# Patent

## Title
Dipole moment term for an electrically small antenna

## Abstract

The present invention relates to a method for designing an electrically small antenna, in one embodiment, within an enclosing volume. In a preferred embodiment, the method comprises the steps of designing the electrically small antenna which has a general cross-sectional contour shape of an oblate spheroid from a top load portion to a stem portion below the top load portion. The oblate spheroid contour shape is represented by an antenna dipole moment algorithm which includes a dipole moment term. The method further comprises the steps of controlling the amplitude of the dipole moment term, including adjusting the amplitude of the dipole moment term to independently change the oblate spheroid contour shape, resulting in a change to the electric field outside the enclosing volume and a change to the electric field inside the enclosing volume.

## Background
CROSS-REFERENCE TO RELATED PATENT APPLICATIONS
This patent application is a continuation-in-part of patent applications (1) Ser. No. 11/959,715, filed Dec. 19, 2007, now U.S. Pat. No. 8,121,821 entitled “Quasi-Static Design Approach For Low Q Factor Electrically Small Antennas” (NC 098,163) and (2) Ser. No. 12/854,955, filed Aug. 12, 2010, entitled “Analytic Antenna Design for a Dipole Antenna” (NC 100,570), both of which are assigned to the same assignee as the present application, and the details of which are hereby incorporated by reference.


FEDERALLY-SPONSORED RESEARCH AND DEVELOPMENT
This invention was developed with funds from the United States Department of the Navy. Licensing inquiries may be directed to Office of Research and Technical Applications, Space and Naval Warfare Systems Center, San Diego, Code 72120, San Diego, Calif., 92152; telephone (619)553-5118; email: ssc_pac_t2@navy.mil.


BACKGROUND
1. Field
This invention relates to antennas. More particularly, this invention relates to an analytic antenna design for a dipole antenna by eliminating or reducing antenna pattern nulls and impedance anti-resonances
2. Background
In 1947, H. A. Wheeler published formulas for qualifying the antenna radiation Q of electrically small antennas in terms of the antenna's physical size. Shortly thereafter, L. J. Chu published additional formulas and theories on this subject, also using the antenna's size as a metric. As the Q of an antenna is inversely related to the antenna's frequency response, of significant interest to antenna engineers is the theoretical Q limits imposed by Wheeler and Chu for given antenna dimensions. Numerous scientists have attempted to corroborate and expand on these dimension-based formulas in terms of actual measurements of scale models, computational models, or revised formulas based on electrical parameters, with some degree of success. However, these approaches have typically yielded solutions that are either very cumbersome, or do not provide an elegant approach for designing low Q antennas.
SUMMARY
The present invention relates to a method for designing an electrically small antenna, in one embodiment, within an enclosing volume. In a preferred embodiment, the method comprises the steps of designing the electrically small antenna which has a general cross-sectional contour shape of an oblate spheroid from a top load portion to a stem portion below the top load portion. The oblate spheroid contour shape is represented by an antenna dipole moment algorithm which includes a dipole moment term. The method further comprises the steps of controlling the amplitude of the dipole moment term, including adjusting the amplitude of the dipole moment term to independently change the oblate spheroid contour shape, resulting in a change to the electric field outside the enclosing volume and a change to the electric field inside the enclosing volume.



BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is an illustration of the contour of an exemplary ACD antenna according to an embodiment of this invention.
FIG. 2 is an illustration of the contour of an exemplary P1 charge antenna, according to an embodiment of this invention.
FIG. 3 is a graph of the computed resistance of an exemplary P1 charge antenna, according to an embodiment of this invention.
FIG. 4 is a graph of the computed reactance of an exemplary P1 charge antenna, according to an embodiment of this invention.
FIG. 5 is a graph of the computed charge of an exemplary P1 charge antenna, according to an embodiment of this invention.
FIG. 6 is a graph comparing experimental measurements of an exemplary P1 charge antenna to computer simulated/predicted values, according to an embodiment of this invention.
FIG. 7 is an illustration of the contour of an exemplary (1−α)P0+α point charge antenna, according to an embodiment of this invention.
FIG. 8 is a graph of the Q/QChu ratio as a function of α using an optimal κ, for an exemplary (1−α)P0+α point charge antenna, according to an embodiment of this invention.
FIG. 9 is an illustration of the contour of an exemplary (1−α)P0+α point charge antenna, according to an embodiment of this invention.
FIG. 10 is a graph of the Q/QChu ratio as a function of α and κ, for an exemplary P1+αP3 charge antenna, according to an embodiment of this invention.
FIG. 11 is a graph of the resistance as a function of α and κ, for an exemplary P1+αP3 charge antenna, according to an embodiment of this invention.
FIG. 12 is a graph of the capacitance as a function of α and κ, for an exemplary P1+αP3 charge antenna, according to an embodiment of this invention.
FIG. 13 is an illustration of the contour of an exemplary P1+αP3 charge antenna, according to an embodiment of this invention.
FIG. 14 is an illustration of the contour of an exemplary P1+αP3 charge antenna, having the local minimum Q solution for α=−1.835 and κ=0.88, according to an embodiment of this invention.
FIG. 15 is a graph of the Q/QChu ratio as a function of α and κ, for α=0.9, for an exemplary (1−α)P0+α oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 16 is a graph of the impedance and resistance for an exemplary (1−α)P0+α oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 17 is a graph of the Q behavior, for an exemplary (1−α)P0+α oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 18 is graph comparing measurements of an exemplary (1−α)P0+α oblate spheroid charge antenna, to numerically derived values, according to an embodiment of this invention.
FIG. 19 is an illustration of the contour of an exemplary (1−α)P0+α oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 20 is an illustration of the contour of an exemplary (1−α)P1+α oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 21 is a graph of the Q/QChu ratio for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 22 is a graph of the β distribution as a function of κ and α, for a minimum Q for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 23 is an illustration of the contour of an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 24 is a graph of the Q/QChu ratio as a function of α and κ, for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 25 is a graph of the β distribution as a function of κ and α, for a minimum Q for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 26 is an illustration of the contour of an exemplary conical enclosing volume (1−α)Po +a oblate spheroid +fi 2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 27 is a 3-D rendering of a wire model of an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge antenna, according to an embodiment of this invention.
FIG. 28 is an illustration of the contour of an exemplary dipole antenna, according to an embodiment of this invention.
FIG. 29 is a flow chart illustrating an exemplary process, according to an embodiment of this invention.
FIG. 30 shows a plot of two eigenmodes.
FIG. 31 shows a plot of resistance and reactance.
FIG. 32 shows a plot of radial distance from the center of the antenna.
FIG. 33 shows a plot of potential as a function of impedance.
FIG. 34 shows a Smith chart between 40 MHz and 6000 MHz.
FIG. 35 shows a plot of resistance, reactance and Q between 0.4 MHz and 2 MHz.
FIG. 36 shows a plot of resistance, reactance and Q between 0.4 MHz and 10 MHz.
FIG. 37 shows an antenna cross-section view of radial distance and height for different moments with τ=0.9 within an enclosing sphere.
FIG. 38 shows a view of the electric field as a function of distance from the top of the antenna.
FIG. 39 shows the Q/QChu ratio (Y-axis) for an oblate spheroid antenna (OSA).
FIG. 40-42 show side views of an oblate spheroid antenna (OSA) for a monopole (FIG. 40), monopole-dipole (FIG. 41) and monopole-dipole-quadrupole (FIG. 42), respectively.



DETAILED DESCRIPTION
The claimed subject matter is now described with reference to the drawings, wherein like reference numerals are used to refer to like elements throughout. In the following description, for purposes of explanation, numerous specific details are set forth in order to provide a thorough understanding of the claimed subject matter. It may be evident, however, that such subject matter may be practiced without these specific details.
Quasi-Static, Low Q Derivation
The radiation Q of an antenna is defined as the ratio of the average stored energy to the average radiated energy, and is usually understood as a metric for qualifying the radiation efficiency of the antenna. This important relationship can be expressed as:

Q=ωEs/Er,  (1)

where Er is the radiated energy and Es is the stored energy.

For an electrically small antenna, the antenna behaves very much like a capacitor, exhibiting quasi-static field behavior, and consequently the instantaneous stored energy Es can be expressed in terms of the well known quasi-static relationship:




  
    
      
        
          
            E
            s
          
          =
          
            
              1
              2
            
            ⁢
            
              Vq
              peak
            
          
        
        ,
      
    
    
      
        (
        2
        )
      
    
  




where V is the voltage and qpeak is the charge. For a time-harmonic current signal I(t)=Ipeak sin(ωt), Eq. 2 can be recast as





  
    
      
        
          
            E
            s
          
          =
          
            
              I
              peak
              2
            
            
              2
              ⁢
              C
              ⁢
              
                
              
              ⁢
              
                ω
                2
              
            
          
        
        ,
      
    
    
      
        (
        3
        )
      
    
  




where C is the capacitance, ω is the radian frequency (2πf). If I(t) is expressed as an rms current (Irms), which is equivalent to an average static current, then the average stored energy Es in a capacitor can be expressed as:





  
    
      
        
          E
          s
        
        =
        
          
            
              I
              
                r
                ⁢
                
                  
                
                ⁢
                m
                ⁢
                
                  
                
                ⁢
                s
              
              2
            
            
              C
              ⁢
              
                
              
              ⁢
              
                ω
                2
              
            
          
          .
        
      
    
    
      
        (
        4
        )
      
    
  




And the radiated energy Er in the capacitor can be expressed as:

Er=RrIrms2,  (5)

where Rr is the resistance in terms of real radiation.

By using the earlier definition of Q (Eq. 1) and combining the above expressions, the following relationship can be developed:




  
    
      
        Q
        =
        
          
            1
            
              ω
              ⁢
              
                
              
              ⁢
              
                CR
                r
              
            
          
          .
        
      
    
    
      
        
          (
          6
          )
        
        .
      
    
  




The above Eq. 6 expresses the Q for a quasi-static field about an electrically small antenna, in terms of electrical parameters of the antenna. However, to be able to arrive at the antenna's electrical parameters as expressed in Eq. 6, the significance of the antenna's shape and its affect on charge distribution must be developed.
It is well understood that the charge on a perfectly conducting antenna is distributed about the surface of the antenna. Since there is an intrinsic relationship between the electric field generated by a source charge, the electric field on the surface of the antenna can be expressed as a function of the distribution of the source charges. For the quasi-static case, this relationship can generally be expressed as:




  
    
      
        
          Φ
          ⁢
          
            
              ∫
              
                ∫
                ∫
              
            
            v
          
          ⁢
          
            q
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              
                r
                dist
              
            
          
          ⁢
          
            d
            v
          
        
        ,
      
    
    
      
        (
        7
        )
      
    
  




where Φ is the electric field potential, q is the charge, ε is the permittivity, and rdist is the distance between the observer and the charge distribution.

For a perfectly conducting surface, the electric field potential is constant, therefore, a solution can be developed for Eq. 7 describing the exact charge distribution on the antenna surface. Therefore, given a simple shape, the charge distribution and attendant antenna parameters can usually be found using a direct, closed form solution. However, due to the complex shapes of most antennas, an indirect solution using numerical techniques must often be employed. As one of ordinary skill in the art can appreciate, numerous researchers have devoted their studies to solving Eq. 7 for assorted geometries and therefore there are a plethora of publications dedicated to this subject matter. These prior art approaches predominately pre-define a shape and thereafter derive a solution from the shape.
For some embodiments of the exemplary methods and apparatuses demonstrated herein, the charge distribution is pre-defined and thereafter the shape of the antenna is generated, based on the corresponding solution. To provide a degree(s) of freedom in the shaping of the antenna, a factor κ, which operates as dimension variable, is inserted into the defining equation(s). From a solution of the defining equation(s), the respective Q-related parameters can be generated to form an expression for Q. Since κ is a variable in the Q expression, it can operate as a solution for a given value of Q. Therefore, varying values of κ will result in varying values of Q. If the Q expression is minimized, then the corresponding value of κ will define a minimum Q antenna. Accordingly, antennas with minimum or otherwise low Q values can be designed by this exemplary method. The application of the exemplary method described will be demonstrated by example below, as derived from differing charge distributions. It should be noted that by use of the exemplary methods and apparatuses described herein, several novel antennas having very low Q values have been developed.
ACD
The Asymptotic Conical Dipole (ACD), sometimes referred to as a Hertzian monopole, is developed, having a constant charge distribution along its vertical axis. The antenna has a length L, and is given a total charge q running from z=−L/2 to z=+L/2. The electric potential Φ for this antenna is defined from the following cylindrical equation:




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              q
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                L
              
            
            ⁢
            
              ln
              ⁡
              
                (
                
                  
                    1
                    +
                    τ
                  
                  
                    1
                    -
                    τ
                  
                
                )
              
            
          
        
        ,
      
    
    
      
        (
        8
        )
      
    
  




where ε0=free space permittivity,





  
    τ
    =
    
      L
      
        
          R
          1
        
        +
        
          R
          2
        
      
    
  
  ,




R1=√{square root over ((z−L/2)2+ρ2)}, and R2=√{square root over ((z+L/2)2+ρ2)}.

On the z-axis, this equation simplifies to




  τ
  =
  
    L
    
      
      
        2
        ⁢
        z
      
      
    
  




for |z|>|L/2|. If h is the actual height of the antenna, we can devise a dimensional variable κ such that κ=L/h. Therefore, the charge distribution per unit length is q/κh. Using this substitution and incorporating image theory, Eq. 8 can be rewritten as:





  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              q
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              ln
              ⁡
              
                (
                
                  
                    
                      (
                      
                        1
                        +
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                  
                    
                      (
                      
                        1
                        -
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        +
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                
                )
              
            
          
        
        ,
      
    
    
      
        (
        9
        )
      
    
  




where





  
    τ
    m
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      h
    
    
      
        R
        f
      
      +
      
        R
        t
      
    
  




is the monopole,





  
    τ
    i
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      h
    
    
      
        R
        f
      
      +
      
        R
        b
      
    
  




is for the image monopole, Rt=√{square root over ((z−κh)2+ρp2)} is the distance from ρ, z to the top of the monopole, Rf=√{square root over (z2+ρ2)} is the distance from ρ, z to the feed point and Rb=√{square root over ((z+κh)2+ρ2)} is the distance from ρ, z to the bottom of the image monopole. It is noted that Eq. 9 expresses the electric potential with the dimensional variable κ embedded within. The capacitance can be calculated by first evaluating the above Rt, Rf, and Rb equations at the point ρ=0, z=h, which is known to be location of the maximum potential location, to result in Rt=(z−κh), Rf=z, Rb=(z+κh),





  
    
      τ
      m
    
    =
    
      
        κ
        ⁢
        
          
        
        ⁢
        h
      
      
        
          2
          ⁢
          z
        
        -
        
          κ
          ⁢
          
            
          
          ⁢
          h
        
      
    
  
  ,




and





  
    τ
    i
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      h
    
    
      
        2
        ⁢
        z
      
      +
      
        κ
        ⁢
        
          
        
        ⁢
        h
      
    
  




for z≧κh. Substituting these values into Eq. 9 yields





  
    
      
        
          Φ
          ⁡
          
            (
            
              ρ
              ,
              z
            
            )
          
        
        =
        
          
            q
            
              4
              ⁢
              π
              ⁢
              
                
              
              ⁢
              
                ɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              ln
              ⁡
              
                (
                
                  
                    z
                    2
                  
                  
                    
                      (
                      
                        z
                        -
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        z
                        +
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                      
                      )
                    
                  
                
                )
              
            
            .
          
        
      
    
    
      
        (
        10
        )
      
    
  




Further evaluating this expression at ρ=0, z=h results in




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            -
            
              q
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
          
          ⁢
          
            
              ln
              ⁡
              
                (
                
                  1
                  -
                  
                    κ
                    2
                  
                
                )
              
            
            .
          
        
      
    
    
      
        (
        11
        )
      
    
  




Since q=CΦ, then the capacitance is




  
    
      
        
          C
          ⁡
          
            (
            κ
            )
          
        
        =
        
          
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
            
              -
              
                ln
                ⁡
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
            
          
          .
        
      
    
    
      
        (
        12
        )
      
    
  




The effective height he for the line charge distribution is he=κh/2. The radiation resistance Rr is




  
    
      
        
          
            
              R
              r
            
            ⁡
            
              (
              κ
              )
            
          
          =
          
            40
            ⁢
            
              
                (
                
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    hk
                  
                  2
                
                )
              
              2
            
          
        
        ,
      
    
    
      
        (
        13
        )
      
    
  




where k is the wave number (2π/λ). The Q can now be expressed as





  
    
      
        
          Q
          ⁡
          
            (
            κ
            )
          
        
        =
        
          
            
              
                -
                3
              
              ⁢
              
                ln
                ⁡
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
            
            
              
                (
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  hk
                
                )
              
              3
            
          
          .
        
      
    
    
      
        (
        14
        )
      
    
  




It should be noted that the quantities C, Re and Q depend on the wave number k and the height h, which can be fixed as constants. A minimum value of Q can be found by taking the derivative of Eq. 14 and setting it to zero resulting in




  
    
      
        
          
            d
            
              (
              
                
                  -
                  
                    ln
                    ⁡
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                  
                
                /
                
                  κ
                  3
                
              
              )
            
          
          
            d
            κ
          
        
        =
        
          
            
              1
              
                
                  κ
                  4
                
                ⁡
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
            
            ⁡
            
              [
              
                
                  
                    (
                    
                      2
                      -
                      
                        3
                        ⁢
                        
                          ln
                          ⁡
                          
                            (
                            
                              1
                              -
                              
                                κ
                                2
                              
                            
                            )
                          
                        
                      
                    
                    )
                  
                  ⁢
                  
                    κ
                    2
                  
                
                +
                
                  3
                  ⁢
                  
                    ln
                    ⁡
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                  
                
              
              ]
            
          
          =
          0.
        
      
    
    
      
        (
        15
        )
      
    
  




As this is a non-trivial equation, Eq. 15 can be solved by any one of several methods of numerical iteration, that are well known in the art, to finally result in κ=0.7304719548370.
This value of κ can now define the dimensional aspects of an antenna that conforms to a constant line charge distribution as defined above. Additionally, as it was derived from a minimized Q relationship, it also defines the dimensional aspects of an antenna that has a minimum Q. Of course, if a non-minimum solution was sought, then Eq. 14 would be evaluated for different values of κ, to provide a spectrum of Q values. Accordingly, different antenna shapes would result from the different κ's, corresponding to different Q values.
In summary, the antenna parameters for the ACD are:




  Q
  =
  
    5.87
    
      
        (
        kh
        )
      
      3
    
  




he=0.3652h

Rr=5.336(hk)2

C=1.064*10−10h h

Based on the minimum κ value, and for a given k and h, the contour of an exemplary ACD antenna was generated. FIG. 1 is an illustration of a cross-sectional view of the exemplary ACD antenna.
P1 Legendre Polynomial Charge Distribution
Various current distribution functions, in addition to a constant line charge, can be chosen, according to design preference. Accordingly, in this exemplary embodiment, a first order Legendre Polynomial (P1) charge distribution is used on a monopole antenna. The P1 charge distribution is linear and defined on the z axis from z=−L/2 to z=+L/2. The static potential equation for this charge in cylindrical coordinates is




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                ∫
                
                  
                    -
                    L
                  
                  /
                  2
                
                
                  L
                  /
                  2
                
              
              ⁢
              
                
                  
                    
                      P
                      1
                    
                    ⁡
                    
                      (
                      
                        2
                        ⁢
                        
                          x
                          /
                          L
                        
                      
                      )
                    
                  
                  
                    
                      
                        
                          (
                          
                            x
                            -
                            z
                          
                          )
                        
                        2
                      
                      +
                      
                        ρ
                        2
                      
                    
                  
                
                ⁢
                
                  d
                  x
                
              
            
            =
            
              
                
                  2
                  ⁢
                  z
                
                L
              
              ⁢
              
                (
                
                  
                    ln
                    ⁡
                    
                      (
                      
                        
                          1
                          +
                          τ
                        
                        
                          1
                          -
                          τ
                        
                      
                      )
                    
                  
                  -
                  
                    2
                    ⁢
                    τ
                  
                
                )
              
            
          
        
        ,
      
    
    
      
        (
        17
        )
      
    
  




where





  
    τ
    =
    
      L
      
        
          R
          1
        
        +
        
          R
          2
        
      
    
  
  ,




L is the length, R1=√{square root over ((z−L/2)2+ρ2)} and R2=√{square root over ((z+L/2)2+ρ2)}. If h is the height of the antenna, the line charge must be less than h by a factor κ<1. The above Eq. 17 represents the potential field for both the monopole and its image, that is, L=2κh.

Our assumption is that the total charge on the monopole is q. The required scale factor, 2q/κh, that enables this assumption is




  
    
      
        q
        =
        
          
            
              2
              ⁢
              q
            
            
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              ∫
              0
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              
                
                  P
                  1
                
                ⁡
                
                  (
                  
                    
                      z
                      /
                      κ
                    
                    ⁢
                    
                      
                    
                    ⁢
                    h
                  
                  )
                
              
              ⁢
              
                
                  d
                  z
                
                .
              
            
          
        
      
    
    
      
        (
        18
        )
      
    
  




With Eq. 18, the resulting potential field equation becomes




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                2
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              z
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              (
              
                
                  ln
                  ⁡
                  
                    (
                    
                      
                        1
                        +
                        τ
                      
                      
                        1
                        -
                        τ
                      
                    
                    )
                  
                
                -
                
                  2
                  ⁢
                  τ
                
              
              )
            
          
        
        ,
      
    
    
      
        (
        19
        )
      
    
  




where





  
    τ
    =
    
      
        2
        ⁢
        κ
        ⁢
        
          
        
        ⁢
        h
      
      
        
          R
          1
        
        +
        
          R
          2
        
      
    
  
  ,




R1=√{square root over ((z−κh)2+ρ2)} and R2=√{square root over ((z+κh)2+ρ2)}. The maximum potential is known to be at z=h and ρ=0 resulting in the simplified expression





  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              
                κ
                2
              
              ⁢
              h
            
          
          ⁢
          
            
              (
              
                
                  ln
                  ⁡
                  
                    (
                    
                      
                        1
                        +
                        κ
                      
                      
                        1
                        -
                        κ
                      
                    
                    )
                  
                
                -
                
                  2
                  ⁢
                  κ
                
              
              )
            
            .
          
        
      
    
    
      
        (
        20
        )
      
    
  




Using principles described in solving for the ACD antenna, the capacitance becomes




  
    
      
        
          C
          ⁡
          
            (
            κ
            )
          
        
        =
        
          4
          ⁢
          
            πɛ
            0
          
          ⁢
          
            
              
                
                  κ
                  2
                
                ⁢
                h
              
              
                2
                ⁢
                
                  (
                  
                    
                      ln
                      ⁡
                      
                        (
                        
                          
                            1
                            +
                            κ
                          
                          
                            1
                            -
                            κ
                          
                        
                        )
                      
                    
                    -
                    
                      2
                      ⁢
                      κ
                    
                  
                  )
                
              
            
            .
          
        
      
    
    
      
        (
        21
        )
      
    
  




Given that the effective height for the P1 charge distribution is he=2κh/3. The radiation resistance is




  
    
      
        
          
            
              R
              r
            
            ⁡
            
              (
              κ
              )
            
          
          =
          
            40
            ⁢
            
              
                (
                
                  
                    2
                    ⁢
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    hk
                  
                  3
                
                )
              
              2
            
          
        
        ,
        
          

        
        ⁢
        
          and
          ⁢
          
            
          
          ⁢
          Q
          ⁢
          
            
          
          ⁢
          is
        
      
    
    
      
        (
        22
        )
      
    
  
  
    
      
        
          Q
          ⁡
          
            (
            κ
            )
          
        
        =
        
          
            
              27
              ⁢
              
                (
                
                  
                    ln
                    ⁡
                    
                      (
                      
                        
                          1
                          +
                          κ
                        
                        
                          1
                          -
                          κ
                        
                      
                      )
                    
                  
                  -
                  
                    2
                    ⁢
                    κ
                  
                
                )
              
            
            
              8
              ⁢
              
                
                  (
                  hk
                  )
                
                3
              
              ⁢
              
                κ
                4
              
            
          
          .
        
      
    
    
      
        (
        23
        )
      
    
  




Again, Q is found to be formulated in terms of κ, k and h. Eq. 23 can be minimized by taking its derivative and setting it to zero:




  
    
      
        
          
            d
            Q
          
          
            d
            κ
          
        
        =
        
          
            
              27
              
                4
                ⁢
                
                  
                    (
                    kh
                    )
                  
                  3
                
              
            
            ⁢
            
              
                (
                
                  
                    
                      
                        2
                        κ
                      
                      ⁢
                      
                        ln
                        ⁡
                        
                          (
                          
                            
                              1
                              +
                              κ
                            
                            
                              1
                              -
                              κ
                            
                          
                          )
                        
                      
                    
                    -
                    3
                  
                  
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                    ⁢
                    
                      κ
                      5
                    
                  
                
                )
              
              [
              
                
                  κ
                  2
                
                -
                
                  (
                  
                    1
                    -
                    
                      1
                      
                        
                          
                            2
                            κ
                          
                          ⁢
                          
                            ln
                            ⁡
                            
                              (
                              
                                
                                  1
                                  +
                                  κ
                                
                                
                                  1
                                  -
                                  κ
                                
                              
                              )
                            
                          
                        
                        -
                        3
                      
                    
                  
                  )
                
              
              ]
            
          
          =
          0.
        
      
    
    
      
        (
        24
        )
      
    
  




The term in the square bracket can be solved by iteration; the quantity,




  
    
      
        2
        κ
      
      ⁢
      
        ln
        ⁡
        
          (
          
            
              1
              +
              κ
            
            
              1
              -
              κ
            
          
          )
        
      
    
    -
    3
  
  ,




being a slowly changing function. The solution is found to be κ=0.69609037983344.

In summary, the parameters for this exemplary antenna become:




  
    
      
        
          
            
              Q
              =
              
                4.7031
                
                  
                    (
                    hk
                    )
                  
                  3
                
              
            
          
        
        
          
            
              
                h
                e
              
              =
              
                .46406
                ⁢
                h
              
            
          
        
        
          
            
              
                R
                r
              
              =
              
                8.614
                ⁢
                
                  
                    (
                    kh
                    )
                  
                  2
                
              
            
          
        
        
          
            
              C
              =
              
                8.22178
                *
                
                  10
                  
                    -
                    11
                  
                
                ⁢
                h
              
            
          
        
      
    
    
      
        (
        25
        )
      
    
  




The radiation resistance for this exemplary P1 charge antenna is 60% higher than the exemplary constant charge ACD antenna and the Q is 20% lower. This is offset by a 23% decrease in the capacitance of the antenna.
Using a κ value of 0.696 a contour of the exemplary P1 charge antenna was generated. FIG. 2 is a cross-sectional illustration of the exemplary P1charge antenna. In addition to generating the contour, a numerical model was created to simulate the exemplary P1 charge antenna's response using Numerical Electromagnetic Code (NEC) and also Computer Simulation Technology (CST) Microwave Studio®. These simulations were performed using varying values of κ, the CST computed minimum being 0.661. The results of the simulation are shown in FIG. 3, which is a graph showing the computed resistance of the exemplary P1 charge antenna as a function of frequency, for a κ value of 0.661. FIG. 4 is a graph showing the computed reactance of the exemplary P1 charge antenna as a function of frequency, for the same κ value. And FIG. 5 is a graph showing the computed charge distribution of the exemplary P1 charge antenna as a function of frequency, again for the same κ value.
Based on the results tabulated above, a half meter physical model was constructed of wires and experimental measurements were performed. FIG. 6 details the comparison of the physical model measurements to the computer simulated/predicted values. It is noted that the measured and computed values appear to track reasonably well throughout the experiment.
It should be noted that typical antennas are known to exhibit anti-resonance or poor radiation characteristics at frequencies that are one half wavelength the longest dimension of the antenna. Examination of FIG. 6 shows that the response is very smooth throughout the one half wavelength region. Therefore, this exemplary P1 charge antenna does not suffer from typical half wavelength anti-resonance problems.
P0+Point Charge Antenna
Progressing from the exemplary P1 charge antenna, another charge distribution was evaluated. A “bulb” antenna charge distribution was generated, using a constant line charge distribution (P0) with a point charge at the end of the constant line charge distribution. In essence, the bulb antenna charge distribution is the charge distribution of the constant line charge ACD antenna modified with a point charge at the top.
The total charge q is split between the line charge and the point charge, with the point charge being αq and the line charge being (1−α)q/κh, where α is a proportionality constant. Using similar dimensional constraints as imposed in the constant line charge ACD antenna, the expression for the potential field distribution becomes




  
    
      
        
          Φ
          ⁡
          
            (
            
              ρ
              ,
              z
            
            )
          
        
        =
        
          
            
              
                (
                
                  1
                  -
                  α
                
                )
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              ln
              ⁡
              
                (
                
                  
                    
                      (
                      
                        1
                        +
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                  
                    
                      (
                      
                        1
                        -
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        +
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                
                )
              
            
          
          +
          
            
              α
              ⁢
              
                
              
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              
                R
                t
              
            
          
          -
          
            
              
                α
                ⁢
                
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  R
                  b
                
              
            
            .
          
        
      
    
    
      
        (
        26
        )
      
    
  




This potential is known to be maximum at ρ=0 and z=h, resulting in Eq. 26 being recast as




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              
                
                  -
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              ln
              ⁡
              
                (
                
                  1
                  -
                  
                    κ
                    2
                  
                
                )
              
            
          
          +
          
            
              
                2
                ⁢
                ακ
                ⁢
                
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  h
                  ⁡
                  
                    (
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
              
            
            .
          
        
      
    
    
      
        (
        27
        )
      
    
  




The capacitance can be found to be




  
    
      
        
          C
          ⁡
          
            (
            
              κ
              ,
              α
            
            )
          
        
        =
        
          
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
            
              
                
                  -
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                
                ⁢
                
                  ln
                  ⁡
                  
                    (
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
              
              +
              
                
                  2
                  ⁢
                  
                    ακ
                    2
                  
                
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
            
          
          .
        
      
    
    
      
        (
        28
        )
      
    
  




The effective height heff is

heff(κ,α)=(+α)κh/2.  (29)

And the radiation resistance is

Rr(κ,α)=10[κhk(1+α)]2.  (30)

Using free space, the resulting Q expression becomes




  
    
      
        
          Q
          ⁡
          
            (
            
              α
              ,
              κ
            
            )
          
        
        =
        
          
            
              3
              
                
                  
                    (
                    
                      1
                      +
                      α
                    
                    )
                  
                  2
                
                ⁢
                
                  
                    (
                    
                      κ
                      ⁢
                      
                        
                      
                      ⁢
                      kh
                    
                    )
                  
                  3
                
              
            
            ⁡
            
              [
              
                
                  
                    2
                    ⁢
                    
                      ακ
                      2
                    
                  
                  
                    (
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
                -
                
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                  ⁢
                  
                    ln
                    ⁡
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                  
                
              
              ]
            
          
          .
        
      
    
    
      
        (
        31
        )
      
    
  




The minimum value Q is found by numerically evaluating Eq. 31 for a range of a values. The value of κ for the minimum Q is given in the following Table 1.











TABLE 1





α
κ
Minimum Q
R
C



















0.5
0.609
33.77
2.29
41.2 pF


0.6
0.600
32.26
2.53
39.1 pF


0.7
0.593
30.84
2.79
37.0 pF


0.8
0.587
29.52
3.06
35.2 pF


0.9
0.582
28.29
3.35
33.5 pF


0.99
0.578
27.26
3.63
32.2 pF









Using the α value of 0.9 (meaning 90% of the total charge on the point) the “bulb” antenna parameters are




  Q
  =
  
    4.061
    
      
        (
        kh
        )
      
      3
    
  




he=0.5529h.

Rr=12.22*(kh)2

C=6.7*10−11h

Of interesting note is that the Q value is diminishing, as compared to the previous exemplary antennas. FIG. 7 is an illustration of a cross-sectional view of the exemplary (1−α)P0+α point charge (“bulb”) antenna demonstrated herein. Also, as a means of validation, this exemplary antenna model exhibited similar characteristics to a “bulb” antenna modeled by R. M. Kalafus, with the principal exception that Kalafus' antenna uses a finite feed point.
A comparison of the Q performance was also made to Chu's theoretical minimum. Chu expresses the theoretical minimum Q of an antenna as




  
    
      
        
          Q
          Chu
        
        =
        
          
            1
            
              
                (
                kh
                )
              
              3
            
          
          +
          
            
              1
              kh
            
            .
          
        
      
    
    
      
        (
        32
        )
      
    
  




Using Eq. 32, only to the first order, a comparison of the Q/QChu ratio for the above exemplary antenna was made, for varying values of α, as a function of κ. FIG. 8 is a plot of this comparison.
P1+Point Charge Antenna
The above demonstrated exemplary P0+point charge model can be modified by using a P1 feed line rather than a constant feed line. As previously described, the P1 charge distribution is a “linear” charge distribution versus a “constant” charge distribution, as in the case of P0. For this formulation, the net charge q is split between the linear charge and the point charge with the point charge quantity being αq and the line charge quantity being 2(1−α)q/κh. The resulting expression for the potential becomes




  
    
      
        
          Φ
          ⁡
          
            (
            
              ρ
              ,
              z
            
            )
          
        
        =
        
          
            
              
                2
                ⁢
                
                  (
                  
                    1
                    -
                    α
                  
                  )
                
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              z
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              (
              
                
                  ln
                  ⁡
                  
                    (
                    
                      
                        1
                        +
                        τ
                      
                      
                        1
                        -
                        τ
                      
                    
                    )
                  
                
                -
                
                  2
                  ⁢
                  τ
                
              
              )
            
          
          +
          
            
              α
              ⁢
              
                
              
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              
                R
                t
              
            
          
          -
          
            
              
                α
                ⁢
                
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  R
                  b
                
              
            
            .
          
        
      
    
    
      
        (
        33
        )
      
    
  




Eq. 33 can be evaluated at ρ=0 and z=h, resulting in




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              
                2
                ⁢
                
                  (
                  
                    1
                    -
                    α
                  
                  )
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  κ
                  2
                
                ⁢
                h
              
            
            ⁢
            
              (
              
                
                  ln
                  ⁡
                  
                    (
                    
                      
                        1
                        +
                        κ
                      
                      
                        1
                        -
                        κ
                      
                    
                    )
                  
                
                -
                
                  2
                  ⁢
                  κ
                
              
              )
            
          
          +
          
            
              
                2
                ⁢
                ακ
                ⁢
                
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  h
                  ⁡
                  
                    (
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
              
            
            .
          
        
      
    
    
      
        (
        34
        )
      
    
  




The capacitance becomes




  
    
      
        
          C
          ⁡
          
            (
            
              κ
              ,
              α
              ,
              h
            
            )
          
        
        =
        
          
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              
                κ
                2
              
              ⁢
              h
            
            
              
                2
                ⁢
                
                  (
                  
                    1
                    -
                    α
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      ln
                      ⁡
                      
                        (
                        
                          
                            1
                            +
                            κ
                          
                          
                            1
                            -
                            κ
                          
                        
                        )
                      
                    
                    -
                    
                      2
                      ⁢
                      κ
                    
                  
                  )
                
              
              +
              
                
                  2
                  ⁢
                  
                    ακ
                    3
                  
                
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
            
          
          .
        
      
    
    
      
        (
        35
        )
      
    
  




The effective height is

heff=(1−α)2κh/3+ακh.  (36)

The radiation resistance is

Rr=40[κheffk(2⁄3+α/3)]2.  (37)

The Q becomes




  
    
      
        
          Q
          ⁡
          
            (
            
              α
              ,
              κ
              ,
              h
            
            )
          
        
        =
        
          
            
              27
              
                4
                ⁢
                
                  
                    (
                    
                      2
                      +
                      α
                    
                    )
                  
                  2
                
                ⁢
                
                  
                    
                      κ
                      4
                    
                    ⁡
                    
                      (
                      kh
                      )
                    
                  
                  3
                
              
            
            ⁡
            
              [
              
                
                  2
                  ⁢
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                  ⁢
                  
                    (
                    
                      
                        ln
                        ⁡
                        
                          (
                          
                            
                              1
                              +
                              κ
                            
                            
                              1
                              -
                              κ
                            
                          
                          )
                        
                      
                      -
                      
                        2
                        ⁢
                        κ
                      
                    
                    )
                  
                
                +
                
                  
                    2
                    ⁢
                    
                      ακ
                      3
                    
                  
                  
                    (
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
              
              ]
            
          
          .
        
      
    
    
      
        (
        38
        )
      
    
  




Using similar procedures as applied to the previous examples, the minimum value of Q can be found by numerically evaluating Eq. 38 for varying α, κ values. Table 2 below details the resulting evaluation.











TABLE 2





α
κ
Minimum Q
R
C



















0.5
0.611
30.39
2.84
36.8 pF


0.6
0.602
29.85
2.99
35.9 pF


0.7
0.595
29.07
3.15
34.8 pF


0.8
0.588
28.42
3.30
33.9 pF


0.9
0.582
27.78
3.47
33.0 pF


0.99
0.578
27.21
3.63
32.1 pF









For large values of α, the addition of a P1 distribution does little to change the performance of the antenna, as compared to the previously modeled antenna. For the case of α=0.9, this antenna's parameters are




  Q
  =
  
    3.988
    
      
        (
        kh
        )
      
      3
    
  




he=0.5626h.

Rr=12.66*(kh)2

C=6.6*10−11h

By using a P1 distribution rather than a constant distribution, only a slight improvement is demonstrated over the P0 distribution for modest values of α. For large values of α, the values will converge in the limit. FIG. 9 is an illustration of the contour of this exemplary (1−α)P1+α point charge antenna.
P1 and P3 Legendre Polynomial Charge Distribution
It should be understood that Q is defined by the stored energy in the near field and far fields of the antenna. Part of the stored energy is in the higher order moments of these fields. These moments could be reduced by adding a higher order charge distribution to account for these higher order field moments. In the following exemplary antenna, a higher order charge distribution term, Legendre polynomial P3, is added to the charge distribution. It should be noted that other polynomials or basis functions (not being Legendre in form) may be used without departing from the spirit and scope of this invention.
It should be understood that the fields created by differing basis functions, for example, Legendre polynomials—P1, P2, P3, . . . , Pn, the respective line charge distributions operate with independent degrees of freedom with different angular and radial dependence. If using Legendre polynomials, the radial and angular dependence can be shown to be Pn(θ)/rn+1 (where is θ measured from the z axis) for Pn (see Appendix for details). Mathematically, a surface constructed from a sequence of multipole functions should quickly converge.
In the following example, the Legendre polynomial P3 with a coefficient α3 (α3P3) is added to a P1 linear charge distribution. The Q can be further reduced by adding a α5P5 term. This shifts the α3 value some but reduces the error term to P7(θ)/r7. An alternate approach is to algebraically calculate the multipole distribution (on the z axis) for each P2m+1 polynomial and solve the equation for the coefficients α3 and α5 to eliminate the P3(θ)/r4 and P5(θ)/r6 terms in the multipole expansion. (See Appendix for example). In addition, since the shape of the feed point is determined by the line charge distribution, combinations of P1, P3, P5, etc. can be used to parameterize the shape of the feed point and the input impedance for high frequencies.
Based on the above observations, the static potential charge distribution on the z axis from z=−/2 to z=L/2 is




  
    
      
        
          
            
              Ψ
              3
            
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                ∫
                
                  
                    -
                    L
                  
                  /
                  2
                
                
                  L
                  /
                  2
                
              
              ⁢
              
                
                  
                    P
                    3
                  
                  ⁡
                  
                    (
                    
                      2
                      ⁢
                      
                        x
                        /
                        L
                      
                    
                    )
                  
                
                
                  
                    
                      
                        (
                        
                          x
                          -
                          z
                        
                        )
                      
                      2
                    
                    +
                    
                      ρ
                      2
                    
                  
                
              
            
            =
            
              
                
                  20
                  
                    L
                    3
                  
                
                [
                
                  
                    
                      5
                      12
                    
                    ⁢
                    
                      Lz
                      ⁡
                      
                        (
                        
                          
                            R
                            1
                          
                          +
                          
                            R
                            2
                          
                        
                        )
                      
                    
                  
                  -
                  
                    2
                    ⁢
                    z
                    ⁢
                    
                      
                    
                    ⁢
                    
                      τ
                      ⁡
                      
                        (
                        
                          
                            
                              11
                              6
                            
                            ⁢
                            
                              z
                              2
                            
                          
                          +
                          
                            
                              L
                              2
                            
                            12
                          
                          -
                          
                            
                              2
                              3
                            
                            ⁢
                            
                              ρ
                              3
                            
                          
                        
                        )
                      
                    
                  
                  +
                  
                    
                      z
                      ⁡
                      
                        (
                        
                          
                            z
                            2
                          
                          -
                          
                            
                              3
                              2
                            
                            ⁢
                            
                              ρ
                              2
                            
                          
                        
                        )
                      
                    
                    ⁢
                    
                      ln
                      ⁡
                      
                        (
                        
                          
                            1
                            +
                            τ
                          
                          
                            1
                            -
                            τ
                          
                        
                        )
                      
                    
                  
                
                ⁢
                
                  
                
                ]
              
              -
              
                
                  
                    3
                    ⁢
                    z
                  
                  L
                
                ⁢
                
                  (
                  
                    
                      ln
                      ⁡
                      
                        (
                        
                          
                            1
                            +
                            τ
                          
                          
                            1
                            -
                            τ
                          
                        
                        )
                      
                    
                    -
                    
                      2
                      ⁢
                      τ
                    
                  
                  )
                
              
            
          
        
        ,
      
    
    
      
        (
        39
        )
      
    
  




(Note: the above three lines are one equation)

where





  
    τ
    =
    
      L
      
        
          R
          1
        
        +
        
          R
          2
        
      
    
  
  ,




L is the length, R1=√{square root over ((z−L/2)2+ρ2)} and R2=√{square root over ((z+L/2)2+ρ2)}. If h is the height of the antenna, the line charge must be less than this height by a factor κ<1. Eq. 40 represents both the monopole and image: L=2κh. The potential Φ can be re-expressed in terms of fields (Ψ1 and Ψ3) generated from the Legendre polynomials, the expression being





  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                2
                ⁢
                q
              
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁡
            
              [
              
                
                  
                    Ψ
                    1
                  
                  ⁡
                  
                    (
                    
                      ρ
                      ,
                      z
                    
                    )
                  
                
                +
                
                  α
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      Ψ
                      3
                    
                    ⁡
                    
                      (
                      
                        ρ
                        ,
                        z
                      
                      )
                    
                  
                
              
              ]
            
          
        
        ,
      
    
    
      
        (
        40
        )
      
    
  




where α is a coefficient for the contribution of the Ψ3 term.

It should be noted that Eq. 40 can have a negative charge distribution. If the charge distribution is negative near the base of the antenna, the feed point is above ground level. If the charge distribution is negative at the top of the antenna, the feed point is at the base. In latter case, the negative charge distribution must be enclosed by the equal potential surface. If the equipotential surface cuts across the charge distributions the solution is understood to not be a valid solution.
Following procedures previously discussed, the total charge q on the monopole is




  
    
      
        
          q
          Net
        
        =
        
          
            
              2
              ⁢
              q
            
            
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              ∫
              0
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              
                [
                
                  
                    
                      
                        P
                        1
                      
                      ⁡
                      
                        (
                        
                          
                            z
                            /
                            κ
                          
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                        )
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      z
                    
                  
                  +
                  
                    α
                    ⁢
                    
                      
                    
                    ⁢
                    
                      
                        P
                        3
                      
                      ⁡
                      
                        (
                        
                          
                            z
                            /
                            κ
                          
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                        )
                      
                    
                    ⁢
                    
                      d
                      z
                    
                  
                
                ]
              
              .
            
          
        
      
    
    
      
        (
        41
        )
      
    
  




Using a substitution of variables by letting u=z/κh, the charge distribution as a function of the Legendre polynomials can be cast as




  
    
      
        
          q
          Net
        
        =
        
          
            
              2
              ⁢
              q
            
            
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          κ
          ⁢
          
            
          
          ⁢
          
            
              h
              ⁡
              
                [
                
                  
                    
                      ∫
                      0
                      1
                    
                    ⁢
                    
                      
                        
                          P
                          1
                        
                        ⁡
                        
                          (
                          u
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        d
                        u
                      
                    
                  
                  +
                  
                    α
                    ⁢
                    
                      
                        ∫
                        0
                        1
                      
                      ⁢
                      
                        
                          
                            P
                            3
                          
                          ⁡
                          
                            (
                            u
                            )
                          
                        
                        ⁢
                        
                          
                        
                        ⁢
                        
                          d
                          u
                        
                      
                    
                  
                
                ]
              
            
            .
          
        
      
    
    
      
        (
        42
        )
      
    
  




Substituting the functions P1(u) and P3(u), and simplifying yields




  
    
      
        
          q
          Net
        
        =
        
          
            q
            ⁡
            
              (
              
                1
                -
                
                  α
                  4
                
              
              )
            
          
          .
        
      
    
    
      
        (
        43
        )
      
    
  




For α≦2⁄3, the feed point is at the ground and the total charge above the feed point is qNet. For α>2⁄3, the feed point is at




  z
  =
  
    
      
        
          3
          ⁢
          α
        
        -
        2
      
      
        5
        ⁢
        α
      
    
  




and the total charge above the feed point is





  
    q
    fp
  
  =
  
    q
    ⁢
    
      
        ⌊
        
          
            (
            
              1
              -
              
                α
                4
              
            
            )
          
          +
          
            
              
                (
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                )
              
              2
            
            
              20
              ⁢
              α
            
          
        
        ⌋
      
      .
    
  




Derivation of the radiation resistance is discussed in the appendix, resulting in





  
    
      
        
          
            
              
                R
                r
              
              =
              
                
                  40
                  ⁢
                  
                    
                      (
                      
                        
                          2
                          ⁢
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          hk
                        
                        
                          3
                          ⁢
                          
                            (
                            
                              1
                              -
                              
                                α
                                4
                              
                            
                            )
                          
                        
                      
                      )
                    
                    2
                  
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  α
                
                ≤
                
                  2
                  /
                  3
                
              
            
          
        
        
          
            
              
                R
                r
              
              =
              
                
                  40
                  ⁢
                  
                    
                      (
                      
                        
                          2
                          ⁢
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          hk
                        
                        
                          3
                          ⁡
                          
                            [
                            
                              
                                (
                                
                                  1
                                  -
                                  
                                    α
                                    4
                                  
                                
                                )
                              
                              +
                              
                                
                                  
                                    (
                                    
                                      
                                        3
                                        ⁢
                                        α
                                      
                                      -
                                      2
                                    
                                    )
                                  
                                  2
                                
                                
                                  20
                                  ⁢
                                  α
                                
                              
                            
                            ]
                          
                        
                      
                      )
                    
                    2
                  
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  α
                
                >
                
                  2
                  /
                  3.
                
              
            
          
        
      
    
    
      
        (
        45
        )
      
    
  




If α<−1, the combination of P3 and P3 creates a negative charge distribution at the end of the wire. The negative charge distribution could cause the equipotential surface to cut cross the charge distributions which, as mentioned above, would generate an invalid solution. Therefore, the potential must be calculated near the end of the charge distribution to verify that the charge distribution is enclosed within the equipotential surface. This is demonstrated as




  
    
      
        
          
            Ψ
            3
          
          ⁡
          
            (
            
              0
              ,
              z
            
            )
          
        
        =
        
          
            
              
                5
                2
              
              ⁢
              
                
                  
                    (
                    
                      2
                      L
                    
                    )
                  
                  3
                
                ⁡
                
                  [
                  
                    
                      
                        5
                        6
                      
                      ⁢
                      
                        Lz
                        2
                      
                    
                    -
                    
                      
                        z
                        ⁡
                        
                          (
                          
                            L
                            z
                          
                          )
                        
                      
                      ⁢
                      
                        (
                        
                          
                            
                              11
                              6
                            
                            ⁢
                            
                              z
                              2
                            
                          
                          +
                          
                            
                              L
                              2
                            
                            12
                          
                        
                        )
                      
                    
                    +
                    
                      
                        z
                        3
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        ln
                        ⁡
                        
                          (
                          
                            
                              1
                              +
                              
                                
                                  L
                                  /
                                  2
                                
                                ⁢
                                z
                              
                            
                            
                              1
                              -
                              
                                
                                  L
                                  /
                                  2
                                
                                ⁢
                                z
                              
                            
                          
                          )
                        
                      
                    
                  
                  ]
                
              
            
            -
            
              
                
                  
                    3
                    2
                  
                  ⁡
                  
                    [
                    
                      
                        
                          2
                          ⁢
                          z
                        
                        L
                      
                      ⁢
                      
                        (
                        
                          
                            ln
                            ⁡
                            
                              (
                              
                                
                                  1
                                  +
                                  
                                    
                                      L
                                      /
                                      2
                                    
                                    ⁢
                                    z
                                  
                                
                                
                                  1
                                  -
                                  
                                    
                                      L
                                      /
                                      2
                                    
                                    ⁢
                                    z
                                  
                                
                              
                              )
                            
                          
                          -
                          
                            2
                            ⁢
                            
                              L
                              
                                2
                                ⁢
                                z
                              
                            
                          
                        
                        )
                      
                    
                    ]
                  
                
                .
                
                  

                
                ⁢
                Substituting
              
              ⁢
              
                
              
              ⁢
              L
            
          
          =
          
            2
            ⁢
            
              
            
            ⁢
            κ
            ⁢
            
              
            
            ⁢
            h
            ⁢
            
              
            
            ⁢
            recasts
            ⁢
            
              
            
            ⁢
            
              Eq
              .
              
                
              
              ⁢
              46
            
            ⁢
            
              
            
            ⁢
            as
          
        
      
    
    
      
        (
        45
        )
      
    
  
  
    
      
        
          
            Ψ
            3
          
          ⁡
          
            (
            
              0
              ,
              z
            
            )
          
        
        =
        
          
            
              5
              2
            
            ⁢
            
              
                (
                
                  1
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    h
                  
                
                )
              
              3
            
            ⁢
            
              ⌊
              
                
                  
                    5
                    3
                  
                  ⁢
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    hz
                    2
                  
                
                -
                
                  2
                  ⁢
                  
                    z
                    ⁡
                    
                      (
                      
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                        z
                      
                      )
                    
                  
                  ⁢
                  
                    (
                    
                      
                        
                          11
                          6
                        
                        ⁢
                        
                          z
                          2
                        
                      
                      +
                      
                        
                          
                            (
                            
                              κ
                              ⁢
                              
                                
                              
                              ⁢
                              h
                            
                            )
                          
                          2
                        
                        3
                      
                    
                    )
                  
                
                +
                
                  
                    z
                    3
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    ln
                    ⁡
                    
                      (
                      
                        
                          z
                          +
                          
                            κ
                            ⁢
                            
                              
                            
                            ⁢
                            h
                          
                        
                        
                          z
                          -
                          
                            κ
                            ⁢
                            
                              
                            
                            ⁢
                            h
                          
                        
                      
                      )
                    
                  
                
              
              ⌋
            
          
          -
          
            
              
                3
                2
              
              ⁡
              
                [
                
                  
                    z
                    
                      κ
                      ⁢
                      
                        
                      
                      ⁢
                      h
                    
                  
                  ⁢
                  
                    (
                    
                      
                        ln
                        ⁡
                        
                          (
                          
                            
                              z
                              +
                              
                                κ
                                ⁢
                                
                                  
                                
                                ⁢
                                h
                              
                            
                            
                              z
                              -
                              
                                κ
                                ⁢
                                
                                  
                                
                                ⁢
                                h
                              
                            
                          
                          )
                        
                      
                      -
                      
                        2
                        ⁢
                        
                          
                            κ
                            ⁢
                            
                              
                            
                            ⁢
                            h
                          
                          z
                        
                      
                    
                    )
                  
                
                ]
              
            
            .
          
        
      
    
    
      
        (
        46
        )
      
    
  




Now with




  
    Φ
    max
  
  =
  
    
      
        2
        ⁢
        q
      
      
        κ
        ⁢
        
          
        
        ⁢
        h
      
    
    ⁢
    
      (
      
        
          Ψ
          1
        
        +
        
          αΨ
          3
        
      
      )
    
  




on the interval κh<z≦1, this value must be larger than a calculated Φmax on the surface of an enclosing sphere. In this situation, the equation for the capacitance and Q on the sphere cannot be written and, therefore, the problem must be numerically solved. FIG. 10 is a plot of the Q/QChu ratio as a function of the parameters κ and α. The upper boundary, in the right hand corner of FIG. 10, is caused by the potential becoming negative on the enclosing sphere which means the antenna is not inside the enclosing sphere. The lower boundary, in FIG. 10, is caused by the potential surface crossing the charge distribution at the top. When α=0, the Eq. 46 reduces to the previous solution for the sole P1 charge distribution antenna, as is expected.

At 50 MHz, the solution to Eq. 46 has two minimums: an absolute minimum at Q=28.706 corresponding to Q/QChu4.1207 at κ=0.6513 and α=0.8600, with R=3.1939 and C=34.718 pF; and a local minimum at Q=31.879 corresponding to Q/QChu=4.576 at κ=0.88, α=−1.835 , with R=1.7737 and C=56.294 pF. These tabulations are shown in FIGS. 11-12. FIG. 11 is a plot of the resistance as a function of α and κ, for an exemplary P1+αP3 charge antenna. And FIG. 12 is a plot of the capacitance as a function of α and κ, for an exemplary P1+αP3 charge antenna. As can be seen from these FIGS., various Q solutions can be found.
FIG. 13 is an illustration of the contour of an exemplary P1+αP3 charge antenna for a global minimum Q of 28.706, using an α value of 0.8600 and a κ value of 0.6513. This exemplary P1+αP3 charge antenna is shaped similar to a volcano and smoke antenna and has a Q of 4.1027 times Chu's theoretical result. The antenna parameters for this design are




  Q
  =
  
    4.1207
    
      
        (
        kh
        )
      
      3
    
  




he=0.53967h

Rr=11.65(kh)2

C=6.943*10−11h

The radiation resistance for this antenna is 35% higher than the single P1 charge distribution antenna and Q is 12% lower. This is offset by 15% decrease in the capacitance of the antenna.
Of some interest is the shape of the local minimum Q solution. FIG. 14 illustrates the cross-sectional view of the of the local minimum solution Q=31.879 for an α value of −1.835 and a κ value of 0.88. It is worthy to note, for this exemplary P1+αP3 charge antenna, the mathematical artifact of having a negative charge is seen in the closed cavity at the top of FIG. 14.
P0 with Oblate Spheroid
From the above examples, it is apparent that the point charge at the end of constant line charge (P0) antenna reduces the Q of the antenna. This point charge is equivalent to giving a spherical end cap to the P0 charge distribution. However, in the above configurations, the top of the spherical volume was not filled by the antenna. If the spherical volume is filled by the antenna, then this will reduce the charge surface density, which is known to increase the capacitance, which would decrease Q.
However, a better choice for the end cap would be an oblate spheroid. The eccentricity can be continuously varied from 0, a sphere, to 1, a disk. It should be noted that this antenna is not the same as a disk loaded monopole, which is well known in the antenna community. In the disk loaded monopole the charge is on the bottom of the disk with minimal charge on the top surface. For the oblate spheroid loaded P0 charge antenna, the oblate spheroid has equal charge on both upper and lower surfaces. When the image is added, the effective charge density on the top surface will be reduced and the effective charge density on the bottom surface will be increased.
The potential for an oblate spheroid with a charge q is derived with oblate spheroidal coordinates, the details being given in the appendix. The cylindrical coordinates can be expressed as oblate spheroid coordinates as

z=a sin h u sin ν

ρ=a cos h u cos ν.  (47)

(Note: a parameterizes the eccentricity variable in the oblate spheroidal domain and α is the charge proportionality variable.) The inversion of the above equations from one domain to the other is given in the appendix. The potential can be expressed as a function of u only:




  
    
      
        
          Φ
          ⁡
          
            (
            u
            )
          
        
        =
        
          
            q
            
              4
              ⁢
              
                πɛ
                0
              
            
          
          ⁢
          
            
              (
              
                
                  π
                  2
                
                -
                
                  arc
                  ⁢
                  
                    
                  
                  ⁢
                  
                    tan
                    ⁡
                    
                      (
                      
                        sinh
                        ⁢
                        
                          
                        
                        ⁢
                        u
                      
                      )
                    
                  
                
              
              )
            
            .
          
        
      
    
    
      
        (
        48
        )
      
    
  




Using procedures similar to those discussed above, the potential field generated from a constant line charge terminated with an oblate spheroidal point charge is




  
    
      
        
          Φ
          =
          
            
              
                
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                  ⁢
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  h
                
              
              ⁢
              
                
              
              ⁢
              
                ln
                ⁡
                
                  (
                  
                    
                      
                        (
                        
                          1
                          +
                          
                            τ
                            m
                          
                        
                        )
                      
                      ⁢
                      
                        (
                        
                          1
                          -
                          
                            τ
                            i
                          
                        
                        )
                      
                    
                    
                      
                        (
                        
                          1
                          -
                          
                            τ
                            m
                          
                        
                        )
                      
                      ⁢
                      
                        (
                        
                          1
                          +
                          
                            τ
                            i
                          
                        
                        )
                      
                    
                  
                  )
                
              
            
            +
            
              
                
                  α
                  ⁢
                  
                    
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                
              
              ⁡
              
                [
                
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            b
                          
                        
                        )
                      
                    
                  
                  -
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            t
                          
                        
                        )
                      
                    
                  
                
                ]
              
            
          
        
        ,
      
    
    
      
        (
        49
        )
      
    
  




where the parameter a plays a role indirectly via u, and ut is for the top of the spheroid and ub is for the bottom of the spheroid in the spheroidal coordinate system.

The maximum potential is not at the top of the oblate spheroid, thus Eq. 49 is numerically evaluated on the surface of the enclosing sphere of the antenna. FIG. 15 illustrates a graph of the values of Q/QChu as a function of a and κ, where α=0.9. For varying values of a, a unique solution can be found for Eq. 49. Each of these solutions can be optimized to result in a κ and a value.
A tabulation of these values is shown in Table 3 below.












TABLE 3





α
κ
a
Minimum Q
R
C




















0.5
0.691
0.722
19.421
2.94
55.7 pF


0.6
0.6985
0.703
17.455
3.42
53.3 pF


0.65
0.707
0.6905
16.786
3.73
50.9 pF


0.675
0.706
0.671
16.523
3.83
50.3 pF


0.7
0.709
0.6575
16.305
3.98
49.0 pF


0.725
0.709
0.649
16.129
4.10
48.2 pF


0.75
0.709
0.647
15.962
4.22
47.2 pF


0.8
0.707
0.643
15.636
4.44
45.8 pF


0.9
0.705
0.637
15.008
4.91
43.16 pF 


0.99
0.702
0.633
14.475
5.35
41.1 pF









For the case of 90% of the charge on the surface of the oblate sphereoid, the antenna parameters are




  Q
  =
  
    2.1544`
    
      
        (
        kh
        )
      
      3
    
  




he=0.670h

Rr=17.9*(kh)2

C=8.632*10−11h

The resulting Q is found to be almost half of the result for the “bulb” antenna, which utilized a constant line charge with a point charge termination (i.e., spherical end cap). The Q is 47% lower; R and C are 46% and 30% higher, respectively, than the bulb antenna. This exemplary (1−α)P0+α oblate spheroid charge antenna was numerical modeled using α=0.90. The exemplary numerical model was generated using a cylindrical antenna with a continuous surface and with a 1⁄2 half meter height. FIG. 16 is a graph of the impedance and resistance of the modeled antenna, as a function of frequency. FIG. 17 is a graph of the Q behavior, as a function of the frequency. Based on the results of these numerical models, an exemplary 1⁄2 meter antenna was physically built using 16 wires to approximate a solid surface. The exemplary antenna was evaluated over a frequency range from 40 MHz to over 200 MHz. The measurement results of the antenna demonstrated a very low Q and well behaved impedance profile.
FIG. 18 is a plot of the comparison of the exemplary experimental model to the exemplary numerical models. FIG. 19 is an illustration of the contour for an exemplary (1−α)P0+α oblate spheroid charge antenna.
P1 with Oblate Spheroid
The same above exemplary technique can be expanded upon using a P1 line charge instead of a P0 line charge, with an oblate spheroid end charge. Using the P1 line charge, the expression for the maximum potential field becomes




  
    
      
        
          Φmax
          ⁡
          
            (
            sphere
            )
          
        
        =
        
          
            
              
                2
                ⁢
                
                  (
                  
                    1
                    -
                    α
                  
                  )
                
                ⁢
                qz
              
              
                4
                ⁢
                
                  
                    
                      πɛ
                      0
                    
                    ⁡
                    
                      (
                      
                        κ
                        ⁢
                        
                          
                        
                        ⁢
                        h
                      
                      )
                    
                  
                  2
                
              
            
            ⁢
            
              (
              
                
                  ln
                  ⁡
                  
                    (
                    
                      
                        1
                        +
                        τ
                      
                      
                        1
                        -
                        τ
                      
                    
                    )
                  
                
                -
                
                  2
                  ⁢
                  τ
                
              
              )
            
          
          +
          
            
              
                
                  α
                  ⁢
                  
                    
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                
              
              ⁡
              
                [
                
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            b
                          
                        
                        )
                      
                    
                  
                  -
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            t
                          
                        
                        )
                      
                    
                  
                
                ]
              
            
            .
          
        
      
    
    
      
        (
        50
        )
      
    
  




The results of this procedure is shown in Table 4 below.












TABLE 4





α
κ
a
Minimum Q
R
C




















0.5
0.6777
0.7217
18.95
3.50
48.0 pF


0.6
0.6720
0.7270
17.14
3.72
49.9 pF


0.61
0.6767
0.7362
16.80
3.80
49.86 pF 


0.62
0.6772
0.7349
16.64
3.84
49.87 pF 


0.7
0.7000
0.7000
15.60
4.35
46.9 pF


0.8
0.7049
0.6516
14.95
4.75
44.9 pF


0.9
0.7026
0.6394
14.69
5.06
42.8 pF


0.99
0.6990
0.636
14.45
5.32
41.4 pF









Using a 90% charge on the oblate spheroid, the antenna parameters become




  Q
  =
  
    2.109
    
      
        (
        kh
        )
      
      3
    
  




he=0.6792h.

Rr=18.46*(kh)2

C=8.56*10−11h

For h=1⁄2 m and 50 MHz, with α=0.9, the minimum Q=14.69 which is 2.109 times Chu's theoretical result. FIG. 20 illustrates a cross-sectional view of an exemplary (1−α)P1+α oblate spheroid charge antenna, using the parameters designed above.
P0 with Oblate Spheroid and 2nd Harmonic on Oblate Spheroid
The oblate spheroid solution can be further improved by redistributing the charge to modify the potential curves and therefore the shape of the surface. This can be calculated by solving the scalar potential equations for the oblate coordinate system. The scalar potential equation in this coordinate system can be expressed as




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                u
                ,
                v
              
              )
            
          
          =
          
            
              ∑
              
                n
                =
                0
              
              ∞
            
            ⁢
            
              
            
            ⁢
            
              
                a
                n
              
              ⁢
              
                
                  K
                  n
                
                ⁡
                
                  (
                  iu
                  )
                
              
              ⁢
              
                
                  P
                  n
                
                ⁡
                
                  (
                  v
                  )
                
              
            
          
        
        ,
      
    
    
      
        (
        51
        )
      
    
  




where Pn are Legendre polynomials, i=√{square root over (−1)}, and the Kn(iu) terms are the far-field dependence functions. Eq. 51 can be tackled using separation of variables in the oblate coordinate system with the solutions for the u dependence obtained from Arfkin's classic text—“Numerical Methods for Physicists,” 2nd Edition, Academic Press, Ohio, 1970.

Eq. 51 with K0(iu), P0(ν) was used in the oblate spheroidal calculation. The second term K1(iu)P1(ν) is an odd function and would change the shape of the oblate spheroid in a non-symmetric manner so that the assumption about the effective height would not be valid. Therefore, it is not used. The third term K2(iu)P2(ν) is symmetric which makes it suitable for our purposes. In Eq. 51, these first and third terms will be used to modify the oblate spheroid's shape. For reference, the first few values of K are




  
    
      
        
          
            K
            0
          
          ⁡
          
            (
            iu
            )
          
        
        =
        
          i
          ⁢
          
            
          
          ⁢
          arc
          ⁢
          
            
          
          ⁢
          
            cot
            ⁡
            
              (
              u
              )
            
          
        
      
    
  
  
    
      
        
          
            K
            1
          
          ⁡
          
            (
            iu
            )
          
        
        =
        
          
            u
            ⁢
            
              
            
            ⁢
            arc
            ⁢
            
              
            
            ⁢
            
              cot
              ⁡
              
                (
                u
                )
              
            
          
          -
          1
        
      
    
  
  
    
      
        
          
            K
            2
          
          ⁡
          
            (
            iu
            )
          
        
        =
        
          
            i
            ⁢
            
              1
              2
            
            ⁢
            
              (
              
                
                  3
                  ⁢
                  
                    u
                    2
                  
                
                +
                1
              
              )
            
            ⁢
            arc
            ⁢
            
              
            
            ⁢
            
              cot
              ⁡
              
                (
                u
                )
              
            
          
          -
          
            i
            ⁢
            
              3
              2
            
            ⁢
            
              u
              .
            
          
        
      
    
  




The general form of these terms being




  
    
      
        
          
            
              K
              n
            
            ⁡
            
              (
              iu
              )
            
          
          =
          
            
              
                -
                
                  
                    iP
                    n
                  
                  ⁡
                  
                    (
                    iu
                    )
                  
                
              
              ⁢
              arc
              ⁢
              
                
              
              ⁢
              
                cos
                ⁡
                
                  (
                  u
                  )
                
              
            
            -
            
              
                
                  
                    2
                    ⁢
                    n
                  
                  -
                  1
                
                n
              
              ⁢
              
                
                  P
                  
                    n
                    -
                    1
                  
                
                ⁡
                
                  (
                  iu
                  )
                
              
            
            -
            
              
                
                  (
                  
                    
                      2
                      ⁢
                      n
                    
                    -
                    5
                  
                  )
                
                
                  3
                  ⁢
                  
                    (
                    
                      n
                      -
                      1
                    
                    )
                  
                
              
              ⁢
              
                
                  
                    P
                    
                      n
                      -
                      3
                    
                  
                  ⁡
                  
                    (
                    iu
                    )
                  
                
                --
              
              ⁢
              
                
              
              ⁢
              ...
            
          
        
        ⁢
        
          
        
        ,
      
    
    
      
        (
        52
        )
      
    
  




where Pn are Legendre polynomials and the sum terminates at the first singular term. For simplicity only the P0(ν) and P2(ν) polynomials were used as they redistribute the charges equally on both sides of the oblate spheroid. None of the high order Pn(ν), n≠0 polynomials change the net charge on the oblate spheriod. Based on this decomposition, the potential field expression becomes





  
    
      
        
          
            Φ
            
              Ob
              ⁢
              
                
              
              ⁢
              2
            
          
          =
          
            
              
                
                  
                    (
                    
                      1
                      -
                      α
                    
                    )
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                  ⁢
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  h
                
              
              ⁢
              
                
              
              ⁢
              ln
              ⁢
              
                
              
              ⁢
              
                (
                
                  
                    
                      (
                      
                        1
                        +
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                  
                    
                      (
                      
                        1
                        -
                        
                          τ
                          m
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        +
                        
                          τ
                          i
                        
                      
                      )
                    
                  
                
                )
              
            
            +
            
              
                
                  α
                  ⁢
                  
                    
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                
              
              ⁡
              
                [
                
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            b
                          
                        
                        )
                      
                    
                  
                  -
                  
                    arc
                    ⁢
                    
                      
                    
                    ⁢
                    
                      tan
                      ⁡
                      
                        (
                        
                          sinh
                          ⁢
                          
                            
                          
                          ⁢
                          
                            u
                            t
                          
                        
                        )
                      
                    
                  
                
                ]
              
            
            +
            
              
                
                  β
                  ⁢
                  
                    
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    
                  
                  ⁢
                  
                    πɛ
                    0
                  
                
              
              ⁡
              
                [
                
                  
                    
                      (
                      
                        
                          
                            (
                            
                              
                                3
                                ⁢
                                
                                  u
                                  t
                                  2
                                
                              
                              +
                              1
                            
                            )
                          
                          ⁢
                          arc
                          ⁢
                          
                            
                          
                          ⁢
                          
                            cot
                            ⁡
                            
                              (
                              
                                u
                                t
                              
                              )
                            
                          
                        
                        -
                        
                          
                            3
                            2
                          
                          ⁢
                          
                            u
                            t
                          
                        
                      
                      )
                    
                    ⁢
                    
                      
                        P
                        2
                      
                      ⁡
                      
                        (
                        
                          v
                          t
                        
                        )
                      
                    
                  
                  -
                  
                    
                      (
                      
                        
                          
                            (
                            
                              
                                3
                                ⁢
                                
                                  u
                                  b
                                  2
                                
                              
                              +
                              1
                            
                            )
                          
                          ⁢
                          arc
                          ⁢
                          
                            
                          
                          ⁢
                          
                            cot
                            ⁡
                            
                              (
                              
                                u
                                b
                              
                              )
                            
                          
                        
                        +
                        
                          
                            3
                            2
                          
                          ⁢
                          
                            u
                            b
                          
                        
                      
                      )
                    
                    ⁢
                    
                      
                        P
                        2
                      
                      ⁡
                      
                        (
                        
                          v
                          b
                        
                        )
                      
                    
                  
                
                ]
              
            
          
        
        ,
      
    
    
      
        (
        53
        )
      
    
  




where β represents the coefficient for the P2 polynomial, νt is for the top oblate spheroid and νb is for the bottom oblate spheroid.

The effective height becomes heff=(1−α)κh/2+ακh and the radiation resistance is Rr=10[κhk(1+α)]2. Using procedures previously described, Q is numerically calculated. It is noted that the addition of the second order term P2 modifies the source charge on the oblate spheroid to where it could be negative at the center of the oblate spheroid or at the edge of the oblate spheroid. A program verified that an equipotential surface encloses all the charge in two areas: at ρ=0 on z axis and at edge of the oblate spheroid.
From the above derivations, the Q/QChu ratio is a function of κ, α, β, and a is plotted in FIG. 21. The parameter β is calculated to obtain the minimum value of Q. From FIG. 21, the minimum Q is found at κ=0.679, a=0.734, and β=−0.321 for a 90% charge distribution on the oblate spheroid. At 50 MHz, the absolute values of the antenna at this minimum is Q=13.311, R=4.563, and C=52.41 pF. The resulting antenna parameters, as a function of k and h become




  Q
  =
  
    1.9107
    
      
        (
        kh
        )
      
      3
    
  




he=0.64505h.

Rr=16.644*(kh)2

C=1.0482*10−10h

FIG. 22 is a plot of the β distribution as a function of κ and a for a minimum Q. FIG. 23 demonstrates the cross-sectional shape of the exemplary (1−α)P0+α oblate spheroid+β second harmonic on oblate spheroid charge antenna. It is interesting to note that the shape of this antenna is practically mushroom-like.
Non-Spherical Enclosing Volume
Using the previous example, the exemplary (1−α)P0+α oblate spheroid+β second harmonic on oblate spheroid charge antenna was constrained to fit within an inverted 45° cone where the angle is measured with respect to the vertical. Mathematically, this is expressed as ρ=1−z. The capacitance is calculated from the largest value of the potential on this cone. It should be noted, however, any enclosing surface may be used to arrive at a correct solution, therefore an appropriate solution is not limited to the use of 45° or a cone, but is used here as just one example of many approaches that may be used. Using the inverted 45° cone, for example, a plot was generated of the Q/QChu ratio as a function of κ, α, β, and a as shown in FIG. 24. The parameter β is calculated to obtain the minimum value of Q. The conical volume is much small than the spherical volume and therefore the Q is understood to be much larger. Using higher resolution parameter steps of 0.001, the antenna parameters were recalculated resulting in Q=34.437 at 50 MHz, Q/QChu=4.9433, R=3.3514 , C=27.58 pF for κ=0.567 , a=0.418, β=−0.442 with 95% of the charge on the top load. The numerical noise at this resolution is about ±0.0005 for the value of Q/QChu. The top load parameter α was used to modify the antenna shape to fit our geometric constraints. As can be seen from the above results, the Q value is significantly less than the first ACD antenna's Q value. FIG. 25 is a plot of the β distribution as a function of κ and a for a minimum Q. FIG. 26 is an illustration of the contour of an exemplary conical enclosing volume (1−α)P0+α oblate spheroid+β second harmonic on oblate spheroid charge antenna.
A model of this antenna can be generated by using wire approximations. In general, this antenna can be approximated using 45° wires for the top portion with a wire tangential to the bottom surface of the antenna. Eight sets of wires can be used to approximate this antenna, with horizontal wires to connect the appropriate vertices. FIG. 27 is a 3-D rendering of a wire model of an exemplary 45 degree (1−α)P0+α oblate spheroid+β second harmonic on oblate spheroid charge antenna. Though this antenna initially appears similar to prior art antennas, the subtle differences in its shape provides a superior performance.
Vertical Dipole
In the above examples, the antennas under study were principally monopole antennas of one form or another. In this embodiment, a dipole antenna is discussed. Of particular concern in electrical small antenna design is the loss caused by ground currents, which is often exacerbated in the case of a monopole antenna. The use of a dipole antenna reduces ground currents. The potential field for a dipole is calculated in principally the same manner as in the monopole case. The dipole is centered at z=h/2 and the image dipole is centered at z=−h/2. The length of the antenna arm is L=κh/2.
The calculation can be divided into two parts. The first part is with the dipole in free space, and the second part is with the image dipole. The top arm of the dipole is excited at a voltage. The other arm of the dipole is excited with the ground of the coaxial cable. The general potential field equation for such a structure is




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              q
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                L
              
            
            ⁢
            
              
            
            ⁢
            ln
            ⁢
            
              
            
            ⁢
            
              (
              
                
                  1
                  +
                  τ
                
                
                  1
                  -
                  τ
                
              
              )
            
          
        
        ,
      
    
    
      
        (
        54
        )
      
    
  




For the free space dipole, the potential field is




  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                2
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              
            
            ⁢
            ln
            ⁢
            
              
            
            ⁢
            
              (
              
                
                  
                    (
                    
                      1
                      +
                      
                        τ
                        t
                      
                    
                    )
                  
                  ⁢
                  
                    (
                    
                      1
                      -
                      
                        τ
                        b
                      
                    
                    )
                  
                
                
                  
                    (
                    
                      1
                      -
                      
                        τ
                        t
                      
                    
                    )
                  
                  ⁢
                  
                    (
                    
                      1
                      +
                      
                        τ
                        b
                      
                    
                    )
                  
                
              
              )
            
          
        
        ,
      
    
    
      
        (
        55
        )
      
    
  




where





  
    τ
    t
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        R
        f
      
      +
      
        R
        t
      
    
  




is for the monopole,





  
    τ
    b
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        R
        f
      
      +
      
        R
        b
      
    
  




is for the image monopole, Rt=√{square root over ((z−κh/2−h/2)2+ρ2)} is the distance from ρ, z to the top of the monopole, Rf=√{square root over ((z−h/2)2+ρ2)} is the distance from ρ, z to the feed point and Rb=√{square root over ((z−h/2+κh/2)2+ρ2)} is the distance from ρ, z to the bottom of the dipole. The capacitance is computed by calculating the potential or voltage at ρ=0 and z=h. The above equations simplify to Rt=(z−h/2−κh/2), Rf=z−h/2, Rb=(z−h/2+κh/2),





  
    
      τ
      m
    
    =
    
      
        κ
        ⁢
        
          
        
        ⁢
        
          h
          /
          2
        
      
      
        
          2
          ⁢
          z
        
        -
        h
        -
        
          κ
          ⁢
          
            
          
          ⁢
          
            h
            /
            2
          
        
      
    
  
  ,




and





  
    τ
    i
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        2
        ⁢
        z
      
      -
      h
      +
      
        κ
        ⁢
        
          
        
        ⁢
        
          h
          /
          2
        
      
    
  




for z≧κh, resulting in





  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              z
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
          
          ⁢
          ln
          ⁢
          
            
          
          ⁢
          
            
              (
              
                
                  
                    (
                    
                      
                        2
                        ⁢
                        z
                      
                      -
                      h
                    
                    )
                  
                  2
                
                
                  
                    (
                    
                      
                        2
                        ⁢
                        z
                      
                      -
                      h
                      -
                      
                        κ
                        ⁢
                        
                          
                        
                        ⁢
                        h
                      
                    
                    )
                  
                  ⁢
                  
                    (
                    
                      
                        2
                        ⁢
                        z
                      
                      -
                      h
                      +
                      
                        κ
                        ⁢
                        
                          
                        
                        ⁢
                        h
                      
                    
                    )
                  
                
              
              )
            
            .
          
        
      
    
    
      
        (
        56
        )
      
    
  




Eq. 57 can be further simplified to yield




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
          
          ⁢
          ln
          ⁢
          
            
          
          ⁢
          
            
              (
              
                1
                
                  (
                  
                    1
                    -
                    
                      κ
                      2
                    
                  
                  )
                
              
              )
            
            .
          
        
      
    
    
      
        (
        57
        )
      
    
  




For the image dipole




  
    τ
    ti
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        R
        f
      
      +
      
        R
        t
      
    
  




is for the monopole,





  
    τ
    bi
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        R
        f
      
      +
      
        R
        b
      
    
  




is for the image dipole, Rt=√{square root over ((z−κh/2+h/2)2+ρ2)} is the distance from ρ, z to the top of the image dipole, Rf=√{square root over ((z+h/2)2+ρ2)} is the distance from ρ, z to the feed point of the image and Rb=√{square root over ((z+h/2+κh/2)2+ρ2)} is the distance from ρ, z to the bottom of the image dipole. The capacitance is computed by calculating the potential or voltage at ρ=0 and z=h. The above equations simplify to Rti=(z+h/2−κh/2), Rfi=z+h/2, Rbi=(z+h/2+κh/2),





  
    
      τ
      ti
    
    =
    
      
        κ
        ⁢
        
          
        
        ⁢
        
          h
          /
          2
        
      
      
        
          2
          ⁢
          z
        
        +
        h
        -
        
          κ
          ⁢
          
            
          
          ⁢
          
            h
            /
            2
          
        
      
    
  
  ,




and





  
    τ
    bi
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        2
        ⁢
        z
      
      +
      h
      +
      
        κ
        ⁢
        
          
        
        ⁢
        
          h
          /
          2
        
      
    
  




for z≧κh. The resulting expression for the potential field for the image dipole becomes





  
    
      
        
          
            Φ
            ⁡
            
              (
              
                0
                ,
                z
              
              )
            
          
          =
          
            
              
                2
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            ⁢
            
              ln
              ⁡
              
                (
                
                  
                    
                      (
                      
                        
                          2
                          ⁢
                          z
                        
                        +
                        h
                      
                      )
                    
                    2
                  
                  
                    
                      (
                      
                        
                          2
                          ⁢
                          z
                        
                        +
                        h
                        -
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        
                          2
                          ⁢
                          z
                        
                        +
                        h
                        +
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                      
                      )
                    
                  
                
                )
              
            
          
        
        ,
      
    
    
      
        (
        58
        )
      
    
  




which can be further simplified to





  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              π
              ⁢
              
                
              
              ⁢
              
                ɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              ln
              ⁡
              
                (
                
                  9
                  
                    (
                    
                      9
                      -
                      
                        κ
                        2
                      
                    
                    )
                  
                
                )
              
            
            .
          
        
      
    
    
      
        (
        59
        )
      
    
  




The combined potential from Eqs. 57 and 59 is




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              π
              ⁢
              
                
              
              ⁢
              
                ɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              ln
              ⁡
              
                (
                
                  
                    1
                    
                      1
                      -
                      
                        κ
                        2
                      
                    
                  
                  ⁢
                  
                    9
                    
                      9
                      -
                      
                        κ
                        2
                      
                    
                  
                
                )
              
            
            .
          
        
      
    
    
      
        (
        60
        )
      
    
  




The capacitance become




  
    
      
        
          C
          ⁡
          
            (
            κ
            )
          
        
        =
        
          
            
              4
              ⁢
              
                πɛ
                0
              
              ⁢
              κ
              ⁢
              
                
              
              ⁢
              h
            
            
              
                -
                2
              
              ⁢
              
                
              
              ⁢
              
                ln
                ⁡
                
                  (
                  
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          
                            κ
                            2
                          
                          /
                          9
                        
                      
                      )
                    
                  
                  )
                
              
            
          
          .
        
      
    
    
      
        (
        61
        )
      
    
  




And the equation for radiation resistance becomes




  
    
      
        
          
            R
            r
          
          =
          
            40
            ⁢
            
              
                
                  k
                  2
                
                [
                
                  
                    
                      ∫
                      0
                      h
                    
                    ⁢
                    
                      
                        zq
                        ⁡
                        
                          (
                          x
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        
                          d
                          3
                        
                        ⁢
                        x
                      
                    
                  
                  
                    
                      ∫
                      fp
                      h
                    
                    ⁢
                    
                      
                        q
                        ⁡
                        
                          (
                          x
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        
                          d
                          3
                        
                        ⁢
                        x
                      
                    
                  
                
                ]
              
              2
            
          
        
        ,
      
    
    
      
        (
        62
        )
      
    
  




where q(z) is





  
    
      
        
          
            
              
                q
                ⁡
                
                  (
                  z
                  )
                
              
              =
              
                
                  0
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  0
                
                ≤
                z
                <
                
                  
                    
                      1
                      -
                      κ
                    
                    2
                  
                  ⁢
                  h
                
              
            
          
        
        
          
            
              
                q
                ⁡
                
                  (
                  z
                  )
                
              
              =
              
                
                  
                    -
                    2
                  
                  ⁢
                  
                    
                      q
                      0
                    
                    /
                    κ
                  
                  ⁢
                  
                    
                  
                  ⁢
                  h
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      1
                      -
                      κ
                    
                    2
                  
                  ⁢
                  h
                
                ≤
                z
                <
                
                  h
                  2
                
              
            
          
        
        
          
            
              
                q
                ⁡
                
                  (
                  z
                  )
                
              
              =
              
                
                  2
                  ⁢
                  
                    
                      q
                      0
                    
                    /
                    κ
                  
                  ⁢
                  
                    
                  
                  ⁢
                  h
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  
                    h
                    2
                  
                
                ≤
                z
                ≤
                
                  
                    
                      1
                      +
                      κ
                    
                    2
                  
                  ⁢
                  h
                
              
            
          
        
        
          
            
              
                q
                ⁡
                
                  (
                  z
                  )
                
              
              =
              
                
                  0
                  ⁢
                  
                    
                  
                  ⁢
                  for
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      1
                      +
                      κ
                    
                    2
                  
                  ⁢
                  h
                
                <
                z
                ≤
                
                  h
                  .
                
              
            
          
        
      
    
    
      
        (
        63
        )
      
    
  




Using a change in variables,




  
    z
    =
    
      x
      +
      
        h
        2
      
    
  
  ,




and performing the following steps:





  
    
      
        
          
            ∫
            0
            h
          
          ⁢
          
            
              zq
              ⁡
              
                (
                z
                )
              
            
            ⁢
            
              
            
            ⁢
            
              d
              z
            
          
        
        =
        
          
            
              ∫
              0
              
                κ
                ⁢
                
                  
                
                ⁢
                
                  h
                  /
                  2
                
              
            
            ⁢
            
              2
              ⁢
              
                
                  q
                  0
                
                /
                κ
              
              ⁢
              
                
              
              ⁢
              
                h
                ⁡
                
                  (
                  
                    
                      h
                      2
                    
                    +
                    x
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                x
              
            
          
          -
          
            
              ∫
              
                
                  -
                  κ
                
                ⁢
                
                  
                
                ⁢
                
                  h
                  /
                  2
                
              
              0
            
            ⁢
            
              2
              ⁢
              
                
                  q
                  0
                
                /
                κ
              
              ⁢
              
                
              
              ⁢
              
                h
                ⁡
                
                  (
                  
                    
                      h
                      2
                    
                    +
                    x
                  
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                x
              
            
          
        
      
    
  
  
    
      
        
          
            ∫
            0
            h
          
          ⁢
          
            
              zq
              ⁡
              
                (
                z
                )
              
            
            ⁢
            
              
            
            ⁢
            
              d
              z
            
          
        
        =
        
          2
          ⁢
          
            
              q
              0
            
            /
            κ
          
          ⁢
          
            
          
          ⁢
          
            h
            ⁡
            
              (
              
                
                  
                    [
                    
                      
                        hx
                        2
                      
                      +
                      
                        
                          x
                          2
                        
                        2
                      
                    
                    ]
                  
                  0
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      h
                      /
                      2
                    
                  
                
                -
                
                  
                    [
                    
                      
                        hx
                        2
                      
                      +
                      
                        
                          x
                          2
                        
                        2
                      
                    
                    ]
                  
                  
                    
                      -
                      κ
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      h
                      /
                      2
                    
                  
                  0
                
              
              )
            
          
        
      
    
  
  
    
      
        
          
            ∫
            0
            h
          
          ⁢
          
            
              zq
              ⁡
              
                (
                z
                )
              
            
            ⁢
            
              
            
            ⁢
            
              d
              z
            
          
        
        =
        
          2
          ⁢
          
            
              q
              0
            
            /
            κ
          
          ⁢
          
            
          
          ⁢
          
            h
            ⁡
            
              (
              
                
                  ⌊
                  
                    
                      
                        κ
                        ⁢
                        
                          
                        
                        ⁢
                        
                          h
                          2
                        
                      
                      4
                    
                    +
                    
                      
                        
                          (
                          
                            κ
                            ⁢
                            
                              
                            
                            ⁢
                            h
                          
                          )
                        
                        2
                      
                      8
                    
                  
                  ⌋
                
                +
                
                  ⌊
                  
                    
                      -
                      
                        
                          κ
                          ⁢
                          
                            
                          
                          ⁢
                          
                            h
                            2
                          
                        
                        4
                      
                    
                    +
                    
                      
                        
                          (
                          
                            κ
                            ⁢
                            
                              
                            
                            ⁢
                            h
                          
                          )
                        
                        2
                      
                      8
                    
                  
                  ⌋
                
              
              )
            
          
        
      
    
  
  
    
      
        
          
            ∫
            0
            h
          
          ⁢
          
            
              zq
              ⁡
              
                (
                z
                )
              
            
            ⁢
            
              
            
            ⁢
            
              d
              z
            
          
        
        =
        
          
            q
            0
          
          ⁢
          κ
          ⁢
          
            
          
          ⁢
          
            h
            /
            2
          
        
      
    
  
  
    
      
        
          
            
              ∫
              fp
              h
            
            ⁢
            
              
                q
                ⁡
                
                  (
                  z
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                d
                z
              
            
          
          =
          
            
              
                ∫
                0
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  
                    h
                    /
                    2
                  
                
              
              ⁢
              
                2
                ⁢
                
                  
                    q
                    0
                  
                  /
                  κ
                
                ⁢
                
                  
                
                ⁢
                h
                ⁢
                
                  
                
                ⁢
                
                  d
                  x
                
              
            
            =
            
              q
              0
            
          
        
        ,
      
    
  




results in Eq. 62 becoming





  
    
      
        
          R
          e
        
        =
        
          40
          ⁢
          
            
              
                (
                
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    hk
                  
                  2
                
                )
              
              2
            
            .
          
        
      
    
    
      
        (
        64
        )
      
    
  




From the derivations shown above, the Q is found to be




  
    
      
        Q
        =
        
          
            
              
                -
                12
              
              ⁢
              
                
              
              ⁢
              
                ln
                ⁡
                
                  (
                  
                    
                      (
                      
                        1
                        -
                        
                          κ
                          2
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          
                            κ
                            2
                          
                          /
                          9
                        
                      
                      )
                    
                  
                  )
                
              
            
            
              
                (
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  hk
                
                )
              
              3
            
          
          .
        
      
    
    
      
        (
        65
        )
      
    
  




The minimum value of Q is at κ=0.7437706. The resulting other antenna parameter values become

Q=12.67326/ka3

heff=0.37188*h.

Rr=5.5319(ka)2

C=4.75458*10−11*h

Based on the above values, an exemplary dipole antenna having a minimum Q was designed. FIG. 28 is a illustration of the contour of an exemplary dipole, using the above parameters. It should be noted that the geometry of this exemplary dipole antenna approximates that of a prolate spheroid centered above the hemisphere of an oblate spheroid.
It should also be noted that, generally speaking, a cylindrically symmetric enclosing volume is the simplest case for monopole or linear-type antennas. However, as alluded above, the enclosing volume can be of very general shapes, in addition to a cylindrically symmetric volume. For example, a rectangular enclosing volume could be used. Various harmonic or non-harmonic basis functions could be used for the top-load. Complex line charge distributions could be used as elements.
Accordingly, by choosing a different enclosing volume shape, and tailoring basis functions to match that shape, it is possible to design a “flatter” antenna with a low Q. Flat antennas are of particular interest in the field of printed circuit card antennas or conformal antennas. For example, a possible design approach for a printed circuit card antenna is to project a three dimensional cylindrically symmetric antenna to a two dimensional antenna. (The two dimensional antenna is a slice of the three dimensional cylindrically symmetric antenna that includes the axis of rotation of the three dimensional cylindrically symmetric antenna.) Now, a spherical monopole becomes a disk monopole, as a gross approximation.
It should be also understood that a low Q, flatter antenna will have most of the charge on its edge. If one slices the flatter antenna to include the largest area, this will better approximate a printed circuit card antenna. The approach could improve the performance of the wire antenna, which was discussed above. The antenna could be any size and mounted on any object.
FIG. 29 is a flow chart illustrating an exemplary process in accordance with an embodiment of this invention. The exemplary process 100 of FIG. 29 enables one of ordinary skill in the art to design electrically small antennas to have a desired Q as a function of a dimensional variable. By evaluating the Q as a function of the dimensional variable, the electrical parameters and contour(s) of a desired Q factor antenna can be designed. Judicially selecting the dimensional variable's value, antennas of varying Q's can be designed. In addition, by minimizing the Q-dimensional variable relationship, a minimum Q antenna can be arrived at.
The exemplary process 100 begins 10 with designation of an electrically small antenna geometry 15 or a form factor corresponding to the antenna geometry 15. This designation is of general form and may simply be the establishment of a coordinate system that is consistent with the geometry chosen. Based on the type of antenna geometry 15 or form factor chosen, the exemplary process 100 incorporates a suitable dimensional variable 20 or a form factor related parameter. It is not necessary that the dimensional variable 20 or form factor related parameter be of a form that is consistent with the geometry's coordinate system, however, it would be preferred as this would obviate the need for a dimensional transformation. Proceeding from the dimensional variable 20 incorporation, the exemplary process 100 formulates quasi-static field equations 25. The quasi-static field equations 25 entail the application of Maxwell's formulas to result in a potential or scalar field equation(s) that relates the field intensities at field point(s) as a function of the source charge distribution.
Next, from the quasi-static field equation formulation 25 (and if the charge distribution is not already included), the exemplary process 100 impresses a given charge distribution 30 within the quasi-static field equation formulation 25. The given charge distribution 30 may be a single “type” charge distribution or of multiple “types” such as from differing basis functions. The charge distribution 30 may be different for different parts of the antenna geometry 15.
It should be appreciated that depending on the complexity of the charge distribution 30 and the quasi-static field formulation 25, the order of the exemplary process 100 for these respective steps may be reversed, without departing from the spirit and scope of this invention. Accordingly, the exemplary process 100 may first impress a given charge distribution 30 in accordance with the antenna geometry 15 or form factor chosen, and then proceed to generate the quasi-static field equations 25, without loss of generality.
Presuming that the charge distribution 30 is performed subsequent to the quasi-static field equation formulation 25, the exemplary process 100 proceeds onto a dimensional variable-reduced expression 35 of the quasi-static field equation formulation 25. The dimensional variable-reduced expression 35 may simply be a reduced algebraic expression or a form that relates the Q of the antenna geometry in terms of the dimensional variable (form factor related parameter) or may take the form of another antenna parameter, such as the capacitance, for example. The capacitance is expressible in terms of the charge distribution and the dimensional variable (form factor related parameter). If expressed in terms of the capacitance, this dimensional variable-reduced expression 35 can be used to generate a Q expression, by appropriate use of a corresponding radiation resistance calculation.
The Q expression can then be solved 40 using a minimization and/or numerical technique to find a minimum Q or dimensional variable-to-Q correspondence. A plot of the computed capacitance and radiation resistance can be generated to correlate to given or desired Q values. By evaluating Q, the capacitance, the radiation resistance, and selecting the appropriate dimensional value (form factor relater parameter), design decisions can be made to meet the performance requirements of an antenna. The solved equation 40 will provide a value(s) of the dimensional variable that generally operate to define the Q and geometry of a corresponding antenna. The exemplary process 100 then terminates at 45. The results of this exemplary process 100 can then be used to physically build or model an antenna that demonstrates the Q performance chosen.
Now that the details of the above cross-referenced application have been shown and described, the details of an analytic antenna design for a dipole antenna by eliminating or reducing antenna pattern nulls and impedance anti-resonances of the present invention will now be described and shown.
Existing antenna designs typically have anti-resonances in the impedance and nulls in the radiation pattern. An objective is to eliminate (or move to a higher frequency) one or more of the anti-resonances in the impedance and one or more of the nulls in the radiation pattern. The radiation pattern would be a dipole pattern over an extended frequency range. The performance of the algorithm should extend beyond a one wavelength high monopole antennas and potentially extends to ultra wide bandwidth antennas. The cross-referenced patent application entitled “Quasi-Static Design Approach For Low Q Factor Electrically Small Antennas”, or more simply “Quasi-static antenna design algorithm”, can be utilized to perform the designs. The algorithm will calculate the quasi-static Q, radiation resistance and capacitance. This design will not require added resistance to create wide bandwidth antennas.
Howard Stuart shows that the DC capacitance is the sum of the effective capacitances of the eigenmodes of the antenna. It is believed that each eigenmode is uniquely linked to the electrostatic spherical harmonic. The eigenmodes are analogous to resonant modes in a radio frequency cavity that decay via resistive loss in the conducing walls. The antenna eigenmodes decay via radiation loss. The ideal antenna would have a high radiation loss or a small Q for each eigenmode. Each eigenmode has a unique radiation pattern; the first eigenmode is a dipole radiation pattern. The higher order eigenmodes are odd vector spherical harmonics.
The impedance of each eigenmode is equivalent to a series LC with a resistor across the inductor. The antenna impedance is a parallel combination of all of the eigenmode equivalent circuits. Stuart pointed out that the DC capacitance was a sum of the effective capacitances from each eigenmode (At DC, the inductors short the resistors.). Stuart modeled the second eigenmode as a capacitance and neglected the higher order eigenmodes. Data for the second eigenmode was selected to illustrate the concept. The higher order eigenmode were neglected. Table I shows the combined data for a disk loaded monopole.







TABLE 1







Eigenmodes for a disk loaded monopole














Res. Freq.
Type
Q
R0
R
L
C
Vector Spherical Harmonics






DC

8.96


12.0 pF



 


220.4 MHz
Dipole
9.08
8.96
738.7
58.7 nH
8.88 pF




  
    
      3
      
        8
        ⁢
        
          
        
        ⁢
        π
      
    
  
  ⁢
  
    sin
    ⁡
    
      (
      θ
      )
    
  






 


 1500 MHz
Octipole
3 est
11 est
105 est
3.61 nH
3.12 pF




  
    
      21
      
        64
        ⁢
        
          
        
        ⁢
        π
      
    
  
  ⁡
  
    [
    
      
        5
        ⁢
        
          
            sin
            3
          
          ⁡
          
            (
            θ
            )
          
        
      
      -
      
        4
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        θ
      
    
    ]
  













The radiation pattern at any given frequency is a linear combination of the eigenmode radiation patterns weighted with the current flow in each mode. FIG. 32 shows the magnitude of the current flow




  
    
    
      I
      1
    
    
  
  =
  
    
      V
      0
    
    
      
      
        Z
        1
      
      
    
  




and





  
    
    
      I
      2
    
    
  
  =
  
    
      V
      0
    
    
      
      
        Z
        2
      
      
    
  




for eigenmode 1 and 2, respectively. The impedance for eigenmode 1 is





  
    Z
    1
  
  =
  
    
      -
      
        j
        
          ω
          ⁢
          
            
          
          ⁢
          
            C
            1
          
        
      
    
    +
    
      
        j
        ⁢
        
          
        
        ⁢
        ω
        ⁢
        
          
        
        ⁢
        
          L
          1
        
        ⁢
        
          R
          1
        
      
      
        
          j
          ⁢
          
            
          
          ⁢
          ω
          ⁢
          
            
          
          ⁢
          
            L
            1
          
        
        +
        
          R
          1
        
      
    
  




where the values of C1=8.88 pF, R1=738.7 and L1=58.7 nH.





  
    Z
    1
  
  =
  
    
      
        -
        
          j
          
            ω
            ⁢
            
              
            
            ⁢
            
              C
              1
            
          
        
      
      +
      
        
          
            j
            ⁢
            
              
            
            ⁢
            ω
            ⁢
            
              
            
            ⁢
            
              L
              1
            
            ⁢
            
              R
              1
            
          
          
            
              j
              ⁢
              
                
              
              ⁢
              ω
              ⁢
              
                
              
              ⁢
              
                L
                1
              
            
            +
            
              R
              1
            
          
        
        ⁢
        
          
            
              
                -
                j
              
              ⁢
              
                
              
              ⁢
              ω
              ⁢
              
                
              
              ⁢
              
                L
                1
              
            
            +
            
              R
              1
            
          
          
            
              
                -
                j
              
              ⁢
              
                
              
              ⁢
              ω
              ⁢
              
                
              
              ⁢
              
                L
                1
              
            
            +
            
              R
              1
            
          
        
      
    
    =
    
      
        
          
            ω
            2
          
          ⁢
          
            L
            1
            2
          
          ⁢
          
            R
            1
          
        
        
          
            
              (
              
                ω
                ⁢
                
                  
                
                ⁢
                
                  L
                  1
                
              
              )
            
            2
          
          +
          
            R
            1
            2
          
        
      
      -
      
        j
        
          ω
          ⁢
          
            
          
          ⁢
          
            C
            1
          
        
      
      +
      
        
          j
          ⁢
          
            
          
          ⁢
          ω
          ⁢
          
            
          
          ⁢
          
            L
            1
          
          ⁢
          
            R
            1
            2
          
        
        
          
            
              (
              
                ω
                ⁢
                
                  
                
                ⁢
                
                  L
                  1
                
              
              )
            
            2
          
          +
          
            R
            1
            2
          
        
      
    
  




resonance frequency for eigenmode 1 occurs when





  
    1
    
      
        ω
        1
      
      ⁢
      
        C
        1
      
    
  
  =
  
    
      
        
          
            ω
            1
          
          ⁢
          
            L
            1
          
          ⁢
          
            R
            1
            2
          
        
        
          
            
              (
              
                
                  ω
                  1
                
                ⁢
                
                  L
                  1
                
              
              )
            
            2
          
          +
          
            R
            1
            2
          
        
      
      ⁢
      
        
      
      ⁢
      or
      ⁢
      
        
      
      ⁢
      
        ω
        1
        2
      
    
    =
    
      
        
          R
          1
          2
        
        
          (
          
            
              
                C
                1
              
              ⁢
              
                L
                1
              
              ⁢
              
                R
                1
                2
              
            
            -
            
              L
              1
              2
            
          
          )
        
      
      .
    
  




The anti-resonance is caused by the cancellation of the combined eigenmode currents (the reactive part) at about 400 MHz. The eigenmodes are in parallel and have the same V0. The impedance is calculated from the equation




  
    
      
        
          I
          0
        
        =
        
          
            
              V
              0
            
            
              Z
              0
            
          
          =
          
            
              I
              1
            
            +
            
              I
              2
            
          
        
      
    
  
  
    
      
        
          I
          1
        
        =
        
          
            
              
                V
                0
              
              
                Z
                1
              
            
            ⁢
            
              
            
            ⁢
            and
            ⁢
            
              
            
            ⁢
            
              I
              2
            
          
          =
          
            
              V
              0
            
            
              Z
              2
            
          
        
      
    
  




The above equation is




  
    
      V
      0
    
    
      Z
      0
    
  
  =
  
    
      
        
          V
          0
        
        
          Z
          1
        
      
      +
      
        
          
            V
            0
          
          
            Z
            2
          
        
        ⁢
        
          
        
        ⁢
        or
        ⁢
        
          
        
        ⁢
        
          Z
          0
        
      
    
    =
    
      
        
          
            Z
            1
          
          *
          
            Z
            2
          
        
        
          
            Z
            1
          
          +
          
            Z
            2
          
        
      
      .
    
  




FIG. 32 shows a plot of the impedance. The cancellation of the reactive current gives a high input resistance 400 Ω and high negative slope in the reactance (anti-resonance). The high impedance values will create a high reflection from the antenna input (assuming a 50 Ω source impedance). Antenna matching an antenna at anti-resonance is very difficult.
The following is an analysis of Stuart's data. At DC both the first and second eigenmode contributes to the radiation pattern. Table 2 (column 3) shows the quasi-static current flow at the feed point. Each vector spherical harmonic reduces to a unique electrostatic spherical harmonic in the quasi-static solution. It is believed that each eigenmode contributes a unique electrostatic spherical harmonic to the quasi-static solution. Thus the quasi-static solution contains information on the high frequency eigenmodes of the antenna.







TABLE 2







Quasi-static solution for a disk loaded monopole











ω << ω1 < ω2
Type
Current
C
Spherical Harmonics





DC


12.0 pF



Eigenmode 1
Dipole
0.74 * I 0
8.88 pF




  
    
      3
      
        4
        ⁢
        
          
        
        ⁢
        π
      
    
  
  ⁢
  
    cos
    ⁡
    
      (
      θ
      )
    
  






 


Eigenmode 2
Octipole
0.26 * I 0
3.12 pF




  
    
      7
      
        4
        ⁢
        
          
        
        ⁢
        π
      
    
  
  ⁡
  
    [
    
      
        
          5
          2
        
        ⁢
        
          
            cos
            3
          
          ⁡
          
            (
            θ
            )
          
        
      
      -
      
        
          3
          2
        
        ⁢
        cos
        ⁢
        
          
        
        ⁢
        θ
      
    
    ]
  













The potential for a charge distribution is




  
    ψ
    ⁡
    
      (
      
        r
        ,
        θ
        ,
        φ
      
      )
    
  
  =
  
    
      1
      
        4
        ⁢
        
          
        
        ⁢
        
          πɛ
          0
        
      
    
    ⁢
    
      ∫
      
        ∫
        
          ∫
          
            
              q
              ⁡
              
                (
                
                  
                    r
                    ′
                  
                  ,
                  
                    θ
                    ′
                  
                  ,
                  
                    φ
                    ′
                  
                
                )
              
            
            ⁢
            
              1
              
                
                
                  r
                  -
                  
                    r
                    ′
                  
                
                
              
            
            ⁢
            sin
            ⁢
            
              
            
            ⁢
            θ
            ⁢
            
              d
              
                r
                ′
              
            
            ⁢
            
              d
              
                θ
                ′
              
            
            ⁢
            
              d
              
                φ
                ′
              
            
          
        
      
    
  




|r−r′| is the distance between the two vectors r and r4 The general potential can be expanded as the sum of





  
    ψ
    ⁡
    
      (
      
        r
        ,
        θ
        ,
        φ
      
      )
    
  
  =
  
    
      
        ∑
        
          l
          =
          1
        
        ∞
      
      ⁢
      
        
      
      ⁢
      
        
          ∑
          
            m
            =
            0
          
          
            m
            =
            l
          
        
        ⁢
        
          
        
        ⁢
        
          
            
              A
              lm
            
            
              r
              
                l
                +
                1
              
            
          
          ⁢
          
            
              Y
              lm
            
            ⁡
            
              (
              
                θ
                ,
                φ
              
              )
            
          
          ⁢
          
            
          
          ⁢
          for
          ⁢
          
            
          
          ⁢
          r
        
      
    
    >
    a
  




if the charge density is zero in this region, r>a. On this surface the ψ(a,θ,φ) has a maximum value ψmax. An equi-potential surface can be define with the potential ψmax.

If this surface encloses all of the charge inside the sphere than the surface is identical to a perfect conductor. In addition, ψ(r,θ,φ) is odd about θ=π/2 (image charge distribution). The image charge distribution has a potential −ψmax. Typically, the feedpoint(s) is in the plane θ=π/2. The general solution could give isolated perfect conductors; implicit assumption is that isolated conductors are connected by very thin wire with neglible charge. The most practical solutions have one perfect conductor and image.



  
    
      ∫
      
        ψ
        ⁢
        
          (
          
            a
            ,
            θ
            ,
            φ
          
          )
        
        ⁢
        
          
            Y
            
              
                l
                ′
              
              ⁢
              
                m
                ′
              
            
          
          ⁡
          
            (
            
              θ
              ,
              φ
            
            )
          
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        θ
        ⁢
        
          d
          θ
        
        ⁢
        
          
        
        ⁢
        
          d
          φ
        
      
    
    =
    
      
        ∑
        
          l
          =
          1
        
        ∞
      
      ⁢
      
        
      
      ⁢
      
        
          ∑
          
            m
            =
            0
          
          
            m
            =
            l
          
        
        ⁢
        
          
        
        ⁢
        
          
            
              A
              lm
            
            
              a
              
                l
                +
                1
              
            
          
          ⁢
          
            ∫
            
              
                
                  Y
                  lm
                
                ⁡
                
                  (
                  
                    θ
                    ,
                    φ
                  
                  )
                
              
              ⁢
              
                
                  Y
                  
                    
                      l
                      ′
                    
                    ⁢
                    
                      m
                      ′
                    
                  
                
                ⁡
                
                  (
                  
                    θ
                    ,
                    φ
                  
                  )
                
              
              ⁢
              sin
              ⁢
              
                
              
              ⁢
              θ
              ⁢
              
                d
                θ
              
              ⁢
              
                d
                φ
              
            
          
        
      
    
  




  
    
      ∫
      
        
          
            Y
            lm
          
          ⁡
          
            (
            
              θ
              ,
              φ
            
            )
          
        
        ⁢
        
          
            Y
            
              
                l
                ′
              
              ⁢
              
                m
                ′
              
            
          
          ⁡
          
            (
            
              θ
              ,
              φ
            
            )
          
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        θ
        ⁢
        
          d
          θφ
        
      
    
    =
    
      
        δ
        
          ll
          ′
        
      
      ⁢
      
        δ
        
          mm
          ′
        
      
    
  




  
    
      A
      lm
    
    =
    
      
        a
        
          l
          +
          1
        
      
      ⁢
      
        ∫
        
          
            ψ
            ⁡
            
              (
              
                a
                ,
                θ
                ,
                φ
              
              )
            
          
          ⁢
          
            
              Y
              lm
            
            ⁡
            
              (
              
                θ
                ,
                φ
              
              )
            
          
          ⁢
          sin
          ⁢
          
            
          
          ⁢
          θ
          ⁢
          
            d
            θφ
          
        
      
    
  



The terms Alm with m>0 can be eliminated by making ψ(r,θ,φ) is independent of φ.




  
    
      Y
      
        l
        ⁢
        
          
        
        ⁢
        0
      
    
    ⁡
    
      (
      
        θ
        ,
        φ
      
      )
    
  
  =
  
    
      
        
          
            2
            ⁢
            l
          
          +
          1
        
        
          4
          ⁢
          π
        
      
    
    ⁢
    
      
        P
        l
      
      ⁡
      
        (
        
          cos
          ⁢
          
            
          
          ⁢
          θ
        
        )
      
    
  




Since, the function ψ(θ) is odd about θ=π/2; therefore, the even l vanish. The following are Legendre Polynomial for the parameters of a design from the cross-reference patent application entitled “Quasi-static Antenna Design Algorithm”.
The potential ψ(θ) for a monopole, with the parameters in the following Table 3, is expanded with Legendre Polynomials. Column 2 is the Asymptotic Conical Dipole (ACD) with a line charge height κ. with the Legendre Polynomial coefficients. The TopLoad is give in Column (3) with a disk height of κ and radius a with the Legendre Polynomial coefficients. The height of the topload was selected to give a negative a3 coefficient. The two solutions were combined to create an OSA with parameters in column (4). The parameter α is the weight factor for the topload and 1−α is the weight factor for the ACD. The equation for combining the Legendre polynomial coefficients (a1, a3, a5, a7, a9)
an(col.4)=(1−α)*an(col.2)+a*an(col.3)










TABLE 3






ACD
TopLoad
Minimum P3 = 0


















κ
0.45
0.45
0.45


a
0
0.5
.5


α
0
1
0.5425183883422


a1
0.450304278947143
0.900317191185674
0.6730790


a3
0.047277383397228
−0.039866913299692
0


a5
0.008915280356837
−0.054190572656918
−0.025320805315145


a7
0.004719340874386
0.014660689955097
+0.010112705555601


a9
0.004960433018193
0.014316408426800
+0.010036221718240









This is an example; there is a family of solutions with different Q and RRad radiation resistance. The antenna shape for column 4 is shown in FIG. 32. The potential as a function of z=cos(θ) is almost a straight line, as shown in FIG. 33.
This antenna is approximately a 70 degree cone at the feed point. This antenna should be a wideband antenna. The current practice is ‘cut and try’ where the antenna shape is changed manually. In addition, resistance may be added to improve performance.
The “Quasi-static Antenna Design Algorithm” was developed to design minimum Q electrically small antennas. This proposal modifies the code to compute the electrostatic spherical harmonics for the antenna solutions. Design curves would be generated. Eliminating an electrostatic spherical harmonic term should eliminate the corresponding eigenmode in the antenna. The capacitance for that eigenmode should be zero. An anti-resonance will also be eliminate. This would dramatically lower the antenna Q over an extended frequency range. The Smith chart will look like FIG. 34. The impedance will change slowly as a function of frequency. The antenna is will be tunable over a very high frequency range with a very wide bandwidth and low VSWR. The designs would be numerical modeled and physical antennas will be built.
In one embodiment, the “Quasi-static Antenna Design Algorithm” requires only 5 msec. to compute each antenna designs A NEC4 run of a similar antenna (100 unknowns) is about 0.1 sec. Additional complexity at the feed point increases the run time. It would be remarkable if ultra wide bandwidth antennas could be designed with a “Quasi-static Antenna Design Algorithm” referenced above.
A Smith chart of an antenna typically has many anti-resonances. FIG. 34 shows the Smith chart, from 40 MHz to 6000 MHz, for the Linear Charge Distribution antenna (LCD). The LCD has only one anti-resonance at 2400 MHz, as shown at point 100. This anti-resonance is shallow with a 4 to 1 VSWR, as shown by circle 104. In FIG. 34, curve 108 shows changes occurring between 40 MHz and 6000 MHZ. It is observed that loops occur in the region 110 in FIG. 34, where the loops occur at approximately every 1/200 MHz.
The above analysis indicates that the antenna has only one eigenmode in the 40 MHz to 6000 MHz frequency range. This antenna was designed as an electrically small antenna with 16 wires approximating a continuous surface. The next step is to numerical model and manufacture a continuous surface LCD antenna. Electrostatic multipoles, Legendre Polynomial P3, P5, P7, . . . , P2n+1 charge distribution, could be added to eliminate the higher order eigenmodes. The modified designs would be numerical modeled and physical antennas can be manufactured. This antenna could have a dipole antenna pattern over an extended frequency range.
Regarding FIGS. 35 and 36, the following description is based on a 1⁄2 meter antenna, scaling the frequency by 100.
FIG. 35 shows a plot of resistance R, Reactance X, and Q for a linear charge distribution (LCD) antenna 16 wire scale model and calculation between 0.4 MHz and 2 MHz. The scale model wires=0.018 cm diameter (1.8 cm full scale), and the calculated wires=5 cm and 0.313 cm. The resistance R and reactance X units are of the left axis, and the Q is on the right axis.
The measured curve R changes from near 0 ohms at 0.4 MHz to approximately 60 ohms at 2 MHz, and the measured reactance X curve changes from −j100 ohms at 0.4 MHz to nearly +j25 ohms around 1.4-1.6 MHz, then changes to approximately +j15 ohms at 2 MHz. The Q curve shows a change from about 80 at 0.4 MHz to approximately 0 at 2 MHz.
The measured plot R, X and Q curves shown in FIG. 35 compare favorably with the calculated values shown as diamond, circle, square and x symbols shown in the Table in FIG. 35.
Similarly, FIG. 36 shows a plot of measured R, X and Q values between 0.25 MHz and 10 MHz, with the Q value approaching 0 value between 1 and 10 MHz, and the R and X values increasing above 1 MHz. The reactance X is zero at 1 MHZ where the X curve has a positive slope at that point.
The impedance plot shown in FIG. 35 is very unusual. Normally the resistance and reactance increase to several hundred ohms to form an anti-resonance. An anti-resonance has a very high input impedance; which is difficult to match to 50 ohms. It appears that there is a very weak anti-resonance with the reactance approaching zero. Another interesting fact about this antenna is the Q steadily decreases to a value of about 2. Normally one would expect a kink in the curve at resonance. Below resonance the stored energy is capacitive and above resonance the stored energy is inductive. In other antennas design the Q changed slope at resonance with much slower rate of decrease in Q. These two observations were a motivation for extending the impedance measurements to higher frequencies.
The impedance plot FIG. 36 is also very unusual. The reactance has never goes through zero. There is no anti-resonance in this frequency range (100 MHz to 1000 MHz). In a dipole antenna there would be 3 anti-resonances near 300 MHz, 600 MHz, and 900 MHz (for 1⁄2 meter monopole) and resonances near 150 MHz, and 450 MHz. The Q appears to be below 2 from 150 MHz to 1000 MHz. The reactance is roughly linear; it appears that the antenna has extra inductance (123 nH) at the feed point. This may be caused by the very narrow wire at the feed point. Increasing this wire diameter would decrease the inductance and shift the reactance down. Adding extra wires to the antenna shape will also modify the impedance. Using an equation from Has Schantz's book the 16 wire design will have reduced gain at 600 Mhz. The ideal antenna would be a smooth continuous surface and would eliminate the drop in gain.
This antenna was designed as an electrically small antenna for the 50 MHz to 200 MHz band. It is remarkable that this antenna is an ultra wideband antenna with a VSWR<4 to 1 from 100 MHz to 2.4 GHz. It should be possible to greatly improve this design.
Previously, the dipole moment top load term was discussed but not calculated. In the following discussion two methods are presented for calculating the dipole moment top load contribution to the radiation resistance. The minimum Q antenna design can be calculated by using different combination of multipole basis function on the top load. The sequence of minimum Q antenna designs is plotted in FIG. 37 (FIGS. 40, 41, and 41 are three dimensional plots with a cutaway section to show the shape) the designs have decreasing Q and more importantly the antenna shape converges. FIG. 38 shows that the dipole moment term reduces the electric field under the antenna. The addition of the quadruple term gives some additional reduction in Q
Adding the dipole term will reduce the Q but may also reduce the radiation resistance. In that case, the dipole moment term allows the solution to fill the top of the spherical enclosing volume. This leads to a much larger capacitance or reduced stored energy. On the other hand, a top load with two or more multipole top-load basis functions (the dipole and quadruple) gives a solution with a positive dipole contribution (spherical enclosing geometry) and increased radiation resistance. FIG. 39 shows the minimum value of the ratio of Q/QChu for different values of the κ parameter for the monopole moment, monopole & dipole moments and monopole & dipole & quadrupole moments. Adding the dipole moment is a significant improvement in Q; the quadrupole term is more modest improvement in Q.
General Calculation of Radiation Resistance.
The effective height calculation to the dipole term, K1(u)P1(cos ν) will now be described. The K1(u)P1(cos νv) term is discontinuous at u=0 and the charge distribution is finite with zero separation.

K1(0)P1(cos ν)=−K1(0)P1(cos(π−νv))

The dipole moment calculation from the charge distribution is

p=∫xq(x)d3x.

Where p and x are vectors (and bolded). If q(x) is rotational symmetric about the z-axis, the above equation reduces to

pz=∫z=−hz=h∫zq(ρ,z)2πρdρdz

where the charge distribution is located between −h and h (inside an inclosing sphere of radius h). If one assumes that the feed point is at z=0, q(ρ,z)=−q(ρ,−z) and qNet=∫z=0z≦h∫q(ρ,z)2πρdρdz.

The dipole moment is




  
    
      1
      
        q
        Net
      
    
    ⁢
    
      P
      z
    
  
  =
  
    2
    ⁢
    
      
        h
        Eff
      
      .
    
  




This follows directly from the definition of





  
    h
    Eff
  
  =
  
    
      1
      
        q
        Net
      
    
    ⁢
    
      
        ∫
        
          z
          =
          0
        
        
          z
          =
          h
        
      
      ⁢
      
        ∫
        
          
            zq
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          ⁢
          2
          ⁢
          π
          ⁢
          
            
          
          ⁢
          ρ
          ⁢
          
            
          
          ⁢
          
            d
            ρ
          
          ⁢
          
            d
            z
          
        
      
    
  




When the feed point is not at z=0, the calculation of read follows the discussion in the above cross-referenced patent application (see paragraphs 0084-0086 and 00136). The above calculation of the dipole moment would appear give zero, which is incorrect, as the potential has a dipole moment.
The dipole moment can be calculated by expanding the potential (rotational symmetric) with Legendre polynomials.




  
    Φ
    ⁡
    
      (
      
        r
        ,
        θ
      
      )
    
  
  =
  
    
      
        ∑
        
          n
          =
          1
        
        ∞
      
      ⁢
      
        
      
      ⁢
      
        
          
            
              β
              
                n
                ⁢
                
                  
                
                ⁢
                0
              
            
            ⁡
            
              (
              
                a
                r
              
              )
            
          
          
            n
            +
            1
          
        
        ⁢
        
          
            P
            n
          
          ⁡
          
            (
            
              cos
              ⁢
              
                
              
              ⁢
              θ
            
            )
          
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
    
    ≥
    a
  




where r=√{square root over (z2+ρ2)}. The dipole moment is dz=β10a2




  
    
      2
      ⁢
      π
      ⁢
      
        
          ∫
          
            θ
            =
            0
          
          
            θ
            =
            π
          
        
        ⁢
        
          
            Φ
            ⁡
            
              (
              
                r
                ,
                θ
              
              )
            
          
          ⁢
          
            
              P
              1
            
            ⁡
            
              (
              
                cos
                ⁢
                
                  
                
                ⁢
                θ
              
              )
            
          
          ⁢
          sin
          ⁢
          
            
          
          ⁢
          θ
          ⁢
          
            d
            θ
          
        
      
    
    =
    
      
        ∑
        
          n
          =
          1
        
        ∞
      
      ⁢
      
        
      
      ⁢
      
        
          
            
              β
              n
            
            ⁡
            
              (
              
                a
                r
              
              )
            
          
          
            n
            +
            1
          
        
        ⁢
        2
        ⁢
        π
        ⁢
        
          ∫
          
            
              
                P
                n
              
              ⁡
              
                (
                
                  cos
                  ⁢
                  
                    
                  
                  ⁢
                  θ
                
                )
              
            
            ⁢
            
              
                P
                1
              
              ⁡
              
                (
                
                  cos
                  ⁢
                  
                    
                  
                  ⁢
                  θ
                
                )
              
            
            ⁢
            sin
            ⁢
            
              
            
            ⁢
            θ
            ⁢
            
              d
              θ
            
            ⁢
            
              
            
            ⁢
            where
          
        
      
    
  




  
    
      ∫
      
        
          
            P
            n
          
          ⁡
          
            (
            
              cos
              ⁢
              
                
              
              ⁢
              θ
            
            )
          
        
        ⁢
        
          
            P
            1
          
          ⁡
          
            (
            
              cos
              ⁢
              
                
              
              ⁢
              θ
            
            )
          
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        θ
        ⁢
        
          d
          θ
        
      
    
    =
    
      
        
          2
          3
        
        ⁢
        
          
            δ
            
              1
              ⁢
              
                
              
              ⁢
              n
            
          
          .
          
            
          
          ⁢
          
            δ
            11
          
        
      
      =
      
        
          1
          ⁢
          
            
          
          ⁢
          and
          ⁢
          
            
          
          ⁢
          
            δ
            
              1
              ⁢
              
                
              
              ⁢
              n
            
          
        
        =
        
          
            0.
            ⁢
            
              
            
            ⁢
            When
            ⁢
            
              
            
            ⁢
            r
          
          =
          
            
              a
              ⁢
              
                3
                2
              
              ⁢
              
                a
                2
              
              ⁢
              
                
                  ∫
                  
                    θ
                    =
                    0
                  
                  
                    θ
                    =
                    π
                  
                
                ⁢
                
                  
                    Φ
                    ⁡
                    
                      (
                      
                        a
                        ,
                        θ
                      
                      )
                    
                  
                  ⁢
                  
                    
                      P
                      1
                    
                    ⁡
                    
                      (
                      
                        cos
                        ⁢
                        
                          
                        
                        ⁢
                        θ
                      
                      )
                    
                  
                  ⁢
                  sin
                  ⁢
                  
                    
                  
                  ⁢
                  θ
                  ⁢
                  
                    d
                    θ
                  
                
              
            
            =
            
              
                
                  β
                  1
                
                ⁢
                
                  a
                  2
                
                ⁢
                
                  

                
                ⁢
                
                  p
                  z
                
              
              =
              
                
                  3
                  2
                
                ⁢
                
                  a
                  2
                
                ⁢
                
                  
                    ∫
                    
                      θ
                      =
                      0
                    
                    
                      θ
                      =
                      π
                    
                  
                  ⁢
                  
                    
                      Φ
                      ⁡
                      
                        (
                        
                          a
                          ,
                          θ
                        
                        )
                      
                    
                    ⁢
                    
                      
                        P
                        1
                      
                      ⁡
                      
                        (
                        
                          cos
                          ⁢
                          
                            
                          
                          ⁢
                          θ
                        
                        )
                      
                    
                    ⁢
                    sin
                    ⁢
                    
                      
                    
                    ⁢
                    θ
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      θ
                    
                  
                
              
            
          
        
      
    
  



The above equation is an alternate method for calculating the dipole moment discussed in paragraph 0136 of the first cross-referenced patent application. The effective height is calculated from




  
    h
    Eff
  
  =
  
    
      
        3
        ⁢
        
          a
          2
        
      
      
        4
        ⁢
        
          q
          Net
        
      
    
    ⁢
    
      
        ∫
        
          θ
          =
          0
        
        
          θ
          =
          π
        
      
      ⁢
      
        
          Φ
          ⁡
          
            (
            
              a
              ,
              θ
            
            )
          
        
        ⁢
        
          
            P
            1
          
          ⁡
          
            (
            
              cos
              ⁢
              
                
              
              ⁢
              θ
            
            )
          
        
        ⁢
        
          
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        θ
        ⁢
        
          d
          θ
        
      
    
  




The following shows the connection between the charge distribution, potential and dipole moment. Simple example of a point charge, q at d/2 and −q at −d/2.
The dipole moment is pz=qd. The potential is




  
    
      
        
          Φ
          ⁡
          
            (
            
              z
              ,
              ρ
            
            )
          
        
        =
        
          
            q
            
              
                
                  
                    (
                    
                      z
                      -
                      
                        d
                        /
                        2
                      
                    
                    )
                  
                  2
                
                +
                
                  ρ
                  2
                
              
            
          
          -
          
            q
            
              
                
                  
                    (
                    
                      z
                      +
                      
                        d
                        /
                        2
                      
                    
                    )
                  
                  2
                
                +
                
                  ρ
                  2
                
              
            
          
        
      
    
  
  
    
      
        
          Φ
          ⁡
          
            (
            
              z
              ,
              0
            
            )
          
        
        =
        
          
            q
            
              
              
                z
                -
                
                  d
                  /
                  2
                
              
              
            
          
          -
          
            q
            
              
              
                (
                
                  z
                  +
                  
                    d
                    /
                    2
                  
                
                )
              
              
            
          
        
      
    
  




Series expansion valid for d/2z<2




  
    Φ
    ⁡
    
      (
      
        z
        ,
        0
      
      )
    
  
  =
  
    
      q
      ⁡
      
        [
        
          
            1
            z
          
          +
          
            
              1
              z
            
            ⁢
            
              (
              
                d
                
                  2
                  ⁢
                  z
                
              
              )
            
          
          +
          
            
              1
              z
            
            ⁢
            
              
                (
                
                  d
                  
                    2
                    ⁢
                    
                      
                    
                    ⁢
                    z
                  
                
                )
              
              2
            
          
          +
          
            
              1
              z
            
            ⁢
            
              
                
                  (
                  
                    d
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
              ++
            
            ⁢
            
              1
              z
            
            ⁢
            
              
                (
                
                  d
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              n
            
          
        
        ]
      
    
    -
    
      q
      ⁢
      
        ⌊
        
          
            1
            z
          
          -
          
            
              1
              z
            
            ⁢
            
              (
              
                d
                
                  2
                  ⁢
                  z
                
              
              )
            
          
          +
          
            
              1
              z
            
            ⁢
            
              
                (
                
                  d
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              2
            
          
          -
          
            
              1
              z
            
            ⁢
            
              
                
                  (
                  
                    d
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
              ++
            
            ⁢
            
              
                
                  (
                  
                    -
                    1
                  
                  )
                
                n
              
              z
            
            ⁢
            
              
                (
                
                  d
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              n
            
          
        
        ⌋
      
    
  




Since only odd n contributes, we write n=2m+1




  
    
      
        
          Φ
          ⁡
          
            (
            
              z
              ,
              0
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            z
          
          ⁢
          
            ⌊
            
              
                (
                
                  d
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              +
              
                
                  
                    
                      (
                      
                        d
                        
                          2
                          ⁢
                          z
                        
                      
                      )
                    
                    3
                  
                  ++
                
                ⁢
                
                  
                    (
                    
                      d
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
            
            ⌋
          
        
      
    
  
  
    
      
        
          Φ
          ⁡
          
            (
            
              r
              ,
              θ
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            r
          
          ⁢
          
            ⌊
            
              
                
                  (
                  
                    d
                    
                      2
                      ⁢
                      r
                    
                  
                  )
                
                ⁢
                
                  
                    P
                    1
                  
                  ⁡
                  
                    (
                    θ
                    )
                  
                
              
              +
              
                
                  
                    (
                    
                      d
                      
                        2
                        ⁢
                        r
                      
                    
                    )
                  
                  3
                
                ⁢
                
                  
                    
                      P
                      3
                    
                    ⁡
                    
                      (
                      θ
                      )
                    
                  
                  ++
                
                ⁢
                
                  
                    (
                    
                      d
                      
                        2
                        ⁢
                        r
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
                ⁢
                
                  
                    P
                    
                      
                        2
                        ⁢
                        m
                      
                      +
                      1
                    
                  
                  ⁡
                  
                    (
                    θ
                    )
                  
                
              
            
            ⌋
          
        
      
    
  
  
    
      
        
          Φ
          ⁡
          
            (
            
              r
              ,
              θ
            
            )
          
        
        =
        
          ⌊
          
            
              
                (
                
                  qd
                  
                    r
                    2
                  
                
                )
              
              ⁢
              
                
                  P
                  1
                
                ⁡
                
                  (
                  θ
                  )
                
              
            
            +
            
              
                
                  2
                  ⁢
                  q
                
                
                  r
                  4
                
              
              ⁢
              
                
                  (
                  
                    d
                    2
                  
                  )
                
                3
              
              ⁢
              
                
                  
                    P
                    3
                  
                  ⁡
                  
                    (
                    θ
                    )
                  
                
                ++
              
              ⁢
              
                
                  2
                  ⁢
                  q
                
                
                  r
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
              ⁢
              
                
                  (
                  
                    d
                    2
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
              ⁢
              
                
                  P
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
                ⁡
                
                  (
                  θ
                  )
                
              
            
          
          ⌋
        
      
    
  




The dipole term allows the radiation resistance to be increased. Also, the dipole term allows the filling of the top of the enclosing volume with a conductor (see antenna shape shown in FIG. 37).
The monopole term changes thickness of the top and bottom with the same sign while the dipole term changes the thickness of the top and bottom with a different sign. Also, the dipole term allows the electric field to be modified between the antenna and the corresponding image antenna, FIG. 38.
The dipole moment components can also be added (by way of vector addition) through superposition to achieve the changes to the electric fields.
FIG. 37 shows an antenna cross-section view of radial distance and height for different moments with τ=0.9 within an enclosing sphere. The cross-section view shows the monopole term with Q/QChu=2.154 (identified as “Monopole” in FIG. 37), the monopole-dipole term with Q/QChu=2.002 (identified as “Dipole” in FIG. 37), and the monopole-dipole-quadruple term with Q/QChu=1.958 (identified as “Quadra pole” in FIG. 37).
FIG. 38 shows a view of the electric field as a function of distance from the top of the antenna, with τ=0.9, showing the monopole; monopole-dipole term; and monopole-dipole-quadruple term.
FIG. 39 shows the Q/QChu ratio (Y-axis) for an oblate spheroid antenna (OSA) when compared with the parameter κ, where curve represents the OSA, curve represents OSA with dipole, curve represents OSA with dipole and octapole, and curve represents a disk loaded monopole.
FIGS. 40-42 show side views of an oblate spheroid antenna (OSA) for a monopole (FIG. 40), monopole-dipole (FIG. 41) and monopole-dipole-quadrupole (FIG. 42), respectively.
The contours shown in FIGS. 40-42 could be implemented as a solid surface in one embodiment, or implemented as an array of vertical wires configured into the contours as shown, as another example.
What has been described above includes examples of one or more embodiments. It is, of course, not possible to describe every conceivable combination of components or methodologies for purposes of describing the aforementioned embodiments. It will, therefore, be understood that many additional changes in the details, materials, steps and arrangement of parts, which have been herein described and illustrated to explain the nature of the invention, may be made by those skilled in the art within the principal and scope of the invention as expressed in the appended claims.
APPENDIX
Calculation of Potential for P3




  
    
      ∫
      
        
          -
          L
        
        /
        2
      
      
        L
        /
        2
      
    
    ⁢
    
      
        
          
            P
            3
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        
          
            
              
                (
                
                  x
                  -
                  z
                
                )
              
              2
            
            +
            
              ρ
              2
            
          
        
      
      ⁢
      
        d
        x
      
    
  
  =
  
    
      
        ∫
        
          
            -
            L
          
          /
          2
        
        
          L
          /
          2
        
      
      ⁢
      
        
          
            
              
                5
                2
              
              ⁢
              
                
                  (
                  
                    2
                    ⁢
                    
                      x
                      /
                      L
                    
                  
                  )
                
                3
              
            
            -
            
              
                3
                2
              
              ⁢
              
                (
                
                  2
                  ⁢
                  
                    x
                    /
                    L
                  
                
                )
              
            
          
          
            
              
                
                  (
                  
                    x
                    -
                    z
                  
                  )
                
                2
              
              +
              
                ρ
                2
              
            
          
        
        ⁢
        
          d
          
            
              ∫
              
                
                  -
                  L
                
                /
                2
              
              
                L
                /
                2
              
            
            ⁢
            
              
                
                  
                    P
                    3
                  
                  ⁡
                  
                    (
                    
                      2
                      ⁢
                      
                        x
                        /
                        L
                      
                    
                    )
                  
                
                
                  
                    
                      
                        (
                        
                          x
                          -
                          z
                        
                        )
                      
                      2
                    
                    +
                    
                      ρ
                      2
                    
                  
                
              
              ⁢
              
                d
                x
              
            
          
        
      
    
    =
    
      
        
          ∫
          
            
              -
              L
            
            /
            2
          
          
            L
            /
            2
          
        
        ⁢
        
          
            
              
                
                  5
                  2
                
                ⁢
                
                  
                    (
                    
                      2
                      ⁢
                      
                        x
                        /
                        L
                      
                    
                    )
                  
                  3
                
              
              -
              
                
                  3
                  2
                
                ⁢
                
                  (
                  
                    2
                    ⁢
                    
                      x
                      /
                      L
                    
                  
                  )
                
              
            
            
              
                
                  
                    (
                    
                      x
                      -
                      z
                    
                    )
                  
                  2
                
                +
                
                  ρ
                  2
                
              
            
          
          ⁢
          
            d
            x
          
          ⁢
          
            

          
          ⁢
          u
        
      
      =
      
        
          x
          -
          
            z
            ⁢
            
              
            
            ⁢
            and
            ⁢
            
              
            
            ⁢
            
              d
              u
            
          
        
        =
        
          
            d
            x
          
          ⁢
          
            

          
          ⁢
          
            5
            2
          
          ⁢
          
            
          
          ⁢
          
            8
            
              L
              3
            
          
          ⁢
          
            
          
          ⁢
          
            ⌊
            
              
                
                  
                    
                      R
                      1
                      3
                    
                    -
                    
                      R
                      2
                      3
                    
                  
                  3
                
                +
                
                  
                     
                     
                  
                  ⁢
                  
                     
                    
                      
                        
                          
                            3
                            ⁢
                            z
                          
                          2
                        
                        ⁢
                        
                          (
                          
                            
                              L
                              /
                              2
                            
                            -
                            z
                          
                          )
                        
                        ⁢
                        
                          R
                          1
                        
                      
                      +
                      
                        
                          
                            3
                            ⁢
                            z
                          
                          2
                        
                        ⁢
                        
                          (
                          
                            
                              L
                              /
                              2
                            
                            +
                            
                              zR
                              2
                            
                          
                          )
                        
                      
                      +
                      
                        
                          (
                          
                            
                              3
                              ⁢
                              
                                z
                                2
                              
                            
                            -
                            
                              ρ
                              2
                            
                          
                          )
                        
                        ⁢
                        
                          (
                          
                            
                              R
                              1
                            
                            -
                            
                              R
                              2
                            
                          
                          )
                        
                      
                      +
                      
                        
                          (
                          
                            
                              z
                              3
                            
                            -
                            
                              
                                3
                                ⁢
                                z
                                ⁢
                                
                                  
                                
                                ⁢
                                
                                  ρ
                                  2
                                
                              
                              2
                            
                          
                          )
                        
                        ⁢
                        
                          log
                          ⁡
                          
                            (
                            
                              
                                1
                                +
                                τ
                              
                              
                                1
                                -
                                τ
                              
                            
                            )
                          
                        
                      
                    
                    ⌋
                  
                
                -
                
                  
                    3
                    2
                  
                  ⁢
                  
                    
                      ∫
                      
                        
                          -
                          L
                        
                        /
                        2
                      
                      
                        L
                        /
                        2
                      
                    
                    ⁢
                    
                      
                        
                          
                            P
                            1
                          
                          ⁡
                          
                            (
                            
                              2
                              ⁢
                              
                                x
                                /
                                L
                              
                            
                            )
                          
                        
                        
                          
                            
                              
                                (
                                
                                  x
                                  -
                                  z
                                
                                )
                              
                              2
                            
                            +
                            
                              ρ
                              2
                            
                          
                        
                      
                      ⁢
                      
                        d
                        u
                      
                      ⁢
                      
                        

                      
                      ⁢
                      
                        where
                        ⁢
                        
                          
                        
                        ⁢
                        
                          R
                          1
                        
                      
                    
                  
                
              
              =
              
                
                  
                    
                      
                        
                          (
                          
                            
                              L
                              /
                              2
                            
                            -
                            z
                          
                          )
                        
                        2
                      
                      +
                      
                        ρ
                        2
                      
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  and
                  ⁢
                  
                    
                  
                  ⁢
                  
                    R
                    2
                  
                
                =
                
                  
                    
                      
                        
                          
                            
                              (
                              
                                
                                  L
                                  /
                                  2
                                
                                +
                                z
                              
                              )
                            
                            2
                          
                          +
                          
                            ρ
                            2
                          
                        
                      
                      ⁢
                      
                        

                      
                      ⁢
                      
                        5
                        2
                      
                      ⁢
                      
                        8
                        
                          L
                          3
                        
                      
                      ⁢
                      
                        ⌊
                        
                          
                            
                              (
                              
                                
                                  
                                    
                                      R
                                      1
                                      2
                                    
                                    +
                                    
                                      R
                                      2
                                      2
                                    
                                  
                                  6
                                
                                +
                                
                                  
                                    
                                      (
                                      
                                        
                                          R
                                          1
                                        
                                        +
                                        
                                          R
                                          2
                                        
                                      
                                      )
                                    
                                    2
                                  
                                  6
                                
                                +
                                
                                  (
                                  
                                    
                                      
                                        3
                                        ⁢
                                        
                                          z
                                          2
                                        
                                      
                                      2
                                    
                                    -
                                    
                                      ρ
                                      2
                                    
                                  
                                  )
                                
                              
                              )
                            
                            ⁢
                            
                              (
                              
                                
                                  R
                                  1
                                
                                -
                                
                                  R
                                  2
                                
                              
                              )
                            
                          
                          +
                          
                            
                              
                                3
                                ⁢
                                Lz
                              
                              4
                            
                            ⁢
                            
                              (
                              
                                
                                  R
                                  1
                                
                                +
                                
                                  R
                                  2
                                
                              
                              )
                            
                          
                          +
                          
                            
                              (
                              
                                
                                  z
                                  3
                                
                                -
                                
                                  
                                    3
                                    ⁢
                                    z
                                    ⁢
                                    
                                      
                                    
                                    ⁢
                                    
                                      ρ
                                      2
                                    
                                  
                                  2
                                
                              
                              )
                            
                            ⁢
                            
                              log
                              ⁡
                              
                                (
                                
                                  
                                    1
                                    +
                                    τ
                                  
                                  
                                    1
                                    -
                                    τ
                                  
                                
                                )
                              
                            
                          
                        
                        ⌋
                      
                    
                    -
                    
                      
                        3
                        2
                      
                      ⁢
                      
                        
                          ∫
                          
                            
                              -
                              L
                            
                            /
                            2
                          
                          
                            L
                            /
                            2
                          
                        
                        ⁢
                        
                          
                            
                              
                                P
                                1
                              
                              ⁡
                              
                                (
                                
                                  2
                                  ⁢
                                  
                                    x
                                    /
                                    L
                                  
                                
                                )
                              
                            
                            
                              
                                
                                  
                                    (
                                    
                                      x
                                      -
                                      z
                                    
                                    )
                                  
                                  2
                                
                                +
                                
                                  ρ
                                  2
                                
                              
                            
                          
                          ⁢
                          
                            d
                            u
                          
                          ⁢
                          
                            

                          
                          ⁢
                          note
                          ⁢
                          
                            
                          
                          ⁢
                          
                            (
                            
                              
                                R
                                1
                              
                              -
                              
                                R
                                2
                              
                            
                            )
                          
                        
                      
                    
                  
                  =
                  
                    
                      
                        (
                        
                          
                            R
                            1
                            2
                          
                          -
                          
                            R
                            2
                            2
                          
                        
                        )
                      
                      
                        (
                        
                          
                            R
                            1
                          
                          +
                          
                            R
                            2
                          
                        
                        )
                      
                    
                    =
                    
                      
                        
                          
                            
                              
                                -
                                2
                              
                              ⁢
                              Lz
                            
                            
                              (
                              
                                
                                  R
                                  1
                                
                                +
                                
                                  R
                                  2
                                
                              
                              )
                            
                          
                          ⁢
                          
                            
                          
                          ⁢
                          and
                          ⁢
                          
                            
                          
                          ⁢
                          
                            R
                            1
                            2
                          
                        
                        +
                        
                          R
                          2
                          2
                        
                      
                      =
                      
                        
                          2
                          ⁢
                          
                            (
                            
                              
                                z
                                2
                              
                              +
                              
                                
                                  (
                                  
                                    L
                                    /
                                    2
                                  
                                  )
                                
                                2
                              
                              +
                              
                                ρ
                                2
                              
                            
                            )
                          
                          ⁢
                          
                            

                          
                          ⁢
                          
                            5
                            2
                          
                          ⁢
                          
                            
                              8
                              
                                L
                                3
                              
                            
                            ⁡
                            
                              [
                              
                                
                                  
                                    
                                      5
                                      ⁢
                                      L
                                    
                                    12
                                  
                                  ⁢
                                  
                                    z
                                    ⁡
                                    
                                      (
                                      
                                        
                                          R
                                          1
                                        
                                        +
                                        
                                          R
                                          2
                                        
                                      
                                      )
                                    
                                  
                                
                                -
                                
                                  2
                                  ⁢
                                  z
                                  ⁢
                                  
                                    
                                  
                                  ⁢
                                  
                                    τ
                                    ⁡
                                    
                                      (
                                      
                                        
                                          
                                            11
                                            6
                                          
                                          ⁢
                                          
                                            z
                                            2
                                          
                                        
                                        +
                                        
                                          
                                            L
                                            2
                                          
                                          12
                                        
                                        -
                                        
                                          
                                            2
                                            3
                                          
                                          ⁢
                                          
                                            ρ
                                            2
                                          
                                        
                                      
                                      )
                                    
                                  
                                
                                +
                                
                                  
                                    (
                                    
                                      
                                        z
                                        3
                                      
                                      -
                                      
                                        
                                          3
                                          2
                                        
                                        ⁢
                                        z
                                        ⁢
                                        
                                          
                                        
                                        ⁢
                                        
                                          ρ
                                          2
                                        
                                      
                                    
                                    )
                                  
                                  ⁢
                                  
                                    log
                                    ⁡
                                    
                                      (
                                      
                                        
                                          1
                                          +
                                          τ
                                        
                                        
                                          1
                                          -
                                          τ
                                        
                                      
                                      )
                                    
                                  
                                
                              
                              ]
                            
                          
                        
                        -
                        
                          
                            3
                            2
                          
                          ⁡
                          
                            [
                            
                              
                                
                                  2
                                  ⁢
                                  z
                                
                                L
                              
                              ⁢
                              
                                (
                                
                                  
                                    ln
                                    ⁡
                                    
                                      (
                                      
                                        
                                          1
                                          +
                                          τ
                                        
                                        
                                          1
                                          -
                                          τ
                                        
                                      
                                      )
                                    
                                  
                                  -
                                  
                                    2
                                    ⁢
                                    τ
                                  
                                
                                )
                              
                            
                            ]
                          
                        
                      
                    
                  
                
              
            
          
        
      
    
  




Expansion of Multipoles and Analytic Calculations of Pure Dipole Solution
The calculation of the potential on the z axis allows the multipole expansion of the potential to be calculated. For P1(u), the potential on the z axis is




  
    
      Ψ
      1
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      ∫
      
        
          -
          L
        
        /
        2
      
      
        L
        /
        2
      
    
    ⁢
    
      
        
          
            P
            1
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        
          
            
              (
              
                x
                -
                z
              
              )
            
            2
          
        
      
      ⁢
      
        d
        x
      
    
  




Expanding this in a series




  
    
      Ψ
      1
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      1
      z
    
    ⁢
    
      
        ∫
        
          
            -
            L
          
          /
          2
        
        
          L
          /
          2
        
      
      ⁢
      
        
          
            P
            1
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        ⁢
        
          ⌊
          
            1
            +
            
              x
              z
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              2
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              3
            
            +
            
              
                
                  
                    (
                    
                      x
                      z
                    
                    )
                  
                  4
                
                ++
              
              ⁢
              
                
                  (
                  
                    x
                    z
                  
                  )
                
                n
              
            
          
          ⌋
        
        ⁢
        
          d
          x
        
      
    
  




where each term





  (
  
    1
    
      z
      
        m
        +
        1
      
    
  
  )




in the far field has an Pm(cos θ) angular dependence. Changing variables to u=2x/L and integrating term by term




  
    
      
        Ψ
        1
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁢
      
        
          ∫
          
            -
            1
          
          1
        
        ⁢
        
          
            
              P
              1
            
            ⁡
            
              (
              u
              )
            
          
          ⁢
          
            ⌊
            
              
                (
                
                  Lu
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              +
              
                
                  
                    
                      (
                      
                        Lu
                        
                          2
                          ⁢
                          z
                        
                      
                      )
                    
                    3
                  
                  ++
                
                ⁢
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
            
            ⌋
          
          ⁢
          
            d
            u
          
        
      
    
  




  
    
      
        Ψ
        1
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁢
      
        
          ∫
          
            -
            1
          
          1
        
        ⁢
        
          
            ⌊
            
              
                
                  Lu
                  2
                
                
                  2
                  ⁢
                  z
                
              
              +
              
                
                  
                    (
                    
                      L
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  3
                
                ⁢
                
                  u
                  4
                
              
              +
              
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  5
                
                ⁢
                
                  u
                  6
                
              
              +
              
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
                ⁢
                
                  u
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    2
                  
                
              
            
            ⌋
          
          ⁢
          
            d
            u
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        
          Ψ
          1
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      =
      
        
          L
          
            2
            ⁢
            z
          
        
        ⁡
        
          [
          
            
              
                2
                3
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                1
              
            
            +
            
              
                2
                5
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
            
            +
            ...
            +
            
              
                2
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
            
          
          ]
        
      
    
  



For P3 (u), the potential on the z axis is




  
    
      Ψ
      3
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      ∫
      
        
          -
          L
        
        /
        2
      
      
        L
        /
        2
      
    
    ⁢
    
      
        
          
            P
            3
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        
          
            
              (
              
                x
                -
                z
              
              )
            
            2
          
        
      
      ⁢
      
        d
        x
      
    
  




Expanding this in a series




  
    
      Ψ
      3
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      1
      z
    
    ⁢
    
      
        ∫
        
          
            -
            L
          
          /
          2
        
        
          L
          /
          2
        
      
      ⁢
      
        
          
            P
            3
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        ⁢
        
          ⌊
          
            1
            +
            
              x
              z
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              2
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              3
            
            +
            
              
                
                  
                    (
                    
                      x
                      z
                    
                    )
                  
                  4
                
                ++
              
              ⁢
              
                
                  (
                  
                    x
                    z
                  
                  )
                
                n
              
            
          
          ⌋
        
        ⁢
        
          d
          x
        
      
    
  




where each term





  (
  
    1
    
      
        z
        m
      
          
  
  )




in the far field has an Pm(cos θ) angular dependence. Changing variables to u=2x/L and integrating term by term




  
    
      
        
          Ψ
          3
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      =
      
        
          L
          
            2
            ⁢
            z
          
        
        ⁢
        
          
            ∫
            
              -
              1
            
            1
          
          ⁢
          
            
              
                P
                3
              
              ⁡
              
                (
                u
                )
              
            
            ⁢
            
              ⌊
              
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  3
                
                +
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  3
                
                +
                
                  
                    (
                    
                      Lu
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
              ⌋
            
            ⁢
            
              d
              u
            
          
        
      
    
    ⁢
    
      
    
  




  
    
      
        Ψ
        3
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁡
      
        [
        
          
            
              4
              35
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              3
            
          
          +
          
            
              8
              63
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              5
            
          
          +
          ...
          +
          
            
              
                4
                ⁢
                m
              
              
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    5
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
              
            
            ⁢
            
              
                (
                
                  Lu
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              
                
                  2
                  ⁢
                  m
                
                +
                1
              
            
          
        
        ]
      
    
  



The Ψ1(0,z) and Ψ3(0,z) can be combined with a −7/2 coefficient to cancel the




  
    (
    
      L
      
        2
        ⁢
        z
      
    
    )
  
  3




term in the far field.




  
    
      
        
          Ψ
          1
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      -
      
        
          7
          2
        
        ⁢
        
          
            Ψ
            3
          
          ⁡
          
            (
            
              0
              ,
              z
            
            )
          
        
      
    
    =
    
      
        
          L
          
            2
            ⁢
            z
          
        
        ⁡
        
          [
          
            
              
                2
                3
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                1
              
            
            +
            
              
                2
                5
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
            
            +
            ...
            +
            
              
                2
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
              
              ⁢
              
                
                  (
                  
                    L
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
            
          
          ]
        
      
      +
      
        -
        
          
            
              7
              ⁢
              L
            
            
              4
              ⁢
              z
            
          
          ⁡
          
            [
            
              
                
                  4
                  35
                
                ⁢
                
                  
                    (
                    
                      L
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  3
                
              
              +
              
                
                  8
                  63
                
                ⁢
                
                  
                    (
                    
                      L
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  5
                
              
              +
              ...
              +
              
                
                  
                    4
                    ⁢
                    m
                  
                  
                    
                      (
                      
                        
                          2
                          ⁢
                          m
                        
                        +
                        5
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        
                          2
                          ⁢
                          m
                        
                        +
                        3
                      
                      )
                    
                  
                
                ⁢
                
                  
                    (
                    
                      L
                      
                        2
                        ⁢
                        z
                      
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
            
            ]
          
        
      
    
  




  
    
      
        
          Ψ
          1
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      -
      
        
          7
          2
        
        ⁢
        
          
            Ψ
            3
          
          ⁡
          
            (
            
              0
              ,
              z
            
            )
          
        
      
    
    =
    
      
        ⌊
        
          
            
              2
              3
            
            ⁢
            
              L
              
                
                  (
                  
                    2
                    ⁢
                    z
                  
                  )
                
                2
              
            
          
          +
          
            
              2
              ⁢
              
                L
                3
              
            
            
              5
              ⁢
              
                
                  (
                  
                    2
                    ⁢
                    z
                  
                  )
                
                4
              
            
          
          +
          ...
          +
          
            
              2
              ⁢
              
                L
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
            
            
              
                (
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  3
                
                )
              
              ⁢
              
                
                  (
                  
                    2
                    ⁢
                    z
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  2
                
              
            
          
        
        ⌋
      
      -
      
        
          7
          2
        
        ⁢
        
          ⌊
          
            
              
                4
                ⁢
                
                  L
                  3
                
              
              
                35
                ⁢
                
                  
                    (
                    
                      2
                      ⁢
                      z
                    
                    )
                  
                  4
                
              
            
            +
            
              
                8
                ⁢
                
                  L
                  5
                
              
              
                63
                ⁢
                
                  
                    (
                    
                      2
                      ⁢
                      z
                    
                    )
                  
                  6
                
              
            
            +
            ...
            +
            
              
                4
                ⁢
                
                  
                
                ⁢
                
                  mL
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    1
                  
                
              
              
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    5
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
                ⁢
                
                  
                    (
                    
                      2
                      ⁢
                      z
                    
                    )
                  
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    2
                  
                
              
            
          
          ⌋
        
      
    
  



(Note: the above two lines are one equation)




  
    
      
        
          Ψ
          1
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      -
      
        
          7
          2
        
        ⁢
        
          Ψ
          3
        
        ⁢
        
          (
          
            0
            ,
            z
          
          )
        
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁡
      
        [
        
          
            
              2
              3
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              1
            
          
          -
          
            
              10
              63
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              5
            
          
          +
          ...
          +
          
            
              
                2
                ⁢
                
                  (
                  
                    5
                    -
                    
                      5
                      ⁢
                      m
                    
                  
                  )
                
              
              
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    5
                  
                  )
                
              
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              
                
                  2
                  ⁢
                  m
                
                +
                1
              
            
          
        
        ]
      
    
  




  
    
      
        
          Ψ
          1
        
        ⁡
        
          (
          
            0
            ,
            z
          
          )
        
      
      -
      
        
          7
          2
        
        ⁢
        
          Ψ
          3
        
        ⁢
        
          (
          
            0
            ,
            z
          
          )
        
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁡
      
        [
        
          
            
              2
              3
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              1
            
          
          -
          
            
              10
              63
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              5
            
          
          +
          ...
          +
          
            
              
                
                  -
                  10
                
                ⁢
                
                  (
                  
                    m
                    -
                    1
                  
                  )
                
              
              
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    5
                  
                  )
                
              
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              
                
                  2
                  ⁢
                  m
                
                +
                1
              
            
          
        
        ]
      
    
  



Additional terms, Ψ5(0,z), could be used to cancel the




  
    (
    
      L
      
        2
        ⁢
        z
      
    
    )
  
  5




term. By using a succession of line sources P1, P3, P5 and P7 an antenna with a pure dipole far field can be generated. The error in the far field is Ψ9(ρ,z).

Radiation Resistance
The effective height for a positively charged distribution is normally calculated with the equation




  
    h
    eff
  
  =
  
    
      ∫
      0
      
        d
        /
        2
      
    
    ⁢
    
      x
      ⁢
      
        
          q
          ⁡
          
            (
            x
            )
          
        
        
          q
          Net
        
      
      ⁢
      
        
          d
          x
        
        .
      
    
  




where the feedpoint is at the ground plane. In the general, the definition must proceed from fundamental concepts. From Jackson the power radiated from a dipole is

P=10c2k4|p|2

where p is the dipole moment

p=∫xq(x)d3x

for a cylindrical symmetric dipole, this reduces to

pz=∫−d/2d/2zq(x)d3x

where by px=0 and py=0. The radiation resistance is defined in terms of the total power radiated as





  P
  =
  
    
      R
      Rad
    
    ⁢
    
      
        I
        fp
        2
      
      2
    
  




where the Ifp is the peak current (





  
    (
    
      
        1
        2
      
      ⁢
      
        
      
      ⁢
      converts
      ⁢
      
        
      
      ⁢
      peak
      ⁢
      
        
      
      ⁢
      to
      ⁢
      
        
      
      ⁢
      rms
    
    )
  
  .




converts peak to rms). In our model, negative charge below the feedpoint is part of the image current. This current is calculated from the continuity equation.

Ifp=iω∫fpd/2q(x)d3x

P=10c2k4|p|2

If we limit the antenna to a symmetric dipole, this equation reduces to



  
    P
    =
    
      10
      ⁢
      
        
      
      ⁢
      
        c
        2
      
      ⁢
      
        k
        4
      
      ⁢
      
        
          
          
            2
            ⁢
            
              
                ∫
                0
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  zq
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
          
          
        
        2
      
    
  




  
    P
    =
    
      10
      ⁢
      
        
      
      ⁢
      
        c
        2
      
      ⁢
      
        k
        4
      
      ⁢
      
        
          
            
              
              
                I
                fp
              
              
            
            2
          
          [
          
            
              2
              ⁢
              
                
                  ∫
                  0
                  
                    d
                    /
                    2
                  
                
                ⁢
                
                  
                    zq
                    ⁡
                    
                      (
                      x
                      )
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      d
                      3
                    
                    ⁢
                    x
                  
                
              
            
            
              I
              fp
            
          
          ]
        
        2
      
    
  




  
    P
    =
    
      10
      ⁢
      
        
      
      ⁢
      
        c
        2
      
      ⁢
      
        
          
            k
            4
          
          [
          
            
              2
              ⁢
              
                
                  ∫
                  0
                  
                    d
                    /
                    2
                  
                
                ⁢
                
                  
                    zq
                    ⁡
                    
                      (
                      x
                      )
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      d
                      3
                    
                    ⁢
                    x
                  
                
              
            
            
              ω
              ⁢
              
                
                  ∫
                  fp
                  
                    d
                    /
                    2
                  
                
                ⁢
                
                  
                    q
                    ⁡
                    
                      (
                      x
                      )
                    
                  
                  ⁢
                  
                    
                  
                  ⁢
                  
                    
                      d
                      3
                    
                    ⁢
                    x
                  
                
              
            
          
          ]
        
        2
      
      ⁢
      
        
          
          
            I
            fp
          
          
        
        2
      
    
  




  
    P
    =
    
      40
      ⁢
      
        
      
      ⁢
      
        
          
            k
            2
          
          [
          
            
              
                ∫
                0
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  zq
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
            
              
                ∫
                fp
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  q
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
          
          ]
        
        2
      
      ⁢
      
        
          
          
            I
            fp
          
          
        
        2
      
    
  




  
    P
    =
    
      80
      ⁢
      
        
      
      ⁢
      
        
          
            k
            2
          
          [
          
            
              
                ∫
                0
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  zq
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
            
              
                ∫
                fp
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  q
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
          
          ]
        
        2
      
      ⁢
      
        
          I
          fp
          2
        
        2
      
    
  




  
    P
    =
    
      
        R
        Rad
      
      ⁢
      
        
          I
          fp
          2
        
        2
      
    
  



If the feedpoint current is the same as the net charge



  
    
      R
      
        Rad
        -
        Dipole
      
    
    =
    
      80
      ⁢
      
        
          
            k
            2
          
          [
          
            
              
                ∫
                0
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  zq
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
            
              
                ∫
                fp
                
                  d
                  /
                  2
                
              
              ⁢
              
                
                  q
                  ⁡
                  
                    (
                    x
                    )
                  
                
                ⁢
                
                  
                
                ⁢
                
                  
                    d
                    3
                  
                  ⁢
                  x
                
              
            
          
          ]
        
        2
      
    
  




  
    
      R
      
        Rad
        -
        Monopole
      
    
    =
    
      40
      ⁢
      
        
          k
          2
        
        [
        
          
            
              ∫
              0
              
                d
                /
                2
              
            
            ⁢
            
              
                zq
                ⁡
                
                  (
                  x
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                
                  d
                  3
                
                ⁢
                x
              
            
          
          
            
              ∫
              fp
              
                d
                /
                2
              
            
            ⁢
            
              
                q
                ⁡
                
                  (
                  x
                  )
                
              
              ⁢
              
                
              
              ⁢
              
                
                  d
                  3
                
                ⁢
                x
              
            
          
        
        ]
      
    
  



The quantity in the square bracket is heff. This equation can be reduces to the usual results, when q(x)=1 and q(x)=δ(z−d/2).
The calculation for




  
    q
    ⁡
    
      (
      x
      )
    
  
  =
  
    
      
        2
        ⁢
        q
      
      
        κ
        ⁢
        
          
        
        ⁢
        h
      
    
    ⁢
    
      (
      
        
          
            P
            1
          
          ⁡
          
            (
            
              x
              
                κ
                ⁢
                
                  
                
                ⁢
                h
              
            
            )
          
        
        +
        
          a
          ⁢
          
            
          
          ⁢
          
            
              P
              3
            
            ⁡
            
              (
              
                x
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  h
                
              
              )
            
          
        
      
      )
    
  




will be discussed in detail below.





  
    p
    z
  
  =
  
    
      ∫
      
        
          -
          κ
        
        ⁢
        
          
        
        ⁢
        h
      
      
        κ
        ⁢
        
          
        
        ⁢
        h
      
    
    ⁢
    
      x
      ⁢
      
        
          2
          ⁢
          
            
          
          ⁢
          q
        
        
          κ
          ⁢
          
            
          
          ⁢
          h
        
      
      ⁢
      
        
      
      ⁢
      
        (
        
          
            
              P
              1
            
            ⁡
            
              (
              
                x
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  h
                
              
              )
            
          
          +
          
            α
            ⁢
            
              
            
            ⁢
            
              
                P
                3
              
              ⁡
              
                (
                
                  x
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    h
                  
                
                )
              
            
          
        
        )
      
      ⁢
      
        d
        x
      
    
  




changing variables





  u
  =
  
    x
    
      κ
      ⁢
      
        
      
      ⁢
      h
    
  




pz=2qκh∫−11u(P1(u)+αP3(u))du

Since u=P1(u) and ∫P1(u)P3(u)du=0

pz=2qκh∫−11u2du





  
    p
    z
  
  =
  
    
      2
      ⁢
      q
      ⁢
      
        
      
      ⁢
      κ
      ⁢
      
        
      
      ⁢
      
        
          h
          ⁡
          
            [
            
              
                u
                2
              
              2
            
            ]
          
        
        
          -
          1
        
        1
      
    
    =
    
      2
      ⁢
      
        
      
      ⁢
      q
      ⁢
      
        
      
      ⁢
      κ
      ⁢
      
        
      
      ⁢
      h
    
  




The same variable change is made for



  
    
      q
      fp
    
    =
    
      
        ∫
        
          -
          fp
        
        
          κ
          ⁢
          
            
          
          ⁢
          h
        
      
      ⁢
      
        
          
            2
            ⁢
            q
          
          
            κ
            ⁢
            
              
            
            ⁢
            h
          
        
        ⁢
        
          
        
        ⁢
        
          (
          
            
              
                P
                1
              
              ⁡
              
                (
                
                  x
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    h
                  
                
                )
              
            
            +
            
              α
              ⁢
              
                
              
              ⁢
              
                
                  P
                  3
                
                ⁡
                
                  (
                  
                    x
                    
                      κ
                      ⁢
                      
                        
                      
                      ⁢
                      h
                    
                  
                  )
                
              
            
          
          )
        
        ⁢
        
          d
          x
        
      
    
  




  
    
      q
      fp
    
    =
    
      2
      ⁢
      q
      ⁢
      
        
          ∫
          
            -
            fp
          
          1
        
        ⁢
        
          
            (
            
              
                
                  P
                  1
                
                ⁡
                
                  (
                  u
                  )
                
              
              +
              
                α
                ⁢
                
                  
                
                ⁢
                
                  
                    P
                    3
                  
                  ⁡
                  
                    (
                    u
                    )
                  
                
              
            
            )
          
          ⁢
          
            
          
          ⁢
          
            d
            x
          
        
      
    
  



This following calculation are performed with the



  
    
      
        q
        Net
      
      ⁡
      
        (
        u
        )
      
    
    =
    
      
        2
        ⁢
        
          q
          ⁡
          
            [
            
              u
              +
              
                
                  α
                  2
                
                ⁢
                
                  (
                  
                    
                      5
                      ⁢
                      
                        u
                        3
                      
                    
                    -
                    
                      3
                      ⁢
                      u
                    
                  
                  )
                
              
            
            ]
          
        
      
      =
      0
    
  




  
    
      qu
      ⁡
      
        [
        
          2
          +
          
            α
            ⁡
            
              (
              
                
                  5
                  ⁢
                  
                    u
                    2
                  
                
                -
                3
              
              )
            
          
        
        ]
      
    
    =
    0
  




  
    
      u
      ⁡
      
        [
        
          2
          +
          
            α5
            ⁢
            
              
            
            ⁢
            
              u
              2
            
          
          -
          
            3
            ⁢
            α
          
        
        ]
      
    
    =
    0
  




  
    
      u
      2
    
    =
    
      
        
          
            3
            ⁢
            α
          
          -
          2
        
        
          5
          ⁢
          α
        
      
    
  



For α≦2⁄3 the feedpoint is at the ground. Note: α>2⁄3



  
    
      
        u
        fp
      
      <
      
        
          1
          
            5
          
        
        ⁢
        
          I
          
            fp
            -
            Peak
          
        
      
    
    =
    
      
        ω
        ⁢
        
          
        
        ⁢
        
          
            q
            ⁡
            
              (
              
                1
                -
                
                  α
                  4
                
              
              )
            
          
          .
          
            

          
          ⁢
          
            q
            fp
          
        
      
      =
      
        2
        ⁢
        
          q
          0
        
        ⁢
        
          
            ∫
            
              u
              fp
            
            1
          
          ⁢
          
            
              [
              
                u
                +
                
                  
                    α
                    2
                  
                  ⁢
                  
                    (
                    
                      
                        5
                        ⁢
                        
                          u
                          3
                        
                      
                      -
                      
                        3
                        ⁢
                        u
                      
                    
                    )
                  
                
              
              ]
            
            ⁢
            
              
            
            ⁢
            
              d
              u
            
          
        
      
    
  




  
    
      q
      fp
    
    =
    
      
        
          q
          0
        
        ⁡
        
          [
          
            
              u
              2
            
            +
            
              α
              ⁡
              
                (
                
                  
                    
                      5
                      4
                    
                    ⁢
                    
                      u
                      4
                    
                  
                  -
                  
                    
                      3
                      2
                    
                    ⁢
                    
                      u
                      2
                    
                  
                
                )
              
            
          
          ]
        
      
      
        u
        fp
      
      1
    
  




  
    
      q
      fp
    
    =
    
      
        
          q
          0
        
        ⁡
        
          [
          
            
              u
              2
            
            ⁡
            
              (
              
                1
                +
                
                  
                    5
                    4
                  
                  ⁢
                  α
                  ⁢
                  
                    
                  
                  ⁢
                  
                    u
                    2
                  
                
                -
                
                  
                    3
                    2
                  
                  ⁢
                  α
                
              
              )
            
          
          ]
        
      
      
        u
        fp
      
      1
    
  




  
    
      q
      fp
    
    =
    
      
        
          q
          0
        
        ⁡
        
          [
          
            
              u
              2
            
            ⁡
            
              (
              
                1
                -
                
                  
                    3
                    ⁢
                    α
                  
                  2
                
                +
                
                  
                    
                      5
                      ⁢
                      α
                    
                    4
                  
                  ⁢
                  
                    u
                    2
                  
                
              
              )
            
          
          ]
        
      
      
        u
        fp
      
      1
    
  




  
    
      q
      fp
    
    =
    
      
        q
        0
      
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                
                  3
                  ⁢
                  α
                
                2
              
              +
              
                
                  5
                  ⁢
                  α
                
                4
              
            
            )
          
          -
          
            
              (
              
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                
                  5
                  ⁢
                  α
                
              
              )
            
            ⁢
            
              (
              
                1
                -
                
                  
                    3
                    ⁢
                    α
                  
                  2
                
                +
                
                  
                    
                      5
                      ⁢
                      α
                    
                    4
                  
                  ⁢
                  
                    
                      
                        3
                        ⁢
                        α
                      
                      -
                      2
                    
                    
                      5
                      ⁢
                      α
                    
                  
                
              
              )
            
          
        
        ]
      
    
  




  
    
      q
      fp
    
    =
    
      
        q
        0
      
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                
                  3
                  ⁢
                  α
                
                2
              
              +
              
                
                  5
                  ⁢
                  α
                
                4
              
            
            )
          
          -
          
            
              (
              
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                
                  5
                  ⁢
                  α
                
              
              )
            
            ⁢
            
              (
              
                1
                -
                
                  
                    3
                    ⁢
                    α
                  
                  2
                
                +
                
                  
                    
                      3
                      ⁢
                      α
                    
                    -
                    2
                  
                  4
                
              
              )
            
          
        
        ]
      
    
  




  
    
      q
      fp
    
    =
    
      
        q
        0
      
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                α
                4
              
            
            )
          
          -
          
            
              (
              
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                
                  5
                  ⁢
                  α
                
              
              )
            
            ⁢
            
              (
              
                
                  3
                  ⁢
                  α
                
                -
                2
              
              )
            
            ⁢
            
              (
              
                
                  1
                  4
                
                -
                
                  1
                  2
                
              
              )
            
          
        
        ]
      
    
  




  
    
      q
      fp
    
    =
    
      
        q
        0
      
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                α
                4
              
            
            )
          
          +
          
            
              (
              
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                
                  20
                  ⁢
                  α
                
              
              )
            
            ⁢
            
              (
              
                
                  3
                  ⁢
                  α
                
                -
                2
              
              )
            
          
        
        ]
      
    
  




  
    
      q
      fp
    
    =
    
      
        q
        0
      
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                α
                4
              
            
            )
          
          +
          
            
              
                (
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                )
              
              2
            
            
              20
              ⁢
              α
            
          
        
        ]
      
    
  



The following is a verification that P3 has no dipole moment




  
    
      Ψ
      3
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      ∫
      
        
          -
          L
        
        /
        2
      
      
        L
        /
        2
      
    
    ⁢
    
      
        
          
            P
            3
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        
          
            
              (
              
                x
                -
                z
              
              )
            
            2
          
        
      
      ⁢
      
        
      
      ⁢
      
        
          d
          x
        
        .
      
    
  




Expanding this in a series




  
    
      Ψ
      3
    
    ⁡
    
      (
      
        0
        ,
        z
      
      )
    
  
  =
  
    
      1
      z
    
    ⁢
    
      
        ∫
        
          
            -
            L
          
          /
          2
        
        
          L
          /
          2
        
      
      ⁢
      
        
          
            P
            3
          
          ⁡
          
            (
            
              2
              ⁢
              
                x
                /
                L
              
            
            )
          
        
        ⁢
        
          ⌊
          
            1
            +
            
              x
              z
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              2
            
            +
            
              
                (
                
                  x
                  z
                
                )
              
              3
            
            +
            
              
                
                  
                    (
                    
                      x
                      z
                    
                    )
                  
                  4
                
                ++
              
              ⁢
              
                
                  (
                  
                    x
                    z
                  
                  )
                
                n
              
            
          
          ⌋
        
        ⁢
        
          
        
        ⁢
        
          d
          x
        
      
    
  




where each term





  (
  
    1
    
      z
      
        m
        +
        1
      
    
  
  )




in the far field has an Pm(cos θ) angular dependence. Changing variables to u=2x/L and integrating term by term




  
    
      
        Ψ
        3
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁢
      
        
          ∫
          
            -
            1
          
          1
        
        ⁢
        
          
            
              P
              3
            
            ⁡
            
              (
              u
              )
            
          
          ⁢
          
            ⌊
            
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
              +
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
              +
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
            
            ⌋
          
          ⁢
          
            
          
          ⁢
          
            d
            u
          
        
      
    
  




  
    
      
        Ψ
        3
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁢
      
        
          ∫
          
            -
            1
          
          1
        
        ⁢
        
          
            
              P
              3
            
            ⁡
            
              (
              u
              )
            
          
          ⁢
          
            ⌊
            
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                3
              
              +
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                5
              
              +
              
                
                  (
                  
                    Lu
                    
                      2
                      ⁢
                      z
                    
                  
                  )
                
                
                  
                    2
                    ⁢
                    m
                  
                  +
                  1
                
              
            
            ⌋
          
          ⁢
          
            d
            u
          
        
      
    
  




  
    
      
        Ψ
        3
      
      ⁡
      
        (
        
          0
          ,
          z
        
        )
      
    
    =
    
      
        L
        
          2
          ⁢
          z
        
      
      ⁢
      
        ⌊
        
          
            
              4
              35
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              3
            
          
          +
          
            
              8
              63
            
            ⁢
            
              
                (
                
                  L
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              5
            
          
          +
          ...
          +
          
            
              
                4
                ⁢
                m
              
              
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    5
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      2
                      ⁢
                      m
                    
                    +
                    3
                  
                  )
                
              
            
            ⁢
            
              
                (
                
                  Lu
                  
                    2
                    ⁢
                    z
                  
                
                )
              
              
                
                  2
                  ⁢
                  m
                
                +
                1
              
            
          
        
        ⌋
      
    
  



The third order Legendre polynomial does not contribute to the diople field



  
    
      p
      z
    
    =
    
      
        
          
            2
            ⁢
            q
          
          
            κ
            ⁢
            
              
            
            ⁢
            h
          
        
        ⁢
        
          
            ∫
            
              
                -
                κ
              
              ⁢
              
                
              
              ⁢
              h
            
            
              κ
              ⁢
              
                
              
              ⁢
              h
            
          
          ⁢
          
            
              x
              ⁡
              
                [
                
                  
                    
                      
                        P
                        1
                      
                      ⁡
                      
                        (
                        
                          
                            x
                            /
                            κ
                          
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                        )
                      
                    
                    ⁢
                    
                      
                    
                    ⁢
                    
                      d
                      x
                    
                  
                  +
                  
                    α
                    ⁢
                    
                      
                    
                    ⁢
                    
                      
                        P
                        3
                      
                      ⁡
                      
                        (
                        
                          
                            x
                            /
                            κ
                          
                          ⁢
                          
                            
                          
                          ⁢
                          h
                        
                        )
                      
                    
                  
                
                ]
              
            
            ⁢
            
              d
              x
            
          
        
      
      =
      
        
          2
          ⁢
          q
          ⁢
          
            
          
          ⁢
          κ
          ⁢
          
            
          
          ⁢
          h
          ⁢
          
            
              ∫
              
                -
                1
              
              1
            
            ⁢
            
              
                u
                ⁡
                
                  [
                  
                    
                      
                        
                          P
                          1
                        
                        ⁡
                        
                          (
                          u
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        d
                        x
                      
                    
                    +
                    
                      α
                      ⁢
                      
                        
                      
                      ⁢
                      
                        
                          P
                          3
                        
                        ⁡
                        
                          (
                          u
                          )
                        
                      
                    
                  
                  ]
                
              
              ⁢
              
                
                  d
                  u
                
                .
                
                  

                
                ⁢
                
                  p
                  z
                
              
            
          
        
        =
        
          2
          ⁢
          q
          ⁢
          
            
          
          ⁢
          κ
          ⁢
          
            
          
          ⁢
          h
          ⁢
          
            
              ∫
              
                -
                1
              
              1
            
            ⁢
            
              
                u
                ⁡
                
                  [
                  
                    
                      
                        
                          P
                          1
                        
                        ⁡
                        
                          (
                          u
                          )
                        
                      
                      ⁢
                      
                        
                      
                      ⁢
                      
                        d
                        x
                      
                    
                    +
                    
                      α
                      ⁢
                      
                        
                      
                      ⁢
                      
                        
                          P
                          3
                        
                        ⁡
                        
                          (
                          u
                          )
                        
                      
                    
                  
                  ]
                
              
              ⁢
              
                d
                u
              
            
          
        
      
    
  




  
    
      p
      z
    
    =
    
      
        2
        ⁢
        
          
        
        ⁢
        q
        ⁢
        
          
        
        ⁢
        κ
        ⁢
        
          
        
        ⁢
        
          
            h
            ⁡
            
              [
              
                
                  u
                  3
                
                3
              
              ]
            
          
          
            -
            1
          
          1
        
      
      =
      
        4
        ⁢
        
          
        
        ⁢
        q
        ⁢
        
          
        
        ⁢
        κ
        ⁢
        
          
        
        ⁢
        
          h
          /
          3
        
      
    
  



Inversion of Oblate Spheroid Coordinates
An oblate spheroid has two limiting cases a sphere and a disk.

x=a cos h u cos ν cos φ

y=a cos h u cos ν sin φ

z=a sin h u sin ν

If one uses oblate spheroid coordinates with rotational symmetry on the z-axis.



  
    ρ
    =
    
      a
      ⁢
      
        
      
      ⁢
      cosh
      ⁢
      
        
      
      ⁢
      u
      ⁢
      
        
      
      ⁢
      cos
      ⁢
      
        
      
      ⁢
      v
    
  




  
    
      ρ
      2
    
    =
    
      
        
          a
          2
        
        ⁡
        
          (
          
            1
            +
            
              
                sinh
                2
              
              ⁢
              u
            
          
          )
        
      
      ⁢
      
        cos
        2
      
      ⁢
      v
    
  




  
    
      ρ
      2
    
    =
    
      
        
          a
          2
        
        ⁡
        
          (
          
            1
            +
            
              
                (
                
                  z
                  
                    a
                    ⁢
                    
                      
                    
                    ⁢
                    sin
                    ⁢
                    
                      
                    
                    ⁢
                    v
                  
                
                )
              
              2
            
          
          )
        
      
      ⁢
      
        cos
        2
      
      ⁢
      v
    
  




  
    
      
        ρ
        2
      
      ⁢
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        (
        
          
            
              a
              2
            
            ⁢
            
              sin
              2
            
            ⁢
            v
          
          +
          
            z
            2
          
        
        )
      
      ⁢
      
        cos
        2
      
      ⁢
      v
    
  




  
    
      
        ρ
        2
      
      +
      
        z
        2
      
    
    =
    
      a
      ⁡
      
        (
        
          
            
              sinh
              2
            
            ⁢
            u
            ⁢
            
              
            
            ⁢
            
              sin
              2
            
            ⁢
            v
          
          +
          
            
              (
              
                1
                +
                
                  
                    sinh
                    2
                  
                  ⁢
                  u
                
              
              )
            
            ⁢
            
              cos
              2
            
            ⁢
            v
          
        
        )
      
    
  




  
    
      
        (
        
          R
          a
        
        )
      
      2
    
    =
    
      (
      
        
          
            sinh
            2
          
          ⁢
          u
        
        ⁢
        
          
        
        +
        
          
            cos
            2
          
          ⁢
          v
        
      
      )
    
  




  
    
      z
      
        a
        ⁢
        
          
        
        ⁢
        sinh
        ⁢
        
          
        
        ⁢
        u
      
    
    =
    
      sin
      ⁢
      
        
      
      ⁢
      v
    
  




  
    
      ρ
      
        a
        ⁢
        
          
        
        ⁢
        cosh
        ⁢
        
          
        
        ⁢
        u
      
    
    =
    
      
        
          cos
          ⁢
          
            
          
          ⁢
          
            
              v
              ⁢
              
                

              
              (
              
                z
                
                  a
                  ⁢
                  
                    
                  
                  ⁢
                  sinh
                  ⁢
                  
                    
                  
                  ⁢
                  u
                
              
              )
            
            2
          
        
        +
        
          
            (
            
              ρ
              
                a
                ⁢
                
                  
                
                ⁢
                cosh
                ⁢
                
                  
                
                ⁢
                u
              
            
            )
          
          2
        
      
      =
      1
    
  




  
    
      
        
          z
          ′2
        
        ⁡
        
          (
          
            1
            +
            
              
                sinh
                2
              
              ⁢
              u
            
          
          )
        
      
      +
      
        
          ρ
          ′2
        
        ⁢
        
          sinh
          2
        
        ⁢
        u
      
    
    =
    
      
        sinh
        2
      
      ⁢
      
        u
        ⁡
        
          (
          
            1
            +
            
              
                sinh
                2
              
              ⁢
              u
            
          
          )
        
      
    
  




  
    
      
        sinh
        2
      
      ⁢
      
        u
        ⁡
        
          (
          
            1
            +
            
              
                sinh
                2
              
              ⁢
              u
            
          
          )
        
      
    
    -
    
      
        z
        ′2
      
      ⁡
      
        (
        
          1
          +
          
            
              sinh
              2
            
            ⁢
            u
          
        
        )
      
    
    -
    
      
        ρ
        ′2
      
      ⁢
      
        sinh
        2
      
      ⁢
      u
    
  




  
    
      
        
          sinh
          4
        
        ⁢
        u
      
      +
      
        
          sinh
          2
        
        ⁢
        
          u
          ⁡
          
            (
            
              1
              -
              
                z
                2
              
              -
              
                ρ
                2
              
            
            )
          
        
        ⁢
        
          sinh
          2
        
        ⁢
        u
      
      -
      
        z
        2
      
    
    =
    0
  




  
    
      
        
          sinh
          4
        
        ⁢
        u
      
      -
      
        
          (
          
            
              z
              ′2
            
            +
            
              ρ
              ′2
            
            -
            1
          
          )
        
        ⁢
        
          sinh
          2
        
        ⁢
        u
      
      -
      
        z
        ′2
      
    
    =
    0
  




  
    
      
        
          sinh
          4
        
        ⁢
        u
      
      -
      
        
          (
          
            
              z
              ′2
            
            +
            
              ρ
              ′2
            
            -
            1
          
          )
        
        ⁢
        
          sinh
          2
        
        ⁢
        u
      
      +
      
        
          
            (
            
              
                z
                ′2
              
              +
              
                ρ
                ′2
              
              -
              1
            
            )
          
          2
        
        4
      
      -
      
        
          
            (
            
              
                z
                ′2
              
              +
              
                ρ
                ′2
              
              -
              1
            
            )
          
          2
        
        4
      
      -
      
        z
        ′2
      
    
    =
    0
  




  
    
      
        
          (
          
            
              
                sinh
                2
              
              ⁢
              u
            
            -
            
              
                (
                
                  
                    z
                    ′2
                  
                  +
                  
                    ρ
                    ′2
                  
                  -
                  1
                
                )
              
              2
            
          
          )
        
        2
      
      -
      
        
          
            (
            
              
                z
                ′2
              
              +
              
                ρ
                ′2
              
              -
              1
            
            )
          
          2
        
        4
      
      -
      
        z
        ′2
      
    
    =
    0
  




  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          (
          
            
              z
              ′2
            
            +
            
              ρ
              ′2
            
            -
            1
          
          )
        
        2
      
      +
      
        
          
            
              
                (
                
                  
                    z
                    ′2
                  
                  +
                  
                    ρ
                    ′2
                  
                  -
                  1
                
                )
              
              2
            
            +
            
              4
              ⁢
              
                z
                ′2
              
            
          
        
        2
      
    
  



There are three algebraically equivalent solutions. The first case is for r<a



  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          
            2
            ⁢
            
              z
              2
            
          
          
            
              (
              
                
                  a
                  2
                
                -
                
                  r
                  2
                
              
              )
            
            +
            
              
                
                  
                    (
                    
                      
                        a
                        2
                      
                      -
                      
                        r
                        2
                      
                    
                    )
                  
                  2
                
                +
                
                  4
                  ⁢
                  
                    a
                    2
                  
                  ⁢
                  
                    z
                    2
                  
                
              
            
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
      <
      a
    
  




  
    
      if
      ⁢
      
        
      
      ⁢
      z
    
    =
    0
  




  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          
            
              ρ
              2
            
            -
            
              a
              2
            
          
          
            a
            2
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        z
      
      =
      
        
          0
          ⁢
          
            
          
          ⁢
          and
          ⁢
          
            
          
          ⁢
          ρ
        
        >
        a
      
    
  




  
    
      if
      ⁢
      
        
      
      ⁢
      ρ
    
    =
    0
  




  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          
            (
            
              z
              a
            
            )
          
          2
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        ρ
      
      =
      0
    
  




  
    
      The
      ⁢
      
        
      
      ⁢
      second
      ⁢
      
        
      
      ⁢
      case
      ⁢
      
        
      
      ⁢
      r
    
    =
    a
  




  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          z
          a
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
      =
      a
    
  




  
    
      The
      ⁢
      
        
      
      ⁢
      third
      ⁢
      
        
      
      ⁢
      case
      ⁢
      
        
      
      ⁢
      r
    
    >
    a
  




  
    
      
        sinh
        2
      
      ⁢
      u
    
    =
    
      
        
          1
          
            2
            ⁢
            
              a
              2
            
          
        
        ⁢
        
          (
          
            
              (
              
                
                  r
                  2
                
                -
                
                  a
                  2
                
              
              )
            
            +
            
              
                
                  
                    (
                    
                      
                        r
                        ′2
                      
                      -
                      
                        a
                        2
                      
                    
                    )
                  
                  2
                
                +
                
                  4
                  ⁢
                  
                    a
                    2
                  
                  ⁢
                  
                    z
                    ′2
                  
                
              
            
          
          )
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
      >
      a
    
  




  
    
      z
      
        a
        ⁢
        
          
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        v
      
    
    =
    
      
        sinh
        ⁢
        
          
        
        ⁢
        u
        ⁢
        
          

        
        ⁢
        
          ρ
          
            a
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
          
        
      
      =
      
        
          
            cosh
            ⁢
            
              
            
            ⁢
            
              
                u
                ⁢
                
                  

                
                (
                
                  ρ
                  
                    a
                    ⁢
                    
                      
                    
                    ⁢
                    cos
                    ⁢
                    
                      
                    
                    ⁢
                    v
                  
                
                )
              
              2
            
          
          -
          
            
              (
              
                z
                
                  a
                  ⁢
                  
                    
                  
                  ⁢
                  sin
                  ⁢
                  
                    
                  
                  ⁢
                  v
                
              
              )
            
            2
          
        
        =
        
          
            1
            ⁢
            
              

            
            ⁢
            
              
                
                  (
                  
                    
                      ρ
                      ′
                    
                    ⁢
                    sin
                    ⁢
                    
                      
                    
                    ⁢
                    v
                  
                  )
                
                2
              
              -
              
                
                  (
                  
                    
                      z
                      ′
                    
                    ⁢
                    cos
                    ⁢
                    
                      
                    
                    ⁢
                    v
                  
                  )
                
                2
              
            
          
          =
          
            
              
                cos
                2
              
              ⁢
              v
              ⁢
              
                
              
              ⁢
              
                sin
                2
              
              ⁢
              v
              ⁢
              
                

              
              ⁢
              
                
                  
                    ρ
                    ′2
                  
                  ⁢
                  
                    sin
                    2
                  
                  ⁢
                  v
                
                -
                
                  
                    z
                    ′2
                  
                  ⁡
                  
                    (
                    
                      1
                      -
                      
                        
                          sin
                          2
                        
                        ⁢
                        v
                      
                    
                    )
                  
                
              
            
            =
            
              
                
                  (
                  
                    1
                    -
                    
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                  
                  )
                
                ⁢
                
                  sin
                  2
                
                ⁢
                v
                ⁢
                
                  

                
                ⁢
                
                  
                    
                      ρ
                      ′2
                    
                    ⁢
                    
                      sin
                      2
                    
                    ⁢
                    v
                  
                  -
                  
                    
                      z
                      ′2
                    
                    ⁡
                    
                      (
                      
                        1
                        -
                        
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                      
                      )
                    
                  
                  -
                  
                    
                      (
                      
                        1
                        -
                        
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                      
                      )
                    
                    ⁢
                    
                      sin
                      2
                    
                    ⁢
                    v
                  
                
              
              =
              
                
                  0
                  ⁢
                  
                    

                  
                  ⁢
                  
                    
                      
                        ρ
                        ′2
                      
                      ⁢
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                    +
                    
                      
                        z
                        ′2
                      
                      ⁡
                      
                        (
                        
                          
                            
                              sin
                              2
                            
                            ⁢
                            v
                          
                          -
                          1
                        
                        )
                      
                    
                    +
                    
                      
                        (
                        
                          
                            
                              sin
                              2
                            
                            ⁢
                            v
                          
                          -
                          1
                        
                        )
                      
                      ⁢
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                  
                
                =
                
                  
                    0
                    ⁢
                    
                      

                    
                    ⁢
                    
                      
                        
                          sin
                          4
                        
                        ⁢
                        v
                      
                      +
                      
                        
                          (
                          
                            
                              ρ
                              ′2
                            
                            +
                            
                              z
                              ′2
                            
                            -
                            1
                          
                          )
                        
                        ⁢
                        
                          sin
                          2
                        
                        ⁢
                        v
                      
                      -
                      
                        z
                        ′2
                      
                    
                  
                  =
                  
                    
                      0
                      ⁢
                      
                        

                      
                      ⁢
                      
                        
                          
                            sin
                            4
                          
                          ⁢
                          v
                        
                        +
                        
                          
                            (
                            
                              
                                ρ
                                ′2
                              
                              +
                              
                                z
                                ′2
                              
                              -
                              1
                            
                            )
                          
                          ⁢
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                        +
                        
                          
                            
                              (
                              
                                
                                  ρ
                                  ′2
                                
                                +
                                
                                  z
                                  ′2
                                
                                -
                                1
                              
                              )
                            
                            2
                          
                          4
                        
                        -
                        
                          
                            
                              (
                              
                                
                                  ρ
                                  ′2
                                
                                +
                                
                                  z
                                  ′2
                                
                                -
                                1
                              
                              )
                            
                            2
                          
                          4
                        
                        -
                        
                          z
                          ′2
                        
                      
                    
                    =
                    
                      
                        0
                        ⁢
                        
                          

                        
                        ⁢
                        
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                      
                      =
                      
                        
                          
                            -
                            
                              1
                              2
                            
                          
                          ⁢
                          
                            (
                            
                              
                                ρ
                                ′2
                              
                              +
                              
                                z
                                ′2
                              
                              -
                              1
                            
                            )
                          
                        
                        +
                        
                          
                            1
                            2
                          
                          ⁢
                          
                            
                              
                                
                                  (
                                  
                                    
                                      ρ
                                      ′2
                                    
                                    +
                                    
                                      z
                                      ′2
                                    
                                    -
                                    1
                                  
                                  )
                                
                                2
                              
                              +
                              
                                4
                                ⁢
                                
                                  z
                                  ′2
                                
                              
                            
                          
                        
                      
                    
                  
                
              
            
          
        
      
    
  




  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          -
          
            1
            2
          
        
        ⁢
        
          (
          
            
              r
              ′2
            
            -
            1
          
          )
        
      
      +
      
        
          1
          2
        
        ⁢
        
          
            
              
                (
                
                  
                    r
                    ′2
                  
                  -
                  1
                
                )
              
              2
            
            +
            
              4
              ⁢
              
                z
                ′2
              
            
          
        
      
    
  



There are three algebraically equivalent solution. The first case is for r<a



  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          
            1
            
              2
              ⁢
              
                a
                2
              
            
          
          ⁢
          
            (
            
              
                a
                2
              
              -
              
                r
                2
              
            
            )
          
        
        +
        
          
            1
            
              2
              ⁢
              
                a
                2
              
            
          
          ⁢
          
            
              
                
                  (
                  
                    
                      a
                      2
                    
                    -
                    
                      r
                      2
                    
                  
                  )
                
                2
              
              +
              
                4
                ⁢
                
                  a
                  2
                
                ⁢
                
                  z
                  2
                
              
            
          
          ⁢
          
            
          
          ⁢
          for
          ⁢
          
            
          
          ⁢
          r
        
      
      <
      a
    
  




  
    
      if
      ⁢
      
        
      
      ⁢
      z
    
    =
    0
  




  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          1
          
            a
            2
          
        
        ⁢
        
          (
          
            
              a
              2
            
            -
            
              ρ
              2
            
          
          )
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        z
      
      =
      0
    
  




  
    
      if
      ⁢
      
        
      
      ⁢
      ρ
    
    =
    0
  




  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          
            z
            2
          
          
            a
            2
          
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        ρ
      
      =
      0
    
  




  
    
      The
      ⁢
      
        
          
        
        ⁢
        
          
        
      
      ⁢
      second
      ⁢
      
        
          
        
        ⁢
        
          
        
      
      ⁢
      case
      ⁢
      
        
      
      ⁢
      r
    
    =
    a
  




  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          
            z
            2
          
          
            a
            2
          
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
      =
      a
    
  




  
    The
    ⁢
    
      
    
    ⁢
    third
    ⁢
    
      
    
    ⁢
    case
    ⁢
    
      
    
    ⁢
    r
    ⁢
    
      
    
    ⁢
        ⁢
    
      
    
    ⁢
    a
  




  
    
      
        sin
        2
      
      ⁢
      v
    
    =
    
      
        
          
            2
            ⁢
            
              z
              2
            
          
          
            
              (
              
                
                  r
                  2
                
                -
                
                  a
                  2
                
              
              )
            
            +
            
              
                
                  
                    (
                    
                      
                        r
                        2
                      
                      -
                      
                        a
                        2
                      
                    
                    )
                  
                  2
                
                +
                
                  4
                  ⁢
                  
                    a
                    2
                  
                  ⁢
                  
                    z
                    2
                  
                
              
            
          
        
        ⁢
        
          
        
        ⁢
        for
        ⁢
        
          
        
        ⁢
        r
      
      >
      a
    
  




  
    
      
        ⌊
        
          
            
              (
              
                3
                +
                
                  9
                  ⁢
                  
                    ξ
                    2
                  
                
              
              )
            
            ⁢
            
              arccot
              ⁡
              
                (
                ξ
                )
              
            
          
          -
          
            3
            ⁢
            
              ξ
              ⁡
              
                (
                
                  1
                  +
                  
                    ξ
                    2
                  
                
                )
              
            
            ⁢
            
              1
              
                
                  ξ
                  2
                
                +
                1
              
            
          
          -
          
            3
            ⁢
            ξ
          
          -
          
            3
            ⁢
            ξ
          
        
        ⌋
      
      -
      
        3
        ⁡
        
          [
          
            
              
                (
                
                  
                    3
                    ⁢
                    
                      ξ
                      2
                    
                  
                  +
                  1
                
                )
              
              ⁢
              
                arccot
                ⁡
                
                  (
                  ξ
                  )
                
              
            
            -
            
              3
              ⁢
              ξ
            
          
          ]
        
      
    
    =
    0
  



Calculation of the Oblate Spheroid Potential

hu=a(sin h2u+sin 2ν)1/2

hν=a(sin h2u+sin2ν)1/2

hφ=a cos h u cos ν




  
    
      
        ∇
        2
      
      ⁢
      Ψ
    
    =
    
      
        
          1
          
            
              h
              u
            
            ⁢
            
              h
              v
            
            ⁢
            
              h
              φ
            
          
        
        ⁡
        
          [
          
            
              
                ∂
                u
              
              ⁢
              
                
                  
                    h
                    v
                  
                  ⁢
                  
                    h
                    φ
                  
                
                
                  h
                  u
                
              
            
            ⁢
            
              
                ∂
                u
              
              ⁢
              
                +
                
                  
                    ∂
                    v
                  
                  ⁢
                  
                    
                      
                        h
                        u
                      
                      ⁢
                      
                        h
                        φ
                      
                    
                    
                      h
                      v
                    
                  
                
              
            
            ⁢
            
              
                ∂
                v
              
              ⁢
              
                +
                
                  
                    ∂
                    φ
                  
                  ⁢
                  
                    
                      
                        h
                        v
                      
                      ⁢
                      
                        h
                        u
                      
                    
                    
                      h
                      φ
                    
                  
                
              
            
            ⁢
            
              ∂
              φ
            
          
          ]
        
      
      ⁢
      Ψ
    
  




  
    
      
        ∇
        2
      
      ⁢
      Ψ
    
    =
    
      
        
          1
          
            
              h
              u
            
            ⁢
            
              h
              v
            
            ⁢
            
              h
              φ
            
          
        
        ⁡
        
          [
          
            
              
                ∂
                u
              
              ⁢
              a
            
            ⁢
            
              
            
            ⁢
            cosh
            ⁢
            
              
            
            ⁢
            u
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
            ⁢
            
              
                ∂
                u
              
              ⁢
              
                +
                
                  
                    ∂
                    v
                  
                  ⁢
                  a
                
              
            
            ⁢
            
              
            
            ⁢
            cosh
            ⁢
            
              
            
            ⁢
            u
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
            ⁢
            
              
            
            ⁢
            
              
                ∂
                v
              
              ⁢
              
                +
                
                  
                    ∂
                    φ
                  
                  ⁢
                  
                    
                      
                        a
                        ⁢
                        
                          
                        
                        ⁢
                        
                          sinh
                          2
                        
                        ⁢
                        u
                      
                      +
                      
                        
                          sin
                          2
                        
                        ⁢
                        v
                      
                    
                    
                      cosh
                      ⁢
                      
                        
                      
                      ⁢
                      u
                      ⁢
                      
                        
                      
                      ⁢
                      cos
                      ⁢
                      
                        
                      
                      ⁢
                      v
                    
                  
                
              
            
            ⁢
            
              ∂
              φ
            
          
          ]
        
      
      ⁢
      Ψ
    
  




  
    
      
        ∇
        2
      
      ⁢
      Ψ
    
    =
    
      
        1
        
          
            a
            3
          
          ⁢
          cosh
          ⁢
          
            
          
          ⁢
          u
          ⁢
          
            
          
          ⁢
          cos
          ⁢
          
            
          
          ⁢
          
            v
            ⁡
            
              (
              
                
                  
                    sinh
                    2
                  
                  ⁢
                  u
                
                +
                
                  
                    sin
                    2
                  
                  ⁢
                  v
                
              
              )
            
          
        
      
      ⁢
      
         
        
          
            
              [
              
                a
                ⁢
                
                  
                
                ⁢
                cos
                ⁢
                
                  
                
                ⁢
                v
                ⁢
                
                  
                    ∂
                    u
                  
                  ⁢
                  cosh
                
                ⁢
                
                  
                
                ⁢
                u
                ⁢
                
                  
                    ∂
                    u
                  
                  ⁢
                  
                    +
                    a
                  
                
                ⁢
                
                  
                
                ⁢
                cosh
                ⁢
                
                  
                
                ⁢
                u
                ⁢
                
                  
                    ∂
                    v
                  
                  ⁢
                  cos
                
                ⁢
                
                  
                
                ⁢
                v
                ⁢
                
                  
                    ∂
                    v
                  
                  ⁢
                  
                    +
                    
                      
                        
                          a
                          ⁢
                          
                            
                          
                          ⁢
                          
                            sinh
                            2
                          
                          ⁢
                          u
                        
                        +
                        
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                      
                      
                        
                          
                            cosh
                            ⁢
                            
                              
                            
                          
                          u
                        
                        ⁢
                        
                          
                            cos
                            ⁢
                            
                              
                            
                          
                          v
                        
                      
                    
                  
                
                ⁢
                
                  
                    ∂
                    φ
                  
                  ⁢
                  
                    ∂
                    φ
                  
                
              
              ]
            
            ⁢
            Ψ
            ⁢
            
              

            
            ⁢
            
              
                ∇
                2
              
              ⁢
              Ψ
            
          
          =
          
            
              ⌊
              
                
                  1
                  
                    
                      a
                      2
                    
                    ⁢
                    cosh
                    ⁢
                    
                      
                    
                    ⁢
                    
                      u
                      ⁡
                      
                        (
                        
                          
                            
                              sinh
                              2
                            
                            ⁢
                            u
                          
                          +
                          
                            
                              sin
                              2
                            
                            ⁢
                            v
                          
                        
                        )
                      
                    
                  
                
                ⁢
                
                  
                    ∂
                    u
                  
                  ⁢
                  cosh
                
                ⁢
                
                  
                
                ⁢
                u
                ⁢
                
                  
                    ∂
                    u
                  
                  ⁢
                  
                    +
                    
                      1
                      
                        
                          a
                          2
                        
                        ⁢
                        cos
                        ⁢
                        
                          
                        
                        ⁢
                        
                          v
                          ⁡
                          
                            (
                            
                              
                                
                                  sinh
                                  2
                                
                                ⁢
                                u
                              
                              +
                              
                                
                                  sin
                                  2
                                
                                ⁢
                                v
                              
                            
                            )
                          
                        
                      
                    
                  
                
                ⁢
                
                  
                    ∂
                    v
                  
                  ⁢
                  cos
                
                ⁢
                
                  
                
                ⁢
                v
                ⁢
                
                  
                    ∂
                    v
                  
                  ⁢
                  
                    +
                    
                      1
                      
                        
                          a
                          2
                        
                        ⁢
                        
                          cosh
                          2
                        
                        ⁢
                        u
                        ⁢
                        
                          
                        
                        ⁢
                        
                          cos
                          2
                        
                        ⁢
                        v
                      
                    
                  
                
                ⁢
                
                  
                    ∂
                    φ
                  
                  ⁢
                  
                    ∂
                    φ
                  
                
              
              ⌋
            
            ⁢
            Ψ
          
        
      
    
  



If we assume that constant u forms equipotential surfaces the Ψ(u)




  
    
      ∇
      2
    
    ⁢
    Ψ
  
  =
  
    
      
        [
        
          
            1
            
              
                a
                2
              
              ⁢
              cosh
              ⁢
              
                
              
              ⁢
              
                u
                ⁡
                
                  (
                  
                    
                      
                        sinh
                        2
                      
                      ⁢
                      u
                    
                    +
                    
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                  
                  )
                
              
            
          
          ⁢
          
            
              ∂
              u
            
            ⁢
            cosh
          
          ⁢
          
            
          
          ⁢
          u
          ⁢
          
            ∂
            u
          
        
        ]
      
      ⁢
      Ψ
    
    =
    0
  




the solution is cos h u∂uΨ=K





  
    Ψ
    ⁡
    
      (
      u
      )
    
  
  =
  
    
      ∫
      
        
          K
          
            cosh
            ⁢
            
              
            
            ⁢
            u
          
        
        ⁢
        
          d
          u
        
      
    
    =
    
      
        K
        ⁢
        
          
        
        ⁢
        
          arctan
          ⁡
          
            (
            
              sinh
              ⁢
              
                
              
              ⁢
              u
            
            )
          
        
      
      +
      C
    
  




the solution is





  
    Ψ
    ⁡
    
      (
      u
      )
    
  
  =
  
    K
    ⁡
    
      (
      
        
          π
          2
        
        -
        
          arctan
          ⁡
          
            (
            
              sinh
              ⁢
              
                
              
              ⁢
              u
            
            )
          
        
      
      )
    
  




The constant K is related to the total charge.




  
    E
    u
  
  =
  
    
      1
      
        
          a
          ⁡
          
            (
            
              
                
                  sinh
                  2
                
                ⁢
                u
              
              +
              
                
                  sin
                  2
                
                ⁢
                v
              
            
            )
          
        
        
          1
          /
          2
        
      
    
    ⁢
    
      
        ∂
        u
      
      ⁢
      Ψ
    
  




q=ε0∫EudA

where dA=hνhφdvdφ=a2(sin h2u+sin2 ν)1/2 cos h u cos νdvdφ



  
    q
    =
    
      
        ɛ
        0
      
      ⁢
      
        ∫
        
          ∫
          
            
              1
              
                
                  a
                  ⁡
                  
                    (
                    
                      
                        
                          sinh
                          2
                        
                        ⁢
                        u
                      
                      +
                      
                        
                          sin
                          2
                        
                        ⁢
                        v
                      
                    
                    )
                  
                
                
                  1
                  /
                  2
                
              
            
            ⁢
            
              K
              
                a
                ⁢
                
                  
                
                ⁢
                cosh
                ⁢
                
                  
                
                ⁢
                u
              
            
            ⁢
            
              
                
                  a
                  2
                
                ⁡
                
                  (
                  
                    
                      
                        sinh
                        2
                      
                      ⁢
                      u
                    
                    +
                    
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                  
                  )
                
              
              
                1
                /
                2
              
            
            ⁢
            cosh
            ⁢
            
              
            
            ⁢
            u
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
            ⁢
            
              d
              v
            
            ⁢
            
              d
              φ
            
          
        
      
    
  




  
    q
    =
    
      
        ɛ
        0
      
      ⁢
      
        
          ∫
          
            
              -
              π
            
            /
            2
          
          
            π
            /
            2
          
        
        ⁢
        
          
            ∫
            0
            
              2
              ⁢
              π
            
          
          ⁢
          
            a
            ⁢
            
              
            
            ⁢
            K
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
            ⁢
            
              
            
            ⁢
            
              d
              v
            
            ⁢
            
              
            
            ⁢
            
              d
              φ
            
          
        
      
    
  




  
    q
    =
    
      
        
          ɛ
          0
        
        ⁢
        
          
            ∫
            
              
                -
                π
              
              /
              2
            
            
              π
              /
              2
            
          
          ⁢
          
            2
            ⁢
            π
            ⁢
            
              
            
            ⁢
            a
            ⁢
            
              
            
            ⁢
            K
            ⁢
            
              
            
            ⁢
            cos
            ⁢
            
              
            
            ⁢
            v
            ⁢
            
              
            
            ⁢
            
              d
              v
            
          
        
      
      =
      
        
          
            ɛ
            0
          
          ⁢
          2
          ⁢
          π
          ⁢
          
            
          
          ⁢
          K
          ⁢
          
            
          
          ⁢
          
            
              a
              ⁡
              
                [
                
                  sin
                  ⁢
                  
                    
                  
                  ⁢
                  v
                
                ]
              
            
            
              
                -
                π
              
              /
              2
            
            
              π
              /
              2
            
          
        
        =
        
          
            ɛ
            0
          
          ⁢
          4
          ⁢
          π
          ⁢
          
            
          
          ⁢
          a
          ⁢
          
            
          
          ⁢
          K
        
      
    
  




  
    K
    =
    
      
        
          q
          /
          4
        
        ⁢
        
          
        
        ⁢
        π
        ⁢
        
          
        
        ⁢
        a
        ⁢
        
          
        
        ⁢
        
          ɛ
          0
        
        ⁢
        
          Ψ
          ⁡
          
            (
            u
            )
          
        
      
      =
      
        
          q
          
            
              4
              ⁢
              
                
              
              ⁢
              π
              ⁢
              
                
              
              ⁢
              a
              ⁢
              
                
              
              ⁢
              
                ɛ
                0
              
            
            ⁢
            
              
            
          
        
        ⁢
        
          (
          
            
              π
              2
            
            -
            
              arctan
              ⁡
              
                (
                
                  sinh
                  ⁢
                  
                    
                  
                  ⁢
                  u
                
                )
              
            
          
          )
        
      
    
  



The potential on the surface is



  
    
      Ψ
      ⁡
      
        (
        0
        )
      
    
    =
    
      q
      
        8
        ⁢
        
          
        
        ⁢
        a
        ⁢
        
          
        
        ⁢
        
          ɛ
          0
        
      
    
  




  
    C
    =
    
      8
      ⁢
      
        
      
      ⁢
      a
      ⁢
      
        
      
      ⁢
      
        ɛ
        0
      
    
  




  
    
      E
      u
    
    =
    
      
        1
        
          
            a
            ⁡
            
              (
              
                
                  
                    sinh
                    2
                  
                  ⁢
                  u
                
                +
                
                  
                    sin
                    2
                  
                  ⁢
                  v
                
              
              )
            
          
          
            1
            /
            2
          
        
      
      ⁢
      
        q
        
          4
          ⁢
          
            
          
          ⁢
          π
          ⁢
          
            
          
          ⁢
          a
          ⁢
          
            
          
          ⁢
          
            ɛ
            0
          
          ⁢
          cosh
          ⁢
          
            
          
          ⁢
          u
        
      
    
  




  
    
      
        E
        u
      
      ⁡
      
        (
        v
        )
      
    
    =
    
      q
      
        
          a
          2
        
        ⁢
        4
        ⁢
        π
        ⁢
        
          
        
        ⁢
        
          ɛ
          0
        
        ⁢
        sin
        ⁢
        
          
        
        ⁢
        v
      
    
  




  
    
      where
      ⁢
      
        
      
      ⁢
      ρ
    
    =
    
      a
      ⁢
      
        
      
      ⁢
      cos
      ⁢
      
        
      
      ⁢
      v
    
  




  
    
      
        E
        u
      
      ⁡
      
        (
        ρ
        )
      
    
    =
    
      q
      
        4
        ⁢
        
          
        
        ⁢
        π
        ⁢
        
          
        
        ⁢
        a
        ⁢
        
          
        
        ⁢
        
          ɛ
          0
        
        ⁢
        
          
            (
            
              
                a
                2
              
              -
              
                ρ
                2
              
            
            )
          
        
      
    
  




  
    σ
    =
    
      
        
          ɛ
          0
        
        ⁢
        
          
            E
            u
          
          ⁡
          
            (
            ρ
            )
          
        
      
      =
      
        q
        
          4
          ⁢
          
            
          
          ⁢
          π
          ⁢
          
            
          
          ⁢
          a
          ⁢
          
            
              (
              
                
                  a
                  2
                
                -
                
                  ρ
                  2
                
              
              )
            
          
        
      
    
  




  
    σ
    =
    
      
        
          ɛ
          0
        
        ⁢
        
          
            E
            u
          
          ⁡
          
            (
            ρ
            )
          
        
      
      =
      
        q
        
          4
          ⁢
          
            
          
          ⁢
          π
          ⁢
          
            
          
          ⁢
          a
          ⁢
          
            
              (
              
                
                  a
                  2
                
                -
                
                  ρ
                  2
                
              
              )
            
          
        
      
    
  



General Solution of Laplace's Equation




  
    
      
        
          
            ∇
            2
          
          ⁢
          Ψ
        
        =
        
          ⌊
          
            
              
                1
                
                  
                    a
                    2
                  
                  ⁢
                  cosh
                  ⁢
                  
                    
                  
                  ⁢
                  
                    u
                    ⁡
                    
                      (
                      
                        
                          
                            sin
                            2
                          
                          ⁢
                          u
                        
                        +
                        
                          
                            sin
                            2
                          
                          ⁢
                          v
                        
                      
                      )
                    
                  
                
              
              ⁢
              
                
                  ∂
                  u
                
                ⁢
                
                   
                  
                    cosh
                    ⁢
                    
                      
                    
                    ⁢
                    u
                    ⁢
                    
                      
                    
                    ⁢
                    
                      
                        ∂
                        u
                      
                      ⁢
                      
                        +
                        
                          1
                          
                            
                              a
                              2
                            
                            ⁢
                            cos
                            ⁢
                            
                              
                            
                            ⁢
                            
                              v
                              ⁡
                              
                                (
                                
                                  
                                    
                                      sinh
                                      2
                                    
                                    ⁢
                                    u
                                  
                                  +
                                  
                                    
                                      sin
                                      2
                                    
                                    ⁢
                                    v
                                  
                                
                                )
                              
                            
                          
                        
                      
                    
                    ⁢
                    
                      
                        ∂
                        v
                      
                      ⁢
                      cos
                    
                    ⁢
                    
                      
                    
                    ⁢
                    v
                    ⁢
                    
                      
                        ∂
                        v
                      
                      ⁢
                      
                        +
                        
                          1
                          
                            
                              a
                              2
                            
                            ⁢
                            
                              cosh
                              2
                            
                            ⁢
                            u
                            ⁢
                            
                              
                            
                            ⁢
                            
                              cos
                              2
                            
                            ⁢
                            v
                          
                        
                      
                    
                    ⁢
                    
                      
                        ∂
                        φ
                      
                      ⁢
                      
                        ∂
                        φ
                      
                    
                  
                  ⌋
                
              
              ⁢
              
                 
                 
              
              ⁢
              Ψ
              ⁢
              
                
              
              ⁢
              
                
                  a
                  2
                
                ⁡
                
                  (
                  
                    
                      
                        sinh
                        2
                      
                      ⁢
                      u
                    
                    +
                    
                      
                        sin
                        2
                      
                      ⁢
                      v
                    
                  
                  )
                
              
              ⁢
              
                
                  ∇
                  2
                
                ⁢
                Ψ
              
            
            =
            
              
                
                  ⌊
                  
                    
                      1
                      
                        cosh
                        ⁢
                        
                          
                        
                        ⁢
                        u
                      
                    
                    ⁢
                    
                      
                        ∂
                        u
                      
                      ⁢
                      cosh
                    
                    ⁢
                    
                      
                    
                    ⁢
                    u
                    ⁢
                    
                      
                        ∂
                        u
                      
                      ⁢
                      
                        +
                        
                          1
                          
                            cos
                            ⁢
                            
                              
                            
                            ⁢
                            v
                          
                        
                      
                    
                    ⁢
                    
                      
                        ∂
                        v
                      
                      ⁢
                      cos
                    
                    ⁢
                    
                      
                    
                    ⁢
                    v
                    ⁢
                    
                      
                        ∂
                        v
                      
                      ⁢
                      
                        +
                        
                          
                            (
                            
                              
                                
                                  sinh
                                  2
                                
                                ⁢
                                u
                              
                              +
                              
                                
                                  sin
                                  2
                                
                                ⁢
                                v
                              
                            
                            )
                          
                          
                            
                              
                                cosh
                                ⁢
                                
                                  
                                
                              
                              2
                            
                            ⁢
                            u
                            ⁢
                            
                              
                            
                            ⁢
                            
                              cos
                              2
                            
                            ⁢
                            v
                          
                        
                      
                    
                    ⁢
                    
                      
                        ∂
                        φ
                      
                      ⁢
                      
                        ∂
                        φ
                      
                    
                  
                  ⌋
                
                ⁢
                Ψ
              
              =
              0
            
          
        
      
    
    
      
        
      
    
  




In the cylindrical symmetry case ∂φΨ=0 and separation of variables Ψ=Φ(ν)K(u) The above equation reduces to



  
    
      [
      
        
          l
          ⁡
          
            (
            
              l
              +
              1
            
            )
          
        
        +
        
          
            1
            
              cos
              ⁢
              
                
              
              ⁢
              v
            
          
          ⁢
          
            
              ∂
              v
            
            ⁢
            cos
          
          ⁢
          
            
          
          ⁢
          v
          ⁢
          
            ∂
            v
          
        
      
      ]
    
    ⁢
    
      Φ
      ⁡
      
        (
        v
        )
      
    
  




  
    
      and
      ⁢
      
        

      
      [
      
        
          1
          
            cosh
            ⁢
            
              
            
            ⁢
            u
          
        
        ⁢
        
          
            ∂
            u
          
          ⁢
          cosh
        
        ⁢
        
          
        
        ⁢
        u
        ⁢
        
          
            ∂
            u
          
          ⁢
          
            -
            
              l
              ⁡
              
                (
                
                  l
                  +
                  1
                
                )
              
            
          
        
      
      ]
    
    ⁢
    
      K
      ⁡
      
        (
        u
        )
      
    
  



The solution in ν is the standard Legendre Polynomials

Φ(ν)=Pl(ν)

and

K(u)=Pl(iu)

In our case, the second




  
    
      K
      2
    
    ⁡
    
      (
      
        i
        ⁢
        
          
        
        ⁢
        u
      
      )
    
  
  =
  
    
      i
      ⁢
      
        1
        2
      
      ⁢
      
        (
        
          
            3
            ⁢
            
              u
              2
            
          
          +
          1
        
        )
      
      ⁢
      
        arccot
        ⁡
        
          (
          u
          )
        
      
    
    -
    
      i
      ⁢
      
        3
        2
      
      ⁢
      u
    
  




The verifying the solution in u is the



  
    
      
        K
        2
      
      ⁡
      
        (
        
          i
          ⁢
          
            
          
          ⁢
          u
        
        )
      
    
    =
    
      
        i
        ⁢
        
          1
          2
        
        ⁢
        
          (
          
            
              3
              ⁢
              
                u
                2
              
            
            +
            1
          
          )
        
        ⁢
        
          arccot
          ⁡
          
            (
            u
            )
          
        
      
      -
      
        i
        ⁢
        
          3
          2
        
        ⁢
        u
      
    
  




  
    
      
        
          ∂
          ξ
        
        ⁢
        
          (
          
            1
            +
            
              ξ
              2
            
          
          )
        
      
      ⁢
      
        
          ∂
          ξ
        
        ⁢
        
          [
          
            
              
                1
                2
              
              ⁢
              
                (
                
                  
                    3
                    ⁢
                    
                      ξ
                      2
                    
                  
                  +
                  1
                
                )
              
              ⁢
              
                arccot
                ⁡
                
                  (
                  ξ
                  )
                
              
            
            -
            
              
                3
                2
              
              ⁢
              ξ
            
          
          ]
        
      
    
    -
    
      6
      ⁡
      
        [
        
          
            
              1
              2
            
            ⁢
            
              (
              
                
                  3
                  ⁢
                  
                    ξ
                    2
                  
                
                +
                1
              
              )
            
            ⁢
            
              arccot
              ⁡
              
                (
                ξ
                )
              
            
          
          -
          
            
              3
              2
            
            ⁢
            ξ
          
        
        ]
      
    
  




  
    
      
        
          
            ∂
            ξ
          
          ⁢
          
            (
            
              1
              +
              
                ξ
                2
              
            
            )
          
        
        ⁢
        
          ⌊
          
            
              3
              ⁢
              
                ξarccot
                ⁡
                
                  (
                  ξ
                  )
                
              
            
            -
            
              
                1
                2
              
              ⁢
              
                (
                
                  
                    3
                    ⁢
                    
                      ξ
                      2
                    
                  
                  +
                  1
                
                )
              
              ⁢
              
                1
                
                  
                    ξ
                    2
                  
                  +
                  1
                
              
            
            -
            
              3
              2
            
          
          ⌋
        
      
      -
      
        
          6
          ⁡
          
            [
            
              
                
                  1
                  2
                
                ⁢
                
                  (
                  
                    
                      3
                      ⁢
                      
                        ξ
                        2
                      
                    
                    +
                    1
                  
                  )
                
                ⁢
                
                  arccot
                  ⁡
                  
                    (
                    ξ
                    )
                  
                
              
              -
              
                
                  3
                  2
                
                ⁢
                ξ
              
            
            ]
          
        
        ⁢
        
          

        
        ⁢
        
          
            ∂
            ξ
          
          ⁢
          
            [
            
              
                3
                ⁢
                
                  ξ
                  ⁡
                  
                    (
                    
                      1
                      +
                      
                        ξ
                        2
                      
                    
                    )
                  
                
                ⁢
                
                  arccot
                  ⁡
                  
                    (
                    ξ
                    )
                  
                
              
              -
              
                
                  1
                  2
                
                ⁢
                
                  (
                  
                    
                      3
                      ⁢
                      
                        ξ
                        2
                      
                    
                    +
                    1
                  
                  )
                
              
              -
              
                
                  3
                  2
                
                ⁢
                
                  (
                  
                    1
                    +
                    
                      ξ
                      2
                    
                  
                  )
                
              
            
            ]
          
        
      
      -
      
        
          3
          ⁡
          
            [
            
              
                
                  (
                  
                    
                      3
                      ⁢
                      
                        ξ
                        2
                      
                    
                    +
                    1
                  
                  )
                
                ⁢
                
                  arccot
                  ⁡
                  
                    (
                    ξ
                    )
                  
                
              
              -
              
                3
                ⁢
                ξ
              
            
            ]
          
        
        ⁢
        
          

        
        ⁢
        
          ⌊
          
            
              
                (
                
                  3
                  +
                  
                    9
                    ⁢
                    
                      ξ
                      2
                    
                  
                
                )
              
              ⁢
              
                
              
              ⁢
              
                arccot
                ⁡
                
                  (
                  ξ
                  )
                
              
            
            -
            
              3
              ⁢
              
                ξ
                ⁡
                
                  (
                  
                    1
                    +
                    
                      ξ
                      2
                    
                  
                  )
                
              
              ⁢
              
                1
                
                  
                    ξ
                    2
                  
                  +
                  1
                
              
            
            -
            
              3
              ⁢
              ξ
            
            -
            
              3
              ⁢
              ξ
            
          
          ⌋
        
      
      -
      
        3
        ⁡
        
          [
          
            
              
                (
                
                  
                    3
                    ⁢
                    
                      ξ
                      2
                    
                  
                  +
                  1
                
                )
              
              ⁢
              
                arccot
                (
                )
              
            
            -
            
              3
              ⁢
              ξ
            
          
          ]
        
      
    
    =
    0

## Claims

1. A method for designing an electrically small antenna within an enclosing volume comprising:
designing the electrically small antenna having a general cross-sectional contour shape of an oblate spheroid from a top load portion to a stem portion below the top load portion, the oblate spheroid contour shape represented by an antenna dipole moment algorithm which includes a dipole moment term;
calculating the amplitude of the dipole moment term, including adjusting the amplitude of the dipole moment term to independently change the oblate spheroid contour shape, resulting in a change to the electric field outside the enclosing volume and a change to the electric field inside the enclosing volume.

2. The method of claim 1 where the independent change includes increasing the electric field outside the enclosing volume and decreasing the electric field inside the enclosing volume.

3. The method of claim 1 where the independent change includes decreasing the electric field outside the enclosing volume and increasing the electric field inside the enclosing volume.

4. The method of claim 1 where the independent change includes changing the radiation resistance.

5. The method of claim 4 where the radiation resistance is increased.

6. The method of claim 4 where the radiation resistance is decreased.