# Patent

## Title
Quasi-static design approach for low Q factor electrically small antennas

## Abstract

Methods and apparatuses are described that enable the design of electrically small antennas in terms of their quality factor (Q) performance or other antenna parameter. A desired charge distribution is defined and thereafter the shape of the antenna is generated, based on the corresponding solution of a quasi-static field approximation. A degree of freedom in the shaping of the antenna is incorporated into the quasi-static field approximation, via a dimensional variable. This expression has a solution set containing the minimum Q factor. By selection of an appropriate value of the dimensional variable, antennas with minimum or otherwise tailored Q values can be quickly designed.

## Background
FEDERALLY-SPONSORED RESEARCH AND DEVELOPMENT
The Quasi-Static Design Approach for Low Q Factor Electrically Small Antennas was developed with funds from the United States Government. Licensing inquiries may be directed to Office of Research and Technical Applications, Space and Naval Warfare Systems Center, San Diego, Code 2112, San Diego, Calif., 92152; telephone (619)553-2778; email: T2@spawar.navy.mil, reference NC 98163.


BACKGROUND
1. Field
The Quasi-Static Design Approach for Low Q Factor Electrically Small Antennas generally relates to antennas and more particularly relates to the design of electrically small antennas using a quasi-static/low Q-factor approach.
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.
Therefore, there has been a long-standing need in the antenna community for simpler methods and systems for the design of electrically small antennas, based on a low Q criterion.
SUMMARY
The foregoing needs addressed by the embodiments of the quasi-static design approach for low Q factor electrically small antennas, wherein methods and systems are provided that in some embodiments electrically small antennas are modeled using a quasi-static approach, where a Q formulation is developed and exploited.
In accordance with one embodiment, a method for designing an electrically small antenna is provided by designating a general form factor for an antenna, defining a charge distribution in terms of at least one form factor related parameter, computing a capacitance in terms of the charge distribution and the form factor related parameter, generating a quasi-static quality factor (Q) based on the computed capacitance and a corresponding radiation resistance, evaluating values of the generated quasi-static Q, the computed capacitance, and the radiation resistance to arrive at a desired antenna performance, and generating a desired antenna geometry using the form factor related parameter for the desired antenna performance.
In accordance with another embodiment, an electrically small antenna is provided, wherein the surface contour of the antenna has a desired Q performance, the Q performance being generated by designating a general form factor for an antenna, defining a charge distribution in terms of at least one form factor related parameter, computing a capacitance in terms of the charge distribution and the form factor related parameter, generating a quasi-static quality factor (Q) based on the computed capacitance and a corresponding radiation resistance, evaluating values of the generated quasi-static Q, the computed capacitance, and the radiation resistance to arrive at a desired antenna performance, and generating a desired antenna geometry using the form factor related parameter for the desired antenna performance.



BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is an illustration of the contour of an exemplary charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 2 is an illustration of the contour of an exemplary P1 charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 3 is a graph of the computed resistance of an exemplary P1 charge distribution antenna, with h=1⁄2, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 4 is a graph of the computed reactance of an exemplary P1 charge distribution antenna, with h=1⁄2, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 5 is a graph of the Q of the P1 charge distribution antenna, with h=1⁄2, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 6 is a graph comparing experimental measurements of an exemplary P1 charge distribution antenna, with h=1⁄2, to computer simulated/predicted values, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 7 is an illustration of the contour of an exemplary (1−α)P0+α point charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 8 is a graph of the Q/QChu ratio as a function of α and κ, for an exemplary (1−α)P0+α point charge distribution antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 9 is an illustration of the contour of an exemplary (1−α)P0+α point charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 10 is a graph of the Q/QChu ratio as a function of α and κ, for an exemplary P1+αP3 charge distribution antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 11 is a graph of the resistance, at 50 MHz and h=1⁄2 m, as a function of α and κ, for an exemplary P1+αP3 charge antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 12 is a graph of the capacitance, at h=1⁄2 m, as a function of α and κ, for an exemplary P1+αP3 charge distribution antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 13 is an illustration of the contour of an exemplary P1+αP3 charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 14 is an illustration of the contour of an exemplary P1+αP3 charge distribution antenna, having the local minimum Q solution for α=−1.835 and κ=0.88 with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 15 is a graph of the Q/QChu ratio as a function of α and κ, for α=0.9 and h=1, for an exemplary (1−α)P0+α oblate distribution spheroid charge antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 16 is a graph of the impedance for an exemplary (1−α)P0+α oblate spheroid charge antenna, with h=1⁄2 m, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 17 is a graph of the Q behavior, for an exemplary (1−α)P0+α oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 18 is graph comparing measurements of an exemplary (1−α)P0+α oblate spheroid charge distribution antenna, with h=1⁄2, to numerically derived values, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 19 is an illustration of the contour of an exemplary (1−α)P0+α oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 20 is an illustration of the contour of an exemplary (1−α)P1+α oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 21 is a graph of the Q/QChu ratio for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 22 is a graph of the β distribution as a function of κ and a, for a minimum Q for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 23 is an illustration of the contour of an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
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 distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 25 is a graph of the β distribution as a function of κ and a, for a minimum Q for an exemplary (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge distribution antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 26 is an illustration of the contour of an exemplary conical enclosing volume (1−α)P0+α oblate spheroid+β2nd harmonic oblate spheroid charge distribution antenna, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
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 distribution antenna enclosed in a 1 m cone, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 28 is an illustration of the contour of an exemplary dipole antenna, with h=1, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.
FIG. 29 is a flow chart illustrating an exemplary process, according to an embodiment of the quasi-static design approach for low Q factor electrically small antennas.



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 peak stored energy to the radiated energy, and is usually understood as a metric for qualifying the bandwidth of the antenna. This important relationship can be expressed as:

Q=ωEs/Erad,   (1)

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





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




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

Erad=RradIrms2,   (5)

where Rrad 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
                rad
              
            
          
          .
        
      
    
    
      
        (
        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, ε0 is the free space 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. A constant line charge 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 equation in cylindrical coordinates:




  
    
      
        
          
            Ψ
            ⁡
            
              (
              
                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 for the monopole,





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




is for the image monopole, Rt=√{square root over ((z−κh)2+ρ2)} 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 Φ 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 on the sphere (h is the radius), 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
              ⁢
              
                
              
              ⁢
              
                πɛ
                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 heff for the line charge distribution is heff=κh/2 . The radiation resistance Rrad is





  
    
      
        
          
            
              R
              rad
            
            ⁡
            
              (
              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
      
    
  




  
    
      h
      eff
    
    =
    
      .3652
      ⁢
      h
    
  




  
    
      R
      rad
    
    =
    
      5.336
      ⁢
      
        
          (
          hk
          )
        
        2
      
    
  




  
    C
    =
    
      1.064
      *
      
        10
        
          -
          10
        
      
      ⁢
      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 distribution in cylindrical coordinates is




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




where





  
    τ
    =
    
      
        2
        ⁢
        
          
        
        ⁢
        κ
        ⁢
        
          
        
        ⁢
        h
      
      
        
          R
          t
        
        +
        
          R
          b
        
      
    
  
  ,




Rt=√{square root over ((z−κh)2+ρ2)} is the distance to the top of the monopole and Rb=√{square root over ((z+κh)2+ρ2)} is the distance to the bottom of the image. The maximum potential is known to be at z=h and ρ=0 on the sphere 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 heff=2κh/3. The radiation resistance is





  
    
      
        
          
            
              R
              rad
            
            ⁡
            
              (
              κ
              )
            
          
          =
          
            40
            ⁢
            
              
            
            ⁢
            
              
                (
                
                  
                    2
                    ⁢
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    hk
                  
                  3
                
                )
              
              2
            
          
        
        ,
      
    
    
      
        (
        22
        )
      
    
  




and Q is





  
    
      
        
          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. 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
                eff
              
              =
              
                .46406
                ⁢
                
                  
                
                ⁢
                h
              
            
          
        
        
          
            
              
                R
                rad
              
              =
              
                8.614
                ⁢
                
                  
                    (
                    kh
                    )
                  
                  2
                
              
            
          
        
        
          
            
              C
              =
              
                8.22178
                ⁢
                
                  
                
                ⋆
                
                  
                    10
                    
                      -
                      11
                    
                  
                  ⁢
                  h
                
              
            
          
        
      
    
    
      
        (
        25
        )
      
    
  




The radiation resistance for this exemplary P1 charge distribution antenna is 60% higher than the exemplary constant charge distribution 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 distribution antenna was generated. FIG. 2 is a cross-sectional illustration of the exemplary P1 charge distribution antenna. In addition to generating the contour, a numerical model was created to simulate the exemplary P1 charge distribution 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 distribution 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 distribution 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 distribution 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 distribution antenna does not suffer from typical half wavelength anti-resonance problems.
P0+Point Charge Distribution Antenna
Progressing from the exemplary P1 charge distribution 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 distribution ACD antenna modified with a point charge at the top.
The total charge q is split between the line charge distribution and the point charge, with the point charge being αq and the line charge distribution being (1−α)q/κh, where α is a proportionality constant. Using similar dimensional constraints as imposed in the constant line charge distribution ACD antenna, the expression for the potential field becomes




  
    
      
        
          Φ
          ⁡
          
            (
            
              ρ
              ,
              z
            
            )
          
        
        =
        
          
            
              
                
                  (
                  
                    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
              
              ⁢
              
                R
                t
              
            
          
          -
          
            
              
                α
                ⁢
                
                  
                
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                
                  R
                  b
                
              
            
            .
          
        
      
    
    
      
        (
        26
        )
      
    
  




This potential is known to be maximum at ρ=0 and z=h on the sphere, 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)=(1+α)κh/2.   (29)

And the radiation resistance is

Rrad(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 α values. At 50 MHz and h=1⁄2, the value of κ for the minimum Q is given in the following Table 1.











TABLE 1





α
κ
Minimum Q
R rad
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
          
        
      
    
  
  
    
      
        
          h
          eff
        
        =
        
          0.5529
          ⁢
          
            
          
          ⁢
          h
        
      
    
  
  
    
      
        
          R
          rad
        
        =
        
          12.22
          *
          
            
              (
              kh
              )
            
            2
          
        
      
    
  
  
    
      
        C
        =
        
          6.7
          ⁢
          
            
          
          ⋆
          
            
              10
              
                -
                11
              
            
            ⁢
            
              h
              .
            
          
        
      
    
  




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 distribution (“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
        )
      
    
  




where h is the radius of the sphere enclosing the antenna.

Using Eq. 32, only to the first order term,




  
    1
    
      
        (
        kh
        )
      
      3
    
  
  ,




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 Distribution Antenna
The above demonstrated exemplary P0+point charge distribution 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 distribution 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
                  
                  )
                
                ⁢
                q
              
              
                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 on the sphere, 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
            
            )
          
        
        =
        
          
            
              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

Rrad=40[κheffk(2⁄3+α/3)]2.   (37)

The Q becomes





  
    
      
        
          Q
          ⁡
          
            (
            
              α
              ,
              κ
            
            )
          
        
        =
        
          
            
              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 κ values with a fixed α. At 50 MHz and h=1⁄2, table 2 below details the resulting evaluation.












TABLE 2





α
κ
Minimum Q
R rad
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
      
    
  




  
    
      h
      eff
    
    =
    
      0.5626
      ⁢
      h
    
  




  
    
      R
      rad
    
    =
    
      12.66
      *
      
        
          (
          kh
          )
        
        2
      
    
  




  
    C
    =
    
      6.6
      *
      
        10
        
          -
          11
        
      
      ⁢
      
        h
        .
      
    
  



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 distribution 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 the embodiments.
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. The quasi-static fields Pn(θ)/rn+1 are terms in the vector spherical harmonics.
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 may shift 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=−L/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
                      ⁢
                      
                        zτ
                        ⁡
                        
                          (
                          
                            
                              
                                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 distribution 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 Eq. 17 and Ψ3) generated from the Legendre polynomials, the expression being





  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                2
                ⁢
                q
              
              
                4
                ⁢
                
                  πɛ
                  0
                
                ⁢
                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 qNet 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
            rad
          
          =
          
            
              40
              ⁢
              
                
                  (
                  
                    
                      2
                      ⁢
                      κ
                      ⁢
                      
                        
                      
                      ⁢
                      hk
                    
                    
                      3
                      ⁢
                      
                        (
                        
                          1
                          -
                          
                            α
                            4
                          
                        
                        )
                      
                    
                  
                  )
                
                2
              
              ⁢
              
                
              
              ⁢
              for
              ⁢
              
                
              
              ⁢
              α
            
            ≤
            
              2
              /
              3
            
          
        
        ⁢
        
          

        
        ⁢
        
          
            R
            rad
          
          =
          
            
              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
                          
                        
                      
                    
                    )
                  
                
                ]
              
            
            .
          
        
      
    
    
      
        (
        45
        )
      
    
  




Substituting L=2κh recasts Eq. 46 as





  
    
      
        
          
            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
      -
      z
    
  
  =
  
    
      
        2
        ⁢
        q
      
      
        4
        ⁢
        
          πɛ
          0
        
        ⁢
        κ
        ⁢
        
          
        
        ⁢
        h
      
    
    ⁢
    
      (
      
        
          Ψ
          1
        
        +
        
          αΨ
          3
        
      
      )
    
  




on the interval κh<z≦1, this value must be larger than a calculated Φmax-sphere 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 and h=1⁄2, the solution to Eq. 46 has two minimums: an absolute minimum at Q=28.706 corresponding to Q/QChu=4.1207 at κ=0.6513 and α=0.8600, with Rrad=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 Rrad=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 distribution antenna. And FIG. 12 is a plot of the capacitance as a function of α and κ, for an exemplary P1+αP3 charge distribution 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 distribution 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 distribution 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
      
    
  




  
    
      h
      eff
    
    =
    
      .53967
      ⁢
      h
    
  




  
    
      R
      rad
    
    =
    
      11.65
      ⁢
      
        
          (
          kh
          )
        
        2
      
    
  




  
    C
    =
    
      6.943
      *
      
        10
        
          -
          11
        
      
      ⁢
      h
    
  



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+α P3 charge distribution antenna, the mathematical artifact of having a negative charge distribution 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 distribution (Po) antenna reduces the Q of the antenna. This point charge is equivalent to giving a spherical end cap to the Po 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 surface charge 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 Po charge distribution 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=αsin h u cos v

ρ=αcos h u sin v.   (47)

(Note: α 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 and α:





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




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





  
    
      
        
          
            Φ
            ⁡
            
              (
              
                ρ
                ,
                z
              
              )
            
          
          =
          
            
              
                
                  
                    (
                    
                      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
                  
                  ⁢
                  a
                
              
              ⁡
              
                [
                
                  
                    arctan
                    ⁡
                    
                      (
                      
                        sinh
                        ⁢
                        
                          
                        
                        ⁢
                        
                          u
                          b
                        
                      
                      )
                    
                  
                  -
                  
                    arctan
                    ⁡
                    
                      (
                      
                        sinh
                        ⁢
                        
                          
                        
                        ⁢
                        
                          u
                          t
                        
                      
                      )
                    
                  
                
                ]
              
            
          
        
        ,
      
    
    
      
        (
        49
        )
      
    
  




where the parameter α also 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 α and κ, where α=0.9. For varying values of α, a unique solution can be found for Eq. 49. Each of these solutions can be optimized to result in a κ and α value. A tabulation of these values at 50 MHz and h=1⁄2 is shown in Table 3 below.












TABLE 3





α
κ
a
Minimum Q
R rad
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 spheroid, the antenna parameters are




  
    
      
        Q
        =
        
          2.1544
          
            
              (
              kh
              )
            
            3
          
        
      
    
  
  
    
      
        
          h
          eff
        
        =
        
          0.670
          ⁢
          
            
          
          ⁢
          h
        
      
    
  
  
    
      
        
          R
          rad
        
        =
        
          17.9
          *
          
            
              (
              kh
              )
            
            2
          
        
      
    
  
  
    
      
        C
        =
        
          8.632
          *
          
            10
            
              -
              11
            
          
          ⁢
          h
        
      
    
  




The resulting Q is found to be almost half of the result for the “bulb” antenna, which utilized a constant line charge distribution 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 distribution 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 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 distribution antenna.
P1 with Oblate Spheroid
The same above exemplary technique can be expanded upon using a P1 line charge distribution instead of a Po line charge distribution, with an oblate spheroid end charge distribution. Using the P1 line charge distribution, the expression for the maximum potential field becomes




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




At 50 MHz and h=1⁄2, the results of this procedure is shown in Table 4 below.












TABLE 4





α
κ
a
Minimum Q
R rad
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
          
        
      
    
  
  
    
      
        
          h
          eff
        
        =
        
          0.6792
          ⁢
          
            
          
          ⁢
          h
        
      
    
  
  
    
      
        
          R
          rad
        
        =
        
          18.46
          *
          
            
              (
              kh
              )
            
            2
          
        
      
    
  
  
    
      
        C
        =
        
          9.192
          *
          
            10
            
              -
              11
            
          
          ⁢
          h
        
      
    
  




For h=1⁄2m 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 distribution 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
                
                ⁡
                
                  (
                  
                    i
                    ⁢
                    
                      
                    
                    ⁢
                    u
                  
                  )
                
              
              ⁢
              
                
                  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), Po(v) was used in the oblate spheroidal calculation. The second term K1(iu)P1(v) 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(v) 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 Kn are



  
    
      
        K
        0
      
      ⁡
      
        (
        
          i
          ⁢
          
            
          
          ⁢
          u
        
        )
      
    
    =
    
      i
      ⁢
      
        
      
      ⁢
      
        arccot
        ⁡
        
          (
          u
          )
        
      
    
  




  
    
      
        K
        1
      
      ⁡
      
        (
        
          i
          ⁢
          
            
          
          ⁢
          u
        
        )
      
    
    =
    
      
        
          u
          ⁢
          arccot
        
        ⁡
        
          (
          u
          )
        
      
      -
      1
    
  




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



The general form of these terms being





  
    
      
        
          
            
              K
              n
            
            ⁡
            
              (
              
                i
                ⁢
                
                  
                
                ⁢
                u
              
              )
            
          
          =
          
            
              
                -
                i
              
              ⁢
              
                
              
              ⁢
              
                
                  P
                  n
                
                ⁡
                
                  (
                  
                    i
                    ⁢
                    
                      
                    
                    ⁢
                    u
                  
                  )
                
              
              ⁢
              
                arccos
                ⁡
                
                  (
                  u
                  )
                
              
            
            -
            
              
                
                  
                    2
                    ⁢
                    n
                  
                  -
                  1
                
                n
              
              ⁢
              
                
                  P
                  
                    n
                    -
                    1
                  
                
                ⁡
                
                  (
                  
                    i
                    ⁢
                    
                      
                    
                    ⁢
                    u
                  
                  )
                
              
            
            -
            
              
                
                  (
                  
                    
                      2
                      ⁢
                      n
                    
                    -
                    5
                  
                  )
                
                
                  3
                  ⁢
                  
                    (
                    
                      n
                      -
                      1
                    
                    )
                  
                
              
              ⁢
              
                
                  
                    P
                    
                      n
                      -
                      3
                    
                  
                  ⁡
                  
                    (
                    
                      i
                      ⁢
                      
                        
                      
                      ⁢
                      u
                    
                    )
                  
                
                --
              
              ⁢
              
                
              
              ⁢
              ...
            
          
        
        ⁢
        
          
        
        ,
      
    
    
      
        (
        52
        )
      
    
  




where Pn are Legendre polynomials and the sum terminates at the first negative order term.

For simplicity only the P0(v)and P2(v) polynomials were used as they redistribute the charges equally on both sides of the oblate spheroid. None of the high order Pn(v), n≠0 polynomials change the net charge on the oblate spheroid. The terms Kn(iu)Pn(v) are a multipole expansion of the scalar potential. 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
                  
                  ⁢
                  a
                
              
              ⁡
              
                [
                
                  
                    arctan
                    ⁡
                    
                      (
                      
                        sinh
                        ⁢
                        
                          
                        
                        ⁢
                        
                          u
                          b
                        
                      
                      )
                    
                  
                  -
                  
                    
                  
                  ⁢
                  
                    arctan
                    ⁡
                    
                      (
                      
                        sinh
                        ⁢
                        
                          
                        
                        ⁢
                        
                          u
                          t
                        
                      
                      )
                    
                  
                
                ]
              
            
            +
            
              
                
                  β
                  ⁢
                  
                    
                  
                  ⁢
                  q
                
                
                  4
                  ⁢
                  
                    πɛ
                    0
                  
                  ⁢
                  a
                
              
              ⁡
              
                [
                
                  
                    
                      (
                      
                        
                          
                            (
                            
                              
                                3
                                ⁢
                                
                                  u
                                  t
                                  2
                                
                              
                              +
                              1
                            
                            )
                          
                          ⁢
                          arccot
                          ⁢
                          
                            
                          
                          ⁢
                          
                            (
                            
                              u
                              t
                            
                            )
                          
                        
                        -
                        
                          
                            3
                            2
                          
                          ⁢
                          
                            u
                            t
                          
                        
                      
                      )
                    
                    ⁢
                    
                      
                        P
                        2
                      
                      ⁡
                      
                        (
                        
                          v
                          t
                        
                        )
                      
                    
                  
                  -
                  
                    
                      (
                      
                        
                          
                            (
                            
                              
                                3
                                ⁢
                                
                                  u
                                  b
                                  2
                                
                              
                              +
                              1
                            
                            )
                          
                          ⁢
                          
                            arccot
                            (
                            
                              u
                              b
                            
                            )
                          
                        
                        +
                        
                          
                            3
                            2
                          
                          ⁢
                          
                            u
                            b
                          
                        
                      
                      )
                    
                    ⁢
                    
                      
                        P
                        2
                      
                      ⁡
                      
                        (
                        
                          v
                          b
                        
                        )
                      
                    
                  
                
                ]
              
            
          
        
        ,
      
    
    
      
        (
        53
        )
      
    
  




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

The effective height becomes heff=(1−α)κh/2+ακh and the radiation resistance is Rrad=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 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=0.698, α=0.706, and β=0.357 for a 90% charge on the oblate spheroid. At 50 MHz and h=1⁄2, the absolute values of the antenna at this minimum is Q=14.363, Rrad=4.822, and C=45.96 pF. The resulting antenna parameters, as a function of k and h become




  
    
      
        Q
        =
        
          2.0617
          
            
              (
              kh
              )
            
            3
          
        
      
    
  
  
    
      
        
          h
          e
        
        =
        
          0.6631
          ⁢
          
            
          
          ⁢
          h
        
      
    
  
  
    
      
        
          R
          rad
        
        =
        
          17.588
          *
          
            
              (
              kh
              )
            
            2
          
        
      
    
  
  
    
      
        C
        =
        
          1.0482
          *
          
            10
            
              -
              10
            
          
          ⁢
          
            h
            .
          
        
      
    
  




FIG. 22 is a plot of the β distribution as a function of κ and α 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 distribution 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 distribution 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 α 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 0.0005, the antenna parameters were recalculated resulting in Q=37.386 at 50 MHz and h=1⁄2, Q/QChu=5.3667, Rrad=3.4287, C=24.832 pF for κ=0.5735 , α=0.4235 , β=0.615 with 95% of the charge on the top load. 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 α 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 distribution 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 distribution 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




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




For the free space dipole, the potential field is





  
    
      
        
          
            
              Φ
              D
            
            ⁡
            
              (
              
                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
      
    
  




if for the top dipole arm,





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




is for the bottom dipole arm, Rt=√{square root over ((z−κh/2−h/2)2+ρ2)} is the distance from ρ,z to the top of the dipole, 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
        td
      
      =
      
        
          κ
          ⁢
          
            
          
          ⁢
          
            h
            /
            2
          
        
        
          
            2
            ⁢
            z
          
          -
          h
          -
          
            κ
            ⁢
            
              
            
            ⁢
            
              h
              /
              2
            
          
        
      
    
    ,
    
      
        and
        ⁢
        
          
        
        ⁢
        
          τ
          bd
        
      
      =
      
        
          κ
          ⁢
          
            
          
          ⁢
          
            h
            /
            2
          
        
        
          
            2
            ⁢
            z
          
          -
          h
          +
          
            κ
            ⁢
            
              
            
            ⁢
            
              h
              /
              2
            
          
        
      
    
  
  ⁢
  
    
  




for z≧(1+κ)h/2, resulting in





  
    
      
        
          
            Φ
            D
          
          ⁡
          
            (
            
              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
            D
          
          ⁡
          
            (
            
              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
        fi
      
      +
      
        R
        ti
      
    
  




is for the top arm of the image dipole,





  
    τ
    bi
  
  =
  
    
      κ
      ⁢
      
        
      
      ⁢
      
        h
        /
        2
      
    
    
      
        R
        fi
      
      +
      
        R
        bi
      
    
  




is for the bottom arm of the image dipole, Rti=√{square root over ((z−κh/2+h/2)2+ρ2)} is the distance from ρ,z to the top of the image dipole, Rfi=√{square root over ((z+h/2)2+ρ2)} is the distance from ρ,z to the feed point of the image dipole and Rbi=√{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 on the sphere. 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≧−(1−κ)h/2. The resulting expression for the potential field for the image dipole becomes





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




which can be further simplified to





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




The combined potential from Eqs. 57 and 59 is




  
    
      
        
          Φ
          ⁡
          
            (
            
              0
              ,
              h
            
            )
          
        
        =
        
          
            
              2
              ⁢
              q
            
            
              4
              ⁢
              
                πɛ
                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
            rad
          
          =
          
            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
        )
      
    
  




Eq. 62 and Eq. 63 treat the dipole as a monopole with the feed point at the center of the antenna.

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
          rad
        
        =
        
          40
          ⁢
          
            
              
                (
                
                  
                    κ
                    ⁢
                    
                      
                    
                    ⁢
                    h
                    ⁢
                    
                      
                    
                    ⁢
                    k
                  
                  2
                
                )
              
              2
            
            .
          
        
      
    
    
      
        (
        64
        )
      
    
  




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




  
    
      
        Q
        =
        
          
            
              
                -
                6
              
              ⁢
              
                ln
                ⁡
                
                  (
                  
                    
                      (
                      
                        1
                        -
                        
                          
                            κ
                            ⁢
                            
                              
                            
                          
                          2
                        
                      
                      )
                    
                    ⁢
                    
                      (
                      
                        1
                        -
                        
                          
                            
                              κ
                              ⁢
                              
                                
                              
                            
                            2
                          
                          /
                          9
                        
                      
                      )
                    
                  
                  )
                
              
            
            
              
                (
                
                  κ
                  ⁢
                  
                    
                  
                  ⁢
                  h
                  ⁢
                  
                    
                  
                  ⁢
                  k
                
                )
              
              3
            
          
          .
        
      
    
    
      
        (
        65
        )
      
    
  




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

    
    
        Q=12.67326/kα3 
        heff=0.37188*h
        Rrad=5.5319(kα)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 an embodiment of the quasi-static design approach for low Q factor electrically small antennas. 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 the embodiments of the quasi-static design approach for low Q factor electrically small antennas. 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.
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 quasi-static design approach for low Q factor electrically small antennas, may be made by those skilled in the art within the principal and scope of the quasi-static design approach for low Q factor electrically small antennas 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 du = dx


 






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


 






  
    
      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
                      
                    
                    )
                  
                  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 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
                
              
            
          
          ]
        
      
    
  




(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 case, the definition must proceed from fundamental concepts. From J. D. Jackson “Classical Electrodynamics”, second edition, John Wiley & Sons, 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). In our model, negative charge below the feed point 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
    =
    
      10
      ⁢
      
        
          
            k
            2
          
          ⁡
          
            [
            
              
                
                
                  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 radiation resistance is




  
    
      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
                  
                
              
            
            ]
          
        
        2
      
    
  



The quantity in the square bracket is heff. This equation can be reduces to the usual results, when q(x)=1 for 0<x≦d/2 or q(x)=δ(z−d/2).

The calculation for





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




p2=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
                3
              
              3
            
            ]
          
        
        
          -
          1
        
        1
      
    
    =
    
      
        4
        3
      
      ⁢
      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
                u
              
            
          
        
      
    
  
  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          2
          ⁢
          q
          ⁢
          
            
              ∫
              
                u
                fp
              
              1
            
            ⁢
            
              
                [
                
                  u
                  +
                  
                    
                      α
                      2
                    
                    ⁢
                    
                      (
                      
                        
                          5
                          ⁢
                          
                            u
                            3
                          
                        
                        -
                        
                          3
                          ⁢
                          u
                        
                      
                      )
                    
                  
                
                ]
              
              ⁢
              
                
              
              ⁢
              
                d
                u
              
            
          
        
      
    
  
  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          
            q
            ⁡
            
              [
              
                
                  u
                  2
                
                +
                
                  α
                  ⁡
                  
                    (
                    
                      
                        
                          5
                          4
                        
                        ⁢
                        
                          u
                          4
                        
                      
                      -
                      
                        
                          3
                          2
                        
                        ⁢
                        
                          u
                          2
                        
                      
                    
                    )
                  
                
              
              ]
            
          
          
            u
            fp
          
          1
        
      
    
  
  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          
            q
            ⁡
            
              [
              
                
                  u
                  2
                
                ⁡
                
                  (
                  
                    1
                    -
                    
                      
                        3
                        ⁢
                        α
                      
                      2
                    
                    +
                    
                      
                        
                          5
                          ⁢
                          α
                        
                        4
                      
                      ⁢
                      
                        u
                        2
                      
                    
                  
                  )
                
              
              ]
            
          
          
            u
            fp
          
          1
        
      
    
  




There are two cases, the first case is the feed point at ground ufp=0





  
    q
    fp
  
  =
  
    q
    ⁡
    
      (
      
        1
        -
        
          α
          4
        
      
      )
    
  




The second case is feed point is the feed point ufp>0. This is found by




  
    
      2
      ⁢
      
        q
        ⁡
        
          (
          
            
              
                P
                1
              
              ⁡
              
                (
                u
                )
              
            
            +
            
              α
              ⁢
              
                
              
              ⁢
              
                
                  P
                  3
                
                ⁡
                
                  (
                  u
                  )
                
              
            
          
          )
        
      
    
    =
    0
  




  
    
      q
      ⁡
      
        (
        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
      fp
    
    =
    
      
        
          
            3
            ⁢
            α
          
          -
          2
        
        
          5
          ⁢
          α
        
      
    
  



The charge above the feed point is





  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          q
          ⁡
          
            [
            
              
                (
                
                  1
                  -
                  
                    α
                    4
                  
                
                )
              
              +
              
                
                  
                    (
                    
                      
                        3
                        ⁢
                        α
                      
                      -
                      2
                    
                    )
                  
                  2
                
                
                  20
                  ⁢
                  α
                
              
            
            ]
          
        
      
    
  
  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          
            q
            ⁡
            
              [
              
                
                  u
                  2
                
                ⁡
                
                  (
                  
                    1
                    -
                    
                      
                        3
                        ⁢
                        α
                      
                      2
                    
                    +
                    
                      
                        
                          5
                          ⁢
                          α
                        
                        4
                      
                      ⁢
                      
                        u
                        2
                      
                    
                  
                  )
                
              
              ]
            
          
          
            u
            fp
          
          1
        
      
    
  
  
    
      
        
          q
          fp
        
        =
        
        ⁢
        
          q
          ⁡
          
            [
            
              
                (
                
                  1
                  -
                  
                    α
                    4
                  
                
                )
              
              +
              
                
                  (
                  
                    
                      
                        3
                        ⁢
                        α
                      
                      -
                      2
                    
                    
                      20
                      ⁢
                      α
                    
                  
                  )
                
                ⁢
                
                  (
                  
                    
                      3
                      ⁢
                      α
                    
                    -
                    2
                  
                  )
                
              
            
            ]
          
        
      
    
  




For α>2⁄3, the effective height is





  
    h
    eff
  
  =
  
    
      2
      ⁢
      κ
      ⁢
      
        
      
      ⁢
      h
    
    
      3
      ⁡
      
        [
        
          
            (
            
              1
              -
              
                α
                4
              
            
            )
          
          +
          
            
              
                (
                
                  
                    3
                    ⁢
                    α
                  
                  -
                  2
                
                )
              
              2
            
            
              20
              ⁢
              α
            
          
        
        ]
      
    
  




For α2⁄3, the effective height is





  
    h
    eff
  
  =
  
    
      2
      ⁢
      κ
      ⁢
      
        
      
      ⁢
      h
    
    
      3
      ⁢
      
        (
        
          1
          -
          
            α
            4
          
        
        )
      
    
  




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
                -
                2
              
              )
            
            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 dipole field

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


    
    
        x=α cos h u sin v cos φ
        y=α cos h u sin v sin φ
        z=α sin h u cos v

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

    
    





  
    ρ
    =
    
      a
      ⁢
      
        
      
      ⁢
      cosh
      ⁢
      
        
      
      ⁢
      u
      ⁢
      
        
      
      ⁢
      sin
      ⁢
      
        
      
      ⁢
      v
    
  




  
    
      z
      
        a
        ⁢
        
          
        
        ⁢
        sinh
        ⁢
        
          
        
        ⁢
        u
      
    
    =
    
      cos
      ⁢
      
        
      
      ⁢
      v
    
  




  
    
      ρ
      
        a
        ⁢
        
          
        
        ⁢
        cosh
        ⁢
        
          
        
        ⁢
        u
      
    
    =
    
      
        
          sin
          ⁢
          
            
          
          ⁢
          
            v
            ⁢
            
              

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




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




  
    
      where
      ⁢
      
        
      
      ⁢
      
        z
        ′
      
    
    =
    
      
        
          z
          a
        
        ⁢
        
          
        
        ⁢
        and
        ⁢
        
          
        
        ⁢
        
          ρ
          ′
        
      
      =
      
        ρ
        a
      
    
  




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




  
    
      
        
          sinh
          4
        
        ⁢
        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 √{square root over (ρ2+z2)}=r<α




  
    
      
        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=α





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




The third case r>α





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




The solution for v is





  
    z
    
      a
      ⁢
      
        
      
      ⁢
      sinh
      ⁢
      
        
      
      ⁢
      u
    
  
  =
  
    cos
    ⁢
    
      
    
    ⁢
    v
  




when z>0. For z=0 and ρ≦α

ρ=α sin v





  
    sin
    ⁢
    
      
    
    ⁢
    v
  
  =
  
    ρ
    a
  




sin v=1

for z=0 and ρ>α

Calculation of the Oblate Spheroid Potential

hu=α(sin h2 u+sin2 v)1/2 

hv=α(sin h2 u+sin2 v)1/2 

hφ=α cos h u cos v 




  
    
      
        
          
            
              ∇
              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=hvh100 dvdφ=α2(sin h2u+sin2 v)1/2 cos h u cos vdvdφ





  
    
      
        
          
            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=ε0∫−π/2π/2∫02παK cos vdvdφ

q=ε0∫−π/2π/22παK cos vdv=ε02πKα[sin v]−π/2π/2=ε04παK

K=q/4παε0 





  
    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
        ⁢
        
          πɛ
          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
                ⁡
                
                  (
                  
                    
                      
                        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
    
  




  
    
      
        
          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(v)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 v is the standard Legendre Polynomials

Φ(v)=Pl(v)

and

K(u)=Pl(iu)

In our case, the second





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




Verifying that K2(iu) is the solution.




  
    
      
        K
        2
      
      ⁡
      
        (
        iu
        )
      
    
    =
    
      
        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
            ⁢
            ξ
            ⁢
            
              
            
            ⁢
            
              arccot
              ⁡
              
                (
                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
                ⁡
                
                  (
                  ξ
                  )
                
              
            
            -
            
              3
              ⁢
              ξ
            
          
          ]
        
      
    
    =
    0

## Claims

1. A method for designing an electrically small antenna, comprising:
designating a general form factor for an antenna;
defining a charge distribution in terms of at least one form factor related parameter;
computing a capacitance in terms of the charge distribution and the form factor related parameter;
generating a quasi-static quality factor (Q) based on the computed capacitance and a corresponding radiation resistance;
evaluating values of the generated quasi-static Q, the computed capacitance, and the radiation resistance to arrive at an antenna performance; and
generating an antenna geometry using the form factor related parameter for the antenna performance.

2. The method according to claim 1, wherein designating the general form factor further comprises establishing a cylindrical coordinate geometry.

3. The method according to claim 1, wherein designating the general form factor further comprises establishing a spherical coordinate geometry.

4. The method according to claim 1, wherein the charge distribution is selected from a set of Legendre polynomials.

5. The method according to claim 1, wherein the charge distribution is selected from at least one of a set of orthogonal basis functions.

6. The method according to claim 1, wherein the charge distribution comprises at least a point charge.

7. The method according to claim 1, wherein the charge distribution comprises at least an oblate spheroidal charge.

8. The method according to claim 1, further comprising validating that the charge distribution is not outside a surface of the antenna geometry.

9. The method according to claim 1, wherein a profile of the charge distribution takes the shape of a profile of the general form factor of the antenna.

10. The method according to claim 1, further comprising adding another degree of freedom for the form factor related parameter to obtain a Q performance.

11. The method according to claim 1, wherein evaluating the Q value is based on a minimum Q value.

12. An electrically small antenna, comprising:
a surface contour of the antenna having a Q performance, the Q performance being generated according to claim 1.

13. The electrically small antenna according to claim 12, wherein the surface contour is an asymptotic conical dipole.

14. The electrically small antenna according to claim 12, wherein the surface contour approximates a shape of a “bulb”.

15. The electrically small antenna according to claim 12, wherein the surface contour approximates a shape of a “volcano and smoke”.

16. The electrically small antenna according to claim 12, wherein the surface contour approximates a shape of a “mushroom”.

17. The electrically small antenna according to claim 12, wherein the surface contour approximates a shape of an oblate spheroid cap.

18. The electrically small antenna according to claim 12, wherein the surface contour approximates a shape of an oblate spheroid cap with a discontinuous slope.

19. The electrically small antenna according to claim 12, wherein the surface contour is of a dipole antenna having an approximate shape of a prolate spheroid centered above and connected to a hemisphere of an oblate spheroid that is connected to ground.

20. The electrically small antenna according to claim 12, wherein the surface contour is approximated using a wire frame model.

21. The electrically small antenna according to claim 12, wherein a shape of a feed point is parameterized by the charge distribution.