Abstract:
Embodiments of the present invention relate to circuits to be inserted in a signal path between a signal generator and a test port of a vector network analyzer (VNA), where the circuits enable the VNA to make scattering parameter measurements for a load (RL) connected to the test port when the signal generator generates signals having frequencies that are below a low frequency limit (e.g., 2 MHz) of an actual dual directional coupler of the VNA. Embodiments of the present invention are also directed to a VNA that includes such circuits.

Description:
FIELD OF THE INVENTION 
   Embodiments of the present invention relate to vector network analyzers (VNAs). More particularly, embodiments of the present invention relate to circuits that enable a VNA to make measurements over a large frequency range (e.g., DC to 8 GHz). 
   BACKGROUND 
     FIG. 1  is a high level diagram of a conventional vector network analyzer (VNA)  100  which is shown as including a signal generator  102  (also known as a voltage source) having a source impedance RS, a forward coupler  104 , a reverse coupler  106 , one or more test port  108 , a receive/detector  110  and a processor/display  120 . The signal generator  102  produces signals that are transmitted to a test device, shown as a resistive load RL, which is connected to one or more test port  108  of the VNA. The resistive load RL, also known as a device under test (DUT), can be, e.g., an open, a short, or anything in-between. 
   The forward coupler, which includes a resistive bridge having an impedance R F , provides a forward signal (also known as an incident signal) to a first input of the receiver/detector  110 . The reverse coupler  108 , which includes a resistive bridge having an impedance R R , provides a reverse signal (also known as a reflected signal) to a second input of the receiver/detector  110 . The forward signal is often referred to as signal VF hereafter. Similarly, the reverse signal is often referred to as signal VR hereafter. In the arrangement shown, the forward coupler  104  and the reverse coupler  106  together form what is known as a dual directional coupler. Preferably, the impedances of the source, the load and the couplers  104  and  106  all match (i.e., preferably R S =R L =R F =R R ). A typical value for R S , R L , R F  and R R  is 50 ohms, or 75 ohms, but other values may be used. 
   The receiver/detector  110 , which is tuned to the frequency of the signal generator  102 , likely includes components such as a local oscillator (LO), band-pass filters (BPFs) and a synchronous detector or digital-signal processor (DSP). The LO is used to mix received RF signals down to lower intermediate frequency (IF) signals and baseband signals. The BPFs are used to filter out undesired harmonics and noise from the IF and baseband signals. The synchronous detector or DSP is used to extract magnitude and phase information from the baseband signals, which can be, e.g., about 200 KHz signals. More specifically, the synchronous detector or DSP can converts the VF and VR signals into digital signals indicative of real and imaginary components of the VF signal and real and imaginary components of the VR signal. The processor/display  120  formats the reflection and transmission data in ways that make it possible to interpret results of measurements. 
   Conventional forward and reverse couplers  104  and  106  typically have a low frequency limit of approximately 1 or 2 MHz. This is primarily due to the balun used to create the single ended to differential signal for the resistive bridge of the coupler. More specifically, a conventional coupler typically includes a coaxial transmission line surrounded by Ferrite beads of about a quarter inch in length. A coupler with one quarter inch Ferrite bead will enable use of the coupler down to about 2 MHz. To get down to about 1 MHz, two such quarter inch Ferrite beads are needed; to get down to about 0.5 MHz, four such quarter inch ferrite beads are required; and so on. Accordingly, it can be appreciated that it would be prohibitive to produce a coupler useful down to about DC, using the above described methodology. Nevertheless, there is a desire to measure S-parameters down to DC. Accordingly, there is a need for couplers that are useful down to DC. 
   SUMMARY 
   Embodiments of the present invention are directed to circuits to be inserted in a signal path between a signal generator and a test port of a vector network analyzer (VNA), where the circuits enable the VNA to make scattering parameter measurements for a load (RL) connected to the test port when the signal generator generates signals having frequencies that are below a low frequency limit (e.g., 2 MHz) of an actual dual directional coupler of the VNA. Embodiments of the present invention are also directed to a VNA that includes such circuits. 
   In accordance with an embodiment of the present invention, the circuit includes an attenuator (e.g., a pi-attenuator) that is inserted in the signal path between the signal generator and the test port. The circuit also includes a monitoring circuit connected to the attenuator to monitor voltages at two nodes along the signal path that are formed by the attenuator. Additionally, the circuit include a dual directional coupler emulation circuit, made up of op-amps and resistors, connected to the monitoring circuit to emulate a dual directional coupler at frequencies that are below the low frequency limit of an actual dual directional coupler. Outputs of the dual directional coupler emulation circuit are indicative of signals that are incident on and reflected from the load (RL), and thus can be used to measure scattering parameters of the load (RL). 
   In accordance with an embodiment of the present invention, the dual directional coupler emulation circuit includes a first circuit portion that emulates a forward coupler, as well as a second circuit portion that emulates a reverse coupler. 
   In accordance with an embodiment of the present invention, when the circuit is inserted in a VNA that includes an actual dual directional coupler, the circuit is preferably inserted in the signal path of the VNA such that it is between the signal generator and the actual dual direction coupler of the VNA. 
   Further and alternative embodiments, and the features, aspects, and advantages of the present invention will become more apparent from the detailed description set forth below, the drawings and the claims. 

   
     BRIEF DESCRIPTION OF THE DRAWINGS 
       FIG. 1  is a high level block diagram of a conventional vector network analyzer (VNA) that can not make measurements below the low frequency limit of the dual directional coupler of the VNA. 
       FIG. 2  is a high level diagram of a VNA that, in accordance with embodiments of the present invention, that can make measurements below the low frequency limit of the actual dual directional coupler of the VNA. 
       FIG. 3  is a circuit diagram that is useful for explaining the known characteristics of a pi-attenuator. 
       FIG. 4  is a circuit diagram that is useful for explaining how op-amps can be used to emulate characteristics of a reverse coupler and a forward coupler at relatively low frequencies (i.e., frequencies in the range of DC to a few MHz), in accordance with embodiments of the present invention. 
       FIG. 5  is diagram that is used to illustrate the characteristics of ideal forward and reverse couplers, and to explain how embodiments of the present invention can be used to emulate such ideal couplers. 
       FIG. 6  illustrates details of a circuit to emulate a dual directional coupler, in accordance with embodiments of the present invention. 
       FIG. 7  illustrates how the embodiments of the present invention can be used together with an actual dual directional coupler to thereby provide a VNA capable of making measurements over a larger bandwidth than a typical VNA. 
   

   DETAILED DESCRIPTION 
     FIG. 2  is a high level diagram of a VNA  200  that, in accordance with embodiments of the present invention, can make measurement from DC to about 8 GHz. As shown in  FIG. 2 , embodiments of the present invention allow the use of conventional forward and reverse couplers  106  and  108  to be used for their intended frequency coverage from the low MHz (e.g., 2 MHz) to the mid GHz (e.g., 8 GHz) range. The addition of a low value (e.g., &lt;1 dB) attenuator  202  placed in the source signal path  103  before the couplers  106  and  108  allows a convenient monitoring point to derive DC to 2 MHz forward and reverse voltage parameters. More specifically, embodiments of the present invention allow S-parameters to be measured down to DC, or at least close thereto. Preferably, embodiments of the present use no switches in the source signal path  103 , which would cause additional losses as well as switching uncertainties. 
   In accordance with an embodiment of the present invention, the attenuator  202 , which is shown as including three resistors (i.e., R 1 , R 2  and R 3 ), is a pi-attenuator, which is given that name because the configuration of resistors R 1 , R 2  and R 3  resemble the symbol pi (i.e., π). Characteristics of the attenuator  202  are described with reference to  FIG. 3 , and Equations 1-3. The insertion loss of the attenuator is preferably low, e.g., below 1 dB. However, embodiments of the present invention also encompass those situations where the attenuator has a greater insertion loss. Further, while the use of a pi-attenuator is preferred, use of other attenuators are within the scope of the present invention. For example, it is possible to string two T-attenuators together to form two shunt arms. Five resistors (instead of three) would be used to form the two T-attenuators, and the math would differ due to the five resistors. For the following discussion, it is assumed that a pi-attenuator is used. 
   Equation 1 below is an equation for the voltage (VA) in  FIG. 3 , which is the voltage at node (N 1 ) in  FIG. 2 . In this and further equations, Vinc is the “incident” voltage generated by the voltage source  102 . 
   
     
       
         
           
             
               
                 VA 
                 = 
                 
                   Vinc 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       + 
                       
                         RL 
                         ⁡ 
                         
                           ( 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               2 
                               · 
                               
                                 + 
                                 R 
                               
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           ) 
                         
                       
                     
                     
                       
                         
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                               ⁢ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       3 
                                       · 
                                       RS 
                                     
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       2 
                                       · 
                                       R 
                                     
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 RS 
                               
                             
                             + 
                           
                         
                       
                       
                         
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   1 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                       + 
                                       RS 
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   RS 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                     
                                     ) 
                                   
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   1 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 2 below is an equation for the voltage (VB) in  FIG. 3 , which is the voltage at node (N 2 ) in  FIG. 2 . 
   
     
       
         
           
             
               
                 VB 
                 = 
                 
                   VA 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     
                       RL 
                       · 
                       
                         R 
                         ⁢ 
                         3 
                       
                     
                     
                       
                         
                           R 
                           ⁢ 
                           2 
                         
                         · 
                         
                           R 
                           ⁢ 
                           3 
                         
                       
                       + 
                       
                         RL 
                         ⁡ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   2 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equations 1 and 2 results in Equation 3 shown below. 
   
     
       
         
           
             
               
                 VB 
                 = 
                 
                   Vinc 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         1 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         2 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                     
                       
                         
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                               ⁢ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       3 
                                       · 
                                       RS 
                                     
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       2 
                                       · 
                                       R 
                                     
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 RS 
                               
                             
                             + 
                           
                         
                       
                       
                         
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   1 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                       + 
                                       RS 
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   RS 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                     
                                     ) 
                                   
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   3 
                 
                 ) 
               
             
           
         
       
     
   
     FIG. 4  illustrates how a pair of op-amps A 1  and A 2  can be used to monitor the voltages VA and VB, which as mentioned above, are the voltages at nodes N 1  and N 2 . As can be seen, the non-inverting (+) input of the op-amp A 1  is connected to ground; the inverting (−) input of the op-amp A 1 , which is the summing input of the op-amp A 1 , is connected to one of the arms of the attenuator  202 ; and a feedback resistor equal to the resistor R 1  is connected between the inverting (−) input and the output of the op-amp A 1 . Similarly, the non-inverting (+) input of the op-amp A 2  is connected to ground; the inverting (−) input of the op-amp A 2 , which is the summing input of the op-amp A 2 , is connected to the other one of the arms of the attenuator  202 ; and a feedback resistor equal to the resistor R 3  is connected between the inverting (−) input and the output of the op-amp A 2 . This arrangement will result in a voltage −VA at the output of the op-amp A 1  (i.e., Node N 5 ), and a voltage −VB at the output of op-amp A 2  (i.e., Node N 6 ). In other words, in the arrangement shown, the op-amps A 1  and A 2  are use used to monitor the voltages VA and VB at nodes N 1  and N 2  of the source signal path  103 . This is only good within the bandwidth of the op-amps A 1  and A 2 , which is likely from DC to about 5 MHz. So for a limited frequency range (e.g., from DC to about 2 MHz), the outputs of the op-amps A 1  and A 2  are equal in magnitude (but opposite in sign) to the voltages at nodes N 1  and N 4 . 
     FIG. 4  also shows how a differential amplifier A 3  can be used to convert voltages VA and VB to the form VO=Kp(VP)−Kn(VN). More specifically, Equation 4 below is an equation for VO of circuit  414 . 
   
     
       
         
           
             
               
                 VO 
                 = 
                 
                   
                     Vp 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                     ⁢ 
                     
                       ( 
                       
                         1 
                         + 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             5 
                           
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             4 
                           
                         
                       
                       ) 
                     
                   
                   - 
                   
                     Vn 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         5 
                       
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         4 
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   4 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 5 below assumes R 4 =R 5 . 
   
     
       
         
           
             
               
                 VO 
                 = 
                 
                   
                     Vp 
                     ⁢ 
                     
                       
                         2 
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   - 
                   Vn 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   5 
                 
                 ) 
               
             
           
         
       
     
   
   The dual directional coupler of the present invention includes a reverse coupler portion and a forward coupler portion. The reverse coupler case is first discussed, followed by a discussion of the forward coupler case. 
   Reverse Coupler Case 
   The reverse coupler case should have the characteristics of a reflectometer in that VO=0 when RL=Ro (i.e., the output is zero when the characteristic impedance of the coupler equals the load impedance). Equation 6 below illustrates this for the reverse coupler case.
 
− VA ( K )−(− VB )=0   (EQ. 6)
 
   Equation 7 below solves for K.
 
 K=VB/VA    (EQ. 7)
 
   Letting Vp=VA and Vn=VB, allows Equation 5 to be rewritten as Equation 8 below. 
   
     
       
         
           
             
               
                 VO 
                 = 
                 
                   
                     VA 
                     ⁢ 
                     
                       
                         2 
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   - 
                   VB 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   8 
                 
                 ) 
               
             
           
         
       
     
   
   Now, assuming RL=Ro=50 ohms, and that VO=0 when RL=Ro, Equation 8 can be rewritten as Equation 9 below. 
   
     
       
         
           
             
               
                 0 
                 = 
                 
                   
                     VA 
                     ⁢ 
                     
                       
                         2 
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   - 
                   VB 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   9 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equation 9 with Equation 2 results in Equation 10 below. 
   
     
       
         
           
             
               
                 
                   
                     2 
                     ⁢ 
                     R 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     7 
                   
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       6 
                     
                     + 
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       7 
                     
                   
                 
                 = 
                 
                   
                     
                       RL 
                       · 
                       R 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     3 
                   
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         2 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                     + 
                     
                       RL 
                       ⁡ 
                       
                         ( 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             2 
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                         
                         ) 
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   10 
                 
                 ) 
               
             
           
         
       
     
   
   Solving for R 7  results in Equation 11 below. 
   
     
       
         
           
             
               
                 
                   R 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   7 
                 
                 = 
                 
                   
                     R 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       3 
                       · 
                       R 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       6 
                       · 
                       RL 
                     
                   
                   
                     
                       2 
                       ⁢ 
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       2 
                       ⁢ 
                       
                         ( 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           + 
                           RL 
                         
                         ) 
                       
                     
                     + 
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         3 
                         · 
                         RL 
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   11 
                 
                 ) 
               
             
           
         
       
     
   
   In other words, by selecting a value for R 7  that satisfies Equation 11, then the circuit  414 , shown in  FIG. 4 , will emulate a reverse coupler. Since this is the reverse coupler case, the term R 7   R  will be used, with the R subscript signifying the reverse coupler case. 
   The pi attenuator  202  should be designed to have low insertion loss, so that minimum power is wasted. This is especially important at microwave frequencies where power is at a premium. Accordingly, it is preferred that the pi attenuator have an attenuation of less than 1 dB. For the following discussion, a pi attenuator with a value of 0.869 dB is selected, e.g., which results when R 1  and R 3  are 1 kilo-ohms. This is a good value for the wide range of op-amps used in developing −VA and −VB. Also assume that R 2  is approximately 5 ohms. More specifically, letting R 1 =R 3 =1K, and R 2 =5.11, the actual insertion loss=0.885 dB, and Zo=50.088 ohms. For an ideal insertion loss of 0.869 and Z 0 =50, R 1  and R 3  would be 1K and R 2  would be 5.012531. For the following examples, it is assumed that R 1 =R 2 =1K and R 2 =5.11 and RL=50. It is also assumed that R 6 =511 ohms. 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           7 
                           R 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             3 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             6 
                             · 
                             RL 
                           
                         
                         
                           
                             
                               2 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             2 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                                 + 
                                 RL 
                               
                               ) 
                             
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               3 
                               · 
                               RL 
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                       ⁢ 
                       
                         
                           1 
                           ⁢ 
                           
                             K 
                             ⁡ 
                             
                               ( 
                               511 
                               ) 
                             
                           
                           ⁢ 
                           
                             ( 
                             50 
                             ) 
                           
                         
                         
                           
                             2 
                             ⁢ 
                             
                               ( 
                               5.11 
                               ) 
                             
                             ⁢ 
                             
                               ( 
                               
                                 
                                   1 
                                   ⁢ 
                                   K 
                                 
                                 + 
                                 50 
                               
                               ) 
                             
                           
                           + 
                           
                             1 
                             ⁢ 
                             
                               K 
                               ⁡ 
                               
                                 ( 
                                 5 
                                 ) 
                               
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       = 
                         
                       ⁢ 
                       420.7077111 
                     
                   
                 
               
             
             
               
                   
               
             
           
         
       
     
   
   VO is now determined for RL other than 50 ohms. 
   Equation 9 can be rewritten as Equation 12 shown below. 
   
     
       
         
           
             
               
                 VB 
                 = 
                 
                   VA 
                   ⁢ 
                   
                     
                       2 
                       ⁢ 
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       7 
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         6 
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   12 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equations 8 and 12 leads to Equation 13 shown below. 
   
     
       
         
           
             
               
                 VO 
                 = 
                 
                   VA 
                   ⁡ 
                   
                     ( 
                     
                       
                         
                           2 
                           ⁢ 
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             6 
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             7 
                           
                         
                       
                       - 
                       
                         
                           
                             RL 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               2 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     ) 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   13 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equations 1 and 13 leads to Equation 14 (EQ. 14) shown below. 
   
     
       
         
           
             
               
                 
                   VO 
                   R 
                 
                 = 
                 
                   Vinc 
                   ( 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       + 
                       
                         RL 
                         ⁡ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 1 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 1 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                     
                     
                       
                         
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               1 
                               ⁢ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       3 
                                       · 
                                       RS 
                                     
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       2 
                                       · 
                                       R 
                                     
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 RS 
                               
                             
                             + 
                           
                         
                       
                       
                         
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   1 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                       + 
                                       RS 
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   RS 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         2 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                     
                                     ) 
                                   
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                   ) 
                 
               
             
           
           
             
               
                 ( 
                 
                   
                     
                       2 
                       ⁢ 
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       7 
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         6 
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                     
                   
                   - 
                   
                     
                       
                         RL 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       + 
                       
                         RL 
                         ⁡ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
                 ) 
               
             
           
         
       
     
   
   Assuming R 1 =R 3 =1K, R 2 =5.11, RS=50, R 6 =511 and R 7 =420.7077111, then Equation 14 becomes: 
   
     
       
         
           
             VO 
             R 
           
           = 
           
             
               Vinc 
               ⁡ 
               
                 ( 
                 
                   
                     50 
                     - 
                     RL 
                   
                   
                     49.85704395 
                     + 
                     RL 
                   
                 
                 ) 
               
             
             ⁢ 
             
               ( 
               0.10202053 
               ) 
             
           
         
       
     
   
   Table 1 below illustrates VO for various values for RL, assuming Vinc=2V. 
   
     
       
             
             
             
             
             
           
             
             
             
             
             
           
         
             
                 
               TABLE 1 
             
             
                 
                 
             
             
                 
               RL 
               VO(mV) 
               dBV 
               Phase(deg) 
             
             
                 
                 
             
           
           
             
                 
             
           
        
         
             
                 
                0 
               +167.475 
               −15.521 
               0 
             
             
                 
                0.5 
               +164.154 
               −15.645 
               0 
             
             
                 
                5.0 
               +136.989 
               −17.266 
               0 
             
             
                 
               25 
               +55.772 
               −25.072 
               0 
             
             
                 
               45 
               +8.803 
               −41.108 
               0 
             
             
                 
               50 
               0 
               −180.637 
               0 
             
             
                 
                 55.5555 
               −8.801 
               −41.109 
               −180 
             
             
                 
               100  
               −55.718 
               −25.080 
               −180 
             
             
                 
               500  
               −136.669 
               −17.287 
               −180 
             
             
                 
               5000  
               −163.694 
               −15.719 
               −180 
             
             
                 
               Infinity 
               −166.996 
               −15.546 
               −180 
             
             
                 
                 
             
           
        
       
     
   
   In Table 1, the phase of VO is 180 degrees out of phase with the standard definition of rho, where rho=(Z−Zo)/(Z+Zo), for rho&lt;Zo, phase=180; for rho&gt;Zo, phase=0. Table 1 is used to illustrate that circuit  414 , with appropriate values for the resistors R 4 , R 5 , R 6  and R 7 , as defined above in Equation 11, will indeed act as a reverse coupler, where RL is a SHORT (i.e., RL=0), an OPEN (i.e., RL=infinity), or any value there-between. 
   The reverse coupler case expressed in Equation 14 above can be rewritten as Equation 15 (EQ. 15) shown below. 
   
     
       
         
           
             
               
                 
                   VO 
                   R 
                 
                 = 
                   
                 ⁢ 
                 
                   - 
                   
                     Vinc 
                     ( 
                     
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         + 
                         
                           RL 
                           ⁡ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                       
                         
                           
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 1 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         3 
                                         · 
                                         RS 
                                       
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         2 
                                         · 
                                         R 
                                       
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       3 
                                     
                                   
                                   ) 
                                 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   2 
                                   · 
                                   RS 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               RL 
                               ⁡ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     1 
                                     ⁢ 
                                     
                                       ( 
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           2 
                                         
                                         + 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           3 
                                         
                                         + 
                                         RS 
                                       
                                       ) 
                                     
                                   
                                   + 
                                   
                                     RS 
                                     ⁡ 
                                     
                                       ( 
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           2 
                                         
                                         + 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           3 
                                         
                                       
                                       ) 
                                     
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                     ) 
                   
                 
               
             
           
           
             
               
                   
                 ⁢ 
                 
                   ( 
                   
                     
                       
                         
                           RL 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         + 
                         
                           RL 
                           ⁡ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                     
                     - 
                     
                       
                         2 
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   ) 
                 
               
             
           
         
       
     
   
   The inversion can be addressed in post processing. 
   Forward Coupler Case 
   For a forward coupler case, the circuit  414  should have the characteristics VO=(Vinc)K for all values of RL. An equation is now determined that produces a constant output VO for all values of RL from 0 to infinity. 
   Equation 16 below illustrates the forward coupler case, with the subscript “o” Indicating an OPEN, and the subscript “s” indicating a SHORT. 
   
     
       
         
           
             
               
                 
                   - 
                   
                     
                       ( 
                       
                         KVBo 
                         - 
                         VAo 
                       
                       ) 
                     
                     OPEN 
                   
                 
                 = 
                 
                   - 
                   
                     
                       ( 
                       
                         KVBs 
                         - 
                         VAs 
                       
                       ) 
                     
                     SHORT 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   16 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 16 can be rewritten as Equation 17 below. 
   
     
       
         
           
             
               
                 K 
                 = 
                 
                   
                     VAo 
                     - 
                     VAs 
                   
                   
                     VBo 
                     - 
                     VBs 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   17 
                 
                 ) 
               
             
           
         
       
     
   
   Equations 18 and 19 shown below are for VA and VB for OPEN. 
   
     
       
         
           
             
               
                 VAo 
                 = 
                 
                   Vinc 
                   ⁢ 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         2 
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                             + 
                             RS 
                           
                           ) 
                         
                       
                       + 
                       
                         RS 
                         ⁡ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   18 
                 
                 ) 
               
             
           
           
             
               
                 VBo 
                 = 
                 
                   Vinc 
                   ⁢ 
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         1 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                             + 
                             RS 
                           
                           ) 
                         
                       
                       + 
                       
                         RS 
                         ⁡ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   19 
                 
                 ) 
               
             
           
         
       
     
   
   Equations 20 and 21 shown below are for VA and VB for SHORT. 
   
     
       
         
           
             
               
                 VAs 
                 = 
                 
                   Vinc 
                   ⁢ 
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         1 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         2 
                         · 
                         R 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 3 
                                 · 
                                 RS 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           RS 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   20 
                 
                 ) 
               
             
           
           
             
               
                 VBs 
                 = 
                 0 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   21 
                 
                 ) 
               
             
           
         
       
     
   
   Equations 17 and 21 can be combined for SHORT, as shown in Equation 22 below. 
   
     
       
         
           
             
               
                 K 
                 = 
                 
                   
                     VAo 
                     - 
                     VAs 
                   
                   VBo 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   22 
                 
                 ) 
               
             
           
         
       
     
   
   Combining equations 18, 19 and 22 leads to Equation 23, shown below. 
   
     
       
         
           
             
               
                 K 
                 = 
                   
                 ⁢ 
                 
                   
                     
                       Vinc 
                       ⁢ 
                       
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             2 
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                         
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                                 + 
                                 RS 
                               
                               ) 
                             
                           
                           + 
                           
                             RS 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     - 
                     
                       
 
                     
                     ⁢ 
                     
                       Vinc 
                       ⁢ 
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     3 
                                     · 
                                     RS 
                                   
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     2 
                                     · 
                                     R 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                           + 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               2 
                               · 
                               RS 
                             
                           
                         
                       
                     
                   
                   
                     Vinc 
                     ⁢ 
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           1 
                           ⁢ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                               + 
                               RS 
                             
                             ) 
                           
                         
                         + 
                         
                           RS 
                           ⁡ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   23 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 23 can be simplified to Equation 24 shown below. 
   
     
       
         
           
             
               
                 K 
                 = 
                 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         2 
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                     
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                   
                   - 
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       2 
                       ⁢ 
                       
                         ( 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                                 + 
                                 RS 
                               
                               ) 
                             
                           
                           + 
                           
                             RS 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 3 
                                 · 
                                 RS 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           RS 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   24 
                 
                 ) 
               
             
           
         
       
     
   
   Letting Vp=VB and Vn=VA, allows Equation 5 to be rewritten as Equation 25 below. 
   
     
       
         
           
             
               
                 VO 
                 = 
                 
                   
                     VB 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         2 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   - 
                   VA 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   25 
                 
                 ) 
               
             
           
         
       
     
   
   From  FIG. 4 , Equation 5 can be rewritten as Equation 26 below.
 
 VO =( VB ) K−VA    (EQ. 26)
 
   Combining equations 25 and 26 leads to Equation 27 below. 
   
     
       
         
           
             
               
                 
                   
                     VB 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       
                         2 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         7 
                       
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           6 
                         
                         + 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           7 
                         
                       
                     
                   
                   - 
                   
                     V 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     A 
                   
                 
                 = 
                 
                   
                     
                       ( 
                       VB 
                       ) 
                     
                     ⁢ 
                     K 
                   
                   - 
                   VA 
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   27 
                 
                 ) 
               
             
           
         
       
     
   
   Solving for K leads to Equation 28 below. 
   
     
       
         
           
             
               
                 K 
                 = 
                 
                   
                     2 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     R 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     7 
                   
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       6 
                     
                     + 
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       7 
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   28 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equations 24 and 28 (i.e., Let K=K) leads to Equation 29 below. 
   
     
       
         
           
             
               
                 
                   
                     2 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     R 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     7 
                   
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       6 
                     
                     + 
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       7 
                     
                   
                 
                 = 
                 
                   
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         2 
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                     
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       3 
                     
                   
                   - 
                   
                     
                       R 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       2 
                       ⁢ 
                       
                         ( 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             1 
                             ⁢ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                                 + 
                                 RS 
                               
                               ) 
                             
                           
                           + 
                           
                             RS 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         1 
                         ⁢ 
                         
                           ( 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 3 
                                 · 
                                 RS 
                               
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                           
                           ) 
                         
                       
                       + 
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           RS 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   29 
                 
                 ) 
               
             
           
         
       
     
   
   Solving for R 7   F  results in Equation 30 below. 
   
     
       
         
           
             
               
                 
                   R 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     7 
                     F 
                   
                 
                 = 
                 
                   R 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   6 
                   ⁢ 
                   
                     ( 
                     
                       
                         
                           
                             
                               
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       2 
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       3 
                                     
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             
                                               3 
                                               · 
                                               RS 
                                             
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             
                                               2 
                                               · 
                                               R 
                                             
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         2 
                                         · 
                                         R 
                                       
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       S 
                                     
                                   
                                   ) 
                                 
                               
                               - 
                             
                           
                         
                         
                           
                             
                               ( 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   2 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             S 
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       RS 
                                       ⁡ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                       
                       
                         
                           
                             
                               
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       3 
                                     
                                     - 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       2 
                                     
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             
                                               3 
                                               · 
                                               RS 
                                             
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             
                                               2 
                                               · 
                                               R 
                                             
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         2 
                                         · 
                                         R 
                                       
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       S 
                                     
                                   
                                   ) 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               ( 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   2 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             S 
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       RS 
                                       ⁡ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                       
                     
                     ) 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   30 
                 
                 ) 
               
             
           
         
       
     
   
   In other words, by selecting a value for R 7  that satisfies Equation 30, then the circuit  414 , shown in  FIG. 4 , will emulate a forward coupler. Since this is the forward coupler case, the term R 7   F  will be used, with the F subscript signifying the reverse coupler case. 
   Still assuming that R 1 =R 2 =1K, R 2 =5.11, R 6 =511 and RS=50, then R 7   F =420.3065369. 
   Combining Equations 26 and 28 with Equations 1 and 3 leads to Equation 31 below. 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         VO 
                         F 
                       
                       = 
                       
                         
                           
                             Vinc 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   2 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   7 
                                 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     6 
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     7 
                                   
                                 
                               
                               ) 
                             
                           
                           ⁢ 
                           
                             ( 
                             
                               
                                 
                                   RL 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                               
                                 
                                   
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         1 
                                         ⁢ 
                                         
                                           ( 
                                           
                                             
                                               R 
                                               ⁢ 
                                               
                                                   
                                               
                                               ⁢ 
                                               
                                                 3 
                                                 · 
                                                 RS 
                                               
                                             
                                             + 
                                             
                                               R 
                                               ⁢ 
                                               
                                                   
                                               
                                               ⁢ 
                                               
                                                 2 
                                                 · 
                                                 R 
                                               
                                               ⁢ 
                                               
                                                   
                                               
                                               ⁢ 
                                               3 
                                             
                                           
                                           ) 
                                         
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           2 
                                           · 
                                           RS 
                                         
                                       
                                       + 
                                     
                                   
                                 
                                 
                                   
                                     
                                       RL 
                                       ( 
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           1 
                                           ⁢ 
                                           
                                             ( 
                                             
                                               
                                                 R 
                                                 ⁢ 
                                                 
                                                     
                                                 
                                                 ⁢ 
                                                 2 
                                               
                                               + 
                                               
                                                 R 
                                                 ⁢ 
                                                 
                                                     
                                                 
                                                 ⁢ 
                                                 3 
                                               
                                               + 
                                               RS 
                                             
                                             ) 
                                           
                                         
                                         + 
                                       
                                     
                                   
                                 
                                 
                                   
                                     
                                       
                                         RS 
                                         ⁢ 
                                         
                                           ( 
                                           
                                             
                                               R 
                                               ⁢ 
                                               
                                                   
                                               
                                               ⁢ 
                                               2 
                                             
                                             + 
                                             
                                               R 
                                               ⁢ 
                                               
                                                   
                                               
                                               ⁢ 
                                               3 
                                             
                                           
                                           ) 
                                         
                                       
                                       ) 
                                     
                                   
                                 
                               
                             
                             ) 
                           
                         
                         - 
                       
                     
                   
                 
                 
                   
                     
                       
                         - 
                         Vinc 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               2 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           + 
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     1 
                                     · 
                                     R 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     1 
                                     · 
                                     R 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                 
                               
                               ) 
                             
                           
                         
                         
                           
                             
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   1 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           3 
                                           · 
                                           RS 
                                         
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           2 
                                           · 
                                           R 
                                         
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     2 
                                     · 
                                     RS 
                                   
                                 
                                 + 
                               
                             
                           
                           
                             
                               
                                 RL 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                           + 
                                           RS 
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       RS 
                                       ⁡ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   31 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 31 can be simplified to Equation 32, shown below. 
   
     
       
         
           
             
               
                 
                   VO 
                   F 
                 
                 = 
                 
                   Vinc 
                   ⁡ 
                   
                     ( 
                     
                       
                         
                           RL 
                           · 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         7 
                                       
                                       - 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         6 
                                       
                                     
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         6 
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         7 
                                       
                                     
                                   
                                   ) 
                                 
                               
                               - 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                       
                       
                         
                           
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 1 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         3 
                                         · 
                                         RS 
                                       
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         2 
                                         · 
                                         R 
                                       
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       3 
                                     
                                   
                                   ) 
                                 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   2 
                                   · 
                                   RS 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               RL 
                               ⁡ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     1 
                                     ⁢ 
                                     
                                       ( 
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           2 
                                         
                                         + 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           3 
                                         
                                         + 
                                         RS 
                                       
                                       ) 
                                     
                                   
                                   + 
                                   
                                     RS 
                                     ⁡ 
                                     
                                       ( 
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           2 
                                         
                                         + 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           3 
                                         
                                       
                                       ) 
                                     
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                     ) 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   32 
                 
                 ) 
               
             
           
         
       
     
   
   Assuming R 1 =R 2 =1K, R 2 =5.11, RS=50, R 6 =511 and R 7   F =420.3065369, then Equation 32 becomes: 
   
     
       
         
           
             
               
                 
                   VO 
                   F 
                 
                 = 
                 
                   - 
                   
                     Vinc 
                     ⁡ 
                     
                       ( 
                       
                         
                           5.11 
                           + 
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               0.10249304 
                               ) 
                             
                           
                         
                         
                           55.1102555 
                           + 
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               1.1053655 
                               ) 
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
             
           
           
             
               
                 = 
                 
                   
                     - 
                     
                       Vinc 
                       ⁡ 
                       
                         ( 
                         
                           
                             49.85704395 
                             + 
                             RL 
                           
                           
                             49.85704395 
                             + 
                             RL 
                           
                         
                         ) 
                       
                     
                   
                   ⁢ 
                   
                     ( 
                     0.09272321 
                     ) 
                   
                 
               
             
           
           
             
               
                 = 
                 
                   - 
                   
                     Vinc 
                     ⁡ 
                     
                       ( 
                       0.09272321 
                       ) 
                     
                   
                 
               
             
           
         
       
     
   
   Table 2 below illustrates the VO for values for RL from 0 to infinity, assuming Vinc=2V. 
   
     
       
             
             
             
             
             
           
             
             
             
             
             
           
         
             
                 
               TABLE 2 
             
             
                 
                 
             
             
                 
               RL 
               VO(mV) 
               dBV 
               Phase(deg) 
             
             
                 
                 
             
           
           
             
                 
             
           
        
         
             
                 
               0 to inf 
               −185.446 
               −14.636 
               −180 
             
             
                 
                 
             
           
        
       
     
   
   In Table 2, the phase of VO is 180 degrees out of phase with the standard definition of Vinc. Table 2 is used to illustrate that circuit  414 , with appropriate values for the resistors R 4 , R 5 , R 6  and R 7 , as defined above in Equation 30, will indeed act as a forward coupler, when RL is a SHORT (i.e., RL=0), when RL is an OPEN (i.e., RL=infinity), or any value there-between. 
   Since the phase of VO F  is 180 degrees from the source Vinc, the forward coupler case expressed in Equation 32 above can be rewritten as Equation 33 shown below. 
   
     
       
         
           
             
               
                 
                   VO 
                   F 
                 
                 = 
                 
                   - 
                   
                     Vinc 
                     ⁡ 
                     
                       ( 
                       
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               2 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           - 
                           
                             RL 
                             ⁡ 
                             
                               ( 
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     1 
                                     · 
                                     R 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   3 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           7 
                                         
                                         - 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           6 
                                         
                                       
                                       
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           6 
                                         
                                         + 
                                         
                                           R 
                                           ⁢ 
                                           
                                               
                                           
                                           ⁢ 
                                           7 
                                         
                                       
                                     
                                     ) 
                                   
                                 
                                 - 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     1 
                                     · 
                                     R 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   2 
                                 
                               
                               ) 
                             
                           
                         
                         
                           
                             
                               
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   1 
                                   ⁢ 
                                   
                                     ( 
                                     
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           3 
                                           · 
                                           RS 
                                         
                                       
                                       + 
                                       
                                         R 
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         
                                           2 
                                           · 
                                           R 
                                         
                                         ⁢ 
                                         
                                             
                                         
                                         ⁢ 
                                         3 
                                       
                                     
                                     ) 
                                   
                                 
                                 + 
                                 
                                   R 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     2 
                                     · 
                                     RS 
                                   
                                 
                                 + 
                               
                             
                           
                           
                             
                               
                                 RL 
                                 ⁡ 
                                 
                                   ( 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       1 
                                       ⁢ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                           + 
                                           RS 
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       RS 
                                       ⁡ 
                                       
                                         ( 
                                         
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             2 
                                           
                                           + 
                                           
                                             R 
                                             ⁢ 
                                             
                                                 
                                             
                                             ⁢ 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   33 
                 
                 ) 
               
             
           
         
       
     
   
   The inversion can be addressed in post processing. 
   Scattering Parameters 
   A reflectometer can be described by scattering parameters, where scattering parameter S 11  represents the value rho. More specifically, S 11  (which is referred to as the “forward reflection” coefficient) is equal to the signal leaving port  1  relative to the signal being injected into port  1 , or simply S 11 =Reflected Voltage/Incident Voltage. Accordingly, the use of Reverse and Forward Couplers can be combined to produce S 11  by dividing VO R  by VO F , as shown in Equation 34 below. 
   
     
       
         
           
             
               
                 
                   S 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   11 
                 
                 = 
                 
                   
                     VO 
                     R 
                   
                   
                     VO 
                     F 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   34 
                 
                 ) 
               
             
           
         
       
     
   
   Combining Equation 34 with Equations 15 and 33, and letting D=R 1 (R 3 ·RS+R 2 ·R 3 )+R 2 ·RS+RL(R 1 (R 2 +R 3 +RS)+(RS(R 2 +R 3 )), leads to Equation 35 below. 
   
     
       
         
           
             
               
                 
                   S 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   11 
                 
                 = 
                 
                   
                     
                       
                         - 
                         Vinc 
                       
                       D 
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         + 
                         
                           RL 
                           ⁡ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                       ) 
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           
                             
                               RL 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                             + 
                             
                               RL 
                               ⁡ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     2 
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                         - 
                         
                           
                             2 
                             ⁢ 
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               7 
                               R 
                             
                           
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               6 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 7 
                                 R 
                               
                             
                           
                         
                       
                       ) 
                     
                   
                   
                     
                       
                         - 
                         Vinc 
                       
                       D 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             
                                 
                             
                             ⁢ 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             
                                 
                             
                             ⁢ 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         - 
                         
                             
                         
                         ⁢ 
                         
                           RL 
                           ( 
                           
                               
                           
                           ⁢ 
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 1 
                                 · 
                                 
                                     
                                 
                                 ⁢ 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 ( 
                                 
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         7 
                                         F 
                                       
                                     
                                     - 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       6 
                                     
                                   
                                   
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       6 
                                     
                                     + 
                                     
                                       R 
                                       ⁢ 
                                       
                                           
                                       
                                       ⁢ 
                                       
                                         7 
                                         F 
                                       
                                     
                                   
                                 
                                 ) 
                               
                             
                             - 
                             
                                 
                             
                             ⁢ 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 1 
                                 · 
                                 
                                     
                                 
                                 ⁢ 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               2 
                             
                           
                           ⁢ 
                           
                               
                           
                           ) 
                         
                       
                       ⁢ 
                       
                           
                       
                       ) 
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   35 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 35 can be simplified to Equation 36 shown below. 
   
     
       
         
           
             
               
                 
                   S 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   11 
                 
                 = 
                 
                   
                     
                       ( 
                       
                         
                           R 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             1 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             2 
                             · 
                             R 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           3 
                         
                         + 
                         
                           RL 
                           ⁡ 
                           
                             ( 
                             
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 2 
                               
                               + 
                               
                                 R 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   1 
                                   · 
                                   R 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                       ) 
                     
                     ⁢ 
                     
                       ( 
                       
                         
                           
                             
                               RL 
                               · 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                           
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 2 
                                 · 
                                 R 
                               
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               3 
                             
                             + 
                             
                               RL 
                               ⁡ 
                               
                                 ( 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     2 
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                         - 
                         
                           
                             2 
                             ⁢ 
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               7 
                               R 
                             
                           
                           
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               6 
                             
                             + 
                             
                               R 
                               ⁢ 
                               
                                   
                               
                               ⁢ 
                               
                                 7 
                                 R 
                               
                             
                           
                         
                       
                       ) 
                     
                   
                   
                     ( 
                     
                       
                         R 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           1 
                           · 
                           
                               
                           
                           ⁢ 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           2 
                           · 
                           
                               
                           
                           ⁢ 
                           R 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         3 
                       
                       - 
                       
                           
                       
                       ⁢ 
                       
                         RL 
                         ( 
                         
                             
                         
                         ⁢ 
                         
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               
                                   
                               
                               ⁢ 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             3 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               ( 
                               
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       7 
                                       F 
                                     
                                   
                                   - 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     6 
                                   
                                 
                                 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     6 
                                   
                                   + 
                                   
                                     R 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       7 
                                       F 
                                     
                                   
                                 
                               
                               ) 
                             
                           
                           - 
                           
                               
                           
                           ⁢ 
                           
                             R 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               1 
                               · 
                               
                                   
                               
                               ⁢ 
                               R 
                             
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             2 
                           
                         
                         ⁢ 
                         
                             
                         
                         ) 
                       
                     
                     ⁢ 
                     
                         
                     
                     ) 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   36 
                 
                 ) 
               
             
           
         
       
     
   
   It can be appreciated from Equation 36 that S 11  has no dependency on the Source Voltage (Vinc) or the Source Impedance (RS). Referring to  FIG. 4 , since the pi attenuator (made up of resistors R 1 , R 2  and R 3 ) and the amplifiers (A 1  and A 2 ) are used for both forward and reverse couplers, only one RF element is needed, which reduces the Through Pass Loss for the high frequency mode, e.g., greater than 2 MHz. 
   Assuming R 1 =R 3 =1K, R 2 =5.11, RS=50, R 6 =511, R 7   R =420.7077111 and R 7   F =420.3065369, then Equation 36 becomes: 
   
     
       
         
           
             S 
             ⁢ 
             
                 
             
             ⁢ 
             11 
           
           = 
           
             
               ( 
               
                 
                   RL 
                   - 
                   50 
                 
                 
                   RL 
                   + 
                   49.85704395 
                 
               
               ) 
             
             ⁢ 
             
               ( 
               0.90050742 
               ) 
             
           
         
       
     
   
   Table 3 below illustrates S 11  for values for RL from 0 to infinity. 
   
     
       
             
             
             
             
             
           
             
             
             
             
             
           
         
             
                 
               TABLE 3 
             
             
                 
                 
             
             
                 
               RL 
               S11 
               20 Log[S11] 
               Phase(deg) 
             
             
                 
                 
             
           
           
             
                 
             
           
        
         
             
                 
               0 
               −0.90770 
               0.84111 
               −180 
             
             
                 
               0.5 
               −0.88930 
               −1.01904 
               −180 
             
             
                 
               5.0 
               −0.74060 
               −2.60830 
               −180 
             
             
                 
               25 
               −0.30100 
               −10.4286 
               −180 
             
             
                 
               45 
               −0.04749 
               −26.4677 
               −180 
             
             
                 
               50 
               0 
               −inf 
               −180 
             
             
                 
               55.5555 
               +0.04748 
               −26.4689 
               0 
             
             
                 
               100 
               +0.30053 
               −10.4423 
               0 
             
             
                 
               500 
               +0.73701 
               −2.65055 
               0 
             
             
                 
               5000 
               +0.88271 
               −1.08369 
               0 
             
             
                 
               Inf 
               +0.90051 
               −0.91025 
               0 
             
             
                 
                 
             
           
        
       
     
   
   The results for the example show a constant multiplier 0.90050742 and a slight difference from 50 Ohms in the denominator. The constant multiplier is due to the attenuator through Insertion Loss, and will always represent that value. The denominator error is due to the approximate values used for R 1 , R 2  and R 3  for the 50 Ohm attenuator values. These approximations also affect the ideal Insertion Loss value. Both errors are removed when a standard Open/Short/Load (OSL) calibration is made. Table 4 below illustrates that results that would be expected for an ideal case, where for an ideal reflectometer 
   
     
       
         
           
             S 
             ⁢ 
             
                 
             
             ⁢ 
             11 
           
           = 
           
             
               
                 RL 
                 - 
                 50 
               
               
                 RL 
                 + 
                 50 
               
             
             . 
           
         
       
     
   
   
     
       
             
             
             
             
             
           
             
             
             
             
             
           
         
             
                 
               TABLE 4 
             
             
                 
                 
             
             
                 
               RL 
               S11 
               20 Log[S11] 
               Phase(deg) 
             
             
                 
                 
             
           
           
             
                 
             
           
        
         
             
                 
               0 
               −1.00000 
               0 
               −180 
             
             
                 
               0.5 
               −0.98020 
               −0.1737 
               −180 
             
             
                 
               5.0 
               −0.81818 
               −1.7430 
               −180 
             
             
                 
               25 
               −0.33333 
               −9.5424 
               −180 
             
             
                 
               45 
               −0.05263 
               −25.5751 
               −180 
             
             
                 
               50 
               0 
               −inf 
               −180 
             
             
                 
               55.5555 
               +0.05263 
               −25.5751 
               0 
             
             
                 
               100 
               +0.33333 
               −9.5424 
               0 
             
             
                 
               500 
               +0.81818 
               −1.7430 
               0 
             
             
                 
               5000 
               +0.98020 
               −0.1737 
               0 
             
             
                 
               Inf 
               +1.00000 
               0 
               0 
             
             
                 
                 
             
           
        
       
     
   
   A comparison of Tables 3 and 4 shows that S 11  parameters measured using embodiments of the present invention are close to what is ideal, but with some errors. But, as mentioned above, such errors can be removed using a standard Open/Short/Load (OSL) calibration. 
     FIG. 5  illustrates a practical reflectometer using forward and reverse couplers  104  and  106  with insertion loss and coupling factor values IL and CPL in dB. From  FIG. 5  it can be seen that PF and PR can be represented, respectively, by Equations 37 and 38 below. 
   
     
       
         
           
             
               
                 PF 
                 = 
                 
                   PINC 
                   - 
                   
                     CPL 
                     F 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   37 
                 
                 ) 
               
             
           
           
             
               
                 PR 
                 = 
                 
                   PINC 
                   - 
                   
                     IL 
                     F 
                   
                   - 
                   
                     IL 
                     R 
                   
                   - 
                   
                     CPL 
                     R 
                   
                   + 
                   
                     20 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     Log 
                     ⁢ 
                     
                        
                       
                         
                           RL 
                           - 
                           50 
                         
                         
                           RL 
                           + 
                           50 
                         
                       
                        
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   38 
                 
                 ) 
               
             
           
         
       
     
   
   Equation 39, below, shows scattering parameter S 11  in terms of PF and PR.
 
 S 11 dB   =PR−PF    (EQ. 39)
 
   Combining Equation 39 with Equations 37 and 38 leads to Equation 40 below. 
   
     
       
         
           
             
               
                 
                   S 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     11 
                     dB 
                   
                 
                 = 
                 
                   ( 
                   
                     PINC 
                     - 
                     
                       IL 
                       F 
                     
                     - 
                     
                       IL 
                       R 
                     
                     - 
                     
                       CPL 
                       R 
                     
                     + 
                     
                       20 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       Log 
                       ⁢ 
                       
                          
                         
                           
                             RL 
                             - 
                             50 
                           
                           
                             RL 
                             + 
                             50 
                           
                         
                          
                       
                     
                     - 
                     
                       ( 
                       
                         PINC 
                         - 
                         
                           CPL 
                           F 
                         
                       
                       ) 
                     
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   40 
                 
                 ) 
               
             
           
         
       
     
   
   Letting CPL F =CPL R  and IL F =IL R , leads to Equation 41 below. 
   
     
       
         
           
             
               
                 
                   S 
                   ⁢ 
                   
                       
                   
                   ⁢ 
                   
                     11 
                     dB 
                   
                 
                 = 
                 
                   
                     20 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     Log 
                     ⁢ 
                     
                        
                       
                         
                           RL 
                           - 
                           50 
                         
                         
                           RL 
                           + 
                           50 
                         
                       
                        
                     
                   
                   - 
                   
                     IL 
                     R 
                   
                   - 
                   
                     IL 
                     F 
                   
                 
               
             
             
               
                 ( 
                 
                   EQ 
                   . 
                   
                       
                   
                   ⁢ 
                   41 
                 
                 ) 
               
             
           
         
       
     
   
   Letting CPL F =CPL R =14.436 dB and IL F =IL R =0.910254/2=0.455127 dB, then Equation 41 becomes: 
   
     
       
         
           
             S 
             ⁢ 
             
                 
             
             ⁢ 
             
               11 
               dB 
             
           
           = 
           
             
               20 
               ⁢ 
               
                   
               
               ⁢ 
               log 
               ⁢ 
               
                  
                 
                   
                     RL 
                     - 
                     50 
                   
                   
                     RL 
                     + 
                     50 
                   
                 
                  
               
             
             - 
             
               0.90050742 
               ⁢ 
               
                   
               
               ⁢ 
               
                 dB 
                 . 
               
             
           
         
       
     
   
   Converting this to linear leads to: 
   
     
       
         
           
             S 
             ⁢ 
             
                 
             
             ⁢ 
             11 
           
           = 
           
             
               ( 
               
                 
                   RL 
                   - 
                   50 
                 
                 
                   RL 
                   + 
                   50 
                 
               
               ) 
             
             ⁢ 
             
               
                 ( 
                 0.90050742 
                 ) 
               
               . 
             
           
         
       
     
   
   This ideal example shows close agreement with measurements that can be obtained using embodiments of the present invention. 
     FIG. 6  illustrates details of the circuit  205  to emulate a dual directional coupler, in accordance with embodiments of the present invention. Resistors R 4   F -R 7   F  on the left emulate a forward coupler, while resistor R 4   R -R 7   R  on the right emulate a reverse coupler. 
     FIG. 7  illustrates how the embodiments of the present invention can be used together with a conventional dual directional coupler to thereby provide a VNA  200  capable of making measurements from the DC to multi-GHz range. In this arrangement, the circuit  205  to emulate a dual directional coupler is used when the frequency of the incident signal Vinc produced by the signal source  102  is within a first range (e.g., DC-2 MHz); and the conventional dual directional coupler made up of actual forward and reverse couplers  104  and  106  are used when the frequency of the incident signal Vinc produced by the signal source  102  is within a higher second range (e.g., 2 MHz-8 GHz). 
   Referring back to  FIG. 2 , a detector/receiver  210  that accepts the VF and VR signals from the circuit  205 , as well as from couplers  104  and  106 , preferably includes a baseband converter for each pair of signals. For example, a baseband converter  212  may use both upconversion and downconversion to provide baseband signals (e.g., of about 200 KHz) to a synchronous detector or DSP  216 . A further baseband converter  214  will likely only used downconversion to provide baseband signals of the same frequency (e.g., of about 200 KHz) to the synchronous detector or DSP  216 . In accordance with an embodiment, through the use of switches S 1  and S 2 , only one of the baseband converters  212  and  214  provides the baseband forward and reverse signals to the synchronous detector or DSP  216 . More specifically, when the frequency of the incident signal Vinc produced by the signal source  102  is within the first range (e.g., DC-2 MHz), the switches S 1  and S 2  provide the output of the baseband converter  212  to the synchronous detector or DSP  216 . When the frequency of the incident signal Vinc produced by the signal source  102  is within the second higher range (e.g., 2 MHz to 8 GHz), the switches S 1  and S 2  provide the output of the baseband converter  214  to the synchronous detector or DSP  216 . In this manner, the VNA can make measurements over a larger bandwidth (e.g., DC to 8 GHz). 
   While various embodiments of the present invention have been described above, it should be understood that they have been presented by way of example, and not limitation. It will be apparent to persons skilled in the relevant art that various changes in form and detail can be made therein without departing from the spirit and scope of the invention. 
   The present invention has been described above with the aid of functional building blocks illustrating the performance of specified functions and relationships thereof. The boundaries of these functional building blocks have often been arbitrarily defined herein for the convenience of the description. Alternate boundaries can be defined so long as the specified functions and relationships thereof are appropriately performed. Any such alternate boundaries are thus within the scope and spirit of the invention. 
   The breadth and scope of the present invention should not be limited by any of the above-described exemplary embodiments.