Abstract:
In one embodiment, a method comprises storing parameters that are related to switch error correction terms of a vector network analyzer (VNA), and applying a calibration process of a TRL group of calibration processes to the VNA to generate calibration measurements, wherein the calibration process generates calibration measurements, calculates a switch error correction matrix using the stored parameters and a subset of the calibration measurements, and applies the switch error correction matrix to calibration measurements before solving for eight-systematic error terms associated with the calibration process.

Description:
TECHNICAL FIELD 
   The present application is generally related to calibration of a vector network analyzer (VNA). 
   BACKGROUND 
     FIG. 1  depicts VNA  100  according to a conventional design. As shown in  FIG. 1 , VNA  100  comprises switch  101  to switch between VNA ports A through D to establish a path to a single reference receiver  102 . When the RF signal path switch  101  changes position, the termination of the test port also changes. The change in termination causes the source match term to be different from the load match term. In a multiport VNA (more than two test ports), it is possible that each test port has multiple different load match terms. During calibration, each of the load match terms needs to be characterized and treated in the same manner as a two-port VNA. The difference between source and load match is referred to as the “switch error.” The standard 12-term VNA error model (or multiport equivalent) derives the load match term from the through connection. 
   The TRL family or group of calibrations is based on the 8-term error model which only factors in a single match term at each testport. The TRL family usually requires two receivers for each test port to factor out any port match variation through additional measurements and to obtain the necessary data to determine the VNA&#39;s systematic error terms. The TRL family of calibrations includes all calibration algorithms that are based on having a constant match defined for each test port independent of switching. The TRL family includes but is not limited to TRL, TRM, LRL, LRM, and Unknown Thru algorithms. Subsequent references to TRL calibration as used herein applies to the entire family of calibration and not to a specific algorithm. Because two receiver requirement, the traditional TRL calibration method cannot be applied to VNA  100 . To address this issue, a two-tier calibration process has been developed. In the two-tier process, the short, open, load, through (SOLT) calibration method is initially performed to obtain the switch error correction terms. After the initial calibration and with error correction turned “on,” the second tier is performed by applying the TRL calibration process. The multi-tier calibration essentially doubles the amount of time required to calibrate a VNA as compared to the calibration time associated with a VNA having a reference receiver for each port. 
   SUMMARY 
   Some representative embodiments enable calibration procedures to be applied to a VNA that only possesses a single reference receiver in an efficient manner. In some representative embodiments, a SOLT calibration is performed as an initial calibration tier. The SOLT calibration is used to calculate parameters associated with the switch error correction terms of the VNA. The calculated parameters are stored. Upon subsequent re-calibration of the VNA, some representative embodiments omit the necessity of repeating the SOLT calibration tier. In particular, some representative embodiments proceed directly to the TRL calibration process. A switch error correction matrix is calculated using the stored parameters and the measurement data from the TRL calibration process. The switch error correction matrix is applied to the measurement data and the error terms are calculated. 

   
     BRIEF DESCRIPTION OF THE DRAWINGS 
       FIG. 1  depicts a single reference receiver VNA according to a conventional design. 
       FIGS. 2 and 3  depict forward and reverse eight-error term models. 
       FIG. 4  depicts a flowchart including calibration operations performed on a VNA having only one reference receiver according to one representative embodiment. 
       FIG. 5  depicts a VNA according to one representative embodiment. 
   

   DETAILED DESCRIPTION 
   To facilitate the discussion of the mathematical basis associated with some representative embodiments, the 8-term error model  200  is shown in  FIG. 2 . Given a VNA with two receivers per test port that can measure the incident and reflected signals, when the switch is placed in the forward position and power is applied to the port- 1  side, the following equations are obtained:
 
 b   0   f   =S   11m   a   0   f   +S   12m   a   3   f  
 
 b   3   f   =S   12m   a   0   f   +S   22m   a   3   f   (1)
 
     FIG. 3  depicts the eight-term error model  300  where power is applied to the port- 2  side. When power the switch is placed in the reverse side and power is applied to the port- 2  side, the following equations are obtained:
   b   3   r   =S   21m   a   0   r   +S   22m   a   3   r      b   0   r   =S   11m   a   0   r   +S   12m   a   0   r   (2) 
   Because the system is not perfect, caused by L 1  and L 2 , the a 3   f  and a 0   r  terms are not zero and must be addressed. Re-arranging equations (1) and (2) into the following form: 
                           b   0   f       a   0   f       =       ⁢       S     11   ⁢   m       +       S     12   ⁢   m       ⁢       a   3   f       a   0   f                           b   0   r       a   3   r       =       ⁢         S     11   ⁢   m       ⁢       a   0   r       a   3   r         +     S     12   ⁢   m                         b   3   f       a   0   f       =       ⁢       S     21   ⁢   m       +       S     22   ⁢   m       ⁢       a   3   f       a   0   f                           b   3   r       a   3   r       =       ⁢         S     21   ⁢   m       ⁢       a   0   r       a   3   r         +     S     22   ⁢   m                       [     S   R     ]     =       ⁢       [           ⁢             b   0   f       a   0   f               b   0   r       a   3   r                   b   3   f       a   0   f               b   3   r       a   3   r             ]     =     [           ⁢           S     11   ⁢   R             S     12   ⁢   R                 S     21   ⁢   R             S     22   ⁢   R             ⁢           ]                     (   3   )               
Let
 
                           [     S   m     ]     =       ⁢     [           ⁢           S     11   ⁢   m             S     12   ⁢   m                 S     21   ⁢   m             S     22   ⁢   m             ⁢           ]       ;                 [     M   sc     ]     =       ⁢       [           ⁢         1           a   0   r       a   3   r                   a   3   f       a   0   f           1         ⁢           ]     =     [           ⁢         1         L   r               L   f         1         ⁢           ]                     (   4   )               
Then
   [S   R   ]=[S   m   ]*[M   sc ] and [ S   m   ]=[S   R   ]*[M   sc ] −1   (5) 
   Solving equation (5) results as follows: 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         S 
                         
                           11 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               0 
                               f 
                             
                             
                               a 
                               0 
                               f 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             ⁢ 
                             
                               
                                 a 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                           
                         
                         Δ 
                       
                     
                   
                 
                 
                   
                     
                       
                         S 
                         
                           12 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               0 
                               r 
                             
                             
                               a 
                               3 
                               r 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 0 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               
                                 a 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                           
                         
                         Δ 
                       
                     
                   
                 
                 
                   
                     
                       
                         S 
                         
                           21 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               3 
                               f 
                             
                             
                               a 
                               0 
                               f 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 3 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             ⁢ 
                             
                               
                                 a 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                           
                         
                         Δ 
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           S 
                           
                             22 
                             ⁢ 
                             m 
                           
                         
                         = 
                           
                         ⁢ 
                         
                           
                             
                               
                                 b 
                                 3 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             - 
                             
                               
                                 
                                   b 
                                   3 
                                   f 
                                 
                                 
                                   a 
                                   0 
                                   f 
                                 
                               
                               ⁢ 
                               
                                 
                                   a 
                                   0 
                                   r 
                                 
                                 
                                   a 
                                   3 
                                   r 
                                 
                               
                             
                           
                           Δ 
                         
                       
                       , 
                       
                         
                           where 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           Δ 
                         
                         = 
                         
                           1 
                           - 
                           
                             
                               
                                 a 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               
                                 a 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 6 
                 ) 
               
             
           
         
       
     
   
   It is noted that two extra measurements a 3   f  and a 0   r  are used to solve [S m ]. However, for a VNA with only one reference receiver, these terms cannot be determined. 
   To solve the switch error correction terms from a different perspective, the L f  and L r  terms can be rewritten as follows: 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           L 
                           r 
                         
                         = 
                           
                         ⁢ 
                         
                           
                             
                               a 
                               0 
                               r 
                             
                             
                               a 
                               3 
                               r 
                             
                           
                           = 
                           
                             
                               
                                 
                                   b 
                                   0 
                                   r 
                                 
                                 
                                   a 
                                   3 
                                   r 
                                 
                               
                               ⁢ 
                               
                                 
                                   a 
                                   0 
                                   r 
                                 
                                 
                                   b 
                                   0 
                                   r 
                                 
                               
                             
                             = 
                             
                               
                                 S 
                                 
                                   12 
                                   ⁢ 
                                   m 
                                 
                               
                               ⁢ 
                               
                                 L 
                                 1 
                               
                             
                           
                         
                       
                       ; 
                     
                   
                 
                 
                   
                     
                       
                         L 
                         1 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           a 
                           0 
                           r 
                         
                         
                           b 
                           0 
                           r 
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           L 
                           f 
                         
                         = 
                           
                         ⁢ 
                         
                           
                             
                               a 
                               3 
                               f 
                             
                             
                               a 
                               0 
                               f 
                             
                           
                           = 
                           
                             
                               
                                 
                                   b 
                                   3 
                                   f 
                                 
                                 
                                   a 
                                   0 
                                   f 
                                 
                               
                               ⁢ 
                               
                                 
                                   a 
                                   3 
                                   f 
                                 
                                 
                                   b 
                                   3 
                                   f 
                                 
                               
                             
                             = 
                             
                               
                                 S 
                                 
                                   21 
                                   ⁢ 
                                   m 
                                 
                               
                               ⁢ 
                               
                                 L 
                                 2 
                               
                             
                           
                         
                       
                       ; 
                     
                   
                 
                 
                   
                     
                       
                         L 
                         2 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           a 
                           3 
                           f 
                         
                         
                           b 
                           3 
                           f 
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 7 
                 ) 
               
             
           
         
       
     
   
   Equation (7) can be substituted back into equation (6) as follows (this method is equivalent to adding an imperfect termination to the respective error box): 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         S 
                         
                           11 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               0 
                               f 
                             
                             
                               a 
                               0 
                               f 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             ⁢ 
                             
                               
                                 b 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               L 
                               2 
                             
                           
                         
                         
                           Δ 
                           ′ 
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         S 
                         
                           12 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               0 
                               r 
                             
                             
                               a 
                               3 
                               r 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 0 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               
                                 b 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             ⁢ 
                             
                               L 
                               2 
                             
                           
                         
                         
                           Δ 
                           ′ 
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         S 
                         
                           21 
                           ⁢ 
                           m 
                         
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             
                               b 
                               3 
                               f 
                             
                             
                               a 
                               0 
                               f 
                             
                           
                           - 
                           
                             
                               
                                 b 
                                 3 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             ⁢ 
                             
                               
                                 b 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               L 
                               2 
                             
                           
                         
                         
                           Δ 
                           ′ 
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         
                           S 
                           
                             22 
                             ⁢ 
                             m 
                           
                         
                         = 
                           
                         ⁢ 
                         
                           
                             
                               
                                 b 
                                 3 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                             - 
                             
                               
                                 
                                   b 
                                   3 
                                   f 
                                 
                                 
                                   a 
                                   0 
                                   f 
                                 
                               
                               ⁢ 
                               
                                 
                                   b 
                                   0 
                                   r 
                                 
                                 
                                   a 
                                   3 
                                   r 
                                 
                               
                               ⁢ 
                               
                                 L 
                                 1 
                               
                             
                           
                           
                             Δ 
                             ′ 
                           
                         
                       
                       , 
                       
                         
                           where 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             Δ 
                             ′ 
                           
                         
                         = 
                         
                           1 
                           - 
                           
                             
                               L 
                               1 
                             
                             ⁢ 
                             
                               L 
                               2 
                             
                             ⁢ 
                             
                               
                                 b 
                                 3 
                                 f 
                               
                               
                                 a 
                                 0 
                                 f 
                               
                             
                             ⁢ 
                             
                               
                                 b 
                                 0 
                                 r 
                               
                               
                                 a 
                                 3 
                                 r 
                               
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 8 
                 ) 
               
             
           
         
       
     
   
   If L 1  and L 2  can be determined and saved, the values can be retrieved and used in equation (6). From  FIGS. 2 and 3 , the following equations may be derived: 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         E 
                         LF 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           e 
                           22 
                         
                         + 
                         
                           
                             
                               e 
                               32 
                             
                             ⁢ 
                             
                               e 
                               33 
                             
                             ⁢ 
                             
                               L 
                               2 
                             
                           
                           
                             1 
                             - 
                             
                               
                                 e 
                                 33 
                               
                               ⁢ 
                               
                                 L 
                                 2 
                               
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         E 
                         LR 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           e 
                           11 
                         
                         + 
                         
                           
                             
                               e 
                               10 
                             
                             ⁢ 
                             
                               e 
                               01 
                             
                             ⁢ 
                             
                               L 
                               1 
                             
                           
                           
                             1 
                             - 
                             
                               
                                 e 
                                 00 
                               
                               ⁢ 
                               
                                 L 
                                 1 
                               
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 9 
                 ) 
               
             
           
         
       
     
   
   Solving for L 2  and L 1  gives: 
   
     
       
         
           
             
               
                 
                   
                     
                       
                         L 
                         2 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             E 
                             LF 
                           
                           - 
                           
                             e 
                             22 
                           
                         
                         
                           
                             
                               e 
                               32 
                             
                             ⁢ 
                             
                               e 
                               23 
                             
                           
                           + 
                           
                             
                               e 
                               33 
                             
                             ⁡ 
                             
                               ( 
                               
                                 
                                   E 
                                   LF 
                                 
                                 - 
                                 
                                   e 
                                   22 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
                 
                   
                     
                       
                         L 
                         1 
                       
                       = 
                         
                       ⁢ 
                       
                         
                           
                             E 
                             LR 
                           
                           - 
                           
                             e 
                             11 
                           
                         
                         
                           
                             
                               e 
                               10 
                             
                             ⁢ 
                             
                               e 
                               01 
                             
                           
                           + 
                           
                             
                               e 
                               00 
                             
                             ⁡ 
                             
                               ( 
                               
                                 
                                   E 
                                   LR 
                                 
                                 - 
                                 
                                   e 
                                   11 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
               
             
             
               
                 ( 
                 10 
                 ) 
               
             
           
         
       
     
   
   Comparing equation (8) to equation (6), it is seen that:
 
 a   3   f   =b   3   f   L   2  and  a   0   r   =b   0   r   L   1   (11)
 
   Based upon the preceding mathematical derivations, it is observed that the parameters L 1  and L 2  can be determined using a calibration process based upon the twelve term error model (e.g., the SOLT method). In VNAs with dual reflectometers at each port, a 3   f  and a 0   r  are directly measured. In VNA systems without dual reflectometers, the terms a 3   f  and a 0   r  cannot be determined directly by calibration methods that use the eight term error model. However, if a VNA is sufficiently stable with time and temperature, it may be assumed that the parameters L 1  and L 2  will remain relatively constant (at least over the “short” term). Accordingly, instead of directly measuring the terms a 3   f  and a 0   r  upon each calibration, these terms can be calculated from the terms b 3   f  and b 0   r  and the parameters L 1  and L 2 . 
   In view of the ability to accurately estimate the terms a 3   f  and a 0   r  using stored parameters, an efficient method of calibrating a VNA having only one reference receiver can be achieved as shown in  FIG. 4 . The portions of the flowchart of  FIG. 4  may be implemented using suitable logic (e.g., software or integrated circuitry) on a VNA or an associated processing platform (e.g., a personal computer). This is also applicable to a VNA with more than one reference receiver. Some reference receiver designs do not measure the a 3   f  and a 0   r  terms correctly. 
   In step  401 , a calibration method (e.g., the SOLT method) based upon the twelve error term model is applied to a VNA. In step  402 , the parameters L 1  and L 2 , as defined above, are calculated and stored. 
   In step  403 , a TRL calibration is performed to generate calibration data. In step  404 , a switch error correction matrix is formed using the calculated terms a 3   f  and a 0   r . In step  405 , the switch error correction matrix is formed and, in step  406 , the switch error correction matrix applied to the calibration data. In step  407 , the corrected calibration data is then used to calculate the eight systematic error terms using standard TRL techniques. In step  408 , device testing occurs using the calibrated VNA. 
   In some embodiments, the SOLT calibration process is repeated from time to time to maintain long term accuracy of the stored parameters L 1  and L 2 . Additionally, the stored parameters L 1  and L 2  preferably include sufficient data points to reduce interpolation errors. 
     FIG. 5  depicts VNA  300  according to one representative embodiment. VNA  500  that includes a minimum of one reference receiver. VNA  500  comprises SOLT calibration software  502  that calculates parameters L 1  and L 2  and stores the parameters in data file  501 . VNA  500  further includes TRL calibration software  304  that calculates terms a 3   f  and a 0   r  using data file  501 , generates a switch error correction matrix, and calculates eight systematic error terms using switch error corrected calibration data. Preferably, TRL calibration software  304  performs interpolation of parameters L 1  and L 2  for frequencies not explicitly represented in data file  501 . Although software is shown in  FIG. 5  to perform calibration operations, other suitable logic could alternatively be employed such as integrated circuitry.