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VSWR - RETURN LOSS - G  CONVERSIONS

 Return Loss Voltage Standing Wave Ratio Reflection Coefficient RL = -20*log|G| dB VSWR = G = RL(dB) VSWR G RL(dB) VSWR G RL(dB) VSWR G RL(dB) VSWR G 46.0 1.01 0.00498 26.0 1.11 0.0521 17.7 1.30 0.130 8.0 2.32 0.398 40.0 1.02 0.00990 25.0 1.12 0.0566 17.0 1.33 0.141 7.0 2.61 0.445 37.0 1.03 0.0148 24.0 1.13 0.0610 16.0 1.38 0.158 6.02 3.01 0.500 34.0 1.04 0.0196 23.5 1.14 0.0654 15.0 1.43 0.178 5.0 3.56 0.562 32.0 1.05 0.0244 23.0 1.15 0.0698 14.0 1.50 0.200 4.0 4.42 0.631 30.4 1.06 0.0291 22.0 1.17 0.0783 13.0 1.58 0.224 3.01 5.85 0.707 29.0 1.07 0.0338 21.5 1.18 0.0826 12.0 1.67 0.250 2.0 8.72 0.794 28.0 1.08 0.0385 20.7 1.20 0.0909 11.0 1.78 0.282 1.0 17.39 0.891 27.0 1.09 0.0431 20.0 1.22 0.100 10.0 1.92 0.316 0.5 34.75 0.944 26.4 1.10 0.0476 19.0 1.25 0.112 9.0 2.10 0.355 0.0 Infinity 1.00

 VSWR REDUCTION BY MATCHED ATTENUATOR By inserting a matched (nominal system impedance) attenuator in front of a mismatched load impedance, the mismatch "seen" at the input of the attenuator is improved by an amount equal to twice the value of attenuator. The explanation is simple. Return loss is determined by the portion of the input signal that is reflected at the load (due to impedance mismatch) and returned to the source. A perfect load impedance (complex conjugate of the source impedance) would absorb 100% of the incident signal and therefore reflect 0% of it back to the source (return loss of Ą dB). For the sake of illustration, assume that the load is an open (or short) circuit, where 0% of the incident signal is absorbed by the load and 100% is reflected back to the source. The reflected signal would therefore have a return loss of 0 dB. Insert a 3 dB  attenuator in front of the load. Now the incident signal is referenced to the input of the attenuator. As signal at the input of the attenuator will experience a 3 dB reduction in power by the time it reaches the load. That 3 dB less power will be 100% reflected by the load and experience another 3 dB reduction in power by the time is returns back to the input, for a total loss of 6 dB. The same principle applies for a load anywhere(§) between zero and infinite load impedance (short and open circuits, respectively). Calculate the improved VSWR as follows: Convert the load VSWR to load return lossper the following equation:                       RLLOAD=20*log dB Add twice the attenuation value to RLLOAD: RLNEW=RLLOAD + 2*ATTEN dB Convert back to VSWR per the following equation: VSWR= Of course, the method can be reversed to predict the attenuator required to improve a load VSWR by a predetermined amount. To do so, calculate the desired return loss and subtract the known load return loss. Divide the answer by two to get the attenuator value needed.  § Actually, the attenuator is only rated for its specified attenuation level when it is connected between two nominal impedances. Therefore, the attenuator will either have to be designed to closely match the two impedances at its input and output (source and load, respectively), or an adjustment will need to be made in the specified attenuation value to compensate for the mismatched load impedance.

VSWR MISMATCH ERRORS

Both amplitude and phase errors are introduced when mismatched impedances are present at an electrical interface. The result is ripple across the frequency band (since the VSWR of each interface typically varies with frequency), as well as a portion of the incident power being reflected back to the source. Amplitude Error Phase Error
eA = 20 * log (1 ± |GA * GB|) dB ef = 180 / p * GA * GB
Resultant MIN and MAX Cascaded VSWR
VSWRMAX = SA * SB
VSWRMIN = SA / SB

 where SA = larger of the two VSWRs SB = smaller of the two VSWRs

Example

VSWRA = 2.5:1  -->  SA = 2.5
VSWRB = 2.0:1  -->  SB = 2.0
VSWRMAX = 2.5 * 2.0 = 5.0  = 5.0:1
VSWRMIN = 2.5 / 2.0 = 1.25  = 1.25:1