This article is demonstration of measurement of Transmission Loss in a section of coaxial transmission line. The scenario is chosen to expose the experiment to some of the things that complicate such measurements.

The very popular nanoVNA-H will be used to make the measurements.

The scenario:

- nanoVNA fully calibrated from 1-5MHz using a 200mm length coax lead on Port 2 (nanoVNA CH1);
- 35m of CCS RG6/U (close to an electrical quarter wavelength); and
- f=1.65MHz (close to a quarter wavelength.

Above is a block diagram of the test configuration. nanoVNA measurements are wrt 50Ω, so \(P=\frac{V^2}{50}\) and \(V=\sqrt{50P}\).

The DUT was swept one way from 1.5-1.8MHz and saved as a .s2p file.

1656000 0.385966661 -6.813504102 0.815628707 -85.534776748 1.0 0.0 1.0 0.0

Above is a record from the .s2p file in MA format.

Above is calculation of Zin=111.2-j11.97Ω from s11 from the .s2p file. Also calculated MismatchLoss=0.701dB. This method of calculating MismatchLoss is only correct if either source or load impedance is purely real, which is true in this case. A more general calculation uses Kurokawa's expression.

Above, Kurokawa's 1/(1-|s|^2) is the correct MismatchLoss in the more general case. They reconcile, as we expect.

Converting the magnitude of s21 from the .s2p file, we get |s21|=-1.77dB.

## So, a bit of accounting is in order

Let's review some meanings of terms (in the 50Ω VNA context):

- \(TransmissionLoss=\frac{PowerIn}{PowerOut}=\frac{P_1}{P_2}\);
- \(s11=\frac{V_{1r}}{V_{1i}}\);
- \(InputMismatchLoss=\frac{P_{1i}}{P_{1}}=\frac1{(1-|s11|^2)}\); and
- \(s21=\frac{V_{2i}}{V_{1i}}\).

It is assumed that Zin of VNA Port 2 is 50+j0Ω, and that therefore P2r=0. Error in Zin of VNA Port 2 flows into the results. A 10dB attenuator is fitted to Port 2 prior to calibration to improve accuracy of Zin.

With the quantities expressed in dB, we can derive that \(TransmissionLoss=-|s21|-InputMismatchLoss\).

In the example given above, \(TransmissionLoss=-|s21|-InputMismatchLoss \\=1.77-0.701=1.07\: dB\)

Note that this is not the Matched Line Loss, the 75Ω line was terminated by the 50Ω VNA port so there is a standing wave.

You might be tempted to apply (Smith 1995)'s expression for loss due to standing waves \(\frac{Loss_{mismatched}}{Loss_{matched}}=\frac{1+S^2}{2S}\) but this scenario does not satisfy his conditions for it to apply, a much misused expression.

## References

- Smith, P. 1995. Electronic applications of the Smith chart 2nd ed. Noble Publishing Tucker.

## Conclusions

The Transmission Loss of a line section may not be directly given by any measures displayed by a VNA, it may take some interpretation and some accounting for elements that can be measured.