In a video titled PDN Measurements with a VNA Rolf Ostergaard shows how to transform the magnitude of the insertion loss S21 to an impedance value. This is the formula he uses:

$$ Z = 25\Omega \times 10^{\frac{S_{21}[dB]}{20}} $$

Then, in the comments, someone asks why he uses \$25\Omega\$ when the port impedances are \$50\Omega\$. The author's response is:

Try doing the math. Remember, 0dB means the receiver sees half the generator amplitude.

I don't get it:

Why does 0 dB mean that the receiver sees half the generator amplitude?


1 Answer 1


Why does 0 dB mean that the receiver sees half the generator amplitude?

Because this is the actual circuit:


simulate this circuit – Schematic created using CircuitLab

In order to match the impedance the generator will have a 50 Ohms output impedance and the load will also be 50 Ohms. Without that the coaxial cable will not be properly matched and you will get signal reflections which cause distortion.

This is an ideal system where S21 = 0 dB, consider that the reference. Any deviations from that will show as S21 not being zero dB.

  • \$\begingroup\$ I guess it all depends on what is meant by the term "generator output". If I have a signal generator or vna who's output impedance is specified to be 50 ohms, and if I take that output to be the generator's output, then the receiver (load) would see the full generator's output voltage if S21 of the cable were 0 dB. \$\endgroup\$
    – SteveSh
    Commented Sep 13, 2021 at 18:41
  • \$\begingroup\$ Todd Hubing's Power Bus Decoupling on Multilayer Printed Circuit Boards (p.161) made everything click into place for me together with your answer. \$\endgroup\$
    – pfabri
    Commented Sep 14, 2021 at 14:21

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