0
\$\begingroup\$

I realized recently that the Touchstone "DB" S-parameter format (.s1p, etc) uses the phase angle of \$\angle \Gamma\$ instead of \$\angle S_{11}\$ and wondered why it didn't use the phase angle of the complex S11 measurement. .

Given the complex variables:

  • \$\Gamma = \frac{Z_{load}-Z_0}{Z_{load}+Z_0}\$
  • \$S_{11} = 20*{clog}_{10}(\Gamma) \$

and phase angle function \$\angle c = {carg}(z) = {atan}(\frac{b}{a})\$ where \$z = a+bi\$,

It seems strange to me that the "real" part would be in DB but the phase part would be from \$\Gamma\$ which is outside of the clog() function; perhaps the meaning of the phase angle in \$\Gamma\$ is lost after going through clog()?

Questions:

  • Is there a meaningful difference between \$\angle S_{11}\$ and \$\angle \Gamma\$?
    • Is \$\angle S_{11}\$ the phase angle of the reflected wave?
    • If so, then what does \$\angle \Gamma\$ mean: what is useful about the phase angle of the reflection coefficient \$\Gamma\$
    • How is that be more useful than \$\angle S_{11}\$ in an RF data format?
  • When would you use one over the other?

Please don't answer "because that is convention" or "because of a Touchstone file format design decision" because I'm trying to understand the difference in between \$\angle S_{11}\$ and \$\angle \Gamma\$ in a meaningful way. It may come down to a design decision, but when would the phase angle of one vs the other be used?

\$\endgroup\$
0

1 Answer 1

2
\$\begingroup\$

It seems strange to me that the "real" part would be in DB but the phase part would be from Γ which is outside of the clog() function; perhaps the meaning of the phase angle in Γ is lost after going through clog()?

This is a misinterpretation of the formula. If you were to actually keep only the real part, your conversion would simply be wrong. Γ = 0 + j1 is an S11 of 0 dB @ 90 degrees, not an S11 of -infinity dB as you suggest.

The value wasn't changed, its magnitude was just represented on a logarithmic scale. Converting "this amplifier has 14 dB voltage gain with a phase shift of 4°" to "this amplifier's H(ω) is 4.9997 + 0.3496i" is the same idea.

Is there a meaningful difference between ∠S11 and ∠Γ?

No. It's the same numerical value. S11 is commonly represented in log-scale, and gamma is often represented in linear scale measurement. For example, I may choose to give this value of a circuit as 0.7071 + j0.7071 (i.e. 1.000∠45°), or as "0 dB, angle of 45°"

Is ∠S11 the phase angle of the reflected wave?

Yes. If the returning voltage wave is in-phase with the incident voltage wave (interfering constructively at the point of measurement), then S11 has an angle of 0. If the returning current wave is in-phase with the incident current wave (interfering constructively at the point of measurement) then S11 has an angle of 180 degrees.

If so, then what does ∠Γ mean: what is useful about the phase angle of the reflection coefficient Γ

Gamma is S11. See previous answer.

How is that be more useful than ∠S11 in an RF data format?

It's the same angle.

When would you use one over the other?

Same remarks about numerical precision that I've given in the past. For the same reason, you may choose between phase-magnitude or real-imaginary representation of gamma/S11 as you see fit for your numerical precision needs.

\$\endgroup\$
3
  • \$\begingroup\$ So in the touchstone DB format, is the real column a magnitude as |S11| or just the real component as real(S11)? \$\endgroup\$
    – KJ7LNW
    Mar 14, 2022 at 20:47
  • \$\begingroup\$ @KJ7LNW I do not know the touchstone DB format in particular, but reporting the log-scale of only the real part of S11 is fairly nonsensical. If you were to report just the real part, you immediately create needless ambiguities and complexity -- what numerical gamma would "-infinity dB, 90 degrees phase" mean? (0+j0.1)? (0+j1)? (0+j99999999)? Applying the same guidance as before - if you have a doubt that you can't resolve, hook up your VNA to the DUT and see what it reports in the different formats. \$\endgroup\$
    – nanofarad
    Mar 14, 2022 at 21:01
  • \$\begingroup\$ Where Γ is complex, clog10 is complex log10, and log10 is real log10: it turns out that real(20*clog10( Γ )) == 20*log10(cabs(Γ)) holds true with tiny a floating point error: creal(cs11)=-6.4614700088017338 20*log10(cabs(cgamma))=-6.4614700088017347 diff=-8.8817841970012523e-16. These must be algebraically equivalent. \$\endgroup\$
    – KJ7LNW
    Mar 14, 2022 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.