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I have done a measurement with a vector analyzer in which I have connected a certain coaxial cable to the two port of the instrument.

Precisely, I have got a measure of the phase of S21 of this type:

enter image description here

You may see that the phase of S21 is a periodic signal with respect to frequency. Why?

I know that in a transmission line S21 = exp(-jβl), and so there is a linear dependence of it with respect to the frequency, not a periodic one,

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    \$\begingroup\$ The phase is not a periodic signal. It looks like one if you were mistakenly thinking you were viewing an oscilloscope. \$\endgroup\$
    – Andy aka
    Oct 31, 2019 at 17:37
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    \$\begingroup\$ The phase is continuous and linear, the network analyzer just wraps it to +/- 180° so it can be displayed nicely. \$\endgroup\$
    – Captainj2001
    Oct 31, 2019 at 18:16

3 Answers 3

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What you're seeing is called "wrapped phase". The phase angle wraps around a circle, like a clock, and the plot only shows the angle as a value between -180 and 180 degrees. This makes the graph much more compact, especially if many multiples of 360 degrees would be shown. It could have chosen to show the angles between 0 and 360 degrees instead, or any other pair of numbers 360 degrees apart.

You can manipulate the data to unwrap the phase and see a linear plot in your graph by tracking when it crosses the -180 to 180 threshold and adding or subtracting 360 degrees as necessary.

But be careful with unwrapping, since all phase measurements are relative to some reference point. As an example, your chart shows 1GHz at about 125 degrees, and at approximately 1.4GHz it's about 125 - 360 = -235 degrees. But that's just relative to the first measurement of your graph. In this case, since you know you measured a transmission line, you can use the slope of the phase change to extrapolate down to DC and unwrap from there to get absolute phase angles for your plot. But in general, you should avoid extrapolating unless you have a model/equation for how the phase will behave outside of the graph.

See this page for more information: https://en.wikipedia.org/wiki/Instantaneous_phase_and_frequency

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  • \$\begingroup\$ This answer is inadequate. But since you don't have graph data all the way to DC you can't really determine the absolute phase with this graph alone" **is false My 1st answer was a brainfaht \$\endgroup\$
    – Tony Stewart EE75
    Oct 31, 2019 at 23:10
  • \$\begingroup\$ I stand corrected on that point. In this case you can use the rate of phase change to extrapolate down to 0Hz and figure out how many times the phase must have cycled. But in general, you should be careful with unwrapping phase unless you have model for the system that lets you determine the absolute values. \$\endgroup\$
    – remcycles
    Oct 31, 2019 at 23:18
  • \$\begingroup\$ as in my answer t= df/dϕ which can be LP filter + delay line + fixed length \$\endgroup\$
    – Tony Stewart EE75
    Oct 31, 2019 at 23:20
  • \$\begingroup\$ As for the answer being inadequate, I don't agree. I think it's clear that the primary question of the OP is about the phase wrapping leading to an unexpected graph. There are no requests to calculate the velocity coefficient of the coax or any other parameter. \$\endgroup\$
    – remcycles
    Oct 31, 2019 at 23:20
  • \$\begingroup\$ Perhaps you can correct the Op's terminology "S21 is a periodic signal", no its linear swept signal with a recursive null phase and total delay= a fixed slope. Assuming no LPF and just the cable , the f1 can be computed \$\endgroup\$
    – Tony Stewart EE75
    Oct 31, 2019 at 23:22
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It is linear, but there are only 360 degrees in a circle. Negative 180 degrees is exactly the same as positive 180 degrees.

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Your measurement indicates that the slope or derivative is constant, which refers to a constant time delay of a fixed cable length.

\$\text{For 1 phase cycle of 360 deg., df/dϕ = 280 MHz/cyc. = 3.57 ns}\$ \$\text{This is the recursive time delay of 1 wavelength, for which you show well above the 1st cycle. }\$

Although your graph has poor resolution, it appears to line up well.

Based on the slope the 1st cycle of recursive phase shift can be determined.

f= [MHz] 280, 560, 840, 1120, 1400, 1680, 1960, 2240, 2520, 2800,

The frequency at 0 deg is highlighted in bold in your restricted plot, while the 1st delay of 360 deg is computed @ f1=280 MHz is the recursive slope of the s21 phase response.

From this f1 = 280 MHz, the length of the cable can be estimated est. from some coax at v=0.6c

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