0
\$\begingroup\$

I'm trying to become reacquainted with microwave circuits and how to analyze them, especially with a VNA. Just to get some practice I'm using Keysite ADS simulator. To get the feel of it I used the circuit below as a test, and analyzed it. It's a simple band-pass, the resonates at 5GHz.
Band pass

My issue that I'm having is interpreting the Smith chart results. By my understanding at 5GHz, the input to the filer, S11 should be 0, with maybe a phase change, and also the reflection coefficient should be 0. In the Smith chart below, at resonance, 5GHz, it shows S11 as being ( 1 + i0 ). This doesn't seem to make sense to me.
Looking at the other graphs of S11 and S21, they seem to makes sense, where at 5GHz S21 goes to 0 dB and S11 goes to very low, meaning %100 of the signal from port 1 goes to port 2 as you would expect. The Smith chart doesn't seem to show the same thing. I would expect it to show 0 at resonance.
So what is the meaning of the S11 plot on the smith chart? Or, is the Smith chart not appropriate for plotting S parameters? I figured that with S parameters being complex they would normally be plotted on the smith chart, but what I'm seeing doesn't seem to make sense to me. Or is the issue just with the way that ADS plots S parameters? Would I expect the same plot on an actual VNA on the Smith chart display.

BTW, I'm also including magnitude and phase charts of S11 to show that those reading are behaving as expected.

enter image description here

S11 and S22 below, with marker at 5Ghz resonance.

S11 and S12

S11 Magitude.

enter image description here

S11 Phase/Magnitude

enter image description here

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

By my understanding at 5GHz, the input to the filer, S11 should be 0, with maybe a phase change,

You're right that the \$S_{11}\$ should be 0.

But a complex number with magnitude zero doesn't have a defined phase.

and also the reflection coefficient should be 0.

\$S_{11}\$ and the forward reflection coefficient are two names for the same thing, so this is correct (but redundant).

In the Smith chart below, at resonance, 5GHz, it shows S11 as being ( 1 + i0 ). This doesn't seem to make sense to me.

The smith chart is showing the input impedance (not the reflection coefficient) to be \$Z_0 (1 + 0 i)\$.

This should be what you expect. When the input is matched to the characteristic impedance \$Z_0\$ is when the reflection coefficient goes to 0.

Notice the marker is showing \$S_{11}\$ to have magnitude 0.004 and angle -89.762 degrees, which is very close to 0, as you expect.

So what is the meaning of the S11 plot on the smith chart?

The values along the curves on the Smith chart are the input impedance, not \$S_{11}\$.

To read \$S_{11}\$, you need to imagine an ordinary set of polar coordinates overlayed on top of the Smith chart, with the origin at the center of the chart and magnitude 1 corresponding to the outer edge of the chart.

If you want to read \$S_{11}\$ off the chart instead of \$Z_{in}\$ in ADS, you should use a polar plot instead of a Smith chart.

Or, is the Smith chart not appropriate for plotting S parameters? I figured that with S parameters being complex they would normally be plotted on the smith chart

The Smith chart is useful for plotting \$S_{11}\$ or \$S_{22}\$. It doesn't really tell you anything useful if you plot \$S_{21}\$ or \$S_{12}\$ on it.


The whole point of the Smith chart is to visualize the transformation between reflection coefficient and input impedance. You can use a ruler and protractor to plot measured reflection coefficient values on the chart, and then the curves on the chart will tell you the corresponding input impedance values.

But you need to be clear in your head what you are measuring. When you do the S-parameter simulation in ADS, it calculates the S-parameters at each frequency. The input impedance of the DUT can then be derived from the S-parameters, and that is what the Smith chart is helping you do. Assuming you don't want to just add an equation to calculate $$Z_{in} = \frac{1+S_{11}}{1-S_{11}}Z_0.$$

\$\endgroup\$

Your Answer

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

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