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I'm trying to become re-acquainted with S parameters, VNA, and transmission lines for some projects at work. I'm using Agilent, Advanced Design Simulator, to practice on the computer and to get the basic theory down and try to really get the feeling of it. I'm simulating some basic circuits and using the built in VNA to look at the results. Below are the results of a simple band stop filter, resonant at 5Ghz. There the Smtih chart represtation of S11 and S21 next to a magnitude and phase plot of each.
My question is, how do I interpret S21 on the Smith chart? S11 makes sense to me. At 5Ghz on the Smith chart, the impedance, Z is at about infinity, so the reflection coefficient, Γ = 1. So all energy from 1 goes back to 1, which the magnitude plot agrees with.
For S21, the magnitude plot makes sence, because it goes to 0, meaning no energy from port 1 is appearing at port 2. However, I don't understand why the Smith chart would show 1 + j0 at the 5 Ghz point.
It looks like for S11, the chart plots the impedance, from which the reflection coefficient is calculated, that represents S11. It seems to me that the impedance looking back from port 2 would be the same.
Unless it is meant to mean the impedance from port 2 to port 1, which in that case it seems to indicate that the ports are matched, in which case I would expect maximum port transfer, which of course is not the case.

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It looks like for S11, the chart plots the impedance, from which the reflection coefficient is calculated,

The way the Smith chart works, if you took the lines of the Smith chart away, and put ordinary polar coordinate lines on the chart instead, these lines would give you the reflection coefficient.

If you then go back to the Smith chart lines instead of the polar coordinate lines, you have the impedance.

In paper and pencil days, you'd use a compass and protractor (or angle markings printed around the perimeter of the chart) to plot the reflection coefficient onto a pre-printed Smith chart. Then the Smith chart lines would give you the impedance without having to do any tedious calculations.

So really what's going on is ADS is plotting the reflection coefficient against lines that aren't shown on the chart, and you're reading the impedance off the lines that are on the chart.

For S21, the magnitude plot makes sence, because it goes to 0, meaning no energy from port 1 is appearing at port 2. However, I don't understand why the Smith chart would show 1 + j0 at the 5 Ghz point.

The Smith chart isn't really meaningful for transmission paramters (\$S_{21}\$ or \$S_{12}\$). Just plot these parameters on an ordinary polar plot, not a Smith chart.

The reason the chart shows "1+j0" at the point where the thing you plotted on it goes to zero is because if you were plotting reflections, then the point where the reflection goes to zero is the point where the input impedance of the DUT is matched (\$Z = (1+j0)\times Z_0\$).

But if you're plotting a transmission parameter, the only thing that's special about this point is that transmission is 0, which you would see more clearly on a polar chart than on a Smith chart.

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  • \$\begingroup\$ Thanks. That all makes sense. \$\endgroup\$ – Frank Aug 13 at 15:12
  • \$\begingroup\$ What is the meaning though, of Z = 1 + J0 in the S21 chart though? It makes sense that transmission coefficient is zero as that's where the band stop is resonant. But impedance wise, "Z = 1 + j0", it doesn't seem to make sense to me because from port 1 the impedance goes toward infinity at that frequency, and the circuit is the same looking from port 1 or port 2. \$\endgroup\$ – Frank Aug 13 at 15:18
  • \$\begingroup\$ @Frank, the chart is scaled for a \$Z_0\$ of 1. If you are actually working in a 50-ohm system, then "1+j0" on the Smith chart means an impedance of 50 ohms. \$\endgroup\$ – The Photon Aug 13 at 15:41
  • \$\begingroup\$ But if you're talking about \$S_{21}\$, it doesn't mean anything. The Smith chart is not meaningful for transmission parameters. \$\endgroup\$ – The Photon Aug 13 at 15:42
  • \$\begingroup\$ Thanks. Just trying to clear that up! \$\endgroup\$ – Frank Aug 13 at 15:51

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