How to compute S21 when port 1 and port 2 reference impedance (Zo) are different

Question

The attenuator given in the figure is a 6 dB attenuator when both port 1 and 2 have a reference impedance of 50 Ω.

Then, their S-parameter matrix (when port 1 and 2 have 50 Ω reference impedance) would look like $$S=\begin{bmatrix}S_{11} && S_{12} \\ S_{21} && S_{22}\end{bmatrix}=\begin{bmatrix}0 && 0.5 \\ 0.5 && 0\end{bmatrix}$$

Now, I changed the reference impedance of port 2 to 100 Ω as shown in the figure. Based on some of readings from Pozar's microwave textbook, the following is what I thought it should be:

Let $$Z_{o1}=50 \Omega $$ and $$Z_{o2}=100 \Omega$$, respectively. Then $$Z_{in,1} = R_1 + R_2 \parallel (R_1 + Z_{o2})$$ $$Z_{in,2} = R_1 + R_2 \parallel (R_1 + Z_{o1})$$ $$\Gamma_S = \frac{Z_{o1} - Z_{in,1}}{Z_{o1} + Z_{in,1}} = \frac{V_1^+}{V_1^-}$$ $$\Gamma_L = \frac{Z_{o2} - Z_{in,2}}{Z_{o2} + Z_{in,2}} = \frac{V_2^+}{V_2^-}$$

Then we have: $$V_1^- = S_{11}V_1^+ + S_{12}V_2^+$$ $$V_2^- = S_{21}V_1^+ + S_{22}V_2^+$$ Since $$\Gamma_L = \frac{V_2^+}{V_2^-} $$, we have: $$V_2^- = S_{21}V_1^+ + S_{22}\Gamma_LV_2^-$$ $$S_{21}'=\frac{V_2^-}{V_1^+} = \frac{S_{21}}{1-S_{22}\Gamma_L}$$ and $$\Gamma_S = \frac{V_1^+}{V_1^-}$$ $$V_1^- = S_{11}V_1^+ + S_{12}V_2^+ = S_{11}\Gamma_SV_1^- + S_{12}V_2^+ $$ $$V_1^- = \frac{S_{12}}{1-S_{11}\Gamma_S}\cdot V_2^+ $$ $$V_2^- = S_{21}\Gamma_SV_1^- + S_{22}V_2^+ = \left(\frac{S_{21}\Gamma_SS_{12}}{1-S_{11}\Gamma_S} + S_{22}\right)V_2^+$$ $$S_{22}'=\frac{V_2^-}{V_2^+}=\frac{S_{21}\Gamma_SS_{12}}{1-S_{11}\Gamma_S} + S_{22}$$

Based on the above computation, I was getting $$S_{11}'= -21.54\text{ dB,} \quad \quad S_{21}'= -6\text{ dB} \quad \text{, and}\quad S_{22}'=-33.543\text{ dB}$$

Compared to the ADS simulation results, S_{11}' is correct but S_{22}' and S_{21}' are wildly wrong. The simulation results say: $$S_{11}'= -21.54\text{ dB,} \quad \quad S_{21}'= -6.512\text{ dB} \quad \text{, and}\quad S_{22}'=-9.542\text{ dB}$$

In addition, intuitively speaking, I thought S_{22}' should be $$S_{22}' = \frac{Z_{in,2} - Z_{o2}}{Z_{in,2} + Z_{o2}} \neq \frac{S_{21}\Gamma_SS_{12}}{1-S_{11}\Gamma_S} + S_{22}$$ For this problem, when I use the middle equation I get the correct S_{22}.

In summary, I would like to know

Q1. How to correctly compute \$S_{21}\$ in this scenario where the reference impedances of port 1 and 2 are different

Q2. If the reference impedances of port 1 and 2 are different, which equation is correct for \$S_{22}'\$?

I spent more than 6 hours today but couldn't come up with the right answer. I would sincerely appreciate it if anyone can answer my question above.

Thanks!

Answer 1

In general, assuming that the reference impedances in consideration are real, S-Parameters have the following definition: $$\frac{V_{r1}}{\sqrt{Z_1}} = S_{11}\frac{V_{i1}}{\sqrt{Z_1}}+S_{12}\frac{V_{i2}}{\sqrt{Z_2}}$$ $$\frac{V_{r2}}{\sqrt{Z_2}} = S_{21}\frac{V_{i1}}{\sqrt{Z_1}}+S_{22}\frac{V_{i2}}{\sqrt{Z_2}}$$ Here, \$Z_1, Z_2\$ is reference impedance at port 1 and 2 respectively.
If reference impedance is real, then the power waves can be viewed as travelling voltage waves scaled by the appropriate reference impedance. Thus, $$S_{11} = \frac{V_{r1}}{V_{i1}}\vert_{V_{i2}=0}$$ $$S_{21} = \sqrt{\frac{Z_1}{Z_2}}\frac{V_{r2}}{V_{i1}}\vert_{V_{i2}=0}$$

So, for your example, $$ Z_{in1} = 16.641+ \frac{116.641 \cdot 66.931}{116.641 + 66.931} = 59.138$$ $$S_{11} = \frac{V_{r1}}{V_{i1}} = \Gamma_1 = 20*log10(\frac{59.138-50}{59.138-50}) = -21.54dB$$ Assuming, voltage at input node is \$V_x\$ and output node is \$V_y\$: $$V_{x} = V_{i1}+V_{r1} = (1+\Gamma_1)V_{i1} = 1.083V_{i1}$$ $$V_{y} = V_{i2}+V_{r2} = V_{r2}$$ Because of matched load, there is no incident wave at node 2. $$V_{y} = V_{r2} = (V_x - \frac{V_x}{59.138}.16.614)\frac{100}{100+16.614} = 0.616V_x$$ $$\frac{V_{r2}}{V_{i1}} = 0.616\cdot1.083 = 0.667$$ $$S_{21} = \sqrt{\frac{50}{100}}\cdot 0.667 = -6.52dB$$ $$S_{22} = \Gamma_2 = \lvert{\frac{50-100}{50+100}}\rvert = -9.54dB$$ This is same as what the simulator shows you.

Answer 2

You show this in your question and, it cannot be correct: -

So, moving on from this despite your downvote...

How to correctly compute S_{21} in this scenario where reference impedance of port 1 and 2 are different

Why don't you use the standard formulas for a T attenuator: -

Where: K is the impedance factor from below, and Z1 is the larger of the source/load impedances and Z2 is the smaller of the source/load impedances.

Unfortunately, the K formula is incorrect but, I've amended that in my addendum at the bottom of this answer.

Images stolen from this electronics-tutorials.

I might add that the derivations are a little bit tricky if you are thinking of doing them yourself.

If the reference impedance of port 1 and 2 are different, which equation is correct for S22'?

If you use the formulas above correctly (noting the error I found) then S22 will be the required output impedance for the load.

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Addendum that corrects the K formula

Micro-cap set-up for a 4:1 attenuation (12.0412 dB) with RIN = 50 Ω and RL = 100 Ω: -

So, with a 1 volt source having a 50 Ω source resistance, the voltage appearing at the left of R1 is, as expected, 0.5 volts. And, with an attenuation of 4:1 (12.0412 dB), as expected, the output voltage is 125 mV. The website having the original pictures neglected the \$10\cdot LOG(RIN/RL)\$ term when calculating K. For equal input and output resistors and 4:1 dB attenuation you get this: -

As you can probably see R1 = R2 = 30 Ω and, R3 = 26.6667 Ω. If you crunched the numbers (do you really need to given that micro-cap is boss), you'll get exactly what is written on the tin.

For 2:1 attenuation: -

Hey, it's given me something to do on a bleak Sunday afternoon!