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I am trying to solve for the scattering parameters of a quarter wave transformer. The example that I am given is a transformer with characteristic impedance 100 Ohms, and the reference impedance (which I understand is the characteristic impedance of the lines conected to both ends of the quarter-wave transformer) is 50 Ohms.

I have been able to arrive to S11 = 0.6, and this result is correct according to the solutions. But my result for S21 (and therefore S12) is incorrect. Can someone explain to me how to calculate the parameters for this network, and why my solution is incorrect?enter image description here

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  • \$\begingroup\$ Please show your calculation that leads to 0.8; otherwise we have no idea what you'd need help with! \$\endgroup\$ Commented Sep 12, 2023 at 7:14
  • \$\begingroup\$ @MarcusMüller my calculations are wrong, I have already seen the mistake I did. But I am still unable to solve for the S12 (and S21) parameters. Unfortunately, I only have the final result, but not the full solution. What I am asking is how to solve this problem. \$\endgroup\$
    – ff18192021
    Commented Sep 12, 2023 at 13:32
  • \$\begingroup\$ again, not able to help you without knowing what your misconceptions are – voting to close this question as "not focused enough", as with any other homework question without own calculations shown. \$\endgroup\$ Commented Sep 12, 2023 at 13:45
  • \$\begingroup\$ @MarcusMüller I have uploaded an image of the problem and my incorrect solution. Do you need anything else? Kindly remove that vote as well. \$\endgroup\$
    – ff18192021
    Commented Sep 12, 2023 at 16:11
  • \$\begingroup\$ Thank you very much! I removed the vote, as it is no longer appropriate \$\endgroup\$ Commented Sep 12, 2023 at 22:31

3 Answers 3

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This is my solution. I think the mistake you made is that you wrote the voltage at the left line with respect to the reference system. Remember that V1+ and V1- are the complex amplitudes at the left-end of the quarter-wave transformer, not at z=0.

enter image description here

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I find your working hard to follow, but just a few hints:

If you are going to derive this from basic wave amplitudes, you need to work in the impedance of the 100 ohm line (as this is where the standing wave is). So in the 100 ohm line at the RH end:

\$\Gamma = \frac{50 - 100}{50+100} = -1/3\$

you can then propagate this to the LH end using the fact that the wave amplitudes translate as \$e^{-j\beta l}\$

Using \$V = V_+(x) + V_-(x)\$ and \$I = I_+(x) + I_-(x)\$, not forgetting that \$I_+(x) = \frac{V_+(x)}{100}\$ and \$I_-(x) = -\frac{V_-(x)}{100}\$ on the 100 ohm line, you can compute the input voltage and current into the line at the LH end which you can relate to the incident wave.

Using this you should get \$S_{21} = -0.8j\$ if I have done it correctly.

Alternatively, using the unitary property of the S matrix,

enter image description here

and using the fact that \$S_{11}=0.6\$ is real and equal to \$S_{22}\$, you can deduce that \$S_{21} = \pm j0.8\$, I'm not sure there is an easy way to deduce the sign.

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OK, the first thing is that this is a passive device, so reflecting 0.6 of the power and letting through 0.8 of the power would mean you get 1.4 of the power you put in. That should have definitely raised some concerns!

A shortcut here is to see that a transmission line is lossless; so, the total power in needs to be the total power out, i.e., \$|S_{11}|+|S_{21}|=1\$.

You can of course also just fully play out the impedance transformation; the 50 Ω of your \$Z_I (=Z_0/2)\$ get transformed to \$Z_0^2/Z_I=(2Z_I)^2/Z_I = 4 Z_I = 2 Z_0 = 200\,Ω =: Z_{\text{transf}}\$. You then arrive at the same value for the transmission coefficient, \$T=1-\frac{Z_{\text{transf}}-Z_I}{Z_{\text{transf}}+Z_I} = 0.4\$.

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    \$\begingroup\$ Don't you mean \$|S_{11}|^2+|S_{21}|^2=1\$ \$\endgroup\$
    – Tesla23
    Commented Sep 12, 2023 at 23:43
  • \$\begingroup\$ @Tesla23 don't think so, S parameters are powers already \$\endgroup\$ Commented Sep 12, 2023 at 23:52
  • \$\begingroup\$ @MarcusMüller I think the other user is right. According to multiple reference books, for example Pozar's, the expression he mentioned is the correct one for lossless networks. This makes sense since S parameters are a relation between voltage waves. Also, the correct answer is 0.8, which satisfies this condition. \$\endgroup\$
    – ff18192021
    Commented Sep 12, 2023 at 23:58
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    \$\begingroup\$ @MarcusMüller Of course that one could just calculate S11 and deduce S21 or S12 by a power conservation argument, given that the network is lossless. However, this strategy fails when the number of ports is bigger than 2 or when the network is lossless. Finally, I do not see where is my mistake. \$\endgroup\$
    – ff18192021
    Commented Sep 12, 2023 at 23:59
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    \$\begingroup\$ S parameters are not powers, they are ratios of wave amplitudes. \$|S_{11}|^2+|S_{21}|^2 = 1\$ is correct. See any textbook. e.g. link p71 \$\endgroup\$
    – Tesla23
    Commented Sep 13, 2023 at 0:33

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