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I have the following question regarding directional couplers. Below is the question. enter image description here

My solution was to multiply the scattering matrices of both couplers as in the following

\begin{pmatrix}0&-\frac{i}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\ \:\:-\frac{i}{\sqrt{2}}&0&0&-\frac{1}{\sqrt{2}}\\ \:\:-\frac{1}{\sqrt{2}}&0&0&-\frac{i}{\sqrt{2}}\\ \:\:\:\:0&-\frac{1}{\sqrt{2}}&-\frac{i}{\sqrt{2}}&0\end{pmatrix}\begin{pmatrix}0&-\frac{i}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\ \:\:\:-\frac{i}{\sqrt{2}}&0&0&-\frac{1}{\sqrt{2}}\\ \:\:\:-\frac{1}{\sqrt{2}}&0&0&-\frac{i}{\sqrt{2}}\\ \:\:\:\:\:0&-\frac{1}{\sqrt{2}}&-\frac{i}{\sqrt{2}}&0\end{pmatrix}

Then I get the following matrix

\begin{pmatrix}0&0&0&i\\ \:0&0&i&0\\ \:0&i&0&0\\ \:i&0&0&0\end{pmatrix}

So does that mean that the resulting phase and amplitude for port 2' and 3' relative to port 1 are zero! Please help me with this as am really doubting that this is correct.

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    \$\begingroup\$ You can't just multiply the S parameters. If you want to multiply transform the matrices to ABCD or T parameters. \$\endgroup\$
    – Mike
    Nov 24, 2017 at 20:16
  • \$\begingroup\$ @Mike, can you elaborate more that please. Do you mean that I have to transform to ABCD, get the total ABCD matrix using multiplication, and then transfer back to S parameters? \$\endgroup\$
    – JordenSH
    Nov 24, 2017 at 22:03
  • \$\begingroup\$ Exactly. But this will only work if port 1 is connected to port 1,which is not your case. I think you'd have to rearrange the ports of one of the matrices. \$\endgroup\$
    – Mike
    Nov 25, 2017 at 9:05

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