I have the following question regarding directional couplers. Below is the question. 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.

• You can't just multiply the S parameters. If you want to multiply transform the matrices to ABCD or T parameters.
– Mike
Nov 24, 2017 at 20:16
• @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? Nov 24, 2017 at 22:03
• 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.
– Mike
Nov 25, 2017 at 9:05