3 dB Power Divider Used as Power Combiner S Matrix Explanation

Question

I'm trying to understand the operation of an ideal RF power combiner.

Below is the scattering matrix of such device. (i.e. Wilkinson Power Divider)

\$[S] = {\frac {-j}{\sqrt {2}}}{\begin{bmatrix} 0 & 1 & 1 \\1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}}\$

I know how the division works but I'm confused as to how power combination works.

If we were to send a wave from port 2 with an amplitude of A, and in same phase a wave with amplitude of B from port 3 we should get A+B in port 1 since the waves can't go anywhere else in the device.

But i fail to explain this with scattering parameters. \$[b] = [S].[a]\$

In our case \$ [a] = {\begin{bmatrix} 0 \\A \\ B \end{bmatrix}}\$

If we do the matrix scalar multiplication we get:

\$ [b] = {\begin{bmatrix} {\frac {-j}{\sqrt {2}} (A+B)} \\0 \\ 0 \end{bmatrix}}\$

So we got \$ \frac {-j}{\sqrt {2}}(A+B) \$ at port 1 instead of \$ (A+B)\$.

Where did the rest go? Same thing goes for magic tee as well. Thank you for your time.

Answer 1

The sum port of a Wilkinson is a superposition of the S-Matrix. Your answer is correct.

\$ [S] = {\frac {-j}{\sqrt {2}}}\begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \$

The sum port is a super position of signals at port 2 (A) and 3 (B):

\$ s = {\frac {-j}{\sqrt {2}}}A + {\frac {-j}{\sqrt {2}}}B = {\frac {-j}{\sqrt {2}}}(A+B)\$

If A = B this would reduce to

\$ 2{\frac {-j}{\sqrt {2}}}A\$

Which is twice the power of A:

\$ |2{\frac {-j}{\sqrt {2}}}A|^2 = 2A^2\$

Therefore, no power is lost.

Please note: this only true for correlated signals. If random signals are used in A and B it becomes the sum not super-position.