# 3 dB Power Divider Used as Power Combiner S Matrix Explanation

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.

• 1/sqrt(2)*(A + B) is correct. If A = B, you'll have 2/sqrt(2)*A which is twice the power of A. It didn't go anywhere. Nov 13, 2023 at 15:39
• @Jason If we apply A from port 2 and 3 at the same time the total input is 2A in your case. Then we got A*sqrt2 at the output. 2A to 1.414*A therefore something is missing. Nov 13, 2023 at 15:42
• I don't follow what you're saying. If you apply signal A to both port 2 and 3, port 1 will be 1/sqrt(2)*(A + B)=2A/sqrt(2) which is still twice the power of A. Nov 13, 2023 at 15:45
• @Jason If we apply waves in same phase both with amplitude of A volts. You are agreeing that the output will be 2A divided by sqrt of 2. Which equals to A times sqrt of 2. But our total applied input is 2*A which is not the same thing as A*sqrt2 Our input of 2*A does not equal to output of 1.414*A Nov 13, 2023 at 15:47
• Yes. Conservation of energy. |2*A/sqrt(2)|^2 = 2A^2 which is TWICE the power of A. If the answer were 2A then |2*A|^2 = 4A^2 which is 4x the power of A. Nov 13, 2023 at 15:50

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.