1
\$\begingroup\$

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.

\$\endgroup\$
8
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jason
    Nov 13, 2023 at 15:39
  • \$\begingroup\$ @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. \$\endgroup\$
    – Exclose
    Nov 13, 2023 at 15:42
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jason
    Nov 13, 2023 at 15:45
  • \$\begingroup\$ @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 \$\endgroup\$
    – Exclose
    Nov 13, 2023 at 15:47
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Jason
    Nov 13, 2023 at 15:50

1 Answer 1

1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.