0
\$\begingroup\$

I'm having a hard time understanding what is actually achievable with an RF combiner.

Let's consider this scenario, here port 1 of the RF combiner has a sinusoid at f1=100Hz and port 2 has a sinusoid f2= 1000Hz, each with 0 dBm in a 50 ohm system (e.g. each having Vpk-pk=0.632V).

In matlab I took these two signals and added them and we get the following: enter image description here

The first question is, can a combiner actually produce a similar output (minus some small insertion losses)? I'm confused because per this application note from minicircuits, https://www.minicircuits.com/app/PWR2-4.pdf (see top of page 2) " If two signals at different RF frequencies are being added; then each signal will appear at the S port with a 3 dB loss. The internal resistor absorbs the 3 dB power loss for each signal."

Certainly it makes sense to me that if you added two in-phase, same frequency sinusoids in the combiner then your output wave would be 3 dB higher. But I don't understand whether adding non-coherent sinusoids MUST result in a 3dB reduction in each signal for ANY type of combiner or if the application note is referring only to a specific type of combiner? I've seen some notes about Wilkinson combiners having this 3dB loss here since the non-coherent waves won't cancel across the resistor.

If this has to do with the specific type of combiner, can anyone point me towards one that would allow adding of non-coherent sinusoids without this loss?

\$\endgroup\$
  • \$\begingroup\$ The importyant thing is you're working in a matched system, with isolation between the sources. This requires the internal resistor, which creates the loss for incoherent adding. Do away with this resistor, and analyse what the new coupler does. \$\endgroup\$ – Neil_UK Aug 9 '18 at 8:21
1
\$\begingroup\$

Since the power is split into 2 paths and it is bidirectional, you lose 3dB in both directions, combining and splitting + 0.5dB typ. loss.

The advantage of this hybrid transformer method is the > 30dB of isolation between twin ports.

schematic

simulate this circuit – Schematic created using CircuitLab

The same behaviour is true for a Wilkinson and waveguide combiner.

\$\endgroup\$
  • \$\begingroup\$ So would you say a passive, non-coherent combiner without the 3dB loss would violate the law of physics? In other words, realizing the idealized output from my matlab plot is impossible? \$\endgroup\$ – user67081 Aug 9 '18 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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