0
\$\begingroup\$

When measuring a 4-way reactive power splitter the S-parameters at the frequency of interest(400 MHz) looks like this:

enter image description here

S11: -34 dB
S22: -3dB
S21 = S12 = -6dB

With IN being port 1 and OUT 1 being port 2. The other outputs were terminated with 50 ohm as shown for simplicity. Return loss and gain were the same.

But why is S21 and S12 equal? Shouldn't there be a large difference because of the terrible return loss on the output port?

I also took the splitter and fed it with a signal generator and power meters to measure power going in, reflected power and transferred power, like this:

enter image description here

In this setup i measured this:

Forward power: 36 dBm
Reflected power: 2 dBm
Output power: 30 dBm

Then i reversed the DUT, so RF input at the OUT 1, and measuring output power at the IN port:

Forward power: 36 dBm
Reflected power: 33 dBm
Output power: 29.7 dBm

Obviously comparing these measurement results to the S21 and S12 gain values, it make sense, but i do not understand why the output power is equal when the return loss is so much worse when using OUT 1 as input.
What am i missing? Or did i make a mistake in my measurement setup?

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

But why is S21 and S12 equal? Shouldn't there be a large difference because of the terrible return loss on the output port?

This property is called reciprocity. Any passive network that doesn't contain certain non-reciprocal materials will have this property.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks. It makes sense to me, but what doesnt make sense is why the output power is the same whether how you connect it, when the RL is so terrible at the output. Should i have split this question in 2? \$\endgroup\$
    – Linkyyy
    Jan 3 at 18:21
  • \$\begingroup\$ I didn't say anything in my post because I'm not sure what would be a convinincing explanation. But consider that whatever causes poor RL for the reverse connection will also cause reflections when a signal is trying to exit the device in the forward connection...possibly causing more absorption internal to the device instead of reflections back to the generator that contribute to S11...but without knowing the actual internal design of the splitter I don't know how to show this in general, aside from the general rule that passive networks are reciprocal. \$\endgroup\$
    – The Photon
    Jan 3 at 20:55

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.