Why the reflection coefficient is 0 at resonant frequency?

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Kaku Yichikun

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Hi,
I cannot explain why the reflection coefficient becomes zero when frequency is the resonant frequency of a microwave cavity resonator. Can you give me an explanation from the view of physics? Thanks.
 

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I almost agree, however Γ and "R" in your calculation are complex quantities. It's right that the filter can be considered as series short in resonance, except for a possible phase shift. You can ignore phase shift if you look for magnitude only.
A general equivalent circuit for the filter could have a pi or T form.
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For clarification, you are asking about reflection coefficient, not S11[dB]?
Γ = 0 means perfect matching, no reflections.

What is your test setup, cavity resonator is one- or two-port device?
 

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FvM said:
For clarification, you are asking about reflection coefficient, not S11[dB]?
Γ = 0 means perfect matching, no reflections.

What is your test setup, cavity resonator is one- or two-port device?
Click to expand...
Thanks. I am asking about Γ. I can image Γ = 0 when the impendence is perfectly matched. But I was told Γ = 0 when resonance occurred from a lecture. I want to consider a two-port device like this (a stepped discontinuity in a circular waveguide):
 

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Thanks for clarification. To get Γ = 0 in resonance, the other port must be terminated with matched impedance too (Γ = 0). In resonance, the filter acts like a piece of waveguide, copying port 2 load impedance to port 1.
 

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Kaku Yichikun said:
Hi,
I cannot explain why the reflection coefficient becomes zero when frequency is the resonant frequency of a microwave cavity resonator. Can you give me an explanation from the view of physics? Thanks.
Click to expand...

It has a minimum, but is not always zero. For zero reflection all energy must be coupled to the resonator and dissipated in the resonator -- that can be achieved by good design of the coupling structure.

edit: This refers to 1-port resonators. I missed your second post and assumed you are talking about 1-port resonators.
 

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FvM said:
Thanks for clarification. To get Γ = 0 in resonance, the other port must be terminated with matched impedance too (Γ = 0). In resonance, the filter acts like a piece of waveguide, copying port 2 load impedance to port 1.
Click to expand...
Thanks for you explanation. I guess I misunderstood the concept. When I want to evaluate the reflection coefficient at interface between region I and II, it seems I need to consider the region II and III as a while but not only the region II.
The equivalent circuit might like this.

So if the resonance occurs, the impedance of the region II is R and that of the whole circled region is R+Z0. Γ is minimum and Γ = 0 when considering R=0. Is it right？
 

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Thanks for you explanation. I guess I misunderstood the concept. When I want to evaluate the reflection coefficient at interface between region I and II, it seems I need to consider the region II and III as a while but not only the region II. View attachment 177774
The equivalent circuit might like this. View attachment 177773

So if the resonance occurs, the impedance of the region II is R and that of the whole circled region is R+Z0. Γ is minimum and Γ = 0 when considering R=0. Is it right？
 

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I almost agree, however Γ and "R" in your calculation are complex quantities. It's right that the filter can be considered as series short in resonance, except for a possible phase shift. You can ignore phase shift if you look for magnitude only.
A general equivalent circuit for the filter could have a pi or T form.
 

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Solution
 

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Because at resonance, all of the energy circulates within
the tank and (at steady state) neither enters nor leaves
(component losses aside)?

Now, if you had a amplitude-modulated waveform you
might well see losses which you might interpret as
reflection / return loss (or gain, if the amplitude is
fading instead of ramping up) as source amplitude
moves relative to "tank amplitude" in real time?
 

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Thanks for your reply. I made the following model in COMSOL.

It shows at the resonance frequency, |S11| is minimum while |S21| is maximum. It seems the energy will transmit (leave) the cavity?

 

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