Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Why the reflection coefficient is 0 at resonant frequency?

Status
Not open for further replies.

Kaku Yichikun

Newbie level 5
Joined
May 10, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
98
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.
 

Solution
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.
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?
 

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?
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):
1659269190830.png
 

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.
 

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.

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.
 

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.
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.
1659352429158.png

The equivalent circuit might like this.
1659352078627.png


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?
 

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.
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?
 

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.
 

Solution
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
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.
Thanks a lot for your kind reply. I am a beginner. It helps me a lot. I would like to only consider the magnitude at this stage.
 

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?
 

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?
Thanks for your reply. I made the following model in COMSOL.
1659421794700.png


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

1659421667655.png
1659421737490.png
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top