Continue to Site

# CAVITY REAONATOR........reflected signal

Status
Not open for further replies.

#### rafalizou

##### Junior Member level 1
There is a cavity resonator coupled to a hybrid. If we send a signal(f+f0) just above its resonant frequency(f) to the cavity resonator what will be the output (or reflected signal) back to the hybrid.

Will it be f0....?

A resonator is a linear component, it can't return other signal frequencies than you send to it.

rafalizou

### rafalizou

Points: 2
Dear Sir, I am working on a accelerometer, and there we are using a cavity resonator.
Any minute change in the dimension of the cavity (due to acceleration) will change its resonant frequency(say f0) by small amount say (f1).
So now the original signal coupled to the resonator(by a hybrid) at (f0) will not be able to cause formation of a standing wave.
Hence there will be some signal coupled back to the hybrid.

A small change in resonance frequency surely won't supress a standing wave. I don't know, how you couple the resonator. Generally, the magnitude and phase of the reflected wave (s11) will change.

I guess, a straightforward way to measure the resonance frequency variation would be an oscillator tuned by the cavity resonator.

rafalizou

### rafalizou

Points: 2
A small change in resonance frequency surely won't supress a standing wave. I don't know, how you couple the resonator. Generally, the magnitude and phase of the reflected wave (s11) will change.

I guess, a straightforward way to measure the resonance frequency variation would be an oscillator tuned by the cavity resonator.

Dear Sir,
That would be correct (oscillator tuned by cavity resonator)
But I want to theortically determine that if, a resonator cavity excited by a signal of frequency greater than the natural resonant frequency by amount f1.... what will be the signal (in terms of frequency) (if any) coupled back or reflected back..?

and now if the input signal (centered at the natural resonant frequency) of the cavity resonator is frequency modulated by a low frequency signal say f1 .. what will be the output signal.???

I think I understand what you are after, - some measurable effect from the deformation of a cavity due to acceleration. Because the cavity will be loaded by your various circuits I am not sure that it will be very sensitive. two things come to mind, the first is to use the "flank" of the resonant circuit, which may fall at 20db/octave, so a frequency change of 1% would lead to a level change of 1%. The other is to have a second cavity and you cancel the two outputs of the cavities at the resting frequency hopefully as one cavity deforms, the balance will go and a large output will result.
Frank

rafalizou

### rafalizou

Points: 2
If you have an antenna that can intercept the RF signal and couple it into your
1-port resonator, then you will have a phase shift in the reflected signal. The phase shift will be in one direction if your interrogator frequency is above the resonant frequency, and the phase shift will be in the other direction if your interrogator frequency is below the resonant frequency.

If there are no other strong reflections from the environment, you should be able to measure that (differential) phase difference.

In order to excite a frequency change, you would need some nonlinear element in addition to the resonator (a diode or a FET transistor) and a strong enough received signal to exercise that non-linearity.

What you might want to do instead is to sweep the interrogating frequency, and look for the resonance frequency in the returned energy (either phase or amplitude). You might want a "retro-reflective" array to be able to send a strong signal back to the interrogator passively.

Rich

Last edited by a moderator:

I think I understand what you are after, - some measurable effect from the deformation of a cavity due to acceleration. Because the cavity will be loaded by your various circuits I am not sure that it will be very sensitive. two things come to mind, the first is to use the "flank" of the resonant circuit, which may fall at 20db/octave, so a frequency change of 1% would lead to a level change of 1%. The other is to have a second cavity and you cancel the two outputs of the cavities at the resting frequency hopefully as one cavity deforms, the balance will go and a large output will result.
Frank

Dear Sir, we want to detect the acceleration as u told.. using it as a accelerometer.
and also we are using double cavities....
But I could not get what will be the output of a single cavity ?
I mean what is the output of a normal cavity resonator, when standing waves are formed in it ?