cavity stabilized oscilator functionality question

Thread Starter

yef smith

Joined Aug 2, 2020
756
Hello, I am trying to understand the logic of a schematics shown bellow taken from the attched article.
double-balanced mixer shown bellow is turned into a phase detector when one signal is shifter 90 degrees with respects to the other.
We have our noisy YIG source (f0+noise) it gets into two routes :
1. going into the cavity oscilator producing pure f0
2.shifted by 90 degrees, so (f0+noise)+90 degree shift

we supress the main tone returned pure signal froms oscilator

So as i see it we have inside our phase detector two signals:
1.returned cavity oscilator signal with supressed carried(so basicly noise)
2.original YIG source (f0+noise) with 90 degree shift
What result do we expect? we compare pure noise to f0+noise with a shift.
they are totally different .
I cant see the physical logic in this comparison.
Thanks.

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Tesla23

Joined May 10, 2009
542
From the paper:

1685657383271.png

so you can see that the reflected signal, extracted by the circulator, changes phase as the frequency moves from the resonant frequency of the cavity. Feeding this to the the DBM they make a frequency discriminator.
 

Thread Starter

yef smith

Joined Aug 2, 2020
756
Hello Tesla23, Could you please help me simulate this effect.
I have a double balance mixer.
What input should i give it and what output should i expect so i could see this effect?
Thanks.
 

Tesla23

Joined May 10, 2009
542
Given that the reflection coefficient phase varies rapidly with frequency offset, use the DBM as a phase detector to measure the phase difference between the incident wave and the reflected one. You have a block diagram that shows how to do it.
 

Thread Starter

yef smith

Joined Aug 2, 2020
756
Hello Tesla23,I have found this explanation taken from the manual shown in the link bellow.
It basicly says that if we put W_LO=W_rf then we get a pahse detecor as shown in the equation bellow.
When we supress carrier then W_LO=0(noise)
but if we look at the schematics shown bellow ,the inputs of the DBM having on one side a supressed carrier
While the other does not have a supressed carrier(by the manual they both has to be with carriers)

by the teory bellow we the have to be both with the same carriers to make it into a phase detector.
what is the intuition behind this ?
Thanks.

https://www.minicircuits.com/app/AN41-001.pdf
1685821969753.png1685821832913.png
 

Thread Starter

yef smith

Joined Aug 2, 2020
756
Hello Tesla23,A simulation was made that showdedhow a DBM mixer can be used as phase detector.
But as you can see both of my inputs have the same amplitude.
In our original curcuit, both inputs dont have the same tone amplitude.
What is the purpose of supressing tone amplitude on one input of the DBM phase detector?
The DBM will work perfectly fine when both tones have the same amplitude.
Thanks.
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