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Noise in an ideal mixer

by diemilio
Tags: ideal, mixer, noise
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diemilio
#1
Feb15-14, 06:17 AM
P: 14
Hello everyone,

I have a question regarding the way white noise behaves when passed through an ideal mixer.

Picture the system shown in the attached image. Assume the mixer is ideal and that the high frequency contents of the multiplication are filtered out. Therefore the output of the system will be A*B/2. My question is, how does wide-band white noise behave when passed through the mixer?

In several textbooks, I've seen that in this system the signal-to-noise ratio remains unchanged from input to output. However, I am confused because I would think that in the case of white noise, it would be possible to express the signal as a composition of in-phase and quadrature components with equal power spectral densities, thus part of the noise should be, phase-filtered.

Is this an unreasonable assumption? Isn't the phase of white noise also white, thus 0 and 90 components are equally probable?

What am I missing here? Why is the SNR unchanged?

Regards,

diemilio
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meBigGuy
#2
Feb15-14, 05:07 PM
P: 1,074
I don't understand "in-phase and quadrature components with equal power spectral densities". Why do you think they would be equal? (asking, not testing you). I don't think the in-phase and quadrature components are in any way correlated. (I'm not an expert)

Are you thinking that when you mix to zero the resulting pos and neg frequencies add and might cancel? That would only happen to the extent they are correlated.
sophiecentaur
#3
Feb16-14, 05:55 AM
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Quote Quote by diemilio View Post

What am I missing here? Why is the SNR unchanged?

Regards,

diemilio
Why would you expect the SNR to be changed at all (apart from a finite amount of added noise from the mixer)? Every baseband component (signal and noise) will find itself mixed up to a corresponding 'product' component and they should all have the same relationship to each other. Are you looking for something more than that?


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