- #1
diemilio
- 32
- 0
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 0º and 90º components are equally probable?
What am I missing here? Why is the SNR unchanged?
Regards,
diemilio
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 0º and 90º components are equally probable?
What am I missing here? Why is the SNR unchanged?
Regards,
diemilio