sampling theorem question

[alok888]

hi,
A signal is band-limited to 0 to 12 kHz. The signal spectrum is corrupted by additive
noise which is band-limited to 10 to 12 kHz. Theoretically, the minimum rate in kilohertz
at which the noisy signal must be sampled so that the UNCORRUPTED PART of the
signal spectrum can be recovered, is _________ kHz?

how to solve this?

 

[Ascozes]

Considering the nyquist theorem the sampling frequency should be bandwidth multiplied by two. In your case I think, bandwidth: 0-10kHz -> sampling frequency: 20 kHz

 

[alok888]

The correct answer is 22 KHz;but I don't know how.The noise is bandlimited to 10-12 kHz.If we apply bandpass sampling theorem the sampling frequency will be (2*F higher/N) where N is an integer.so sampling frequency is (2*12/12/2)=24/6=4 kHz

 

[KlausST]

Hi,

Tricky question... Or tricky answer.

My initial answer was 12kHz × 2 = 24kHz. So a digital filter is able to attenuate the noise signal from 10 kHz ...12kHz.

But.

If you sample with 22kHz, then noise frequency from 10 kHz ... 11 kHz is sampled correctely and can be filtered out.
Now the tricky thing:
Because of nyquist the noise frequency in the range of 11kHz ...12kHz is mirrored from 11kHz down to 10kHz.
So if a digital filter attenuates the noise at digital side from 10..11kHz, then it automatically attenuates "mirrored" noise of 10..12kHz on the analog side.

Klaus

 

[Andrew8611]

So the minimum frequency to avoid mirroring should be 26 KHz, because mirrored noise will appear on 13 Khz and they will be filtered by a digital filter.

 

[alok888]

I considered 4 sampling frequencies and the spectrums are as shown in the image.