0
$\begingroup$

A basic theorem in communications is the matched filter maximizes the SNR at sampling. I'm a little confused on how this relates to discrete time systems and sampling rate.

Normally if you sample at Nyquist rate it's "good enough" but how does this interact with the discrete time correlator? Does sampling at a higher rate benefit the system? Does up-sampling the signal and passing it through the correlator yield the same benefit?

I understand the idea of Noise being White and the output of white noised passed through the correlator limits its PSD but lets say your analog signal is pre-filtered to remove bands outside of some region and your sample rate meets the Nyquist rate for this bandwidth. Is there any benefit to over-sampling and then discrete time matched filtering?

|improve this question
$\endgroup$
  • $\begingroup$ Just a snide remark: that theorem says that MF maximizes the SNR for a channel where only additive uncorrelated noise happens to the signal. It's not generally true that matched filtering maximizes SNR. $\endgroup$ – Marcus Müller Apr 22 '19 at 8:05
1
$\begingroup$

Normally if you sample at Nyquist rate it's "good enough" but how does this interact with the discrete time correlator?

It interacts very well! See the following:

Does sampling at a higher rate benefit the system?

You mean oversampling.

If you filter your signal to the original signals bandwidth, which is now smaller than the Nyquist rate. You thus get a component that is correlated between samples (which is your signal) and one that isn't (which is your noise). By averaging / downsampling your signal, you increase the variance of your signal component more than that of your noise component – you get an SNR gain.

Does up-sampling the signal and passing it through the correlator yield the same benefit?

No, because upsampling doesn't "uncorrelate" the noise that you already digitize.

Is there any benefit to over-sampling and then discrete time matched filtering?

Yes, as described above, the oversampling allows you to increase your SNR. Think of it this way:

Assume that if critically sampled, your pulse length is $N$; thus, your MF does a correlation over $N$ samples.

You oversample your analog signal by a factor of $f>1$. You interpolate your MF by the same factor, so that the pulse and the filter are now $Nf$ in length.

While the power of the samples stays the same, the amplitude of the signal part in the (digital) correlator output has been scaled by a factor of $f$, which means the power has been scaled by $f^2$.

For the noise component in the correlator output: White Noise samples are uncorrelated, and we know that thus the variance of their sum (which is what a correlator is, weighted sum) is the sum of their variances. Since variance is power for zero-mean noise, that means we get a noise power increase of $f$.

Thus, oversampling by $f$ gets us an SNR increase of $\frac{f^2}f$ :)

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Marcus, are you saying that if I oversample by a factor of one million, I'll get a one million increase in SNR? $\endgroup$ – MBaz Apr 22 '19 at 16:07
  • $\begingroup$ Thanks, as a couple questions. 1. If the AWGN has already passed through a bandpass filter, it's no longer white and the noise is then correlated? So it seems the argument doesn't work. 2. I still don't understand why over-sampling would increase SNR, if you sample at the Nyquist rate for your bandpass filter, shouldn't you be able to reconstruct the signal perfectly? It seems you could then apply the matched filter in the frequency domain with maybe a lower sample rate(but it would need to be longer impulse response approximation). $\endgroup$ – FourierFlux Apr 22 '19 at 18:37
  • $\begingroup$ Also I always wondered, do ADCs take the average value over the interval or the instantaneous value at sampling time? Seems the later would be more prone to noise but I guess that's an ASIC question. $\endgroup$ – FourierFlux Apr 22 '19 at 18:38
  • $\begingroup$ We typically idealize them to do the latter. You'll find that in reality, they inherently do some integration (typically by charging a sample and hold capacitor), or that to be honest things are quite a bit more complex; if you look at sigma-delta converters, you'll find that these, in the process of taking one measurement, have spectral shaping properties. That's a topic worth reading on, by the way – it's an interesting concept :) $\endgroup$ – Marcus Müller Apr 23 '19 at 5:51
  • $\begingroup$ Ok thanks, any additional input on the other two questions? $\endgroup$ – FourierFlux Apr 23 '19 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.