I recently built a simulation to study sampling, the effects of aliasing and the effects of anti-aliasing filters on the sampled signal.

For fundamental frequencies above the sample band it's obvious one sees 'imposters' in the sampled signal. Using an antialiasing filter I can eliminate imposters.

But if I rather impose a broadband noise (actually white noise) signal into the sampler then it doesn't make much difference whether the anti-aliasing filter is present or not. The peak to peak noise is the same in either case. Of course the bandwidth of the noise has changed.

But furthermore I would expect the (imposter) aliased broadband noise outside the sample band to be superimposed on the broadband noise that is genuinely passed in the sample band thus 'piling up' with a larger peak to peak level.

Why doesn't this happen?

I should mention that my simulation time step is in the MHz and my system under study in the 1 kHz range. So the system is virtually in a continuous world.

  • \$\begingroup\$ This is a fantastic question that I have always wondered about myself... \$\endgroup\$ – Matt Young Feb 19 '15 at 21:39
  • \$\begingroup\$ If you measure the noise amplitude on a scope, what amplitude do you see (a) before and (b) after the AA filter? \$\endgroup\$ – Brian Drummond Feb 19 '15 at 21:39
  • \$\begingroup\$ @BrianDrummond That experiment doesn't necessarily address the point of my question. Even a digital scope greatly over-samples and has its own anti-aliasing filters built in. So virtually the scope is 'continuous' and the effects of sampling aren't addressed. \$\endgroup\$ – docscience Feb 19 '15 at 21:52
  • \$\begingroup\$ Why do you say the AA filter does not make a difference? I find it easiest to think of the peak to peak output of the sampler but it also works for RMS. If you input broadband noise of 1MHz BW and 1V pk-pk directly into your 2KHz sampler the output of the sampler will be 1v pk-pk. If you now add the AA filter (brick wall 1KHz BW) and feed that into the sampler the input voltage will be ~30mV pk-pk (30dB att) and the sampler output will now be 30mv p-p still with 500Hz BW. The noise above Nyquist has been aliased into the output band. Kevin \$\endgroup\$ – Kevin White Feb 20 '15 at 2:19

You are correct: after sampling, the aliased noise components do pile up in the frequency band below the Nyquist frequency. The question is just what exactly it is that piles up, and what is its consequence.

In the following I assume that we deal with random noise modeled as a wide-sense stationary (WSS) random process, i.e. a random process for which we can define a power spectrum. If \$N(t)\$ is the noise process and \$R_k= N(kT)\$ is the sampled noise process (with sample period \$T\$), then the power spectrum of \$R_k\$ is an aliased version of the power spectrum of \$N(t)\$:


where \$f_s=1/T\$ is the sampling frequency. Of course, if \$N(t)\$ is band-limited (which is always the case) then only a finite number of shifted power spectra of \$N(t)\$ add up in the band of interest \$[0,f_s/2]\$.

The noise power is given by the integral of the respective power spectrum. In the case of \$N(t)\$ we have to integrate over the whole bandwidth of \$N(t)\$, whereas in the case of the sampled noise \$R_k\$ we have to integrate in the band \$[0,f_s/2]\$. From (1) it becomes clear that in both cases we obtain the same power because either we integrate the original power spectrum \$S_N(f)\$, or we integrate an aliased (i.e., piled up) version in the band \$[0,f_s/2]\$.

Consequently, the noise power does not change after sampling, regardless of the sampling frequency. The sampled noise has the same power as the original continuous-time noise.

So the power of the sampled noise only changes if you change the power of the continuous-time noise, and this can be done by the anti-aliasing filter, because the filter reduces noise band-width and, consequently, noise power. Note that only looking at the peak-to-peak value doesn't say much, because you need to consider the power.


E.A. Lee, D.G. Messerschmitt: Digital Communication, 2nd ed., section 3.2.5 (pp. 64)

| improve this answer | |

The energy represented by the sampled signal is related only to the PDF (probability density function) of the input signal and the sample frequency. The actual bandwidth of the input signal does not affect this.

In other words, when you undersample a wide-bandwidth signal, you get a set of samples that have the same PDF as the original wideband signal, but those samples only have an effective bandwidth of Fs/2. The "excess" energy outside that bandwidth was simply never captured by the sampling process.

If you double the sample rate, you will "capture" twice as much energy.

| improve this answer | |
  • \$\begingroup\$ Are you saying that for a given input noise power, increasing the sampling rate increases the noise power of the sampled noise? \$\endgroup\$ – Matt L. Feb 20 '15 at 9:04
  • \$\begingroup\$ Yes, as long as the noise bandwidth is still greater than or equal to the new sampling bandwidth. \$\endgroup\$ – Dave Tweed Feb 20 '15 at 12:41
  • 1
    \$\begingroup\$ That's not the case. If you model the noise as a (wide-sense) stationary random process, then the sampled noise has the same power as the original continuous-time noise process, regardless of the sampling rate. \$\endgroup\$ – Matt L. Feb 20 '15 at 12:58
  • \$\begingroup\$ @MattL.: On what do you base that assertion? Perhaps you should explain in greater detail in a separate answer. \$\endgroup\$ – Dave Tweed Feb 20 '15 at 13:20
  • \$\begingroup\$ OK, I'll write up an answer as soon as I have more time; might take till tomorrow though. \$\endgroup\$ – Matt L. Feb 20 '15 at 13:57

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.