From a comment in another question I read a statement

"In addition, the initial rate of 48kHz is already performing 6 times oversampling and therefore the subsequent decimation by 6 will not cause any aliasing either."

Could you please explain it? If I would oversample by 2 and then downsample it by 2 without AA-filter I would have stuff folding back to by final Nyquist band. Is this not exactly what is called aliasing?

  • $\begingroup$ it seems @MarcusMüller is already explaining it... Please not that some ideal system may require strictly bandlimited signals and impossible to realise filters to implement. $\endgroup$
    – Fat32
    Jul 25, 2016 at 10:33
  • $\begingroup$ Have a look at this related question and its answers. $\endgroup$
    – Matt L.
    Jul 25, 2016 at 10:46
  • $\begingroup$ Johu may be you need a complete treatment on sampling theory from its beginning... and that would require more than an answer. I suggest you begin by reading sampling from a basic signals and system theory book, then move on to its further applications from any one of a DSP book. And whenever you face a barriering concept, come and ask. $\endgroup$
    – Fat32
    Jul 25, 2016 at 13:30
  • $\begingroup$ @MattL. I did. The quote just does not make any sense to me, as you would not oversample if there was nothing you would aa-filter. In every practical case you do. $\endgroup$
    – Johu
    Jul 25, 2016 at 20:03

1 Answer 1


So the point is, oversampling with a factor of $N$ requires your signal to be confined to at most $\frac1N$ of your Nyquist bandwidth. Otherwise, you're not oversampling.

If the signal is six time oversampled, then there's no loss of information/aliasing when reducing the sample rate by six.

If I would oversample by 2 and then downsample it by 2 without AA-filter I would have stuff folding back to by final Nyquist band.

No, you wouldn't; there's, by definition of oversampling, no energy in the "upper half" of your spectrum.

Now, through discussion in the comments it became clear that "oversampled signal" might be ambiguous:

If you had your signal alone, AA filtering would do nothing, because there's no spectral component that could be folded back (aliased) into the decimated Nyquist band. However, if we're observing noisy signals, you'd have noise energy (and be it just the quantization noise) that is relatively wideband. Thus, in that situation, although the signal of interest is oversampled, the noise is not, and to avoid getting noise aliased into our band of interest, proper AA filtering is necessary.

This is really the difference between mathematically perfect signals and noisy signals; for example, if you generated a digital sine with 100 Samples per period, you could, without incurring any problems, decimate strongly – because a digitally generated signal typically has zero noise.

  • $\begingroup$ My confusion come from "by definition of oversampling, no energy in the "upper half" of your spectrum. Why do I need AA-filter for, if there is no energy above Nyquist? I agree, that my signal has to be in Nyquist band, but it's tail or noise would be also above. $\endgroup$
    – Johu
    Jul 25, 2016 at 9:56
  • $\begingroup$ To make sure there's no energy above nyquist limit. However, if the situation already is that your signal is oversampled, then there's no energy above that limit, and you don't need to AA. $\endgroup$ Jul 25, 2016 at 9:59
  • $\begingroup$ Please don't be offended, I'll try to make a simple analogy: The comment said "if your box is already green, throwing away all but the green channel will not make your box look different", to which your question is "then if I can just throw away everything but the green channel, why do I need green paint", ignoring the fact that by what has been said about the box, it's already green. $\endgroup$ Jul 25, 2016 at 10:01
  • $\begingroup$ I am not offended and you are helpful. My confusion is in terminology only I think. I am still confused about the following: if I digitize with high sampling rate in order to apply digital AA-filter before down sampling to my final bandwidth, then recording the high sampling rate is not called oversampling as I do have energy there? E.g. the box is not yet green. $\endgroup$
    – Johu
    Jul 25, 2016 at 10:41
  • $\begingroup$ If the observed signal spans more than $\frac1N$ of your Nyquist bandwidth, you're not doing $N$-oversampling. But: we might actually have a misunderstanding of what signal means here. I was assuming that you're observing an analog signal that is already confined to $\frac1N$ of your sampling rate alone, but it's possible that you're considering a system where you have more than just that one signal superimposed, and the other signals need to be filtered out first. $\endgroup$ Jul 25, 2016 at 10:43

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