If a signal contains frequencies from $500$ to $1000\textrm{ Hz}$, what is the Nyquist sampling frequency, $2$ times $500$ or $2$ times $1000$?

Twice the highest frequency of the signal, or twice the bandwidth of the signal?

  • $\begingroup$ Twice the highest frequency of the signal.. $\endgroup$ Nov 27, 2016 at 13:02
  • 2
    $\begingroup$ Hi Hana. Before posting questions here kindly search for answer.. $\endgroup$ Nov 27, 2016 at 13:04
  • $\begingroup$ In Discrete Wavelet Decomposition, if we start with a signal containing the frequencies from 0 to 1000, the output of the low pass will be a signal containing 0-500 Hz, and the output of the high pass will be 500-1000 Hz. After that, downsampling is performed to both the output of the low and high pass filters according to the nyquist rate. This means that the Nyquist is twice the signal bandwidth, right?? $\endgroup$
    – Noha
    Nov 27, 2016 at 13:13
  • 1
    $\begingroup$ Possibly very closely related to this question. I would suggest that we keep one of them and make the other a part of it (?). $\endgroup$
    – A_A
    Nov 27, 2016 at 15:17
  • $\begingroup$ SE.DSP wishes you a happy new year 2017, with a kind reminder that your question and its answers may require some action (update, votes, acceptance, etc.) $\endgroup$ Dec 31, 2016 at 18:20

2 Answers 2


Both. For baseband sampling (thus using only low-pass filtering), a sampling rate over over 2x the highest frequency is necessary. For band-pass under-sampling, sampling at a bit over twice the bandwidth will work, as long as you remember the original band and don't fold (alias) the given band onto itself. 1000 Hz sampling will fold and invert a (slightly narrower than) 500 to 1000 Hz signal down to 500 to 0 Hz, and thus the signal won't be aliased with itself, but with an empty frequency band (after band-pass filtering stuff out, if needed). Thus, no information will be lost (aliased), as long as you remember the original frequency band (e.g. it was not baseband). Reconstruction may require a band-pass anti-aliasing filter, rather than a low-pass anti-aliasing filter. Using knowledge of the original band, you can de-invert and upconvert the samples back to the original lower side-band of a 1000 Hz carrier, for example.


The "Nyquist rate" is often defined as:

the minimum rate at which a signal can be sampled without introducing errors

The most conservative option is twice the maximum frequency. A cleverer option is to take advantage of the actual bandwidth or the signal. This resorts to generalized sampling theorems, by Papoulis and Gerchberg, notably. And using the most recent findings, you can go below, using the fact that inside $[500,1000]$ there may be holes.

My take is to go for $1000$ $\text{Hz}$, for rhetorical reasons, as the question is about Nyquist. Would the question have been about "the minimum sampling frequency", I would have said "something" below $500$ $\text{Hz}$, since we do not have enough information.

In the special case of $M$-band multi-band wavelets, or multi-rate filter banks, the context is more specific. $M$ filters, quite often FIR, are designed to allow perfect reconstruction, even if they are not ideal. So each channel, that span $1/M$th of the frequency range, can be subsampled at an additional $1/M$ rate. In the $2$-band case, subsampling both the $[0,500]$ and the $[500,1000]$ bands at the same rate is perfectly valid, as long as the low- and high-pass filters are designed correctly. However, note that after filtering and downsampling, there still exists aliasing in the subbands (esp. due to FIR)), but it is cancelled at the synthesis.

  • $\begingroup$ In Discrete wavelet decomposition, we start with a signal containing the frequencies from 0 to 1000 HZ and sampled with 2000 samples, After first level decomposition, the output of low pass filter(0-500HZ) is downsampled to 1000 samples, and this is reasonable according to Nyquist. The output of the high pass filter (500-1000 HZ) is also downsampled to 1000 samples, although the highest frequency is still 1000, Why? $\endgroup$
    – Noha
    Nov 27, 2016 at 16:37
  • 1
    $\begingroup$ Why? Because, in addition to the samples, you know the band (500-1000) and the spectral bandwidth (500). If you don't know that, you will need a higher sample rate (above 2000). $\endgroup$
    – hotpaw2
    Nov 27, 2016 at 18:02
  • $\begingroup$ @hana I have added details regarding the discrete wavelet transform and subsampling. Here, the bandlimited sampling property is used $\endgroup$ Nov 27, 2016 at 20:00
  • $\begingroup$ @ hotpaw2 If you don't know that, you will need a higher sample rate (above 2000). Why knowing the bandwidth allows me to sample at twice the bandwidth not twice the highest frequency? $\endgroup$
    – Noha
    Nov 27, 2016 at 20:49
  • $\begingroup$ @hana Mostly because downsampling sends you to the baseband, whatever the initial freq range was, and knowing the original range allows you to recast the data to the proper frequency location $\endgroup$ Nov 27, 2016 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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