1
$\begingroup$

I am learning some basic DSP and I have a pretty good intuition as to why sampling creates spectral images of the frequency response at intervals of the sampling frequency (convolution with pulse train).

I am reading about sample rate conversions (downsampling/decimation and upsampling/interpolation) and I can't seem to get intuition as to why downsampling/decimation stretches the frequency response by M (the downsample factor) or why upsampling/interpolation compresses it by I (the upsample factor) and creates more images of the originally sampled datas frequency response

|improve this question
$\endgroup$
  • $\begingroup$ ain't it the other way around? $\endgroup$ – robert bristow-johnson Jul 2 at 2:34
  • $\begingroup$ i dont think so? These are images from the book i'm reading: imgur.com/a/7YiFP63 $\endgroup$ – Taako Jul 2 at 2:36
  • 1
    $\begingroup$ i always think of "downsampling" as meaning reducing the sample rate (meaning that fewer samples represent the same stretch of signal) and "upsampling" as increasing the sample rate (meaning more samples reprenting the same stretch of signal). aliasing can happen when information is lost or destroyed. redundancy is the opposite of that. $\endgroup$ – robert bristow-johnson Jul 2 at 2:39
  • $\begingroup$ your definitions seem to align with what i'm reading. But the book i'm reading clearly talks about down sampling as stretching the frequency response and upsampling shrinking and creating images. If you're interested i'm referencing section 2.2.5 of this free text analog.com/media/en/training-seminars/design-handbooks/… $\endgroup$ – Taako Jul 2 at 2:41
  • $\begingroup$ do you mean fig. 2-13 (b)? that spectrum of the intermediate signal is what happens when there is zero-stuffing done in the up-sampling. low-pass filtering is needed to remove the repeated images. $\endgroup$ – robert bristow-johnson Jul 2 at 2:43
1
$\begingroup$

I try to think of it this way: In the time domain, when downsampling occurs, the signal gets compressed; while on upsampling, the signal gets stretched.

Then, from the Fourier transform we know that time stretching means frequency compression, and vice-versa.

It may not be a rigorous answer, but hope this helps!

|improve this answer
$\endgroup$

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.