We have a signal sampled att 22 kHz that we want to interpolate to 40 kHz.

So we can do this by upsampling by a factor 20, then downsampling by a factor 11. My question regards the choice of cut-off frequencies after the upsampling and the downsampling.

The old Nyquist frequency was at $11$ kHz so it would make sense to me to low-pass frequencies above $11$ kHz after upsampling - we're basically replicating our signal 20 times across the new spectra, right? The correct cut-off is apparently 20 kHz, but I can't quite wrap my head around that fact.

We then downsample by a factor 11 to 40 kHz. Now the correct LP cut-off is 11 kHz. I am seriously confused. Can anyone shed some wisdom here?


Don't be confused; you're doing everything correctly here:

$$\begin{align*} f_{sample,in} &&\overset{\text{interpolate}}{\rightarrow}&& f_{sample,intermediate} &&\overset{\text{decimate}}{\rightarrow}&& f_{sample,out}\\ 22 \mathrm{kHz}&&\overset{\uparrow 20}{\rightarrow}&& 440 \mathrm{kHz} &&\overset{\downarrow 11}{\rightarrow}&& f_{sample,out} \end{align*}$$

  1. To avoid images after interpolating, you will have to suppress all signal above 11kHz, which demands low pass with a cut off of $\frac1{2\cdot20}f_{sample,intermediate}$.
  2. To avoid aliases during decimation, you will have to suppress all signal above 20kHz, which demands a low pass filter with a cut off frequency of $\frac1{2\cdot11}f_{sample,intermediate}$.

Since the filter from 1. implies 2., using one filter is sufficient.

This is a very commonly applied optimization.

| improve this answer | |
  • $\begingroup$ Very nice @Marcus, I have been patiently waiting for the answer as well. Pretty concise. $\endgroup$ – Gilles Jan 19 '16 at 11:27
  • $\begingroup$ Yep, very nice! $\endgroup$ – Benjamin Lindqvist Jan 19 '16 at 14:06

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.