I am a little confused about the use of anti-aliasing filters. As on how I am confused, please consider the following task:

Consider a simple sampling rate conversion system with a conversion rate of 4/3. The system consists of two upsampling blocks, each by 2, and one downsampling block of 3.

system block diagram

The frequency spectrum has periodic repetitions at integer multiples of the sampling frequency. Therefore, upsampling creates additional - unwanted - spectral images. These can be cancelled by an anti-imaging filter:


The spectrum within the blue dotted rectangles should be my output Y_1. This output will be upsampled a second time, which I think results just having 4 spectrum repetitions until 2*pi.

But, how does downsampling work? I am given the following figure from my lecture notes:


So, apparently before downsampling we apply the anti-aliasing filter, which will result in the part of the spectrum within the blue-dotted shapes. Now, we increase Omega by a factor of 2, such that.. I don't know. What exactly happens here ?


1 Answer 1


I think you make this needlessly complicated. Let's do a specific example: let's say we want to go from 48 kHz up by 4/3 to 64 kHz and your signal bandwidth is 20 kHz.

In the first step, you would up-sample by a factor of 4 by inserting 3 zeros between each sample. Your new sample rate is 192 kHz. Your original spectrum is preserved and you get mirror spectra centered around 48 kHz & 96 kHz.

To eliminate this you need a lowpass filter with cutoff of 20 kHz at 192 kHz sample rate and sufficient attenuation at 24 kHz.

Finally you just down sample the result by factor of three by simply throwing away every second and third sample. This works without aliasing, since the lowpass filter has already eliminated all content that would alias.

Regarding your graphs: in general I would avoid drawing anything past the Nyqusit frequency. It's just periodic repetition and doesn't add any useful information and it can get confusing.

  • $\begingroup$ First of all, thx very much for your reply! I'm afraid I still don't fully get it, though. You say: In the first step I add 3 zeros between each sample. Since our initial sampling frequency is 48 kHz, we add zeros at 12, 24 and 36 kHz, is that right? So I would expect additional spectra at these exact frequencies. Instead, you say we have additional spectra at 48 and 96 kHz. But these spectra should have been there just from the original image, I think. Apparently I am terribly wrong, but can you explain why, again? $\endgroup$
    – user503842
    Jan 28, 2019 at 12:43
  • $\begingroup$ You insert zeros in the time domain. So your up-sampled sequence is x[0],0,0,0,x[1],0,0,0,x[2],0,0,0,x[3], .... with a sample rate of 192 kHz. Let's assume your orginal signal x[n] is a sine wave at 10 kHz. Once its upsampled you still have the 10 kHz plus the images: 38 kHz, 58kHz, 86 kHz. (48+-10, 96-10) $\endgroup$
    – Hilmar
    Jan 29, 2019 at 11:47

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