I should say which of the filters from the following image implements an ideal low-pass filter with cutoff frequency $\frac{\pi}{4}$. In the picture, Sa is a zero insertion filter, which inserts a zero sample after each input sample and Sb is a decimator, which extracts every second sample of its input. The first filter is fine and the second one does not represent the desired low-pass (that's the answer on my book).

enter image description here

However, I've not been able to explain why is the second filter wrong. I know that up-sampling compresses the frequency content of a signal and that down-sampling expands it. So for the first filter, we first compress the input band by 2 and conclude that if the original limits were less than or equal to $\frac{\pi}{4}$, nothing gets filtered out (because $\frac{\pi}{4}$ becomes $\frac{\pi}{8}$ after Sa). And then, Sb recovers the original spectrum by multiplying by 2.

I analysed the second filter in a similar way, just noticing that Sb was switched with Sa thus making a multiplication by 2 first. The first Frequency Response would not filter anything from inputs with frequency lower than $\frac{\pi}{4}$ (which becomes $\frac{\pi}{2}$ after Sb). Then, after Sa, we have a division by 2 which restores the original spectrum and anything less than $\frac{\pi}{4}$ can proceed to the output.

What am I doing wrong?


1 Answer 1


If the input sequence is band limited to $\pi/2$, the second alternative is also OK (as you realized with your argument). However, if it not band limited you will get aliasing, so it is not a feasible solution in the general case.

  • $\begingroup$ I didn't mean that band limitation to $\frac{\pi}{2}$ will be able to pass the second filter, because the first stage, Sb, multiplies the band by 2 making it limited to $\pi$, which would get filtered by the first Frequency Response. Did you miss type $\frac{\pi}{2}$ instead of $\frac{\pi}{4}$ or did I get something wrong? $\endgroup$
    – Thiago
    Commented Jun 3, 2014 at 13:23
  • $\begingroup$ I was correct. :-) The problem is that if you have a signal at a frequency, say $0.9\pi$, that one will be $1.8\pi \Rightarrow 0.2\pi$ after Sb due to aliasing and you can not recover the correct signal. As a second reason (after talking to a colleague), the second approach is time-variant. You could try using Noble identities and in that way change order of the up-/downsampler and filters. In the first case you will get upsampling followed by downsampling which can be removed, while in the second you get downsampling follows by upsampling, which is replacing every other sample with zero. $\endgroup$
    – Oscar
    Commented Jun 3, 2014 at 16:04

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