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Let's say I have a discrete-time signal $y[n]$ that is passed through an Ideal Lowpass Filter (impulse response $h[n]$) to get $y_1[n]$. The filter has cutoff frequencies at $\pm \pi/N$ rad.

Now this $y_1[n]$ is downsampled by a factor of $N$ to obtain $z[n]$. I know that $z[n] = y_1[nN] = \sum_{k=-\infty}^{\infty} h[k] \ast y[nN - k]$.

So in order to show this procedure in the frequency domain I can use the fact that convolution in time-domain is equal to multiplication in frequency domain.

As such, the frequency response of $y_1[n]$ in terms of $y[n]$ is simply:

$$Y_1(e^{j\omega}) = H_{LP}(e^{j\omega}) \cdot Y(e^{j\omega})$$

And using the expression for downsampling in frequency domain:

$$Z(e^{j\omega}) = \frac 1N \sum_{k=0}^{N-1} Y_1\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right)$$

Subbing in to the summation should we get this:

$$Z(e^{j\omega}) = \frac 1N \sum_{k=0}^{N-1} H_{LP}\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right) \cdot Y\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right)$$

or this:

$$Z(e^{j\omega}) = \frac{H_{LP}(e^{j\omega})}{N} \sum_{k=0}^{N-1} Y\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right)$$

First of all which one is correct (should $H_{LP}$ be inside the summation?) and if so does this suffice? Or is there a more intuitive way of writing/explaining this procedure?

Edit:

One other minor question. What if we want to show $Z(e^{jω})$ only in terms of $Y(e^{jω})$. Then can we write $H_{LP}$ as $u(ω+π/N)−u(ω−π/N)$ and then shift and scale it inside the summation as follows: $$Z(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} \left(u(\frac{(\omega - 2\pi k)}{N} + \pi/N) - u(\frac{(\omega - 2\pi k)}{N} - \pi/N) \right) \cdot Y\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right)$$

  • May help – OverLordGoldDragon Apr 11 '23 at 10:36
  • @OverLordGoldDragon Thanks but not exactly! The mathematical intuition given is only for the downsampling part and not the entire decimation procedure. –  Apr 11 '23 at 13:12
  • What do you mean by "suffice"? I believe my linked post answers your question completely, if read carefully. It's just my answer applied to $H \cdot Y$ - and no, you can't move it outside the summation. The lowpassing part, in sense of removing aliasing, is covered in "Lossless subsampling criteria", to be paired with the first section's visuals. – OverLordGoldDragon Apr 11 '23 at 13:19
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    My bad I didn't go down to that part. By suffice I meant is it mathematically sufficient which after reading your other answer seems yes. Thanks again! –  Apr 11 '23 at 13:26
  • @OverLordGoldDragon One other minor question. What if we only want to show $Z(e^{j\omega})$ in terms of $Y(e^{j\omega})$. Then can we write $H_{LP}$ as $u(\omega + \pi/N) - u(\omega - \pi/N)$ and then shift and scale it inside the summation as follows: $$Z(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} \left(u(\frac{(\omega - 2\pi k)}{N} + \pi/N) - u(\frac{(\omega - 2\pi k)}{N} - \pi/N) \right) \cdot Y\left(e^{j\frac{(\omega - 2\pi k)}{N}}\right)$$ –  Apr 12 '23 at 15:44
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    Caporal- If what OLGD referred you to answered your question as originally posted, please consider upvoting that and reformulating/rewriting your question here to be your updated question in the comment above (which is really your only question now). You can refer to the other answer for context and that will allow him or others to properly answer here and clean this up. Thanks! – Dan Boschen Apr 13 '23 at 02:45
  • Minor questions are fine, but that's not minor. Or, my answer is simply "yes". Agreed with Dan's suggestion, you can edit that it and I may answer. – OverLordGoldDragon Apr 13 '23 at 12:39
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    @OverLordGoldDragon Done! –  Apr 13 '23 at 17:23
  • You seem to be conflating the discrete-time Fourier transform (DTFT) with its sampled counterpart, the discrete Fourier transform (DFT), in your question. You start with $Y(\omega),H_{LP}(\omega),$ and $Y_1(\omega)$, which are all continuous functions of $\omega$ and suggests that they are DTFTs. However, you then write $Z(e^{j\omega})$ as a finite-sum, which suggests that this is a DFT. Consider checking equation (1) in @OverLordGoldDragon's linked answer. Equation (1) is a DTFT, while the equation below it is a DFT. Best not to confuse the two. – mhdadk Apr 15 '23 at 12:46
  • So, your post is written in terms of DTFT which I've not worked out for this context; if a DFT-based answer for a length $N$ sequence is acceptable, I may provide it, and your could relate the two. -- @mhdadk Strictly, if we're looking at sampling of a continuous function, (1) is actually CFT, see under "Periodizing" - which is where DTFT comes from AFAIK. – OverLordGoldDragon Apr 15 '23 at 12:56
  • @OverLordGoldDragon yes, DTFT is just CFT of a discrete-time signal, or equivalently CFT of a continuous-time signal multiplied by a Dirac comb. – mhdadk Apr 15 '23 at 15:07
  • @mhdadk CFT operates on impulses by integration, DTFT on instants of a function by summation. They're equivalent but not identical in every sense (but for this question yes, it makes no difference). – OverLordGoldDragon Apr 15 '23 at 15:43
  • @OverLordGoldDragon I’m not sure what you mean. How would they not be identical? You can use the sifting property of Dirac deltas to reduce the CFT integration to a summation. – mhdadk Apr 15 '23 at 16:36
  • @mhdadk $Z(e^{j\omega})$ is the downsampled DTFT of $Y_1(e^{j\omega})$ –  Apr 16 '23 at 12:36
  • @mhdadk I can't explain it quickly, see e.g. here – OverLordGoldDragon Apr 21 '23 at 10:55
  • "Proof: bandlimited subsampling" here shows it for finite decimation, but it's for convolution of subsamplings rather than subsampling of convolution, and more complicated. If you study the answer it becomes obvious, but it's much easier to study my very first comment here, where you plug in $X = Y_1$. You just show $X$ is unchanged because we're adding zeros to it. It's same in DTFT. – OverLordGoldDragon Jun 09 '23 at 12:36

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