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

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  • $\begingroup$ May help $\endgroup$ Commented Apr 11, 2023 at 10:36
  • $\begingroup$ @OverLordGoldDragon Thanks but not exactly! The mathematical intuition given is only for the downsampling part and not the entire decimation procedure. $\endgroup$
    – user64710
    Commented Apr 11, 2023 at 13:12
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    $\begingroup$ 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! $\endgroup$
    – user64710
    Commented Apr 11, 2023 at 13:26
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    $\begingroup$ 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! $\endgroup$ Commented Apr 13, 2023 at 2:45
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    $\begingroup$ @OverLordGoldDragon Done! $\endgroup$
    – user64710
    Commented Apr 13, 2023 at 17:23

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