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I'm currently studying FIR filter and am having trouble understand how the following equation works, and it's implication.

$$ y[n] = h[n] * z^n = H(z) \cdot z^n $$

I don't really understand how this works given that

$$ y[n] = h[n] * x[n] $$ $$ Y(z) = H(z) \cdot X(z) $$

It seems to me that $ x[n] = X(z) = z^n $. What's going on here? What domain is $ z^n $ in?

I thought that convolution is done on two functions in the n-domain; however, in the equation $ y[n] = h[n] * z^n $, $ z^n $ does not appear to me to be in the n-domain.

If it is in the n-domain, then I don't understand how the other part works, $ y[n] = H(z) \cdot z^n $, because I thought that then it should be

$$ y[n] = Z^{-1}(\,H(z) \cdot Z(z^n)\,) $$

Thanks.

Edit: Here's the page from the book with the equation. It does seem to me that $z^n$ is used in both the time and frequency domain. Page From Book

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  • $\begingroup$ Please tell us the context of your first displayed equation. Is $y[n]$ the FIR filter output at time $n$? What is the input that produces this $y[n]$? I will assume that $H(z)$ is the transfer function of the FIR filter and $h[n]$ is the FIR response at time $n$. $\endgroup$ – Dilip Sarwate Apr 15 '12 at 12:38
  • $\begingroup$ Thanks for the reply. $y[n]$ is the output of the FIR filter at time $n$. the input that produces $y[n]$ is $x[n]$, and in the case above, $x[n] = z^n$ — I'm not really sure if z is supposed to be constant or not. Your assumptions re $H[n]$ and $h[n]$ are correct. If this is confusing, I can try and post the page from the book I'm basing this question on. I appreciate the help, thanks! $\endgroup$ – user968243 Apr 15 '12 at 13:35
  • $\begingroup$ $x[n] = z^n$ makes no sense. Are you sure it does not say $X(z) = z^n$, or more likely, $X(z) = z^{-n}$? $\endgroup$ – Dilip Sarwate Apr 15 '12 at 16:13
  • $\begingroup$ I've edited my question with book page which 'explains' it, though I don't really understand it. $\endgroup$ – user968243 Apr 16 '12 at 13:01
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    $\begingroup$ I am not surprised that you do not understand it; I don't think I understand it either. The point being made is totally obscured by the poor choice of notation by the authors of the text $\endgroup$ – Dilip Sarwate Apr 16 '12 at 13:23
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Looks like a notation or transcription error. The "z" that's used as the transform variable to calculate the Z-Transform can not show up as an independent variable in the time domain as well. One of them needs a different name.

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I suspect it should be something like the following: $$ y[n] = h[n] * \delta[n + N] \Rightarrow H(z)\cdot z^N $$

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    $\begingroup$ Is the $z$-transform of the sequence $h[0]=1, h[2]=3, h[n] = 0$ otherwise $H(z) = 1 + 3z$ or $1+3z^{-1}$? The response to a unit impulse at $n=N$ should be $h[n-N]$, no? $\endgroup$ – Dilip Sarwate Apr 16 '12 at 1:30
  • $\begingroup$ @DilipSarwate You're right, I had the power inverted. I believe that I have fixed the problem. $\endgroup$ – Jim Clay Apr 16 '12 at 12:16

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