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I have a system which is expressed as following: $$y[n] = x[n] - \alpha y[n-N]$$ where $y[n]$ is the output and $x[n]$ is the input.

When I take the z-transform of both sides, I've found the transfer function as $$H(z)=\frac{1}{1+\alpha z^{-N}}$$.

However, I need to find the impulse response of this system but I couldn't find $h[n]$. Can anybody help me how to solve this problem?

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Cconsider the following pair of signals and their transforms:

If $g[n]$ has a Z-transform $G(z)$ then the expanded (zero stuffed) signal $h[n]$ has the Z-transform $H(z) = G(z^N)$, where

$$ h[n] = \begin{cases} {g[\frac{n}{N}] ~~~,~~~ n = m N \\ 0 ~~~~~~~~~,~~~ \text{otherwise} \tag{1}}\end{cases} $$

Looking at the given Z-transform : $$H(z) = \frac{ 1}{ 1 - \alpha z^{-N} }$$

it can be seen that if $H(z) = G(z^N)$ where

$$G(z)= \frac{ 1}{ 1 - \alpha z^{-1} }$$

and $g[n] = \alpha^n u[n] $.

Then you can simply conlcude that :

$$ h[n] = \alpha^{n/N} ~u[n/N] = \begin{cases} {\alpha^{n/N} ~~ ,~~ n = m N, m = 0,1,.. \\ 0 ~~~~~~~~~,~~ \text{otherwise} }\end{cases} $$

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HINT:

Find the impulse response corresponding to

$$\tilde{H}(z)=\frac{1}{1+\alpha z^{-1}}\tag{1}$$

and then try to figure out what happens to the impulse response if you replace $z$ by $z^N$. (Note that $H(z)=\tilde{H}(z^N)$).

HINT 2:

Answer the following questions for yourself: what is the impulse response corresponding to $H(z)=1+z^{-1}$? And what is the impulse response corresponding to $H(z^N)=1+z^{-N}$? So what does replacing $z$ by $z^N$ do to the impulse response?

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  • $\begingroup$ The impulse response of this function is $(-a)^{n} u(n)$. But I'm asking $z^{-N}$ version. Please can you help me? $\endgroup$
    – Jason
    Commented Oct 30, 2019 at 15:41
  • $\begingroup$ @Jason: I know what you're asking, but as you might know, this is no homework service. I tried to help you but you need to show a bit of effort on your side. Did you actually try to think about what happens when $z$ is replaced by $z^N$? $\endgroup$
    – Matt L.
    Commented Oct 30, 2019 at 15:46
  • $\begingroup$ @MattL. Sir, this is not a homework, just a simple question. I put an effort as you can see. I did everything and stuck at this step. If I replace z with $z^{N}$, does it mean shifting by N? And the result is h[n] = $(-a)^{n-N}u(n-N)$. Am I right? $\endgroup$
    – Jason
    Commented Oct 30, 2019 at 15:50
  • $\begingroup$ @Ben it is z/(z-1). $\endgroup$
    – Jason
    Commented Oct 30, 2019 at 15:51
  • $\begingroup$ Sorry I misread, I thought you meant step function. It's even easier because it's an impulse response. You simply need to find the inverse transform of H(z), then you will have the impulse response for each sample. $\endgroup$
    – Ben
    Commented Oct 30, 2019 at 15:53
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I find it easiest to find the impulse response in such cases by substituting $x(n) = \delta(n)$ and $y(n) = h(n)$ (after all it is an 'impulse' response). That gives you $$h(n) = \delta(n) - \alpha h(n-N)$$ If the system is causal (no negative time indices), $$\begin{aligned} h(0) &= 1 \\ h(1) &= 0 \\ h(2) &= 0 \\ &\vdots \\ h(N) &= -\alpha h(0) = -\alpha\\ h(N+1) &= 0 \\ &\vdots \\ h(2N) &= -\alpha h(N) = \alpha^2\\ \therefore h(kN) &= (-1)^k \alpha^k, \ k = 0,1,2,\ldots \end{aligned}$$ This is just an upsampled version of the sequence $h(k) = (-1)^k \alpha^k$, with an upsampling factor of $N$. For $N=5, \alpha=0.8$, this is what the impulse response looks like for the first $100$ samples

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