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



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 $H(z)=\tilde{H}(z^N)$).


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?

  • $\begingroup$ 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 Oct 30 '19 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. Oct 30 '19 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 Oct 30 '19 at 15:50
  • $\begingroup$ @Ben it is z/(z-1). $\endgroup$ – Jason Oct 30 '19 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 Oct 30 '19 at 15:53

Everything becomes very clear once you cinsder the following pair of signals and 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

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

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 :

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


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