So the question is basically z transform the given system.

$(y[n+2] + 3y[n+1] - 4y[n])=(x[n+2] - 5x[n+1])$

I've to find h[z] first then it's really easy to solve it. So that's what I got so far;

$z^2y(z) + 3zy(z) - 4y(z) = z^2x(z) - 5zx(z)$

$(z^2+3z-4)y(z) = (z^2 - 5z)x(z)$

$H(z) = \frac{y(z)}{x(z)} = \frac{z^2-5z}{z^2+3z-4}$


$H(z) = 1 - \frac{4}{5(z-1)}-\frac{36}{5(z+4)}$

I've to find Z transform pair but I'm stuck here. Thanks in advance!

Here is the z transform table from schaums;


  • $\begingroup$ $z^2+3z-4 = (z-1)(z+4)$ then partial fractions $\endgroup$ – G Cab Jan 13 at 18:08
  • $\begingroup$ I've updated the question. I tried every way to simplify this. Can you check it out? @GCab $\endgroup$ – L4W Jan 13 at 18:17
  • $\begingroup$ Are you studying bilateral $Z$ transform ($z$ ranges from $-\infty$ to $\infty$)? $\endgroup$ – Shubham Johri Jan 13 at 18:34

Using $H(z)=\frac{z^2-5z}{(z+4)(z-1)}=\frac{Az}{z-1}+\frac{Bz}{z+4}+C$ we get $A=-4/5,B=9/5,C=0$. Can you complete?

From your list we see that $Z[-a^nu[-1-n]]=\frac z{z-a}$ and thus$$Z[-u[-1-n]]=\frac z{z-1}\\Z[-(-4)^nu[-1-n]]=\frac z{z+4}$$Thus, by the linearity of the inverse $Z$ transform, we get$$Z^{-1}[H(z)]=AZ^{-1}[z/(z-1)]+BZ^{-1}[z(z+4)]\\=\frac45u[-1-n]-\frac95(-4)^nu[-1-n]$$

  • $\begingroup$ Can you help me out finding the solution please. I know I'm asking too much :( $\endgroup$ – L4W Jan 13 at 18:38
  • $\begingroup$ @L4W Look at the spoiler in my answer $\endgroup$ – Shubham Johri Jan 13 at 18:44
  • $\begingroup$ Thanks a lot, couldn't find out without you! $\endgroup$ – L4W Jan 13 at 18:46
  • $\begingroup$ You are welcome ;) $\endgroup$ – Shubham Johri Jan 13 at 18:46
  • $\begingroup$ By the way isn't $C = 1$ $\endgroup$ – L4W Jan 13 at 18:47

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