# Help me with z transform

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;

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

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

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

then

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

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

Here is the z transform table from schaums; • $z^2+3z-4 = (z-1)(z+4)$ then partial fractions – G Cab Jan 13 at 18:08
• I've updated the question. I tried every way to simplify this. Can you check it out? @GCab – L4W Jan 13 at 18:17
• Are you studying bilateral $Z$ transform ($z$ ranges from $-\infty$ to $\infty$)? – 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$$Z^{-1}[H(z)]=AZ^{-1}[z/(z-1)]+BZ^{-1}[z(z+4)]\\=\frac45u[-1-n]-\frac95(-4)^nu[-1-n]$$
• By the way isn't $C = 1$ – L4W Jan 13 at 18:47