# How to find $h[n]$ system response of this equation?

$2y[n-2]-2y[n-3]-4y[n-4]=x[n]-10x[n-1]-4x[n-2] + 4x[n-3]$ is the system that I'm looking for the response.

I transformed this system via using Z transform:

$$\frac{Y(z)}{X(z)}=H(z)=\frac{z^4 - 10z^3 - 4z^2 + 4z}{2( z^2 -z -2 )}$$

$$H(z) = 0.5z^2 -4z + \frac{-5}{z-2} + \frac{-1/3}{z-2} + \frac{1/3}{z+1}$$

How to find ROC to inverse the Z transfom to $h[n]$ ?

• This is a homework type question. What have you tried? We can help you, but we can't just provide solutions to homework questions if you don't show us what you've done and where exactly you're stuck. – Matt L. Jan 7 '18 at 16:38
• Wait, I am preparing a question. I asked this question because I'm working on Fourier transformations where I think Z transform is much more easier than Fourier transform. This is not the question you are thinking like this. If you didn't see my step below the code side, I want to show you that, H(Z) is my answer where I couldn't get the ROC regions correctly. I always frustrate when I ask a question on here, everytime someone blocked my question asking feature. If I can't ask here, where should I go to ask my quesions? @MattL. please consider twice before you -1 to someone. – Bay Jan 7 '18 at 16:49
• First of all, apparently there was someone else who -1'd you (even though I should have done that too). Second, you wrote a long comment, but you didn't take the time to tell us what your problem is. So you have a partial fraction expansion, but why and where are you stuck now? – Matt L. Jan 7 '18 at 16:56

1. get your partial fraction expansion right; right now it does not equal the transfer function $H(z)$, which you calculated correctly.
2. find the poles and zeros of $H(z)$; note that you have $4$ poles, only two of which are finite. The locations of the poles determine the possible ROCs.
3. note that there are two possible ROCs, and, consequently, two different impulse responses $h[n]$ corresponding to the given difference equation.
4. there is no causal impulse response corresponding to the given difference equation because in order to compute the output $y[n-2]$ you need the input values $x[n]$ and $x[n-1]$.