How do I find the poles of this difference equation?

Question

I have an equation:

$$y(n) = 0.634x(n) - 0.634x(n-2) + 0.268y(n-2)$$

I completed a $z$ transform and got: $$ H(z) = \frac{1-0.268z^{-2}}{0.634 - 0.634z^{-2}}$$

What is the next step to find the poles of this equation?

Answer 1

Your transfer function is incorrect. Assuming that $y(n)$ is the output sequence and $x(n)$ is the input sequence, the transfer function is given by

$$H(z)=\frac{Y(z)}{X(z)}=0.634\frac{1-z^{-2}}{1-0.268z^{-2}}\tag{1}$$

The poles are the values of $z$ for which $(1)$ becomes infinity, or, equivalently, for which the denominator of $(1)$ becomes zero. I'm sure you can take it from here.

Answer 2

Apply Z-transform to the equation, gives:

$Y(z)=0.634X(z)-0.634z^{-2}X(z)+0.268z^{-2}Y(z)$

or:

$Y(z)\left(1-0.268z^{-2} \right)=0.634X(z)\left(1-z^{-2} \right)$

Now computing $H(z)=\frac{Y(z)}{X(z)}$,

$H(z)=\frac{Y(z)}{X(z)}=\frac{0.634\left(1-z^{-2}\right)}{1-0.268z^{-2}}$

For finding the poles, you simply need to solve $1-0.268z^{-2}=0$, $z=\pm 0.5177$.