The following question was on a homework assignment, and the solutions have been distributed already. However, I am having trouble reproducing the solutions.
Given the initial conditions $y[-1] = y[-2] = 0$ and the difference equation $$ 8 y[n] - 6 y[n-1] + y[n-2] = x[n] $$ Find $h[n]$.
I was able to find $$ H(z)=\frac{1}{8-6z^{-1}+z^{-2}} $$ but I can't figure out how to find the inverse $\mathcal Z$-transform. The solutions say
$$h[n] = \left(\frac{1}{2^{n+2}}-\frac{1}{2^{2n+3}}\right) U[n]$$
Can anyone shed some light on arriving to the result? Thanks in advance!
You have to do a partial fraction expansion of $H(z)$:
$$H(z)=\frac18\frac{1}{(1-\frac12 z^{-1})(1-\frac14 z^{-1})}=\frac18\left[\frac{2}{1-\frac12 z^{-1}}-\frac{1}{1-\frac14 z^{-1}}\right]\tag{1}$$
Then you can use the $\mathcal{Z}$-transform correspondence
$$\frac{1}{1-az^{-1}}\Longleftrightarrow a^nu[n],\quad |a|<1\tag{2}$$
which yields the given result.