I am doing the following problem for my DSP exam and I am doubting my answers while being stuck on the last part:
Given a causal LTI system and difference equations $y(n)=x(n)+20y(n-1)-100y(n-2),$ find the transfer function using a $\mathcal{Z}-$transform and the impulse response $h(n)$ using an inverse $\mathcal{Z}-$transform and check the stability of the system.
I have started by calculating the transfer function (where each $b_k$ and $a_k$ is the coefficient of the corresponding $x(n-k)$ and $y(n-k)$ terms respectively. $$\mathcal{H}(z)=\displaystyle\frac{\displaystyle \sum_{k=0}^{M}{b_kz^{-k}}}{\displaystyle \sum_{k=0}^{M}{a_kz^{-k}}}=\frac{1}{\displaystyle1-\frac{20}{z}+\frac{100}{z^2}}=\frac{z^2}{(z-10)^2}$$ and I have found that the ROC of this is $|z|<10$. So the impulse response $h(n)$ should be the inverse $\mathcal{Z}-$transform of $\mathcal{H}(z)$. And then I need to calculate the integral on the circle with radius $10$, so I must then calculate the contour integral, but it seems that it is an utter failure and it diverges. Because I reparametrised as $z=10e^{ix},$ for $0\leq x\leq 2\pi$ and got the integral $$\frac{1}{2\pi i}\int_{0}^{2\pi}{\frac{(10e^{ix})^2}{(10e^{ix}-10)^2}\cdot10ie^{ix}\cdot (e^{ix})^{n-1}dx}$$ which is absolutely horrifying. And for stability it must be that this $h(n)$ I'm supposed to find is absolutely integrable. I'm so confused. Thank you in advance.
I need to calculate the integral on the circle with radius 10
. The ROC does not include the circle of radius 10... so the integral will not converge. Usually, people don't do the integral to find the impulse response, they just use a table of $z$-transform pairs like this one. $\endgroup$