A problem on $\mathcal{Z}-$transforms and signal stability

Question

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.

Answer 1

The integral you'll get for the inverse $\mathcal{Z}$-transform is

$$h[n]=\frac{1}{2\pi i}\oint_C\frac{z^2}{(z-10)^2}z^{n-1}dz\tag{1}$$

where the positively oriented closed curve $C$ must lie inside the region of convergence $|z|>10$. Solving the integral $(1)$ is best done via the residue theorem.

However, the transfer function in this example has a standard form, and its inverse transform can be found in most $\mathcal{Z}$-transform tables, such as this one (entry 13). The result is

$$h[n]=(n+1)10^nu[n]\tag{2}$$

Clearly, $h[n]$ grows without bound, and hence, the corresponding system is unstable.

The fact that the system is unstable can of course already be seen from $H(z)$ and the additional information that the system is causal. In that case, stability requires all poles to lie inside the unit circle, which is obviously not the case here.