Inverse Z Transform of $A/(z-z_p)^2$

Question

I'm searching online the inverse z transform of $A/(z-z_p)^2$. I tried to do it and I got:

$$A \space n \space z_p^{n-2} heaviside(n-1)$$

Is it right?

Thank you in advance.

Answer 1

OK, let us check on the Z-Transform you provided:

$$X(z)=\frac{A}{(z-z_p)^2}$$

And we want to find it's inverse, so that $\mathcal{Z}^{-1}\{X(z)\}=x[n]$

What I think should be the best to do here is to find a transform pair in a table that gets the closest to your case.

Such as:

$$n(z_p)^{n-1}u[n-1] \leftrightarrow \frac{1}{z-z_p}$$

Where $u[n-1]$ is the shifted Heaviside Step Function and can be also defined as $1^{n-1}$, $\forall n\geq 1$.

It is easy to see that if:

$$X(z)=\frac{A}{(z-z_p)^2}=A\frac{1}{z-z_p}\frac{1}{z-z_p}$$

Then, throught the Z-transform properties:

$$x[n]=A\left\{\left[(z_p)^{n-1}u[n-1]\right]\ast\left[(z_p)^{n-1}u[n-1]\right]\right\}$$

So, we proceed to solve this convolution:

$$ \begin{align} x[n]&=A\sum_{k=-\infty}^{+\infty}\left[(z_p)^{k-1}u[k-1](z_p)^{n-k-1}u[n-k-1]\right]\\ &=A\sum_{k=-\infty}^{+\infty}\left[(z_p)^{k-1+n-k-1}u[k-1]u[n-k-1]\right]\\ &=A\sum_{k=-\infty}^{+\infty}\left[(z_p)^{n-2}u[k-1]u[n-k-1]\right]\\ &=A(z_p)^{n-2}\sum_{k=-\infty}^{+\infty}\left[u[k-1]u[n-k-1]\right]\\ &=A(z_p)^{n-2}\sum_{k=1}^{+\infty}\left[u[n-k-1]\right]\\ &=A(z_p)^{n-2}\sum_{k=1}^{+\infty}\left[1^{-k+n-1}\right]\\ &=A(z_p)^{n-2}\sum_{k=1}^{+\infty}\left[1^{n-1}1^{-k}\right]\\ &=A(z_p)^{n-2}u[n-1]\sum_{k=1}^{+\infty}\left[1^{-k}\right]\\ &=A(z_p)^{n-2}u[n-1]\sum_{n=1}^{+\infty}\left[1^{-n}\right]\\ &=An(z_p)^{n-2}u[n-1] \end{align} $$

$$x[n]=An(z_p)^{n-2}u[n-1]$$

So Yes, you were correct on your solution!

Please let me know if I made it confusing... I tried to break through all steps. Hope I was able to help :)