What is the inverse z transform of 1/(z-1)^2?

Question

I'd like to know how to calculate the inverse z transform of $\frac{1}{(z-1)^2}$ and the general case $\frac{1}{(z-a)^2}$

Answer 1

A simple way is to recall the convolution property: if $x[n] \leftrightarrow X(z)$, then $x[n] \star y[n] \leftrightarrow X(z) Y(z) $ and, in particular $x[n] \star x[n] \leftrightarrow [X(z)]^2$

Then, we only need to get the anti-transform of $\frac{1}{z-a}= \frac{ z^{-1}}{1-a z^{-1}}$ (piece of cake) and convolve it with itself.

BTW, you need to specify the ROC (region of convergence), or if you want a "causal" sequence - without this, there can be more than one solution.

Answer 2

Hint: find the Maclauren series expansion of $\frac{1}{(z-a)^2}$ by differentiating a power series term-by-term.

Solution: Recall that for a sequence $x[n], n\in\mathbb{Z}$, the $\mathcal{Z}$ transform is defined as

$$ \mathcal{Z}\{x[n]\}(z)=\sum_{n\in\Bbb{Z}}x[n]z^{-n} $$If we can find a power series expansion for $(x-a)^{-2}$, the coefficients of our power series will be the sequence $x[n]$. This will be easier than using definition of the inverse transform.

Now, for any $a\neq 0$, we have

$$ \frac{1}{z-a}=-\frac{1}{a}\frac{1}{1-z/a}=-\frac{1}{a}\sum_{n=0}^\infty\left(\frac{z}{a}\right)^n $$Hence, inside the radius of convergence, we have

$$ (z-a)^{-2}=-\frac{d}{dz}(z-a)^{-1}=\sum_{n=1}^\infty\frac{n}{a^{n+1}}z^{n-1} $$

We now have a power series, so the only thing that remains is to define the appropriate $x[n]$ such that $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$. Note that we have only nonnegative powers of $z$, so if we define

$$ x[n]=\left\{\begin{array}{cc} \frac{-n+1}{a^{-n+2}} & n\leq 0\\ 0 &n>0 \end{array}\right. $$ we have $\mathcal{Z}\{x[n]\}=(z-a)^{-2}$, as you can check.