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What is the inverse Z-transform of $frac{1}{(1-z^{-1})^2}$?

Title says it all. I have a one line solution but can't work out how to get there from tables or first principals.

Thanks!

The Z-transform of a discrete time-domain signal $x = (x[n])$ is just the generating function $$X(z) = sum limits_{n=-infty}^infty x[n] z^{-n} .$$ Conversely, given a signal in the frequency-domain representation, i.e., given $X(z)$, the inverse Z-transform is simply its power series representation.

In order to compute the inverse Z-transform of $X(z) = (1 - z^{-1})^{-2}$, we develop this as a power series in $z^{-1}$. This can be done by writing the Taylor series for the function $g(x) = (1-x)^{-2}$ and plugging in $x = z^{-1}$. Alternatively, we can also use the Newton binomial series formula: $$
(1-x)^{-(beta+1)} = sum_{n=0}^{infty} binom{n+beta}{beta} x^n.
$$
Plugging in $x = z^{-1}$ and $beta=1$, we get
$$
(1-x)^{-2} = sum_{n=0}^{infty} binom{n+1}{1} z^{-n} = sum_{n=0}^{infty} (n+1) z^{-n}.
$$
Comparing with the definition of Z-transform, we conclude that the inverse Z-transform of $X(z)$ is $$
x[n] = begin{cases}
n+1, & ngeqslant 0, \ 0, &n < 0.
end{cases}
$$
Done! $qquad diamond$

Using the properties.

The above procedure is quite straightforward, but a bit tedious. Often it is possible to get the answer quicker by exploiting the properties of the Z-transform. To implement this approach effectively, one needs to be familiar with a table of Z-transforms of common series as well as a dictionary to translate elementary operations between the time and frequency domains. The Wikipedia article on Z-transforms contains such an extensive table of properties; my notation closely follows this page.

Method 1.

Given a signal $y[n]$ with Z-transform $Y(z)$, its accumulation $sum limits_{k=-infty}^{n} y[k]$ has the Z-transform $frac{Y(z)}{1 - z^{-1}}$. Now the delta function $delta[n]$ (a unit spike at the origin and zero everywhere else) has the Z-transform $1$. Therefore, its accumulation given by
$$
u[n] = begin{cases} 1, &n geqslant 0, \ 0, &n < 0, end{cases}
$$
has the Z-transform $(1-z^{-1})^{-1}$.

Applying the accumulation operation once again results in $(1-z^{-1})^{-2}$ in the frequency domain; in the time domain, we have
$$
sum_{k=-infty}^{n} u[k] = sum_{k=-infty}^n [k geqslant 0] = begin{cases} n+1, &n geqslant 0, \ 0, &n < 0, end{cases}
$$
which is our desired inverse Z-transform. $qquad diamond$

Alternate method using differentiation.

We can replace the second accumulation operation with a differentiation operation. We again start from the signal
$$
u[n] = begin{cases} 1, &n geqslant 0, \ 0, &n < 0,end{cases}
$$
whose Z-transform is $U(z) = (1 - z^{-1})^{-1}$. Therefore the Z-transform of $n cdot u[n]$ is given by
$$
- z frac{d U(z)}{dz} = -z cdot frac{(-1)}{(1 - z^{-1})^{2}} cdot frac{1}{z^2} = frac{z^{-1}}{(1-z^{-1})^2}.
$$
Notice that we got what we wanted except for an extra $z^{-1}$. We can get rid of this by shifting the series by $1$ in the time domain: i.e., the Z-transform of $(n+1)cdot u[n+1]$ is $(1-z^{-1})^{-2}$. Thus our inverse Z-transform is given by $(n+1) cdot u[n+1]$. $qquad diamond$

This is the one that you find in the table:

$$
n alpha^{n} u[n] <=> frac{alpha z^{-1}}{(1-alpha z^{-1})^2} ROC: |z|>|alpha|
$$

Then you find the shift property:

$$
x[n-k] <=> z^{-k}X(z) ROC: R_x
$$

What happens if you multiply RHS of transform by z? then, LHS of transform has a shift.

$$
(n+1) alpha^{n+1} u[n+1] <=> frac{alpha}{(1-alpha z^{-1})^2} ROC: |z| > |alpha|
$$

when n=-1, (n+1) term equals zero. and u[n+1] = 1
when n=0, (n+1) term equals one and u[n+1] = 1
now we notice that if we substitute u[n] in place of u[n+1], its the same because (n+1) is zero when n= -1. Thus:

$$
(n+1) alpha^{n+1} u[n] <=> frac{alpha}{(1-alpha z^{-1})^2} ROC: |z| > |alpha|
$$

Then we divide the RHS of transform by alpha which corresponds to dividing by alpha on RHS of transform.

$$
(n+1) alpha^{n} u[n] <=> frac{1}{(1-alpha z^{-1})^2} ROC: |z| > |alpha|
$$