Inverse Z-transform of $\frac{1}{(1-z^{-1})^2}$?

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!

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Is this what it means? Find the coefficients $x_n$ in closed form: $$\frac{1}{(1-z^{-1})^2} = \sum_{n=0}^\infty x_n z^{-n}$$ And if that's not it, tell us what it is! – GEdgar Jan 5 '12 at 1:39

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 $$x[n] = \begin{cases} n+1, & n\geqslant 0, \\ 0, &n < 0. \end{cases}$$ 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 $(n+1) \cdot u[n+1]$. $\qquad \diamond$

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