Lecture 2: Inverse Z-Transforms
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Given the nature of the inverse Z-transform integral, we look for other ways of computing the inverse. The approach is very similar to what we did for Laplace transforms: We break the function down into pieces that we can recognize from the table, then do table lookup. As for Laplace transforms, there will be a variety of properties (delay, convolution, scaling, etc.) that will help us. The most important tool, however, remains the Partial Fraction Expansion. However, we will find it convenient to modify it slightly for our purposes here.
![\begin{example}Find the inverse Z-transform of $F[z] = \frac{z}{z-.2}$.
\par Kno...
...rd to write
\begin{displaymath}f[k] = (.2)^k u[k]
\end{displaymath}\end{example}](img13.png)
![\begin{example}
Find the inverse Z-transform of
\begin{displaymath}F[z] = \frac...
...a[k] + [\frac{3}{2}(2)^k +
\frac{5}{3}(3)^k]u[k]. \end{displaymath}\end{example}](img14.png)
The point is: Computing inverse Z-transforms using PFE is exactly
analogous to computing inverse Laplace transforms, provided
that you form ![$F[z]/z$](img15.png){kind=small}
first.
There is another method of obtaining inverse Z-transforms which is
useful if you only need a few terms. Recall that the Z-transform is
simply a power series. All you need to do is find the coefficients of
the power series. One way to do this is by long polynomial division.
This gives you a numerical expression for as many terms of ![$f[k]$](img4_1.png){kind=small}
as
you choose to compute, not a closed-form mathematical expression.
![\begin{example}
If $F[z] = z/(z-.5)$, find $f[k]$. Work through the first
few.
\end{example}](img16.png){kind=small}
