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You are here: Home Electrical and Computer Engineering Signals and Systems Lecture 2: Properties of the Z-Transform

Lecture 2: Properties of the Z-Transform

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Schedule :: Intro :: Inverse :: Properties :: Solution :: Transfer Functions :: System Realization :: Bilateral :: Frequency Response :: Rubber Sheet Geometry :: FIR Filters

In the descriptions of these properties, take

\begin{displaymath}f[k] \leftrightarrow F[z]. \end{displaymath}
Delay property

This is very analogous to the differentiation property of Laplace transforms, and will similarly allow us to solve differential equations.

\begin{displaymath}\boxed{f[k-1]u[k-1] \leftrightarrow z^{-1} F[z]} \end{displaymath}

So $z^{-1}$ is the delay operator. (As $s$ is the differentiation operator.) Also
\begin{displaymath}\boxed{f[k-1]u[k] \leftrightarrow z^{-1}F[z] + f[-1].}
\end{displaymath}
Note the difference between these two!

This property is used to introduce the initial conditions when we use transforms to solve difference equations.
\begin{proof}For the more general case of shifting by $m$,
\begin{displaymath}\Z...
...row z^{-3}F[z] + z^{-2}f[-1] + z^{-1}f[-2] +
f[-3]} \end{displaymath}\end{proof}

Left Shift (Advance)
Similar to the last property,
\begin{displaymath}\boxed{f[k+m]u[k] \leftrightarrow z^m F[z] - z^m \sum_{k=0}^{m-1} f[k]
z^{-k}} \end{displaymath}


\begin{example}
Find the Z-transform of the sequence
\begin{displaymath}f[k] = ...
...6}\frac{z}{z-1} = \frac{z^6 - 6z + 5}{z^5(z-1)^2}
\end{displaymath}\end{example}

Convolution
Like the convolution property for Laplace transforms, the convolution property for Z-transforms is very important for systems analysis and design. In words: The transform of the convolution is the product of the transforms. This holds for both Laplace and Z-transforms.

If $f_1[k] \leftrightarrow F_1[z]$ and $f_2[k] \leftrightarrow
F_2[z]$ then

\begin{displaymath}f_1[k]* f_2[k] \leftrightarrow F_1[z]F_2[z] \end{displaymath}

where denotes convolution (in this case, discrete-time convolution).
\begin{proof}This is somewhat easier (and more general) to prove for
noncausal ...
... \sum_{r=-\infty}^\infty
f_2[r]z^{-r} = F_1[z]F_2[z].
\end{eqnarray*}\end{proof}

Multiplication by $\gamma^k$
 

\begin{displaymath}\gamma^k f[k]u[k] \leftrightarrow F[z/\gamma] \end{displaymath}

Multiplication by $k$
 

\begin{displaymath}kf[k]u[k] \leftrightarrow -z \frac{d}{dz} F[z]
\end{displaymath}

Initial Value theorem
For a causal $f[k]$ ,

\begin{displaymath}f[0] = \lim_{z\rightarrow \infty} F[z] \end{displaymath}

Final Value theorem
If $F[z]$ has no poles outside the unit circle (i.e. it is stable),
\begin{displaymath}\lim_{n\rightarrow \infty} f[n] = \lim_{z\rightarrow 1}(z-1)F(z).
\end{displaymath}
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 17). Lecture 2: Properties of the Z-Transform. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/node3_1.html. This work is licensed under a Creative Commons License Creative Commons License