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- The z-transform
- Z-transform properties
As with other signal transforms, the z-transform has a number of significant properties, including ways in which changes to a signal in one domain impacts its representation in another domain. To examine these properties, we will again use the notation of a z-transform "pair." If we have
$X(z)=\sum\limits_{n=-\infty}^{\infty}x[n]z^{-n}$,
then we express the relationship between $x[n]$ and its z-transform $X(z)$ as
$x[n]\leftrightarrow X(z)$
The DTFT is a special case of the z-transform. If we have $x[n]\leftrightarrow X(z)$ and the ROC of $X(z)$ contains the unit circle, then the DTFT of $x[n]$, $X(e^{j\omega})$, is simply the z-transform $X(z)$ evaluated on the unit circle $z=e^{j\omega}$:
$X(e^{j\omega})=X(z)|_{z=e^{j\omega}}$
Given that the z-transform is simply an infinite sum, if follows then that it will be a linear operator. If we have $x_1[n]\leftrightarrow X_1(z)$ and $x_2[n]\leftrightarrow X_2(z)$, and $\alpha_1,\alpha_2\in C$, then:
$\alpha_1 x_1[n]+\alpha_2 x_2[n]\leftrightarrow \alpha_1 X_1(z)+\alpha_2 X_2(z)$However, the ROC of the new transform is the intersection of those of $X_1(x)$ and $X_2(z)$: $ROC_{\alpha_1 X_1+\alpha_2 X_2}=ROC_{X_1}\bigcap ROC_{X_2}$
If $x[n]$ and $X(z)$ are a z-transform pair, then:$x[n-m]\leftrightarrow z^{-m}X(z)$and the ROC remains the same as for $X(z)$ (with the possible addition/removal of a pole at zero and/or zero at infinity):
$\begin{align*}\sum_{n=-\infty}^{\infty} x[n-m]\, z^{-n}&= \sum_{r=-\infty}^{\infty} x[r]\, z^{-(r+m)}\\&=\sum_{r=-\infty}^{\infty} x[r]\, z^{-r} \, z^{-m} \\&=z^{-m} \sum_{r=-\infty}^{\infty} x[r]\, z^{-r}\\&=z^{-m} X(z)
\end{align*}$
If $x[n]$ and $X(z)$ are a z-transform pair, then:$z^n_0 x[n]\leftrightarrow X(\frac{z}{z_0})$To determine the ROC of the new z-transform, substitute $\frac{z}{z_0}$ for $z$ in the original ROC definition, and simplify. For example, if the ROC of $X(z)$ is $|z|\gt 1$, then the new ROC will be $|\frac{z}{z_0}|\gt 1 \rightarrow |z|\gt |z_0|$
Proof:$\begin{align*}
\sum_{n=-\infty}^{\infty} (z_0^n x[n]) z^{-n}&=\sum_{n=-\infty}^{\infty} x[n](z/z_0)^{-n}\\&=X\left(\frac{z}{z_0}\right)\end{align*}$
If $x[n]$ and $X(z)$ are a z-transform pair, then:$x^*[n] \leftrightarrow} X^*(z^*)$Proof:
$\begin{align*}\sum_{n=-\infty}^{\infty} x^*[n] z^{-n}&=\left(\sum_{n=-\infty}^{\infty} x[n] (z^*)^{-n}\right)^*\\&=X^*(z^*)
\end{align*}$
If $x[n]$ and $X(z)$ are a z-transform pair, then:$x[-n]\leftrightarrow}X(z^{-1})$With the ROC inverted: to find the ROC, substitute $\frac{1}{z}$ for $z$ in the original ROC expression and simplify.
Proof:$\begin{align*}
\sum_{n=-\infty}^{\infty} x[-n]z^{-n}&=\sum_{m=-\infty}^{\infty} x[m]z^{m}\\&=\sum_{m=-\infty}^{\infty} x[m] (z^{-1})^{-m}\\&=X(z^{-1})
\end{align*}$
If $x[n]\leftrightarrow X(z)$, $h[n]\leftrightarrow H(z)$, and $y[n]\leftrightarrow Y(z)$ all all z-transform pairs, and $y[n]=x[n]\ast h[n]$, then:
$Y(z)=X(z)H(z)$,with the ROC of $Y(z)$ being the intersection of the ROCs of $X(z)$ and $H(z)$.
$\begin{array}{c|c|c}
x[n]&X(z)&ROC \\
\hline \\\delta[n]&1&all z\\
u[n]&\frac{1}{1-z^{-1}}&|z|\gt 1 \\
\alpha^n u[n]&\frac{1}{1-\alpha z^{-1}}&|z|\gt|\alpha| \\
-\alpha^n u[-n-1]&\frac{1}{1-\alpha z^{-1}}&|z|\lt|\alpha|
\end{array}$
Questions & Answers
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Source:
OpenStax, Discrete-time signals and systems. OpenStax CNX. Oct 07, 2015 Download for free at https://legacy.cnx.org/content/col11868/1.2
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