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Relation to the z Transform

The Laplace transform is used to analyze continuous-time systems. Its discrete-time counterpart is the $ z$ transform:

$\displaystyle X_d(z) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\sum_{n=0}^\infty x_d(nT) z^{-n}
$

If we define $ z=e^{sT}$ , the $ z$ transform becomes proportional to the Laplace transform of a sampled continuous-time signal:

$\displaystyle X_d(e^{sT}) = \sum_{n=0}^\infty x_d(nT) e^{-snT}
$

As the sampling interval $ T$ goes to zero, we have

\begin{eqnarray*}
\lim_{T\to 0} X_d(e^{sT})T &=&
\lim_{T\to 0}
\sum_{n=0}^\infty x_d(t_n)\, e^{-st_n} T \\
&=& \int_{0}^\infty x_d(t)\, e^{-st} dt \mathrel{\stackrel{\mathrm{\Delta}}{=}}X(s)
\end{eqnarray*}

where $ t_n\mathrel{\stackrel{\mathrm{\Delta}}{=}}nT$ and $ \Delta t \mathrel{\stackrel{\mathrm{\Delta}}{=}}t_{n+1} - t_n = T$ .

In summary,

$\textstyle \parbox{0.8\textwidth}{the {\it z} transform\ (times the sampling interval $T$) of a
discrete time signal $x_d(nT)$\ approaches, as $T\to0$, the Laplace Transform\ of
the underlying continuous-time signal $x_d(t)$.}$

Note that the $ z$ plane and $ s$ plane are related by

$\displaystyle \zbox{z = e^{sT}}
$

In particular, the discrete-time frequency axis $ \omega_d \in(-\pi/T,\pi/T)$ and continuous-time frequency axis $ \omega_a \in(-\infty,\infty)$ are related by

$\displaystyle \zbox{e^{j\omega_d T} = e^{j\omega_a T}}
$

For the mapping $ z=e^{sT}$ from the $ s$ plane to the $ z$ plane to be invertible, it is necessary that $ X(j\omega_a )$ be zero for all $ \vert\omega_a \vert\geq \pi/T$ . If this is true, we say $ x(t)$ is bandlimited below half the sampling rate. As is well known, this condition is necessary to prevent aliasing when sampling the continuous-time signal $ x(t)$ at the rate $ f_s=1/T$ to produce $ x(nT)$ , $ n=0,1,2,\ldots\,$


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``The Laplace Transform'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2020-06-27 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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