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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) \isdef \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

$\displaystyle \lim_{T\to 0}
X_d(e^{sT})T =
\lim_{\Delta t\to 0}
\sum_{n=0}^\infty x_d(t_n) e^{-st_n} \Delta t
= \int_{0}^\infty x_d(t) e^{-st} dt
\isdef X(s)

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

In summary,

$\textstyle \parbox{0.8\textwidth}{the {\it z} transform\ (times the sampling in...
...o0$, the Laplace transform\ of
the underlying continuous-time signal $x_d(t)$.}$

Note that the $ z$ plane and $ s$ plane are generally 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 to 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\,$ (see [84, Appendix G]).

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