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Consider a continuous time signal <math>x(t)</math>. Its one sided Laplace transform is defined as :

<math>L\{x(t)\} \equiv X(s) \equiv \int_0^{\infty}{x(t)e^{-st}dt}

</math>

If the continuous time signal <math>x(t)</math> is uniformly sampled with a train of impulses to get a discrete time signal <math>x^{*}(k) = x(kT</math>), then it can be represented as :

<math>x^{*}(k) = x(kT) = \sum_{k=0}^{\infty}{x(t).\delta(t-kT)}</math>

where <math>T</math> is the sampling interval.

Now the Laplace transform of the sampled signal (discrete time) is called [[Star_transform]] and is given by :

<math> \begin{array}{l l l l l l}

L\{x^{*}(k)\} & = & X^{*}(s) & = & \int_0^{\infty}{\sum_{k=0}^{\infty}{x(t).\delta(t-kT)} e^{-st}dt} \\

& = & \sum_{k=0}^{\infty}{x(kT).e^{-kTs}}, & & \text{by sifting property} \\

& = & \sum_{k=0}^{\infty}{x^{*}(k).z^{-k}}, z = e^{sT} \\

\left. L\{x^{*}(k)\}\right|_{s = \frac{\ln{(z)}}{T}} & = & \left.X^{*}(s)\right|_{s = \frac{\ln{(z)}}{T}} & = & Z\{x^{*}(k)\}

\end{array} </math>

It can be seen that the [[Laplace_Transform]] of an impulse sampled signal is the called the [[Star_transform]] and is the same as the [[Z_Transform]] of the corresponding sequence when <math>s = \frac{\ln{(z)}}{T}</math>.

<ref name=ogata_dtcs>{{cite book|last=Ogata|first=Katsuhiko|title=Discrete-Time Control Systems|publisher=Pearson Education|___location=India|isbn=81-7808-335-3|pages=75-77}}</ref>

<references/>