User:Cplusplusboy/sandbox

Consider a continuous time signal $x(t)$. Its one sided Laplace transform is defined as : $L\{x(t)\}\equiv X(s)\equiv \int _{0}^{\infty }{x(t)e^{-st}dt}$

If the continuous time signal $x(t)$ is uniformly sampled with a train of impulses to get a discrete time signal $x^{*}(k)=x(kT$), then it can be represented as : $x^{*}(k)=x(kT)=\sum _{k=0}^{\infty }{x(t).\delta (t-kT)}$ where $T$ is the sampling interval.

Now the Laplace transform of the sampled signal (discrete time) is called Star_transform and is given by : ${\begin{array}{l l l l l l}L\{x^{*}(k)\}&=&X^{*}(s)&=&\int _{0}^{\infty }{\sum _{k=0}^{\infty }{x(t).\delta (t-kT)}e^{-st}dt}\\&=&\sum _{k=0}^{\infty }{x(kT).e^{-kTs}},&&{\text{by sifting property}}\\&=&\sum _{k=0}^{\infty }{x^{*}(k).z^{-k}},z=e^{sT}\\\left.L\{x^{*}(k)\}\right|_{s={\frac {\ln {(z)}}{T}}}&=&\left.X^{*}(s)\right|_{s={\frac {\ln {(z)}}{T}}}&=&Z\{x^{*}(k)\}\end{array}}$

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 Failed to parse (syntax error): {\displaystyle s = \frac{\ln{(z)}}{T}}} .