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 starred transform and is given by : Failed to parse (syntax error): {\displaystyle 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}} \Failed to parse (syntax error): {\displaystyle left. L\{x^{*}(k)\}\right|_{s = \frac{\ln{(z)}}{T}} = \left.X^{*}(s)\right|_{s = \frac{\ln{(z)}}{T}} = Z\{x^{*}(k)\}} It can be seen that the Laplace_Transform of an impulse sampled signal is the called the Starred_Transform and is the same as the Z_Transform of the corresponding sequence when {{{1}}}{T}}}.