Question

Laplace transform of
$x\left(t\right)={\sum }_{k=0}^{\infty }\delta (t-KT)$ is

• A. $\frac{1}{1-e^{sT}}$
• B. ${\sum }_{K=-\infty }^{\infty }(e^{sT}{)}^k$
• C. $\frac{1}{1+e^{-sT}}$
• D. $\frac{1}{1-e^{-sT}}$

Answer 1

The correct option is D $\frac{1}{1-e^{-sT}}$

$X\left(s\right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-st}dt={\int }_{-\infty }^{\infty }\left({\sum }_{k=0}^{\infty }\delta \right(t-KT\left)\right)e^{-st}dt$

$={\sum }_{k=0}^{\infty }\left({\int }_{-\infty }^{\infty }\delta \right(t-KT\left)e^{-st}dt\right)={\sum }_{k=0}^{\infty }{e}^{-sKT}$

$={\sum }_{K=0}^{\infty }(e^{-sT}{)}^K=\frac{1}{1-e^{-sT}},Re[s]>0$
Hence it is right sided signal.