Answer to Question #117355 in Math for Akshay

Question #117355
Find the Laplace transform of f(t)=e^-t,where f(t+1)=f(t)
1
Expert's answer
2020-05-21T16:28:24-0400

If f(t+T)=f(t),f(t+T)=f(t), then

F(s)=L(f(t))=0Testf(t)dt1esTF(s)=L(f(t))={\displaystyle\int_{0}^T e^{-st}f(t)dt\over 1-e^{-sT}}

f(t+1)=f(t), f(t)=etf(t+1)=f(t),\ f(t)=e^{-t}




F(s)=L(f(t))=01estetdt1es(1)=F(s)=L(f(t))={\displaystyle\int_{0}^1 e^{-st}e^{-t}dt\over 1-e^{-s(1)}}=

=1(1e1)(s+1)[e(s+1)t]10==-{1\over (1-e^{-1})(s+1)}\bigg[e^{-(s+1)t}\bigg]\begin{matrix} 1\\ 0 \end{matrix}=

=1e(s+1)(1e1)(s+1)={1-e^{-(s+1)}\over (1-e^{-1})(s+1)}


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