For the function f(t)=e^t g(t)=e^{-2t} 0leq t < infty compute in two different ways: a) By directly evaluating the integral in the defination of f cdot g b) By computing L^{-1}left{F(s)G(s)right} text{ where } F(s)=Lleft{f(t)right} text{ and } G(s)=Lleft{g(t)right}

For the function f(t)=e^t g(t)=e^{-2t} 0leq t < infty compute in two different ways: a) By directly evaluating the integral in the defination of f cdot g b) By computing L^{-1}left{F(s)G(s)right} text{ where } F(s)=Lleft{f(t)right} text{ and } G(s)=Lleft{g(t)right}

Question
Laplace transform
asked 2021-02-24
For the function
\(f(t)=e^t\)
\(g(t)=e^{-2t}\)
\(0\leq t < \infty\)
compute in two different ways:
a) By directly evaluating the integral in the defination of \(f \cdot g\)
b) By computing \(L^{-1}\left\{F(s)G(s)\right\} \text{ where } F(s)=L\left\{f(t)\right\} \text{ and } G(s)=L\left\{g(t)\right\}\)

Answers (1)

2021-02-25
\(\text{Step 1 }\)
\(\text{ (a) }\)
\(\text{Formula used: }\)
\(\text{Let f and g be two functions. The convolution of f with g is defined as } (f*g)(t)=\int_0^t f(w)g(t-w)dw\)
\(\text{Calculation: }\)
\(\text{Here given that } f(t)=e^t \text{ and } g(t)=e^{-2t}\)
\(\text{Then the convolution of f with g is given below:}\)
\((f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau\)
\(=\int_0^t (e^{w})(e^{-2(t-w)})dw\)
\(\int_0^t (e^{3w})(e^{-2w})dw\)
\(=\frac{1}{3}e^{-2t}\left[e^{3w}\right]_0^t\)
\(=\frac{1}{3}e^{-2t}\left[e^{3t}-1\right]\)
\(\text{Step 2}\)
\(\text{ (b) }\)
\(\text{Here, the Laplace transform of f is }L(f(t))=L(e^t)=\frac{1}{s-1}\)
\(\text{The Laplace transform of g is } L(g(t))=L(e^{-2t})=\frac{1}{s+2}\)
\(\text{Now, find the inverse of the convolution of f with g as follows: }\)
\((f*g)(t)=L^{-1}\left\{F(s)G(s)\right\}\)
\(=L^{-1}\left\{\frac{1}{s-1} \times \frac{1}{s+2}\right\}\)
\(=L^{-1}\left\{\frac{1}{s-1}\right\} * L^{-1}\left\{\frac{1}{s+2}\right\}\)
\(=e^t * e^{-2t}\)
0

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