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

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