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}

ediculeN 2021-02-24 Answered
For the function
f(t)=et
g(t)=e2t
0t<
compute in two different ways:
a) By directly evaluating the integral in the defination of fg
b) By computing L1{F(s)G(s)} where F(s)=L{f(t)} and G(s)=L{g(t)}
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Expert Answer

SabadisO
Answered 2021-02-25 Author has 108 answers
Step 1 
 (a) 
Formula used: 
Let f and g be two functions. The convolution of f with g is defined as (fg)(t)=0tf(w)g(tw)dw
Calculation: 
Here given that f(t)=et and g(t)=e2t
Then the convolution of f with g is given below:
(fg)(t)=0tf(τ)g(tτ)dτ
=0t(ew)(e2(tw))dw
0t(e3w)(e2w)dw
=13e2t[e3w]0t
=13e2t[e3t1]
Step 2
 (b) 
Here, the Laplace transform of f is L(f(t))=L(et)=1s1
The Laplace transform of g is L(g(t))=L(e2t)=1s+2
Now, find the inverse of the convolution of f with g as follows: 
(fg)(t)=L1{F(s)G(s)}
=L1{1s1×1s+2}
=L1{1s1}L1{1s+2}
=ete2t
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