Find the inverse Laplace transform of the given function by using the convolution theorem. {F}{left({s}right)}=frac{s}{{{left({s}+{1}right)}{left({s}^{2}+{4}right)}}}

Tammy Todd 2021-01-13 Answered
Find the inverse Laplace transform of the given function by using the convolution theorem. F(s)=s(s+1)(s2+4)
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Expert Answer

Viktor Wiley
Answered 2021-01-14 Author has 84 answers

Step 1
the given function is F(s)=s(s+1)(s2+4)
Let G(s)=1s+1  and  H(s)=ss2+4
Since L1{1sa}=eat  and  L1{1s2+a2}=cosat
step 2
That implies,
L1{G(s)}=L1{1s+1}=et and
L{H(s)}=L1{1s2+4}=cos2t
g(t)=et  and  h(t)=cos2t
Now, F(s)=s(s+1)(s2+4)
Step 3
By using the convolution theorem,
If F(s)=L{f(t)}  and  G(s)=L{g(t)} both exists for s>a0 ,then
H(s)=F(s)G(s)
=L{h(t)},s>a
Where h(t)=0tf(tτ)g(τ)dτ=0tf(τ)g(tτ)dτ
Step 4
That implies,
F(s)=G(s)H(s)
=s(s+1)(s2+4)
=L{f(t)}
=L{0tg(tτ)h(τ)dτ}
Thus , F(s)=L{0tg(tτ)h(τ)dτ}
Step 5
Taking the inverse Laplace transform,
L1{F(s)}=0tg(tτ)h(τ)dτ
=0te(tτ)cos2τdτ
Therefore ,

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