Find Laplace transform for frac{s+8}{s^2+4s+5}

ruigE 2021-01-31 Answered
Find Laplace transform for
s+8s2+4s+5
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Expert Answer

Brittany Patton
Answered 2021-02-01 Author has 100 answers

Step 1
Formula Used:
L{eatsinbt}=b(sa)2+b2
L{eatcosbt}=sa(sa)2+b2
Step 2
Consider F(s)=s+8s2+5s+5
=s+8(s+2)2+1
=s+2+6(s+2)2+1
=(s+2)(s+2)2+1+6(s+2)2+1
Step 3
Consider F(s)=(s+2)(s+2)2+1+6(s+2)2+1
We now apply Inverse Laplace Transform, we get
L1{F(s)}=L1{(s+2)(s+2)2+1}+6L1{1(s+2)2+1}
 f(t)=e2tcost+6e2tsint
=e2t(cost+6sint)
Step 4
ANSWER
: The inverse Laplace Transform of F(s) is
f(t)=e2t(cost+6sint)

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