Find inverse Laplace transform, F(s)=\frac{10}{s^3+4s^2+9s+10}

illusiia

illusiia

Answered question

2021-09-08

Find inverse Laplace transform
F(s)=10s3+4s2+9s+10

Answer & Explanation

wheezym

wheezym

Skilled2021-09-09Added 103 answers

Step 1
F(s)=10s3+4s2+9s+10
Now using partial fraction of 10s3+4s2+9s+10
s3+4s2+9s+10=(s+2)(s2+2s+5)
10(s2)(s2+2s+5)=a0s+2+a2s+a1s2+2s+5
10=a0(s2+2s+5)+(a2s+a1)(s+2)
10=s2(a2+2)+s(a1+2a2+4)+(2a1+10)s˙(3)
Now equate coefficients
10=10+2a1,4+2a2+a1=0
2+a2=0
a2=2,a1=0,a0=2(4)
from 2 ,4
10(s+2)(s2+2s+5)=2s+22ss2+2s+5(5)
taking inverse laplace both sides we get
L1{10(s+2)(s2+2s+5)}=L1{2s+2}2L1{ss2+2s+5}
from (6)
L1=L1{2s+2}=2L1{1s+2}=2e2t   (7)
L2=L1{ss2+2s+5}=L1{s+11s2+2s+1+4}=L1{s+11(s+1)2+4}
=L1{s+11(s+1)

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