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91Ó°ÊÓ

Determine the inverse Laplace transform of the given function.

2s-1s2-4s+6

Short Answer

Expert verified

The solution is

L−12(s−2)+3(s−2)2+(2)2(t)=2e2tcos2t+32e2tsin2t

Step by step solution

01

Given Information

The given function value in s domain is2s-1s2-4s+6

02

Use partial fractions

Factorize the denominator of the given function as

s2−4s+6=s2−4s+4+2=s-12+22

The function becomes.2s-1(s−2)2+(2)2Decompose the function as:

2s-1(s−2)2+(2)2=2(s−2)+3(s−2)2+(2)2=2(s−2)(s−2)2+(2)2+322(s−2)2+(2)2

Take inverse Laplace transform using L−1s−a(s−a)2+b2(t)=eatcosbtand L−1b(s−a)2+b2(t)=eatsinbtas:

L−12(s−2)+3(s−2)2+(2)2(t)=2e2tcos2t+32e2tsin2t

Therefore,the required inverse Laplace transform is:

L−12(s−2)+3(s−2)2+(2)2(t)=2e2tcos2t+32e2tsin2t

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