Find the inverse Laplace transforms of FS S2 5 S 6S 4 S

Find the inverse Laplace transforms of F_(S) = S^2 + 5 S + 6/(S + 4) (S + 1)^2 F_(S) = 8(S + 1) (S + 3)/S(S + 4) (S + 1) F_(S) = 5 S^2 +7 S + 29/S (S^2 + 4 S + 29) F_(S) = 5/S(S + 1) (S^2 + 6 S + 10)

Solution

1) f(s)=[(s^2+5s+6)/[(s+4)(s+1)^2]]=a/(s+4)+b/(s+1)+c/(s+1)^2

s^2+5s+6=a(s+1^2)+b(s+4)(s+1)+c(s+4)

s^2+5s+6=s^2(a+b)+s(2a+5b+c)+(a+4b+4c)

a+b=0

2a+5b+c=5

a+4b+4c=6

solve above equations

a=-1.5

b=1.5

c=0.33

f(t)=-1.5e^(-4t)+1.5e(-t)+0.33.t.e(-t)

2) f(s)=[8(s+1)(s+3)]/[s(s+1)(s+4)]=a/s+b/(s+1)+c/(s+4)

8(s^2+4s+3)=a(s^2+5s+4)+b(s^2+4s)+c(s^2+s)

a+b+c=8

5a+4b+c=32

4a=24

solve above equations

a=6

b=0

c=2

f(t)=6+2*e^(-4t)