Question

FindL-1[s+3/(s2+6s+13)2].​

Answer 1

Answer:

Answer:

L^-1[s/[(s-b)^2+a^2] ^2] = e^bt [t sin at]/2a EQ 1

L^-1[1/ [(s-b)^2+b^2]] = e^bt[ sin at -at cos at]/2a^3 , EQ 2

L^-1[(s^2 -6s +13 )^2]=L^-1[[(s-3)^2 +2^2]^2] ,

L^-1[s/[(s-3)^2+2^2] = e^3t[t sin 2t]/4 , EQ 1

L^-1[-6/[(s-3)^2+2^2]=-6e^3t[ sin 2t -2t cos 2t]/16 , EQ 2

F[t] = EQ 1+ EQ 2

F[t] = [e^3t/16][(4t-6) sin 2t + 12t cos 2t }