Is this the correct inverse laplace transform

[Ready2GoXtr]

Homework Statement

Well i did all my work but the forum logged me out and when i submitted it I lost it all
because it wanted me to login again >_<
anyways i need to find the inverse laplace transform of
F(s)=[12*s^2 - (28+2*sqrt(3))*s + 26 + 6*sqrt(3)]/[(s-3)(s^2-2s+2)]

Homework Equations

The Attempt at a Solution

F(s)=K1/(s-3) + K2/(s-1-j) + K2^*/(s-1+j) Note * is for conjugate
K1= [12(3)^2 - (28+2*sqrt(3))*(3) + 26 + 6*sqrt(3)]/[(3)^2-2(3)+2]=50/5=10
K1=10
K2=[12(1+j)^2 - (28+2*sqrt(3))*(1+j) + 26 + 6*sqrt(3)]/[(1+j+3)*(1+j-1+j)]
=-1.0231-.3603j
K2=-1.0231-.3603j
K2^*=-1.0231+.3603j


F(s)=10/(s-3) + (-1.0231-.3603j)/(s-1-j) + (-1.0231+.3603j)/(s-1+j)

note: (-1.0231-.3603j)=1.0846$$\angle$$19.4=1.0846*exp(j*19.4)
note: (-1.0231+.3603j)=1.0846$$\angle$$-19.4=1.0846*exp(-j*19.4)
doing the inverse laplace transform
I think the second exponent for the last two terms come from translation in frequency operation for laplace transforms but i could be wrong
it says
for f(t): exp(-a*t)*f(t)
for F(s):F(s+a)

f(t)= 10*exp(3*t) + 1.0846*exp(j*19.4)*exp(-(-1-j)t) + 1.0846*exp(-j*19.4)*exp(-(-1+j)t)

if this is correct how do i get rid of the j's here?

 

[vela]

When you calculated K2, you accidentally changed (s-3) to (s+3) in the denominator.

You can use e^jθ=cos θ + j sin θ to eliminate the j's, but you might find it easier and a lot less tedious to avoid them in the first place. When you do the partial fractions expansion, use the form

$$F(s) = \frac{A}{s-3} + \frac{B s + C}{s^2-2s+2}$$

The constants A, B, and C will all be real. To find the inverse Laplace transform of the second term, express it in the form

$$\frac{B s + C}{s^2-2s+2} = \frac{B(s-1)}{(s-1)^2+1} + \frac{B+C}{(s-1)^2+1}$$

and use the shifting property you mentioned.