0

# Simple inverse Laplace transform?

Find the inverse Laplace transform of (4s-6)/(s^2-2s+2).

### 1 Answer by Expert Tutors

Maurizio T. | Statistics Ph.D and CFA charterholder with a true passion to teach.Statistics Ph.D and CFA charterholder wi...
5.0 5.0 (327 lesson ratings) (327)
0

Sun, I am afraid that this is not a "simple" inverse Laplace transform problem. it is not terribly difficult, but it requires quite a bit of calculations.....

and it requires that you use complex numbers. Note that the solutions of s2-2s+2 = 0 are s = 1-i and s = 1+i. Hence you can decompose your assigned expresssion as

(4s-6)/[(s-(1-i))*(s-(1+i))]

and then you need to go from here using partial fractions, but in this case you will have to use complex numbers. If I am not mistaken, you get:

4s+6/(s^2-2s+2) = (2-5i)/(s-(1+i)) + (2+5i)/(s-(1-i))

and then you go from here.. But you need to be careful as you will be using complex numbers (technically you are closer to using Fourier transforms rather than Laplace transforms)...