# Inverse Laplace : with Convolution

1. Nov 15, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
Use the convolution property to obtain the inverse Laplace transform of
F(s)= e-3s * ((3s+15)/s2+25)

2. Relevant equations

3. The attempt at a solution
= (3*(s/s2+52) + 15*(1/s2+52)) *e-3s

Using table of Laplace:
3*(s/s2+52) = 3*cos(5*t) = T7
15*(1/s2+52) = 15/5*sin(5*t) =T18
e-3s = s(t-a) = s(t-3) (T16)

=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)

Convolution : int 0 to t (f(u)*g(t-u))du
integral 0 to t ((3*cos(5*u)+15/5*sin(5*u))*s((t-u)-3)) du

Something is wrong in my work here, and I cant find where. :([/SUP]

2. Nov 15, 2015

### LCKurtz

None of the above equalities are true; you have functions of s "equal" to functions of t.

And you are apparently using s for two different things. With 156 posts, it is past time for you to learn to use latex. It isn't worth the effort to decipher what you have written.

3. Nov 16, 2015

### vela

Staff Emeritus
Is the asterisk supposed to denote multiplication or convolution?

You need to use parentheses as appropriate. Assuming the asterisk is supposed to be multiplication, what you wrote means
$$F(s) = e^{-3s}\left(\frac 3s + \frac{15}{s^2} + 25\right),$$ which is quite different from what you intended.

Is there some reason you're not simplifying 15/5?

Why do you think something's wrong?

Besides the problems LCKurtz pointed out with what you've written, did you notice that when you wrote "=(3*cos(5*t)+15/5*sin(5*t))*s(t-3)," you never said what it was equal to and that you're using * to mean both multiplication and convolution? You should make an effort to be less sloppy, especially when you're trying to communicate with others.