# Thread: Tricky Inverse Laplace! Convolution perhaps?

1. ## Tricky Inverse Laplace! Convolution perhaps?

Hello All,

I have been working at the following for a while, trying to find the inverse of it, my working to date is shown below, although I suspect I may be approaching it wrong.

$\displaystyle F(s) = \frac{3e^{-3s}}{s(s+3)}$

My train of thought so far is as follows;

By using the convolution theorem and treating
$\displaystyle \mathcal{L}^{-1}(F(s)G(s)) = \int^{t}_{0} f(t)g(\tau-t)dt$

$\displaystyle G(s) = \frac{3}{s}$
$\displaystyle F(s) = \frac{e^{-3s}}{s+3}$
from identity

$\displaystyle \mathcal {L}[H(t-a)y(t-a)]=e^{-as}Y(s)$

let $\displaystyle Y(s)=\frac{1}{s+3}$
therefore
$\displaystyle {L}[H(t-3)(e^{-3t}-3)]=\frac{e^{-3s}}{s+3}$

$\displaystyle {L}^{-1}\left\{\frac{3e^{-3s}}{s\left(s+3\right)}\right\}= 3\int_0^t H(t-3)(e^{-3t}-3)dt$

Any solutions or help would be appreciated, or pointers as to any possible errors.

Thanks, Alex.

2. Originally Posted by arickard1988
Hello All,

I have been working at the following for a while, trying to find the inverse of it, my working to date is shown below, although I suspect I may be approaching it wrong.

$\displaystyle F(s) = \frac{3e^{-3s}}{s(s+3)}$

My train of thought so far is as follows;

By using the convolution theorem and treating
$\displaystyle \mathcal{L}^{-1}(F(s)G(s)) = \int^{t}_{0} f(t)g(\tau-t)dt$

$\displaystyle G(s) = \frac{3}{s}$
$\displaystyle F(s) = \frac{e^{-3s}}{s+3}$
from identity

$\displaystyle \mathcal {L}[H(t-a)y(t-a)]=e^{-as}Y(s)$

let $\displaystyle Y(s)=\frac{1}{s+3}$
therefore
$\displaystyle {L}[H(t-3)(e^{-3t}-3)]=\frac{e^{-3s}}{s+3}$

$\displaystyle {L}^{-1}\left\{\frac{3e^{-3s}}{s\left(s+3\right)}\right\}= 3\int_0^t H(t-3)(e^{-3t}-3)dt$

Any solutions or help would be appreciated, or pointers as to any possible errors.

Thanks, Alex.
Can't you just use $\displaystyle LT^{-1}\left[ e^{-as} G(s) \right] = H(t - a) g(t - a)$ where $\displaystyle g(t) = LT^{-1} [G(s)]$.

3. fantastic,

cool, using that;

$\displaystyle \mathcal{L}^{-1}(F(s))=f(t)=3H(t-3)[1-e^{-(t-3)}]$

However, using Matlab to give an answer, it gives something different, but similar; the following:

$\displaystyle f(t)=-3H(t-1)(\frac{e^{3-3t}}{\frac{8}{3}})$

I am not quite sure where I am making an error, please give a pointer.

Thanks, Alex.

4. Originally Posted by arickard1988
fantastic,

cool, using that;

$\displaystyle \mathcal{L}^{-1}(F(s))=f(t)=3H(t-3)[1-e^{-(t-3)}]$

However, using Matlab to give an answer, it gives something different, but similar; the following:

$\displaystyle f(t)=-3H(t-1)(\frac{e^{3-3t}}{\frac{8}{3}})$

I am not quite sure where I am making an error, please give a pointer.

Thanks, Alex.
You will need to show your working because I get $\displaystyle H(t - 3) \left(1 - e^{-3(t-3)}\right)$ (and this is the correct answer ).

Your Matlab result is wrong (probbaly because of a data entry error).

5. Fantastic,

Thanku, I have looked over the syntax and corrected it. I simply had the factor of 3 in the wrong place, al sorted now.

Alex