# Inverse Laplace transform problem 1

• Jan 16th 2011, 03:17 AM
chthon
Inverse Laplace transform problem 1
Hello,

I got the following differential equation :
$\displaystyle E_2'(t)+2{}E_2(t) = E_1(t) + E_1'(t)$

Transforming this into a Laplace equation and computing the transfer function, I get :
$\displaystyle H(s) = {{s+1} \over {s+2}}$

Where I get into trouble is to obtain the inverse transformation. It seems probably silly. I do not have any problem solving more complex equations using partial fractions, but trying, this does not seem to apply here :
$\displaystyle {{s+1} \over {s+2}} = {A \over {s+2}}$

What am I missing here?

Thanks,

Jurgen
• Jan 16th 2011, 04:06 AM
FernandoRevilla
Take into account that:

$\displaystyle \displaystyle\lim_{s \to{+}\infty}{\dfrac{s+1}{s+2}}\neq 0$

Fernando Revilla
• Jan 16th 2011, 04:41 AM
chisigma
... and also that is...

$\displaystyle \displaystyle \frac{s+1}{s+2}= 1-\frac{1}{s+2}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 16th 2011, 05:04 AM
FernandoRevilla
... and also we deduce that

$\displaystyle \mathcal{L}^{-1}\left\{{ \dfrac{s+1}{s+2} }\right\}$

does not exist. :)

Fernando Revilla
• Jan 16th 2011, 06:48 AM
chthon
Reply : Inverse Laplace transform problem 1
Why not? I have controlled the answer above yours, which says
$\displaystyle \displaystyle \frac{s+1}{s+2}= 1-\frac{1}{s+2}$

which then translates to

$\displaystyle \delta{(t)}-e^{-2t}$
• Jan 16th 2011, 07:56 AM
chisigma
It is important to observe that $\displaystyle H(s)$ is a Transfer function, i.e. the ratio of the L-transforms of the 'output signal' $\displaystyle e_{2} (t)$ and the 'input signal' $\displaystyle e_{1} (t)$ . In this case the $\displaystyle H(s)$ is a [simple] first order 'high pass filter'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 16th 2011, 08:14 AM
FernandoRevilla
Quote:

Originally Posted by chthon
$\displaystyle \delta{(t)}-e^{-2t}$

$\displaystyle \dfrac{s+1}{s+2}$

into simple fractions to find its inverse as a piecewise continuous and exponential order function not as a distribution. For that reason I talked about the non zero limit.

Fernando Revilla