# Thread: Inverse Laplace of this function

1. ## Inverse Laplace of this function

Hi,

I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

Thanks for any help.

2. $\displaystyle (s+a)e^{-(a+b)t}$

3. Originally Posted by algorithm
Hi,

I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

Thanks for any help.
What exactly are you trying to find the inverse transform of? What's the X(s) doing there?

If the problem is to find the inverse transform of
$\displaystyle \frac{s + a}{s + a + b}$

Note that
$\displaystyle \frac{s + a}{s + a + b} = 1 - \frac{b}{s + a + b}$

The second term is simple enough. You may need to look up the inverse transform of 1. It is $\displaystyle \delta (t)$.

-Dan

4. Hi

Thanks.

Originally Posted by lilaziz1
$\displaystyle (s+a)e^{-(a+b)t}$
Should the "s" be "n(t)"?

Originally Posted by topsquark
What exactly are you trying to find the inverse transform of? What's the X(s) doing there?
X(s) is an arbitrary function. I want to see how it is affected in the time domain by this multiplication. For example, if I had "sX(s)", then in the time domain it would be d/dt of x(t).

5. Originally Posted by algorithm
Hi,

I've looked through the transform tables, and haven't been able to figure out how to find the inverse:

Thanks for any help.
If that is the cases you can use the convolution theorem to bring this back to the time domain. Using Topsquark's idea gives

$\displaystyle X(s)\frac{s+a}{s+a+b}=X(s)\frac{s+a+b-b}{s+a+b}=X(s)-X(s)\frac{b}{s+a+b}$

So now taking the inverse transform gives

$\displaystyle x(t) - \int_{0}^{t}e^{-(a+b)(t-\tau)}X(\tau)d\tau=x(t) - be^{-(a+b)t}\int_{0}^{t}e^{(a+b)\tau}X(\tau)d\tau$