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