# Inverse Laplace Transform

• Dec 13th 2008, 07:35 PM
petition Edgecombe
Inverse Laplace Transform
Find the IVL of $\displaystyle \frac{(s-2)e^{-s}}{s^2-4s+3}$.

I am not understanding how my textbook gets the answer $\displaystyle u_1(t)e^{2(t-1)}cosh(t-1)$

I end up with

$\displaystyle e^{-s}\left(\frac{1}{2(s-1)}+\frac{1}{2(s-3)}\right)$

$\displaystyle =u_1(t)\left(\frac{1}{2}e^{(t-1)}+\frac{1}{2}e^{3(t-1)}\right)$
• Dec 13th 2008, 08:01 PM
mr fantastic
Quote:

Originally Posted by petition Edgecombe
Find the IVL of $\displaystyle \frac{(s-2)e^{-s}}{s^2-4s+3}$.

I am not understanding how my textbook gets the answer $\displaystyle u_1(t)e^{2(t-1)}cosh(t-1)$

I end up with

$\displaystyle e^{-s}\left(\frac{1}{2(s-1)}+\frac{1}{2(s-3)}\right)$

$\displaystyle =u_1(t)\left(\frac{1}{2}e^{(t-1)}+\frac{1}{2}e^{3(t-1)}\right)$

$\displaystyle = u_1(t) e^{2(t-1)} \left(\frac{1}{2}e^{-(t-1)}+\frac{1}{2}e^{(t-1)}\right) \, ....$
• Dec 13th 2008, 08:31 PM
petition Edgecombe
OK, so my answer wasn't wrong, it just wasn't simplified. I've got to remember my hyperbolic functions.