## The problem

Find the Laplace Transform of the following: $x(t)=e^{-2t}u(t)$

## Solution

Now, from other classes we have become used to simply looking up a simple problem like this in a Laplace table. This may or may not be valid on exam 3 and the final, so we have to know how to solve the problem using the definition of a Laplace Transform.

$X(s)=\int_{-\infty}^{+\infty}x(t)e^{-st}dt$

     $=\int_{0}^{\infty}e^{-2t}e^{-st}dt$

     $=\int_{0}^{\infty}e^{-(2+s)t}dt$

     $=\int_{0}^{\infty}e^{-(2+a+j\omega)t}dt$


We can now see that if if $2+a\leq 0$, the integral diverges and if $2+a>0$:

$X(s)=\frac{e^{-(2+a)t}e^{-j\omega t}}{-(2+s)} \bigg| _0^{\infty}$

     $=0-\frac{1}{-2-s}$
$=\frac{1}{2+s}$


## Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.