The problem

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


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.

Buyue Zhang