# Convolve exponential functions

• Oct 29th 2012, 01:59 PM
trojsi
Convolve exponential functions
hi,
I read about convolution and the popular examples of rectangular pulses and exponentials with square functions but
unfortunately I did not find any help on convolving two exponential functions.

I need to convolve: $x(t) = e^{2t} [u(t-1)-u(t-4)]$ and $x(t) = 2e^{-t} [u(t+1)-u(t-5)]$

Is the solution much different from the rectangular pulses derivation?

I sketched each signal to identify the t 'limits' of each one due to the unit steps. What is the closed form solution exactly?

I would really appreciate any guidance thanks.
• Oct 29th 2012, 08:29 PM
chiro
Re: Convolve exponential functions
Hey trojsi.

Did you try and setup the convolution integral directly and just use that to evaluate the actual integral? Are you just trying to find the convolution or the frequencies for something like a Laplace Transform?
• Oct 29th 2012, 10:55 PM
trojsi
Re: Convolve exponential functions
Hi chiro,
I need to apply convolution to the pair of signals and sketch the output. Then I also have to apply the commutative property of convolution.
• Oct 29th 2012, 11:28 PM
chiro
Re: Convolve exponential functions
Well the convolution theorem is given by Integral (0 to t) f(t-u)g(u)du. Did you compute this integral and if so what did you get?
• Oct 30th 2012, 01:48 AM
trojsi
Re: Convolve exponential functions
I need to understand how to decide which limits to take for the integral. Attached find my sketches till now.Attachment 25465
• Oct 30th 2012, 10:39 AM
trojsi
Re: Convolve exponential functions
attached is the updated version with all the cases if I understood well the proper method. I am confused how to the Integral limits for each integral.Attachment 25470
• Oct 30th 2012, 04:02 PM
chiro
Re: Convolve exponential functions
The standard convolution operation has limits from 0 to t. If your signal is 0 in particular regions, just split up the integral to get the non-zero parts and note where the limits are, but to start off with, your standard convolution integral of two general functions (or signals) is Integral (0 to t) f(t-u)g(u)du or Integral (0 to t) g(t-u)f(u)du.

If the signal is largely missing across chunks of the real line, then just remember that Integral (a to c) = Integral (a to b) + Integral (b to c) where a < b < c and if one region is zero then it can disappear.