# Convolution theorem

1. Sep 29, 2014

### kidsasd987

I am having a hard time to understand why convolution integral gives the area overlaps of the two signal functions.
if we use

http://en.wikipedia.org/wiki/Convolution#mediaviewer/File:Comparison_convolution_correlation.svg

for convolution, it is pretty obvious that one of the functions gives always 1 and therefore we just need to find another function's integral (area) and that area represents the area overlaps.

What I feel is though, this seems one special coincidence that convolution of the two function gives the overlapped area of the two functions.

if we use some arbitrary two functions, lets say

x(t)=e^2t

and

h(t)=e^(-7t+2)

then convolution is

integral from negative infity to infinity x(τ)*(t-τ)
= integral from negative infinity to infinity e^2τ*e^(-7t-7τ+2)

we are finding integral of product of the two functions, and it will likely be greater than the overlapped are of the two functions.

Is the convolution 'always' the overlapped area of two signal function

2. Sep 30, 2014

### olivermsun

No, it is not the overlapped area of two signal functions.

The picture is just depicting the unit "top hat" function and the ramp-shaped function overlapping with different offsets. The convolution at each offset is the integral over of the point-wise product of the two functions with that offset.

3. Oct 2, 2014

### analogdesign

Do you understand the principle of Superposition? One way to visualize convolution is to imagine taking your single and splitting it up into however many impulses. Then the convolution is the sum of the response of all the different impulses with the other signal or system. If you're doing a convolution in discrete math it is done in just this way. I found it more intuitive.