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## Main Question or Discussion Point

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^2

and

h(t)=e^(-7

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

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^2

*t*and

h(t)=e^(-7

*t*+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