Time reversal of a signal in convolution

[Ershadh]

While performing convolution of two signals why do we time reverse one of the signals? What would happen if we do not time reverse one of these?

 

[Prashanth.vinnakota]

This is answered in another domain.

Any function or signal can be expressed in time or frequency domain. When you do the multiplication operation between two such signals it is same as doing the convolution operation in the other domain.

Consider two signals f(t) and g(t) in time domain. If you need the convolution of both i.e f(t)*g(t) you will be doing the integral of f(p)g(p-t) which is the definition of the convolution.

Instead you can do the multiplication of F(w) and G(w) which are the fourier transforms of the signals f(t) and g(t) respectively.

Answer to your question: Take the Inverse Fourier Transform of F(w)G(w) which is exactly the result of f(t)*g(t).
To maintain spacial equivalence in two domains we do the (time reversal + shift) operation on one signal and do the integration.

 

[permute]

if a signal is not reversed, you will have "correlation".

The reverse is explained as follows:
y(t) = integral(-inf to inf) of x(tau) h(t-tau) dtau
this represents sweeping over all of time, and determine the effect of the input at each instant in time on the output at time "t". x(tau) is the value of the input at time tau. (t-tau) is the difference in time between now "t", and tau. h(t-tau) describes how much an impulse would affect the output after this interval -- it would be the effect of the system on an input at time tau on the output at time "t".

causal systems will have impulses defined on 0 to inf for this reason -- if (t-tau) in the above is negative, and h(t-tau) is not 0, then it means an input in the "future" affects the current input.

Keep in mind though, that this only sounds bad when "t" corresponds to "time". convolution as an operator can be used on images, in which case pixels to both the right and left will be known.