Domain of convolution vs. domain of Fourier transforms

[skynelson]

TL;DR Summary

Convolving two signals, g and h, of lengths X and Y respectively, results in a signal with length X+Y-1. How can the length of an output signal of convolution be different from the input signals , given the contents of the Convolution Theorem? Thank you!

Convolving two signals, g and h, of lengths X and Y respectively, results in a signal with length X+Y-1. But through convolution theorem, g*h = F^{-1}{ F{g} F{h} }, where F and F^{-1} is the Fourier transform and its inverse, respectively. The Fourier transform is unitary, so the output signal is the same length as the input signal for that operation. The prescribed pointwise multiplication also requires signals of the same length (I believe the smaller will be padded to match the larger if needed).

How can the length of an output signal of convolution be different from the input signals , due to the Convolution Theorem?

 

[martinbn]

Fourier is unitary for the ##L^2## norm, not for the length of the support. Consider this, the Fourier transform of a function of compact support will not have compact support.

 

[jasonRF]

You aren't being careful enough with your statement of the convolution theorem. Recall that the Discrete Fourier Transform (DFT) assumes that the input signal is one period of an infinitely long periodic signal. So using the DFT computes the so-called circular convolution of the two periodic signals. If you want the linear convolution (which has the length you are stating) then you will need to zero-pad both signals so that you do not get any of the artifacts from the periodic extension of the signals. At the end of the day you should find that in order to compute the linear convolution via the DFT, you need to zero pad both signals so that they each have length## \geq X+Y-1##.

jason

EDIT: I am obviously assuming you are dealing with discrete-time signals because otherwise your length question doesn't make sense. If my assumption is incorrect then your question might not make sense. In any case, it is usually best to give such details in your question.

 

[skynelson]

Thank you for your helpful reply. Indeed, this clears things up. Yes, I am referring to a discrete transform. Much appreciated!