I am trying to understand where does the convolution fit in Discrete Fourier Transform. I know that convolution is producing a third signal from two other signals. I also know that DFT transforms one signal from time to frequency domain.

So, what is the relationship between the two (if any)?

  • $\begingroup$ In Fourier space convolution is a simple multiplication, so to make convolution of two signals with equal length you should make their FFT, multiply them, make inverse FFT of product. Absolutely correlated signals will give you a large peak in zero shift value of convolution. $\endgroup$
    – Eddy_Em
    Apr 3, 2013 at 7:28

1 Answer 1


FFT and DFT are a faster way to implement convolution. Since convolution in time domain corresponds to multiplication in the Fourier domain, you replace the sliding-window based implementation of convolution (in time domain) by a pointwise multiplication (in Fourier domain). This avoids you an extra inner loop over the input data and saves time.

With respect to boundary conditions, there is a minor difference with finite-length signals. Using the convolution theorem to compute a convolution product in the Fourier domain implicitely assumes that the input data is periodic, i.e., when you reach the right end of the signal you re-enter by its leftmost part. This is also known as circular convolution. If you need to have a convolution of explicitely non-periodic signals, you can add zeros at the beginning and end of your signal to avoid these border effects.

  • $\begingroup$ It should be mentioned that DFT implements the circular convolution unless zero padding is applied. $\endgroup$
    – Deve
    Apr 3, 2013 at 12:07
  • $\begingroup$ @Deve you're right. I have edited my answer, thanks. $\endgroup$
    – sansuiso
    Apr 3, 2013 at 13:50

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