Maybe it's a trivial question, but I couldn't find any explanation for this.

According to the convolution theorem, in the continues case we add normalization factor, i.e. $$ \mathcal F\left\{h\star g\right\} = \sqrt{2\pi}\cdot \mathcal F\left\{h\right\}\cdot \mathcal F\left\{g\right\} $$

Do we still need to add a normalization factor also in the discrete case? If we do, what should its value be?


2 Answers 2


The choice of the normalization factor is just a matter of convention. Note that the specific correspondence between convolution in the time domain and multiplication in the frequency domain with a scaling of $\sqrt{2\pi}$, as shown in your question, applies only to the unitary definition of the Fourier transform with angular frequency as the independent variable in the frequency domain:

$$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-j\omega t}d\omega\tag{1}$$

If we use the non-unitary definition (as is common in the field of signal processing)

$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}d\omega\tag{2}$$

then the correspondence is simply

$$\mathcal{F}\{(f\star g)(t)\}=F(\omega)G(\omega)\tag{3}$$

The same is true if we use frequency instead of angular frequency.

For the discrete-time Fourier transform (DTFT), we have exactly the same choices.

For the discrete Fourier transform (DFT), where we deal with samples in both domains, the same holds for the normalization factor $N$, where $N$ is the DFT length. In the field of signal processing, the DFT is commonly defined as a non-unitary transform:

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{4}$$

The inverse transform is then given by

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{5}$$

The unitary DFT definition is

$$X[k]=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{6}$$

and the corresponding inverse transform is

$$x[n]=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{7}$$

  • $\begingroup$ Thanls! so just to be sure I understand, if I have $x,y \in \mathbb{R}^N$ and $x^F_l, y^F_l $are the $l$-th Fourier coefficients of $x,y$. Than $(x*y)_l = x^F_l \cdot y^F_l$ or $(x*y)_l = \sqrt{N} x^F_l \cdot y^F_l $ ? I use unitary definition of the Fourier transform, and a circular convolution $\endgroup$
    – Nave Tseva
    Commented Jan 3, 2021 at 13:21
  • 2
    $\begingroup$ @NaveTseva: For the unitary definition of the DFT you get a factor of $\sqrt{N}$ for the convolution. $\endgroup$
    – Matt L.
    Commented Jan 3, 2021 at 13:28

There is no normalization factor in either the continuous or the discrete case. Please check your source. A normalization factor ONLY occurs for the Fourier Series with $2\pi$ periodic functions and circular convolution.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.