The DFT is the most important discrete transform, used to perform Fourier analysis in many practical applications.^[2] In digital signal processing, the function is any quantity or signal that varies over time, such as the pressure of a sound wave, a radio signal, or daily temperature readings, sampled over a finite time interval (often defined by a window function^[3]). In image processing, the samples can be the values of pixels along a row or column of a raster image. The DFT is also used to efficiently solve partial differential equations, and to perform other operations such as convolutions or multiplying large integers.


Since it deals with a finite amount of data, it can be implemented in computers by numerical algorithms or even dedicated hardware. These implementations usually employ efficient fast Fourier transform (FFT) algorithms;^[4] so much so that the terms "FFT" and "DFT" are often used interchangeably. Prior to its current usage, the "FFT" initialism may have also been used for the ambiguous term "finite Fourier transform".


The DFT has many applications, including purely mathematical ones with no physical interpretation. But physically it can be related to signal processing as a discrete version (i.e. samples) of the discrete-time Fourier transform (DTFT), which is a continuous and periodic function. The DFT computes N equally-spaced samples of one cycle of the DTFT. (see Fig.2 and § Sampling the DTFT)

Example[edit]

This example demonstrates how to apply the DFT to a sequence of length ${\displaystyle N=4}$ and the input vector


$${\displaystyle \mathbf {x} ={\begin{pmatrix}x_{0}\\x_{1}\\x_{2}\\x_{3}\end{pmatrix}}={\begin{pmatrix}1\\2-i\\-i\\-1+2i\end{pmatrix}}.}$$


Calculating the DFT of ${\displaystyle \mathbf {x} }$ using Eq.1


$${\displaystyle {\begin{aligned}X_{0}&=e^{-i2\pi 0\cdot 0/4}\cdot 1+e^{-i2\pi 0\cdot 1/4}\cdot (2-i)+e^{-i2\pi 0\cdot 2/4}\cdot (-i)+e^{-i2\pi 0\cdot 3/4}\cdot (-1+2i)=2\\X_{1}&=e^{-i2\pi 1\cdot 0/4}\cdot 1+e^{-i2\pi 1\cdot 1/4}\cdot (2-i)+e^{-i2\pi 1\cdot 2/4}\cdot (-i)+e^{-i2\pi 1\cdot 3/4}\cdot (-1+2i)=-2-2i\\X_{2}&=e^{-i2\pi 2\cdot 0/4}\cdot 1+e^{-i2\pi 2\cdot 1/4}\cdot (2-i)+e^{-i2\pi 2\cdot 2/4}\cdot (-i)+e^{-i2\pi 2\cdot 3/4}\cdot (-1+2i)=-2i\\X_{3}&=e^{-i2\pi 3\cdot 0/4}\cdot 1+e^{-i2\pi 3\cdot 1/4}\cdot (2-i)+e^{-i2\pi 3\cdot 2/4}\cdot (-i)+e^{-i2\pi 3\cdot 3/4}\cdot (-1+2i)=4+4i\end{aligned}}}$$


results in $${\displaystyle \mathbf {X} ={\begin{pmatrix}X_{0}\\X_{1}\\X_{2}\\X_{3}\end{pmatrix}}={\begin{pmatrix}2\\-2-2i\\-2i\\4+4i\end{pmatrix}}.}$$