The Discrete Fourier Transform (DFT) is a mathematical function, and the Fast Fourier Transform (FFT) is an algorithm for computing that function. Since the DFT is almost always computed via the FFT, the distinction between the two is sometimes lost.

It is often not necessary to distinguish between the two. In my previous post, I used the two terms interchangeably because the distinction didn’t matter in that context.

Here I will make a distinction between the DFT and the FFT; I’ll use **DFT** to refer to the DFT as it is defined in [1], and **FFT** for the DFT as computed in NumPy‘s FFT routine. The differences are due to varying conventions; this is often an issue.

Suppose we have a function *f* on an interval [-*A*/2, *A*/2] and we sample *f* at *N* points where *N* is an even integer.

Define

for *n* running from –*N*/2 + 1 to *N*/2.

DFT of the sequence {*f*_{n}} is defined in [1] as the sequence {*F*_{k}} where

Now suppose the function *f* that we sampled has a Fourier series

The Fourier coefficients *c*_{k} relate to the DFT output *F*_{k} according to the **Discrete Poisson Summation Formula** [1]:

This means that the *c*_{k} simply equal the *F*_{k} if *f* has no frequency components higher than *N*/2 Hz because all the terms in the infinite sum above are zero. That is, if *f* is band-limited and we have sampled at a frequency higher than the Nyquist frequency, then we can simply read the Fourier coefficients off the DFT.

However, when *f* is not band-limited, or when it is band-limited but we have not sampled it finely enough, we will have aliasing. Our Fourier coefficients will differ from our DFT output by an error term involving high-frequency Fourier series coefficients.

In application it’s usually too much to hope that your function is exactly band-limited, but it may be approximately band-limited. The Fourier coefficients of smooth functions eventually decay rapidly (see details here) and so the error in approximating Fourier series coefficients by DFT terms is small.

## Computing a DFT with the FFT

We defined the DFT of the sequence {*f*_{n}} above to be the sequence {*F*_{k}} where

and *k* runs from –*N*/2 + 1 to *N*/2.

NumPy, on the other hand, defines the DFT of the sequence {*a*_{n}} to be the sequence {*A*_{k}} where

and *k* runs from 0 to *N*-1.

Relative to the definition in the previous post, the NumPy definition difference in three ways:

- The normalization factor 1/
*N*is missing. - The input indices are numbered differently.
- The output is arranged differently.

The first difference is trivial to overcome: simply divide the FFT output by *N*.

The second difference is easy to deal with if we think of the inputs to the FFT being samples from a periodic function, which they usually are. The *f*_{k} come from sampling a periodic function *f* over an interval [-*A*/2, *A*/2]. If we sample the same function over [0, *A*] we get the *a*_{n}. We have

If we extend the *f*s and the *a*s periodically past their original ranges of definition, then they all agree. But since we start our sampling in difference places, our samples would be listed in different orders if we stored them by increasing index.

Something similar occurs with the output.

For 0 ≤ *k* ≤ *N*/2,

*F*_{k} = *A*_{k}.

But for *N* < *k* < *N*,

*F*_{k} = *A*_{N–k}.

## Example

We’ll use a band-limited function in our example so that we find the Fourier coefficients exactly.

*f*(*x*) = 7 cos(6π*x*) – 5 sin(4π*x*)

We compute the FFT as follows.

import numpy as np def f(x): return 7*np.sin(3*2*np.pi*x) - 5*np.cos(2*2*np.pi*x) N = 8 delta = 1/N x = [f(n*delta) for n in range(N)] print( np.fft.fft(x)/N )

The output is

[0, 0, -2.5, -3.5i, 0, 3.5i, -2.5, 0]

This says *F*_{2} and *F*_{-2} = 5/2, and so the coefficient of cos(2·2π*x*) is 5.

Also *F*_{3} = -7/2 and *F*_{-3} = 7/2, so the coefficient of cos(3·2π*x*) is 7. These results follow from Euler’s equation

exp(*i*θ) = cos(θ) + *i* sin(θ)

[1] William L. Briggs and Van Emden Henson. The DFT: An Owner’s Manual for the Discrete Fourier Transform. SIAM 1995.

Well, I definitely got my answer to my previous comment. :-)

Consider the periodic function ()=cos(3), ()=sin(2), ℎ()=sin(2)sin(3), with a period of 2. Use the MATLAB functions of fft, ifft, and fftshift (to have the mean at zero) to calculate the Fourier modes of these functions.

a) Plot the f(x), g(x), h(x) functions in physical domain, f(x) vs. x.

b) Plot the magnitudes of Fourier modes of f(x), g(x), h(x) as a function of wave number, |̂| vs.

Can you please tell me how do I solve this problem in MATLAB? I saw the above example but I cannot understand a few terms so can you please help me out?