# fft

Fast Fourier transform

## Syntax

Y = fft(X)
Y = fft(X,n)
Y = fft(X,n,dim)

## Description

example

Y = fft(X) computes the discrete Fourier transform (DFT) of X using a fast Fourier transform (FFT) algorithm. If X is a vector, then fft(X) returns the Fourier transform of the vector.If X is a matrix, then fft(X) treats the columns of X as vectors and returns the Fourier transform of each column.If X is a multidimensional array, then fft(X) treats the values along the first array dimension whose size does not equal 1 as vectors and returns the Fourier transform of each vector. 

example

Y = fft(X,n) returns the n-point DFT. If no value is specified, Y is the same size as X. If X is a vector and the length of X is less than n, then X is padded with trailing zeros to length n.If X is a vector and the length of X is greater than n, then X is truncated to length n.If X is a matrix, then each column is treated as in the vector case.If X is a multidimensional array, then the first array dimension whose size does not equal 1 is treated as in the vector case. 

example

Y = fft(X,n,dim) returns the Fourier transform along the dimension dim. For example, if X is a matrix, then fft(X,n,2) returns the n-point Fourier transform of each row.

## Examples

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Use Fourier transforms to find the frequency components of a signal buried in noise.

Specify the parameters of a signal with a sampling frequency of 1 kHz and a signal duration of 1.5 seconds.

Fs = 1000; % Sampling frequency T = 1/Fs; % Sampling period L = 1500; % Length of signal t = (0:L-1)*T; % Time vector

Form a signal containing a 50 Hz sinusoid of amplitude 0.7 and a 120 Hz sinusoid of amplitude 1.

S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);

Corrupt the signal with zero-mean white noise with a variance of 4.

X = S + 2*randn(size(t));

Plot the noisy signal in the time domain. It is difficult to identify the frequency components by looking at the signal X(t).

plot(1000*t(1:50),X(1:50)) title("Signal Corrupted with Zero-Mean Random Noise") xlabel("t (milliseconds)") ylabel("X(t)")

Compute the Fourier transform of the signal.

Y = fft(X);

Compute the two-sided spectrum P2. Then compute the single-sided spectrum P1 based on P2 and the even-valued signal length L.

P2 = abs(Y/L); P1 = P2(1:L/2+1); P1(2:end-1) = 2*P1(2:end-1);

Define the frequency domain f and plot the single-sided amplitude spectrum P1. The amplitudes are not exactly at 0.7 and 1, as expected, because of the added noise. On average, longer signals produce better frequency approximations.

f = Fs*(0:(L/2))/L; plot(f,P1) title("Single-Sided Amplitude Spectrum of X(t)") xlabel("f (Hz)") ylabel("|P1(f)|")

Now, take the Fourier transform of the original, uncorrupted signal and retrieve the exact amplitudes, 0.7 and 1.0.

Y = fft(S); P2 = abs(Y/L); P1 = P2(1:L/2+1); P1(2:end-1) = 2*P1(2:end-1); plot(f,P1) title("Single-Sided Amplitude Spectrum of S(t)") xlabel("f (Hz)") ylabel("|P1(f)|")

Convert a Gaussian pulse from the time domain to the frequency domain.

Specify the parameters of a signal with a sampling frequency of 44.1 kHz and a signal duration of 1 ms. Create a Gaussian pulse with a standard deviation of 0.1 ms.

Fs = 44100; % Sampling frequency T = 1/Fs; % Sampling period t = -0.5:T:0.5; % Time vector L = length(t); % Signal length X = 1/(0.4*sqrt(2*pi))*(exp(-t.^2/(2*(0.1*1e-3)^2)));

Plot the pulse in the time domain.

plot(t,X) title("Gaussian Pulse in Time Domain") xlabel("Time (t)") ylabel("X(t)") axis([-1e-3 1e-3 0 1.1]) 

The execution time of fft depends on the length of the transform. Transform lengths that have only small prime factors result in significantly faster execution time than those that have large prime factors.

In this example, the signal length L is 44,101, which is a very large prime number. To improve the performance of fft, identify an input length that is the next power of 2 from the original signal length. Calling fft with this input length pads the pulse X with trailing zeros to the specified transform length.

n = 2^nextpow2(L);

Convert the Gaussian pulse to the frequency domain.

Y = fft(X,n);

Define the frequency domain and plot the unique frequencies.

f = Fs*(0:(n/2))/n; P = abs(Y/n).^2; plot(f,P(1:n/2+1)) title("Gaussian Pulse in Frequency Domain") xlabel("f (Hz)") ylabel("|P(f)|^2")

Compare cosine waves in the time domain and the frequency domain.

Specify the parameters of a signal with a sampling frequency of 1 kHz and a signal duration of 1 second.

Fs = 1000; % Sampling frequency T = 1/Fs; % Sampling period L = 1000; % Length of signal t = (0:L-1)*T; % Time vector

Create a matrix where each row represents a cosine wave with scaled frequency. The result, X, is a 3-by-1000 matrix. The first row has a wave frequency of 50, the second row has a wave frequency of 150, and the third row has a wave frequency of 300.

x1 = cos(2*pi*50*t); % First row wave x2 = cos(2*pi*150*t); % Second row wave x3 = cos(2*pi*300*t); % Third row wave X = [x1; x2; x3];

Plot the first 100 entries from each row of X in a single figure in order and compare their frequencies.

for i = 1:3 subplot(3,1,i) plot(t(1:100),X(i,1:100)) title("Row " + num2str(i) + " in the Time Domain") end

Specify the dim argument to use fft along the rows of X, that is, for each signal.

dim = 2;

Compute the Fourier transform of the signals.

Y = fft(X,L,dim);

Calculate the double-sided spectrum and single-sided spectrum of each signal.

P2 = abs(Y/L); P1 = P2(:,1:L/2+1); P1(:,2:end-1) = 2*P1(:,2:end-1);

In the frequency domain, plot the single-sided amplitude spectrum for each row in a single figure.

for i=1:3 subplot(3,1,i) plot(0:(Fs/L):(Fs/2-Fs/L),P1(i,1:L/2)) title("Row " + num2str(i) + " in the Frequency Domain") end

Create a signal that consists of two sinusoids of frequencies 15 Hz and 40 Hz. The first sinusoid is a cosine wave with phase $-\pi /4$, and the second is a cosine wave with phase $\pi /2$. Sample the signal at 100 Hz for 1 s.

Fs = 100; t = 0:1/Fs:1-1/Fs; x = cos(2*pi*15*t - pi/4) + cos(2*pi*40*t + pi/2);

Compute the Fourier transform of the signal. Plot the magnitude of the transform as a function of frequency.

y = fft(x); z = fftshift(y); ly = length(y); f = (-ly/2:ly/2-1)/ly*Fs; stem(f,abs(z)) title("Double-Sided Amplitude Spectrum of x(t)") xlabel("Frequency (Hz)") ylabel("|y|") grid

Compute the phase of the transform, removing small-magnitude transform values. Plot the phase as a function of frequency.

tol = 1e-6; z(abs(z) < tol) = 0; theta = angle(z); stem(f,theta/pi) title("Phase Spectrum of x(t)") xlabel("Frequency (Hz)") ylabel("Phase/\pi") grid

Interpolate the Fourier transform of a signal by padding with zeros.

Specify the parameters of a signal with a sampling frequency of 80 Hz and a signal duration of 0.8 s.

Fs = 80; T = 1/Fs; L = 65; t = (0:L-1)*T;

Create a superposition of a 2 Hz sinusoidal signal and its higher harmonics. The signal contains a 2 Hz cosine wave, a 4 Hz cosine wave, and a 6 Hz sine wave.

X = 3*cos(2*pi*2*t) + 2*cos(2*pi*4*t) + sin(2*pi*6*t);

Plot the signal in the time domain.

plot(t,X) title("Signal superposition in time domain") xlabel("t (ms)") ylabel("X(t)")

Compute the Fourier transform of the signal.

Y = fft(X);

Compute the single-sided amplitude spectrum of the signal.

f = Fs*(0:(L-1)/2)/L; P2 = abs(Y/L); P1 = P2(1:(L+1)/2); P1(2:end) = 2*P1(2:end);

In the frequency domain, plot the single-sided spectrum. Because the time sampling of the signal is quite short, the frequency resolution of the Fourier transform is not precise enough to show the peak frequency near 4 Hz.

plot(f,P1,"-o") title("Single-Sided Spectrum of Original Signal") xlabel("f (Hz)") ylabel("|P1(f)|")

To better assess the peak frequencies, you can increase the length of the analysis window by padding the original signal with zeros. This method automatically interpolates the Fourier transform of the signal with a more precise frequency resolution.

Identify a new input length that is the next power of 2 from the original signal length. Pad the signal X with trailing zeros to extend its length. Compute the Fourier transform of the zero-padded signal.

n = 2^nextpow2(L); Y = fft(X,n);

Compute the single-sided amplitude spectrum of the padded signal. Because the signal length n increased from 65 to 128, the frequency resolution becomes Fs/n, which is 0.625 Hz.

f = Fs*(0:(n/2))/n; P2 = abs(Y/L); P1 = P2(1:n/2+1); P1(2:end-1) = 2*P1(2:end-1);

Plot the single-sided spectrum of the padded signal. This new spectrum shows the peak frequencies near 2 Hz, 4 Hz, and 6 Hz within the frequency resolution of 0.625 Hz.

plot(f,P1,"-o") title("Single-Sided Spectrum of Padded Signal") xlabel("f (Hz)") ylabel("|P1(f)|")

## Input Arguments

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Input array, specified as a vector, matrix, or multidimensional array.

If X is an empty 0-by-0 matrix, then fft(X) returns an empty 0-by-0 matrix.

Data Types: double | single | int8 | int16 | int32 | uint8 | uint16 | uint32 | logical
Complex Number Support: Yes

Transform length, specified as [] or a nonnegative integer scalar. Specifying a positive integer scalar for the transform length can improve the performance of fft. The length is typically specified as a power of 2 or a value that can be factored into a product of small prime numbers (prime factors not greater than 7). If n is less than the length of the signal, then fft ignores the remaining signal values past the nth entry and returns the truncated result. If n is 0, then fft returns an empty matrix.

Example: n = 2^nextpow2(size(X,1))

Data Types: double | single | int8 | int16 | int32 | uint8 | uint16 | uint32 | logical

Dimension to operate along, specified as a positive integer scalar. If you do not specify the dimension, then the default is the first array dimension of size greater than 1.

• fft(X,[],1) operates along the columns of X and returns the Fourier transform of each column.

• fft(X,[],2) operates along the rows of X and returns the Fourier transform of each row.

If dim is greater than ndims(X), then fft(X,[],dim) returns X. When n is specified, fft(X,n,dim) pads or truncates X to length n along dimension dim.

Data Types: double | single | int8 | int16 | int32 | uint8 | uint16 | uint32 | logical

## Output Arguments

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Frequency domain representation returned as a vector, matrix, or multidimensional array.

If X is of type single, then fft natively computes in single precision, and Y is also of type single. Otherwise, Y is returned as type double.

The size of Y is as follows:

• For Y = fft(X) or Y = fft(X,[],dim), the size of Y is equal to the size of X.

• For Y = fft(X,n,dim), the value of size(Y,dim) is equal to n, while the size of all other dimensions remains as in X.

If X is real, then Y is conjugate symmetric, and the number of unique points in Y is ceil((n+1)/2).

Data Types: double | single

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### Discrete Fourier Transform of Vector

Y = fft(X) and X = ifft(Y) implement the Fourier transform and inverse Fourier transform, respectively. For X and Y of length n, these transforms are defined as follows:

$\begin{array}{l}Y\left(k\right)=\sum _{j=1}^{n}X\left(j\right)\text{\hspace{0.17em}}{W}_{n}^{\left(j-1\right)\text{​}\left(k-1\right)}\\ X\left(j\right)=\frac{1}{n}\sum _{k=1}^{n}Y\left(k\right)\text{\hspace{0.17em}}{W}_{n}{}^{-\left(j-1\right)\text{​}\left(k-1\right)},\end{array}$

where

${W}_{n}={e}^{\left(-2\pi i\right)/n}$

is one of n roots of unity.

## Tips

• The execution time of fft depends on the length of the transform. Transform lengths that have only small prime factors (not greater than 7) result in significantly faster execution time than those that are prime or have large prime factors.

• For most values of n, real-input DFTs require roughly half the computation time of complex-input DFTs. However, when n has large prime factors, there is little or no speed difference.

• You can potentially increase the speed of fft using the utility function fftw. This function controls the optimization of the algorithm used to compute an FFT of a particular size and dimension.

## Algorithms

The FFT functions (fft, fft2, fftn, ifft, ifft2, ifftn) are based on a library called FFTW [1] [2].

## References

[2] Frigo, M., and S. G. Johnson. “FFTW: An Adaptive Software Architecture for the FFT.” Proceedings of the International Conference on Acoustics, Speech, and Signal Processing. Vol. 3, 1998, pp. 1381-1384.

## Version History

Introduced before R2006a