离散傅立叶变换

 Discrete Fourier transform

Syntax

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

Definition

The functions X = fft(x) and x = ifft(X) implement the transform and inverse transform pair given for vectors of length http://www.mathworks.com/access/helpdesk/help/techdoc/ref/math_f19.gif by:

http://www.mathworks.com/access/helpdesk/help/techdoc/ref/math_fa.gif

where

http://www.mathworks.com/access/helpdesk/help/techdoc/ref/math_f2.gif

is an http://www.mathworks.com/access/helpdesk/help/techdoc/ref/math_f36.gifth root of unity.

Description

Y = fft(X) returns the discrete Fourier transform (DFT) of vector X, computed with a fast Fourier transform (FFT) algorithm.

If X is a matrix, fft returns the Fourier transform of each column of the matrix.

If X is a multidimensional array, fft operates on the first nonsingleton dimension.

Y = fft(X,n) returns the n-point DFT. If the length of X is less than n, X is padded with trailing zeros to length n. If the length of X is greater than n, the sequence X is truncated. When X is a matrix, the length of the columns are adjusted in the same manner.

Y = fft(X,[],dim) and Y = fft(X,n,dim) applies the FFT operation across the dimension dim.

Examples

A common use of Fourier transforms is to find the frequency components of a signal buried in a noisy time domain signal. Consider data sampled at 1000 Hz. Form a signal containing a 50 Hz sinusoid of amplitude 0.7 and 120 Hz sinusoid of amplitude 1 and corrupt it with some zero-mean random noise:

Fs = 1000;                    % Sampling frequency


T = 1/Fs;                     % Sample time


L = 1000;                     % Length of signal


t = (0:L-1)*T;                % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid


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


y = x + 2*randn(size(t));        % Sinusoids plus noise


plot(Fs*t(1:50),y(1:50))


title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)')

http://www.mathworks.com/access/helpdesk/help/techdoc/ref/new_fft1.gif

It is difficult to identify the frequency components by looking at the original signal. Converting to the frequency domain, the discrete Fourier transform of the noisy signal y is found by taking the fast Fourier transform (FFT):

NFFT = 2^nextpow2(L); % Next power of 2 from length of y


Y = fft(y,NFFT)/L;


f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2))) 


title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')

http://www.mathworks.com/access/helpdesk/help/techdoc/ref/new_fft2.gif

The main reason the amplitudes are not exactly at 0.7 and 1 is because of the noise. Several executions of this code (including recomputation of y) will produce different approximations to 0.7 and 1. The other reason is that you have a finite length signal. Increasing L from 1000 to 10000 in the example above will produce much better approximations on average.