MATLAB —— 信号处理工具箱之fft的介绍和相关案例分析

mytomorrow · 发表于 2019-11-20 10:16

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x

目录

Syntax

Description

   Y = fft(X)

   Y = fft(X,n)

   Y = fft（X，n，dim）

Examples

   Noisy Signal

Syntax

Y = fft(X)

Y = fft(X,n)

Y = fft(X,n,dim)

Description

% j: d2 w2 a4 S9 }Y = fft(X)

Y = fft(X) 使用fast Fourier transform(FFT)算法计算信号X的离散傅里叶变换：

如果 X 是一个向量，那么 fft(X) 返回向量的傅里叶变换；
如果 X 是一个矩阵，则 fft(X) 视X的列为向量，然后返回每列的傅里叶变换；
如果X是多维数组，则fft（X）将沿大小不等于1的第一个数组维度的值视为向量，并返回每个向量的傅里叶变换。- u5 A! ?, B% K& v5 L- N

Y = fft(X,n)( j4 _) |( x1 A* R' [

Y = fft(X,n) 返回 n 点 DFT。如果未指定任何值，则Y与X的大小相同。

如果X是向量并且X的长度小于n，则用尾随零填充X到长度n。
如果X是向量并且X的长度大于n，则X被截断为长度n。
如果X是矩阵，那么每个列都被视为向量情况。
如果X是多维数组，则大小不等于1的第一个数组维度将被视为向量的情况。- F2 K* }, C/ t5 V" E- w

Y = fft（X，n，dim）

Y = fft（X，n，dim）沿维度dim返回傅立叶变换。例如，如果X是矩阵，则fft（X，n，2）返回每行的n点傅立叶变换。

Examples

$ [" G9 A7 F! i: BNoisy Signal

使用傅立叶变换来查找隐藏在噪声中的信号的频率分量。

指定采样频率为1 kHz且信号持续时间为1.5秒的信号参数。

clc
clear
close all
% 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).
figure();
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.
figure();
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);
figure();
plot(f,P1)
title('Single-Sided Amplitude Spectrum of S(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
' S+ P: }& z. N( D; _