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% t/ r; T& K9 Q, h5 W# r6 a目录) [2 U5 w" |& E. N9 Z5 B
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Y = fft(X)- R [) c7 N6 a
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Y = fft(X,n)
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Examples
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1 u- e& s! q4 ~$ I- Y2 AY = fft(X)& b# g; D$ M S1 M" x
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Y = fft(X)
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( U' M2 E3 i1 Y: t; A5 M( nY = fft(X) 使用fast Fourier transform(FFT)算法计算信号X的离散傅里叶变换:+ c7 N4 i9 i5 f; ]( U" w& t, }( l* X
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- 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。
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Y = fft(X,n) \2 s$ s6 m4 l4 V, s' u: R! M
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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。
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% y; o/ { z* m# J, H) C% p% t- 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。! B9 {) Y% t4 `9 k" k2 o
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Y = fft(X,n,dim)
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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。
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Examples
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Noisy Signal
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使用傅立叶变换来查找隐藏在噪声中的信号的频率分量。! w/ i( T0 E2 Q$ a# P3 ?
& l) x+ ~4 j" ^: g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。4 H- H" h7 M- ]/ d
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- 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)|')
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figure(1)是加上零均值的随机噪声后的信号时域图形,通过观察这幅图很难辨别其频率成分。5 x+ |# ]) M7 q6 n4 E
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1 Y3 B n3 T: Z! K$ H2 v7 b7 tfigure(2)是X(t)的单边幅度谱,通过这幅图其实已经能够看出信号的频率成分,分别为50Hz和120Hz,其他的频率成分都会噪声的频率分量。; _' ^. s# V8 U$ T6 _1 k: F' ?
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figure(3)是信号S(t)的单边幅度谱,用作和figure(2)的幅度谱对比,原信号确实只有两个频率成分。
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