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目录2 d* T+ Y( H* l Z1 d4 v
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Syntax! J0 R6 X; i' G
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Description! n) j6 d _- z3 }/ e$ S" J5 m
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Y = fft(X)# J9 f- ^- t; h
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Y = fft(X,n)) j2 f6 R: q c+ L2 o$ m: |0 s& Z% a7 h
* Y' u( B, a% M* J5 j: U Y = fft(X,n,dim)
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Examples8 J! x& W6 w5 S$ y
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Y = fft(X)
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Y = fft(X,n,dim)
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Description1 i9 s% {5 W( t1 ]9 \" {
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Y = fft(X)7 O2 C$ c* L/ T3 k5 d
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Y = fft(X) 使用fast Fourier transform(FFT)算法计算信号X的离散傅里叶变换:
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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。" O% I# e# W k% m
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% {! A" q$ V5 S$ uY = fft(X,n)
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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。) W9 R4 q0 \( }7 \
+ o/ j+ e }8 L. V- 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。
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& z: A+ d) ~6 xY = fft(X,n,dim) ?6 r& u5 m- O+ C( y t6 }& x/ [
, S. w5 @& ~& p( U: y% \( S8 dY = 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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, ^) D2 E' r- U- p+ u' RExamples
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Noisy Signal3 [ d' u9 w# {- c# x5 Q/ |5 p9 L
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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。
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/ N( K+ N: b) g* v3 F+ x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。
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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.
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- 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。
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figure(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。5 a, t+ D: P; T# O. w& Z, h0 c* g
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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。" o: Y& L5 c r" ^* D
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2 k0 ^& Q; |/ \8 j' y: J上面三幅图画到一起:
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