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MATLAB —— 信号处理工具箱之fft的案例分析

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发表于 2019-11-26 14:06 | 只看该作者 |只看大图 回帖奖励 |倒序浏览 |阅读模式

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上篇:
MATLAB —— 信号处理工具箱之fft的介绍和相关案例分析介绍了MATLAB信号处理工具箱中的信号变换 fft 并分析了一个案例,就是被噪声污染了的信号的频谱分析。
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: L/ h) M1 v1 D- O0 {7 V这篇博文继续分析几个小案例:. j$ D+ U+ T  Q' G0 b
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Gaussian Pulse

& O/ \3 {8 T" u5 c3 S2 I4 B这个案例是将高斯脉冲从时域变换到频域,高斯脉冲的信息在下面的程序中都有注释:
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  • clc
  • clear
  • close all
  • % Convert a Gaussian pulse from the time domain to the frequency domain.
  • %
  • % Define signal parameters and a Gaussian pulse, X.
  • Fs = 100;           % Sampling frequency
  • t = -0.5:1/Fs:0.5;  % Time vector
  • L = length(t);      % Signal length
  • X = 1/(4*sqrt(2*pi*0.01))*(exp(-t.^2/(2*0.01)));
  • % Plot the pulse in the time domain.
  • figure();
  • plot(t,X)
  • title('Gaussian Pulse in Time Domain')
  • xlabel('Time (t)')
  • ylabel('X(t)')
  • % To use the fft function to convert the signal to the frequency domain,
  • % first identify a new input length that is the next power of 2 from the original signal length.
  • % This will pad the signal X with trailing zeros in order to improve the peRFormance of fft.
  • 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);
  • figure();
  • plot(f,P(1:n/2+1))
  • title('Gaussian Pulse in Frequency Domain')
  • xlabel('Frequency (f)')
  • ylabel('|P(f)|')

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. H1 r1 e# P' @$ ^7 C! H高斯脉冲在时域的图像:) i5 T: o3 M- R7 [

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高斯脉冲在频域的图像:
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2 D1 b$ }; a. k9 h- m+ MCosine Waves
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这个例子比较简单,就是不同频率的余弦波在时域以及频域的比较:6 l# k( N9 D* Y4 m, N; m
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  • clc
  • clear
  • close all
  • % Compare cosine waves in the time domain and the frequency domain.
  • %
  • % Specify the parameters of a signal with a sampling frequency of 1kHz and a signal duration of 1 second.

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  • 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.
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  • 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

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  • X = [x1; x2; x3];
  • % Plot the first 100 entries from each row of X in a single figure in order and compare their frequencies.
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  • figure();
  • 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

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  • % For algorithm performance purposes, fft allows you to pad the input with trailing zeros.
  • % In this case, pad each row of X with zeros so that the length of each row is the next higher power of 2 from the current length.
  • % Define the new length using the nextpow2 function.
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  • n = 2^nextpow2(L);
  • % Specify the dim argument to use fft along the rows of X, that is, for each signal.

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  • dim = 2;
  • % Compute the Fourier transform of the signals.
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  • Y = fft(X,n,dim);
  • % Calculate the double-sided spectrum and single-sided spectrum of each signal.

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  • P2 = abs(Y/L);
  • P1 = P2(:,1:n/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.

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  • figure();
  • for i=1:3
  •     subplot(3,1,i)
  •     plot(0: (Fs/n): (Fs/2-Fs/n),P1(i,1:n/2))
  •     title(['Row ',num2str(i),' in the Frequency Domain'])
  • end- k* z9 x+ I1 Q) C+ n
           
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下图是频率为50Hz,150Hz以及300Hz的余弦波在时域的图像:5 X- Z6 R8 ]/ k

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下图分别为其fft:
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从频域图中可以清晰的看到它们的频率成分位于何处。$ I+ L9 m: F* p7 n0 q: x3 X4 I8 A
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