MATLAB —— 信号处理工具箱之fft的案例分析

mytomorrow

: \/ h  A) e( A; }& i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。

这篇博文继续分析几个小案例：

Gaussian Pulse
" T% N8 \/ F, S& d; _2 i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：

• 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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高斯脉冲在时域的图像：
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高斯脉冲在频域的图像：
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Cosine Waves
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4 u3 i4 d3 q7 U7 T$ ~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：

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

           

下图是频率为50Hz，150Hz以及300Hz的余弦波在时域的图像：
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下图分别为其fft：
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, ?: `- Y% J7 P" z# c$ \0 e从频域图中可以清晰的看到它们的频率成分位于何处。


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ddny

看看，学习一下