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以下 Matlab 脚本用于过滤由 50 Hz 和 120 Hz 正弦波组成的信号。我将 rad/s 的频率计算为 Fp= (2*PI * 30)/1000=0.184。我保持 fp=0.184 和 fst=0.185，因为我想过滤掉 50 Hz 和 120 Hz。但是当我绘制滤波器输出的 FFT 时，我得到的是 50 Hz 的正弦波。为什么这个 50Hz 的正弦波即使在滤波之后也会出现？

理想情况下，图中不应有任何峰值。

过滤前 在此处输入图像描述

过滤后 在此处输入图像描述 

Fs = 1000;                    % Sampling frequency
T = 1/Fs;    % Sample time
L = 1000;                     % Length of signal
t =(0:L-1)*T;                % Time vector 
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);  % Sum of a 50 Hz sinusoid and a 120 Hz sinusoid %
y = x + 2*randn(size(t));     % Sinusoids plus noise

y = x ;

plot(Fs*t(1:50),y(1:50));title('Signal');xlabel('time (milliseconds)')

%pause;

NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L; f = Fs/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1))) 
title('Single-Sided Amplitude Spectrum of y(t)');xlabel('Frequency(Hz)');ylabel('|Y(f)|')


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%       Now let us see Low Pass Filtering of this signal  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Fp= (2*pi * 30)/1000; %=0.184 %only frequncies less than 30Hz will be passed  

d=fdesign.lowpass('Fp,Fst,Ap,Ast',0.184,0.185,2,60);

designmethods(d);

Hd = design(d,'equiripple'); fvtool(Hd);

Filterd_Output = filter(Hd,y);

NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Filtered_Freq = fft(Filterd_Output,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.
plot(f,2*abs(Filtered_Freq(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of Low Pass Filtered_Output')
xlabel('Frequency (Hz)');ylabel('|Filtered_Freq_Amplitude|')

更新 正如建议的那样，我将原始光谱与过滤后的光谱进行了比较。这解释了我很多。但是有什么办法可以让我进一步降低 50 Hz 的峰值吗？

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1 回答 1

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您错误地指定了过滤器的归一化频率。Matlab 假设频率在 [0, 1] 中，而不是在 [0, pi] 中。

代替

d=fdesign.lowpass('Fp,Fst,Ap,Ast',0.184,0.185,2,60);

和

d=fdesign.lowpass('Fp,Fst,Ap,Ast', 2*30/Fs, 2*35/Fs,2,60);

或者

d=fdesign.lowpass('Fp,Fst,Ap,Ast', 30, 35,2,60, Fs);

它应该按预期工作。

于 2013-02-27T14:11:24.003 回答