我正在制作一个带通滤波器,并基于正弦波创建了一个具有一些不需要频率的信号:
Fs = 8e3; % Sampling Rate
fn = Fs/2; % Nyquist frequency
L = 1e3; % Length of signal
T = 1/Fs; % Sampling period
t = T*linspace(0,L,Fs); % Time domain
% Frequencies in Hz
f1 = 1500;
f2 = 700;
f3 = 2500;
f4 = 3500;
% Signal
x = 6*sin(2*pi*f1*t);
% Noise
noise = 3*sin(2*pi*f2*t)...
+ 2*sin(2*pi*f3*t)...
+ 1*sin(2*pi*f4*t);
x_noise = x + noise;
然后,我创建一个Butterworth带通滤波器:
[b,a] = butter(10,[1000 2000]/fn,'bandpass');
具有带通响应(带有freqz)的时间和频率空间中的信号如下所示:
Fig. 1 Signal with corruption | Fig. 2 Frequency with bandpass response
我会从这里开始想,只是做
xf = filter(b,a,x_noise);
会产生与原始信号非常相似的东西,但是,我得到的是really far from the filtered signal with a high response far from the bandpass。
我在这里做错了什么?
这是完整的代码:
clear all
Fs = 8e3; % Sampling Rate
fn = Fs/2; % Nyquist frequency
L = 1e3; % Length of signal
T = 1/Fs; % Sampling period
t = T*linspace(0,L,Fs); % Time domain
% Frequencies in Hz
f1 = 1500;
f2 = 700;
f3 = 2500;
f4 = 3500;
% Signal
x = 6*sin(2*pi*f1*t);
% Noise
noise = 3*sin(2*pi*f2*t)...
+ 2*sin(2*pi*f3*t)...
+ 1*sin(2*pi*f4*t);
x_noise = x + noise;
subplot(221);
idx = 1:round(length(t)/30);
plot(t(idx),x(idx),t(idx),x_noise(idx));
xlabel('Time (s)'); ylabel('Signal Amplitudes');
legend('Original signal','Noisy signal');
% Frequency space
f = fn*linspace(0,1,L/2+1);
X = fft(x_noise)/L;
[b,a] = butter(10,[1000 2000]/fn,'bandpass');
h = abs(freqz(b,a,floor(L/2+1)));
subplot(222);
plot(f,abs(X(1:L/2+1)),f,h*max(abs(X)));
xlabel('Freq (Hz)'); ylabel('Frequency amplitudes');
legend('Fourier Transform of signal','Filter amplitude response');
% Filtered signal
xf = filter(b,a,x_noise);
subplot(223)
plot(t(idx),xf(idx));
xlabel('Time (s)'); ylabel('Signal Amplitudes');
legend('Filtered signal');
% Filtered in frequency space
Xf = abs(fft(xf)/L);
subplot(224);
plot(f,Xf(1:L/2+1),f,h*5e-6);
xlabel('Freq (Hz)'); ylabel('Frequency amplitudes');
legend('Fourier Transform of filtered signal','Bandpass');
最佳答案
您的时间变量t
是错误的,例如1/(t(2)-t(1))
应该提供Fs
,但没有。
请尝试:
t = T*(0:L-1);