Change frequency level after filter

First grahp í the source data. After I filter it, I got the grahp 2. But as you can see, in the second grahp, the yelow line(I don't know how to call it) is not the same with the 0-0 line(blue). So I want to change the frequency level from 0,5 to 0. How can I do it.

Thanks in advance

Answer 1

[Star Strider]

You seem to be using a low-pass filter. Use a bandpass filter instead, with a low-frequency cutoff that will eliminate the D-C offset (value at 0 Hz), and with the high-frequency cutoff you used in your original low-pass filter.

[vu ngothanh]

How to change it. Could you please to explain more. I'm using code like this

filter(2,0.01,x)

[Star Strider]

Use the filter design functions in the Signal Processing Toolbox to design your filter. The filter you are using with that filter call is actually not doing any filtering.

For example, your filter has b=2, a=0.01 and freqz plots it as:

b = 2;
a = 0.01;
figure(1)
freqz(b, a, 512)

I do not have your signal, and I do not know what frequencies you want your filter to pass, so I cannot be specific about your bandpass filter design.

[vu ngothanh]

here is a part of my input data

   0.000000,6.200694E-6,-1.875651E-6
  3.906250E-5,1.195134E-5,3.528527E-6
  7.812500E-5,6.992699E-6,-1.703543E-6
  0.000117,9.885240E-6,8.637580E-8
  0.000156,7.302614E-6,2.633568E-6
  0.000195,1.484388E-5,9.827665E-6
  0.000234,9.781935E-6,1.359972E-6
  0.000273,9.954110E-6,2.633568E-6
  0.000312,1.470614E-5,2.977783E-6
  0.000352,9.024364E-6,3.617479E-7
  0.000391,8.852189E-6,-3.611039E-7
  0.000430,8.886624E-6,-2.922609E-7
  0.000469,1.050507E-5,7.418159E-6
  0.000508,1.056979E-7,-1.302822E-5
  0.000547,5.133209E-6,-9.689335E-6
  0.000586,1.109046E-5,4.216958E-6
  0.000625,2.688323E-6,-5.248959E-6
  0.000664,6.786089E-6,-1.152799E-6
  0.000703,4.926599E-6,-8.243631E-6
  0.000742,6.889394E-6,-2.922609E-7
  0.000781,5.064339E-6,-4.987900E-7
  0.000820,7.130439E-6,-6.143919E-6
  0.000859,2.102928E-6,-3.699991E-6

This is my code:

 y=load('vibration_019.txt');
  a=y(:,1);
  c=y(:,3);
  b=y(:,2);
  figure(1);
  subplot(2,1,1);
  plot(a,b);    
  subplot(2,1,2);
  [B,A]=butter(2,0.01);
  b1=filter(B,A,b);
  plot(a,b1);

Now, when I fix the code as you said. the result like this

But, I want like this:

[Star Strider]

What is the sampling frequency of your data?

Are:

a=y(:,1);

your sampling times?

[vu ngothanh]

a=y(:,1); %time
b=(y:,2); %value at point 1
b=(y:,2); %value at point 2

[Star Strider]

There is not enough data to test the filter, but there is enough available to design it. I designed a bandpass filter with a lower frequency cutoff of 500 Hz and an upper frequency cutoff of 10 KHz. That should do what you want it to.

The code:

a = y(:,1);
b = y(:,2);
c = y(:,3);
% 
Ts = mean(diff(a));                 % Analyse Signal Spectrum
Fs = 1/Ts;
Fn = Fs/2;
fa = fft(a)/length(a);
fb = fft(b)/length(b);
Fv = linspace(0, 1, fix(length(fa)/2)+1)*Fn;
Iv = 1:length(Fv);
% 
figure(1)                           % Plot Fourier Transformed Signals
subplot(2,1,1)
plot(Fv, abs(fa(Iv)))
grid
subplot(2,1,2)
plot(Fv, abs(fb(Iv)))
grid
% 
Wp = [500 10000]/Fn;                 % Design Bandpass Filter
Ws = Wp.*[0.8  1/0.8];
Rp =  1;
Rs = 10;
[n,Wn] = buttord(Wp, Ws, Rp, Rs);
[b,a] = butter(n,Wn);
[sos,g] = tf2sos(b,a);
% 
figure(2)
freqz(sos, 128, Fs)                 % Check Filter Performance
% 
bfilt = filtfilt(sos, g, b);        % Filter Signals
cfilt = filtfilt(sos, g, c);
% 
figure(3)                           % Plot Filtered Signals
subplot(2,1,1)
plot(a, bfilt)
title('Velocity 1 Filtered')
grid
subplot(2,1,2)
plot(a, cfilt)
title('Velocity 2 Filtered')
grid

The documentation for the filter design is in the Signal Processing Toolbox documentation. It is relatively straightforward.

[vu ngothanh]

Thank you very much. I has just fix it as you say. it work. Thank you one more time

[Star Strider]

My pleasure!