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Filtering Issue

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sanchezb

Civil/Environmental
Feb 5, 2012
10
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CA
I am using fast fourier transforms to convert raw acceleration to displacement. I am able to get extremely close to expected results; however, I am having issues at the beginning of the dataset. I believe there may be an issue with the filter I am using.

Here is the code for the butterworth filter I am using
Code:
[B,A] = butter(5,0.5/(Fs/2),'high');

I apply it to the raw data and after conversion.

Here are the results, I move the device +/- 10cm, +/- 7.5cm, and +/- 5cm as seen. The green line represents actual movement and the blue line represents the dataset produced through the conversion algorithm.

wJBc0.png


Any suggestions for better filtering techniques? I can provide the code and some data upon request.
 
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Here is most of the code

Code:
clear
clc
home

load ACCfilter
load ElapsedTime

%definitions

Fs = 32;
t = ElapsedTime; % sampling range
yno = ACCfilter;

[B,A] = butter(5,0.5/(Fs/2),'high');
y = filter(B,A,yno);

%plot in time domain

subplot(2, 1, 1); %
plot(t, y,'r'); grid on % plot with grid
title('Acceleration vs Time')
xlabel('Time (s)'); % time expressed in seconds
hold on
plot(t,yno,'g')
%FFT section

Y = fft(y); % compute Fourier transform enter
n = length(y); %
Amp = abs(Y )/n; % absolute value and normalize


%if n is even the fft will be symmetric
%the first n/2 + 1 points will be unique and the rest are symmetrically
%redundant. Point 1 is the DC component. Point n/2 +1 is the nyquist
%component

%1 2 3 4       5           6  7  8
%4 3 2 1      0          -1 -2 -3


% if n is odd the nyquist component is not evaluated. The number of unique
% points is (n+1)/2

%1 2 3 4 5  6  7  8  9
%5 4 3 2 1 -1 -2 -3 -4


%create the frequency axis

NumUniquePts=ceil((n+1)/2);

% the positive frequencies   

freq=(0:NumUniquePts-1)*Fs/(n-1);

%mirror the positive frequencies but make them negative then adjust some things
%to look right

freq2= -1*freq(end:-1:1);
if mod (n,2)==0
    freq2(1)=[];
end

freq3=[freq, freq2];
freq3';
size(freq3);

if mod(n,2)==1
freq3(n+1)=[];
end
if mod(n,2)==0
    freq3(n)=[];
end

%store freq3 in a column

freq3c = freq3';

%plot acceleration in frequency domain

subplot(2, 1, 2);
plot(freq, Amp(1 : NumUniquePts),'b.'); grid on % plot amplitude spectrum enter
xlabel('Frequency (Hz)'); % 1 Herz = number of cycles per second enter
ylabel('Acceleration Amplitude'); % amplitude as function of frequency enter


%Omega Arithmetic to convert acceleration to velocity
%uses complete frequency vector from above
V=Y./(2*pi*freq3c*1i);
%result=[freq3',Y',G']   %use this line to look at freq3, Y, and G next to each other 


% Get rid of infinities resulting from divisions by zero to prepare for
% ifft
V(1)=[1];
if mod(n,2)==0
    V(n)=[1];
end



%go back to time domain with the velocity

inversedVno=ifft(V);

[B,A] = butter(5,0.5/(Fs/2),'high');
inversedV = filter(B,A,inversedVno);


figure
plot(t,real(inversedV),'b')
hold on
title('Velocity vs Time'); % amplitude as function of time
xlabel('Time (s)'); % time expressed in seconds
ylabel('FFT Velocity');

% FsdirV=100; %sampling rate
% dtdirV = 1/Fs; %
% etdirV = 3; % end of the interval
% tdirV = 0 : dt : et; % sampling range
% ydirV = -cos(4*pi*t)/(2*pi);% define the signal
% plot(t,ydirV,'g')


%Omega Arithmetic to convert velocity to displacement
D=V./(2*pi*freq3c*1i);

% Get rid of infinities resulting from divisions by zero to prepare for
% ifft
D(1)=[1];
if mod(n,2)==0
    D(n)=[1];
end

%go back to time domain with the displacement

inversedDno=ifft(D);

[B,A] = butter(5,0.5/(Fs/2),'high');
inversedD = filter(B,A,inversedDno);
 
Use a FIR filter, Matlab fir2 maybe, since IIR has nonlinear group delay you will see different time shifts for different frequencies. You can then 'shift' the resulting filtered data by just defining the x-label for plotting purposes or by filtering using filtfilt.
 
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