Remove peak at 0 hz of fft

Hi everybody,

I want to calculate the fft of my data, but I get a large peak at 0 Hz. I have read all the answers to the previous questions. Mostly recommended using x=x-mean(x) or x=detrend(x). I have tried that but nothing works.

Another suggestion was to look at the diff(x) and diff(diff(x)). I tried that and both give a noisy straight line.

I dont now what else to do so can somebody help me?

Thanks in advance!

x=load('data.txt');
% diffx=diff(x);
% diffdiffx=diff(diff(x));

xm=x-mean(x);
xd=detrend(x);
y=detrend(xm,'constant');

N=length(x);

fs=1000;
Ts=1/fs;
f=(0:1/N:1/2-1/N)*fs;

% figure
% plot(x)
% hold on
% plot(diffx)
% plot(diffdiffx)
% hold off
% legend('x','diffx','diffdiffx')

xhat=fft(x,N);
xhat(1)=0;
figure
plot(f,abs(xhat(1:N/2)));
title('Normal data')

xhat=fft(xm,N);
xhat(1)=0;
figure
plot(f,abs(xhat(1:N/2)));
title('Removed mean data')

xhat=fft(xd,N);
xhat(1)=0;
figure
plot(f,abs(xhat(1:N/2)));
title('Detrend data')

xhat=fft(y,N);
xhat(1)=0;
figure
plot(f,abs(xhat(1:N/2)));
title('Removed mean and detrend data')

[S,fm]=periodogram(x,[],N,fs,'power');
figure
plot(fm,10*log10(S))
title('Periodogram')

[S,fm]=periodogram(xm,[],N,fs,'power'); 
figure
plot(fm,10*log10(S))
title('Periodogram removed mean')

[S,fm]=periodogram(xd,[],N,fs,'power'); 
figure
plot(fm,10*log10(S))
title('Periodogram detrend data')

[S,fm]=periodogram(y,[],N,fs,'power'); 
figure
plot(fm,10*log10(S))
title('Periodogram removed mean and detrend data')

[Sw,fw]=pwelch(x,ones(N,1),0,N,fs,'power');
figure
plot(fw,Sw)
title('Welch method')

[Sw,fw]=pwelch(xm,ones(N,1),0,N,fs,'power'); 
figure
plot(fw,Sw)
title('Welch method removed mean')

[Sw,fw]=pwelch(xd,ones(N,1),0,N,fs,'power'); 
figure
plot(fw,Sw)
title('Welch method detrend data')

[Sw,fw]=pwelch(y,ones(N,1),0,N,fs,'power'); 
figure
plot(fw,Sw)
title('Welch method removed mean and detrend data')

[Mehmed Saad]

apply a hpf

[Emmy]

Applying a highpass filter did not solve the problem. There still is a peak from the point of the high pass filter. And it doesn't matter at which point I take the high pass filter. Even when I look in the normal data, where the peak ends, there still is a peak after the high pass filter.

Answer 1

[David Goodmanson]

Hi Emmy

It's all in how you view things. Literally. Here is what happens when using x - mean(x).

fs = 1000;
N = 30000;
figure(1)
plot(x)
xm = x - mean(x);
y = fftshift(abs(fft(xm)));    % shift zero frequency to the center
f = (-N/2:N/2-1)*fs/N;         % make f array match
y(15001) = 1;                  % adjust 0 Hz value in fft
figure(2)
semilogy(f,y)
figure(3)
semilogy(f,y)
xlim([-50 50])

Both x and y have 30000 elements. Ordinarily, y(1) would be the element corresponding to zero frequency. For plotting purposes, fftshift puts that element at the center of the array, element 15001.

Since the y values have a large range, it makes sense to look at abs(y) on a semilog plot. Note that since y is the fft of a real function, abs(y) is symmetric about f==0.

After subtracting the mean, the DC value should be zero. Because of 'numerical error' it is small but nonzero. For plot purposes I arbitrarily set that value to 1 so as to not have a large negative spike in the semilog plot.

In figure 2, you can see some lower frequency stuff rising above the noise. In the zoomed-in figure 3. there is something interesting happening between about 17-20 Hz. If you zoom in a little bit in figure 3 you will see no peak at f=0, with the largest values at one freq point on either side, falling off from there.

[Image Analyst]

Adding plots for visualization purposes:

x=load('data.txt');
fs = 1000;
N = 30000;
subplot(3, 1, 1);
plot(x)
grid on;
title('x', 'FontSize', 20);
xm = x - mean(x);
y = fftshift(abs(fft(xm)));    % shift zero frequency to the center
f = (-N/2:N/2-1)*fs/N;         % make f array match
y(15001) = 1;                  % adjust 0 Hz value in fft
subplot(3, 1, 2);
semilogy(f,y)
grid on;
title('FT', 'FontSize', 20);
subplot(3, 1, 3);
semilogy(f,y)
xlim([-50 50])
grid on;
title('Zoomed FT', 'FontSize', 20);

Answer 2

[Mehmed Saad]

FFT of X

FFT of X High pass filtered