Sharp peak at 0hz for FFT

Hi,

I am using the FFT example and I am getting a sharp peak at 0hz. I have tried removing the DC component by subtracting the mean, however I still get this large peak. The signal is a 600hz signal, I am sampling at 5khz. Does anyone have any suggestions on how to remove this?

Here is my code:

Fs = 5000;            % Sampling frequency                    
T = 1/Fs;             % Sampling period       
L = length(Y);             % Length of signal
t = (0:L-1)*T;        % Time vector

t2 = flipud(rot90(t));
y2 = Y + 2*randn(size(t2));

y3 = fft(y2);
y3 = y3 - mean(y3);

y3 = detrend(y3,'constant');

P2 = abs(y3/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);

f = Fs*(0:(L/2))/L;

%Plot time-domain signal
subplot(3,1,1);
plot(t, Y);
ylabel('Amplitude'); xlabel('Time (secs)');
axis tight;
title('Noisy Input Signal');

subplot(3,1,2);
plot(1000*t(1:1000),y2(1:1000))
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('t (milliseconds)')
ylabel('X(t)')

subplot(3,1,3);
plot(f,P1) 
title('Single-Sided Amplitude Spectrum of X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')

Thanks!

Mike

[Walter Roberson]

Your third plot has the x axis starting about -500 or so. The command you used is plot(f,P1) . You set f = Fs*(0:(L/2))/L; which cannot have any negative values, and you do not do any xlim(). So it is not clear why you would have negative x values?

This tends to cast doubt that what you plotted is the same as what is in your code, so until this is cleared up it does not seem worth while to debug the code you posted.

Answer 1

[Ameer Hamza]

You need to subtract the mean value from time domain signal, not the frequency domain signal. Add the line

y2 = Y + 2*randn(size(t2));
y2 = y2 - mean(y2);

and delete the line

y3 = y3 - mean(y3);

[Michael Ratterman]

That was it, thanks so much!

[Ameer Hamza]

You are welcome.