MATLAB: Is there an error with the example for FFT in the MATLAB 7.6 (R2008a) documentation

MATLAB

The example code for FFT in the documentation is not correct. While the example follows the correct general method for finding the frequency components of a signal, to ensure mathematically correct results, the code should be changed from:

f = Fs/2*linspace(0,1,NFFT/2);
plot(f,2*abs(Y(1:NFFT/2)))

f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))

Best Answer

This change has been incorporated into the documentation in Release 2009a (R2009a). For previous releases, read below for any additional information:

This is a bug in the documentation for FFT in MATLAB 7.6 (R2008a)

When using the example, use the corrected code:

f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))

Related Solutions

MATLAB: Does the example in FFT divide the FFT result by the length of the signal and multiply the magnitude of FFT by 2 when displaying the single side spectrum in MATLAB 7.8 (R2009a)

Since the calculated FFT will be symmetric such that the magnitude of only the first half points are unique (and the rest are symmetrically redundant), the main idea is to display only half of the FFT spectrum. The following code illustrates this point:

NumUniquePts = ceil((NFFT+1)/2); 
FFTX=FFTX(1:NumUniquePts);
MX=abs(FFTX);  % take magnitude of X
MX=MX*2;    % multiply by 2 to take into account the fact that we threw out second half of FFTX above

However, the DC component of the signal is unique and should not be multiplied by 2:

MX(1)=MX(1)/2; % FFTX(1) is the DC component of x for both ODD and EVEN NFFT.

Also, for odd NFFT, the Nyquist frequency component is not evaluated, and the number of unique points is (NFFT+1)/2.

If NFFT is even, then the FFT will be symmetric such that the first (1+NFFT/2) points are unique, and FFTX(1+NFFT/2) is the Nyquist frequency component of x, which is unique. The rest are symmetrically redundant.

if ~rem(NFFT,2)  % when NFFT is even, FFTX(1+NFFT/2) is the Nyquist frequency component of x, and needs to make sure it is unique.
    MX(length(MX))=MX(length(MX))/2;
end

Try the following short example to see this:

a = [1 2 3 4 5]; A = fft(a)  % 5-point FFT
b = [1 2 3 4 5 6]; B = fft(b) % 6-point FFT

You can see that A(1) (or B(1)) is the DC component of a (or b). Also, note that B(4) is the Nyquist frequency component of b (even NFFT), and A does not have such component. The divisions by 2 are done to take care of these unique components.

MATLAB: Plotting results of DFT loop

Hi, it is difficult to provide an exact solution without your original data. However, when I executed your code using random data, at last iteration it is trying to access beyond the last element. After modifying the loop parameter, the code should work.

clc; 
clear all; 
DifDr=randi(100,525158,5); % Demo data 
  
ss=300000:6000:525158; 
k=length(ss); 
for i=1:k-1 
va(:,i)= DifDr(ss(i):ss(i+1),1); %  
    y = va; 
    [L , ~]= size(y); 
    Fs=10;  
    NFFT = 2^nextpow2(L); % Next power of 2 from length of y 
    f = Fs/2*linspace(0,1,NFFT/2+1); 
    Y(:,i) = fft(y(:,i),NFFT)/L; 
    % Plot single-sided amplitude spectrum. 
    fg=figure('Renderer', 'painters', 'Position', [100 100 700 500]); 
    plot(f,2*abs(Y(1:NFFT/2+1,i))); % semilogx(f,2*abs(Y(1:NFFT/2+1)))% 
end

Best Answer

Related Solutions

MATLAB: Does the example in FFT divide the FFT result by the length of the signal and multiply the magnitude of FFT by 2 when displaying the single side spectrum in MATLAB 7.8 (R2009a)

MATLAB: Plotting results of DFT loop

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