MATLAB: Fft importance of time sampled

fast fourierfftfourierMATLABsample

I guessed that the longer one samples, the most exact result with fft. And so it seems to do with frequency, but not at all with amplitude. The amplitude decreases the longer the time sampled. Did I do something wrong ? Is this reasonable? Thanks to any help
% T=[.01 .02 .1 .2 1];
% T=[1:10]
T=[1 5 10];
n=10; % maximal number of harmonic
freq = 5;
Fs = 150e4; % Sampling frequency (frecuencia de muestreo)
nfft = 1e7; % Length of FFT % number of fft bins
tic
for m=1:length(T) 
      t = 0:1/Fs:T(m); % Time vector of 1 second
      x = cos(2*pi*t*freq); % sine wave of f Hz.
      x = x-mean(x);% restamos de la fcn 'x' su DC offset
      figure(1);
      subplot(length(T),1,m);
      plot(t,x,'.');
      % Take fft, padding with zeros so that length(X) is equal to nfft
      X = fft(x,nfft);
      % FFT is symmetric, throw away second half
      X = X(1:nfft/2);
      % Scaling is done here using the number of samples: length(x)/2
      X=X/(length(x)/2);
      % Take the magnitude of fft of x
      mx = abs(X);
      % Frequency vector
      f_fft = (0:nfft/2-1)*Fs/nfft;
      f_Axis=f_fft(1:nfft/2);
      % picos de amplitud
      pks= findpeaks(mx(1:nfft/2));
      n=min(n,length(pks));
      pks_sort=sort(pks,'descend');% vector de picos
      for k =  1:n
          locs=find(mx(1:nfft/2)==pks_sort(k));
          f(k)=f_Axis(locs);
      end
      % bar plot freq y magnitude
      figure
      uds='Amp';
      pks_sort=pks_sort(:); % to make it column array for text of bar plot

      f=f(:); % to make it column array for text of bar plot
      subplot(1,2,1)
      bar(pks_sort(1:n),'r');
      ylabel({sprintf('Amplitude (%s)'...
          ,uds)},'fontweight','bold','fontsize',16);
      text(1:n,pks_sort(1:n),num2str(pks_sort(1:n),'%.2f'),... 
          'HorizontalAlignment','center',...
          'VerticalAlignment','bottom')
      subplot(1,2,2)
      bar(f(1:n),'b');
      ylabel('Frequency (Hz)','fontweight','bold','fontsize',16);
      text(1:n,f(1:n),num2str(f(1:n),'%.2f'),...
          'HorizontalAlignment','center',...
          'VerticalAlignment','bottom')
      bar3d_pks_sort(m,:)=pks_sort(1:n);
      bar3d_f(m,:)=f(:);
  end
figure;
col(1,:)=[0 0 1]; col(2,:)=[0 .5 0];
plot(bar3d_pks_sort(:,1),'-+','Color',col(1,:),'Linewidth',2);
h1 = gca;
h2 = axes('Position',get(h1,'Position'));
plot(bar3d_f(:,1),'-o','Color',col(2,:),'Linewidth',2);
set(h1,'YColor',col(1,:))
set(h2,'YAxisLocation','right','Color','none','XTickLabel',[],'YColor',col(2,:))
set(h2,'YAxisLocation','right','Color','none','XTickLabel',[])

Best Answer

You have to normalise the Fourier transform by the length of the time-domain vector.
So this line should be:
      Fn = Fs/2;                                                          % Nyquist Frequency
      % Take fft, padding with zeros so that length(X) is equal to nfft
      X = fft(x,nfft)/nfft;
... and the following line is then:
      % FFT is symmetric, throw away second half
      Fv = linspace(0, 1, fix(X/2)+1)*Fn;                                  % Frequency Vector (For Plots, &c.)  
      Iv = 1:length(Fv);                                                   % Index Vector
      X = X(1:nfft(Iv))*2;
The current documentation is a bit confusing. Much more straightforward description is in the R2015a fft documentation.
I didn’t go through the rest of your code, but that should correct some of it.
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