Hi - Please why I don't get the excat amplitude in the spectrum of two added sine waves xt= 0.7*sin(2*​pi*50*t)+s​in(2*pi*12​0*t) , I mean i don't get excatly 0.7 for 50 Hz - nor excatly 1 for 120 Hz

% generate sin wave
% FFT on N=2^nextpow2(L) 
clear;
clc;
Fs=1000
dt=1/Fs
t=0:dt:1.5-dt;
xt= 0.7*sin(2*pi*50*t)+sin(2*pi*120*t);
L=length(xt)
N=2^nextpow2(L)
% xt=1.7*sin(2*pi*50*t) + 2.3*sin(2*pi*1200*t)+randn(size(xt));
subplot(2,1,1)
plot(t,xt)
grid on
grid minor
X=fft(xt,N)/L; % Always divide by L
subplot(2,1,2)
a=2*abs(X(1:N/2+1)); % Take N/2+1 points 
grid on
grid minor
f=linspace(0,Fs/2,N/2+1); % make N/2+1 points from 0 to Fs/2 
plot(f,a)
grid on
grid minor
ylim([0 1.2])
title('Single-Sided Amplitude Spectrum of x(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')

Answer 1

[Ameer Hamza]

It is happening because your signal is 1500 samples whereas you are trying to calculate 2048 point fft. MATLAB pad extra zeros to the end of your signal, it will introduce extra frequency components and the FFT changes. Try replacing the following line in your code

N = L; % instead of N = 2^nextpow2(L)

Answer 2

[dpb]

To amplify on Ameer's answer...

>> [p, fr]=findpeaks(a,f,'minpeakheight',0.25)
p =
  0.6097    0.9503
fr =
 49.8047  120.1172
>>

NB: With the frequency space of Fmax/2048 the frequency binning isn't precisely that of the input frequencies and so there's discretization/binning roundoff/noise introduced. That "drains off" energy into neighboring bins so that the all the energy isn't in the single-frequency bin as it is in the example case when the FFT is done over the L bins and the frequency bins do precisely match the signal frequencies.

To see this,

>> hAx=subplot(2,1,2);
>> hAx.YScale='log';
>> ylim([1E-5 1])

NB: the spread in the two peaks; to get closer approximation the total energy you would need to integrate over the peak regions.

Contrast this image to that of the one-sided noiseless PSD shown in the documentation example for FFT of the same signal. It would be good for the doc to illustrate this effect as well.

NB3: In the one-sided energy normalization you used

a=2*abs(X(1:N/2+1)); % Take N/2+1 points

This is slightly in error as there is only a single DC and Fmax element in the returned FFT; the above will double those, too. If there's a significant DC component in particular, this may make a noticeable distortion in the one-sided PSD.

Use the form shown in the example of

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

instead to get the positive frequency components first, then scale all except DC and Fmax. The reason for the above should be documented as well, granted, instead of leaving it for the user to try to discern the magic.

ADDENDUM

I went ahead and added the L-length result to the plot and then scaled to show the lower range of the peaks more clearly...you can see how when the bins align with the clean input signal frequencies all the power is in one bin, the energy on either side is of the magnitude of the floating-point roundoff.

>> [pk,loc]=findpeaks(P1,f1,'NPeaks',2,'MinPeakHeight',0.25)
pk =
  0.7000    1.0000
loc =
  50   120
>>

[Mohamad]

Thanks a lot for the valuable answers . So when computing FFT for a signal , no need to make : fft(xt,2^nextpow2(length(xt)) ? I have to make fft(xt,length(xt)) , right ? When I have to use N=nextpowe2(length(xt)) in computing fft ?

[dpb]

" When [do] I have to use N=nextpowe2(length(xt)) in computing fft?"

You never "have to"; the power-of-two is really mostly an historical remnant from the time when computing power wasn't nearly what it is today and the efficiencies of the algorithm coded specifically for those series lengths was important in being able to markedly speed up computation of the DFT.

With current processors, while still more efficient, the speed at which the algorithm runs is so quick that unless one is doing realtime control work or some other unusual application where time becomes critical, there's very little reason to not simply use the length of the input signal other than if there is a need for specifically interpolating to a given frequency spacing that doesn't match the sample rate.

Note, however, that just as the effect of augmenting the signal by zeros such that the frequency binning doesn't precisely match the pure frequencies in the signal in your example, changing the frequency in the signal for a given sample length will have the same effect. To see this, take the original example again with the 1500 sampled points but change the signal frequencies to, say, 51 and 119 Hz, respectively. You'll see very similar occurrence.

The point is, virtually never in the real world will the frequency content of a signal be either pure with no noise whatsoever, or absolutely stable in frequency or centered at the precise frequency of a bin in the frequency domain. The moral is, therefore, to always expect that you will have to integrate the peaks to get a true estimate of total power and that, unlike in this case where the two frequencies are separated sufficiently as to be essentially non-overlapping, in the real world you may not be able to separate them out so easily, one from another.

To further illustrate it all has to do with where the bin centers are with respect to the input signal frequency, I redid your example except doubled the FFT length from 2048 to 4096; the next yet power-of-two. This results in

>> [pk,loc]=findpeaks(P1,f,'NPeaks',2,'MinPeakHeight',0.25)
pk =
  0.6926    0.9503
loc =
 50.0488  120.1172
>>

Notice that the 50 Hz peak is now much closer to 0.7 because the frequency bin is only off by 0.0488 Hz but the 120 Hz peak nearest bin is still the same location as was at N=2048 and it has the same magnitude as before. You can further experiment to see how the choices for sampling and computing the FFT affect the results.

ADDENDUM

Actually, other than compute time, there is reason to use longer N which is to improve resolution. If one were to double N yet again to 8192, the results are

>> [pk,loc]=findpeaks(P1,f,'NPeaks',2,'MinPeakHeight',0.25)
pk =
  0.6926    0.9999
loc =
 50.0488  119.9951
>>

which is getting you pretty close to theoretical.

[Mohamad]

Thanks a lot for the valuable answers

[dpb]

Glad to help -- a few more votes or acceptances never hurts... :)

Altho that isn't real motivation but appreciate knowing is of some value.

Q? Did you try the suggested example of using the base length FFT on the sample waveform with a slight change in frequency?