My snipet code have 3 harmonics, second's amplitude is maximum, but abs(fft(X)) returns first as maximum. I use MATLAB R2022a.

In Xn first harmonic's amplitude is 170, second's 220, and third's 150. But abs(fft(Xn) returns maximum first's. Thanks in advance. It is executing in R2022a version.

SampFreq = 16000;
Segm = 1:2048;
Pitch = 45;
FirstHarmAngles = Pitch*2*pi/SampFreq*Segm+1.9*pi;
SinFirstHarmAngles = sin(FirstHarmAngles);
SecondHarmAngles = Pitch*2*2*pi/SampFreq*Segm+2.9*pi;
SinSecondHarmAngles = sin(SecondHarmAngles);
ThirdHarmAngles = Pitch*3*2*pi/SampFreq*Segm+0.3*pi;
SinThirdHarmAngles = sin(ThirdHarmAngles);
Xn = 170*SinFirstHarmAngles+220*SinSecondHarmAngles+150*...
SinThirdHarmAngles;
ABS = abs(fft(Xn))

ABS = 1×2048

1.0e+05 * 0.0359 0.0414 0.0581 0.0907 0.1634 0.4385 1.6552 0.4040 0.3079 0.3324 0.4876 1.3711 1.5150 0.5278 0.3604 0.3266 0.4070 1.4252 0.4674 0.1809 0.1132 0.0852 0.0703 0.0610 0.0546 0.0499 0.0461 0.0431 0.0405 0.0383

plot(ABS(1:50))

Answer 1

[Jeffrey Clark]

@Georges Theodosiou, you don't have enough samples to easily interpret the abs(fft), if you change Segm = 1:2048 to use 16384 you get

Your plot from 2048 samples shows that the second peak contains the energy over a wider frequency range. Instead of increasing the number of samples this can be captured by increasing the fft n-point DFT by adding 16384 as the DFT size e.g., ABS = (abs(fft(Xn,16384))) which results in

Please see Fast Fourier transform - MATLAB fft (mathworks.com) and its examples for more information.

[Georges Theodosiou]

Mr. Jeffrey Clark, please let me express you my sincere gratitude for your instructive answer. Regards.

[Georges Theodosiou]

@Jeffrey Clark I need that segment be at most 1/8 of a second represented by 16000 samples. Then I leave Segm = 1:2048, and change fft(Xn) to fft(Xn,16384). Regards.

[Georges Theodosiou]

@Jeffrey Clark, please let me ask you the time your dsp kit processor executes 16384 points FFT. Regards.

[Jeffrey Clark]

@Georges Theodosiou, I don't have a dsp, my intel cpu for 16384 points takes 2.83e-04 seconds using MATLAB's fft.

If the fft time for you is too long there is a way to estimate the actual peak from your 2048 fft, which only works if you know that the frequency of the signals being measured are stable and not chirped (as is the case for your test data). In essance you compute the area under the fft curve where the phase adjacent to the peaks' phase are related using unwrap(angle(fft)).

[Georges Theodosiou]

@Jeffrey Clark. Dear gentlemen, please le me express you my sincere gratitude for your kindness to help me, and tell you I'm beginner in DSP and MATLAB, also homeless (in France, though Greek). I get internet access in municipal Libraries and social Centers. My project, metrw-pitch, demands that user could see on screen his voice's pitch at least 8 times per second. Regards.