My snipet code have 3 harmonics, second's amplitude is maximum, but abs(fft(X)) returns first as maximum. I use MATLAB R2022a.
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Georges Theodosiou 2022 年 9 月 14 日
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*...
ABS = abs(fft(Xn))
Jeffrey Clark 2022 年 9 月 14 日
@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.