why fft spectrum for f1=sin(2*pi*10*t) is so much different from f2=sin(2*pi*9*t)?

when I plot p=fft(f1) vs freq, I get perfect peak at 10Hz. When I plot f=fft(f2) vs freq, at get a poor peak at 9Hz. Why?

My program look like this: N=1000; T=3; t=[0:N-1]/N; t=t*T; f1=sin(2*pi*10*t); f2=sin(2*pi*9*t); p1=abs(fft(f1))/(N/2); p2=abs(fft(f2))/(N/2); p1=p1(1:N/2); p2=p2(1:N/2); freq=[0:N/2-1]/T; plot(freq,p1); plot(freq,p2); I expect the first chart getts a perfect peak at 10Hz, indeed, I got it. I do not get a good peak for the second at 9Hz. Why? Thank you!

Answer 1

[Youssef Khmou]

hi,

Nothing wrong with the code, it is true, the two spectra have an amplitude of 1. Perhaps you have drawn the two graphs in the same figure, so that the second graph has replaced the first, draw figures separated or in the same figure using "hold on" command: i re-post the code:

      N=1000;
     T=3; 
     t=[0:N-1]/N;
     t=t*T;
     f1=sin(2*pi*10*t); 
     f2=sin(2*pi*9*t);
     p1=abs(fft(f1))/(N/2);
     p2=abs(fft(f2))/(N/2); 
     p1=p1(1:N/2);
     p2=p2(1:N/2);
     freq=[0:N/2-1]/T;
     plot(freq,p1); 
     hold on
     plot(freq,p2,'r');