Prashant Kumar answered . 2024-04-23 12:03:10

y = sin(2*pi*(0:9)/10)    % correct   but   y = sin(2*pi*(0:10)/10)      % incorrect 

The code below gives two peaks, one at +160 kHz and one at -160 kHz for any number of fft points (above a minimum number), any number of complete oscillations in the time window and any amplitude. The abs(amplitude) of each peak is A/2, consistent with cos(w*t) = (1/2)(exp(i*w*t)+exp(-i*w*t)). If you want, as is common for  plotting purposes, you could ignore the negative freqencies and double the height of the positive frequency peak.

f0 = 160e3;
ncyc = 23;     % number of cycles in time window
A = 6;         % amplitude
psi = pi/6;    % arbitrary phase angle
N = 1299;      % need N >2*ncyc to avoid aliasing
t0 = 1/f0;
T = ncyc*t0;   % time window
delt = T/N;
t = (0:N-1)*delt;
y = A*cos(2*pi*f0*t+psi);
figure(1)
plot(t,y);grid on

delf = 1/(delt*N);    % golden rule for fft
if rem(N,2) ==0
  f = (-N/2:N/2-1)*delf;
else
  f = (-(N-1)/2:(N-1)/2)*delf;   
end
z = fftshift(fft(y))/N;     % fftshift to put f=0 in the middle
figure(2)
stem(f,abs(z),'o-');grid on

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