I can't place the fft curve at the signal frequency exactly. There's been a slight deviation of the fft peak at that frequency

%program to find the fft of sine function of frequency 1GHz
%Available information , frquency=1GHz
clear all
close all
clc
fmax=1e9;                 %maximun frequency of the signal
fs=60*fmax; 
ts=1/fs;                  %time increment
T=10e-9;                  %total time signal exists 
TA=(-T/2:ts:T/2);                %time axis
N=numel(TA);
f=1e9;                     %frequency of the signal
Y=sin(2*pi*f*TA);          %sine function
plot(TA,Y);

FA=(-N/2:N/2-1)*fs/N;      %frequency axis
FFT=fft(Y);                %fourier transform of the signal
figure;
plot(FA,fftshift(abs(FFT)));
xlim([-2e9 2e9]);
xline(-1e9);

As shown below there's a slight deviation at the peak of the fft curve, I can't figure why there is deviation at the peak

Isn't the peak supposed to be at exactly at 1GHz.

xline(1e9);

Answer 1

[Paul]

fmax=1e9;                 %maximun frequency of the signal
fs=60*fmax; 
ts=1/fs;                  %time increment
T=10e-9;                  %total time signal exists 
TA=(-T/2:ts:T/2);                %time axis
N=numel(TA)

N = 601

f=1e9;                     %frequency of the signal
Y=sin(2*pi*f*TA);          %sine function
plot(TA,Y);

This expression for FA is incorrect when N is odd (601)

FA=(-N/2:N/2-1)*fs/N;      %frequency axis

The correct expression for N odd is

FA = (-(N-1)/2:(N-1)/2)/N*fs;

A general expression that works for N odd or even is

FA = ((0:N-1) - floor(N/2))/N*fs;

Use ifftshift on Y prior to the call to fft if the phase of the result is important

FFT=fft(Y);                %fourier transform of the signal
figure;
plot(FA,fftshift(abs(FFT)),'-o');
xlim([-2e9 2e9]);
xline(-1e9);

The small-but-nonzero points around the peaks arise because the discrete-time period of the sampled signal is P = 60, but the number of points in the data is 601, which is not an integer multiple of the period.

[Yogesh]

Thank you!!..