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

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Yogesh
Yogesh 2024-5-5
回答： Paul 2024-5-5
Ran in:
clear all
close all
clc
L=10;
n=1.45;
c=2.9979e8;
dt=6e-12;
T=10*2*L*n/c;
fmax = 1e9;
fs=80*fmax;
TA=-T/2:dt:T/2;
%t = (-T/2/dt:1:T/2/dt)*dt;
Nt=round(T/dt);
vsine = 1;
phi = vsine*sin(2*pi*fmax*TA);
EL1t=1.274e7*exp(1i*phi);
plot(TA,(EL1t));
Warning: Imaginary parts of complex X and/or Y arguments ignored.
%FA = ((0:Nt-1)-floor(Nt/2))/Nt*fs;
FA = (-Nt/2:Nt/2-1)/Nt*fs*2;
FP=fft(phi);
%fs=1/dt/Nt;
figure;
Z=plot(FA,fftshift(abs(fft(EL1t/Nt))));
xline(1e9);
xlim([-2e9 2e9]);
As per the logic , each peak should have been at the multiples of 1GHz like the first peak should have been at 1GHz and the next at 2GHz and so on.
But evidently there has been a slight devition.

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回答（1 个）

Paul
Paul 2024-5-5
The code has
fs = 80*fmax;
but the TA vector is space by dt, so fs should be
fs = 1/dt;

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