why this fourrier transform is not good?
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
this script should calculate the fourier transform of a single sine with frequency of 1KHz and for some reason i can't see the pulses in the frequency domain at the -1000 and 1000, please help
clear all
close all
clc
% Downsampling
f_max=1e3;
fs=4e3;
t=linspace(0,1/f_max,fs);
y=sin(2*pi*f_max*t);
frequency_domain=linspace(-fs/2,fs/2,length(t));
figure;
col=4; row=2;
subplot(col,row,1)
plot(t,y)
title('\bf The Original signal')
ylabel('Amplitude')
xlabel('Time')
grid on
subplot(col,row,3)
stem(t,y)
title('\bf The sampled signal')
ylabel('Amplitude')
xlabel('Time')
grid on
% sampled signal in the frequency domain
subplot(col,row,4)
Y=fft(y)/length(y);
Y=fftshift(Y);
plot(frequency_domain,abs(Y))
title('\bf The Original signal in the frequency domain')
ylabel('Magnitude')
xlabel('frequency')
grid on
xlim([-2e3 2e3])
0 commentaires
Réponse acceptée
Geoff Hayes
le 16 Juil 2016
Kobi - why is your time domain initialized as
t=linspace(0,1/f_max,fs);
Aren't you interested in the full second since fs is the sampling frequency (number of samples per second)? Instead, try
t = linspace(0,1-1/f_max,fs);
and see what happens.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Parametric Spectral Estimation dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!