why this fourrier transform is not good?

1 次查看（过去 30 天）
Kobi
Kobi 2016-7-15
评论： Kobi 2016-7-16
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])

请先登录，再回答此问题。

采纳的回答

Geoff Hayes
Geoff Hayes 2016-7-16
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.
  1 个评论
Kobi
Kobi 2016-7-16
if i change the script as you said, i can't sample the signal properly, i get something like a sawtooth wave

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Parametric Spectral Estimation 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by