Frequency domain to Time domain Using IFFT : Symmetric figure problem
1 回表示 (過去 30 日間)
古いコメントを表示
Hi...
I want to convert the frequency domain data to time domain data...
But.. there is some problems...
clc
clear
close all
% Signal Generating
Fs = 3000;
Ts = 1/Fs;
t = 0 : Ts : 1-Ts;
f1 = 30;f2 = 70; f3 = 100;
w1 = 2* pi * f1; w2 = 2* pi * f2; w3 = 2* pi * f3;
theta1 = 0; theta2 = 0; theta3 = 0;
y = 1*exp(-1*t).*sin(w1.*t + theta1)+ ...
2*exp(-2*t).*sin(w2.*t + theta2)+ ...
3*exp(-3*t).*sin(w3.*t + theta3);
plot(t,y)
%% FFT & IFFT
[Frequency, Amplitude] = FFT_hg(t, y);
figure;
plot(Frequency, abs(Amplitude));
xlim([0 150])
B = ifft(Amplitude/2)*length(y);
t1 = (0:(1-Ts)/(length(B)-1):1-Ts);
figure;
B = real(B);
plot(t1,B)
grid on
From the above code...
Shape of figure(3) looks like symmetric...
I don't know, why the figure(1) and figure(3) are different...
Can you help me?
function [Frequency, Amplitude] = FFT_hg(Time, T_Amplitude)
fl=(length(Time)-1)/(max(Time)-min(Time));
L=length(T_Amplitude); %
LN=ceil(L/2);
A = fft(T_Amplitude);
X2=abs(2*A/L); %
Amplitude = X2(1:LN); %
f=fl*(0:L)/L;
Frequency=f(1:LN);
end
0 件のコメント
回答 (1 件)
Pat Gipper
2021 年 1 月 12 日
Your function produced a vector of complex numbers called "A". The original sequence will be reproduced using ifft(A). But the function went further by removing all the phase information by taking the absolute value of A.
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Fourier Analysis and Filtering についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!