Filter löschen
Filter löschen

FFT vs Table Fourier Pairs

3 Ansichten (letzte 30 Tage)
Eugene
Eugene am 5 Feb. 2012
Beantwortet: Sadiq Nasser am 8 Jan. 2021
Dear friends, I done FFTs of good known functions from Fourier Transform pairs table. The result of FFT doesn't match analytic transform in most cases by magnitude value, and in example 4 also by envelopment form. Code examples below. Where I wrong, please ?
fs=1000; % sampling rate
dt=1/fs; % time step
Sec=5; % Signal length
N=Sec*fs; % number of samples
n=(0:N-1)-N/2;
tn=n*dt; % Time axis
df = fs/N; % Frequency resolution
fk=n*df; % Frecuency axis [Hz]
wk=fk*2*pi; % Angular frequency axis [Rad/sec]
%--------------
% Example 1
% Time domain
alfa=7;
ex1=exp(-alfa.*abs(tn));
% FFT result
Ex1a=fftshift(fft(ifftshift(ex1)));
% Analytics Fourier transform
Ex1b=(2*alfa)./((alfa^2)+(wk.^2));
figure(1);
subplot(3,1,1);
plot(tn,ex1);
grid;
title('e^{-a|t|}');
subplot(3,1,2);
plot(wk,real(Ex1a));
grid;
title('Ex1a');
subplot(3,1,3);
plot(wk,real(Ex1b));
grid;
title('2a/(a^2+w^2)');
%--------------
% Example 2
% Time domain
sigma=0.5;
ex2=exp(-(tn.^2)/(2*(sigma^2)));
% FFT result
Ex2a=fftshift(fft(ifftshift(ex2)));
% Analytics Fourier transform
Ex2b=sigma*sqrt(2*pi)*exp(-(sigma^2).*(wk.^2)/2);
figure(2);
subplot(3,1,1);
plot(tn,ex2);
grid;
title('e^{-t^2/2s^2}');
subplot(3,1,2);
plot(wk,real(Ex2a));
grid;
title('Ex2a');
subplot(3,1,3);
plot(wk,real(Ex2b));
grid;
title('s*sqrt(2pi)*e^{-s^2w^2/2}');
%--------------
% Example 3
% Time domain
ex3=1i./(pi.*tn);
% FFT result
Ex3a=fftshift(fft(ifftshift(ex3)));
% Analytics Fourier transform
Ex3b=-1*(wk<0)+1*(wk>0);
figure(3);
subplot(3,1,1);
plot(tn,abs(ex3));
grid;
title('-i/(t*pi)');
subplot(3,1,2);
plot(wk,real(Ex3a));
grid;
title('Ex3a');
subplot(3,1,3);
plot(wk,real(Ex3b));
grid;
title('(w<0)=>-1;(w>0)=>1');
%--------------
% Example 4
% Time domain
ex4=cos(alfa.*(tn.^2));
% FFT result
Ex4a=fftshift(fft(ifftshift(ex4)));
% Analytics Fourier transform
Ex4b=sqrt(pi/alfa)*cos((wk.^2)/4/alfa-pi/4);
figure(4);
subplot(3,1,1);
plot(tn,ex4);
grid;
title('cos(a*t^2)');
subplot(3,1,2);
plot(wk,abs(Ex4a));
grid;
title('Ex4a');
subplot(3,1,3);
plot(wk,abs(Ex4b));
grid;
title('sqrt(pi/a)*cos(w^2/4a-pi/4)');
%--------------
% Example 5
% Time domain
ex5=2*cos(2*pi*4*tn)+3*cos(2*pi*7*tn);
% FFT result
Ex5a=fftshift(fft(ifftshift(ex5)));
figure(5);
subplot(2,1,1);
plot(tn,ex5);
grid;
title('2cos(4wt)+3cos(7wt)');
subplot(2,1,2);
plot(wk,abs(Ex5a));
grid;
title('Delta(-14pi)+Delta(-8pi)+Delta(8pi)+Delta(14pi)');

Akzeptierte Antwort

Dr. Seis
Dr. Seis am 5 Feb. 2012
Multiply your fft result by "dt".
  1 Kommentar
Eugene
Eugene am 8 Feb. 2012
Dear Elige,
thank you a LOT

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (3)

Sadiq Nasser
Sadiq Nasser am 8 Jan. 2021
f(t)=cos⁡(10πt-3)e^(-j10ωt)

Sadiq Nasser
Sadiq Nasser am 8 Jan. 2021
f(t)=coc^2(10πt-3)e^(-i10ωt)

Sadiq Nasser
Sadiq Nasser am 8 Jan. 2021
f(t)=coc^2(10πt-3)e^(-i10ωt)

Kategorien

Mehr zu Fourier Analysis and Filtering finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by