Different results when compearing DFT from fft to the "real" fourier transformation
    9 Ansichten (letzte 30 Tage)
  
       Ältere Kommentare anzeigen
    
    Lillebror Sagmen-Andersson
 am 18 Mai 2022
  
    
    
    
    
    Bearbeitet: Paul
      
      
 am 21 Mai 2022
            
I'm trying to compare the DFT computed from matlabs fft, to the "real" fourier transformation for the signal that can derived from the figure above. However I can't seem to get the the signals to match. 
I do not have access to syms for this. 
It is known that T = 3, and a = 1.  EDIT : T_0 = 3, || T =/= 3
I have tried doing it likes this:
x=@(t) exp(-a.*t).*((t>=0)&(t<=3)); % a = 1
T=T_0/N_0; %T_0 & N_0 may be choosen to achive T = 3. 
omega=linspace(-pi/T,pi/T,4097);
X = (1-exp(-(a+1i*omega)*T))./(a+1i*omega);
t=(0:T:T*(N_0-1))';
xf=T*x(t); 
xf(1)=T*(x(T_0)+1)/2; 
X_r=fft(xf); 
r=[-N_0/2:N_0/2-1]'; 
omega_r=r*2*pi/T_0;
then using 
subplot(211);
plot(omega,abs(X),'r',omega_r,fftshift(abs(X_r)),'bo');
xlabel('\omega');ylabel('|X(\omega)|')
%and
legend('True FT',strcat('DFT with T_0 = ',num2str(T_0),' , N_0 = ',num2str(N_0)));
subplot(212);
plot(omega,angle(X),'r',omega_r,fftshift(angle(X_r)),'bo');
xlabel('\omega'); ylabel('\angle X(\omega)')
legend('True FT',strcat('DFT with T_0 = ', num2str(T_0),' , N_0 = ',num2str(N_0)));
However as you can see the plots do not agree, and I don't see whats is going wrong. 

0 Kommentare
Akzeptierte Antwort
Weitere Antworten (3)
  Paul
      
      
 am 20 Mai 2022
        
      Bearbeitet: Paul
      
      
 am 21 Mai 2022
  
      Hi Lillebror,
Referring only to the code in the original question, it looks like there is a mix up between the variables used to define the duration of the signal and the sample period to generate the samples of the same, but it's hard to say because the entire code is not posted in the question. I think you're looking for something like this.
Define the signal of interest
T = 3; % T defines the duration of the signal
a = 1;
xfunc = @(t) exp(-a.*t).*((t >= 0) & (t <= T)); % a = 1, T = 3
Pick a sampling period Ts. Though not really necessary, choose Ts such that T/Ts is a nice integer
N = 31;
t = linspace(0,T,N);
Ts = t(2); % Ts is the sampling period
Define the frequency vector for the CTFT of x(t) and then compute the CTFT
omega_c = linspace(-pi/Ts,pi/Ts,4097);
XCTFT = (1 - exp(-(a + 1i*omega_c)*T)) ./ (a + 1i*omega_c);
Samples of x(t) 
x = xfunc(t);
Adjust the endpoint to account for the effect of impulse sampling at discontinuities. I think this is similar to your adjustment of xf(1).
x(1) = x(1)/2;
x(end) = x(end)/2;
Compute the DFT
XDFT = fft(x);
Do the fftshift and get the associated frequency vector for N odd
XDFT = fftshift(XDFT);
omega_n = (-(N-1)/2 : (N-1)/2)*2*pi/N/Ts;
Compare, note the scaling by Ts on the DFT
figure
subplot(211);hold on
plot(omega_c,abs(XCTFT));
stem(omega_n,abs(Ts*XDFT));
subplot(212);hold on
plot(omega_c,180/pi*angle(XCTFT));
stem(omega_n,180/pi*angle(T*XDFT));
xlabel('rad/sec')
0 Kommentare
  Michal
      
 am 18 Mai 2022
        Can you clarify the purpose of this operation (why multiply by T?):
t=(0:T:T*(N_0-1))';
xf=T*x(t) 
I'm not following your operations with the time axis, sorry
0 Kommentare
Siehe auch
Kategorien
				Mehr zu Install Products 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!

