Using IFFT for obtaining time response of measured freq response
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Hello, I have numerical frequency response data (G(s=j2*pi*f), f) for a system. I want to obtain the impulse response in time domain. I should be able to obtain this using the IFFT function. But the scaling is not clear to me. Here is the code I am using. What am I missing? Thanks. Amit.
clc; close all;
fmax = 1e6;
L = 1024;
fdelta = 2*fmax/(L-1)
fgrid = [-fmax:fdelta:fmax]; %fmax = fs/2 => deltaT = 1/(2*fmax)
tau = 1/(2*pi*5e4)
% Y(s) = 1/(1+s.tau) Given Freq Domain data
Y = 1./(1+j*2*pi*fgrid*tau);
y_t = ifft((Y));
Tdelta = 1/(2*fmax);
t = [0:Tdelta:Tdelta*(length(y_t)-1)];
% compare with known time domain function
figure; plot(t, y_t, 'bx', t, exp(-t/tau), 'r');
grid; axis([0 1e-4 0 1])
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Respuestas (4)
Rick Rosson
el 28 de Dic. de 2011
Please try:
Fs = 2e6;
L = 1024;
dF = Fs/L;
f = (-Fs/2:dF:Fs/2-dF)';
s = j*2*pi*f;
Fc = 50e3;
alpha = 2*pi*Fc;
tau = 1/alpha;
Y = alpha./(s+alpha);
dt = 1/Fs;
t = dt*(0:L-1)';
y = L*ifft(ifftshift(Y));
x = exp(-t/tau);
HTH.
Rick
2 comentarios
Rick Rosson
el 28 de Dic. de 2011
Do you have access to either the Control Systems Toolbox or the Signal Processing Toolbox? If so, which one (or both)?
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Rick Rosson
el 29 de Dic. de 2011
I modified the code I posted earlier to correct the scaling factor:
Fs = 2e6;
L = 1024;
dF = Fs/L;
f = (-Fs/2:dF:Fs/2-dF)';
s = j*2*pi*f;
Fc = 50e3;
alpha = 2*pi*Fc;
tau = 1/alpha;
Y = alpha./(s+alpha);
dt = 1/Fs;
t = dt*(0:L-1)';
G = 2*pi;
y = G*abs(ifft(ifftshift(Y)));
x = exp(-t/tau);
figure;
plot(t,x,t,y);
HTH.
Rick
1 comentario
Greg Heath
el 30 de Dic. de 2011
FFTing and IFFTing is tricky business because you have to be perfectly clear about original assumptions. For example, time domain signals contain N measurements at a sampling frequency
Fs. If the signal is real the corresponding nonnegative frequency spectrum measurements have a spacing df = Fs/N and
have either the range
f = df*[0:N/2], Neven
fmax = Fs/2
L = length(f) = N/2+1
with L odd when N is even
or
f = df*[0:(N-1)/2], N odd
fmax = Fs/2 - df/2
L = (N+1)/2
with L even when N is odd
Now since you seem to have L = 1024 measurements, my conclusion is that
N = 2*L + 1 = 2049 is odd
and
Fs = 2*N*fmax/(N-1) = 2.001e6
Hope this helps.
Greg
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