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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Réponses (4)
Rick Rosson
le 28 Déc 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
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Rick Rosson
le 28 Déc 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
le 29 Déc 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
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Greg Heath
le 30 Déc 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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