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Fourier transform of an impulse response function

22 vues (au cours des 30 derniers jours)
Jason Soukas
Jason Soukas le 30 Août 2023
Modifié(e) : Paul le 30 Août 2023
Hello,
It seems that I have trouble calculating an integral (Fourier transform). Here is my code:
syms t tau w f u
% SYSTEM
h = piecewise(t>=0, 0.01005*(exp(-t))*sin(9.949874371*t), t<0, 0);
%% Autocorrelation of h
h1 = subs(h,t,t-tau/2);
h2 = subs(h,t,t+tau/2);
%%
hh = h1*h2;
hh_simple = simplify(hh);
%% Fourier transform of the autocorrelation function
hh_test = hh_simple*exp(-j*2*pi*f*tau);
f3 = int(hh_test,tau,[-2*t 2*t]);
"h" is the impulse response function of a 1-dof system, "hh" its autocorrelation function and "f3" is the Fourier transform of the autocorrelation function.
"f3" is a function of frequency f and time t. When I assign some values to time "t", I plot the function f3 which is now of one variable, f. Seems like my plotted function has singularities and is not smooth. I was expecting a smooth curve with 1 peak (at the natural frequency of the system). Could someone please tell me if there is something wrong with it? Is there any problem with the integration?
Here is f3 at 3 time instants.
  2 commentaires
Paul
Paul le 30 Août 2023
Hi Jason,
What's the source for that definiton of the autocorrelation function?
Jason Soukas
Jason Soukas le 30 Août 2023
Modifié(e) : Jason Soukas le 30 Août 2023
Hello, thank you for the response!
The source is 3-4 papers I found, I attach a screenshot of one at the specific point you asked me for. I followed the formulas exactly, with the constraint that t>=0 (and as a result t-τ/2 >= 0 and t+τ/2 >= 0, which leads to -2t <= τ <= 2t ).

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Réponses (1)

Paul
Paul le 30 Août 2023
Modifié(e) : Paul le 30 Août 2023
Hi Jason,
syms t tau f real
% SYSTEM
I'd use heaviside, but piecewise works too.
h = piecewise(t>=0, 0.01005*(exp(-t))*sin(9.949874371*t), t<0, 0);
%h = 0.01005*(exp(-t))*sin(9.949874371*t)*heaviside(t);
%% Autocorrelation of h
h1 = subs(h,t,t-tau/2);
h2 = subs(h,t,t+tau/2);
%%
hh = h1*h2;
hh_simple = simplify(hh)
hh_simple = 
Safer to use sym(pi) here
%% Fourier transform of the autocorrelation function
hh_test = hh_simple*exp(-1j*2*sym(pi)*f*tau)
hh_test = 
f3 = int(hh_test,tau,[-2*t 2*t]);
%f3 = int(hh_test,tau,-inf,inf)
Plot f3 for some values of t
figure
fplot(subs(f3,t,1))
xline(9.949874371/2/pi,'r')
xlabel('f')
figure
fplot(subs(f3,t,5))
xline(9.949874371/2/pi,'r')
xlabel('f')

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