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From the basics of the signal theory the impulse response function (IRF) can be calculate like (L=Laplace transformation): L[x(t)]= X(s) , L[y(t)]= Y(s) and transfer function H(s)= Y(s)/X(s), then IRF is inverse Laplace transformation of transfer function h(t)=L−1[H(s)]. By convolution of input signal (stimulus) and IRF h(t) can be calculate output (response) of the system y(t)=h(t)*x(t).

I tried to solve my problem firstly with Fourier transformations. Since Gaussian is analytical function, its Fourier transformation can be found analytically,and than can be found impulse response of the system. I did some calculations but discrepancy of result convolution is too high(y(t)=h(t)*x(t) compare to original response y(t)). Possibly i did some mistakes in calculations or it have to be solved by Laplace, but Laplace transformation of Gaussian result in error functions and algebraic solution is very complicated.

stimulus: f(t)=(A/(s Sqrt[2 Pi])) Exp[(-(t-c)^2)/(2 s^2)]

response: g(t)=(B/(r Sqrt[2 Pi])) Exp[(-(t-d)^2)/(2 r^2)]

where:

A- Aera of stimulus B- Aera of response c- Center of stimulus d- Center of response s- Sigma of stimulus r- Sigma of response

than impulse response is:

h(t)=F-1{F[g(t)]/F[f(t)])}= B/A Sqrt[2]/(E^((t+c-2d)^2/(4 (r^2 + s^2))) Pi Sqrt[r^2 + s^2])

Plese, can someone calculate it properly or even calculate it with Laplace transformations?

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  • $\begingroup$ Is this a question related to the software Mathematica or is this related to the mathematical problem you are trying to solve? If it is Mathematica maybe you can show the code that you have tried till now. $\endgroup$ – Dunlop Feb 19 at 14:43
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can someone calculate it properly or even calculate it with Laplace transformations?

Your question is little hard to follow. You mentioned FourierTransform and then mention LaplaceTransform.

Using LaplaceTransform, since $Y(s)=X(s)H(s)$ then $H(s)=\frac{Y(s)}{X(s)}$ so you could find $h(t)$ as follows (I called your $f(t)$ as $x(t)$ and your $g(t)$ as $y(t)$ as these are the common notation and used $z$ instead of $s$ for Laplace since you already have $s$ used in your equations.

ClearAll[f, t, g, c, d, r, A, B, z];
x[t_] := (A/(s Sqrt[2 Pi])) Exp[(-t - c^2)/(2 s^2)];
y[t_] := (B/(r Sqrt[2 Pi])) Exp[(-t - d^2)/(2 r^2)];
H0 = LaplaceTransform[y[t], t, z]/LaplaceTransform[x[t], t, z];
h = InverseLaplaceTransform[H0, z, t]

$$ \frac{B s e^{\frac{c^2}{2 s^2}-\frac{d^2}{2 r^2}} \left(\delta (t)+\frac{\left(r^2-s^2\right) e^{-\frac{t}{2 r^2}}}{2 r^2 s^2}\right)}{A r} $$

If this is not what you meant, then will delete this answer.

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  • $\begingroup$ Thank You Nasser for your answer. Firstly i am sorry for my confusing terminology and secondly I am sorry for that, I forgot brackets in exponent of input functions, this is why You get such a nice result. In real it have to be like this: ClearAll[f, t, g, c, d, r, A, B, z]; x[t_] := (A/(s Sqrt[2 Pi])) Exp[(-(t - c)^2)/(2 s^2)]; y[t_] := (B/(r Sqrt[2 Pi])) Exp[(-(t - d)^2)/(2 r^2)]; H0 = LaplaceTransform[y[t], t, z]/LaplaceTransform[x[t], t, z]; h = InverseLaplaceTransform[H0, z, t] $\endgroup$ – Antonín Opíchal Feb 20 at 14:02
  • $\begingroup$ I will correct it in my Question also $\endgroup$ – Antonín Opíchal Feb 20 at 14:03
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Thank You Nasser for your answer. Firstly i am sorry for my confusing terminology and secondly I am sorry for that, I forgot brackets in exponent of input functions, this is why You get such a nice result. In real it have to be like this:

ClearAll[f, t, g, c, d, r, A, B, z];
x[t_] := (A/(s Sqrt[2 Pi])) Exp[(-(t - c)^2)/(2 s^2)];
y[t_] := (B/(r Sqrt[2 Pi])) Exp[(-(t - d)^2)/(2 r^2)];
H0 = LaplaceTransform[y[t], t, z]/LaplaceTransform[x[t], t, z];
h = InverseLaplaceTransform[H0, z, t]

Result is not satisfying

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