Impulse response function h(t) of Gauss functions as a stimulus and response

Question

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?

Answer 1

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.

Answer 2

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]