I understand its usage and why is it important because It transforms differential equations to algebraic ones.. But I can't get the physical meaning of the new form of the equation and the meaning of this transformation.. and also what does it mean to change the domain of the function ?
Physical meaning behind Frequency domain?
3 Answers
instead of considering all the data as it happens in the times domain; chronologically,
consider the frequency domain ; how often the various values the occur over the entire time interval
In the frequency domain, you essentially restrict yourself to functions of the form $ x(t) = e^{\alpha t} \text{ where } \alpha \in \mathbb{C} $ Now, the derivative of such a function is just a scaled version of the same function because $ x'(t) = \alpha e^{\alpha t} $
Thus, if you restrict yourself to such functions, and if the system is linear with constant coefficients, then a differential equation simply becomes an algebraic equation for $\alpha$ once you divide by $e^{\alpha t}$. That algebraic equation thus characterizes all the solutions of the original differential equation which have the form $e^{\alpha t}$.
Now, $\Re(e^{\alpha t})$ is an exponentially damped sinusoidal wave where the real part of $\alpha$ determines the damping and the imaginary part the frequency. You may thus view the transformed equation as an equation which, instead of characterizing an a priori completely unknown function (as an ODE does), actually characterizes the frequencies and dampings of the solutions of the ODE.
Laplace transformation in its usual form is applied to onset-processes (OP's for short) $f:\ {\mathbb R}_{\geq 0}\to{\mathbb C}\ ,\qquad t\mapsto f(t)\quad(t\geq0)\ .$ The result ${\cal L}f$, defined by ${\cal L}f(s):=\int_0^\infty f(t)e^{-s\, t}\ dt\ ,$ has no physical interpretation whatever; so nobody has ever looked at the graph of an ${\cal L}f$. But the transformation $f\mapsto{\cal L}f$ has interesting formal properties which make it useful in applications: Differentiation with respect to $t$ is transformed into multiplication with $s$, etcetera. Above all ${\cal L}$ is injective. This implies that knowing the Laplace transform of some unknown OP $f$ it is in principle possible to get back $f$.
As a rule, Laplace transformation and its inverse is not applied to data, but to finite analytic expressions, using a set of rules and catalogues.
Contrasting this, Fourier transformation is applied to time signals (TS's) $f:\ {\mathbb R}\to{\mathbb C},\qquad t\mapsto f(t)\quad(-\infty
Fourier transform is applied to "abstract" functions in theoretical considerations, to "analytic expressions" in many applications where "solutions in finite terms" are desired, but also in a large extent to data coming from sampled TS's.