What is the physical meaning of Laplace operator?
What is the physical meaning of Laplace operator?
(OP)
According to some dynamics book, it says the 2nd differential equation can be transformed to just a simple algbraic equation by applying Laplace transformation and I know how it works and what is the beneit of doing that in terms of the ease of caluculation. However what I want to know is the physical meaning of Laplace transform and how it is related to tranfer function. I want to know someting that is not explained in basic dynamics book about Laplace.





RE: What is the physical meaning of Laplace operator?
transforms represent a one-to-one mapping between the time and frequency response of linear systems.
the frequency response is algebraic as opposed to solving differential equations in time, and is simpler to work with.
RE: What is the physical meaning of Laplace operator?
mx''+cx'+kx = f(t)
algebraic Laplace equation
ms^2X(s) + cX(s)+ kX(s) = F(s)
assumed x(0)=x'(0)=0.
Laplace transfer function:
H(s) = X(s)/F(s) = 1 / [ms^2+cs+k]
Fourier transfer function: substitute s=jw
H(w) = X(w)/F(w) = 1 / [-mw^2+jcw+k]
Physical meaning of the fourier transfer function (function of frequency): gives complex amplification factor of the system for sinusoidal input.
RE: What is the physical meaning of Laplace operator?
It is a mathematical tool - like complex arithmetic.
Would you ask the question, "What is the physical meaning of complex arithmetic?"?
(maybe you would?!?!)
M
--
Dr Michael F Platten