Register to reply

Physical meaning of Laplace transform

by pixel01
Tags: laplace, meaning, physical, transform
Share this thread:
pixel01
#1
Feb11-07, 02:04 PM
P: 691
Hi every body,

I am not so sure this kind of question is suitable here. In styding maths, I have some difficulties with the Laplace transform. The formulas are OK, I can apply them, but I still do not understand the physics meaning behind that. Can anyone can explain to me a little bit more abou this. Thanks
Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
ranger
#2
Feb11-07, 02:13 PM
PF Gold
ranger's Avatar
P: 1,684
By "physics meaning", do you mean like applications of Lt?
http://en.wikipedia.org/wiki/Laplace...s_and_theorems

Its also used a lot in circuit analysis.
arunma
#3
Feb11-07, 02:34 PM
P: 906
Quote Quote by pixel01 View Post
Hi every body,

I am not so sure this kind of question is suitable here. In styding maths, I have some difficulties with the Laplace transform. The formulas are OK, I can apply them, but I still do not understand the physics meaning behind that. Can anyone can explain to me a little bit more abou this. Thanks
Well I have a math degree, and I still couldn't tell you what the physical meaning of the Laplace Transform is. I don't think there is any. But maybe this will help: the Laplace Transform transforms a function into an entirely new variable space (traditionally called s). Since derivatives of all orders are eliminated, this turns a calculus problem into an algebra problem.

I realize this answer is insufficient, but I'm not sure there exists any good answer to your question. We physics people often get the idea that math was invented by physicists and for physicists, and that every mathematical idea has some physical basis. But I can't think of any obvious physical analogy for an integral transform.

rbj
#4
Feb11-07, 02:34 PM
P: 2,251
Physical meaning of Laplace transform

Quote Quote by pixel01 View Post
Hi every body,

I am not so sure this kind of question is suitable here. In styding maths, I have some difficulties with the Laplace transform. The formulas are OK, I can apply them, but I still do not understand the physics meaning behind that. Can anyone can explain to me a little bit more abou this. Thanks
are you good with the Fourier Transform? the Laplace Transform is an "analytical extension" of the Fourier Transform. and the Fourier Transform can be thought of the limiting case of the Fourier Series where the original [itex]x(t)[/itex] is truncated and periodically extended so that:

[tex] x_T(t) = x_T(t-nT) \ \ \ \forall \ \ \mathrm{integer} \ \ n [/tex]

and

[tex] x_T(t) = x(t) \ \ \ \mathrm{for} \ \ |t|< \frac{T}{2} [/tex]

and, in the limit, [itex] T \rightarrow \infty [/itex].

if you have this Fourier Series down solid, and then get an understanding of how the Fourier Transform generalizes the Fourier Series (to non-periodic [itex]x(t)[/itex]), then generalizing further to the (bilateral) Laplace Transform is a cake walk.
marcusl
#5
Feb11-07, 06:28 PM
Sci Advisor
PF Gold
P: 2,081
Because
[tex]s=\sigma + i\omega[/tex]
the Laplace kernel exp(-st) is more general than the Fourier kernel. Laplace transforms can readily handle dissipative systems; [tex]\omega[/tex] gives the angular frequency, and [tex]\sigma[/tex] the time constant, of a damped sinusoid.

The fundamental Fourier transform pair is [tex]\delta(t)[/tex] an infinite impulse excitation, and 1, indicating that the spectrum contains sinusoids of all frequencies. The fundamental Laplace transform pair is H(t), the Heaviside step function, and 1/s, its spectrum of damped sinusoids. Note that the spectrum is weighted towards low frequencies (1/abs(s) goes to zero as abs(s) goes to infinity), as one would expect for an excitation that begins but is never turned off.
planish
#6
Feb12-07, 04:19 AM
P: 33
Quote Quote by marcusl View Post
Because
[tex]s=\sigma + i\omega[/tex]
the Laplace kernel exp(-st) is more general than the Fourier kernel. Laplace transforms can readily handle dissipative systems; [tex]\omega[/tex] gives the angular frequency, and [tex]\sigma[/tex] the time constant, of a damped sinusoid.
I barely remember any of it, but in the dim past I recall having to learn about LaPlace transforms for a course in "Feedback and Control Systems". We would be figuring out the critical damping for things like robot arms, where the object was to get the arm into some new position as quickly as possible, yet slow it down as it approaches the end position in such a way that it doesn't overshoot. Then build the electronic circuit that controlled the servomotor. What a nightmare that was.
pixel01
#7
Feb13-07, 03:13 PM
P: 691
Thank you all for reply the thread. I think I ve got to read more about this and hope come back to this topic not very long.
EugP
#8
Jul22-07, 06:33 PM
P: 109
Quote Quote by pixel01 View Post
Hi every body,

I am not so sure this kind of question is suitable here. In styding maths, I have some difficulties with the Laplace transform. The formulas are OK, I can apply them, but I still do not understand the physics meaning behind that. Can anyone can explain to me a little bit more abou this. Thanks
From what I know, the Laplace transform lets you transform something from time domain to frequency domain. Doing this allows you to look at the behavior of that something (signal, circuit, etc.) in terms of freqency.

I hope this helped.
rbj
#9
Jul22-07, 10:53 PM
P: 2,251
i think the thread is pretty stale, but i thought the OP's question was more about whether the Laplace Transform of a function (say a function of time) that represents a physical quantity that varies in time (a physical "process", let's call it "x(t)"), if the value of the X(s) for some complex value of s represents a physical quantity that is some component of "x(t)". it can be answered more clearly if they OP was asking about the continuous Fourier Transform, of which L.T. is a generalization of that. i suppose there is an icky answer to the question by considering the closed-form expression of the inverse Laplace Transform with some fixed values of [itex]\mathrm{Re} \left\{ s \right\} [/itex]. but it's an icky answer.
Micko
#10
Jul23-07, 06:05 AM
P: 43
Well, in my opinion, there is no clear physical meaning of Laplace transform. Laplace transform is introduced as a utility tool to solve differential eqations more easily. As you know every dynamic behavior can be described as a set of differential equations. System's behavior is often described with set of linear differential equations with constant coefficients (linear systems or approximation of nonlinear systems).
Also Laplace transform is used to describe linear systems (relationship between input and output) in control theory.
On the other hand,
Fourier transform have physical meaning. If you apply Fourier transform to some signal x(t), you'll get that its frequency domain, i.e. picture of signal in frequency domain (base and higher harmonics).
AlephZero
#11
Jul23-07, 09:50 AM
Engineering
Sci Advisor
HW Helper
Thanks
P: 7,177
It's easy to see the "physical meaning" of time-domain differential equations, because you have been using algebraic variables to represent time, velocity, etc for many years already.

The s-plane (and the z-plane for digital signals) are just as "physically real", once you get familiar with them. Having different ways of looking at the same physical system is very valuable. If you use the s-plane for long enough, it becomes just as "real" as Euclidean space or Relativistic space-time.

Micko, The Fourier transformation is a bit simpler than the Laplace, but I don't see how you can claim that the frequency domain has a physical interpretation, but the s-plane does not. The imaginary axis in the s-plane IS the frequency domain!

However since you said "base and higher harmonics", perhaps you are only thinking about Fourier transformations of periodic functions? Or, you are confusing "Fourier Analysis" of periodic functions with "the Fourier Transform", which can be also used with non-periodic functions?
Micko
#12
Jul24-07, 12:27 AM
P: 43
Well, I'm not confusing Fourier transforms for periodic and aperiodic signals. I wasn't enough precise.
If signal is periodic, then Fourier transform becomes Fourier series (base and higher harmonics). For me, Fourier transform in contrast to Laplace has physical meaning because frequency response methods of analyzing systems are relying on Fourier transform. When system is described with Fourier transform it's easier to see how it is behaved (for example as low pass filter).
If for example I do Fourier transform of periodic (say rectangular) signal I can immediately see how this picture looks like in frequency domain, i.e. what are amplitudes of harmonics and similar...
abj
#13
Nov9-09, 02:34 PM
P: 1
The Laplace transform have physical meaning.

The Fourier transform analyzes the signal in terms of sinosoids, but the Laplace transform analyzes the signal in terms of sinousoids and exponentials.

Traveling along a vertical line in the s-plane reveal frequency content of the signal weighted by exponential function with exponent defined by the constant real axe value.

Traveling along a horisontal line reveal the exponential content of the signal weighted by sinousoids function with frequency defined by the constant imaginary axe value.

In particular, traveling along the imaginaly axe reveal frequency content of the signal weighted by 1 which is equivalent to the Fourier transform of the signal.

abj


Register to reply

Related Discussions
Physical Meaning Quantum Physics 4
Help on physical meaning General Math 6
Physical meaning of a fourier transform? General Physics 6
Finding an inverse Laplace Transform for a function - solving IVPs with Laplace Calculus & Beyond Homework 2
What is the physical meaning of this eqn Advanced Physics Homework 2