# Physical meaning of Laplace transform

by pixel01
Tags: laplace, meaning, physical, transform
 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
 PF Patron P: 1,686 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.
P: 907
 Quote by pixel01 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.

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## Physical meaning of Laplace transform

 Quote by pixel01 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 $x(t)$ is truncated and periodically extended so that:

$$x_T(t) = x_T(t-nT) \ \ \ \forall \ \ \mathrm{integer} \ \ n$$

and

$$x_T(t) = x(t) \ \ \ \mathrm{for} \ \ |t|< \frac{T}{2}$$

and, in the limit, $T \rightarrow \infty$.

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 $x(t)$), then generalizing further to the (bilateral) Laplace Transform is a cake walk.
 PF Patron Sci Advisor P: 1,998 Because $$s=\sigma + i\omega$$ the Laplace kernel exp(-st) is more general than the Fourier kernel. Laplace transforms can readily handle dissipative systems; $$\omega$$ gives the angular frequency, and $$\sigma$$ the time constant, of a damped sinusoid. The fundamental Fourier transform pair is $$\delta(t)$$ 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.
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 Quote by marcusl Because $$s=\sigma + i\omega$$ the Laplace kernel exp(-st) is more general than the Fourier kernel. Laplace transforms can readily handle dissipative systems; $$\omega$$ gives the angular frequency, and $$\sigma$$ 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.
 P: 2,246 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 $\mathrm{Re} \left\{ s \right\}$. but it's an icky answer.