where does s domain in laplace physically exist..? I can solve the maths part of it?


by aditya23456
Tags: domain, exist, laplace, maths, physically, solve
aditya23456
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#1
Apr22-12, 02:58 AM
P: 111
I need a physical explaination of s domain..Is s-domain a higher dimensional plane..?
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aditya23456
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#2
Apr27-12, 03:21 AM
P: 111
Is my queston valid??
Ali, Ahmed
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#3
Apr27-12, 08:44 AM
P: 21
- I can't say whether your question is entirely valid or not because the truth is I don't precisely know the answer to your question.

- What I do know however, is that in modelling physical systems, we call the s-domain the "frequency domain" whereas the t-domain would be called the "time-domain." We use the s-domain to model the frequency characteristics of a system (the response of the system to a sinusoidal input).

- In mathematics, I just see the s-domain as a tool to bring a calculus problem into the realm of pure algebra.

- Your question is intriguing though, and I would like to hear a more solid answer!

sri sharan
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#4
Apr27-12, 02:16 PM
P: 32

where does s domain in laplace physically exist..? I can solve the maths part of it?


No, Its not a higher dimensional plane whatever you may mean by that. It is just a mathematical technique in which you write it is a function of s
Hechima
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#5
Jun14-12, 11:58 AM
P: 10
Here's one way of looking at it that I believe is useful:

Consider the function
[itex] f(t) = A cos(\omega_0t) = Real\{A e^{i \omega_0 t}\} [/itex]

It is periodic, with period [itex]T = \frac{2π}{\omega_0}[/itex]

Now consider
[itex] g(t) = e^{-\alpha t}[/itex]

It is transient, with a time constant [itex]\tau = \frac{1}{\alpha}[/itex]

So, what about something like the following?
[itex] h(t) = A e^{-\alpha t} cos(\omega_0t) [/itex]

It oscillates, but it isn't actually periodic because it also decays. The spacing between the local maxima doesn't even match up with the period of the pure cosine (check this yourself). So how can we keep going in the frequency domain?

Well, [itex] h(t) = Real\{A e^{(-\alpha + i \omega_0) t}\} [/itex]

Then you can say it has a sort of "generalized frequency" of [itex] s = -\alpha + i \omega_0 [/itex]

So, in some sense the Laplace transform variable represents a point in the complex plane. The real part is representative of a transient decay (or growth if positive), and the imaginary part represents oscillation frequency. The units of 's' are reciprocal time, so it can be regarded as a generalization of a time constant or frequency variable.

This approach is used extensively in classical feedback and control theory.

Here's an example where a function in the 's' domain is analyzed by plotting it's poles and zeros in the complex 's' plane.
http://cnx.org/content/m10112/latest/
algebrat
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#6
Jun14-12, 12:59 PM
P: 428
I don't really think physically that it is the frequency domain. In a sense, mathematically, because it can be related to the fourier expansion, where we are mapping more literally to the frequency domain. For the laplace, I would say we're more mapping to the decaying exponential response domain. If each ping that f gives at each x results in a response of a decaying exponential e^(-sx)=1...0 as x=0...∞.

We get contributions from each s value? As s gets larger, that's a faster decaying exponential. If we assign the coefficient 1/s to each e^(-sx), then we get the inverse laplace of 1/s which is a constant. So conceptually, we have Ʃ(1/s)e^(-sx) over all s gives us f(x)=1. I'm being grossly negligent and incorrect here, because the inverse laplace integral form is a complex residue problem along a vertical line in the complex plane, but just wanted to add some intuition like you get in fourier version.

While I am pretty sure my attempt is false, my guess is that it is more correct than saying that laplace maps to the frequency domain.
Vargo
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#7
Jun14-12, 07:50 PM
P: 350
By dimensional analysis, you have e^(-st), so s should have units inverse to t. In other words, it is still frequency. And of course, if you set s = i omega, you have exactly the frequency of the fourier transform. So you can say that s is in the complex angular frequency domain.

See http://en.wikipedia.org/wiki/Laplace_transform

This sentence in particular might help put meaning to the Laplace transform:
"The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments."
algebrat
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#8
Jun15-12, 02:01 AM
P: 428
Quote Quote by Vargo View Post
By dimensional analysis, you have e^(-st), so s should have units inverse to t. In other words, it is still frequency.
"[In other words, 1 over seconds is always frequency]"? Only for periodic motion, which the decaying exponentials are not. The factor s is more like the time to decrease from 1 to 1/e. Now we're getting somewhere, thanks for giving me something to bite into.

And of course, if you set s = i omega, you have exactly the frequency of the fourier transform. So you can say that s is in the complex angular frequency domain.
At which point physical interpretation continues to fall apart. This is not a good analogy, though I wouldn't assume that examining the strange inverse laplace transform would not help develop the intuition somehow, but if I had to guess, I doubt it.

See http://en.wikipedia.org/wiki/Laplace_transform

This sentence in particular might help put meaning to the Laplace transform:
"The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments."
Yes, I've seen that statement. It is not given much support or explanation, perhaps it should be flagged as such on wikipedia. Perhaps my statement above is a half-baked attempt in the correct direction. I think I've done a google search before and not found much on the connection between moments and laplace transform.

Maybe s is the "moment" of time time it takes for response to decay from 1 to 1/e? Just kidding, but sort of.
Vargo
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#9
Jun15-12, 08:23 AM
P: 350
The connection to moments is simple and explicit. The Laplace transform is (almost) the generating function for the moments of f(x). The nth "Moment" in this context means

$$ \int_0^\infty x^n f(x)\, dx.$$

The integrals are only over half the line because that is the relevant "support" of f in the context of the Laplace transform.

See http://en.wikipedia.org/wiki/Moment-generating_function to see the connection in the context of probability theory.


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