Procedure for inverse Laplace transformation to calculate periodic switch-on processes

Hi, The calculation of switch-on processes according to Laplace often involves the problem of transforming back into the time domain.

The inverse Lapace transformation is almost impossible to solve analytically, especially in real lines with excitation with hormonal functions.

FFT-based, numerical methods are suitable for the numerical inverse Laplace transformation.

One of the oldest numerical methods is the Koizuni algorithm.

The numerical, inverse Laplace transformation for

w/(p^2+w^2) with w=2*Pi*50 Hz

is shown in this Mathematica-example.

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Regards :)

Reply to
Leo Baumann
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The old-fashioned one-sided Laplace transform isn't too useful anymore. It's suited to initial-value problems, as long as you don't mind having to look up the inverse transforms in books.

The modern fashion is to use two-sided Fourier/Laplace transforms, with or without a Heaviside unit step function to turn it into a one-sided problem. Whether or not to substitute s for (j omega) is down to personal taste. One important reason for this is that it's possible to pick functions that have the right behaviour but are easier to calculate with.

I generally stick to Fourier myself. (See Bracewell for more.)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Reply to
Phil Hobbs

Am 27.01.2021 um 03:56 schrieb Phil Hobbs:

It is not possible to analytically solve harmonic switch-on processes on real lines with Fourier or Laplace. Most real lines are distorting because the Heaviside condition is not met.

The problem is the inverse transformation.

In the case of real lines, you have no choice but to use the numerical Koizuni algorithm, for example.

There are other FFT-based algorithms for harmonic functions in the time domain, but I've had good experiences with Koizumi, when a analytic calculation was impossible, either with Fourier or Laplace.

Fourier can only be used for the steady state.

:)

Reply to
Leo Baumann

Here an example of a cosinus switch-on to a real 100 km long 380 kV-4-bunch overhead line:

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regards

:)

Reply to
Leo Baumann

Why not use LTspice to get the turnon, steady state and frequency response?

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Reply to
Steve Wilson

You're cracked. (But we already knew that, don't be discouraged.) ;)

Fourier and Laplace are mathematically exactly equivalent apart from needing that Heaviside unit step to make a one-sided transform.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Am 27.01.2021 um 12:14 schrieb Phil Hobbs:

You can use Fourier to calculate switch-on processes, but You need to transform the stimulus into frequency-area, and later transform the results of multiplication with U2/U0(w) back into time-area.-

But tell me, how You want to traform a equitation like this?

(Z2*Cosh[Sqrt[(Rs + p*Ls)*(Gs + p*Cs)]*(l - x)] + Z0*Sinh[Sqrt[(Rs + p*Ls)*(Gs + p*Cs)]*(l - x)])/((Z1 + Z2)* Cosh[Sqrt[(Rs + p*Ls)*(Gs + p*Cs)]*l] + (Z0 + Z1*Z2/Z0)* Sinh[Sqrt[(Rs + p*Ls)*(Gs + p*Cs)]*l])

lol

:)

Reply to
Leo Baumann

This is what I would call the best "On Topic" pissing contest on S.E.D for a long time

Reply to
Brent Locher

You can use Fourier transform pretty straightforwardly IIRC to solve the problem of e.g. a DC step being applied to one end of a long transmission line. It becomes equivalent to a diffusion equation, DC current "diffuses" into a long line of this type.

Reply to
bitrex

This isn't the 18th century anymore. Time to get back to numerically solving the circuit differential equations. You know those things from which the transforms are derived...

Reply to
Fred Bloggs

You probably don't. You have to construct a boundary-value problem in both time and space for the particular solution you're actually looking for, not the infinity of solutions you get by just bunging the Laplace transforms of the stimulus and the transfer function together and then trying to invert that.

An infinite half-line driven by e.g. a sine switched on at 0 that has series resistance and shunt conductance has bounded behavior both in time and in the dimension of x and you have to enforce what they is.

See also e.g.:

Reply to
bitrex

Am 27.01.2021 um 15:53 schrieb Fred Bloggs:

For solving 100 km of such a line in time area You need about a ladder network with 1000 equivalent extended Pi-networks - I have tried it

Reply to
Leo Baumann

Sometimes it's productive in these type of problems where there are two independent variables that have boundary conditions in both dimensions, and they can't be easily uncoupled by exploiting a symmetry/linearity,is to apply a sequence of transforms in series, first Laplace to the time variable and then Fourier to the space variable, or vice versa whatever makes it easiest, solve with boundary conditions on the second step, transform back, solve with boundary conditions on what you get from that, transform back.

Reply to
bitrex

Right, and I estimate the chances someone hasn't solved the telegraph equation on a long cable driven by a switched-sine starting at 0 analytically in a book somewhere to be about 0%, it's why they called it the telegraph equation there was a great deal of interest in solutions to it long before digital computers showed up. Trouble is finding where that solution is at.

Reply to
bitrex

I'm sure there's some kind of non-linear extension to the concept of circuits terminated in their characteristic impedance that can be used without solving for every millimeter of infinitesimal bit of length.

Reply to
Fred Bloggs

Am 27.01.2021 um 10:36 schrieb Steve Wilson:

lol

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:)

Reply to
Leo Baumann

It's a lead-pipe sinh. (*)

(I'm sure you picked that ugly mess completely at random.)

Cheers

Phil Hobbs

(*) pronounced 'cinch'.

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Reply to
Phil Hobbs

When I say that Fourier and Laplace are mathematically equivalent, what I mean is that the one becomes the other by a very simple change of variable, s = j omega (s = -i omega for physicists), and multiplying by the Heaviside unit step.

'Tain't rocket surgery, even slightly.

Splitting out the unit step as a multiplicative function allows one to freely use the full power of the simple Fourier theorems and analytic continuations, and hack together useful approximations for a very wide range of cases.

Bracewell is a very good read on that sort of stuff. I took his course back in 1985 iirc, and it was an excellent use of three hours a week plus problem sets. (He was a pretty good guy, despite being Australian--he wasn't out to kill anybody as far as I know. May God hold him in memory eternal.)

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Heaviside has a lot of things named after him. The transatlatic cable thing is one, but there's also the Heaviside Layer and the Heaviside unit step, among many others. Smart guy, if not necessarily the most clubbable.

The unit step function is H(t) = {0 (t 0)}

with various approaches to the removable singularity at 0, commonly setting it to 0.5. It's a set of measure zero, so it doesn't affect any of the relevant integrals.

Cheers

Phil Hobbs

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Reply to
Phil Hobbs

Numerical solutions have a far lower information density than a well-chosen analytical approximation.

Cheers

Phil Hobbs (Who uses both routinely--horses for courses)

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Reply to
Phil Hobbs

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