| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I have a Laplace transform problem which is a bit tricky.
There are two possible approaches. Perhaps there might be some
suggestions forthcoming as to how this particular problem can be
solved.
I would like to know if there is an inverse Laplace tranform of
either of the following two general functions.
-1
L { f(g(s)) } = ???
-1 { � }
L { a [f(s)] } = ???
{ e }
Thanks,
Lewis.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 978.1 | don't know of any | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Sun Dec 04 1988 12:34 | 9 |
I assume you want the results expressed in terms of
-1 -1
L { f(s) } and L { g(s) }. I don't recall any such
general formula for Laplace transforms and doubt that one
exists. I only vaguely recall a connection between
multiplying by x on one side and differentiating on the
other.
Dan
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| 978.2 | Try a different tack | AUSSIE::BENHAM | Lewis, CSS NSG-S Engineering | Mon Dec 12 1988 17:42 | 29 |
OK, seems like no go on those ones in .0. Let's try a different
approach. How about the following.
�
[ ] �
f(s) = [ ( a + bs ) ( c + ds ) ] = g( s )
[ ]
Find g(s) ?
Also I have managed to find a couple of Laplace tranform tables.
One result in particular is:
{ � }
{ -as }
-1 { e } ( a )
L { ----- } = erfc(-----)
{ s } ( � )
( 2t )
Could anyone provide a proof of this result and thus in combination
with the result for the top expression, find a solution to my problem.
Thanks,
Lewis.
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