View Full Version : Laplace inverse transform
eljose79
Sep25-04, 09:13 AM
Let,s suppose we want to get the inverse Laplace transform of a function f(s) numerically,we should calculate the integral from (c-i8,c+i8) of exp(st)f(s) my question is what c we should choose for calculating the integral?..wouldn,t depend the integral of the value of c..where could i find the proof that the Inverse Laplace transform does not depend on the value of c chosen?..
Another question,let,s take the Laplace transform in 2 dimensions hten how would we define the Laplace inverse transform?..
Feynman
Sep25-04, 11:28 AM
you can use the complexe analysis and résidue theorem
eljose79
Sep26-04, 05:40 AM
but what would happen if i try solving it numerically for example we should calculate the integral over all R of exp(ixt)f(c+ix)exp(ct) and this would be equal to our inverse Laplace transform, the problem is what c would i choose?..thanx.
Feynman
Sep28-04, 01:01 AM
So you have a complex integral (the integral of the inverse Laplace is a line integral in complex analysis) .
It is impossible to calculate numerically these integral , because it is a line complex integral and not a simple integral like the reel integral , for this (if you know the complex analysis) their is an important theorem to calculate the value of the complex integral it is : Residue theorem, so use this theorem
eljose79
Sep28-04, 03:23 PM
But making the change of variable c+iu the integral becomes simply a Fourier inverse transform (is a integral on the real plane of exp(iu)f(c+iu)exp(ct)) so if we can have a real integral and should be able to compute it numerically.
Feynman
Sep29-04, 01:57 AM
It is wrong
When you change the variale c+iu=v , so u change an COMPLEX variable and not a real variable!
eljose79
Sep29-04, 07:44 AM
but the Laplace inverse transform is not just a special case of fourier transform?
Feynman
Sep30-04, 08:58 AM
Ya
the Laplace inverse transform is not just a special case of fourier transform
But when we calculate the Laplace we use the complexe variables and not the real variables
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