Why exp(-st) in the Laplace Transform? [Archive]

Eric Erpelding

Nov23-04, 03:08 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nDoes anyone have an explaination why the kernel function exp(-st) was\nused in the definition of the Laplace transform?\n\nIs there a physical meaning to the use of this function?\n\nI know s is a complex frequency, how can we visualize what the Laplace\ntransform is doing?\n\n--Eric Erpelding\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Does anyone have an explaination why the kernel function \exp(-st) was
used in the definition of the Laplace transform?

Is there a physical meaning to the use of this function?

I know s is a complex frequency, how can we visualize what the Laplace
transform is doing?

--Eric Erpelding

Han de Bruijn

Nov24-04, 01:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eric Erpelding wrote:\n\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n\nIn a nutshell, the exp(-st) comes from the fact that convolution\nintegrals can be written as integrals over an exp(t.d/dx) operator.\nBut you should read the whole story (= heuristics):\n\nhttp://hdebruijn.soo.dto.tudelft.nl/jaar2004/uitboek.pdf\n\nEspecially read the subsection "Laplace and Statistics". Hope it helps\nin providing some background understanding.\n\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n\nIf s is complex - but it doesn\'t have to be - then what you probably\nwant is a visualization of the complex Fourier transform. I don\'t have\nit at hand, though.\n\nHan de Bruijn\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eric Erpelding wrote:

> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?

In a nutshell, the \exp(-st) comes from the fact that convolution
integrals can be written as integrals over an \exp(t.d/dx) operator.
But you should read the whole story (= heuristics):

http://hdebruijn.soo.dto.tudelft.nl/jaar2004/uitboek.pdf

Especially read the subsection "Laplace and Statistics". Hope it helps
in providing some background understanding.

> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?

If s is complex - but it doesn't have to be - then what you probably
want is a visualization of the complex Fourier transform. I don't have
it at hand, though.

Han de Bruijn

Zdislav V. Kovarik

Nov24-04, 01:47 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Tue, 23 Nov 2004, Eric Erpelding wrote:\n\n>\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n>\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n>\n> --Eric Erpelding\n\nWrite s=u+iv, then you have Fourier transform (dual variable v) of the\ndamped function exp(-ut)f(t), depending on the real parameter u. It is\nassumed that f(t) vanishes for all t<0.\n\nAnd the physical, or signal-processing meaning of Fourier transform is\nwell-known.\n\nHope it helps, ZVK(Slavek).\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 23 Nov 2004, Eric Erpelding wrote:

>
> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?
>
> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?
>
> --Eric Erpelding

Write s=u+iv, then you have Fourier transform (dual variable v) of the
damped function \exp(-ut)f(t), depending on the real parameter u. It is
assumed that f(t) vanishes for all t<0.

And the physical, or signal-processing meaning of Fourier transform is
well-known.

Hope it helps, ZVK(Slavek).

ueb

Nov24-04, 01:48 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eric Erpelding wrote:\n\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n\n> Is there a physical meaning to the use of this function?\n\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n\nBe s = \\sigma + j \\omega .\nWith \\sigma > 0 , the factor exp(-\\sigma t) makes each finite\nfunction of time converge. Thus, you can integrate up to each time\nvalue. You get a quite exact image (and the exact origin at back-\ntransformation).\n\nUlrich Bruchholz\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eric Erpelding wrote:

> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?

> Is there a physical meaning to the use of this function?

> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?

Be s = \sigma + j \omega .
With \sigma > , the factor \exp(-\sigma t) makes each finite
function of time converge. Thus, you can integrate up to each time
value. You get a quite exact image (and the exact origin at back-
transformation).

Ulrich Bruchholz

David C. Ullrich

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOn Wed, 24 Nov 2004 07:47:49 +0000 (UTC), "Zdislav V. Kovarik"\n<kovarik@mcmaster.ca> wrote:\n\n>On Tue, 23 Nov 2004, Eric Erpelding wrote:\n>\n>>\n>> Does anyone have an explaination why the kernel function exp(-st) was\n>> used in the definition of the Laplace transform?\n>>\n>> Is there a physical meaning to the use of this function?\n>>\n>> I know s is a complex frequency, how can we visualize what the Laplace\n>> transform is doing?\n>>\n>> --Eric Erpelding\n>\n>Write s=u+iv, then you have Fourier transform (dual variable v) of the\n>damped function exp(-ut)f(t), depending on the real parameter u. It is\n>assumed that f(t) vanishes for all t<0.\n>\n>And the physical, or signal-processing meaning of Fourier transform is\n>well-known.\n>\n>Hope it helps, ZVK(Slavek).\n\nDoesn\'t help me much...\n\nI was assuming that s was real; then v = 0 and we just have the\nmean of exp(-ut)f(t), or the FT at the origin. What does that\nmean mean, in terms of f?\n\n************************\n\nDavid C. Ullrich\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 24 Nov 2004 07:47:49 +0000 (UTC), "Zdislav V. Kovarik"
<kovarik@mcmaster.ca> wrote:

>On Tue, 23 Nov 2004, Eric Erpelding wrote:
>
>>
>> Does anyone have an explaination why the kernel function \exp(-st) was
>> used in the definition of the Laplace transform?
>>
>> Is there a physical meaning to the use of this function?
>>
>> I know s is a complex frequency, how can we visualize what the Laplace
>> transform is doing?
>>
>> --Eric Erpelding
>
>Write s=u+iv, then you have Fourier transform (dual variable v) of the
>damped function \exp(-ut)f(t), depending on the real parameter u. It is
>assumed that f(t) vanishes for all t<0.
>
>And the physical, or signal-processing meaning of Fourier transform is
>well-known.
>
>Hope it helps, ZVK(Slavek).

Doesn't help me much...

I was assuming that s was real; then v = and we just have the
mean of \exp(-ut)f(t), or the FT at the origin. What does that
mean mean, in terms of f?

************************

David C. Ullrich

robert bristow-johnson

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nin article d4e9eb01.0411221145.1ba28f9d@posting.google.com, Eric Erpelding\nat e_erpelding@yahoo.com wrote on 11/23/2004 04:08:\n\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n\nit\'s a good fundamental question. may i suggest posting it to comp.dsp and\nletting those folks whack at it?\n\ni dunno what Laplace was doing, but i see the Laplace Transform as a\ngeneralization of the Fourier Transform:\n\n\n+inf\nx(t) = integral{ X(f) exp(+i*2*pi*f*t) df}\n-inf\n\n\n+inf\nX(f) = integral{ x(t) exp(-i*2*pi*f*t) dt}\n-inf\n\n\nand the Fourier Transform as a generalization (maybe "limit" is a better\ndescription) of the Fourier series:\n\n+inf\nx(t) = SUM{ X(n/T)/T * exp(+i*2*pi*n/T*t) }\nn=-inf\n\nt0+T\nX(f)/T = 1/T * integral{ x(t) exp(-i*2*pi*n/T*t) dt}\nt0\n\nwhere x(t+T) = x(t) for all t and t0 can be any real number.\n\nlet the period T go to infinity (so you can describe a non-periodic function\nas a sum of sinusoids) and the summation becomes a Reimann integral for the\nF.T. integral on top. at least if x(t) and X(f) are decent functions.\nmathematicians will be happy to pick apart this explanation but this is how\nNeanderthal electrical engineers look at it.\n\nthe generalization of the (bilateral) Laplace Transform is that of\nsubstituting s for i*2*pi*f and possibly allowing s to have a non-zero real\npart to make the integrals converge.\n\n--\n\nr b-j rbj@audioimagination.com\n\n"Imagination is more important than knowledge."\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>in article d4e9eb01.0411221145.1ba28f9d@posting.google.com, Eric Erpelding
at e_{erpelding}@yahoo.com wrote on 11/23/2004 04:08:

> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?

it's a good fundamental question. may i suggest posting it to comp.dsp and
letting those folks whack at it?

i dunno what Laplace was doing, but i see the Laplace Transform as a
generalization of the Fourier Transform:

+infx(t) =[/itex] integral{ X(f) \exp(+i*2*\pi*f*t) df}
-inf+infX(f) = integral{ x(t) \exp(-i*2*\pi*f*t) dt}
[itex]-inf

and the Fourier Transform as a generalization (maybe "limit" is a better
description) of the Fourier series:

+infx(t) = SUM{ X(n/T)/T * \exp(+i*2*\pi*n/T*t) }n=-inft0+TX(f)/T = 1/T * integral{ x(t) \exp(-i*2*\pi*n/T*t) dt}
t0

where x(t+T) = x(t) for all t and t0 can be any real number.

let the period T go to infinity (so you can describe a non-periodic function
as a sum of sinusoids) and the summation becomes a Reimann integral for the
F.T. integral on top. at least if x(t) and X(f) are decent functions.
mathematicians will be happy to pick apart this explanation but this is how
Neanderthal electrical engineers look at it.

the generalization of the (bilateral) Laplace Transform is that of
substituting s for i*2*\pi*f and possibly allowing s to have a non-zero real
part to make the integrals converge.

--

r b-j rbj@audioimagination.com

"Imagination is more important than knowledge."

Gerard Westendorp

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEric Erpelding wrote:\n\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n>\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n\n\nThe way I think of it, is that the Laplace transform is really\na Fourier transform of a signal f(t) that has been secretly\nmultiplied by an exponential exp(-at)) before it was Fourier\ntransformed.\nThis secrtet exponetial is related to s and w by\n\ns = a + iw\n\n\nBy doing this, you can "tame" an exponentially growing signal\nbefore you apply Fourier to it.\n\nGerard\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eric Erpelding wrote:

> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?
>
> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?

The way I think of it, is that the Laplace transform is really
a Fourier transform of a signal f(t) that has been secretly
multiplied by an exponential \exp(-at)) before it was Fourier
transformed.
This secrtet exponetial is related to s and w by

s = a + iw

By doing this, you can "tame" an exponentially growing signal
before you apply Fourier to it.

Gerard

Dirk Van de moortel

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Eric Erpelding" <e_erpelding@yahoo.com> wrote in message news:d4e9eb01.0411221145.1ba28f9d@posting.google.c om...\n>\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n>\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n\nFWIW, I\'m not sure about this at all, but historically, perhaps\nthis is what happened...\nWhen you look at s as a real, and at positive functions f(t),\nyou can look at exp(-st) as a thing that "drags" the function f(t)\n"down" in such a way that the total area under the graph is\nfinite for all s > a for some real a.\nThe greater the value of s, the smaller the area. The Laplace\ntransform F(s) of f(t) is the area as a function of s.\n\nI know, it\'s not much, but it might help :-)\n\nDirk Vdm\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Eric Erpelding" <e_{erpelding}@yahoo.com> wrote in message news:d4e9eb01.0411221145.1ba28f9d@posting.google.c om...
>
> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?
>
> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?

FWIW, I'm not sure about this at all, but historically, perhaps
this is what happened...
When you look at s as a real, and at positive functions f(t),
you can look at \exp(-st) as a thing that "drags" the function f(t)
"down" in such a way that the total area under the graph is
finite for all s > a for some real a.
The greater the value of s, the smaller the area. The Laplace
transform F(s) of f(t) is the area as a function of s.

I know, it's not much, but it might help :-)

Dirk Vdm

Arnold Neumaier

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEric Erpelding wrote:\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n>\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n\nIt enables one to repalce differentiation by multiplication.\nThat explains the exponential, apart from a factor. If you take a purely\nimaginary factor - the other important case - you get the fourier transform.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eric Erpelding wrote:
> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?
>
> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?

It enables one to repalce differentiation by multiplication.
That explains the exponential, apart from a factor. If you take a purely
imaginary factor - the other important case - you get the fourier transform.

Arnold Neumaier

David C. Ullrich

Nov24-04, 09:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nOn Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_erpelding@yahoo.com (Eric\nErpelding) wrote:\n\n>\n>Does anyone have an explaination why the kernel function exp(-st) was\n>used in the definition of the Laplace transform?\n>\n>Is there a physical meaning to the use of this function?\n>\n>I know s is a complex frequency, how can we visualize what the Laplace\n>transform is doing?\n\nI believe the answer is no, there\'s no "interpretation" of what the\nLaplace transform really means, analogous to the way one interprets,\nsay, the Fourier transform. The kernels are not orthogonal in any\nobvious sense...\n\nIt\'s hard to know for certain that no is the correct answer, of\ncourse. But I\'ve never seen an answer of the sort I think you\'re\nlooking for, I\'ve thought about it and never found one, and\nI once posted more or less the same question to sci.math, and\nhad a few smart people agree the answer was no.\n\n>--Eric Erpelding\n\n\n************************\n\nDavid C. Ullrich\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_{erpelding}@yahoo.com (Eric
Erpelding) wrote:

>
>Does anyone have an explaination why the kernel function \exp(-st) was
>used in the definition of the Laplace transform?
>
>Is there a physical meaning to the use of this function?
>
>I know s is a complex frequency, how can we visualize what the Laplace
>transform is doing?

I believe the answer is no, there's no "interpretation" of what the
Laplace transform really means, analogous to the way one interprets,
say, the Fourier transform. The kernels are not orthogonal in any
obvious sense...

It's hard to know for certain that no is the correct answer, of
course. But I've never seen an answer of the sort I think you're
looking for, I've thought about it and never found one, and
I once posted more or less the same question to sci.math, and
had a few smart people agree the answer was no.

>--Eric Erpelding

************************

David C. Ullrich

John C. Polasek

Nov24-04, 09:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_erpelding@yahoo.com (Eric\nErpelding) wrote:\n\n>\n>Does anyone have an explaination why the kernel function exp(-st) was\n>used in the definition of the Laplace transform?\n>\n>Is there a physical meaning to the use of this function?\n>\n>I know s is a complex frequency, how can we visualize what the Laplace\n>transform is doing?\n>\n>--Eric Erpelding\nFor all time derivative and time integral operations and all analytic\nfunctions of t, there is a transformed equivalent function LT of s\nthat is self-consistent and algebraic. Derivatives are represented by\ns and integrals by 1/s, the unit step function 1/s and impulse by\n1.For example if the transform is 1/(s + a) then 1/s(1/s+a) is the\nintegral.\n\nThe original functions become disguised, but after the desired\nalgebraic operations, the whole thing can be unrolled by the inverse\nLT.\n\nThe Laplace transform makes it possible to show block diagrams\ncontaining "s factories" able to cascade operations by simple\nmultiplication.\n\nAccompanying this work should be a sturdy table of transforms and\ntheir inverses. Formal study of LT can be hairy, but for engineering\nwork you can usually get by with the rudiments.\n\nThe Laplace transform actually maps the function onto the s plane =\nsigma + j omega. If you limit yourself to sigma = 0, you stay on the\nfrequency axis and get the Fourier transform.\n\nIs that any help?\n\nMr. Dual Space\nIf you have something to say, write an equation.\nIf you have nothing to say, write an essay\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_{erpelding}@yahoo.com (Eric
Erpelding) wrote:

>
>Does anyone have an explaination why the kernel function \exp(-st) was
>used in the definition of the Laplace transform?
>
>Is there a physical meaning to the use of this function?
>
>I know s is a complex frequency, how can we visualize what the Laplace
>transform is doing?
>
>--Eric Erpelding
For all time derivative and time integral operations and all analytic
functions of t, there is a transformed equivalent function LT of s
that is self-consistent and algebraic. Derivatives are represented by
s and integrals by 1/s, the unit step function 1/s and impulse by
1.For example if the transform is 1/(s + a) then 1/s(1/s+a) is the
integral.

The original functions become disguised, but after the desired
algebraic operations, the whole thing can be unrolled by the inverse
LT.

The Laplace transform makes it possible to show block diagrams
containing "s factories" able to cascade operations by simple
multiplication.

Accompanying this work should be a sturdy table of transforms and
their inverses. Formal study of LT can be hairy, but for engineering
work you can usually get by with the rudiments.

The Laplace transform actually maps the function onto the s plane =
\sigma + j \omega. If you limit yourself to \sigma = 0, you stay on the
frequency axis and get the Fourier transform.

Is that any help?

Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay

Eckard Blumschein

Nov25-04, 03:37 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Comparing all answers so far, I wondered why nobody referred to books\nlike B. Girod, R. Rabenstein, A. Stenger. Signals and Systems.\nChichester: Wiley 2001. They decided to introduce LT first, then FT.\nAlso they convincingly visualize LT.\n\nNonetheless I appreciate the simplicity of the answer by Eric Erpelding:\n> The way I think of it, is that the Laplace transform is really\n> a Fourier transform of a signal f(t) that has been secretly\n> multiplied by an exponential exp(-at)) before it was Fourier\n> transformed.\n> By doing this, you can "tame" an exponentially growing signal\n> before you apply Fourier to it.\n\nThis way LT ensures convergence corresponding to the natural decay of\nall real life.\n\n>> Does anyone have an explaination why the kernel function exp(-st) was\n>> used in the definition of the Laplace transform?\n\nLT seems to have been used first by Denis Poisson (1781-1840) in 1815.\nHowever, not even Heaviside was aware of it and reinvented nearly the\nsame method decades later.\n\n>> Is there a physical meaning to the use of this function?\n\nAs does FT, LT performs convolution.\nLT is seemingly attractive as a unilateral transform because real-world\nsignals are typically one-sided. However, LT is restricted to a\ndeterministic analysis. So it only fits to so future, not to past.\nSo-called causal signals are actually deterministically predicted ones.\n\n\n\n\n\n\n> By doing this, you can "tame" an exponentially growing signal\n> before you apply Fourier to it.\n\nWhat about the physical meaning of the\n\n>\n> Gerard\n>\n>\n>\n>\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Comparing all answers so far, I wondered why nobody referred to books
like B. Girod, R. Rabenstein, A. Stenger. Signals and Systems.
Chichester: Wiley 2001. They decided to introduce LT first, then FT.
Also they convincingly visualize LT.

Nonetheless I appreciate the simplicity of the answer by Eric Erpelding:
> The way I think of it, is that the Laplace transform is really
> a Fourier transform of a signal f(t) that has been secretly
> multiplied by an exponential \exp(-at)) before it was Fourier
> transformed.
> By doing this, you can "tame" an exponentially growing signal
> before you apply Fourier to it.

This way LT ensures convergence corresponding to the natural decay of
all real life.

>> Does anyone have an explaination why the kernel function \exp(-st) was
>> used in the definition of the Laplace transform?

LT seems to have been used first by Denis Poisson (1781-1840) in 1815.
However, not even Heaviside was aware of it and reinvented nearly the
same method decades later.

>> Is there a physical meaning to the use of this function?

As does FT, LT performs convolution.
LT is seemingly attractive as a unilateral transform because real-world
signals are typically one-sided. However, LT is restricted to a
deterministic analysis. So it only fits to so future, not to past.
So-called causal signals are actually deterministically predicted ones.

> By doing this, you can "tame" an exponentially growing signal
> before you apply Fourier to it.

What about the physical meaning of the

>
> Gerard
>
>
>
>

Dan Heyman

Nov25-04, 03:38 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>e_erpelding@yahoo.com (Eric Erpelding) wrote in message news:<d4e9eb01.0411221145.1ba28f9d@posting.google. com>...\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n>\n> I know s is a complex frequency, how can we visualize what the Laplace\n> transform is doing?\n>\n> --Eric Erpelding\n\nHere\'s an economic interpretation of the LT. Let T be a positive\nrandom variable with density function f(). Then E[exp(-sT)] is the LT\nof f(). Suppose you will receive a dollar at time T. The expected\npresent value with (real)continuous interest rate s is the expection\nas shown. This interpretation is useful in operations research.\n\nDan Heyman\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>e_{erpelding}@yahoo.com (Eric Erpelding) wrote in message news:<d4e9eb01.0411221145.1ba28f9d@posting.google.com>...
> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?
>
> I know s is a complex frequency, how can we visualize what the Laplace
> transform is doing?
>
> --Eric Erpelding

Here's an economic interpretation of the LT. Let T be a positive
random variable with density function f(). Then E[\exp(-sT)] is the LT
of f(). Suppose you will receive a dollar at time T. The expected
present value with (real)continuous interest rate s is the expection
as shown. This interpretation is useful in operations research.

Dan Heyman

Eckard Blumschein

Nov26-04, 01:17 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>B. Girod, R. Rabenstein, A. Stenger. Signals and Systems. Chichester:\nWiley 2001 first introduce and visualize LT, then FT as a special case\nof it. Decide yourself whether or not this somewhat demanding approach\nis the pedagogically best one.\n\nPierre-Simon, marquis de Laplace (1749-1825) first of all embodies\ndeterminism. The famous integral transform carries his name, but seems\nto have been used first by Denis Poisson (1781-1840) in 1815. Heaviside\n(1850-1925) was not aware of it when he nearly reinvented a slightly\nmodified calculus with the operator pt =3D d/dt instead of st. Laplace di=\nd\nnot trust in the work by Jean Baptiste Fourier (1768-1830) and rejected\nhis famous paper which was nonetheless published with several years delay.\n\nDeterminism is the wrong belief that anything can be predicted in\nadvance by calculation. Correspondingly, one-sided LT assumes that\nanything starts at t=3D0. Pertaining signals are misleadingly called\ncausal signals. The name deterministic signals would be more correct.\nOne-sidedness of LT and the option to consider initial values make LT\nattractive for the first glance. In order to find a fan of LT you might\nGoogle for Kastrup. As far as I can judge, FT is much more in use than\nLT, mainly because of difficulties with inversion.\n\nIn reality, all physical quantities are one-sided because future merely\nexists in imagination. LT is, however, not suited for performing a\nfrequency analysis of such quantities only existing within past, not\nwithin future. Real-valued cosine transform is, in principle, adequate\nfor that task. In other words: LT belongs to deterministic prediction of\nprocesses to come. CT belongs to causal analysis of the past. Causality\nis hidden in influences from what already happened. The laws of physics\nare symmetrical with respect to time. the caveat =91in principle=92 refer=\ns\nto time t in the kernel of LT. In case of CT in IR+, time has to be\nreplaced by elapsed time. LT generally assumes an event-related time\nscale. CT makes sense with an observer-bound time-scale permanently\nsliding relatively to the event-related one.\n\nHaving dealt with LT, Ernst Terhardt of MMK Munich copied the trick of\nmaking each finite function to converge, as does any real process.\nHowever, he finds fault with the widespread practice to start\nintegration a little bit left from zero. In this case, definition of LT\nis not to blame for the fallacy. Instead, notion of real numbers should\nbe revised.\n\nWhat about the minus sign of the imaginary part of s or omega in LT or\nFT, respectively, this is an arbitrarily chosen option, enforced by\ntransformation into the inevitably redundant complex representation.\n(The only way out of that dilemma would be use of the observer-bound\ntime-scale that has a natural zero to rely on.) Physicists still prefer\nthe opposite sign of imaginary part. This discrepancy in sign does not\nmatter because inverse transform returns the correct input. In order to\nbridge the gap between imaginable reality and visualization of complex\nrepresentations like LT or FT, one might get aware of their half-reality\nin the sense the other half of reality is just economized. They are\nexhibiting just one out of two phasors, one of which rotates clockwise,\nthe other one anticlockwise. Omission of the half picture paradoxically\nresults in redundant doubling of data. Being a complex transform, at\nleast the bilateral LT should possibly lead to ambiguous and even\nnon-causal results, as does FT.\n\nEckard Blumschein\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>B. Girod, R. Rabenstein, A. Stenger. Signals and Systems. Chichester:
Wiley 2001 first introduce and visualize LT, then FT as a special case
of it. Decide yourself whether or not this somewhat demanding approach
is the pedagogically best one.

Pierre-Simon, marquis de Laplace (1749-1825) first of all embodies
determinism. The famous integral transform carries his name, but seems
to have been used first by Denis Poisson (1781-1840) in 1815. Heaviside
(1850-1925) was not aware of it when he nearly reinvented a slightly
modified calculus with the operator pt =3D d/dt instead of st. Laplace di=
d
not trust in the work by Jean Baptiste Fourier (1768-1830) and rejected
his famous paper which was nonetheless published with several years delay.

Determinism is the wrong belief that anything can be predicted in
advance by calculation. Correspondingly, one-sided LT assumes that
anything starts at t=3D0. Pertaining signals are misleadingly called
causal signals. The name deterministic signals would be more correct.
One-sidedness of LT and the option to consider initial values make LT
attractive for the first glance. In order to find a fan of LT you might
Google for Kastrup. As far as I can judge, FT is much more in use than
LT, mainly because of difficulties with inversion.

In reality, all physical quantities are one-sided because future merely
exists in imagination. LT is, however, not suited for performing a
frequency analysis of such quantities only existing within past, not
within future. Real-valued cosine transform is, in principle, adequate
for that task. In other words: LT belongs to deterministic prediction of
processes to come. CT belongs to causal analysis of the past. Causality
is hidden in influences from what already happened. The laws of physics
are symmetrical with respect to time. the caveat =91in principle=92 refer=
s
to time t in the kernel of LT. In case of CT in IR+, time has to be
replaced by elapsed time. LT generally assumes an event-related time
scale. CT makes sense with an observer-bound time-scale permanently
sliding relatively to the event-related one.

Having dealt with LT, Ernst Terhardt of MMK Munich copied the trick of
making each finite function to converge, as does any real process.
However, he finds fault with the widespread practice to start
integration a little bit left from zero. In this case, definition of LT
is not to blame for the fallacy. Instead, notion of real numbers should
be revised.

What about the minus sign of the imaginary part of s or \omega in LT or
FT, respectively, this is an arbitrarily chosen option, enforced by
transformation into the inevitably redundant complex representation.
(The only way out of that dilemma would be use of the observer-bound
time-scale that has a natural zero to rely on.) Physicists still prefer
the opposite sign of imaginary part. This discrepancy in sign does not
matter because inverse transform returns the correct input. In order to
bridge the gap between imaginable reality and visualization of complex
representations like LT or FT, one might get aware of their half-reality
in the sense the other half of reality is just economized. They are
exhibiting just one out of two phasors, one of which rotates clockwise,
the other one anticlockwise. Omission of the half picture paradoxically
results in redundant doubling of data. Being a complex transform, at
least the bilateral LT should possibly lead to ambiguous and even
non-causal results, as does FT.

Eckard Blumschein

Paul Arendt

Dec1-04, 11:05 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>e_erpelding@yahoo.com (Eric Erpelding) wrote:\n\n> Does anyone have an explaination why the kernel function exp(-st) was\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n\nHere is a group-theoretic answer.\n\nThe existence (and utility) of both the Laplace and Fourier transforms\ncome from the symmetry of the real line under translations (t -> t + a,\nwhere "a" is some constant shift). This is a one-dimensional Lie\ngroup, so it is Abelian (commutative), and its unitary irreducible\nrepresentations are therefore one-dimensional. These are the Fourier\ntransform kernels, exp(-i w t), for w any real number. Under the\ntranslation operation (t -> t + a), these get multiplied by the\n(unitary) number exp(-i w a).\n\nIn other words, when you take the Fourier transform of a well-behaved\ncomplex function, you are splitting it into functions which transform\nin a simple way under translations! This splits the infinite-\ndimensional space of functions over the real line (t) into an infinite\nnumber of one-dimensional subspaces (the Fourier kernels) which are\ndense (in "nice" functions), and which transform trivially\n(multiplication by a constant) under translations. If you draw the\nFourier kernels as their graph in the complex plane versus t, they are\nconstant-pitch helices. Moving a constant-pitch helix by a constant is\nequivalent to rotating it!\n\nIf you take the Laplace transform, you are splitting a well-behaved\nfunction into its *real* irreducible representations: exponentials!\nShifting an exponential (exp(-st)) by a constant (t -> t + a) is\nequivalent to multiplying it by a real number (exp(-sa)). Taking the\nLaplace transform is splitting the infinite-dimensional space of\nwell-behaved functions into an infinite number of one-dimensional\nsubspaces which are dense in well-behaved functions, and which each\nbehave nicely under translations.\n\nSo, both the Laplace and Fourier transforms are sort of splitting\nfunctions into the "normal modes" of the group of translations of\nthe real line.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>e_{erpelding}@yahoo.com (Eric Erpelding) wrote:

> Does anyone have an explaination why the kernel function \exp(-st) was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?

Here is a group-theoretic answer.

The existence (and utility) of both the Laplace and Fourier transforms
come from the symmetry of the real line under translations (t -> t + a,
where "a" is some constant shift). This is a one-dimensional Lie
group, so it is Abelian (commutative), and its unitary irreducible
representations are therefore one-dimensional. These are the Fourier
transform kernels, \exp(-i w t), for w any real number. Under the
translation operation (t -> t + a), these get multiplied by the
(unitary) number \exp(-i w a).

In other words, when you take the Fourier transform of a well-behaved
complex function, you are splitting it into functions which transform
in a simple way under translations! This splits the infinite-
dimensional space of functions over the real line (t) into an infinite
number of one-dimensional subspaces (the Fourier kernels) which are
dense (in "nice" functions), and which transform trivially
(multiplication by a constant) under translations. If you draw the
Fourier kernels as their graph in the complex plane versus t, they are
constant-pitch helices. Moving a constant-pitch helix by a constant is
equivalent to rotating it!

If you take the Laplace transform, you are splitting a well-behaved
function into its *real* irreducible representations: exponentials!
Shifting an exponential (\exp(-st)) by a constant (t -> t + a) is
equivalent to multiplying it by a real number (\exp(-sa)). Taking the
Laplace transform is splitting the infinite-dimensional space of
well-behaved functions into an infinite number of one-dimensional
subspaces which are dense in well-behaved functions, and which each
behave nicely under translations.

So, both the Laplace and Fourier transforms are sort of splitting
functions into the "normal modes" of the group of translations of
the real line.

T. Monroe

Dec3-04, 04:53 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message\nnews:r1d6q0lc1hdebg5pq17h3jaudkh4s0141u@4 ax.com...\n>\n>\n>\n> On Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_erpelding@yahoo.com (Eric\n> Erpelding) wrote:\n>\n>>\n>>Does anyone have an explaination why the kernel function exp(-st) was\n>>used in the definition of the Laplace transform?\n>>\n>>Is there a physical meaning to the use of this function?\n>>\n>>I know s is a complex frequency, how can we visualize what the Laplace\n>>transform is doing?\n>\n> I believe the answer is no, there\'s no "interpretation" of what the\n> Laplace transform really means, analogous to the way one interprets,\n> say, the Fourier transform. The kernels are not orthogonal in any\n> obvious sense...\n>\n> It\'s hard to know for certain that no is the correct answer, of\n> course. But I\'ve never seen an answer of the sort I think you\'re\n> looking for, I\'ve thought about it and never found one, and\n> I once posted more or less the same question to sci.math, and\n> had a few smart people agree the answer was no.\n>\n\nArthur Mattuck, an MIT professor, \'\'interprets\'\' the Laplace transform as a\ncontinuous analog of a power series expansion of a function. You can watch\na video of him lecturing on this point at\n\nhttp://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring2004/VideoLectures/\n\nClick on Lecture 19.\n\nT. Monroe\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
news:r1d6q0lc1hdebg5pq17h3jaudkh4s0141u@4ax.com...
>
>
>
> On Tue, 23 Nov 2004 09:08:18 +0000 (UTC), e_{erpelding}@yahoo.com (Eric
> Erpelding) wrote:
>
>>
>>Does anyone have an explaination why the kernel function \exp(-st) was
>>used in the definition of the Laplace transform?
>>
>>Is there a physical meaning to the use of this function?
>>
>>I know s is a complex frequency, how can we visualize what the Laplace
>>transform is doing?
>
> I believe the answer is no, there's no "interpretation" of what the
> Laplace transform really means, analogous to the way one interprets,
> say, the Fourier transform. The kernels are not orthogonal in any
> obvious sense...
>
> It's hard to know for certain that no is the correct answer, of
> course. But I've never seen an answer of the sort I think you're
> looking for, I've thought about it and never found one, and
> I once posted more or less the same question to sci.math, and
> had a few smart people agree the answer was no.
>

Arthur Mattuck, an MIT professor, ''interprets'' the Laplace transform as a
continuous analog of a power series expansion of a function. You can watch
a video of him lecturing on this point at

http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring2004/VideoLectures/

Click on Lecture 19.

T. Monroe

Boo

Dec6-04, 07:16 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring2004/VideoLectures/\n>\n> Click on Lecture 19.\nNote: this is realmedia only.\n\n--\nBoo\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring2004/VideoLectures/
>
> Click on Lecture 19.
Note: this is realmedia only.

--
Boo

EDWARD HYMAN

Dec16-04, 08:08 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Paul,\n\nI am not sure that I comprehend all of your commentary, however I do =\nnote one of your statements\n\nIf you take the Laplace transform, ...\n\nShifting an exponential (exp(-st)) by a constant (t -> t + a) is\nequivalent to multiplying it by a real number (exp(-sa)).\n\n\nYour statement is not quite valid, in that, for the case of a Laplace =\ntransform, the transform variable s has both real and imaginary =\ncomponents, thus exp(-sa) has both real and imaginary components as =\nwell. As in the Fourier transform, you would integrate over the =\nimaginary component. It is the real component, however, which =\ndistinguishes the Laplace Transform from the Fourier Transform. That =\nis, the real component of our multiplier will tend to attenuate the =\nfunction. An example of the use of a Laplace transform would be to =\ntransform the unit step function U(t). Note that it\'s Laplace Transform =\nis determined to be 1/s, under the restriction re(s) > 0. The Fourier =\nTransform of U(t) does not exist.\n\nEd\n\n--\nEdward Hyman\nEdwardH@email.uophx.edu\nOther EMail: e.hyman@worldnet.att.net\n\n"Paul Arendt" <parendt@nmt.edu> wrote in message =\nnews:2e2051b7.0411301530.2a0e7dd4@posting.googl e.com...\ne_erpelding@yahoo.com (Eric Erpelding) wrote:\n\n> Does anyone have an explaination why the kernel function exp(-st) =\nwas\n> used in the definition of the Laplace transform?\n>\n> Is there a physical meaning to the use of this function?\n\nHere is a group-theoretic answer.\n\nThe existence (and utility) of both the Laplace and Fourier transforms\ncome from the symmetry of the real line under translations (t -> t + a,\nwhere "a" is some constant shift). This is a one-dimensional Lie\ngroup, so it is Abelian (commutative), and its unitary irreducible\nrepresentations are therefore one-dimensional. These are the Fourier\ntransform kernels, exp(-i w t), for w any real number. Under the\ntranslation operation (t -> t + a), these get multiplied by the\n(unitary) number exp(-i w a).\n\nIn other words, when you take the Fourier transform of a well-behaved\ncomplex function, you are splitting it into functions which transform\nin a simple way under translations! This splits the infinite-\ndimensional space of functions over the real line (t) into an infinite\nnumber of one-dimensional subspaces (the Fourier kernels) which are\ndense (in "nice" functions), and which transform trivially\n(multiplication by a constant) under translations. If you draw the\nFourier kernels as their graph in the complex plane versus t, they are\nconstant-pitch helices. Moving a constant-pitch helix by a constant is\nequivalent to rotating it!\n\nIf you take the Laplace transform, you are splitting a well-behaved\nfunction into its *real* irreducible representations: exponentials!\nShifting an exponential (exp(-st)) by a constant (t -> t + a) is\nequivalent to multiplying it by a real number (exp(-sa)). Taking the\nLaplace transform is splitting the infinite-dimensional space of\nwell-behaved functions into an infinite number of one-dimensional\nsubspaces which are dense in well-behaved functions, and which each\nbehave nicely under translations.\n\nSo, both the Laplace and Fourier transforms are sort of splitting\nfunctions into the "normal modes" of the group of translations of\nthe real line.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul,

I am not sure that I comprehend all of your commentary, however I do =
note one of your statements

If you take the Laplace transform, ...

Shifting an exponential (\exp(-st)) by a constant (t -> t + a) is
equivalent to multiplying it by a real number (\exp(-sa)).

Your statement is not quite valid, in that, for the case of a Laplace =
transform, the transform variable s has both real and imaginary =
components, thus \exp(-sa) has both real and imaginary components as =
well. As in the Fourier transform, you would integrate over the =
imaginary component. It is the real component, however, which =
distinguishes the Laplace Transform from the Fourier Transform. That =
is, the real component of our multiplier will tend to attenuate the =
function. An example of the use of a Laplace transform would be to =
transform the unit step function U(t). Note that it's Laplace Transform =
is determined to be 1/s, under the restriction re(s) > . The Fourier =
Transform of U(t) does not exist.

Ed

--
Edward Hyman
EdwardH@email.uophx.edu
Other EMail: e.hyman@worldnet.att.net

"Paul Arendt" <parendt@nmt.edu> wrote in message =
news:2e2051b7.0411301530.2a0e7dd4@posting.google.c om...
e_{erpelding}@yahoo.com (Eric Erpelding) wrote:

> Does anyone have an explaination why the kernel function \exp(-st) =
was
> used in the definition of the Laplace transform?
>
> Is there a physical meaning to the use of this function?

Here is a group-theoretic answer.

The existence (and utility) of both the Laplace and Fourier transforms
come from the symmetry of the real line under translations (t -> t + a,
where "a" is some constant shift). This is a one-dimensional Lie
group, so it is Abelian (commutative), and its unitary irreducible
representations are therefore one-dimensional. These are the Fourier
transform kernels, \exp(-i w t), for w any real number. Under the
translation operation (t -> t + a), these get multiplied by the
(unitary) number \exp(-i w a).

In other words, when you take the Fourier transform of a well-behaved
complex function, you are splitting it into functions which transform
in a simple way under translations! This splits the infinite-
dimensional space of functions over the real line (t) into an infinite
number of one-dimensional subspaces (the Fourier kernels) which are
dense (in "nice" functions), and which transform trivially
(multiplication by a constant) under translations. If you draw the
Fourier kernels as their graph in the complex plane versus t, they are
constant-pitch helices. Moving a constant-pitch helix by a constant is
equivalent to rotating it!

If you take the Laplace transform, you are splitting a well-behaved
function into its *real* irreducible representations: exponentials!
Shifting an exponential (\exp(-st)) by a constant (t -> t + a) is
equivalent to multiplying it by a real number (\exp(-sa)). Taking the
Laplace transform is splitting the infinite-dimensional space of
well-behaved functions into an infinite number of one-dimensional
subspaces which are dense in well-behaved functions, and which each
behave nicely under translations.

So, both the Laplace and Fourier transforms are sort of splitting
functions into the "normal modes" of the group of translations of
the real line.

Aaron Denney

Dec19-04, 07:19 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2004-12-16, EDWARD HYMAN <e.hyman@worldnet.att.net> wrote:\n> Paul,\n>\n> I am not sure that I comprehend all of your commentary, however I do\n> note one of your statements\n>\n> If you take the Laplace transform, ...\n>\n> Shifting an exponential (exp(-st)) by a constant (t -> t + a) is\n> equivalent to multiplying it by a real number (exp(-sa)).\n>\n>\n> Your statement is not quite valid, in that, for the case of a Laplace\n> transform, the transform variable s has both real and imaginary\n> components, thus exp(-sa) has both real and imaginary components as\n> well.\n\nTerminology issue. In many fields a "Laplace transform" means\nexclusively real-valued exponent. If it\'s complex than it\'s a\n"complex Laplace transform" or even "generalized Laplace transform",\nthough that\'s also used for other extensions, such as adding discrete\nportions.\n\n> As in the Fourier transform, you would integrate over the\n> imaginary component. It is the real component, however, which\n> distinguishes the Laplace Transform from the Fourier Transform. That\n> is, the real component of our multiplier will tend to attenuate the\n> function. An example of the use of a Laplace transform would be to\n> transform the unit step function U(t). Note that it\'s Laplace Transform\n> is determined to be 1/s, under the restriction re(s) > 0. The Fourier\n> Transform of U(t) does not exist.\n\nWell, that depends on what you mean by "exist". It sure does, if\nyou allow distributions, rather than just functions as Fourier\ntransforms.\n\nUnsurprisingly, the non-singular portion is -i/k, or what you get\nfor 1/s, when you replace s with (ik). The singular portion is just\na delta distribution, caused by the (1/2) offset that is the even\nportion of the unit step.\n\nOne way to do this is to take a series of function that converge\nto the unit step but whose Fourier transforms are functions --\ne.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving\nthe delta distribution noticed above. Write signum(x) as the\nlimit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0.\n\nThe fourier transform of these will be -1/(a - ik) + 1/(a + ik) =\n(-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0\n= 2/ki, but we need to divide by 2 to get the step function.\n\nPhysicists often just use it via formal manipulation in integrals, which\nwhile not normally made rigorous, does give workable results, along\nthe lines of:\n\nThe unit step is just the integral of the delta function.\n\nIntegrating a function in the real domain can be done by dividing the\nfunction in the Fourier domain by ik.\n\nAgain this gives the non-singular portion as 1/ik.\n\nI have left out constant multiplicative factors, as they\'ll depend\non one\'s convention.\n\n--\nAaron Denney\n-><-\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2004-12-16, EDWARD HYMAN <e.hyman@worldnet.att.net> wrote:
> Paul,
>
> I am not sure that I comprehend all of your commentary, however I do
> note one of your statements
>
> If you take the Laplace transform, ...
>
> Shifting an exponential (\exp(-st)) by a constant (t -> t + a) is
> equivalent to multiplying it by a real number (\exp(-sa)).
>
>
> Your statement is not quite valid, in that, for the case of a Laplace
> transform, the transform variable s has both real and imaginary
> components, thus \exp(-sa) has both real and imaginary components as
> well.

Terminology issue. In many fields a "Laplace transform" means
exclusively real-valued exponent. If it's complex than it's a
"complex Laplace transform" or even "generalized Laplace transform",
though that's also used for other extensions, such as adding discrete
portions.

> As in the Fourier transform, you would integrate over the
> imaginary component. It is the real component, however, which
> distinguishes the Laplace Transform from the Fourier Transform. That
> is, the real component of our multiplier will tend to attenuate the
> function. An example of the use of a Laplace transform would be to
> transform the unit step function U(t). Note that it's Laplace Transform
> is determined to be 1/s, under the restriction re(s) > . The Fourier
> Transform of U(t) does not exist.

Well, that depends on what you mean by "exist". It sure does, if
you allow distributions, rather than just functions as Fourier
transforms.

Unsurprisingly, the non-singular portion is -i/k, or what you get
for 1/s, when you replace s with (ik). The singular portion is just
a \delta distribution, caused by the (1/2) offset that is the even
portion of the unit step.

One way to do this is to take a series of function that converge
to the unit step but whose Fourier transforms are functions --
e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving
the \delta distribution noticed above. Write signum(x) as the
limit a -> of f_a(t) = e^{-at} for t > 0, -e^{at} for t < .

The fourier transform of these will be -1/(a - ik) + 1/(a + ik) =(-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a ->= 2/ki, but we need to divide by 2 to get the step function.

Physicists often just use it via formal manipulation in integrals, which
while not normally made rigorous, does give workable results, along
the lines of:

The unit step is just the integral of the \delta function.

Integrating a function in the real domain can be done by dividing the
function in the Fourier domain by ik.

Again this gives the non-singular portion as 1/ik.

I have left out constant multiplicative factors, as they'll depend
on one's convention.

--
Aaron Denney
-><-

EDWARD HYMAN

Dec20-04, 10:04 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Aaron,\n\nYour point is well made. Indeed, the Laplace transform of U(t) is\ndetermined by the real part of s, the transform variable corresponding\nto t. So, we would have\n\nLaplace transform of U(t) is 1/s for re(s) > 0\nFourier transform of U(t) is pi*delta(k) + 1/ik for the case re(s) =\n0. s, in this case, being equal to ik.\n\nAs you point out, each piece of the signum function which you define\nis Fourier transformable. Note that your selection of\nexp(-at) for the U(t) portion is, in essence, adding the positive real\ncomponent a to the Fourier transform variable ik. That is,\nlim a->0 of Fourier transform of exp(-at)U(t) is identical to the limit\nre(s)->0 of Laplace transform of U(t), having re(s) > 0.\nA similar argument is made for the U(-t) portion. One could therefore\nargue that the Fourier transform of U(t) exists, since\nits Laplace transform exists in the limiting sense as re(s)->0.\n\nEd\n\n--\nEdward Hyman\nEdwardH@email.uophx.edu\nOther EMail: e.hyman@worldnet.att.net\n\n"Aaron Denney" <wnoise@ofb.net> wrote in message\nnews:slrncsa9rg.350.wnoise@ofb.net...\nOn 2004-12-16, EDWARD HYMAN <e.hyman@worldnet.att.net> wrote:\n> Paul,\n>\n> I am not sure that I comprehend all of your commentary, however\nI do\n> note one of your statements\n>\n> If you take the Laplace transform, ...\n>\n> Shifting an exponential (exp(-st)) by a constant (t -> t + a) is\n> equivalent to multiplying it by a real number (exp(-sa)).\n>\n>\n> Your statement is not quite valid, in that, for the case of a\nLaplace\n> transform, the transform variable s has both real and imaginary\n> components, thus exp(-sa) has both real and imaginary components as\n> well.\n\nTerminology issue. In many fields a "Laplace transform" means\nexclusively real-valued exponent. If it\'s complex than it\'s a\n"complex Laplace transform" or even "generalized Laplace transform",\nthough that\'s also used for other extensions, such as adding discrete\nportions.\n\n> As in the Fourier transform, you would integrate over the\n> imaginary component. It is the real component, however, which\n> distinguishes the Laplace Transform from the Fourier Transform.\nThat\n> is, the real component of our multiplier will tend to attenuate the\n> function. An example of the use of a Laplace transform would be to\n> transform the unit step function U(t). Note that it\'s Laplace\nTransform\n> is determined to be 1/s, under the restriction re(s) > 0. The\nFourier\n> Transform of U(t) does not exist.\n\nWell, that depends on what you mean by "exist". It sure does, if\nyou allow distributions, rather than just functions as Fourier\ntransforms.\n\nUnsurprisingly, the non-singular portion is -i/k, or what you get\nfor 1/s, when you replace s with (ik). The singular portion is just\na delta distribution, caused by the (1/2) offset that is the even\nportion of the unit step.\n\nOne way to do this is to take a series of function that converge\nto the unit step but whose Fourier transforms are functions --\ne.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving\nthe delta distribution noticed above. Write signum(x) as the\nlimit a -> 0 of f_a(t) = e^{-at} for t > 0, -e^{at} for t < 0.\n\nThe fourier transform of these will be -1/(a - ik) + 1/(a + ik)\n(-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a -> 0\n= 2/ki, but we need to divide by 2 to get the step function.\n\nPhysicists often just use it via formal manipulation in integrals,\nwhich\nwhile not normally made rigorous, does give workable results, along\nthe lines of:\n\nThe unit step is just the integral of the delta function.\n\nIntegrating a function in the real domain can be done by dividing\nthe\nfunction in the Fourier domain by ik.\n\nAgain this gives the non-singular portion as 1/ik.\n\nI have left out constant multiplicative factors, as they\'ll depend\non one\'s convention.\n\n--\nAaron Denney\n-><-\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Aaron,

Your point is well made. Indeed, the Laplace transform of U(t) is
determined by the real part of s, the transform variable corresponding
to t. So, we would have

Laplace transform of U(t) is 1/s for re(s) >
Fourier transform of U(t) is \pi*\delta(k) + 1/ik for the case re(s) =
. s, in this case, being equal to ik.

As you point out, each piece of the signum function which you define
is Fourier transformable. Note that your selection of
\exp(-at) for the U(t) portion is, in essence, adding the positive real
component a to the Fourier transform variable ik. That is,
lim a->0 of Fourier transform of \exp(-at)U(t) is identical to the limit
re(s)->0 of Laplace transform of U(t), having re(s) > .
A similar argument is made for the U(-t) portion. One could therefore
argue that the Fourier transform of U(t) exists, since
its Laplace transform exists in the limiting sense as re(s)->0.

Ed

--
Edward Hyman
EdwardH@email.uophx.edu
Other EMail: e.hyman@worldnet.att.net

"Aaron Denney" <wnoise@ofb.net> wrote in message
news:slrncsa9rg.350.wnoise@ofb.net...
On 2004-12-16, EDWARD HYMAN <e.hyman@worldnet.att.net> wrote:
> Paul,
>
> I am not sure that I comprehend all of your commentary, however
I do
> note one of your statements
>
> If you take the Laplace transform, ...
>
> Shifting an exponential (\exp(-st)) by a constant (t -> t + a) is
> equivalent to multiplying it by a real number (\exp(-sa)).
>
>
> Your statement is not quite valid, in that, for the case of a
Laplace
> transform, the transform variable s has both real and imaginary
> components, thus \exp(-sa) has both real and imaginary components as
> well.

Terminology issue. In many fields a "Laplace transform" means
exclusively real-valued exponent. If it's complex than it's a
"complex Laplace transform" or even "generalized Laplace transform",
though that's also used for other extensions, such as adding discrete
portions.

> As in the Fourier transform, you would integrate over the
> imaginary component. It is the real component, however, which
> distinguishes the Laplace Transform from the Fourier Transform.
That
> is, the real component of our multiplier will tend to attenuate the
> function. An example of the use of a Laplace transform would be to
> transform the unit step function U(t). Note that it's Laplace
Transform
> is determined to be 1/s, under the restriction re(s) > . The
Fourier
> Transform of U(t) does not exist.

Well, that depends on what you mean by "exist". It sure does, if
you allow distributions, rather than just functions as Fourier
transforms.

Unsurprisingly, the non-singular portion is -i/k, or what you get
for 1/s, when you replace s with (ik). The singular portion is just
a \delta distribution, caused by the (1/2) offset that is the even
portion of the unit step.

One way to do this is to take a series of function that converge
to the unit step but whose Fourier transforms are functions --
e.g. rewrite it as (1 + signum(x))/2. Discard the 1/2 as giving
the \delta distribution noticed above. Write signum(x) as the
limit a -> of f_a(t) = e^{-at} for t > 0, -e^{at} for t < .

The fourier transform of these will be -1/(a - ik) + 1/(a + ik)(-(a + ik) + (a-ik)) / (a^2 + k^2) = -2ik / (a^2 + k^2). Lim a ->= 2/ki, but we need to divide by 2 to get the step function.

Physicists often just use it via formal manipulation in integrals,
which
while not normally made rigorous, does give workable results, along
the lines of:

The unit step is just the integral of the \delta function.

Integrating a function in the real domain can be done by dividing
the
function in the Fourier domain by ik.

Again this gives the non-singular portion as 1/ik.

I have left out constant multiplicative factors, as they'll depend
on one's convention.

--
Aaron Denney
-><-

reilly

Jan18-05, 01:20 PM

Why exp(-st) ? Why not? Actually there are circumstances for which the laplace xform can be derived for certain kinds of diferential equations.Once can derive subsidiary differential equations for integral transforms, of which he exponential kernal is one of many. (Ince's treatise on DEs covers this material in great detail.) So, often there can be a strong logic for the choice of an LT, well beyond the cookbook level.

Regards,
Reilly Atkinson

Penny314

Jan27-05, 09:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nDear Reilly,\nTrue, there are other kernels. But the idea behind the exponential in both\nFourier and Laplace transforms is that certain ODE\'s form a semigroup.\nThis was explained clearly by Norbert Wiener in his articles.\nbest\nPenny\n\n> So,\n>often there can be a strong logic for the choice of an LT, well beyond\n>the cookbook level.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dear Reilly,
True, there are other kernels. But the idea behind the exponential in both
Fourier and Laplace transforms is that certain ODE's form a semigroup.
This was explained clearly by Norbert Wiener in his articles.
best
Penny

> So,
>often there can be a strong logic for the choice of an LT, well beyond
>the cookbook level.