View Full Version : Laplace vs. Fourier
Stuart Wilson
Aug27-04, 06:10 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\nBeing a physicist, I am comfortable with using Fourier transforms\nto swap from time to frequency spaces. This involves (to within a\nconstant or so..)taking the integral from negative infinity to\npositive infinity.\n\nF(w) ~ int f(t)e^(iwt) dt\n\nHowever, reading the engineering literature, it seems the Laplace\ntransform is used in a similar way. ( Note that the limits of the\nintegral are now 0 to infinity. )\n\nL(s) ~ int f(t)e^(-st) dt\n\nAlthough I understand that the resulting Laplace transform is a\ncomplex function, how are the two related? Is one `more general\'?\nWhat are the restrictions on the function in the time domain?\n\nThanks in advance,\nStu\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>Being a physicist, I am comfortable with using Fourier transforms
to swap from time to frequency spaces. This involves (to within a
constant or so..)taking the integral from negative infinity to
positive infinity.
F(w) ~ \int f(t)e^(iwt) dt
However, reading the engineering literature, it seems the Laplace
transform is used in a similar way. ( Note that the limits of the
integral are now to infinity. )
L(s) ~ \int f(t)e^(-st) dt
Although I understand that the resulting Laplace transform is a
complex function, how are the two related? Is one `more general'?
What are the restrictions on the function in the time domain?
Thanks in advance,
Stu
<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>repeat after me: \'google is our friend\'... searched for \'laplace vs fourier\ntransform\', only 7800 hits.\nhttp://engineering.dartmouth.edu/~engs092/outlines/30Laplace.pdf\nhttp://www.hep.ph.ic.ac.uk/~hallg/Instrumentation/Lectures/Lecture26.pdf\nhttp://www.ece.ucsb.edu/courses/ECE130/130A_Su04ElSoussi/topics_review.pdf\netc, etc, etc...\n\n"Stuart Wilson" <stuart_wlsn@yahoo.com.au> wrote in message\nnews:e227b6ef.0408270029.1cc9211a@posting .google.com...\n>\n>\n> Being a physicist, I am comfortable with using Fourier transforms\n> to swap from time to frequency spaces. This involves (to within a\n> constant or so..)taking the integral from negative infinity to\n> positive infinity.\n>\n> F(w) ~ int f(t)e^(iwt) dt\n>\n> However, reading the engineering literature, it seems the Laplace\n> transform is used in a similar way. ( Note that the limits of the\n> integral are now 0 to infinity. )\n>\n> L(s) ~ int f(t)e^(-st) dt\n>\n> Although I understand that the resulting Laplace transform is a\n> complex function, how are the two related? Is one `more general\'?\n> What are the restrictions on the function in the time domain?\n>\n> Thanks in advance,\n> Stu\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>repeat after me: 'google is our friend'... searched for 'laplace vs fourier
transform', only 7800 hits.
http://engineering.dartmouth.edu/~engs092/outlines/30Laplace.pdf
http://www.hep.ph.ic.ac.uk/~hallg/Instrumentation/Lectures/Lecture26.pdf
http://www.ece.ucsb.edu/courses/ECE130/130A_Su04ElSoussi/topics_review.pdf
etc, etc, etc...
"Stuart Wilson" <stuart_wlsn@yahoo.com.au> wrote in message
news:e227b6ef.0408270029.1cc9211a@posting.google.c om...
>
>
> Being a physicist, I am comfortable with using Fourier transforms
> to swap from time to frequency spaces. This involves (to within a
> constant or so..)taking the integral from negative infinity to
> positive infinity.
>
> F(w) ~ \int f(t)e^(iwt) dt
>
> However, reading the engineering literature, it seems the Laplace
> transform is used in a similar way. ( Note that the limits of the
> integral are now to infinity. )
>
> L(s) ~ \int f(t)e^(-st) dt
>
> Although I understand that the resulting Laplace transform is a
> complex function, how are the two related? Is one `more general'?
> What are the restrictions on the function in the time domain?
>
> Thanks in advance,
> Stu
William R. Frensley
Aug28-04, 04:27 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>Stuart Wilson wrote:\n> Being a physicist, I am comfortable with using Fourier transforms\n> to swap from time to frequency spaces. This involves (to within a\n> constant or so..)taking the integral from negative infinity to\n> positive infinity.\n>\n> F(w) ~ int f(t)e^(iwt) dt\n>\n> However, reading the engineering literature, it seems the Laplace\n> transform is used in a similar way. ( Note that the limits of the\n> integral are now 0 to infinity. )\n>\n> L(s) ~ int f(t)e^(-st) dt\n>\n> Although I understand that the resulting Laplace transform is a\n> complex function, how are the two related? Is one `more general\'?\n> What are the restrictions on the function in the time domain?\n>\n\nIn a nutshell: The Fourier transform is optimum when dealing\nwith boundary-value problems and the Laplace transform is optimum\nwhen dealing with initial-value problems. Engineers have\nfavored the latter because the systems they deal with clearly do\nnot have a history that goes back to t = - infinity.\n\nBy the way, if you set up a Green\'s function formalism for the\nSchroedinger equation in terms of Laplace, rather than Fourier\ntransforms, you will automatically get the retarded (or causal)\ncase, without having to worry about "adiabatic switching on"\nand the resulting +/- i epsilon terms in the denominator.\n\n- Bill Frensley\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>Stuart Wilson wrote:
> Being a physicist, I am comfortable with using Fourier transforms
> to swap from time to frequency spaces. This involves (to within a
> constant or so..)taking the integral from negative infinity to
> positive infinity.
>
> F(w) ~ \int f(t)e^(iwt) dt
>
> However, reading the engineering literature, it seems the Laplace
> transform is used in a similar way. ( Note that the limits of the
> integral are now to infinity. )
>
> L(s) ~ \int f(t)e^(-st) dt
>
> Although I understand that the resulting Laplace transform is a
> complex function, how are the two related? Is one `more general'?
> What are the restrictions on the function in the time domain?
>
In a nutshell: The Fourier transform is optimum when dealing
with boundary-value problems and the Laplace transform is optimum
when dealing with initial-value problems. Engineers have
favored the latter because the systems they deal with clearly do
not have a history that goes back to t = - infinity.
By the way, if you set up a Green's function formalism for the
Schroedinger equation in terms of Laplace, rather than Fourier
transforms, you will automatically get the retarded (or causal)
case, without having to worry about "adiabatic switching on"
and the resulting +/- i \epsilon terms in the denominator.
- Bill Frensley
robert bristow-johnson
Aug28-04, 04:28 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>this could be a good one for comp.dsp ...\n\nstuart_wlsn@yahoo.com.au (Stuart Wilson) wrote in message news:<e227b6ef.0408270029.1cc9211a@posting.google. com>...\n> Being a physicist, I am comfortable with using Fourier transforms\n> to swap from time to frequency spaces. This involves (to within a\n> constant or so..)\n\nthat constant being 1/(2*pi) in the inverse transform, i presume. if\nyou express frequency in Hz (or kHz or whatever non-radian form), then\nthere is no 1/(2*pi) in the inverse Fourier transform and there is a\nreally nice symmetry between forward and inverse transform. makes the\napplication of the "duality theorem" easy which is, i think, why it is\npopular with us electrical engineers.\n\n> taking the integral from negative infinity to positive infinity.\n>\n> F(w) ~ int f(t)e^(iwt) dt\n>\n> However, reading the engineering literature, it seems the Laplace\n> transform is used in a similar way. ( Note that the limits of the\n> integral are now 0 to infinity. )\n\nactually the most correct definition of the LT has limits of -inf to\n+inf. the form you see commonly in sophmore linear system texts (with\nbottom limit "0-" or -epsilon) is there for pedagogical reasons.\neither the systems analyzed are completely "relaxed" at t=0 (which\nmeans all states are zero) and the derivatives of the input eventually\nare zero at t=0, or there is information of the initial state that\nactually changes the Laplace Transform in comparison to the "textbook"\ntransforms.\n\nafter t=0, it makes no difference if you know all of the initial\nstates of the system H(s) and all of the derivatives of input x(t) at\nt=0 or if you know all of x(t) before t=0 and assume that H(s) was\n"relaxed" at t=-inf. they get you to the same place at t=0.\n\n> L(s) ~ int f(t)e^(-st) dt\n>\n> Although I understand that the resulting Laplace transform is a\n> complex function,\n\nso is the FT.\n\n> how are the two related? Is one `more general\'?\n\nyes, of course. the LT is more general than the FT. s = sigma +\ni*omega and you set sigma to a positive enough value to make the\nintegral converge. for most decent signals (that ain\'t blowing up to\ninfinity), sigma = 0 is good enough and the two are precisely the\nsame.\n\n> What are the restrictions on the function in the time domain?\n\nthe LT or FT integral has to exist (or converge). if x(t) won\'t let\nit do that (no matter what sigma is) then you cannot LT it.\n\nr b-j\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>this could be a good one for comp.dsp ...
stuart_wlsn@yahoo.com.au (Stuart Wilson) wrote in message news:<e227b6ef.0408270029.1cc9211a@posting.google.com>...
> Being a physicist, I am comfortable with using Fourier transforms
> to swap from time to frequency spaces. This involves (to within a
> constant or so..)
that constant being 1/(2*\pi) in the inverse transform, i presume. if
you express frequency in Hz (or kHz or whatever non-radian form), then
there is no 1/(2*\pi) in the inverse Fourier transform and there is a
really nice symmetry between forward and inverse transform. makes the
application of the "duality theorem" easy which is, i think, why it is
popular with us electrical engineers.
> taking the integral from negative infinity to positive infinity.
>
> F(w) ~ \int f(t)e^(iwt) dt
>
> However, reading the engineering literature, it seems the Laplace
> transform is used in a similar way. ( Note that the limits of the
> integral are now to infinity. )
actually the most correct definition of the LT has limits of -inf to
+inf. the form you see commonly in sophmore linear system texts (with
bottom limit "0-" or -\epsilon) is there for pedagogical reasons.
either the systems analyzed are completely "relaxed" at t=0 (which
means all states are zero) and the derivatives of the input eventually
are zero at t=0, or there is information of the initial state that
actually changes the Laplace Transform in comparison to the "textbook"
transforms.
after t=0, it makes no difference if you know all of the initial
states of the system H(s) and all of the derivatives of input x(t) at
t=0 or if you know all of x(t) before t=0 and assume that H(s) was
"relaxed" at t=-inf. they get you to the same place at t=0.
> L(s) ~ \int f(t)e^(-st) dt
>
> Although I understand that the resulting Laplace transform is a
> complex function,
so is the FT.
> how are the two related? Is one `more general'?
yes, of course. the LT is more general than the FT. s = \sigma +i*\omega and you set \sigma to a positive enough value to make the
integral converge. for most decent signals (that ain't blowing up to
infinity), \sigma = is good enough and the two are precisely the
same.
> What are the restrictions on the function in the time domain?
the LT or FT integral has to exist (or converge). if x(t) won't let
it do that (no matter what \sigma is) then you cannot LT it.
r b-j
<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\nStuart Wilson wrote:\n> Being a physicist, I am comfortable with using Fourier transforms\n> to swap from time to frequency spaces. This involves (to within a\n> constant or so..)taking the integral from negative infinity to\n> positive infinity.\n>\n> F(w) ~ int f(t)e^(iwt) dt\n>\n> However, reading the engineering literature, it seems the Laplace\n> transform is used in a similar way. ( Note that the limits of the\n> integral are now 0 to infinity. )\n>\n> L(s) ~ int f(t)e^(-st) dt\n>\n> Although I understand that the resulting Laplace transform is a\n> complex function, how are the two related? Is one `more general\'?\n> What are the restrictions on the function in the time domain?\n>\n> Thanks in advance,\n> Stu\n\nThe Laplace transform is more general since:\n\ns = v + iw\nL(s) ~ int f(t) e^(-iwt)e^(-vt)\n\nso for v = 0 L(s) = F(w).\n\nA lot of the time in engineering the limits are from 0 to infinity\nbecause the value of f(t) for values of t < 0 = 0. In general for the\nLaplace transform the limits should be -infinity to +infinity.\n\nThe Laplace transform is able to converge for a larger number of\nfunctions than the Fourier transform due to the e^(-vt) term.\n\n!Q\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>Stuart Wilson wrote:
> Being a physicist, I am comfortable with using Fourier transforms
> to swap from time to frequency spaces. This involves (to within a
> constant or so..)taking the integral from negative infinity to
> positive infinity.
>
> F(w) ~ \int f(t)e^(iwt) dt
>
> However, reading the engineering literature, it seems the Laplace
> transform is used in a similar way. ( Note that the limits of the
> integral are now to infinity. )
>
> L(s) ~ \int f(t)e^(-st) dt
>
> Although I understand that the resulting Laplace transform is a
> complex function, how are the two related? Is one `more general'?
> What are the restrictions on the function in the time domain?
>
> Thanks in advance,
> Stu
The Laplace transform is more general since:
s = v + iwL(s) ~ \int f(t) e^(-iwt)e^(-vt)
so for v = L(s) = F(w).
A lot of the time in engineering the limits are from to infinity
because the value of f(t) for values of t < = . In general for the
Laplace transform the limits should be -infinity to +infinity.
The Laplace transform is able to converge for a larger number of
functions than the Fourier transform due to the e^(-vt) term.
!Q
<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\nStuart Wilson wrote:\n\n> Being a physicist, I am comfortable with using Fourier transforms\n> to swap from time to frequency spaces. This involves (to within a\n> constant or so..)taking the integral from negative infinity to\n> positive infinity.\n\n> F(w) ~ int f(t)e^(iwt) dt\n\n> However, reading the engineering literature, it seems the Laplace\n> transform is used in a similar way. ( Note that the limits of the\n> integral are now 0 to infinity. )\n\n> L(s) ~ int f(t)e^(-st) dt\n\n> Although I understand that the resulting Laplace transform is a\n> complex function, how are the two related? Is one `more general\'?\n> What are the restrictions on the function in the time domain?\n\nBoth F(w) and L(s) are complex. The difference is in the argument\n\ns = a + iw (a > 0) , i.e.\n\ne^{-st} = e^{-at} \\dot [cos(wt) - i sin(wt)]\n\nThus, the Laplace transform might be `more general\'. Because if\nthe time function begins at t >= 0 (that\'s the only one restriction\non the time function ;) , you can calculate with the (enlarged)\nspectrum. Sets of differential equations become simple algebra.\nYou get the *exact* result after the back transform.\n\nHowever, the Fourier transform is only applicable to functions\nfinite in time (or in a time window that does as though the\nfunction be zero out of the window).\n\nUlrich\nhttp://home.t-online.de/home/Ulrich.Bruchholz/\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>Stuart Wilson wrote:
> Being a physicist, I am comfortable with using Fourier transforms
> to swap from time to frequency spaces. This involves (to within a
> constant or so..)taking the integral from negative infinity to
> positive infinity.
> F(w) ~ \int f(t)e^(iwt) dt
> However, reading the engineering literature, it seems the Laplace
> transform is used in a similar way. ( Note that the limits of the
> integral are now to infinity. )
> L(s) ~ \int f(t)e^(-st) dt
> Although I understand that the resulting Laplace transform is a
> complex function, how are the two related? Is one `more general'?
> What are the restrictions on the function in the time domain?
Both F(w) and L(s) are complex. The difference is in the argument
s = a + iw[/itex] (a > 0) , i.e.
[itex]e^{-st} = e^{-at} \dot [cos(wt) - i sin(wt)]
Thus, the Laplace transform might be `more general'. Because if
the time function begins at t >= (that's the only one restriction
on the time function ;) , you can calculate with the (enlarged)
spectrum. Sets of differential equations become simple algebra.
You get the *exact* result after the back transform.
However, the Fourier transform is only applicable to functions
finite in time (or in a time window that does as though the
function be zero out of the window).
Ulrich
http://home.t-online.de/home/Ulrich.Bruchholz/
Arnold Neumaier
Aug29-04, 12:54 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>robert bristow-johnson wrote:\n\n> actually the most correct definition of the LT has limits of -inf to\n> +inf. the form you see commonly in sophmore linear system texts (with\n> bottom limit "0-" or -epsilon) is there for pedagogical reasons.\n> either the systems analyzed are completely "relaxed" at t=0 (which\n> means all states are zero) and the derivatives of the input eventually\n> are zero at t=0, or there is information of the initial state that\n> actually changes the Laplace Transform in comparison to the "textbook"\n> transforms.\n>\n> after t=0, it makes no difference if you know all of the initial\n> states of the system H(s) and all of the derivatives of input x(t) at\n> t=0 or if you know all of x(t) before t=0 and assume that H(s) was\n> "relaxed" at t=-inf. they get you to the same place at t=0.\n\nhttp://mathworld.wolfram.com/LaplaceTransform.html says:\n\n\'\'The unilateral Laplace transform is almost always what is meant by\n"the" Laplace transform.\'\'\n\n[unilateral = extending the integral from 0 to inf]\n\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>robert bristow-johnson wrote:
> actually the most correct definition of the LT has limits of -inf to
> +inf. the form you see commonly in sophmore linear system texts (with
> bottom limit "0-" or -\epsilon) is there for pedagogical reasons.
> either the systems analyzed are completely "relaxed" at t=0 (which
> means all states are zero) and the derivatives of the input eventually
> are zero at t=0, or there is information of the initial state that
> actually changes the Laplace Transform in comparison to the "textbook"
> transforms.
>
> after t=0, it makes no difference if you know all of the initial
> states of the system H(s) and all of the derivatives of input x(t) at
> t=0 or if you know all of x(t) before t=0 and assume that H(s) was
> "relaxed" at t=-inf. they get you to the same place at t=0.
http://mathworld.wolfram.com/LaplaceTransform.html says:
''The unilateral Laplace transform is almost always what is meant by
"the" Laplace transform.''
[unilateral = extending the integral from to inf]
Arnold Neumaier
robert egri
Aug30-04, 03:57 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\nrbj@surfglobal.net (robert bristow-johnson) wrote in message news:<4cbb922e.0408270948.162fd525@posting.google. com>...\n> this could be a good one for comp.dsp ...\n>\n> stuart_wlsn@yahoo.com.au (Stuart Wilson) wrote in message news:<e227b6ef.0408270029.1cc9211a@posting.google. com>...\n> > Being a physicist, I am comfortable with using Fourier transforms\n[...]\n>\n> > taking the integral from negative infinity to positive infinity.\n> >\n> > F(w) ~ int f(t)e^(iwt) dt\n> >\n> > However, reading the engineering literature, it seems the Laplace\n> > transform is used in a similar way. ( Note that the limits of the\n> > integral are now 0 to infinity. )\n>\n> actually the most correct definition of the LT has limits of -inf to\n> +inf. [..]\n\nThat is not quite right. There is a so-called one-sided LT (that is\nused in electrical engineering almost exclusively), and then there is\nthe two-sided LT you are referring to, rarely used in engineering but\none is not more correct than the other, they are different. The former\nis great for initial value problems like transient analysis, the\nlatter is not very much used by engineers for in their practical but\nsloppy ways it can be subsumed in anything they do with FT.\n\n>\n> after t=0, it makes no difference if you know all of the initial\n> states of the system H(s) and all of the derivatives of input x(t) at\n> t=0 or if you know all of x(t) before t=0 and assume that H(s) was\n> "relaxed" at t=-inf. they get you to the same place at t=0.\n>\n> > L(s) ~ int f(t)e^(-st) dt\n> >\n> > Although I understand that the resulting Laplace transform is a\n> > complex function,\n>\n> so is the FT.\n>\n> > how are the two related? Is one `more general\'?\n>\n> yes, of course. the LT is more general than the FT. s = sigma +\n> i*omega and you set sigma to a positive enough value to make the\n> integral converge. for most decent signals (that ain\'t blowing up to\n> infinity), sigma = 0 is good enough and the two are precisely the\n> same.\n>\n\nWhat is "more" general in this context is matter of taste. The FT\nexists for any finite energy signal, and a scalar product what is\nusually called by engineers the cross-corrleation, can be defined and\nproved that the cross-correlation is the same in time as in frequency\ndomain (Parseval\'s theorem). This gives a Hilbert (vector) space\nstructure for the FT. There is no such resuslt for LT (not even if it\nis two-sided LT) and one could argue that that is the reason why FT is\nso interesting.\n\nFor FT they are all defined on the same lines, time and frequency. The\ndifficulty with LT is that the LT of different functions have\ndifferent regions of holomorphy and that, so-to-speak, has to be\n"sorted" out. For example, here is an application where pure "Fourier"\nmethods do not work, and what makes it work is extension of the\nFourier transform analytically to a one-sided Laplace Transform: this\nis the Wiener-Hopf method of solving semi-infinite one-sided\nconvolution-type integral equation. The two-sided conovolution problem\nis trivially solved by Fourier transform.\n\nThe one-sided convolution equation\n\nx(t) = C int_0^inf{ a(t-s).x(s).ds}\n\ncannot be solved by directly FT becasue the lower limit is not -inf,\nand it cannot be solved by LT becasue the upper limit is not t. The\ntrick to the solution is to massage the poles of the analytically\ncontinued FT to result in a product of a casual and anti-causal\nfunctions. See the details in any book on optimum Wiener (Kolmogorov)\nfiltering of statinary processes.\n\n\n> > What are the restrictions on the function in the time domain?\n>\n> the LT or FT integral has to exist (or converge). if x(t) won\'t let\n> it do that (no matter what sigma is) then you cannot LT it.\n>\n> r b-j\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>rbj@surfglobal.net (robert bristow-johnson) wrote in message news:<4cbb922e.0408270948.162fd525@posting.google.com>...
> this could be a good one for comp.dsp ...
>
> stuart_wlsn@yahoo.com.au (Stuart Wilson) wrote in message news:<e227b6ef.0408270029.1cc9211a@posting.google.com>...
> > Being a physicist, I am comfortable with using Fourier transforms
[...]
>
> > taking the integral from negative infinity to positive infinity.
> >
> > F(w) ~ \int f(t)e^(iwt) dt
> >
> > However, reading the engineering literature, it seems the Laplace
> > transform is used in a similar way. ( Note that the limits of the
> > integral are now to infinity. )
>
> actually the most correct definition of the LT has limits of -inf to
> +inf. [..]
That is not quite right. There is a so-called one-sided LT (that is
used in electrical engineering almost exclusively), and then there is
the two-sided LT you are referring to, rarely used in engineering but
one is not more correct than the other, they are different. The former
is great for initial value problems like transient analysis, the
latter is not very much used by engineers for in their practical but
sloppy ways it can be subsumed in anything they do with FT.
>
> after t=0, it makes no difference if you know all of the initial
> states of the system H(s) and all of the derivatives of input x(t) at
> t=0 or if you know all of x(t) before t=0 and assume that H(s) was
> "relaxed" at t=-inf. they get you to the same place at t=0.
>
> > L(s) ~ \int f(t)e^(-st) dt
> >
> > Although I understand that the resulting Laplace transform is a
> > complex function,
>
> so is the FT.
>
> > how are the two related? Is one `more general'?
>
> yes, of course. the LT is more general than the FT. s = \sigma +
> i*\omega and you set \sigma to a positive enough value to make the
> integral converge. for most decent signals (that ain't blowing up to
> infinity), \sigma = is good enough and the two are precisely the
> same.
>
What is "more" general in this context is matter of taste. The FT
exists for any finite energy signal, and a scalar product what is
usually called by engineers the cross-corrleation, can be defined and
proved that the cross-correlation is the same in time as in frequency
domain (Parseval's theorem). This gives a Hilbert (vector) space
structure for the FT. There is no such resuslt for LT (not even if it
is two-sided LT) and one could argue that that is the reason why FT is
so interesting.
For FT they are all defined on the same lines, time and frequency. The
difficulty with LT is that the LT of different functions have
different regions of holomorphy and that, so-to-speak, has to be
"sorted" out. For example, here is an application where pure "Fourier"
methods do not work, and what makes it work is extension of the
Fourier transform analytically to a one-sided Laplace Transform: this
is the Wiener-Hopf method of solving semi-infinite one-sided
convolution-type integral equation. The two-sided conovolution problem
is trivially solved by Fourier transform.
The one-sided convolution equation
x(t) = C \int_0^inf{ a(t-s).x(s).ds}
cannot be solved by directly FT becasue the lower limit is not -inf,
and it cannot be solved by LT becasue the upper limit is not t. The
trick to the solution is to massage the poles of the analytically
continued FT to result in a product of a casual and anti-causal
functions. See the details in any book on optimum Wiener (Kolmogorov)
filtering of statinary processes.
> > What are the restrictions on the function in the time domain?
>
> the LT or FT integral has to exist (or converge). if x(t) won't let
> it do that (no matter what \sigma is) then you cannot LT it.
>
> r b-j
Arnold Neumaier
Aug30-04, 04:37 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>\n\nWilliam R. Frensley wrote:\n\n> In a nutshell: The Fourier transform is optimum when dealing\n> with boundary-value problems and the Laplace transform is optimum\n> when dealing with initial-value problems. Engineers have\n> favored the latter because the systems they deal with clearly do\n> not have a history that goes back to t = - infinity.\n>\n> By the way, if you set up a Green\'s function formalism for the\n> Schroedinger equation in terms of Laplace, rather than Fourier\n> transforms, you will automatically get the retarded (or causal)\n> case, without having to worry about "adiabatic switching on"\n> and the resulting +/- i epsilon terms in the denominator.\n\n\nBut the Fourier transform is needed to switch between the position\nand the momentum representation. It is a unitary transform, which\nmeans that it preserves the inner product, essential for QM.\nWith the Laplace transform, this is spoilt. Thus I doubt that one\ncan work out a valid QM in terms of the Laplace transform.\n\n\nArnold Neumaier\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>William R. Frensley wrote:
> In a nutshell: The Fourier transform is optimum when dealing
> with boundary-value problems and the Laplace transform is optimum
> when dealing with initial-value problems. Engineers have
> favored the latter because the systems they deal with clearly do
> not have a history that goes back to t = - infinity.
>
> By the way, if you set up a Green's function formalism for the
> Schroedinger equation in terms of Laplace, rather than Fourier
> transforms, you will automatically get the retarded (or causal)
> case, without having to worry about "adiabatic switching on"
> and the resulting +/- i \epsilon terms in the denominator.
But the Fourier transform is needed to switch between the position
and the momentum representation. It is a unitary transform, which
means that it preserves the inner product, essential for QM.
With the Laplace transform, this is spoilt. Thus I doubt that one
can work out a valid QM in terms of the Laplace transform.
Arnold Neumaier
robert bristow-johnson
Aug31-04, 02:40 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<41318A41.4080504@univie.ac.at>...\n> robert bristow-johnson wrote:\n>\n> > actually the most correct definition of the LT has limits of -inf to\n> > +inf. the form you see commonly in sophmore linear system texts (with\n> > bottom limit "0-" or -epsilon) is there for pedagogical reasons.\n> > either the systems analyzed are completely "relaxed" at t=0 (which\n> > means all states are zero) and the derivatives of the input eventually\n> > are zero at t=0, or there is information of the initial state that\n> > actually changes the Laplace Transform in comparison to the "textbook"\n> > transforms.\n> >\n> > after t=0, it makes no difference if you know all of the initial\n> > states of the system H(s) and all of the derivatives of input x(t) at\n> > t=0 or if you know all of x(t) before t=0 and assume that H(s) was\n> > "relaxed" at t=-inf. they get you to the same place at t=0.\n>\n> http://mathworld.wolfram.com/LaplaceTransform.html says:\n>\n> \'\'The unilateral Laplace transform is almost always what is meant by\n> "the" Laplace transform.\'\'\n>\n> [unilateral = extending the integral from 0 to inf]\n\nanother quote from the same website (different page):\n\nhttp://mathworld.wolfram.com/BilateralLaplaceTransform.html\n\n"While some authors use this as the primary definition of "the"\nLaplace transform (Oppenheim et al. 1997), it is much more common for\nthe unilateral Laplace transform to be considered the primary\ndefinition."\n\nanother text that uses the bilateral LT for the primary definition is\nthe old Papoulis "Signal Analysis".\n\nwhile my POV may not be the most popular for undergrad engineering\ntexts, it *is* more fundamental. think about the LT of the derivative\nof x(t) and compare the Unilateral LT to the Bilateral LT. for the\nUnilateral LT:\n\n\nLT{ x\'(t) } = s*X(s) - f(0)\n\nof course, that f(0) is needed because taking the derivative of a\nfunction destroys information about its constant value component. for\nthe Bilateral LT as well as the FT:\n\nLT{ x\'(t) } = s*X(s)\n\nFT{ x\'(t) } = i*w * X(i*w)\n\nwhat happened to the f(0)? it is the result of the first half\nintegral of the Bilateral LT. the Bilateral LT takes care of that in\nits definition and operation. that is why i would agree with\nOppenhiem and Wilskey and Papoulis that it is more basic or\nfundamental and you can, if you like causal systems and signals,\nderive the Unilateral FT and all of its properties from the Bilateral.\n\nno?\n\nr b-j\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<41318A41.4080504@univie.ac.at>...
> robert bristow-johnson wrote:
>
> > actually the most correct definition of the LT has limits of -inf to
> > +inf. the form you see commonly in sophmore linear system texts (with
> > bottom limit "0-" or -\epsilon) is there for pedagogical reasons.
> > either the systems analyzed are completely "relaxed" at t=0 (which
> > means all states are zero) and the derivatives of the input eventually
> > are zero at t=0, or there is information of the initial state that
> > actually changes the Laplace Transform in comparison to the "textbook"
> > transforms.
> >
> > after t=0, it makes no difference if you know all of the initial
> > states of the system H(s) and all of the derivatives of input x(t) at
> > t=0 or if you know all of x(t) before t=0 and assume that H(s) was
> > "relaxed" at t=-inf. they get you to the same place at t=0.
>
> http://mathworld.wolfram.com/LaplaceTransform.html says:
>
> ''The unilateral Laplace transform is almost always what is meant by
> "the" Laplace transform.''
>
> [unilateral = extending the integral from to inf]
another quote from the same website (different page):
http://mathworld.wolfram.com/BilateralLaplaceTransform.html
"While some authors use this as the primary definition of "the"
Laplace transform (Oppenheim et al. 1997), it is much more common for
the unilateral Laplace transform to be considered the primary
definition."
another text that uses the bilateral LT for the primary definition is
the old Papoulis "Signal Analysis".
while my POV may not be the most popular for undergrad engineering
texts, it *is* more fundamental. think about the LT of the derivative
of x(t) and compare the Unilateral LT to the Bilateral LT. for the
Unilateral LT:
LT{ x'(t) } = s*X(s) - f(0)
of course, that f(0) is needed because taking the derivative of a
function destroys information about its constant value component. for
the Bilateral LT as well as the FT:
LT{ x'(t) } = s*X(s)
FT{ x'(t) } = i*w * X(i*w)
what happened to the f(0)? it is the result of the first half
integral of the Bilateral LT. the Bilateral LT takes care of that in
its definition and operation. that is why i would agree with
Oppenhiem and Wilskey and Papoulis that it is more basic or
fundamental and you can, if you like causal systems and signals,
derive the Unilateral FT and all of its properties from the Bilateral.
no?
r b-j
<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>\nAs you have observed "..that f(0) is needed because taking the\nderivative of a function destroys information about its constant value\ncomponent...". The constant (unless it is zero) has no Bi-LT; this\nintegral int_-inf^+inf {exp(-st)dt} does not exist over any finte strip\nin s.\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>As you have observed "..that f(0) is needed because taking the
derivative of a function destroys information about its constant value
component...". The constant (unless it is zero) has no Bi-LT; this
integral \int_-inf^+inf {\exp(-st)dt} does not exist over any finte strip
in s.
da_willem
Sep4-04, 02:05 AM
In the Laplace transform the argument s:= v+wi can be complex. So the Fourier transform is a Laplace transform with v=0, or:
{\cal L}[f(t)]={\cal F} [f(t)e^{-vt}]
So you can use the extra freedom of the real part of s to make the transformation-intergal converge.
The Laplace transform has a region of concergence (values of s) depending on the function transformed but If v=0 is included in the region of convergence the Fourier transform of the same function converges also.
sridhar_n
Sep4-04, 02:06 AM
The Laplace Transforms are a special case of the Fourier Transfoms. While the Fourier Transforms can be used to move from any domain to any other domain, the Laplace Transforms can only be used to move from Time to Frequency and Back.
Sridhar
J. B. Wood
Sep9-04, 02:59 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>In article <sridhar_n.1bzcjm@physicsforums.com>, sridhar_n\n<sridhar_n@email.com> wrote:\n\n> The Laplace Transforms are a special case of the Fourier Transfoms.\n> While the Fourier Transforms can be used to move from any domain to any\n> other domain, the Laplace Transforms can only be used to move from Time\n> to Frequency and Back.\n>\n> Sridhar\n\nHello, and if anything it\'s the other way around. The Fourier transform\nis obtained from the bi-lateral Laplace transform by setting the real part\nof the complex variable s to zero. In electrical engineering terms (where\nsuch transforms are used to characterize the simulus-response of a system\nor plant) this is equivalent to ignoring the impulse response of a system\nand dealing only with the steady-state (frequency) response. Sincerely,\n\nJohn Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil\nNaval Research Laboratory\n4555 Overlook Avenue, SW\nWashington, DC 20375-5337\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 <sridhar_n.1bzcjm@physicsforums.com>, sridhar_n
<sridhar_n@email.com> wrote:
> The Laplace Transforms are a special case of the Fourier Transfoms.
> While the Fourier Transforms can be used to move from any domain to any
> other domain, the Laplace Transforms can only be used to move from Time
> to Frequency and Back.
>
> Sridhar
Hello, and if anything it's the other way around. The Fourier transform
is obtained from the bi-lateral Laplace transform by setting the real part
of the complex variable s to zero. In electrical engineering terms (where
such transforms are used to characterize the simulus-response of a system
or plant) this is equivalent to ignoring the impulse response of a system
and dealing only with the steady-state (frequency) response. Sincerely,
John Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337
<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\nrobert egri <rge11x@netscape.net> writes\n>What is "more" general in this context is matter of taste. The FT\n>exists for any finite energy signal, and a scalar product what is\n>usually called by engineers the cross-corrleation, can be defined and\n>proved that the cross-correlation is the same in time as in frequency\n>domain (Parseval\'s theorem). This gives a Hilbert (vector) space\n>structure for the FT. There is no such resuslt for LT (not even if it\n>is two-sided LT) and one could argue that that is the reason why FT is\n>so interesting.\n>\n>For FT they are all defined on the same lines, time and frequency. The\n>difficulty with LT is that the LT of different functions have\n>different regions of holomorphy and that, so-to-speak, has to be\n>"sorted" out. For example, here is an application where pure "Fourier"\n>methods do not work, and what makes it work is extension of the\n>Fourier transform analytically to a one-sided Laplace Transform: this\n>is the Wiener-Hopf method of solving semi-infinite one-sided\n>convolution-type integral equation. The two-sided conovolution problem\n>is trivially solved by Fourier transform.\n>\n>The one-sided convolution equation\n>\n>x(t) = C int_0^inf{ a(t-s).x(s).ds}\n>\n>cannot be solved by directly FT becasue the lower limit is not -inf,\n>and it cannot be solved by LT becasue the upper limit is not t. The\n>trick to the solution is to massage the poles of the analytically\n>continued FT to result in a product of a casual and anti-causal\n>functions. See the details in any book on optimum Wiener (Kolmogorov)\n>filtering of statinary processes.\n\nThat\'s a very interesting overview.\nI find it interesting because in real life we have (for example) a\nphoton emitted by atom #1 at t_1, and absorbed by a different atom at\nt_2. The wavefunction of the photon thus exists only between t_1 and\nt_2. Of course that\'s not precisely so because both atoms have some\n\'photon-ness\' about their description even before t_1 and after t_2.\n\nCan I attempt to sum it up?\n\nWe have two formulations of a pulse.\n\n1a) We can sum it as a series of sine waves and get the FT.\n1b) We can sum it as a series of pulses and get the LT.\n\n2a) The FT is unphysical because its time domain goes from -inf to +inf\nand this is impossible (eg big bang).\n2b) The LT is unphysical because it confounds energy and momentum.\n\n3) There is a mixed formulation that has fewer problems, although I\nwonder whether it solves them all.\n\n====Oz\'s brain wanders off about here ...\n\nI\'ve always wondered if the problem associated with these is due to the\nover-simplistic model that is used. Both have elemental forms that are\nnon-physical, so its not surprising that the result is non-physical.\n\nI note that there are an infinity of possible waveforms in between the\ntwo used above that can also be summed (using the same form) to describe\nany other waveform.\n\nA real-life pulse always has tails because (in essence) the risetime is\nalways finite. In general real pulses all have the same form. They have\nelements of an FT (sinusoidal tails) and elements of Lt\'s in that they\nare finite in time. One ought in principle to be able to replicate any\ngiven waveform as a sum of pulses of this form (there\'s probably an\ninfinity of possible elemental pulseforms one could choose).\n\nI can see this as having some problems that might make it a tad tricky.\nHaving chosen an appropriate waveform there are three required variables\nto solve. Amplitudes for each (elemental) pulseform, the position in\nspace for each pulseform and the nominal width (ie frequency/how much\nits \'squeezed). For Lt\'s and FT\'s we only have two, amplitude and either\nfrequency or delta-position. I know nothing about gaussian waveforms\n(along with a host of other things) but from gleaning here I get the\nimpression that these are \'quite well understood\', and am somewhat\npuzzled that modelling along these lines isn\'t a standard practice among\nthe heavyweights.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>robert egri <rge11x@netscape.net> writes
>What is "more" general in this context is matter of taste. The FT
>exists for any finite energy signal, and a scalar product what is
>usually called by engineers the cross-corrleation, can be defined and
>proved that the cross-correlation is the same in time as in frequency
>domain (Parseval's theorem). This gives a Hilbert (vector) space
>structure for the FT. There is no such resuslt for LT (not even if it
>is two-sided LT) and one could argue that that is the reason why FT is
>so interesting.
>
>For FT they are all defined on the same lines, time and frequency. The
>difficulty with LT is that the LT of different functions have
>different regions of holomorphy and that, so-to-speak, has to be
>"sorted" out. For example, here is an application where pure "Fourier"
>methods do not work, and what makes it work is extension of the
>Fourier transform analytically to a one-sided Laplace Transform: this
>is the Wiener-Hopf method of solving semi-infinite one-sided
>convolution-type integral equation. The two-sided conovolution problem
>is trivially solved by Fourier transform.
>
>The one-sided convolution equation
>
>x(t) = C \int_0^inf{ a(t-s).x(s).ds}
>
>cannot be solved by directly FT becasue the lower limit is not -inf,
>and it cannot be solved by LT becasue the upper limit is not t. The
>trick to the solution is to massage the poles of the analytically
>continued FT to result in a product of a casual and anti-causal
>functions. See the details in any book on optimum Wiener (Kolmogorov)
>filtering of statinary processes.
That's a very interesting overview.
I find it interesting because in real life we have (for example) a
photon emitted by atom #1 at t_1, and absorbed by a different atom at
t_2. The wavefunction of the photon thus exists only between t_1 and
t_2. Of course that's not precisely so because both atoms have some
'photon-ness' about their description even before t_1 and after t_2.
Can I attempt to sum it up?
We have two formulations of a pulse.
1a) We can sum it as a series of sine waves and get the FT.
1b) We can sum it as a series of pulses and get the LT.
2a) The FT is unphysical because its time domain goes from -inf to +inf
and this is impossible (eg big bang).
2b) The LT is unphysical because it confounds energy and momentum.
3) There is a mixed formulation that has fewer problems, although I
wonder whether it solves them all.
====Oz's brain wanders off about here ...
I've always wondered if the problem associated with these is due to the
over-simplistic model that is used. Both have elemental forms that are
non-physical, so its not surprising that the result is non-physical.
I note that there are an infinity of possible waveforms in between the
two used above that can also be summed (using the same form) to describe
any other waveform.
A real-life pulse always has tails because (in essence) the risetime is
always finite. In general real pulses all have the same form. They have
elements of an FT (sinusoidal tails) and elements of Lt's in that they
are finite in time. One ought in principle to be able to replicate any
given waveform as a sum of pulses of this form (there's probably an
infinity of possible elemental pulseforms one could choose).
I can see this as having some problems that might make it a tad tricky.
Having chosen an appropriate waveform there are three required variables
to solve. Amplitudes for each (elemental) pulseform, the position in
space for each pulseform and the nominal width (ie frequency/how much
its 'squeezed). For Lt's and FT's we only have two, amplitude and either
frequency or \delta-position. I know nothing about gaussian waveforms
(along with a host of other things) but from gleaning here I get the
impression that these are 'quite well understood', and am somewhat
puzzled that modelling along these lines isn't a standard practice among
the heavyweights.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
p.kinsler@imperial.ac.uk
Sep27-04, 03:32 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\nOz <oz@farmeroz.port995.com> wrote:\n> 1a) We can sum it as a series of sine waves and get the FT.\n> 1b) We can sum it as a series of pulses and get the LT.\n\n> 2a) The FT is unphysical because its time domain goes from -inf to +inf\n> and this is impossible (eg big bang).\n> 2b) The LT is unphysical because it confounds energy and momentum.\n\n> 3) There is a mixed formulation that has fewer problems, although I\n> wonder whether it solves them all.\n\nDon\'t get hung up on the fourier/sine or laplace/exponential\ndecomposition. Any suitable set of basis functions will do\n-- although I\'ve hidden a lot of detail inside that "suitable".\n\nFor example, if you are in a confined volume, such as in\na laser resonator, the Gauss-Laguerre modes are a good\nway to decompose the electric and magnetic fields.\n\n--\n---------------------------------+---------------------------------\nDr. Paul Kinsler\nBlackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714\nImperial College London, Dr.Paul.Kinsler@physics.org\nSW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/\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>Oz <oz@farmeroz.port995.com> wrote:
> 1a) We can sum it as a series of sine waves and get the FT.
> 1b) We can sum it as a series of pulses and get the LT.
> 2a) The FT is unphysical because its time domain goes from -inf to +inf
> and this is impossible (eg big bang).
> 2b) The LT is unphysical because it confounds energy and momentum.
> 3) There is a mixed formulation that has fewer problems, although I
> wonder whether it solves them all.
Don't get hung up on the fourier/sine or laplace/exponential
decomposition. Any suitable set of basis functions will do
-- although I've hidden a lot of detail inside that "suitable".
For example, if you are in a confined volume, such as in
a laser resonator, the Gauss-Laguerre modes are a good
way to decompose the electric and magnetic fields.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London, Dr.Paul.Kinsler@physics.org
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
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