Want to find the inverse Laplace transform of the following term: H(s)=1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to +infinity) How to do that? ------------------------------ Making relaxation to the problem, if I have to find only certain sampled values of the inverse Laplace transform of H(s), let's denote it as h(t), I just need to find h(1), h(2), h(3), etc. Is there a short cut for it? Thanks a lot!
does anybody know how to get the inverse laplace transform of this wierd thing?
Started by ●January 31, 2006
Reply by ●January 31, 20062006-01-31
Its been many years since I have studied the transforms in this fashion, but the concepts of final value theorem and initial value theorem come to mind. Perhaps there is some varient thereof that would be beneficial. It seems like a real odd equation to crank through algebraically. The part about having in integral in there also makes me think of residue theory, but it is just a vague recollection. Sorry if I haven't been of much help.
Reply by ●February 1, 20062006-02-01
gino wrote:> Want to find the inverse Laplace transform of the following term: > > H(s)=1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to +infinity) > > How to do that? > > ------------------------------ > > Making relaxation to the problem, if I have to find only certain sampled > values of the inverse Laplace transform of H(s), let's denote it as h(t), > > I just need to find h(1), h(2), h(3), etc. > > Is there a short cut for it? > > Thanks a lot! > >Dino, Let G(s) = sH(s) so that H(s) = (1/s)G(s). Any book on Laplace Transforms will tell you that if g() is the inverse Laplace Transform of G(s) then the inverse Laplace Transform of H(s) is integral over [0,t] of g(). All that remains is to determine g(). Since your integral is essentially the complementary error function, formula 29.3.113 in Abramowitz and Stegun "Handbook of Mathematical Functions" (still indispensible in the 21'st century) is directly applicable, yielding g(x) = sqrt(pi/2) * erf(x/(2k)), k = a/sqrt(2) I suggest you check this yourself to make sure I have the factors correct (free advice cannot be guaranteed). How to numerically evaluate this is another matter but Matlab has the tools to do it. Good luck! Mike
Reply by ●February 2, 20062006-02-02
On Tue, 31 Jan 2006 17:10:16 -0800, "gino" <loseminds@hotmail.com> wrote:>Want to find the inverse Laplace transform of the following term: > >H(s)=3D1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to =+infinity)> >How to do that?[snip] In what follows, define the error function Erf(x) as Erf(x) :=3D (2 / sqrt(pi) ) Int[0..oo] exp(-u^2) du and define the complementary error function Erfc(x) as Erfc(x) :=3D (2 / sqrt(pi) ) Int[x..oo] exp(-u^2) du =46irst, take F(s) =3D exp(a^2 s^2 / 2) Int[as..oo] exp(-u^2 / 2) du Let u =3D sqrt(2) v=20 and define a new constant b =3D a / sqrt(2) Then, if I've done my substitutions correctly, F(s) =3D sqrt(2) exp(b^2 s^2) Int[bs..oo] exp(-v^2) dv F(s) =3D sqrt(pi/2) exp(b^2 s^2) Erfc(bs) Let G(s) =3D (1/s) F(s) The inverse Laplace transform of G(s) is g(t) =3D sqrt(pi/2) Erf[t/(2b)] Re b^2 > 0 and Re s > 0 Now, your problem asks for the inverse of H(s) =3D (1/s) G(s) which would be h(t) =3D sqrt(pi/2) Int[0..t] Erf[u/(2b)] du b=3Da/sqrt(2) HTH =20
Reply by ●February 2, 20062006-02-02
On Thu, 02 Feb 2006 10:17:33 GMT, Anon <spamhole@galaxycentral.gww> wrote:>On Tue, 31 Jan 2006 17:10:16 -0800, "gino" <loseminds@hotmail.com> >wrote: > >>Want to find the inverse Laplace transform of the following term: >> >>H(s)=3D1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to =+infinity)>> >>How to do that? > >[snip] > >In what follows, define the error function Erf(x) as > Erf(x) :=3D (2 / sqrt(pi) ) Int[0..oo] exp(-u^2) duAddendum ... =46irst, correct typo in definition of Erf(x) Erf(x) :=3D (2 / sqrt(pi) ) Int[0..x] exp(-u^2) du> h(t) =3D sqrt(pi/2) Int[0..t] Erf[u/(2b)] du b=3Da/sqrt(2)Since Integral Erf(x) dx =3D x Erf(x) + exp(-x^2) / sqrt(pi), and writing {u/2b} as meaning u/(2b), h(t) =3D sqrt(pi/2) [ {u/2b} Erf{u/2b}=20 + exp(-{u/2b}^2) / sqrt(pi) ] | [0 .. t] =20 =3D sqrt(pi/2) [ {t/2b} Erf{t/2b} + exp(-{t/2b}^2) / sqrt(pi) - 1/sqrt(pi) ] Define a new constant c =3D sqrt(2) / (2a) so that t/(2b) =3D [sqrt(2) t / (2a)] =3D ct h(t) =3D sqrt(pi/2) [ ct Erf(ct) + exp(-(ct)^2) / sqrt(pi) - 1/sqrt(pi) ] h(t) =3D [1/sqrt(2)] [ sqrt(pi) ct Erf(ct) + exp(-(ct)^2) - 1 ] Hopefully someone will correct me if I've gone off the rails. 8-) =20
Reply by ●February 2, 20062006-02-02
In article <abm3u1do6cm669279f3ljqvi3tb13l698k@4ax.com>, Anon <spamhole@galaxycentral.gww> wrote:> On Tue, 31 Jan 2006 17:10:16 -0800, "gino" <loseminds@hotmail.com> > wrote: > > >Want to find the inverse Laplace transform of the following term: > > > >H(s)=1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to +infinity) > > > >How to do that? > > [snip] > > In what follows, define the error function Erf(x) as > Erf(x) := (2 / sqrt(pi) ) Int[0..oo] exp(-u^2) duShouldn't "(2 / sqrt(pi) ) Int[0..oo] exp(-u^2) du" have an 'x' in it somewhere if it is to be a function of 'x'?> > and define the complementary error function Erfc(x) as > Erfc(x) := (2 / sqrt(pi) ) Int[x..oo] exp(-u^2) du > > > First, take > F(s) = exp(a^2 s^2 / 2) Int[as..oo] exp(-u^2 / 2) du > > Let u = sqrt(2) v > and define a new constant b = a / sqrt(2) > > Then, if I've done my substitutions correctly, > F(s) = sqrt(2) exp(b^2 s^2) Int[bs..oo] exp(-v^2) dv > > F(s) = sqrt(pi/2) exp(b^2 s^2) Erfc(bs) > > > Let G(s) = (1/s) F(s) > > The inverse Laplace transform of G(s) is > g(t) = sqrt(pi/2) Erf[t/(2b)] Re b^2 > 0 and Re s > 0 > > > Now, your problem asks for the inverse of H(s) = (1/s) G(s) > which would be > > h(t) = sqrt(pi/2) Int[0..t] Erf[u/(2b)] du b=a/sqrt(2) > > HTH >
Reply by ●February 4, 20062006-02-04
On Thu, 02 Feb 2006 13:44:18 GMT, Anon <spamhole@galaxycentral.gww> wrote:>On Thu, 02 Feb 2006 10:17:33 GMT, Anon <spamhole@galaxycentral.gww> >wrote: > >>On Tue, 31 Jan 2006 17:10:16 -0800, "gino" <loseminds@hotmail.com> >>wrote: >> >>>Want to find the inverse Laplace transform of the following term: >>> >>>H(s)=3D1/s^2*exp(s^2*a^2/2)*integrate(exp(-u^2/2), u from s*a to =+infinity)>>> >>>How to do that? >> >>[snip] >> >>In what follows, define the error function Erf(x) as >> Erf(x) :=3D (2 / sqrt(pi) ) Int[0..oo] exp(-u^2) du > >Addendum ... > >First, correct typo in definition of Erf(x) > Erf(x) :=3D (2 / sqrt(pi) ) Int[0..x] exp(-u^2) du > > >> h(t) =3D sqrt(pi/2) Int[0..t] Erf[u/(2b)] du b=3Da/sqrt(2) > >Since Integral Erf(x) dx =3D x Erf(x) + exp(-x^2) / sqrt(pi), > >and writing {u/2b} as meaning u/(2b), > > h(t) =3D sqrt(pi/2) [ {u/2b} Erf{u/2b}=20 > + exp(-{u/2b}^2) / sqrt(pi) ] | [0 .. t] > =20 > =3D sqrt(pi/2) [ {t/2b} Erf{t/2b} + exp(-{t/2b}^2) / sqrt(pi) > - 1/sqrt(pi) ] > >Define a new constant c =3D sqrt(2) / (2a) so that > t/(2b) =3D [sqrt(2) t / (2a)] =3D ct > > h(t) =3D sqrt(pi/2) [ ct Erf(ct) + exp(-(ct)^2) / sqrt(pi) > - 1/sqrt(pi) ] > > h(t) =3D [1/sqrt(2)] [ sqrt(pi) ct Erf(ct) + exp(-(ct)^2) - 1 ] > > >Hopefully someone will correct me if I've gone off the rails. 8-) >=20Don't know if the OP is still around to care, but I did get off the rails and lost a factor of 1/c in the final result above. =20 Errata (do over) ... h(t) =3D sqrt(pi/2) Int[0..t] Erf[u/(2b)] du, b=3Da/sqrt(2) Let x =3D u/(2b) and define a new constant c =3D 1/(2b) h(t) =3D sqrt(pi/2) (1/c) Int[0..ct] Erf(x) dx Since Int Erf(x) dx =3D x Erf(x) + exp(-x^2) / sqrt(pi), h(t) =3D sqrt(pi/2) (1/c) [ x Erf(x) =20 + exp(-x^2) / sqrt(pi) ] | [0 .. ct] =20 =3D sqrt(pi/2) (1/c) [ ct Erf(ct) + exp(-(ct)^2) / sqrt(pi) - 1/sqrt(pi) ] h(t) =3D [1/sqrt(2)] (1/c) [sqrt(pi) ct Erf(ct) + exp(-(ct)^2) - 1] In summary, given=20 H(s) =3D (1/s^2) exp(s^2*a^2/2) Int[as .. oo] exp(-u^2/2) du, the inverse Laplace transform of H(s) is h(t) =3D a [sqrt(pi) ct Erf(ct) + exp(-(ct)^2) - 1], c =3D sqrt(2) / (2a) Sorry for the previous errors. =20
