Hi all, I have a function f(x) and I take its unilateral Laplace transform to obtain F(z) where z is on the complex plane. However it is complicated and it is possible only to obtain the F(z) numerically. That's to say: f(x) -- (numerical) Laplace Transform -- > F(z) However, I want to find (hopefully theoretically) the poles and hence the region of convergence of F(z), because the subsequent treatment of F(z) requires making sure that F(z)'s poles and ROC are correct. A closed form solution of the poles and region of convergence will give some useful intuition and guidelines... In order to find F(z)'s poles and hence the ROC, I compute f(x)'s asymptotics as x grows large. I found the asymptotic behavior of f(x) is the same as g(x), and g(x) has a convenient closed form of Laplace transform G(z). Will I be able to use the poles of G(z) to infer the poles of the original laplace transform F(z)? Am I on the right track? Thanks a lot!

# Laplace transform and poles and region of convergence...

Started by ●July 4, 2007

Reply by ●July 4, 20072007-07-04

Hi Vista, Not quite sure about the assumption that f(x) and g(x) shares the same pole/ROC if they shares the same asymptotic behavior. Consider the functions exp(-x), exp(-2x), and 0,(x>0). It seems they don't get the same pole. Regards, Is A Bug On Tue, 2007-07-03 at 20:16 -0700, Vista wrote:> Hi all, > > I have a function f(x) and I take its unilateral Laplace transform to obtain > F(z) where z is on the complex plane. However it is complicated and it is > possible only to obtain the F(z) numerically. > > That's to say: > f(x) -- (numerical) Laplace Transform -- > F(z) > > However, I want to find (hopefully theoretically) the poles and hence the > region of convergence of F(z), because the subsequent treatment of F(z) > requires making sure that F(z)'s poles and ROC are correct. A closed form > solution of the poles and region of convergence will give some useful > intuition and guidelines... > > In order to find F(z)'s poles and hence the ROC, I compute f(x)'s > asymptotics as x grows large. > > I found the asymptotic behavior of f(x) is the same as g(x), and g(x) has a > convenient closed form of Laplace transform G(z). Will I be able to use the > poles of G(z) to infer the poles of the original laplace transform F(z)? Am > I on the right track? > > Thanks a lot! > > > >

Reply by ●July 4, 20072007-07-04

Vista wrote:> In order to find F(z)'s poles and hence the ROC, I compute f(x)'s > asymptotics as x grows large. > > I found the asymptotic behavior of f(x) is the same as g(x), and g(x) has a > convenient closed form of Laplace transform G(z). Will I be able to use the > poles of G(z) to infer the poles of the original laplace transform F(z)? Am > I on the right track?If f(x) -> g(x) for large x then it seems the poles of both are going to be the same. But there is a bit of a question here since f(x) and g(x) are of different character. F(x) is actually a discrete function of points (I presume) and g(x) seems to be a continuous analytic function so a proof of shall we say "equivalence" may take some hand- waving. But if you can show that both functions are equivalent for large x I think that allows an assumption that they have the same poles. However since g(x) is only "equivalent" for large x, it can't be substituted for any other assumptions about f(x) including zeros. At least that's the way it seems to me. Benj

Reply by ●July 4, 20072007-07-04

On 7/3/07 8:16 PM, in article f6f3c6$883$1@news.Stanford.EDU, "Vista" <abc@gmai.com> wrote:> Hi all, > > I have a function f(x) and I take its unilateral Laplace transform to obtain > F(z) where z is on the complex plane. However it is complicated and it is > possible only to obtain the F(z) numerically. > > That's to say: > f(x) -- (numerical) Laplace Transform -- > F(z) > > However, I want to find (hopefully theoretically) the poles and hence the > region of convergence of F(z), because the subsequent treatment of F(z) > requires making sure that F(z)'s poles and ROC are correct. A closed form > solution of the poles and region of convergence will give some useful > intuition and guidelines... > > In order to find F(z)'s poles and hence the ROC, I compute f(x)'s > asymptotics as x grows large. > > I found the asymptotic behavior of f(x) is the same as g(x), and g(x) has a > convenient closed form of Laplace transform G(z). Will I be able to use the > poles of G(z) to infer the poles of the original laplace transform F(z)? Am > I on the right track? > > Thanks a lot! > > > >I have some suggestions. 1. Laplace transform methods are used to solve differential equations. Usually, but not exclusively, these are ODEs with constant coefficients. Thus, your f(x) represents some ODE. Solve the ODE numerically without using transform methods. Add some random noise to the solution just to make sure that any unstable poles get excited. 1. Do a numerical integration as per the Cauchy contour integral theorem. That should give you a fairly good value to the number of poles inside the contour even after computational roundoff. Bill -- Support the troops. Impeach Bush. Oh, I forgot about Cheney.

Reply by ●July 4, 20072007-07-04

"Salmon Egg" <salmonegg@sbcglobal.net> wrote in message news:C2B10E4F.85F8C%salmonegg@sbcglobal.net...> On 7/3/07 8:16 PM, in article f6f3c6$883$1@news.Stanford.EDU, "Vista" > <abc@gmai.com> wrote: > >> Hi all, >> >> I have a function f(x) and I take its unilateral Laplace transform to >> obtain >> F(z) where z is on the complex plane. However it is complicated and it is >> possible only to obtain the F(z) numerically. >> >> That's to say: >> f(x) -- (numerical) Laplace Transform -- > F(z) >> >> However, I want to find (hopefully theoretically) the poles and hence the >> region of convergence of F(z), because the subsequent treatment of F(z) >> requires making sure that F(z)'s poles and ROC are correct. A closed form >> solution of the poles and region of convergence will give some useful >> intuition and guidelines... >> >> In order to find F(z)'s poles and hence the ROC, I compute f(x)'s >> asymptotics as x grows large. >> >> I found the asymptotic behavior of f(x) is the same as g(x), and g(x) >> has a >> convenient closed form of Laplace transform G(z). Will I be able to use >> the >> poles of G(z) to infer the poles of the original laplace transform F(z)? >> Am >> I on the right track? >> >> Thanks a lot! >> >> >> >> > I have some suggestions. > > 1. Laplace transform methods are used to solve differential equations. > Usually, but not exclusively, these are ODEs with constant coefficients. > Thus, your f(x) represents some ODE. Solve the ODE numerically without > using > transform methods. Add some random noise to the solution just to make sure > that any unstable poles get excited. >How do I decide the coefficient of the ODEs from a very complicated Laplace transform?> 1. Do a numerical integration as per the Cauchy contour integral theorem. > That should give you a fairly good value to the number of poles inside the > contour even after computational roundoff. >Numerical integral is a much more weird animal...

Reply by ●July 4, 20072007-07-04

"Benj" <bjacoby@iwaynet.net> wrote in message news:1183526707.918541.60890@k79g2000hse.googlegroups.com...> > Vista wrote: >> In order to find F(z)'s poles and hence the ROC, I compute f(x)'s >> asymptotics as x grows large. >> >> I found the asymptotic behavior of f(x) is the same as g(x), and g(x) >> has a >> convenient closed form of Laplace transform G(z). Will I be able to use >> the >> poles of G(z) to infer the poles of the original laplace transform F(z)? >> Am >> I on the right track? > > If f(x) -> g(x) for large x then it seems the poles of both are going > to be the same. But there is a bit of a question here since f(x) and > g(x) are of different character. F(x) is actually a discrete function > of points (I presume) and g(x) seems to be a continuous analytic > function so a proof of shall we say "equivalence" may take some hand- > waving. But if you can show that both functions are equivalent for > large x I think that allows an assumption that they have the same > poles. However since g(x) is only "equivalent" for large x, it can't > be substituted for any other assumptions about f(x) including zeros. > > At least that's the way it seems to me. > > Benj >Mathematically, the "equivalence" of g(x) and f(x) for lare x, is an if and only if condition of the "equivalence" of G(z) and F(z)'s poles behavior... Am I right?

Reply by ●July 4, 20072007-07-04

"isabug" <isabut@gmail.com> wrote in message news:1183526136.6584.6.camel@iwu7...> Hi Vista, > Not quite sure about the assumption that f(x) and g(x) shares the same > pole/ROC if they shares the same asymptotic behavior. > > Consider the functions exp(-x), exp(-2x), and 0,(x>0). It seems they > don't get the same pole. > > Regards, > Is A Bug >But the example you gave are not the same asymptotics.

Reply by ●July 4, 20072007-07-04

"Vista" <abc@gmai.com> writes:> Hi all, > > I have a function f(x) and I take its unilateral Laplace transform to > obtain > F(z) where z is on the complex plane. However it is complicated and it is > possible only to obtain the F(z) numerically. > > That's to say: > f(x) -- (numerical) Laplace Transform -- > F(z) > > However, I want to find (hopefully theoretically) the poles and hence the > region of convergence of F(z), because the subsequent treatment of F(z) > requires making sure that F(z)'s poles and ROC are correct. A closed form > solution of the poles and region of convergence will give some useful > intuition and guidelines... > > In order to find F(z)'s poles and hence the ROC, I compute f(x)'s > asymptotics as x grows large. > > I found the asymptotic behavior of f(x) is the same as g(x), and g(x) has > a > convenient closed form of Laplace transform G(z). Will I be able to use the > > poles of G(z) to infer the poles of the original laplace transform F(z)? Am > > I on the right track?Under some conditions, the rightmost pole(s) of F(z) and G(z) may be the same. Probably not the other poles (if any). For that matter, there's no guarantee that F(z) has poles at all. What may be true is that the abscissa of convergence is the same. For example, this is true if g(x) > 0 for all sufficiently large x and f(x)/g(x) is bounded as x -> infinity. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2