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Series expansion

  1. Nov 17, 2009 #1
    I'm trying to work something on inverse Laplace transform. I need to express a transfer function F(s) to the form
    [tex]F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }[/tex]

    I can easily do it for rational function e.g.
    [tex]\frac{s^3+2s^2+3s+1}{s+4}= \frac{s^{-1} (1+2s^{-1}+3s^{-2}+s^{-3})}{s^{-3}+4s^{-4}}[/tex]

    for some indicial e.g
    [tex]\frac{1}{\sqrt{s^2+s}}=\frac{1}{s}(1+s^{-1})^{-1/2}[/tex]
    Expand using the binomial theorem.

    My problem is how to express irrational functions such as s-3/2 or
    [tex]\frac{e^{-\sqrt{s}}}{s}[/tex]
    to the form of F(s).
     
  2. jcsd
  3. Nov 18, 2009 #2
    [itex]F(s)[/itex] is regular at infinity (in the sense of complex variables) but neither [itex]s^{-3/2}[/itex] nor [itex]e^{-\sqrt{s}}/s[/itex] is.
     
  4. Nov 19, 2009 #3
    That is an interesting observation that you make. If you could please explain more what regular at infinity means. I'm do not know much about complex variables theory. That's why I want to avoid dealing with Bromwich integral.

    My idea is to repeatedly use
    [tex]L^{-1} { \frac{F(s)}{s} }=\int_0^t f(t) dt[/tex]
    for evaluating inverse transform.

    From Laplace transform table I know that [itex]s^{-3/2}[/itex] and [itex]e^{-2\sqrt{s}}/s[/itex] can be inverted. That's why I expect the functions could be express as F(s)
     
  5. Nov 19, 2009 #4
    OK, the function

    [tex]F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }[/tex]

    has the property that [itex]s^kF(s)[/itex] converges to a finite nonzero limit as [itex]s \to \infty[/itex] for some integer [itex] k [/itex] . Neither of the two functions in question has that property.
     
  6. Nov 19, 2009 #5
    There is no standart way for those transforms. You have to proceed from the integral form:
    [tex]\frac{1}{2\pi}PV \int_{-\infty}^{\infty}{exp(st)s^{-3/2}ds}=\frac{1}{2\pi}2\sqrt{t}\int_{0}^{\infty}{exp(z)z^{-1/2-1}dz}=2\sqrt{t}\frac{1}{2\pi}\Gamma[-1/2]=-2\sqrt{\frac{t}{\pi}} [/tex]
    (for t>0)
     
    Last edited: Nov 19, 2009
  7. Nov 20, 2009 #6
    Actually I'm trying to solve pde not inverse Laplace transform per se. You know after we transform the pde wrt time and solve the transformed ode in s-domain, we usually obtain quite complicated F(s). It is this F(s) that I want to invert numerically.

    But from what have you have shown, may be I can learnt something from it. Just need further clarification. Why is your inverse transform for s-3/2 has a negative sign. The entry http://www.vibrationdata.com/Laplace.htm" [Broken] has no such sign. And why is your integral form different from entry 1.2 of the table. Where has the integration end limit [itex]c+i\infty[/itex] gone?
     
    Last edited by a moderator: May 4, 2017
  8. Nov 20, 2009 #7
    So the converse statement is not true for those cases. Thanks g_edgar. I have to sort out this first. May be I will change the form to

    [tex]F(s)=\frac{s^{-1} (a_0 + a_1s^{-0.5} + a_2s^{-1}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }[/tex]

    I will try to figure out how to handle this case and come back again if I have a problem.
     
  9. Nov 26, 2009 #8
    you are right abount the factor in the limits of the integral (missed another i in the denominator ). The difference in sign is probably because i chose the wrong contour
     
  10. Nov 26, 2009 #9
    Apparently it is easier to prove the converse, I am not sure about the inversion formula. Now,
    [tex]\int_{0}^{\infty}t^{\nu}e^{-st}dt=\frac{1}{\nu+1}\int_{0}^{\infty}z^{\nu}e^{-z}dz=\frac{\Gamma(\nu+1)}{s^{\nu+1}}[/tex]
    The correct form of the integral I posted previously is then an alternative definition for [tex]\Gamma[/tex]:
    [tex]\frac{1}{\Gamma(\nu+1)}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{s}}{s^{\nu+1}}[/tex]
    Which I didn't kow
     
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