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Evaluating the function

  1. Sep 24, 2003 #1
    Let be the function

    f(t)=L(-1)1/(exp(-s)-1) where L(-1) means the laplace inverse transform..my doubt is to know what is the value of

    f(0),f(1),f(2)...f(n) being n an integer.
     
  2. jcsd
  3. Sep 24, 2003 #2

    Tom Mattson

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    I'll take a stab at this.

    First, we need to develop the Laplace transform of a periodic function:

    f(t)=f(t+nT) for all integers n.

    Start with the definition of L{f(t)}:

    L{f(t)}=∫0∞e-stf(t)dt

    Let's break this up over intervals of width T:

    L{f(t)}=∫0Te-stf(t)dt+∫T2Te-stf(t)dt+∫2T3Te-stf(t)dt+...

    Now perform substitutions on each integral such that the limits of each integral are [0,T]:

    L{f(t)}=∫0Te-stf(t)dt+∫0Te-s(t+T)f(t+T)dt+∫0Te-s(t+2T)f(t+2T)dt
    +∫0Te-s(t+3T)f(t+3T)dt+...

    Noting that f(t)=f(t+T)=f(t+2T)=f(t+3T)=..., and factoring ∫0Te-stf(t)dt out of each factor yields:

    L{f(t)}=(1+e-sT+e-2sT+e-3sT+...)∫0te-stf(t)dt

    The first factor on the right is a geometric series whose sum is:

    (1-e-sT)-1

    So, I have finally:

    L{f(t)}=(1-e-sT)-1∫0Tf(t)dt

    Now, we get your function if we let f(t)=d(t+n+1/2). That is, a periodic delta function whose period is 1. I used the half integer n+1/2 so that there is only one delta function in each interval. I could just as easily have chosen n+1/3, n+1/4, or whatever. So, it seems that the function is not unique.

    Does that help?

    edit: fixed a variety of bracket errors
     
    Last edited: Sep 24, 2003
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