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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.

- Thread starter eljose79
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- #1

- 214

- 1

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

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^{-st}f(t)dt

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

L{f(t)}=∫_{0}^{T}e^{-st}f(t)dt+∫_{T}^{2T}e^{-st}f(t)dt+∫_{2T}^{3T}e^{-st}f(t)dt+...

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

L{f(t)}=∫_{0}^{T}e^{-st}f(t)dt+∫_{0}^{T}e^{-s(t+T)}f(t+T)dt+∫_{0}^{T}e^{-s(t+2T)}f(t+2T)dt

+∫_{0}^{T}e^{-s(t+3T)}f(t+3T)dt+...

Noting that f(t)=f(t+T)=f(t+2T)=f(t+3T)=..., and factoring ∫_{0}^{T}e^{-st}f(t)dt out of each factor yields:

L{f(t)}=(1+e^{-sT}+e^{-2sT}+e^{-3sT}+...)∫_{0}^{t}e^{-st}f(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}∫_{0}^{T}f(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

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)}=∫

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

L{f(t)}=∫

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

L{f(t)}=∫

+∫

Noting that f(t)=f(t+T)=f(t+2T)=f(t+3T)=..., and factoring ∫

L{f(t)}=(1+e

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

(1-e

So, I have finally:

L{f(t)}=(1-e

Now, we get

Does that help?

edit: fixed a variety of bracket errors

Last edited:

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