How does the inverse Laplace transform work in signal processing?

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Discussion Overview

The discussion centers on the inverse Laplace transform in the context of signal processing, specifically examining the mathematical formulation and its implications. Participants explore the relationship between integration over a complex variable and summation, as well as the conditions under which these expressions hold true.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how a complex integral can be expressed as a sum, highlighting confusion over the definition of \(\Delta s\) in the context of the integral.
  • Another participant suggests that the definition resembles a Riemann integral and proposes that \(\Delta s\) might be related to \(j\Delta y\), indicating a need for clarity on the nature of \(F\).
  • Concerns are raised about the independence of \(c\) in the expression, which some participants find dubious without further information about \(F\), such as whether it is an entire function or its behavior away from the y-axis in the complex plane.
  • A humorous remark is made questioning whether the original poster can return the book, suggesting dissatisfaction with the material presented.

Areas of Agreement / Disagreement

Participants express varying levels of confusion and skepticism regarding the formulation of the inverse Laplace transform, with no consensus reached on the validity of the expression or the interpretation of \(\Delta s\).

Contextual Notes

Participants note the need for additional assumptions about the function \(F\) to fully understand the implications of the integral and its relationship to the sum.

Swapnil
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My book on signal processing says that:

[tex]f(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} F(s) e^{st} ds = \lim_{\Delta s \to 0} \sum_{n = -\infty}^{\infty} \Big[ \frac{F(n\Delta s)\Delta s}{2\pi j} \Big] e^{n\Delta s t}[/tex]

I don't get this. How/Why can you write a integration over a complex variable as the above sum?

edit: I forgot a coefficient [tex]\frac{1}{2\pi j}[/tex] on the LHS. Sorry about that. Its fixed now.
 
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Isn't that definition of a Reimann integral?
 
I understand how you could be confused! It would have to be explained what \Delta s is in the sum since the integral seems to have s lie on a vertical line c + jy in the complex plane (real y). So do we perhaps have \Delta s =
j\Delta y?Supposing the above and not knowing anything else about F, using a Riemann sum, you should have limit as \Delta s -> 0 of sum over n of the following terms:
F(c+n\Delta s)\exp((c+n\Delta s)t)\Delta s
which is the limit as \Delta y-> 0 of the sum of terms

F(c+n j \Delta y)\exp((c+ n j \Delta y)t)j\Delta y

The 2\pi j in the denominator seems something like Cauchy's integral thm., but we need to know quite a bit more about F to get that. (Is F an entire fuction? Does it vanish very rapidly away from the y-axis in the complex plane?)

Generally speaking, the independence of c on the RHS makes the formula dubious.

Can you still get the money back for the book?
 
gammamcc said:
I understand how you could be confused! It would have to be explained what [tex]\Delta s[/tex] is in the sum since the integral seems to have [itex]s[/itex] lie on a vertical line [tex]c + jy[/tex] in the complex plane (real y). So do we perhaps have [tex]\Delta s = <br /> j\Delta y[/tex]?Supposing the above and not knowing anything else about [tex]F[/tex], using a Riemann sum, you should have limit as [tex]\Delta s -> 0[/tex] of sum over [itex]n[/itex] of the following terms:
[tex]F(c+n\Delta s)\exp((c+n\Delta s)t)\Delta s[/tex]
which is the limit as [tex]\Delta y-> 0[/tex] of the sum of terms

[tex]F(c+n j \Delta y)\exp((c+ n j \Delta y)t)j\Delta y[/tex]

The [tex]2\pi j[/tex] in the denominator seems something like Cauchy's integral thm., but we need to know quite a bit more about [tex]F[/tex] to get that. (Is [tex]F[/tex] an entire function? Does it vanish very rapidly away from the y-axis in the complex plane?)

Generally speaking, the independence of [itex]c[/itex] on the RHS makes the formula dubious.

Can you still get the money back for the book?

Edited gammamcc's post to look nice...
 

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