How does the inverse Laplace transform work in signal processing?

In summary, the book on signal processing says that:- f(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} F(s) e^{st} ds- This limit as \Delta s -> 0 of the sum of terms is the limit as \Delta y-> 0 of the sum of terms- 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 function? Does it vanish very rapidly away from the y-axis in the complex plane?)
  • #1
Swapnil
459
6
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.
 
Last edited:
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  • #2
Isn't that definition of a Reimann integral?
 
  • #3
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?
 
  • #4
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 =
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...
 

Related to How does the inverse Laplace transform work in signal processing?

What is the Inverse Laplace Transform?

The Inverse Laplace Transform is a mathematical operation that takes a function expressed in the Laplace domain and converts it back to the time domain.

Why is the Inverse Laplace Transform important?

The Inverse Laplace Transform is important because it allows us to analyze and solve problems in the time domain that are difficult or impossible to solve in the Laplace domain. It is also a key tool in many areas of engineering and science, including control systems, signal processing, and differential equations.

What is the process for finding the Inverse Laplace Transform?

The process for finding the Inverse Laplace Transform involves using a table of known Laplace transforms, algebraic manipulations, and partial fraction decomposition to convert the function from the Laplace domain to the time domain.

Are there any limitations to the Inverse Laplace Transform?

Yes, there are limitations to the Inverse Laplace Transform. The function in the Laplace domain must have a Laplace transform that exists and is unique. It also cannot have any singularities on the imaginary axis.

What are some common applications of the Inverse Laplace Transform?

The Inverse Laplace Transform has many applications in engineering and science, including circuit analysis, signal processing, control systems, and solving differential equations. It is also used in the field of probability and statistics to solve problems involving probability distributions.

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