Laplace Transforms, Region of Convergence

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SUMMARY

The discussion focuses on the concept of the Region of Convergence (ROC) in relation to the Laplace Transform. The ROC is defined as the set of values of \( s \) for which the improper integral \( \int_0^\infty f(t)e^{-st} dt \) converges. It is established that the ROC typically forms a half-plane in the complex plane, determined by the behavior of the function \( f(t) \). For example, for the function \( f(t) = e^{5t} \), the ROC is defined by the condition on the real part of \( s \) that ensures convergence.

PREREQUISITES
  • Understanding of Laplace Transforms
  • Knowledge of improper integrals
  • Familiarity with complex variables
  • Concept of exponential functions and their growth rates
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  • Study the properties of Laplace Transforms in detail
  • Learn about improper integrals and their convergence criteria
  • Explore complex analysis, focusing on half-planes and convergence
  • Investigate specific examples of functions and their Regions of Convergence
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Students and professionals in engineering, mathematics, and physics who are working with Laplace Transforms and need a clear understanding of the Region of Convergence for various functions.

tanky322
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Can anyone explain the region of convergence to me in english? I understand the Laplace transform and can do it with my eyes closed, but I can't figure out how to figure out the ROC. I've scoured the internet, and every definition is vague or just incomprehensible by me.


Thanks!
 
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Do you know anything about complex variables? The region of convergence is just the values of t where
\int_0^\infty f(t)e^{-s t} dt
converges as an improper integral.
That can be difficult to find in general, but in many elementary applications only very well behaved f are considered for example the functions of exponential order.
 
I'm somewhat familiar with complex variables, although not too much. I guess what I am not really sure of, is what exactly converges? The function and e^-st?


Thanks for your reply!
 
What exactly converges? The improper integral. That is, the limit
\lim_{M\to+\infty}\int_0^M f(t)e^{-s t} dt
exists. Generally, the region of convergence is a half-plane: all s to the right of some vertical line in the complex plane.
 
As an example to what g_edgar wrote, consider
f(t) = e^{5 t},
for
t\geq0.

Now calculuate

<br /> \lim_{M\to+\infty}\int_0^M f(t)e^{-s t} dt = \lim_{M\to+\infty}\int_0^M e^{-(s-5) t}.<br />

You should find that the limit only converges if s satisfies some condition. That condition defines the region of convergence. Note that s is complex in general, and the constraint will be on the real part.

Jason
 

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