Computing the Inverse Laplace Transform

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SUMMARY

The discussion focuses on computing the inverse Laplace transform, specifically using the formula \(\ell^{-1} \left\{\frac{F(s)}{s}\right\} = \int^{t}_{0} f(\tau) d\tau\). The user seeks clarity on applying this definition to the function \(F(s) = \frac{1 - e^{-s}}{1 + e^{-s}}\). The solution involves manipulating \(F(s)\) and utilizing a table lookup for inversion, followed by integrating the resulting function to find the inverse transform.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with integration techniques
  • Knowledge of function manipulation and simplification
  • Experience with mathematical tables for transform pairs
NEXT STEPS
  • Study the properties of Laplace transforms in detail
  • Learn techniques for manipulating functions in the context of transforms
  • Explore integration methods relevant to inverse Laplace transforms
  • Review standard Laplace transform tables for common functions
USEFUL FOR

Students in engineering or mathematics, particularly those studying control systems or differential equations, will benefit from this discussion on inverse Laplace transforms.

audifanatic51
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Hi, I recently posted another question about a Laplace transform and now have a question about taking the inverse Laplace transform. Again, my professor did not cover this topic as well as I could have hoped for, and so I am stuck on this problem as well. Again, I understand the idea of Laplace transforms (and their inverses) very well, however actually computing them seems to be another issue. Anyway, here is the problem.

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I saw a definition in the book that seemed like it could be useful, however, when I tried to implement it, things seemed to get ugly. Here's the definition I found:

[itex]\ell^{-1}[/itex] [itex]\left\{\stackrel{F(s)}{s}\right\}[/itex] = [itex]\int^{t}_{0}[/itex] f([itex]\tau[/itex]) d[itex]\tau[/itex]

Please note that due to my typing incompetency, I cannot figure out how to make fractions with the forum tools. That should be F(s)/s in the inverse Laplace transform brackets above.

Any help or step in the right direction would be appreciated. thanks
 
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Since you have that formula, you observe in your problem that F(s)=(1-exp(-s))/(1+exp(-s)), which can be inverted by a little manipulation (and table look up). Then F(s)/s is inverted by integrating f(tau).
 

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