Laplace Transform: Time Scaling Property

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SUMMARY

The Laplace Transform's time scaling property states that if the Laplace transform of x(t) is X(s) with a region of convergence (ROC) R, then the transform of x(at) is (1/|a|)X(s/a) with ROC R/a, as per Oppenheim's "Signal and Systems" (2nd edition). However, there is a debate regarding the ROC, with some arguing it should be aR instead of R/a, particularly in the case of the example x(t)=e-|t|. The consensus leans towards aR being the correct interpretation.

PREREQUISITES
  • Understanding of Laplace Transform and its properties
  • Familiarity with the concept of Region of Convergence (ROC)
  • Knowledge of the function x(t)=e-|t| and its implications
  • Basic principles of signal processing as outlined in "Signal and Systems" by Oppenheim
NEXT STEPS
  • Study the implications of time scaling in Laplace Transforms
  • Examine the differences between ROC interpretations in various examples
  • Learn about the effects of different values of 'a' on the ROC
  • Review additional examples in "Signal and Systems" by Oppenheim for deeper insights
USEFUL FOR

Students and professionals in electrical engineering, particularly those focusing on signal processing and control systems, will benefit from this discussion on the Laplace Transform's time scaling property.

asmani
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Hi all

According to the textbook Signal and Systems by Oppenheim (2nd edition) pages 685 and 686, if the Laplace transform of x(t) is X(s) with ROC (region of convergence) R, then the Laplace transform of x(at) is (1/|a|)X(s/a) with ROC R/a.

Consequently, for a>1, there is a compression in the size of the ROC of X(s) by a factor 1/a.

But I think the ROC must be aR and not R/a, as the example x(t)=e-|t| shows. Which is correct?

Thanks
 
Engineering news on Phys.org
aR is correct. . .
 

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