What is the laplace transform of u(-t)?

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SUMMARY

The Laplace transform of the unit step function u(-t) is determined to be 0 for t > 0, as the function evaluates to 0 in this range. The integral of u(-t) multiplied by e^(-st) results in 0, confirming that the Laplace transform does not yield a non-zero value. The discussion emphasizes that while functions can be zero for t > 0, their Laplace transforms may still be non-zero in other contexts, such as with the unit impulse function. This analysis is specific to one-sided Laplace transforms, as two-sided transforms differ in behavior.

PREREQUISITES
  • Understanding of Laplace transforms and their definitions
  • Familiarity with the unit step function u(t) and its properties
  • Knowledge of integral calculus, particularly improper integrals
  • Concept of one-sided vs. two-sided Laplace transforms
NEXT STEPS
  • Study the properties of the unit step function u(t) and its transformations
  • Learn about the implications of one-sided versus two-sided Laplace transforms
  • Explore the Laplace transform of the unit impulse function and its significance
  • Investigate the behavior of Laplace transforms for piecewise functions
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Students and professionals in engineering, mathematics, and physics who are studying control systems, signal processing, or differential equations will benefit from this discussion.

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Homework Statement


Find the laplace transform of u(-t)


Homework Equations





The Attempt at a Solution


For u(t), the laplace transform of it is 1/s, basically taking the integral of e^-st from 0 to infinity.

In this case, since the unit step function approaches from the negative side, do I just take the integral of e^(-st), but switch the limit (from infinity to 0) for the purpose of transformation, leaving with -1/s?
 
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What is the value of u(-t) for t>0... what does that say about the integral of u(-t)*[e^(-st)]?
 
learningphysics said:
What is the value of u(-t) for t>0... what does that say about the integral of u(-t)*[e^(-st)]?

u(-t) for t > 0 is equal to 0, does that immediately lead to the conclusion that the transformation of u(-t) = 0? What if t < 0?
 
l46kok said:
u(-t) for t > 0 is equal to 0, does that immediately lead to the conclusion that the transformation of u(-t) = 0? What if t < 0?

Yes, though a function can be 0 for t>0, but still have non-zero laplace transform... like the unit impulse function... but that's because the value at 0 is infinite... the laplace transform is defined from 0- (a value slightly less than 0) to infinity to include what happens at t=0... what happens for t<0 doesn't matter... but what happens at zero does matter.

But for u(-t) the value at t=0 is finite... so the integral is 0.

Hope I'm not missing anything... I think this is all true...

This is all for one-sided laplace transforms... two-sided is different...
 
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