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

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Homework Help Overview

The discussion revolves around finding the Laplace transform of the unit step function u(-t), particularly focusing on its behavior for different values of t. Participants explore the implications of the function's definition and its transformation properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the integral of e^(-st) in relation to the Laplace transform and question how the limits of integration should be adjusted for u(-t). There is also inquiry into the value of u(-t) for t > 0 and its implications for the transform.

Discussion Status

The discussion is active, with participants raising questions about the behavior of u(-t) and its implications for the Laplace transform. Some guidance is offered regarding the significance of the value at t = 0 and how it affects the integral, but no consensus has been reached.

Contextual Notes

Participants note that this discussion pertains to one-sided Laplace transforms, indicating a specific context for the analysis of u(-t).

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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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