How Do You Derive Laplace Transforms for Delayed Functions?

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Homework Help Overview

The discussion revolves around deriving the Laplace transform of delayed functions, specifically focusing on the unit step function and its implications in the context of Laplace transforms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the notation used in the problem, particularly the role of T and the unit step function u. Questions are raised about whether T represents the Laplace transform of t and the definition of u as the unit step function.

Discussion Status

Some participants have provided clarifications regarding the definitions of u and T, emphasizing that u is indeed the step function and that T is a parameter affecting the transform. There is an ongoing exploration of the integral that defines the Laplace transform, with some guidance offered on evaluating specific integrals.

Contextual Notes

Participants are navigating the definitions and implications of parameters in the context of the Laplace transform, with an emphasis on understanding the integration process and the specific characteristics of the functions involved.

discombobulated
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Homework Statement



Derive the Laplace transform of the following functions, using first principles

3d) u(t - T) \} = 0, \ t<T \ (= 1, t>T)

3e) f(t) = e^{-a(t-T)}u(t-T)

Homework Equations



see above

The Attempt at a Solution



I know I need to derive the transform using by integration using this:

L(f) = \int^\infty_{0} \mbox{f(t) e^{-st}} \ dt

but I don't understand the notation, is T the laplace transform of t ? Is u the unit step function? so u(s) = 1/s ?
If someone could explain this to me please?
 
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u is a function and T is a value for t, t is the variable.
 
discombobulated said:
but I don't understand the notation, is T the laplace transform of t ? Is u the unit step function? so u(s) = 1/s ?
If someone could explain this to me please?


Yes u is the stepfunction. You must use two useful identities in laplace transform theory.
 
discombobulated said:

Homework Statement



Derive the Laplace transform of the following functions, using first principles

3d) u(t - T) \} = 0, \ t<T \ (= 1, t>T)

3e) f(t) = e^{-a(t-T)}u(t-T)

Homework Equations



see above

The Attempt at a Solution



I know I need to derive the transform using by integration using this:

L(f) = \int^\infty_{0} \mbox{f(t) e^{-st}} \ dt

but I don't understand the notation, is T the laplace transform of t ? Is u the unit step function?
u is *defined* in the problem. yes, it is a step function, but the step does not occur at zero. tell us: where does the step occur?
so u(s) = 1/s ?
no. u depends on T as a parameter, so too will the transform depend on T as a parameter. if T happened to be zero then you *would* be correct, but T is not (necessarily) zero.
If someone could explain this to me please?

your teacher or professor or whoever obviously wants you to evaluate the integral that defines the transform. The first one should be very easy if you know how to integrate an exponential function by itself. I.e., do you know the value of this integral
<br /> \int_a^b e^x<br />

?
 

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