How Do You Derive Laplace Transforms for Delayed Functions?

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 />

?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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