How Do You Solve the Unilateral Laplace Transform for Delayed Functions?

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To solve the unilateral Laplace transform for the given delayed function x(t)=tu(t) - (t-1)u(t-1) - (t-2)u(t-2) + (t-3)u(t-3), recognize that the Laplace transform of delayed functions involves a shift property. The transform of a function f(t-a)u(t-a) can be expressed as L[f(t-a)u(t-a)] = e^{-as}F(s), where F(s) is the Laplace transform of f(t). For the terms (t-1), (t-2), and (t-3), apply this property to account for the delays. The overall transform will combine the results from each component, factoring in the exponential decay due to the delays. Understanding these properties is crucial for accurately computing the Laplace transform of delayed functions.
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Please help me explain how to solve this to find the unilateral laplace transform
x(t)=tu(t) - (t-1)u(t-1) - (t-2)u(t-2) + (t-3)u(t-3)

I know the part tu(t)
as a(t) = u(t) --> 1/s
then b(t) = tu(t) ---> - d/dx a(t) = 1/s^2

But for those (t-1), (t-2) in front u(t), how to solve these?

Please help me, thank you very very much
 
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Look up how a delay affects the Laplace transform. It's usually one of the properties listed for the transform.

Alternatively, you could derive the relationship for yourself from the definition. Start with

L[f(t-a)u(t-a)] = \int_0^\infty f(t-a)u(t-a)e^{-st}\,dt

and use the substitution t'=t-a.
 

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