Laplace of step function

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The Laplace Transform of step functions (Sect. 6.3).

IVP’s with Step Functions – This is the section where the reason for using Laplace transforms really becomes apparent. We will use Laplace transforms to solve IVP’s that


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unit step function Archives

Solution: Using step function notation, f (t) = u(t − 1)(t2 − 2t +2). Completing the square we obtain, t2 − 2t +2 = (t2 − 2t +1) − 1+2 = (t − 1)2 +1. This is a parabola t2 translated to the right by 1 and
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Differential Equations

L [ u ( t − 1)] = e − s s , L [ u ( t − 4)] = e − 4 s s. With each of these Laplace functions found, I then put it all together back in the original equation, and simplified. 7 s − 1 s 2 ∗ e − s s

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Laplace Transform of Functions

2. The unit step function can be shifted and then used to model the switching on and off of another function. 3. The function U (t−a)−U (t−b) is equal to 1 on [a,b) and equal to zero outside