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Hi have to determinate the Laplace Transform of u where u is the solution to this problem: enter image description here

I really have no idea on how to manage this mess.

2 Answers 2

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For any real numbers $a\in\mathbb{R}$, let $H(t-a)$ be the shifted unit step function, i.e. $$ H(t-a) = \begin{cases} \, 1 & \ \ \textrm{ if }t\ge a, \\ \, 0 & \ \ \textrm{ if }t

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    Can you please give me the exact result of L[u](z)? – 2017-02-09
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    I am assuming you know how to do Laplace transform of unit step functions and integrals, but if you don't, let me know so that I can add details. Regardless, you can find the table of Laplace transform here https://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms – 2017-02-09
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    I just added some more details (: – 2017-02-09
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    The solution is not the One given by my teacher, but your answer helped me a lot! – 2017-02-09
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    Can you tell me what is the given solution ? – 2017-02-10
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    It should be one of the follows: http://imgur.com/a/Z2Xvh Assuming s=1 – 2017-02-10
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    Thanks! Unless I misunderstood the function $\chi_[1,2](t)$ (which is an indicator function?), I think my solution is correct. – 2017-02-10
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    The correct One is the First. Anyway the $\chi$ function is 1 for x in 1..2 and 0 otherwise – 2017-02-10
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Well you have an equation in terms of $u$. Slap on an $e^{-st}$ and integrate. We have $$\int_0^\infty u(t)e^{-st}dt - \int_0^\infty e^{-st}\int_t^\infty u(t')dt'dt = \int_0^\infty e^{-st}\chi_{[1,2]}(t)dt.$$ The RHS is $$\int_0^\infty e^{-st}\chi_{[1,2]}(t)dt = \int_1^2e^{-st}dt = \frac{1}{s}(e^{-s}-e^{-2s}) = C(s)/s.$$

The first term on the LHS is the Laplace transform $\mathcal L_u(s).$ The second term can be integrated by parts $$ \int_0^\infty e^{-st}F(t)dt = -\left.F(t)\frac{1}{s}e^{-st}\right|_0^\infty + \int_0^\infty\frac{1}{s}e^{-st}F'(t)dt$$ where $F(t) = \int_t^\infty u(t')dt'.$ We have $$F(0) = \int_0^\infty u(t')dt' = 1$$ by definition, $ \lim_{t\to \infty} F(t)=0$ by the convergence of the integral,and $F'(t) = -u(t)$ by the FTC so we get $$ \int_0^\infty e^{-st}F(t)dt = \frac{1}{s} - \int_0^\infty \frac{1}{s}e^{-st}u(t)dt = \frac{1}{s}-\frac{1}{s}\mathcal L_u(s).$$

Plugging this all into the initial equation, $$ (1+\frac{1}{s})\mathcal L_u(s) - \frac{1}{s} = C(s)/s$$ so we get $$ \mathcal L_u(s) = \frac{1}{1+s}(C(s)+1)$$ where $C(s) = e^{-s}-e^{-2s}.$

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    You should have $e^{-s} - e^{-2s}$ on your $C$. – 2017-02-09
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    @CheeHan Thanks! Wasn't my only error. – 2017-02-09