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anonymous

  • one year ago

what is the laplace transform of t^2u(t-1)?

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  1. DARTHVADER2900
    • one year ago
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    http://openstudy.com/study#/updates/5170a3c3e4b0aca440dfb7fb

  2. anonymous
    • one year ago
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    $$\int_0^\infty e^{-st}t^2u(t-1)\, dt=\int_0^1e^{-st}t^2\cdot 0\, dt+\int_1^\infty e^{-st}t^2\cdot 1\, dt$$

  3. anonymous
    • one year ago
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    we can split the integral into one on \((0,1)\) and another on \((1,\infty)\) because we have that the step function \(u(t-1)=0\) for \(t-1<0\Leftrightarrow t<1\) and \(u(t-1)=1\) for \(t-1>0\Leftrightarrow t>1\)

  4. anonymous
    • one year ago
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    the first integral is identically zero, while the second we evaluate using integration by parts: $$\begin{align*}\int_1^\infty e^{-st} t^2\, dt&=-\frac1s t^2e^{-st}\bigg|_1^\infty+\frac2s\int_1^\infty te^{-st}\, dt\\&=\left[-\frac1s t^2e^{-st}-\frac2{s^2}te^{-st}\right]_1^\infty +\frac2{s^2}\int_1^\infty e^{-st}\, dt\\&=\frac1se^{-st}+\frac2{s^2}e^{-st}+\frac2{s^2}\left[-\frac1se^{-st}\right]_1^\infty\\&=\left(\frac1s+\frac2{s^2}+\frac2{s^3}\right)e^{-s}\end{align*}$$

  5. anonymous
    • one year ago
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    which is the same thing you get after shifting \(t\to t+1\) so that \(t^2\to(t+1)^2=t^2+2t+1\) and pulling out the factor of \(e^{-s}\) from \(e^{-st}\to e^{-s(t+1)}=e^{-s}e^{-st}\), so: $$\mathcal{L}\{t^2 u(t-1)\}=e^{-s}\mathcal{L}\{t^2+2t+1\}=\left(\frac1s+\frac2{s^2}+\frac2{s^3}\right)e^{-s}$$ considering that $$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

  6. IrishBoy123
    • one year ago
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    \(F(p) =\mathcal {L} \{ t^2 \; \; u(t-1)\} \) [\(\bar t = t -1\)] \( F(p) = \mathcal {L} \; \left\{(\bar t+1)^2 \; u(\bar t) \right\}\) \(= \mathcal {L} \; \left\{ ( \bar t^2 + 2 \bar t + 1 \; )u(\bar t) \right\}\) \(= \left( \dfrac{2}{p^3} + \dfrac{2}{p^2} + \dfrac{1}{p} \right) e^{-p}\) |dw:1443657391199:dw|

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