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UnkleRhaukus

  • 3 years ago

Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral \[\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\}\]

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  1. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\\ &=\\ &=\end{align*}\]

  2. wio
    • 3 years ago
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    I'm guessing that this involve integration by parts...

  3. UnkleRhaukus
    • 3 years ago
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    do i have to reverse the order of integration ?

  4. wio
    • 3 years ago
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    Not quite sure though.

  5. UnkleRhaukus
    • 3 years ago
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    @Zarkon

  6. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^ux(u)e^{-pt}\cdot\text dt\cdot\text du\\ ?&=\int\limits_0^\infty x(u)\int\limits_0^ue^{-pt}\cdot\text dt\cdot\text du\\ \end{align*}\]

  7. UnkleRhaukus
    • 3 years ago
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    @mahmit2012

  8. wio
    • 3 years ago
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    Somehow u become a t

  9. UnkleRhaukus
    • 3 years ago
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    |dw:1353224604904:dw|

  10. UnkleRhaukus
    • 3 years ago
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    have to reverse the order of integration correctly ?

  11. wio
    • 3 years ago
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    is u still constant with respect to t then?

  12. UnkleRhaukus
    • 3 years ago
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    i think so

  13. wio
    • 3 years ago
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    Well, if we say that \(X(t)\) is the antiderivative of \(x(t)\)....then \[ X(t) = \int_0^tx(u)\cdot du \]and \(X'(t)=x(t)\) So the antiderivative is going to be dependent on \(t\) and yet the derivative is constant...?

  14. wio
    • 3 years ago
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    Is there anyway to verify the answer you're getting at?

  15. UnkleRhaukus
    • 3 years ago
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    wolfram tells me the answer is \[\frac{\mathcal L\{x(t)\}}{p}=\frac1p\int\limits_0^\infty x(t)e^{-pt}\text dt\]

  16. wio
    • 3 years ago
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    Okay nevermind what I said before, it looks like what you are doing is correct.

  17. wio
    • 3 years ago
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    What are you getting for that inner integral?

  18. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}&=\int\limits_0^\infty x(u)\int\limits_0^ue^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{e^{-pt}}{-p}\Big|_0^u\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{1-e^{-pu}}p\cdot\text du\\\end{align*}\]

  19. UnkleRhaukus
    • 3 years ago
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    i think i have made a mistake but im not sure where

  20. mahmit2012
    • 3 years ago
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    |dw:1353228810868:dw|

  21. mahmit2012
    • 3 years ago
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    |dw:1353228991552:dw|

  22. UnkleRhaukus
    • 3 years ago
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    H(t-u) is the Heaviside unit step function?

  23. mahmit2012
    • 3 years ago
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    yes

  24. mahmit2012
    • 3 years ago
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    |dw:1353229734181:dw|

  25. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\int\limits_0^\infty H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\mathcal L \Big\{H(t-u)\Big\}\cdot\text du\\ &=\int\limits_0^\infty x(u)\frac{e^{-up}}{p}\cdot\text du\\ &=\frac1p\int\limits_0^\infty x(u)e^{-up}\cdot\text du\\ &=\frac{X(p)}p\color{red}\checkmark \\ \end{align*}\] Thankyou mahmit2012!

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