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UnkleRhaukus

  • 2 years ago

find the function whose transforms is given below. \[\frac{e^{-ap}}{p^2},\qquad a>0\]

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  1. UnkleRhaukus
    • 2 years ago
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    \[ \begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{e^{-ap}\times\frac1{p^2}\right\}\\ &= \end{align*}\]

  2. AccessDenied
    • 2 years ago
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    Are you familiar with this theorem about convolution? \( \displaystyle \mathcal{L}^{-1} \left\{ F(s) G(s) \right\}(t) = \mathcal{L}^{-1} \{F(s)\}(t) * \mathcal{L}^{-1} \{G(s)\}(t) \) \( \displaystyle = (f * g)(t) \) \( \displaystyle = \int_{0}^{t} f(t - \tau) g(\tau) \; \textrm{d} \tau \) I found this in my PDF. If we can find the inverse Laplace transform of our two factors individually, then their convolution should give the inverse Laplace transform of their product.

  3. UnkleRhaukus
    • 2 years ago
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    i was trying to apply that but i got stuck

  4. AccessDenied
    • 2 years ago
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    Hmm.. where did you get stuck?

  5. UnkleRhaukus
    • 2 years ago
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    Second Shift Theorem

  6. AccessDenied
    • 2 years ago
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    Hmm... \( \mathcal{L} \{H(t-a) f(t-a) \}(s) = e^{-as} F(s) \) \( H(t-a) f(t-a) = \mathcal{L}^{-1} \{e^{-as} F(s) \} \) \(\displaystyle F(s) = \frac{1}{s^2} \) in this case. Looking at this, I think we simply need the inverse laplace transform of 1/s^2 to find f(t), and then sub t=t-a. I think that's possible w/o convolution... :)

  7. AccessDenied
    • 2 years ago
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    From a table, I find the inverse laplace transform of 1/s^2 is f(t) = t. f(t-a) = (t-a) So the final answer would be... \( \displaystyle \mathcal{L}^{-1} \{\frac{e^{-as}}{s^2}\}(t) = (t-a) H(t-a) \)

  8. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{\frac{e^{-ap}}p\times\frac1{p}\right\}\\ &=\mathcal L ^{-1}\left\{\frac{e^{-ap}}p\right\}\times\mathcal L^{-1} \left\{\frac1{p}\right\}_{t\rightarrow t-a}\\ &=h(t-a)\times t\Big|_{t\rightarrow t-a}\\ &=h(t-a)\ (t-a)\\ \end{align*}\]??

  9. UnkleRhaukus
    • 2 years ago
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    im getting muddled up

  10. UnkleRhaukus
    • 2 years ago
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    \[\begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{{e^{-ap}}\times\frac1{p^2}\right\}\\ &=h(t-a)\times\mathcal L^{-1} \left\{\frac1{p^2}\right\}_{t\rightarrow t-a}\\ &=h(t-a)\times t\Big|_{t\rightarrow t-a}\\ &=h(t-a)(t-a) \end{align*}\] ?

  11. AccessDenied
    • 2 years ago
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    Yes, that is what I got as well. :)

  12. UnkleRhaukus
    • 2 years ago
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    ok i still dont understand this very well

  13. AccessDenied
    • 2 years ago
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    I apologize for not being very much help with the intuition part here. I am not very familiar with it myself. ;(

  14. AccessDenied
    • 2 years ago
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    My current understanding is that this theorem establishes that a factor of e^(-ap) on the laplace transform of a function translates to a shift of 'a' units to the right in the original function; or in reverse, that a shift of 'a'units translates to a factor of e^(-ap) in the transform.

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