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UnkleRhaukus Group Title

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

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

    • one year ago
  2. AccessDenied Group Title
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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.

    • one year ago
  3. UnkleRhaukus Group Title
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    i was trying to apply that but i got stuck

    • one year ago
  4. AccessDenied Group Title
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    Hmm.. where did you get stuck?

    • one year ago
  5. UnkleRhaukus Group Title
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    Second Shift Theorem

    • one year ago
  6. AccessDenied Group Title
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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... :)

    • one year ago
  7. AccessDenied Group Title
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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... \( \displaystyle \mathcal{L}^{-1} \{\frac{e^{-as}}{s^2}\}(t) = (t-a) H(t-a) \)

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

    • one year ago
  9. UnkleRhaukus Group Title
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    im getting muddled up

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

    • one year ago
  11. AccessDenied Group Title
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    Yes, that is what I got as well. :)

    • one year ago
  12. UnkleRhaukus Group Title
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    ok i still dont understand this very well

    • one year ago
  13. AccessDenied Group Title
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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. ;(

    • one year ago
  14. AccessDenied Group Title
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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.

    • one year ago
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