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

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[ \begin{align*} f(t)&=\mathcal L ^{1}\left\{\frac{e^{ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{1}\left\{e^{ap}\times\frac1{p^2}\right\}\\ &= \end{align*}\]
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i was trying to apply that but i got stuck
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Hmm.. where did you get stuck?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
Second Shift Theorem
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Hmm... \( \mathcal{L} \{H(ta) f(ta) \}(s) = e^{as} F(s) \) \( H(ta) f(ta) = \mathcal{L}^{1} \{e^{as} F(s) \} \) \(\displaystyle F(s) = \frac{1}{s^2} \) 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=ta. I think that's possible w/o convolution... :)
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
From a table, I find the inverse laplace transform of 1/s^2 is f(t) = t. f(ta) = (ta) So the final answer would be... \( \displaystyle \mathcal{L}^{1} \{\frac{e^{as}}{s^2}\}(t) = (ta) H(ta) \)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\begin{align*} f(t)&=\mathcal L ^{1}\left\{\frac{e^{ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{1}\left\{\frac{e^{ap}}p\times\frac1{p}\right\}\\ &=\mathcal L ^{1}\left\{\frac{e^{ap}}p\right\}\times\mathcal L^{1} \left\{\frac1{p}\right\}_{t\rightarrow ta}\\ &=h(ta)\times t\Big_{t\rightarrow ta}\\ &=h(ta)\ (ta)\\ \end{align*}\]??
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
im getting muddled up
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\begin{align*} f(t)&=\mathcal L ^{1}\left\{\frac{e^{ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{1}\left\{{e^{ap}}\times\frac1{p^2}\right\}\\ &=h(ta)\times\mathcal L^{1} \left\{\frac1{p^2}\right\}_{t\rightarrow ta}\\ &=h(ta)\times t\Big_{t\rightarrow ta}\\ &=h(ta)(ta) \end{align*}\] ?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Yes, that is what I got as well. :)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
ok i still dont understand this very well
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
I apologize for not being very much help with the intuition part here. I am not very familiar with it myself. ;(
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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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