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

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

AccessDenied Group TitleBest 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.
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i was trying to apply that but i got stuck
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
Hmm.. where did you get stuck?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
Second Shift Theorem
 2 years ago

AccessDenied Group TitleBest 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... :)
 2 years ago

AccessDenied Group TitleBest 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... \( \displaystyle \mathcal{L}^{1} \{\frac{e^{as}}{s^2}\}(t) = (ta) H(ta) \)
 2 years ago

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
im getting muddled up
 2 years ago

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
Yes, that is what I got as well. :)
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
ok i still dont understand this very well
 2 years ago

AccessDenied Group TitleBest 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. ;(
 2 years ago

AccessDenied Group TitleBest 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.
 2 years ago
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