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

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

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Are 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.

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

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

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

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm... \( \mathcal{L} \{H(ta) f(ta) \}(s) = e^{as} F(s) \) \( H(ta) f(ta) = \mathcal{L}^{1} \{e^{as} F(s) \} \) \(\displaystyle F(s) = \frac{1}{s^2} \) 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... :)

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1From 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... \( \displaystyle \mathcal{L}^{1} \{\frac{e^{as}}{s^2}\}(t) = (ta) H(ta) \)

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

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

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

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

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

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

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1My 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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