Question

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

Answer 1

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

Answer 2

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.

Answer 3

i was trying to apply that but i got stuck

Answer 4

Hmm.. where did you get stuck?

Answer 5

Second Shift Theorem

Answer 6

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... :)

Answer 7

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...
\( \displaystyle \mathcal{L}^{-1} \{\frac{e^{-as}}{s^2}\}(t) = (t-a) H(t-a) \)

Answer 8

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

Answer 9

im getting muddled up

Answer 10

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

Answer 11

Yes, that is what I got as well. :)

Answer 12

ok i still dont understand this very well

Answer 13

I apologize for not being very much help with the intuition part here. I am not very familiar with it myself. ;(

Answer 14

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.