Here's the question you clicked on:
UnkleRhaukus
find the function whose transforms is given below. \[\frac{e^{-ap}}{p^2},\qquad a>0\]
\[ \begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{e^{-ap}\times\frac1{p^2}\right\}\\ &= \end{align*}\]
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.
i was trying to apply that but i got stuck
Hmm.. where did you get stuck?
Second Shift Theorem
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... :)
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... \( \displaystyle \mathcal{L}^{-1} \{\frac{e^{-as}}{s^2}\}(t) = (t-a) H(t-a) \)
\[\begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{\frac{e^{-ap}}p\times\frac1{p}\right\}\\ &=\mathcal L ^{-1}\left\{\frac{e^{-ap}}p\right\}\times\mathcal L^{-1} \left\{\frac1{p}\right\}_{t\rightarrow t-a}\\ &=h(t-a)\times t\Big|_{t\rightarrow t-a}\\ &=h(t-a)\ (t-a)\\ \end{align*}\]??
im getting muddled up
\[\begin{align*} f(t)&=\mathcal L ^{-1}\left\{\frac{e^{-ap}}{p^2}\right\},\quad a>0\\ &=\mathcal L ^{-1}\left\{{e^{-ap}}\times\frac1{p^2}\right\}\\ &=h(t-a)\times\mathcal L^{-1} \left\{\frac1{p^2}\right\}_{t\rightarrow t-a}\\ &=h(t-a)\times t\Big|_{t\rightarrow t-a}\\ &=h(t-a)(t-a) \end{align*}\] ?
Yes, that is what I got as well. :)
ok i still dont understand this very well
I apologize for not being very much help with the intuition part here. I am not very familiar with it myself. ;(
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.