## UnkleRhaukus Group Title find the function whose transforms is given below. $\frac{e^{-ap}}{p^2},\qquad a>0$ one year ago one year ago

1. UnkleRhaukus Group Title

\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. AccessDenied Group Title

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.

3. UnkleRhaukus Group Title

i was trying to apply that but i got stuck

4. AccessDenied Group Title

Hmm.. where did you get stuck?

5. UnkleRhaukus Group Title

Second Shift Theorem

6. AccessDenied Group Title

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... :)

7. AccessDenied Group Title

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

8. UnkleRhaukus Group Title

\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*}??

9. UnkleRhaukus Group Title

im getting muddled up

10. UnkleRhaukus Group Title

\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*} ?

11. AccessDenied Group Title

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

12. UnkleRhaukus Group Title

ok i still dont understand this very well

13. AccessDenied Group Title

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

14. AccessDenied Group Title

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.