## UnkleRhaukus 3 years ago find the function whose transforms is given below. $\frac{e^{-ap}}{p^2},\qquad a>0$

1. UnkleRhaukus

\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

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

i was trying to apply that but i got stuck

4. AccessDenied

Hmm.. where did you get stuck?

5. UnkleRhaukus

Second Shift Theorem

6. AccessDenied

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

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

\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

im getting muddled up

10. UnkleRhaukus

\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

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

12. UnkleRhaukus

ok i still dont understand this very well

13. AccessDenied

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

14. AccessDenied

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.