Laplace transformation with $(t+3)^2$

Question

The problem says that: $$F(s) = \frac{-2 \times {(s)}^{-1/2}}{1}$$

I need to find the laplace transformation of ${e}^{-8t} \times f(t) \times ({t+3})^{2} $

I know that if it was $({t})^{2} $ instead of $({t+3})^{2} $, I would have to: 1st) substitute s for s+8 (in my country we say translation, right?), than I would have to find the second derivate of it. I've already done the 1st step (s+8 thing) and got $$F(s) = \frac{-2 \times {(s+8)}^{-1/2}}{1}$$ What am I supposed to do now? Sorry for my english, I hope you understand my doubt.

Answer 1

To solve this question, you need to use a combination of some properties of the Laplace Transform: linearity, frequency shift and frequency differentiation. Let's remember them first.

$$ \newcommand{\tl}[3]{#1 & #2 & #3 \\\hline} \newcommand{\bs}[0]{\,\,\,\,\,} \begin{array}{|c|c|c|} \hline \rule[0]{0pt}{2.5ex} \rule[0]{0pt}{-1.5ex} \tl {{\mathbf{\bs Property \bs}}} {{\mathbf{\bs f(t) \bs}}} {{\mathbf{\bs F(s) \bs}}} \rule[0]{0pt}{2ex} \rule[0]{0pt}{-1ex} \tl {\text{Linearity}} {a\!\cdot\! f(t) + b\!\cdot\! g(t)} {a \!\cdot\! F(s) +b\!\cdot\! G(s)} \rule[0]{0pt}{2ex} \rule[0]{0pt}{-1ex} \tl {\text{Frequency Shift}} {e^{-at}f(t)} {F(s+a)} \rule[0]{0pt}{2ex} \rule[0]{0pt}{-1ex} \tl {\text{Frequency Differentiation}} {t^nf(t)} {(-1)^n \,F^{(n)}(s)} \end{array} \\ \small\text{(Note: a,b}\in \mathbb{R} \, \text{and} \, F^{(n)} \,\text{denotes the} \, \text{n$^{\text{th}}$} \, \text{derivative of F)} $$

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Let $g(t) = e^{-8t} \cdot f(t) \cdot(t+3)^2$. Rewriting the expression, in order to use the above properties, we have:

$$ g(t) =\underbrace{t^2e^{-8t}f(t)}_{\text{(I)}} \,\,+ \,\,\underbrace{6te^{-8t}f(t)}_{\text{(II)}} \,\,+ \,\, \underbrace{9e^{-8t}f(t)}_{\text{(III)}} $$

For $\text{(I)}$, we use frequency shift combined with frequency differentiation. For $\text{(II)}$, we use the previous ones plus linearity. And for $\text{(III)}$ we use frequency shift and linearity. Then $G(s)$ is:

$$ G(s) = (-1)^2 F^{(2)}(s+8) \,\, + \,\, 6\cdot (-1)^1 F^{(1)}(s+8) \,\, + \,\, 9F(s+8) $$

Since $F(s) = -2\cdot (s)^{-1/2}$, we have:

$$ \begin{alignat}{1} G(s) &= -\frac{-3/2}{\sqrt{s+8}\cdot (s+8)^2}-\frac{6}{\sqrt{s+8}\cdot (s+8)} -\frac{18}{\sqrt{s+8}} \\[6pt] &= -\frac{-3/2 \cdot (12s^2+196s+801)}{\sqrt{s+8}\cdot (s+8)^2} \end{alignat} $$