Inverse Laplace of $F(s) = \frac{3s}{(s^2+9)^2}$

Question

Can somebody please show how to go about answering the following;

${\scr L^{-1}}(F(s)) $ where $F(s) = \dfrac{3s}{(s^2+9)^2}$

I know the ${\scr L}\left(\dfrac{3}{s^2+9}\right)=\sin(3t)$ and that $\dfrac{d}{ds}\dfrac{1}{s^2+9}=\dfrac{2s}{(s^2+9)^2}$ but dont quite understand how to apply these two to get a final answer, if somebody could show a step by step that would be very helpful thanks.

Answer 1

Remember the definition of the Laplace transform.

$$\mathscr{L}[f](s) = \int_0^\infty f(t) e^{-st}\,dt$$

is a parameter-dependent integral. These can often be differentiated under the integral sign, and if $f$ has at most exponential growth - $\lvert f(t)\rvert \leqslant C\cdot e^{Kt}$ for some constants $C,\,K$ - you obtain the derivative of $\mathscr{L}[f]$ by differentiating under the integral,

$$\frac{d}{ds} \mathscr{L}[f](s) = \int_0^\infty f(t)\frac{d}{ds} e^{-st}\,dt = - \int_0^\infty t\cdot f(t)e^{-st}\,dt.$$

In your case - you have a sign error, by the way - we have

$$F(s) = \frac{3s}{(s^2+9)^2} = -\frac{1}{2} \frac{d}{ds} \frac{3}{s^2+9} = -\frac{1}{2}\frac{d}{ds} \mathscr{L}[\sin (3t)](s).$$

Since the Laplace transform is linear, and hence its inverse is linear too, and $\sin (3t)$ has at most exponential growth (it is bounded), the above makes determining

$$\mathscr{L}^{-1}\left[\frac{3s}{(s^2+9)^2}\right]$$

easy.

Answer 2

One of the basic properties of the inverse transform is that if ${\scr L}\left(f(t)\right)=F(s)$ then ${\scr L}^{-1}\left(-F'(s)\right)=tf(t)$.

So you have $F(s) = \dfrac{3}{s^2+9}$ and $-F'(s) = \dfrac{6s}{(s^2+9)^2}$ (note you lost a sign in your derivative which gets removed anyway by the negative in front of $F'(s)$). So

$\begin{align} {\scr L}^{-1}\left(\dfrac{3s}{(s^2+9)^2}\right) &= {\scr L}^{-1}\left(\dfrac{1}{2}\dfrac{6s}{(s^2+9)^2}\right) \\ &= \dfrac{1}{2}{\scr L}^{-1}\left(\dfrac{6s}{(s^2+9)^2}\right) \\ &= \dfrac{1}{2}t\sin(3t) \end{align}$

Answer 3

• $$\dfrac{\mathrm d}{\mathrm ds} \left(\dfrac{1}{s^2+9}\right)=\dfrac{-2s}{(s^2+9)^2} \implies \dfrac{\mathrm d}{\mathrm ds}\left(\dfrac{3/2}{s^2+9}\right)=\dfrac{-3s}{(s^2+9)^2}$$
• $$f(t) \overset{\mathscr{L}}\longrightarrow F(s) \qquad \text{and } \qquad t\cdot f(t) \overset{\mathscr{L}}\longrightarrow-F'(s)$$
• $$F(s)=\left(\dfrac{3/2}{s^2+9}\right) \underset{\mathscr{L}}{\overset{\mathscr{L}^{-1}}\leftrightharpoons} f(t)={1 \over 2}\sin(3t)$$
• $$-F'(s)=\dfrac{3s}{(s^2+9)^2} \underset{\mathscr{L}}{\overset{\mathscr{L}^{-1}}\leftrightharpoons} \color{red}{f(t)={t \over 2}\sin(3t)} \qquad \longleftarrow$$