Welcome to our community

Be a part of something great, join today!

Emad's question via email about Inverse Laplace Transform

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I have a question in Laplace transforms and I wonder if you can help me out. What's the inverse Laplace of $\displaystyle \begin{align*} \frac{12\,s}{s^4 + 324} \end{align*}$?
The closest Inverse Laplace Transform from my table is

$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{2\,a\,s\,\omega}{\left( s^2 + \omega ^2 - a^2 \right) ^2 + 4\,a^2\,\omega ^2 } \right\} = \sin{ \left( \omega \, t \right) } \sinh{ \left( a \, t \right) } \end{align*}$

so we would hope that we can write the denominator in this format.

Notice that if we expand out the denominator, we have

$\displaystyle \begin{align*} \left( s^2 + \omega ^2 - a^2 \right) ^2 + 4\,a^2 \, \omega ^2 &= \left( s^2 + \omega ^2 - a^2 \right) \left( s^2 + \omega ^2 - a^2 \right) + 4\,a^2 \,\omega ^2 \\ &= s^4 + \omega ^2 \,s^2 - a^2\,s^2 + \omega ^2 \,s^2 + \omega ^4 - a^2\,\omega ^2 - a^2\,s^2 - a^2\,\omega ^2 + a^4 + 4\,a^2\,\omega ^2 \\ &= s^4 + \left( 2\,\omega ^2 - 2\,a^2 \right) \, s^2 + \omega ^4 + 2\,a^2\,\omega ^2 + a^4 \\ &= s^4 + \left( 2\,\omega ^2 - 2\,a^2 \right) \, s^2 + \left( \omega ^2 + a^2 \right) ^2 \end{align*}$

so equating to the denominator in our problem we have

$\displaystyle \begin{align*} s^4 + 0\,s^2 + 18^2 &\equiv s^4 + \left( 2\,\omega ^2 - 2\,a^2 \right) \,s^2 + \left( \omega ^2 + a^2 \right) ^2 \end{align*}$

We can see that $\displaystyle \begin{align*} 2\,\omega ^2 - 2\,a^2 = 0 \implies \left| \omega \right| = \left| a \right| \end{align*}$ and $\displaystyle \begin{align*} \left| \omega ^2 + a^2 \right| = 18 \end{align*}$. We can see that a possible solution is $\displaystyle \begin{align*} \omega = 3 \end{align*}$ and $\displaystyle \begin{align*} a = 3 \end{align*}$.

However our numerator needs to be $\displaystyle \begin{align*} 2\,a\,s\,\omega = 2 \cdot 3 \cdot s \cdot 3 = 18\,s \end{align*}$, so that means we will write

$\displaystyle \begin{align*} \mathcal{L}^{-1}\,\left\{ \frac{12\,s}{s^2 + 324} \right\} &= \frac{2}{3} \,\mathcal{L}^{-1} \,\left\{ \frac{18\,s}{s^2 + 324} \right\} \\ &= \frac{2}{3}\,\mathcal{L}^{-1}\,\left\{ \frac{18\,s}{\left( s^2 + 3^2 - 3^2 \right) ^2 + 4 \cdot 3^2 \cdot 3^2 } \right\} \\ &= \frac{2}{3}\sin{ \left( 3\,t \right) }\sinh{ \left( 3\,t\right) } \end{align*}$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,180
Rochester, MN
Another option is to note that
$$s^4+324=s^4+(\sqrt{18})^2,$$
and therefore factors as
$$s^4+324=(s^2+6s+18)(s^2-6s+18).$$
Now try partial fractions:
$$\frac{12s}{s^4+324}=\frac{1}{s^2-6s+18}-\frac{1}{s^2+6s+18}.$$
Complete the square, and off you go.