Inverse Laplace transfer second order equation

Question

How should I inverse Laplace transform this function?

$$\frac{(\omega_n)^2}{s^2+bs+c}$$

I don't even know were to start and how should I transform it? A little help would be helpful here.

Answer 1

In such a case the most convenient way to compute the inverse transform is to use standard results from Laplace transform tables. The transform pairs that you need here can be found in most tables:

$$\frac{1}{s^2}\Longleftrightarrow t\cdot u(t)\\ \frac{\omega_0}{s^2+\omega_0^2}\Longleftrightarrow \sin(\omega_0t)\cdot u(t)\\ \frac{\omega_0}{s^2-\omega_0^2}\Longleftrightarrow \sinh(\omega_0t)\cdot u(t)\\ f(t)\Longleftrightarrow F(s)\quad\Rightarrow\quad e^{-\alpha t}f(t)\Longleftrightarrow F(s+\alpha)$$

Now rewrite the function as

$$X(s)=\frac{\omega_n^2}{s^2+bs+c}=\frac{\omega_n^2}{(s+b/2)^2+c-b^2/4}$$

You have to distinguish 3 cases:

• $c>b^2/4$: Define $\omega_0^2=c-b^2/4$.

$$X(s)=\frac{\omega_n^2}{\omega_0}\frac{\omega_0}{(s+b/2)^2+\omega_0^2} \Longleftrightarrow\frac{\omega_n^2}{\omega_0}e^{-bt/2}\sin(\omega_0t)\cdot u(t)$$

• $c<b^2/4$: Define $\omega_0^2=b^2/4-c$.

$$X(s)=\frac{\omega_n^2}{\omega_0}\frac{\omega_0}{(s+b/2)^2-\omega_0^2} \Longleftrightarrow\frac{\omega_n^2}{\omega_0}e^{-bt/2}\sinh(\omega_0t)\cdot u(t)$$

• $c=b^2/4$:

$$X(s)=\omega_n^2\frac{1}{(s+b/2)^2} \Longleftrightarrow\omega_n^2e^{-bt/2}t\cdot u(t)$$