How should I inverse Laplace transform this function?

$$\frac{({\omega}_{n}{)}^{2}}{{s}^{2}+bs+c}$$

$$\frac{({\omega}_{n}{)}^{2}}{{s}^{2}+bs+c}$$

limfne2c
2022-10-17
Answered

How should I inverse Laplace transform this function?

$$\frac{({\omega}_{n}{)}^{2}}{{s}^{2}+bs+c}$$

$$\frac{({\omega}_{n}{)}^{2}}{{s}^{2}+bs+c}$$

You can still ask an expert for help

espava8b

Answered 2022-10-18
Author has **12** answers

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:

$$\frac{1}{{s}^{2}}\u27fat\cdot u(t)\phantom{\rule{0ex}{0ex}}\frac{{\omega}_{0}}{{s}^{2}+{\omega}_{0}^{2}}\u27fa\mathrm{sin}({\omega}_{0}t)\cdot u(t)\phantom{\rule{0ex}{0ex}}\frac{{\omega}_{0}}{{s}^{2}-{\omega}_{0}^{2}}\u27fa\mathrm{sinh}({\omega}_{0}t)\cdot u(t)\phantom{\rule{0ex}{0ex}}f(t)\u27faF(s)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{e}^{-\alpha t}f(t)\u27faF(s+\alpha )$$

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 ${\omega}_{0}^{2}=c-{b}^{2}/4$

$$X(s)=\frac{{\omega}_{n}^{2}}{{\omega}_{0}}\frac{{\omega}_{0}}{(s+b/2{)}^{2}+{\omega}_{0}^{2}}\u27fa\frac{{\omega}_{n}^{2}}{{\omega}_{0}}{e}^{-bt/2}\mathrm{sin}({\omega}_{0}t)\cdot u(t)$$

$c<{b}^{2}/4$: Define ${\omega}_{0}^{2}={b}^{2}/4-c$

$$X(s)=\frac{{\omega}_{n}^{2}}{{\omega}_{0}}\frac{{\omega}_{0}}{(s+b/2{)}^{2}-{\omega}_{0}^{2}}\u27fa\frac{{\omega}_{n}^{2}}{{\omega}_{0}}{e}^{-bt/2}\mathrm{sinh}({\omega}_{0}t)\cdot u(t)$$

$c={b}^{2}/4$

$$X(s)={\omega}_{n}^{2}\frac{1}{(s+b/2{)}^{2}}\u27fa{\omega}_{n}^{2}{e}^{-bt/2}t\cdot u(t)$$

$$\frac{1}{{s}^{2}}\u27fat\cdot u(t)\phantom{\rule{0ex}{0ex}}\frac{{\omega}_{0}}{{s}^{2}+{\omega}_{0}^{2}}\u27fa\mathrm{sin}({\omega}_{0}t)\cdot u(t)\phantom{\rule{0ex}{0ex}}\frac{{\omega}_{0}}{{s}^{2}-{\omega}_{0}^{2}}\u27fa\mathrm{sinh}({\omega}_{0}t)\cdot u(t)\phantom{\rule{0ex}{0ex}}f(t)\u27faF(s)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{e}^{-\alpha t}f(t)\u27faF(s+\alpha )$$

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 ${\omega}_{0}^{2}=c-{b}^{2}/4$

$$X(s)=\frac{{\omega}_{n}^{2}}{{\omega}_{0}}\frac{{\omega}_{0}}{(s+b/2{)}^{2}+{\omega}_{0}^{2}}\u27fa\frac{{\omega}_{n}^{2}}{{\omega}_{0}}{e}^{-bt/2}\mathrm{sin}({\omega}_{0}t)\cdot u(t)$$

$c<{b}^{2}/4$: Define ${\omega}_{0}^{2}={b}^{2}/4-c$

$$X(s)=\frac{{\omega}_{n}^{2}}{{\omega}_{0}}\frac{{\omega}_{0}}{(s+b/2{)}^{2}-{\omega}_{0}^{2}}\u27fa\frac{{\omega}_{n}^{2}}{{\omega}_{0}}{e}^{-bt/2}\mathrm{sinh}({\omega}_{0}t)\cdot u(t)$$

$c={b}^{2}/4$

$$X(s)={\omega}_{n}^{2}\frac{1}{(s+b/2{)}^{2}}\u27fa{\omega}_{n}^{2}{e}^{-bt/2}t\cdot u(t)$$

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