12 Problem A.5.8 Ogata 4edition (Equation)

Laplace.T	Root-locus	Transient response	Frecuency response

12 Problem A.5.8 Ogata 4edition (Equation)

Solve this equation:

$\begin{displaymath} \ddot{x}+2\dot{x}+100x=0 \end{displaymath}$ (1)

Solution:

$\begin{displaymath} \ddot{x}=> s^{2}X(s)-sx(0) \end{displaymath}$ (2)

$\begin{displaymath} \dot{x}=> sX(s)-x(0) \end{displaymath}$ (3)

Substituting the two above equations in the (1)

$\begin{displaymath} s^{2}X(s)-sx(0)+2(sX(s)-x(0))+100X(s)=0 \end{displaymath}$ (4)

$\begin{displaymath} (s^{2}+2s+k)X(s)=(s+2)x(0) \end{displaymath}$ (5)

$\begin{displaymath} X(s)=\frac{s+2}{(s^{2}+2s+100)}x(0) \end{displaymath}$ (6)

We need $\sigma$ , $w_{d}$ , sine and cosine transforms

$\begin{displaymath}\frac{s+2}{(s^{2}+2\sigma s+w_{n}^{2})}x(0)=\frac{s+2}{(s^{2}... ...}+\sigma^{2})}x(0)=\frac{s+2}{((s+\sigma)^{2}+w_{d}^{2})}x(0) \end{displaymath}$

we get $\sigma=1$ , $w_{n}^2=100$ y $w_{d}^{2}= w_{n}^{2}-\sigma^{2}=99$ .

$\begin{displaymath} X(s)=\frac{s+2}{((s+1)^{2}+99)}x(0) \end{displaymath}$ (7)

$\begin{displaymath} X(s)=\frac{s+1}{((s+1)^{2}+99)}x(0)+\frac{1}{((s+1)^{2}+99)}x(0) \end{displaymath}$ (8)

We get $w_{d}=\sqrt{99}=9.95$

$\begin{displaymath} x(t)=e^{-t}\cos{(9.95t)}x(0)+\frac{e^{-t}\sin{(9.95t)}}{9.95} x(0) \end{displaymath}$ (9)

The resulting equation has a period $T=\frac{2\pi}{9.95}$ due to the sine and cosine. Ie, four cicles after, it's $4T=\frac{8\pi}{9.95}$

$\begin{displaymath} x(4T)=e^{-4T}\cos{(9.95\cdot4T)}x(0)+\frac{e^{-4T}\sin{(9.95\cdot4T)}}{9.95} x(0)=e^{-4T}x(0)=0.004 \end{displaymath}$ (10)