Consider a closed-loop system as shown. $G_p(s)=\frac{14.4}{s(1+0.1s)}$ is the plant transfer function and G${}_c$ (s) = 1 is the compensator. For a unit-step input, the output response has damped oscillations. The damped natural frequency is __________ rad / s. (Round off to 2 decimal places.)

Concept:

Time Response of the second-order System:



$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}$

$G\left(s\right)=\frac{{\omega }_n^2}{s\left(s+2\zeta {\omega }_n\right)},H\left(s\right)=1$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{{\omega }_n^2}{s\left(s+2\zeta {\omega }_n\right)}}{1+\frac{{\omega }_n^2}{s\left(s+2\zeta {\omega }_n\right)}}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{[s^2+2\zeta {\omega }_{n}s+{\omega }_n^2]}$

In the above transfer function, the power of 's' is two in the denominator therefore it is called a second-order system.

${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$

Where ω${}_d$ is damped natural frequency in rad/sec.

Calculation:

Given,

$G_p(s)=\frac{14.4}{s(1+0.1s)}$

Gc(s) = 1

H(s) = 1

G(s) = Gc(s)GP(s)

$G(s)=\frac{14.4}{s(1+0.1s)}$

The transfer function of the system will be,

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{\frac{14.4}{s\left(1+0.1s\right)}}{1+\frac{14.4}{s\left(1+0.1s\right)}}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{14.4}{0.1s^2+s+14.4}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{14.4}{0.1[s^2+10s+144]}$

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{144}{[s^2+10s+144]}$ .... (1)

We know that,

$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{[s^2+2\zeta {\omega }_{n}s+{\omega }_n^2]}$ .... (2)

From equation (1) and (2),

${\omega }_n^2=144$

${\omega }_n=12$

$2\zeta {\omega }_n=10$

$\zeta =\frac{10}{24}=\frac{5}{12}$

${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=12\sqrt{1-{\left(\frac{10}{24}\right)}^2}=10.91rad/sec$