Control System Model
Consider a control system's transfer function, which relates a step input to output, is 25/(s^3+0.6s^2+25s). Estimate the natural frequency, steady-state gain, and damping ratio.
Expand Hint
$$$\frac{Y(s)}{R(s)}=\frac{K\omega _{n}^{2}}{s^2+2\zeta\omega_ns+\omega _{n}^{2} }$$$
where
$$K$$
is the steady state gain,
$$\zeta$$
is the damping ratio, and
$$\omega_n$$
is the undamped natural frequency (
$$\zeta=0$$
).
Hint 2
In the second order form, our transfer function can be written as:
$$$\frac{Y(s)}{R(s)}=\frac{25}{s^3+0.6s^2+25s}$$$
One standard second-order control system model is
$$$\frac{Y(s)}{R(s)}=\frac{K\omega _{n}^{2}}{s^2+2\zeta\omega_ns+\omega _{n}^{2} }$$$
where
$$K$$
is the steady state gain,
$$\zeta$$
is the damping ratio, and
$$\omega_n$$
is the undamped natural frequency (
$$\zeta=0$$
),
In the second order form, our transfer function can be written as:
$$$\frac{Y(s)}{R(s)}=\frac{25}{s^3+0.6s^2+25s}=\frac{25}{s(s^2+0.6s+25)}$$$
To find the natural frequency,
$$\omega_n$$
, we know
$$\omega_n^{2}=25$$
, based on the transfer function format. Thus,
$$$\omega_n=\sqrt{25}=5$$$
To find the steady-state gain,
$$K$$
, we know
$$K\omega _{n}^{2}=25$$
, based on the transfer function format. Thus,
$$$K=\frac{25}{\omega _{n}^{2}}=\frac{25}{5^2}=\frac{25}{25}=1$$$
To find the damping ratio,
$$\zeta$$
, we know
$$2\zeta\omega_n=0.6$$
, based on the transfer function format. Thus,
$$$\zeta=\frac{0.6}{2\omega_n}=\frac{0.6}{2(5)}=\frac{0.6}{10}=0.06$$$
$$\omega_n=5$$
,
$$K=1$$
,
$$\zeta=0.06$$
Time Analysis
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