# Control Systems/Examples/Second Order Systems

## Second Order Systems: Examples

### Example 1

A damped control system for aiming a hydrophonic array on a minesweeper vessel has the following open-loop transfer function from the driveshaft to the array.

$G(s)=\frac{K}{Js^2 + K_ds}$

The gain parameter K can be varied. The moment of inertia, J, of the array and the force due to viscous drag of the water, Kd are known constants and given as:

• $J = 9 \, N \, m \, s^2 \, rad^{-1}$
• $K_d = 2 \, N \, m \, s \, rad^{-1}$

1. The system is arranged as a closed loop system with unity feedback. Find the value of K such that, when the input is a unit step, the closed loop response has at most a 50% overshoot (approximately). You may use standard response curves. Should K be greater or less than this value for less overshoot?
2. Find the corresponding time-domain response of the system.
3. The system is now given an input of constant angular velocity, V. For the limiting value of K found above, calculate the maximum value of V such that the array follows the input with at most 5° error.

First, let us draw the block diagram of the system. We know the open-loop transfer function, and that there is unit feedback. Therefore, we have:

The closed-loop gain is given by:

 $H(s)$ $=\frac{G(s)}{1+G(s)}$ $=\frac{\dfrac{K}{Js^2 + K_ds}}{\dfrac{Js^2 + K_ds+K}{Js^2 + K_ds}}$ $=\frac{K}{Js^2 + K_ds+K}$

We now need to express the closed-loop transfer function in the standard second order form.

 $\frac{\omega_n^2}{s^2+2\zeta\omega_n s + \omega_n^2}$ $=\frac{K}{Js^2 + K_ds+K}$ $=\frac{\frac{K}{J}}{s^2 + \frac{K_d}{J}s+\frac{K}{J}}$

We can now express the natural frequency ωn and damping ratio, ζ:

$\omega_n^2 = \frac{K}{J} = \frac{K}{9}$
$2 \zeta \omega_n = \frac{K_d}{J}$
$\zeta = \frac{K_d}{2J} \sqrt{\frac{J}{K}} = \frac{K_d}{2} \sqrt{\frac{1}{KJ}}=\frac{1}{3\sqrt{K}}$

We now look at the standard response curves for second order systems.

We see that for 50% overshoot, we need ζ=0.2 or more.

$\zeta = \frac{1}{3 \sqrt{K}} \le \frac{1}{5}$
$K \ge \frac{25}{9}$

This is the maximum permissible value, thus K should be less than this value for less overshoot. We can now evaluate the natural frequency fully:

$\omega_n = \frac{5}{9}$

The output of the second order system is given by the following equation:

 $y(t)$ $= 1 - \frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \sin \left( \sqrt{1-\zeta^2} \omega_n t + \sin^{-1} \sqrt{1-\zeta^2} \right)$ $= 1 - \frac{1}{\sqrt{0.96}} e^{-t/9} \sin \left( \frac{5\sqrt{0.96}}{9} t + \sin^{-1} \sqrt{0.96} \right)$ $= 1 - 1.02 e^{-t/9} \sin \left( 0.544 t +0.436 \pi \right)$

We can plot the output of this system:

The tracking error signal, E(s), is equal to the output's deviation from the input.

$E(s) = R(s) - Y(s)\,$

Now, we can find the gain from the reference input, R(s) to the error tracking signal:

$\frac{E(s)}{R(s)} = \frac{R(s)-Y(s)}{R(s)}$

The gain from the input to the error tracking signal of a unity feedback system like this is simply $\frac{1}{1+G(s)}$.

 $\frac{E(s)}{R(s)}$ $= \frac{1} { 1 + \frac{K}{9s^2 + 2s}}$ $= \frac{9s^2 + 2s}{9s^2 + 2s + 2.78}$

Now, R(s) is given by the Laplace transform of a ramp of slope V:

$R(s) = \frac{v}{s^2}$

We now use the final value theorem to find the value of E(s) in the steady state:

 $\lim_{t \to \infty}{e(t)}$ $= \lim_{s \to 0}{sE(s)}$ $=\lim_{s \to 0}{s \frac{9s^2 + 2s}{9s^2 + 2s + 2.78} \frac{V}{s^2} }$ $=\frac{2V}{2.78}$

We require this to be less than $5^\circ = \frac{5\pi}{180} \mbox{rad}$

$V \le \frac{2.78}{2}\times \frac{5\pi}{180} = 0.12 \,\mbox{rad/s}$