Control Systems/Examples/Second Order Systems

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:


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

Tasks

 * 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.

Task 1
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:




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

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 &omega;n and damping ratio, &zeta;:


 * $$\omega_n^2 = \frac{K}{J} = \frac{K}{9}$$


 * $$2 \zeta \omega_n = \frac{K_d}{J}$$


 * $$\zeta = \frac{K_d}{2J} \sqrt{\frac{J}{K}} = \frac{K_d}{2} \sqrt{\frac{1}{KJ}}=\frac{1}{3\sqrt{K}}$$

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



We see that for 50% overshoot, we need &zeta;=0.2 or more.


 * $$\zeta = \frac{1}{3 \sqrt{K}} \geq \frac{1}{5}$$


 * $$K \le \frac{25}{9}$$

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:


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

Task 2
The output of the second order system is given by the following equation:




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

We can plot the output of this system:



Task 3
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:




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

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


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