Homework 4: Lag-Lead and PID Compensation¶

Problems¶

Add a lag compensator to the lead compensator design for Question 3 of the “Dominant Poles and lead Compensation” problem sheet in order to give a position error constant $K_P = 20$ .

A process control system has open-loop transfer function

$G_o(s)=\frac{9}{(s+3)^2}.$

A PID compensator $ $D(s) = K_p + T_D s + 1/(T_I s)$ $is placed in cascade with the plant and unity feedback is applied. Write down the new closed-loop transfer function and tune the values of proportional gain$ K_p $, differential time$ T_D $and integral rate$ 1/T_I $required to give a steady-state open-loop gain of 15, zero step-error, rise-time$ t_r \le 200 $ms and peak overshoot$ %OS \le 10%$.

Design a PID compensator for the control system with open-loop transfer function $ $\frac{5}{(s+1)(s+5)}$ $such that the dominant closed-loop poles satisfy$ \zeta = 0.5 $,$ \omega_n = 10 $rad/s and the velocity error constant$ K_v = 25$.

A cancellation compensator is to be designed to achieve dominant closed-loop poles at $s = - 1.5 \pm j2.6$ for the system with open-loop transfer function $ $\frac{K}{s(s+1)}.$ $Determine the compensation required and the loop gain$ K$ of the compensated system. Use the root-locus technique to examine the worst case effect of a 5% cancellation mismatch due to component tolerances.

A control system has open-loop poles at $s = 0$ , $-1$ and $-5$ . Determine the value of the velocity error constant $K_v$ for this system. Use the zero of a lag compensator to cancel the pole at $s = -1$ and position the pole in order to raise the value of $K_v$ by $10$ . Sketch the root-loci for both the compensated and uncompensated systems and comment on the relative stability of each.

Using the plant equation $ $G(s)=\frac{K}{s-1},\;K>0,$ $and a cancellation compensator$ D(s)=\frac{s-1}{s+1} examine the effect on stability of a small error in the compensator zero position.