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.