Question 1 of 6: PI control of a double-integrator servo

Question 1: PI control of a double-integrator servo (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A unity-feedback loop in which a PI controller drives a plant that already contains one integrator, with a disturbance entering negatively at the plant input.

Find. The proportional gain $K_1^{0}$ that meets the 10 % overshoot, the integral gain $K_2^{max}$ at which the loop loses stability, and the two steady-state quantities requested in part (c).

Approach. With $K_2 = 0$ the loop collapses to a textbook second-order system, so the overshoot fixes $\zeta$ and the plant pole fixes $\omega_n$; restoring $K_2$ makes the characteristic polynomial cubic, which Routh–Hurwitz handles; the steady-state values then follow from the final-value theorem applied to the two different transfer functions the block diagram actually defines.

1. Close the loop with the integral action switched off. Setting $K_2 = 0$ makes $C(s) = K_1$, so the loop gain and closed-loop transfer function are $$L(s) = \frac{K_1}{s(s+3)}, \qquad T(s) = \frac{L}{1+L} = \frac{K_1}{s^{2} + 3s + K_1}.$$ This is the canonical form $\omega_n^{2}/(s^{2} + 2\zeta\omega_n s + \omega_n^{2})$ with no closed-loop zero, so the standard overshoot formula applies exactly. Matching coefficients gives $\omega_n = \sqrt{K_1}$ and $2\zeta\omega_n = 3$.
2. Convert the overshoot specification into a damping ratio. For a pole pair with no zero, $$M_p = \exp\!\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^{2}}}\right) \;\Longrightarrow\; \zeta = \frac{|\ln M_p|}{\sqrt{\pi^{2} + \ln^{2} M_p}} = \frac{2.30259}{\sqrt{9.8696 + 5.30190}} = 0.59115 .$$ Ten per cent overshoot is therefore a moderately damped design, a little below the familiar $\zeta = 0.707$.
3. Read the natural frequency off the plant and solve for the gain. The damping term $2\zeta\omega_n$ is fixed at 3 by the plant pole, so $\omega_n$ is not free once $\zeta$ is chosen: $$\omega_n = \frac{3}{2\zeta} = \frac{3}{2(0.59115)} = 2.5374\ \text{rad/s}, \qquad K_1^{0} = \omega_n^{2} = \boxed{K_1^{0} = 6.438}.$$ The resulting closed-loop poles are $-1.5 \pm j2.046$, and a numerical step response peaks at 1.100 — exactly the requested 10 %.
4. Restore the integral term and form the characteristic polynomial. With $C(s) = (K_1 s + K_2)/s$, $$1 + C(s)P(s) = 1 + \frac{K_1 s + K_2}{s^{2}(s+3)} = 0 \;\Longrightarrow\; \Delta(s) = s^{3} + 3s^{2} + K_1 s + K_2 .$$ Two free integrators now sit in the loop, which is what makes part (c) interesting and part (b) necessary.
5. Apply the Routh–Hurwitz test. The array for $s^{3} + 3s^{2} + K_1 s + K_2$ is $$\begin{array}{c|cc} s^{3} & 1 & K_1\\ s^{2} & 3 & K_2\\ s^{1} & \dfrac{3K_1 - K_2}{3} & 0\\ s^{0} & K_2 & \end{array}$$ A stable loop needs every first-column entry positive, i.e. $K_2 \gt 0$ and $3K_1 - K_2 \gt 0$.
6. Evaluate the limit at $K_1 = K_1^{0}$. The binding condition is the $s^{1}$ row: $$K_2 \lt 3K_1^{0} = 3(6.438) \;\Longrightarrow\; \boxed{K_2^{max} = 19.32}.$$ At exactly $K_2^{max}$ the $s^{1}$ row vanishes and the auxiliary polynomial $3s^{2} + K_2^{max} = 0$ gives the sustained-oscillation frequency $\omega_{osc} = \sqrt{K_2^{max}/3} = \sqrt{K_1^{0}} = 2.537$ rad/s. A numerical root solve at that gain returns $s = -3$ and $s = \pm j2.5374$, confirming the pivot was kept.
7. Fix the operating gain and confirm it is stable. Part (c) sets $K_2 = K_2^{max}/2 = 9.658$, comfortably inside the Routh window. The closed-loop roots are $-2.115$ and $-0.443 \pm j2.091$; the complex pair is dominant and only lightly damped ($\zeta \approx 0.21$), which is the price of pushing the integral gain to half its stability limit.
8. Steady-state error to the ramp. With $d = 0$ the error obeys $E(s) = R(s)/[1 + L(s)]$ and $R(s) = 2/s^{2}$. Because $L(s) = (K_1 s + K_2)/[s^{2}(s+3)]$ carries two poles at the origin, the loop is Type 2 and the velocity constant is unbounded: $$K_v = \lim_{s\to 0} sL(s) = \lim_{s\to 0}\frac{K_1 s + K_2}{s(s+3)} = \infty \;\Longrightarrow\; e_{ss} = \frac{2}{K_v} = \boxed{e_{ss}=0}.$$ The ramp is followed with zero asymptotic error — the whole reason for adding $K_2$.
9. Steady-state control effort against the disturbance. Reading the block diagram, the plant sees $u - d$, and with $r = 0$ the error is $e = -y$. Eliminating $y$: $$y = P\,(u-d),\quad u = -C y \;\Longrightarrow\; \frac{u}{d} = \frac{C(s)P(s)}{1 + C(s)P(s)} = T(s).$$ Since $CP$ has a double pole at the origin, $T(0) = 1$, so by the final-value theorem $$u(\infty) = \lim_{s\to 0} s\,T(s)\frac{1}{s} = T(0) = \boxed{u(\infty) = +1}.$$ The physical argument is quicker and worth writing down: the plant contains an integrator, so its input must settle at zero or $y$ would ramp away. The plant input is $u - d$; hence $u(\infty) = d = 1$. The controller ends up exactly cancelling the disturbance, and it does so independently of $K_1$ and $K_2$.