Question 1 of 8: Controller Design by Pole Placement

Question 1: Controller Design by Pole Placement (20 marks, compulsory)

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. Unity-feedback positioning loop of Figure Q1.1, with

Find. The closed-loop transfer function and characteristic equation in terms of $a_0$, $a_1$, $b_1$; the required $\zeta$ and $\omega_n$; the three controller parameters and the third closed-loop pole $-d$; and a judgement on whether the design is lead or lag and whether the extra pole and zero are truly insignificant.

[Figure not reproduced: Figure Q1.1 (redrawn) — unity-feedback positioning loop with the first-order controller ahead of the second-order process. See the official exam paper.]

Approach. Reduce the loop to a single third-order transfer function, convert the two transient specifications into a dominant second-order pole pair, fix $a_0$ from the position-error requirement, and then match the actual characteristic polynomial coefficient-by-coefficient against a desired polynomial built from the dominant pair and one unknown real pole.

1. Close the loop and read off the characteristic equation. For a unity-feedback loop the closed-loop transfer function is $G_{cl}(s) = \dfrac{G_c G_p}{1 + G_c G_p}$. Substituting the two blocks, $$G_{cl}(s) = \frac{30\,(a_1 s + a_0)}{(b_1 s + 1)(s^2 + 10s + 20) + 30\,(a_1 s + a_0)}.$$ Expanding the product in the denominator and collecting powers of $s$ gives the characteristic equation $$\boxed{Q(s) = b_1 s^3 + (1 + 10 b_1)s^2 + (20 b_1 + 10 + 30 a_1)s + (20 + 30 a_0) = 0.}$$ Note the closed-loop zero at $s = -a_0/a_1$ that the controller numerator introduces — part 4c will ask whether it matters.
2. Convert the overshoot specification into a damping ratio. Inverting $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ with $L = \ln(PO/100) = \ln 0.10 = -2.3026$, $$\zeta = \frac{-L}{\sqrt{\pi^2 + L^2}} = \frac{2.3026}{\sqrt{9.8696 + 5.3019}} = \frac{2.3026}{3.8951} = 0.591.$$
3. Convert the settling time into a natural frequency. With the usual four-time-constant estimate $T_{s(\pm 2\%)} = 4/(\zeta\omega_n)$, the product $\zeta\omega_n$ is fixed directly by the specification: $\zeta\omega_n = 4/3 = 1.3333\ \text{s}^{-1}$. Dividing by the damping ratio just found, $$\boxed{\zeta = 0.591, \qquad \omega_n = \frac{1.3333}{0.591} = 2.255\ \text{rad/s}.}$$ The dominant pair therefore sits at $s_{1,2} = -1.333 \pm j\,1.819$.
4. Fix the controller DC gain from the error specification. The uncompensated loop ($G_c = 1$) is Type 0 with position constant $K_{pos,u} = G_p(0) = 30/20 = 1.5$, so $e_{ss,u} = 1/(1 + K_{pos,u}) = 1/2.5 = 0.400$. The requirement $e_{ss,c} = e_{ss,u}/4 = 0.100$ needs $K_{pos,c} = 1/e_{ss,c} - 1 = 9.0$. Since the controller contributes its own DC gain $G_c(0) = a_0$, $K_{pos,c} = a_0 \cdot 1.5$ and $$\boxed{a_0 = \frac{9.0}{1.5} = 6.0.}$$
5. Build the desired characteristic polynomial. The desired third-order polynomial is the dominant quadratic multiplied by one unknown real factor: $$Q_d(s) = (s^2 + 2\zeta\omega_n s + \omega_n^2)(s + d) = (s^2 + 2.6667 s + 5.0872)(s + d).$$ Expanding, $Q_d(s) = s^3 + (2.6667 + d)s^2 + (5.0872 + 2.6667 d)s + 5.0872\,d$.
6. Normalise the actual polynomial and match coefficients. Dividing $Q(s)$ by $b_1$ to make it monic, and using $20 + 30a_0 = 200$, $$\frac{Q(s)}{b_1} = s^3 + \Big(10 + \frac{1}{b_1}\Big)s^2 + \Big(20 + \frac{10 + 30 a_1}{b_1}\Big)s + \frac{200}{b_1}.$$ Matching the constant terms gives $200/b_1 = 5.0872\,d$, i.e. $b_1 = 200/(5.0872\,d)$. Substituting that into the $s^2$ match $10 + 1/b_1 = 2.6667 + d$ leaves a single equation in $d$: $$0.025436\,d + 10 = 2.6667 + d \quad\Longrightarrow\quad 0.97456\,d = 7.3333.$$
7. Solve for the third pole and the two remaining parameters. The last equation gives $d = 7.525$. Back-substituting, $b_1 = 200/(5.0872 \times 7.525) = 5.225$. Finally the $s^1$ match, $20 + (10 + 30a_1)/b_1 = 5.0872 + 2.6667 \times 7.525 = 25.153$, yields $10 + 30a_1 = 5.153 \times 5.225 = 26.93$ and hence $a_1 = 0.564$: $$\boxed{a_1 = 0.564, \quad b_1 = 5.225, \quad d = 7.525.}$$ Substituting all three back into $Q(s)$ and solving numerically returns the roots $-1.333 \pm j\,1.819$ and $-7.525$, confirming the design closes on itself.
8. Write the controller and locate its pole and zero. $$G_c(s) = \frac{0.564\,s + 6.0}{5.225\,s + 1} = 6.0\;\frac{1 + s/10.64}{1 + s/0.1914}.$$ The controller zero lies at $s = -a_0/a_1 = -10.64$ and its pole at $s = -1/b_1 = -0.1914$.
9. Part 4a — lead or lag? The controller pole ($-0.191$) is closer to the origin than the controller zero ($-10.64$). A network whose pole precedes its zero attenuates and retards phase over the intermediate band, so this is a $$\boxed{\text{LAG controller } (\alpha = 0.1914/10.64 = 0.018 \lt 1).}$$ That is exactly what one expects: the design was driven by a fourfold reduction in steady-state error, which is a low-frequency (DC-gain) demand, and lag compensation is the classical way to buy DC gain without disturbing the transient behaviour.
10. Part 4b — is the third pole insignificant? The dominant pair has real part $\sigma = \zeta\omega_n = 1.333$, while the third pole sits at $-7.525$. The separation ratio is $7.525/1.333 = 5.6$. The customary engineering threshold is a factor of five or more, so the third pole is just inside the insignificant region: its transient decays about $5.6$ times faster than the dominant oscillation and contributes only a small, quickly-vanishing correction. The dominant-pole model is acceptable, though not generously so.
11. Part 4c — is the closed-loop zero insignificant? The closed-loop zero introduced by the controller numerator lies at $-10.64$, which is $10.64/1.333 = 8.0$ times further from the imaginary axis than the dominant pair. A zero that remote adds negligible extra overshoot, so it too is insignificant and the realised step response should track the $PO = 10\%$, $T_s = 3$ s specification closely.