Question 4 of 8: Controller Design by Pole Placement (PI + Rate Feedback)

Question 4: Controller Design by Pole Placement (PI + Rate Feedback) (20 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. From Figure Q4.1: a single summing junction driven by $R(s)$ positively and by the rate-feedback path negatively; forward path consisting of the PI controller $G_c(s) = K_p\left(1 + \frac{K_i}{s}\right)$ followed by the process $P(s) = \dfrac{1}{s^2+7s+5}$; and a feedback block $H(s) = K_d s + 1$ carrying $Y(s)$ back to that junction. Specifications: $PO = 5\%$, $T_{s(\pm2\%)} = 0.5$ s, $e_{ss(step)} = 0\%$.

Find. The closed-loop transfer function and characteristic equation in terms of $K_p$, $K_i$ and $K_d$; the damping ratio, natural frequency and DC gain the specifications demand; and the three controller gains that place the poles accordingly with the third pole cancelling the closed-loop zero.

[Figure not reproduced: Figure 4.1 — Redrawn closed loop under PI plus rate feedback. Note the single summing junction and the multiplicative controller form Kp(1 + Ki/s), which places the closed-loop zero at s = -Ki. See the official exam paper.]

Approach. Reduce the single-loop diagram algebraically to get $Y/R$, translate the three specifications into a target second-order pair plus a DC-gain condition, then match the closed-loop characteristic polynomial coefficient by coefficient against the desired cubic formed from that pair and a third pole placed exactly on the closed-loop zero.

1. Reduce the loop. With $E = R - H\,Y$ into the controller, $$Y = P\,G_c\,(R - H\,Y) \;\Longrightarrow\; Y\left(1 + P G_c H\right) = P G_c R$$ so $\dfrac{Y}{R} = \dfrac{PG_c}{1+PG_cH}$. Substituting $G_c = K_p(s+K_i)/s$, $P = 1/(s^2+7s+5)$ and $H = K_ds+1$, then clearing the fractions by multiplying numerator and denominator by $s(s^2+7s+5)$: $$\boxed{\,G_{cl}(s) = \frac{K_p\,(s+K_i)} {s(s^2+7s+5) + K_p(s+K_i)(K_ds+1)}\,}$$
2. Expand the characteristic equation. Multiplying out the second term, $K_p(s+K_i)(K_ds+1) = K_pK_d s^2 + K_p(1+K_iK_d)s + K_pK_i$, so $$\boxed{\,Q(s) = s^3 + (7 + K_pK_d)\,s^2 + (5 + K_p + K_pK_iK_d)\,s + K_pK_i = 0\,}$$ Observe immediately that $G_{cl}(0) = K_pK_i/(K_pK_i) = 1$: the integral action forces unity DC gain for any admissible gains, so the zero steady-state error requirement is met automatically and $$K_{dc} = 1$$
3. Convert the transient specifications. From $PO = 5\%$, $$\zeta = \frac{-\ln(0.05)}{\sqrt{\pi^2+\ln^2(0.05)}} = \boxed{0.690}$$ and from $T_{s(\pm2\%)} = 4/(\zeta\omega_n) = 0.5$ s, $$\zeta\omega_n = \frac{4}{0.5} = 8 \;\Longrightarrow\; \omega_n = \frac{8}{0.690} = \boxed{11.59\ \text{rad/s}}$$ The desired dominant pair is therefore $s = -\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^2} = -8 \pm j8.390$, and the target quadratic factor is $s^2 + 16s + 134.38$.
4. Set up the pole-placement match. The closed-loop zero is at $s = -K_i$, so the third pole must be placed there. The desired characteristic polynomial is $$(s^2 + 2\zeta\omega_n s + \omega_n^2)(s + K_i) = s^3 + (16+K_i)s^2 + (\omega_n^2 + 16K_i)s + \omega_n^2K_i$$ Matching against $Q(s)$ gives three equations: $$7 + K_pK_d = 16 + K_i, \qquad 5 + K_p + K_pK_iK_d = \omega_n^2 + 16K_i, \qquad K_pK_i = \omega_n^2K_i$$
5. Solve for $K_p$. The constant-term equation divides through by $K_i$ immediately: $$K_p = \omega_n^2 = \boxed{134.4}$$
6. Reduce to a quadratic in $K_i$. The $s^2$ equation gives $K_pK_d = 9 + K_i$. Substituting both this and $K_p = \omega_n^2$ into the $s^1$ equation, $$5 + \omega_n^2 + K_i(9+K_i) = \omega_n^2 + 16K_i$$ The $\omega_n^2$ terms cancel — a neat simplification — leaving $$K_i^2 - 7K_i + 5 = 0$$ Notice that the coefficients are precisely the plant's own: for a process $1/(s^2+as+b)$ this matching always collapses to $K_i^2 - aK_i + b = 0$. Solving, $$K_i = \frac{7 \pm \sqrt{49-20}}{2} = \frac{7 \pm \sqrt{29}}{2} \;\Rightarrow\; K_i = 6.193 \ \text{or}\ 0.807$$
7. Choose between the two roots and finish. Both values place the dominant pair exactly and both cancel the third pole exactly, so either is mathematically admissible. Prefer the larger, $K_i = 6.193$: it puts the cancelled mode at $s = -6.19$ rather than at $s = -0.807$, and since no cancellation is ever perfect in practice, the residual mode then decays roughly eight times faster and leaves a far smaller tail on the step response. With $K_d = (9+K_i)/K_p$, $$\boxed{\,K_p = 134.4, \qquad K_i = 6.193, \qquad K_d = \frac{9+6.193}{134.4} = 0.1131\,}$$
8. Verify the placement. Substituting these gains back, $Q(s) = s^3 + 22.193s^2 + 233.47s + 832.18$, whose roots are $-8 \pm j8.390$ and $-6.193$. The real root coincides with the closed-loop zero at $-K_i = -6.193$ to machine precision, so it cancels and the realised transfer function is exactly the desired second-order model with $\zeta = 0.690$, $\omega_n = 11.59$ rad/s and $K_{dc} = 1$.