22-Elec-A2 Systems and Control · May 2018

Question 7 of 8: PD Controller Design by Pole Placement — Rate Feedback versus Derivative Control

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7: PD Controller Design by Pole Placement — Rate Feedback versus Derivative Control (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. Process $G_p(s) = 30/(s^{2}+10s+5)$ in two configurations: Figure Q7.1 puts the gain $K_p$ in the forward path and returns $(1 + T_ds)Y(s)$ around it (proportional plus rate feedback), while Figure Q7.2 puts the whole controller $K_p(T_ds+1)$ in the forward path with unity feedback (proportional plus derivative). Specifications: $PO = 5\%$ and $e_{step(\%)} = 5\%$.

Find. $K_p$ and $T_d$ meeting both specifications; $G_{cl1}(s)$ and its settling time; $G_{cl2}(s)$; and a comparison of the two with respect to overshoot, steady-state error, rise time and settling time.

[Figure not reproduced: Figure Q7.1 (redrawn) — proportional gain in the forward path with a rate feedback branch $T_ds$ summed with the direct unity branch, so the signal returned to the input summer is $(1+T_ds)Y(s)$. See the official exam paper.]

[Figure not reproduced: Figure Q7.2 (redrawn) — the same process under a cascade proportional-plus-derivative controller $K_p(T_ds+1)$ with plain unity feedback. The derivative now acts on the error rather than on the output. See the official exam paper.]

Approach. Derive each closed-loop transfer function, note that both share a denominator, use the step-error specification to fix $K_p$ (which fixes $\omega_n$) and the overshoot specification to fix $\zeta$ (which fixes $T_d$), then compare the two numerators.

1. Closed-loop transfer function for rate feedback. In Figure Q7.1 the forward path is $K_pG_p$ and the feedback is $H(s) = 1 + T_ds$, so $$G_{cl1}(s) = \frac{K_pG_p}{1+K_pG_p(1+T_ds)} = \frac{30K_p}{s^{2}+10s+5+30K_p(1+T_ds)}$$ $$G_{cl1}(s) = \frac{30K_p}{s^{2} + (10 + 30K_pT_d)s + (5 + 30K_p)}$$ This is a pure second-order system — the rate branch contributes only to the damping term, and there is no numerator zero.
2. Fix $K_p$ from the steady-state error. The DC gain is $G_{cl1}(0) = 30K_p/(5+30K_p)$, so the step error is $$e_{step} = 1 - G_{cl1}(0) = \frac{5}{5+30K_p} = 0.05 \;\Longrightarrow\; 5 + 30K_p = 100 \;\Longrightarrow\; \boxed{\;K_p = \frac{95}{30} = 3.1667\;}$$ Because $\omega_n^{2} = 5 + 30K_p$, this immediately gives $$\omega_n = \sqrt{100} = 10.0\ \text{rad/s}$$ a convenient round figure that confirms the intended design.
3. Fix $T_d$ from the overshoot. Inverting the overshoot correlation for $PO = 5\%$, $$\zeta = \frac{-\ln(0.05)}{\sqrt{\pi^{2}+\ln^{2}(0.05)}} = \frac{2.9957}{4.3410} = 0.6901$$ Matching the damping coefficient $2\zeta\omega_n = 10 + 30K_pT_d$: $$2(0.6901)(10) = 13.802 = 10 + 95\,T_d \;\Longrightarrow\; \boxed{\;T_d = \frac{3.802}{95} = 0.04002\ \text{s}\;}$$
4. Substitute and estimate the settling time. With the two settings in place, $$\boxed{\;G_{cl1}(s) = \frac{95}{s^{2} + 13.802\,s + 100}\;}$$ and, since $\zeta\omega_n = 6.901$, $$T_{settle(\pm2\%)} = \frac{4}{\zeta\omega_n} = \frac{4}{6.901} = \boxed{\;0.580\ \text{s}\;}$$ A direct simulation of this transfer function returns an overshoot of 5.00 % and a final value of 0.95, confirming both specifications exactly.
5. Closed-loop transfer function under PD control. In Figure Q7.2 the controller multiplies the error, so it appears in the numerator as well: $$G_{cl2}(s) = \frac{K_p(T_ds+1)G_p}{1+K_p(T_ds+1)G_p} = \frac{30K_p(T_ds+1)}{s^{2}+10s+5+30K_p(T_ds+1)}$$ Substituting the same $K_p = 3.1667$ and $T_d = 0.04002$: $$\boxed{\;G_{cl2}(s) = \frac{3.802\,s + 95}{s^{2} + 13.802\,s + 100}\;}$$
6. Compare the two. The denominators are identical, so both loops have the same poles, the same $\omega_n = 10$ rad/s and the same $\zeta = 0.690$. The single difference is that $G_{cl2}$ carries a numerator zero at $$s = -\frac{1}{T_d} = -24.99$$ and both have the same DC gain $95/100 = 0.95$. The zero sits $24.99/6.901 = 3.6$ times further left than the pole real part, so its influence is real but moderate.
7. Effect on each specification. A left-half-plane zero adds a derivative term $T_d\,\dot{y}$ to the response, lifting the early part of the transient:
   • Steady-state error — unchanged. Both DC gains are 0.95, so $e_{step} = 5\%$ in each case. A zero cannot change the final value.
   • Percent overshoot — increases. Simulation gives 5.00 % for $G_{cl1}$ against 5.60 % for $G_{cl2}$. The rise is modest precisely because the zero is 3.6 times beyond the pole real part; a nearer zero would inflate it sharply.
   • Rise time — decreases. Simulation gives 0.322 s against 0.270 s, a 16 % speed-up: the derivative kick makes the PD loop reach its final value sooner.
   • Settling time — essentially unchanged. Settling is governed by the envelope $e^{-\zeta\omega_n t}$, which depends only on the poles, so both remain near 0.58 s; the slightly larger overshoot of $G_{cl2}$ can push the 2 % crossing marginally later.
   The engineering summary: rate feedback and PD control place identical poles, but PD additionally differentiates the reference. Rate feedback is therefore preferred when a clean, low-overshoot step response matters (and it avoids differentiating a noisy or stepped command), while PD gives a faster rise at the cost of extra overshoot.