Question 4 of 7: PD vs. proportional-plus-rate-feedback control

Question 4 — PD vs. proportional-plus-rate-feedback control [8 + 4 + 8]

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. Plant $\dfrac{30}{s^2+10s+5}$; specs $e_{ss}=5\%$, $\zeta=0.7$.

Find. $G_{cl1}$ and $K_p,T_d$ (A); step-response specs (B); $G_{cl2}$ and comparison (C).

[Figure not reproduced: Figure Q4.1 (redrawn) — forward gain $K_p$ with inner rate ($T_d s$) feedback added to the unity output feedback, so the feedback transfer function is $1+T_d s$. See the official exam paper.]

Part A — Rate-feedback closed loop and controller sizing [8]

Approach. The rate path makes the feedback $H(s)=1+T_d s$; form $G_{cl1}=K_pG/(1+K_pGH)$ and match the resulting second-order denominator to the $e_{ss}$ and $\zeta$ specs.

1. Closed-loop transfer function. With forward $K_p\cdot\dfrac{30}{s^2+10s+5}$ and feedback $1+T_d s$, $$G_{cl1}(s)=\frac{30K_p}{s^2+(10+30K_pT_d)s+(5+30K_p)}.$$
2. Stability restrictions. For a second-order denominator, stability needs both middle and constant coefficients positive: $5+30K_p\gt0$ and $10+30K_pT_d\gt0$, i.e. $K_p\gt0$ and $T_d\gt-\dfrac{1}{3K_p}$ — any $K_p\gt0,\ T_d\ge0$ is safe.
3. Gain from steady-state error. $e_{ss}=1-G_{cl1}(0)=\dfrac{5}{5+30K_p}=0.05\Rightarrow 5+30K_p=100\Rightarrow \boxed{K_p=3.17}$, giving $\omega_n^2=5+30K_p=100$, $\omega_n=10$ rad/s.
4. Rate gain from damping. $2\zeta\omega_n=10+30K_pT_d\Rightarrow 14=10+95T_d\Rightarrow \boxed{T_d=0.0421\ \text{s}}.$
5. Numerical closed loop. $$G_{cl1}(s)=\frac{95}{s^2+14s+100}.$$

Part B — Step-response specifications of $G_{cl1}$ [4]

Approach. Read the specs directly from $\zeta=0.7$, $\omega_n=10$.

1. Percent overshoot. $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}\times100=e^{-2.199/0.714}\times100=\boxed{4.6\%}.$
2. Settling time. $T_{s(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{7}=\boxed{0.57\ \text{s}}.$
3. Steady-state error. $e_{ss}=\boxed{5\%}$ by design (DC gain $0.95$).

Part C — PD control and comparison [8]

[Figure not reproduced: Figure Q4.2 (redrawn) — the same plant under forward PD control $K_p(T_d s+1)$ with plain unity feedback. See the official exam paper.]

Approach. The forward PD zero enters the numerator while the denominator is unchanged; compare the two closed loops term by term.

1. PD closed loop. $G_{cl2}(s)=\dfrac{30K_p(T_d s+1)}{s^2+(10+30K_pT_d)s+(5+30K_p)}$; with the same $K_p,T_d$ $$G_{cl2}(s)=\frac{95(0.0421s+1)}{s^2+14s+100}=\frac{4.0s+95}{s^2+14s+100}.$$
2. The difference. $G_{cl1}$ and $G_{cl2}$ share the identical denominator (same poles, same $\zeta=0.7$, $\omega_n=10$); $G_{cl2}$ adds a finite zero at $s=-1/T_d=-23.75$.
3. Effect on the Part-B specs. $e_{ss}$: unchanged ($G_{cl2}(0)=95/100=0.95$, so $e_{ss}=5\%$). PO: the extra LHP zero raises the overshoot well above $4.6\%$ (the derivative kick), and gives a faster rise/peak time. $T_s$: governed by the same $\zeta\omega_n=7$, so the $\pm2\%$ envelope is essentially the same. In short, rate feedback (Q4.1) avoids the numerator zero and keeps the clean, low-overshoot response, whereas forward PD (Q4.2) is faster but more oscillatory for identical gains.