Question 6 of 8: Lead/Lag Controller Design by Pole Placement

Question 6: Lead/Lag Controller Design by Pole Placement (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. $G(s)=1/(s+0.5)$ (Type 0, first order), unity feedback, series compensator $G_c(s)=(a_1s+a_0)/(b_1s+1)$.

Find. $G_{cl}(s)$ and $Q(s)$; the target second-order model $G_m(s)$; the compensator parameters and its Lead/Lag classification; and a comparison of actual vs. model response.

Approach. Because both $G(s)$ and $G_c(s)$ are first order, the closed loop here is EXACTLY second order (unlike Questions 4–5): match $Q(s)$'s coefficients directly to $s^2+2\zeta\omega_ns+\omega_n^2$ after using the steady-state error spec to fix $a_0$.

1. Part 1) — closed-loop TF and characteristic equation. $$G_{cl}(s)=\frac{G_c G}{1+G_cG}=\frac{a_1s+a_0}{(b_1s+1)(s+0.5)+(a_1s+a_0)},\qquad \boxed{Q(s)=b_1s^2+(0.5b_1+1+a_1)s+(0.5+a_0)=0.}$$
2. Part 2) — target model $G_m(s)$. $PO=10\%\Rightarrow\zeta=\boxed{0.591}$ (as in Q5); $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=2$ s gives $\omega_n=4/(0.591\times2)=\boxed{3.38\ \text{rad/s}}$. For a Type 0 loop, $e_{ss(step)}=1-K_{dc}$, so $e_{ss}=5\%\Rightarrow\boxed{K_{dc}=0.95}$. The target model, in the same $K_{dc}\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)$ form as the formula sheet, is $$\boxed{G_m(s)=\frac{0.95(3.38)^2}{s^2+4.00s+11.45}}.$$
3. Part 3) — compensator parameters and Lead/Lag classification. Matching $G_{cl}(0)=a_0/(0.5+a_0)=K_{dc}=0.95$ gives $\boxed{a_0=9.5}$ directly (this also reproduces the required open-loop position constant $K_{pos}=2a_0=19$, i.e. $e_{ss}=1/(1+19)=5\%$, independently). Matching the $s^0$ coefficient, $0.5+a_0=b_1\omega_n^2$, gives $\boxed{b_1=10/11.45=0.874}$; matching the $s^1$ coefficient, $0.5b_1+1+a_1=2\zeta\omega_nb_1$, gives $\boxed{a_1=2.06}$. The compensator's zero sits at $s=-a_0/a_1=-4.62$ and its pole at $s=-1/b_1=-1.14$: because the ZERO is farther from the origin than the pole ($4.62>1.14$), this is a $\boxed{\text{LAG compensator}}$ (a Lead network would need the pole farther out than the zero).
4. Part 4) — actual response vs. the model. Because $G(s)$ and $G_c(s)$ are BOTH first order, the closed loop's denominator is exactly the second-order polynomial designed in Part 3 — no extra (non-dominant) pole appears, unlike Questions 4–5. However, the closed-loop NUMERATOR is $a_1s+a_0$, not the constant $K_{dc}\omega_n^2$ of the idealized model $G_m(s)$: the actual system carries a real LHP zero at $s=-4.62$ that the pure second-order model does not have. A left-half-plane zero of this kind speeds up the rise time and pushes the overshoot somewhat ABOVE the nominal $10\%$ predicted by $G_m(s)$, even though both share identical poles ($\zeta=0.591$, $\omega_n=3.38$ rad/s).