Question 6 of 8: Lead/Lag controller design by pole placement
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, May 2016, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard ζ–overshoot and ζ–resonant-peak design plots are supplied). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — time response and second-order specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot below are redrawn as inline figures.
Reading the supplied design charts. The percent overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.
Question 6 — Lead/Lag controller design by pole placement [5 + 5 + 7 + 3]
Find. $G_{cl}$, $Q(s)$; $G_m(s)$; the three controller parameters and the controller type.
Part (1) — closed-loop transfer function
Loop. $G_cG=\dfrac{a_1s+a_0}{(b_1s+1)(s+0.5)}$, unit feedback:
$G_{cl}(s)=\dfrac{a_1s+a_0}{b_1s^2+(0.5b_1+1+a_1)s+(0.5+a_0)}$, so $Q(s)=b_1s^2+(0.5b_1+1+a_1)s+(0.5+a_0)=0$.
Part (2) — the model
Damping and frequency. $PO=10\%\Rightarrow\zeta=0.591$; $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=2\Rightarrow\zeta\omega_n=2\Rightarrow\omega_n=3.38$ rad/s ($\omega_n^2=11.45$, $2\zeta\omega_n=4$).
DC gain from error. $e_{ss}=5\%\Rightarrow G_m(0)=0.95$, so $\boxed{G_m(s)=\dfrac{0.95\cdot11.45}{s^2+4s+11.45}=\dfrac{10.88}{s^2+4s+11.45}}$.
Q6.1: controller pole (×) at $-1.15$ sits nearer the origin than the zero (○) at $-4.62$ ⇒ a lag network.
Part (3) — controller parameters and type
Match the $s^0$ / DC-gain (error) condition. $G_{cl}(0)=\dfrac{a_0}{0.5+a_0}=0.95\Rightarrow\boxed{a_0=9.5}$.
Match $\omega_n^2$. $\dfrac{0.5+a_0}{b_1}=11.45\Rightarrow b_1=\dfrac{10}{11.45}=\boxed{0.873}$.
Match $2\zeta\omega_n$. $\dfrac{0.5b_1+1+a_1}{b_1}=4\Rightarrow a_1=3.5b_1-1=\boxed{2.06}$.
Classify. zero at $-a_0/a_1=-4.62$, pole at $-1/b_1=-1.15$. The pole is closer to the origin than the zero, so $G_c$ is a lag controller (large DC gain $a_0=9.5$ to pull the steady-state error down).
Part (4) — actual vs model
The realised closed loop carries an extra zero at $-4.62$ (from the controller numerator) that the pure second-order model $G_m$ omits. Being a few times farther out than the dominant poles, the zero adds a little extra overshoot and a slightly faster rise, but the two responses track closely; steady state, $\zeta$ and $\omega_n$ match by construction.
