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22-Elec-A2 Systems and Control · May 2013

Question 4 of 8: Safe operating range under P vs PI control (Routh–Hurwitz)

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 2013, 3 hours, closed book (approved calculator + one handwritten formula sheet, and a printed short table of Laplace transforms). Eight questions of 20 marks each; Questions 1 and 2 are compulsory and the candidate chooses three of the remaining six (100 marks total). Every question is worked in full below, since the whole set is a study resource.

Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — root locus (Ch. 8), frequency-response / Bode and Nyquist (Ch. 10), stability and Routh–Hurwitz (Ch. 6), PID and cascade compensation (Ch. 9), steady-state error (Ch. 7), state space and pole placement (Ch. 11–12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of the s-domain, dominant-poles modelling and controllability/observability; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, root loci, Bode/Nyquist plots and pole–zero maps below are redrawn as inline figures.

Reading the exam figures. Two questions (Q2 and Q6) require reading a second-order model off a printed step-response curve; the values read from those curves are stated explicitly in the Given block so a reader can reproduce every subsequent number.

Question 4 — Safe operating range under P vs PI control (Routh–Hurwitz) [20]

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. $G(s)=\dfrac{100}{(s+5)(s+7)(s+10)}$, i.e. $(s+5)(s+7)(s+10)=s^3+22s^2+155s+350$. P control: $G_c=K_p$. PI control: $G_c=K_p(s+2)/s$.

Find. Safe $K_p$ range for each scheme and the direction of the gain-margin change.

R(s)+−ControllerC(s)P or PIProcessG(s)Y(s)
Figure Q4.1 — Common unit-feedback configuration; controller is $K_p$ (P) or $K_p(s+2)/s$ (PI).

Proportional control

  1. Characteristic polynomial. $s^3+22s^2+155s+(350+100K_p)=0$.
  2. Routh $s^1$ row. $\dfrac{22(155)-(350+100K_p)}{22}=\dfrac{3060-100K_p}{22}$. Marginal at $K_p=30.6$.
  3. Oscillation frequency. $22s^2+(350+100\cdot30.6)=0\Rightarrow s^2=-155\Rightarrow\omega_{osc}=\sqrt{155}=12.45\ \text{rad/s}$ (matches the given 12.5). Safe range $\boxed{0\lt K_p\lt 30.6}$.

Proportional + Integral control

  1. Characteristic polynomial. $s(s+5)(s+7)(s+10)+100K_p(s+2)=0\Rightarrow s^4+22s^3+155s^2+(350+100K_p)s+200K_p=0$.
  2. Routh array (marginal condition). With $b_1=\dfrac{3060-100K_p}{22}$, the $s^1$ row vanishes when $b_1(350+100K_p)=22(200K_p)$, i.e. $$10000K_p^2-174200K_p-1\,071\,000=0\Rightarrow \boxed{K_p=22.24}.$$
  3. Oscillation frequency. From the $s^2$ auxiliary row at $K_p=22.24$: $b_1=38.0$, $\omega_{osc}=\sqrt{200(22.24)/38.0}=10.82\ \text{rad/s}$ (matches the given 10.7). Safe range $\boxed{0\lt K_p\lt 22.24}$.
Answer to the qualitative question. The PI system goes unstable at a lower gain ($K_{crit}=22.24$) than the P system ($K_{crit}=30.6$). For the same operating gain the gain margin $K_{crit}/K_{op}$ is therefore smaller: adding the integrator (and its extra $-90^\circ$ of low-frequency phase lag, only partly offset by the zero at $-2$) reduces the gain margin. The table asks "Will the PI gain margin increase?" — the answer is NO.
QuantityP controlPI control
Critical gain $K_{crit}$30.622.24
Oscillation frequency $\omega_{osc}$12.45 rad/s10.82 rad/s
Safe range$0\lt K_p\lt 30.6$$0\lt K_p\lt 22.24$
Gain margin under PI vs P$\boxed{\text{decreases (answer: NO, it does not increase)}}$