Question 2 of 8: Routh–Hurwitz and root locus of the same plant (compulsory)
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 2014, 3 hours, closed book (approved calculator plus one signed 8.5 × 11" formula sheet). Questions 1 and 2 are compulsory; answer three of the remaining six (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) — Routh–Hurwitz and stability (Ch. 6), steady-state error and error constants (Ch. 7), root locus (Ch. 8), PID and lag/lead design (Ch. 9–11), frequency response, Nyquist and gain/phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of dominant-poles modelling and Mason’s rule; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode/Nyquist plots and signal-flow graphs below are redrawn as inline figures.
Reading the exam figures. Q1 supplies an open-loop Bode plot and Q6 supplies open-loop Bode plots of the process. Where a transfer function is given exactly in the text (Q1’s system is the same plant as Q2, stated exactly there), every graphical reading is also confirmed analytically and the exact model governs. Q6 is a pure graph-reading problem (no closed form given); its numeric reads are approximate and are flagged accordingly. Note the printed exam mislabels the header of the last two pages “May 2013” — this is the May 2014 sitting throughout.
Question 2 — Routh–Hurwitz and root locus of the same plant (compulsory) [20]
Given. $L(s)=K_p\,G(s)$, $G(s)=\dfrac{(s+40)(s+50)}{(s+2)^2(s+6)}$. Poles $-2,-2,-6$; zeros $-40,-50$; relative degree $n-m=1$.
Find. $K_{crit}$, $\omega_{osc}$, safe range(s), and a labelled root locus.
Open-loop pole–zero map: double pole at $-2$, pole at $-6$, zeros at $-40,-50$.
Part 1 — Routh–Hurwitz (exact $K_{crit}$, $\omega_{osc}$, safe range)
Approach. Form the closed-loop characteristic polynomial, build the Routh array, and set the $s^1$ row to zero; the $s^2$-row auxiliary equation gives $\omega_{osc}$.
Characteristic polynomial. With $(s+2)^2(s+6)=s^3+10s^2+28s+24$ and $(s+40)(s+50)=s^2+90s+2000$, $1+K_pG=0$ gives $$s^3+(10+K_p)s^2+(28+90K_p)s+(24+2000K_p)=0.$$
Marginal stability. The $s^1$ numerator must vanish: $(10+K_p)(28+90K_p)-(24+2000K_p)=0$, i.e. $$90K_p^2-1072K_p+256=0\ \Rightarrow\ \boxed{K_{crit}=0.244\ \text{or}\ 11.67}.$$ Two positive roots — matching the two Nyquist crossings of Q1.
Safe range. The $s^1$ entry is positive outside the two roots (the parabola $90K_p^2-1072K_p+256$ opens upward), and the $s^2,s^0$ entries are positive for all $K_p\gt0$. Hence $\boxed{0\lt K_p\lt0.244\ \text{or}\ K_p\gt11.67}$ — identical to Q1.
Part 2 — Root-locus sketch
Approach. Apply the real-axis rule, count asymptotes ($n-m=1\Rightarrow$ one branch to $\infty$ at $180^\circ$), and mark the numerically-found $j\omega$-axis crossings from Part 1.
Root locus (not to scale). The two branches from the double pole at $-2$ break away into the RHP, cross the imaginary axis at $\pm j7.07$ ($K=0.244$), loop back and re-cross at $\pm j32.8$ ($K=11.67$) returning to the LHP, then terminate on the zeros $-40,-50$; the third branch runs left to $-\infty$.
Real-axis segments. Loci lie where an odd number of real poles/zeros sit to the right: the segment $[-40,-6]$ (three singularities to the right) and $(-\infty,-50]$ (five). The double pole at $-2$ has an even count to its immediate left, so those two branches leave the axis at once.
Break-away (rough). The two branches leaving the double pole at $-2$ depart vertically (a double pole splits at $\pm90^\circ$) and bow into the RHP — this is what makes the low-gain closed loop go unstable. A break-in occurs on the far-left real segment near $s\approx-45$ (between the zeros $-40$ and $-50$ region, where two branches rejoin the axis en route to the zeros). No exact computation is required; the shape is set by the two poles at $-2$ and the two distant zeros.
Imaginary-axis crossings. From Part 1 (or $s=j\omega$ substitution) the locus crosses at $\pm j7.07$ ($K=0.244$) and $\pm j32.8$ ($K=11.67$). Between these gains two branches are in the RHP; outside them all roots are in the LHP.
