22-Mec-A3 System Analysis and Control · May 2016

Question 6 of 6: Counting Right-Half-Plane Roots with Routh's Criterion

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Mec-A3 System Analysis and Control, National Exams May 2016 — 3 hours, closed book (Casio or Sharp approved calculator; semi-log graph paper is the only permitted aid). The rubric states that any four (4) questions constitute a complete paper and that all questions are of equal value, so each question is worth 25 marks of the 100 marked. Because this document is a study resource rather than an exam script, all six questions are solved in full.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Pearson) — Ch. 3–7; N. S. Nise, Control Systems Engineering, 8th ed. (Wiley) — Ch. 4, 6, 8, 10; R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson) — Ch. 5–8; G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson) — Ch. 3–6. A table of Laplace transform pairs is appended to the examination paper (pages 3–4 of the original), together with two sheets of 3-cycle semi-logarithmic graph paper for the Bode question.

Question 6: Counting Right-Half-Plane Roots with Routh's Criterion (25 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. Three characteristic polynomials with numerical coefficients:

Part	Polynomial	Degree	All coefficients positive?
(a)	$s^{4}+8s^{3}+32s^{2}+80s+100$	4	Yes
(b)	$s^{5}+10s^{4}+30s^{3}+80s^{2}+344s+480$	5	Yes
(c)	$s^{4}+2s^{3}+7s^{2}-2s+8$	4	No — the $s^{1}$ coefficient is $-2$

Find. The number of roots with positive real part for each polynomial.

Approach. Build the Routh array for each polynomial and count the sign changes down the first column; by Routh's theorem that count equals the number of roots in the open right half-plane. Part (c) can be pre-judged by the coefficient test, but the array is still needed to say how many.

Part (a)

Construct the array. The first two rows are the alternate coefficients; each later row follows from the two above it:$$\begin{array}{c|ccc}s^{4}&1&32&100\\ s^{3}&8&80&0\\ s^{2}&22&100&\\ s^{1}&43.64&&\\ s^{0}&100&&\end{array}$$with $b_{1}=\dfrac{8(32)-1(80)}{8}=22$, $b_{2}=\dfrac{8(100)-1(0)}{8}=100$, and $c_{1}=\dfrac{22(80)-8(100)}{22}=\dfrac{960}{22}=43.64$.
Count sign changes. The first column is $1,\;8,\;22,\;43.64,\;100$ — all strictly positive, so$$\boxed{\;\text{(a) zero roots with positive real part; the polynomial is stable}\;}$$Factoring confirms it: $s^{4}+8s^{3}+32s^{2}+80s+100=(s^{2}+2s+10)(s^{2}+6s+10)$, whose roots are $-1\pm j3$ and $-3\pm j1$, all comfortably in the left half-plane.

Part (b)

Construct the array for the fifth-order polynomial.$$\begin{array}{c|ccc}s^{5}&1&30&344\\ s^{4}&10&80&480\\ s^{3}&22&296&\\ s^{2}&-54.55&480&\\ s^{1}&489.6&&\\ s^{0}&480&&\end{array}$$The entries follow in the usual way: $\dfrac{10(30)-1(80)}{10}=22$ and $\dfrac{10(344)-1(480)}{10}=296$ on the $s^{3}$ row; then $\dfrac{22(80)-10(296)}{22}=\dfrac{-1200}{22}=-54.55$ on the $s^{2}$ row; then $\dfrac{-54.55(296)-22(480)}{-54.55}=489.6$.
Count sign changes. The first column reads $1,\;10,\;22,\;-54.55,\;489.6,\;480$: it goes positive→negative once and negative→positive once, so there are two sign changes and$$\boxed{\;\text{(b) two roots with positive real part}\;}$$The polynomial in fact factors exactly, $s^{5}+10s^{4}+30s^{3}+80s^{2}+344s+480=(s^{2}-2s+10)(s+2)(s+4)(s+6)$, so the actual roots are $1\pm j3$ (the unstable pair) together with $-2$, $-4$ and $-6$ — two in the right half-plane, exactly as counted. Note that all coefficients being positive was never sufficient; it is only necessary.

Part (c)

Apply the coefficient test first. The $s^{1}$ coefficient is $-2$ while the others are positive. A Hurwitz (all-roots-in-LHP) polynomial cannot have a negative or missing coefficient, so instability is guaranteed before any arithmetic is done. The array is still required to count the offending roots.
Construct the array.$$\begin{array}{c|ccc}s^{4}&1&7&8\\ s^{3}&2&-2&0\\ s^{2}&8&8&\\ s^{1}&-4&&\\ s^{0}&8&&\end{array}$$with $\dfrac{2(7)-1(-2)}{2}=8$, $\dfrac{2(8)-1(0)}{2}=8$, and $\dfrac{8(-2)-2(8)}{8}=\dfrac{-32}{8}=-4$.
Count sign changes. The first column is $1,\;2,\;8,\;-4,\;8$: two sign changes, hence$$\boxed{\;\text{(c) two roots with positive real part}\;}$$Direct root-finding gives $0.3138\pm j0.9771$ in the right half-plane and $-1.3138\pm j2.4228$ in the left, confirming the count.

Taken together the three parts make the examiner's point efficiently: part (a) shows a stable case where the array must be carried to the end, part (b) shows that positive coefficients guarantee nothing for orders above two, and part (c) shows the shortcut of the coefficient test — useful for a verdict, useless for a count.

Part	Routh first column	Sign changes	RHP roots	Actual roots
(a)	$1,\;8,\;22,\;43.64,\;100$	0	0 (stable)	$-1\pm j3$, $-3\pm j1$
(b)	$1,\;10,\;22,\;-54.55,\;489.6,\;480$	2	2	$1\pm j3$, $-2$, $-4$, $-6$
(c)	$1,\;2,\;8,\;-4,\;8$	2	2	$0.314\pm j0.977$, $-1.314\pm j2.423$

Topic. The Routh–Hurwitz criterion as a root-counting tool rather than a yes/no stability test.

Key relations. Array entries $b_{i}=\dfrac{a_{1}a_{2i}-a_{0}a_{2i+1}}{a_{1}}$ formed from the two rows above; number of RHP roots $=$ number of sign changes in the first column; necessary condition for stability: all coefficients present and of the same sign; special cases — a zero in the first column with a non-zero row (replace by $\epsilon$) and an entirely zero row (differentiate the auxiliary polynomial).

Why this works. The Routh array is a systematic Sturm-like sequence: each row is a remainder in a Euclidean-style division of the even and odd parts of the polynomial, and the Cauchy index of the resulting continued fraction counts encirclements of the origin by the polynomial as $s$ traverses the imaginary axis. The practical consequence is that the array delivers the number of unstable roots without factoring anything — which is what makes it viable in a closed-book examination and, historically, what made it viable before computers. It is also the reason the criterion generalises so naturally to a symbolic gain, as in Question 3: the same first-column positivity conditions become inequalities in $K$.

Common pitfall. Stopping at the coefficient test. In part (b) every coefficient is positive, yet the polynomial has two right-half-plane roots; positivity is necessary but sufficient only for first- and second-order polynomials. The other frequent errors are dropping the trailing zero when a row is short (which corrupts the next row) and mis-signing the cross-multiplication — the numerator is always (pivot of the row above)×(next entry two rows up) minus (leading entry two rows up)×(next entry one row up), and it is worth writing the pattern out once at the top of the answer.

Source. Ogata, Modern Control Engineering 5th ed., §5-6 (Routh's stability criterion, including the special cases); Nise, Control Systems Engineering 8th ed., §6.2–6.4; Dorf & Bishop, 13th ed., §6.2.

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