22-Mec-A3 System Analysis and Control · Undated paper
Question 1 of 6: Routh–Hurwitz stability of two characteristic equations
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mec-A3 System Analysis and Control, National Exam, May 2019 sitting. Three hours, closed book, Casio or Sharp approved calculator only. Six questions; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied with the paper. Every one of the six questions is worked below.
Reference texts. Ogata, Modern Control Engineering, 5th ed. (Chs. 5–8); Nise, Control Systems Engineering, 8th ed. (Chs. 4, 6, 8, 10); Dorf & Bishop, Modern Control Systems, 13th ed.; Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed.
Question 1: Routh–Hurwitz stability of two characteristic equations (25 marks)
Given. Two closed-loop characteristic polynomials in the variable $p$, with all coefficients real and positive. The paper instructs that limited (marginal) stability is to be reported as instability, so any root on or to the right of the imaginary axis condemns the system.
Find. For each polynomial, the number of roots in the right half-plane and hence a stable / unstable verdict.
Approach. Build the Routh array for each polynomial and count the sign changes down the first column; each change is one right-half-plane root, and the factorisation of the polynomial is then used as an independent check.
Write the Routh array for (a). The coefficients alternate into the first two rows, $1,\ 8,\ 3$ and $4,\ 8,\ 0$, and every later entry is the usual $2\times2$ determinant divided by the pivot immediately above it:
$$b_1=\frac{4\cdot 8-1\cdot 8}{4}=6,\quad b_2=\frac{4\cdot 3-1\cdot 0}{4}=3,\quad c_1=\frac{6\cdot 8-4\cdot 3}{6}=6,\quad d_1=3$$
Collecting these gives the complete array for part (a):
Row
Column 1
Column 2
Column 3
$p^4$
1
8
3
$p^3$
4
8
0
$p^2$
6
3
$p^1$
6
$p^0$
3
Read the verdict for (a). The first column is $1,\ 4,\ 6,\ 6,\ 3$ — every entry positive, so there are no sign changes and no right-half-plane roots:
$$\boxed{p^4+4p^3+8p^2+8p+3=0 \text{ is STABLE (0 RHP roots)}}$$
Confirm by factorisation. Trial division shows $p=-1$ is a repeated root, and the polynomial factors exactly as
$$p^4+4p^3+8p^2+8p+3=(p+1)^2\,(p^2+2p+3)$$
whose roots are $p=-1$ (twice) and $p=-1\pm j\sqrt{2}$. All four sit strictly in the left half-plane, in agreement with the array. The dominant pair has $\zeta = 1/\sqrt{3}=0.577$ and $\omega_n=\sqrt{3}=1.732$ rad/s, so the response is well damped.
Build the array for (b). Starting from the rows $1,\ 3,\ 1$ and $2,\ 8,\ 4$, the third row already goes negative:
$$b_1=\frac{2\cdot 3-1\cdot 8}{2}=-1,\quad b_2=\frac{2\cdot 1-1\cdot 4}{2}=-1$$
and continuing with these pivots gives $c_1=\dfrac{(-1)(8)-2(-1)}{-1}=6$, $c_2=\dfrac{(-1)(4)-2(0)}{-1}=4$, then $d_1=\dfrac{6(-1)-(-1)(4)}{6}=-\dfrac{1}{3}$ and finally $e_1=4$.
The array for part (b) is therefore
Row
Column 1
Column 2
Column 3
$p^5$
1
3
1
$p^4$
2
8
4
$p^3$
−1
−1
$p^2$
6
4
$p^1$
−1/3
$p^0$
4
Count the sign changes in (b). The first column reads $+1,\ +2,\ -1,\ +6,\ -\tfrac13,\ +4$. The sign flips four times (plus to minus, minus to plus, plus to minus, minus to plus), so
$$\boxed{p^5+2p^4+3p^3+8p^2+p+4=0 \text{ is UNSTABLE, with 4 RHP roots}}$$
Cross-check numerically. Solving the quintic gives roots at approximately $-1.906$, $0.317\pm j1.545$ and $0.636\pm j0.615$: one root in the left half-plane and four in the right, exactly as the array predicted. Note that all five coefficients of the polynomial are positive — a necessary condition for stability that is nowhere near sufficient above second order, which is precisely the trap this part sets.
