Routh–Hurwitz Stability of Three Feedback Loops: 22-Mec-A3 System Analysis and Control, December 2019, Q4

Question 4: Routh–Hurwitz Stability of Three Feedback Loops (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.

Figure 3.0 — the unity-negative-feedback configuration to which all three transfer functions of Question 4 are applied.

Given. The unity-negative-feedback loop of Fig 3.0 with, in turn, $G_a(s)=\dfrac{100}{s\left(s^{2}+8s+24\right)}$, $G_b(s)=\dfrac{3s+1}{s^{2}\left(300s^{2}+600s+50\right)}$ and $G_c(s)=\dfrac{24}{s\left(s+2\right)\left(s+4\right)}$.

Find. Whether each closed loop is stable, and where it is not, how many closed-loop poles lie in the right half plane.

Approach. Form $1+G(s)=0$ in polynomial form for each case, check the necessary condition (all coefficients present and of the same sign), then build the Routh array and count sign changes in its first column.

(a) Form the characteristic polynomial. $$1+\frac{100}{s\left(s^{2}+8s+24\right)}=0 \;\Longrightarrow\; s^{3}+8s^{2}+24s+100=0.$$ All four coefficients are present and positive, so the necessary condition is met and the array must be built.
(a) Build the Routh array. $$\begin{array}{c|cc} s^{3} & 1 & 24\\ s^{2} & 8 & 100\\ s^{1} & \dfrac{8(24)-1(100)}{8}=\dfrac{92}{8}=11.5 & 0\\ s^{0} & 100 & \end{array}$$ The first column is $1,\;8,\;11.5,\;100$ — all positive, no sign changes, so $\boxed{\;\text{(a) is STABLE}\;}$ (roots $s=-6.652$ and $-0.674\pm j3.818$). The margin is not large: the loop would reach the imaginary axis if the numerator gain were raised from 100 to $8\times24=192$.
(b) Form the characteristic polynomial. $$1+\frac{3s+1}{s^{2}\left(300s^{2}+600s+50\right)}=0 \;\Longrightarrow\; 300s^{4}+600s^{3}+50s^{2}+3s+1=0.$$ Again all coefficients are present and positive, so the necessary condition alone decides nothing.
(b) Build the Routh array. $$\begin{array}{c|ccc} s^{4} & 300 & 50 & 1\\ s^{3} & 600 & 3 & 0\\ s^{2} & \dfrac{600(50)-300(3)}{600}=48.5 & \dfrac{600(1)-300(0)}{600}=1 & \\ s^{1} & \dfrac{48.5(3)-600(1)}{48.5}=-9.371 & & \\ s^{0} & 1 & & \end{array}$$ The first column is $300,\;600,\;48.5,\;-9.371,\;1$: two sign changes, therefore $\boxed{\;\text{(b) is UNSTABLE, with 2 RHP poles}\;}$ (the offending pair is $s=+0.0283\pm j0.1073$). Physically the near-free double integrator is destabilised by the very slow plant dynamics; the zero at $s=-1/3$ is too far from the origin to rescue it.
(c) Form the characteristic polynomial. $$1+\frac{24}{s\left(s+2\right)\left(s+4\right)}=0 \;\Longrightarrow\; s^{3}+6s^{2}+8s+24=0.$$
(c) Build the Routh array. $$\begin{array}{c|cc} s^{3} & 1 & 8\\ s^{2} & 6 & 24\\ s^{1} & \dfrac{6(8)-1(24)}{6}=\dfrac{24}{6}=4 & 0\\ s^{0} & 24 & \end{array}$$ All entries positive, no sign changes, so $\boxed{\;\text{(c) is STABLE}\;}$ (roots $s=-5.343$ and $-0.328\pm j2.094$). For a general gain $K$ in place of 24 the $s^{1}$ entry is $(48-K)/6$, so this loop is stable for $0<K<48$ and the printed gain of 24 sits at exactly half the critical value.

Cases (a) and (c) share the classic three-pole shape and are stable because their gains are below the critical values 192 and 48 respectively; case (b) fails not because its gain is large — it is tiny — but because two poles at the origin combined with a heavily damped, very slow quadratic leave the loop with almost no phase to spare.

Question 4 — Routh–Hurwitz results
Case	Characteristic polynomial	Routh first column	Sign changes	Verdict
a	$s^{3}+8s^{2}+24s+100$	$1,\;8,\;11.5,\;100$	0	Stable
b	$300s^{4}+600s^{3}+50s^{2}+3s+1$	$300,\;600,\;48.5,\;-9.371,\;1$	2	Unstable — 2 RHP poles
c	$s^{3}+6s^{2}+8s+24$	$1,\;6,\;4,\;24$	0	Stable

Topic: The Routh–Hurwitz criterion as an algebraic test for right-half-plane roots without factoring the polynomial.

Key relations: Necessary condition — every coefficient present and of the same sign. Sufficient test — the number of RHP roots equals the number of sign changes in the first column of the Routh array. For the cubic $s^{3}+a_2s^{2}+a_1s+a_0$ the condition reduces to $a_2a_1>a_0$ with all coefficients positive.

Why this works: The array is a systematic Euclidean-style reduction of the polynomial; each new row is a linear combination that preserves the root distribution across the imaginary axis, so the sign pattern of the first column counts RHP roots exactly. Because the test is purely algebraic it works symbolically, which is why the same array gives the critical gain ($K<192$ in case (a), $K<48$ in case (c)) as a by-product.

Common pitfall: Concluding stability from the necessary condition alone. All three polynomials here have positive coefficients throughout, yet one of them has two right-half-plane poles. A second trap is dividing a row by a positive constant to simplify (legitimate) versus by a negative one (which corrupts the sign count).

Source: Ogata, Modern Control Engineering, 5th ed., Ch. 5.6; Nise, Control Systems Engineering, 8th ed., Ch. 6.2–6.3; Dorf & Bishop, Modern Control Systems, 13th ed., Ch. 6.

Question 4 of 6: Routh–Hurwitz Stability of Three Feedback Loops

Question 4: Routh–Hurwitz Stability of Three Feedback Loops (25 marks)