22-Mec-A3 System Analysis and Control · December 2017

Question 2 of 6: Routh Stability — Range of Gain and an Open-Loop Unstable Plant

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

Notes on this paper

Paper format. National Exams, December 2017 — 16-Mec-A3, System Analysis and Control. Three hours, closed book; a Casio or Sharp approved calculator and semi-log graph paper are the only permitted aids. Six questions are printed; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied as page 5. All six questions are solved here.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Prentice Hall) — the source of this paper's problem style; N. S. Nise, Control Systems Engineering, 7th ed. (Wiley); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson).

Check: conventions used throughout. All transfer functions are written in the Laplace variable s with zero initial conditions unless a question states otherwise. “Unity feedback” means $H(s)=1$, so the closed-loop transfer function is $T(s)=G(s)/[1+G(s)]$ and the error signal is $E(s)=R(s)-C(s)$. Root-locus gains are the gain $K$ exactly as it appears in the printed $G(s)$.

Question 2: Routh Stability — Range of Gain and an Open-Loop Unstable Plant (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. Part (a): a type-1 loop with adjustable gain $K$ and open-loop poles at $s=0,\,-1,\,-2$. Part (b): a fixed-gain loop with poles at $s=0,\,+1,\,-3/2$ — note the right-half-plane open-loop pole at $s=+1$, which makes this an open-loop unstable plant.

Find. (a) all values of $K$ for which the closed loop is stable, together with the frequency at which the roots cross the imaginary axis; (b) a definite yes/no verdict on the stability of the second loop, supported by the Routh array.

Approach. In both parts form the closed-loop characteristic polynomial $1+G(s)=0$, clear denominators, and apply the Routh–Hurwitz criterion: the loop is stable if and only if every entry in the first column of the array is positive, and the number of sign changes equals the number of closed-loop poles in the right half plane.

Form the characteristic equation for part (a). With unity feedback the closed-loop poles satisfy $1+G(s)=0$, i.e. $$s(s+1)(s+2)+K=0\;\Longrightarrow\;s^3+3s^2+2s+K=0$$ All coefficients must be positive for stability, which already requires $K>0$ — a necessary condition, but for a third-order polynomial not a sufficient one.
Build the Routh array. Writing the coefficients in the standard interleaved pattern: $$\begin{array}{c|cc} s^3 & 1 & 2\\ s^2 & 3 & K\\ s^1 & \dfrac{3(2)-1(K)}{3}=\dfrac{6-K}{3} & 0\\ s^0 & K & \end{array}$$
Impose positivity of the first column. The entries are $1,\;3,\;(6-K)/3,\;K$. The first two are positive automatically; the remaining two require $$\frac{6-K}{3}>0\;\Rightarrow\;K<6,\qquad\text{and}\qquad K>0$$ so that $$\boxed{\;0<K<6\;}$$
Locate the imaginary-axis crossing. At the upper limit $K=6$ the $s^1$ row vanishes, and the auxiliary polynomial is read from the row above it: $$A(s)=3s^2+K=3s^2+6=0\;\Longrightarrow\;s^2=-2\;\Longrightarrow\;s=\pm j\sqrt{2}=\pm j1.4142$$ So at $K=6$ the loop sustains an undamped oscillation at $\omega=1.414\ \text{rad/s}$; the third root is at $s=-3$. Substituting $K=5.9$ and $K=6.1$ into the cubic confirms numerically that the pair crosses from the left half plane into the right half plane as $K$ passes 6.
Now part (b): form its characteristic equation. Expanding the open-loop denominator first, $$s(s-1)(2s+3)=s\left(2s^2+s-3\right)=2s^3+s^2-3s$$ so $1+G(s)=0$ gives $$\boxed{\;2s^3+s^2-3s+10=0\;}$$ The coefficient of $s$ is negative while the others are positive. Because a necessary condition for stability is that all coefficients of the characteristic polynomial share the same sign, the system can be declared unstable at this point — but the array tells us how unstable.
Complete the Routh array for part (b). $$\begin{array}{c|cc} s^3 & 2 & -3\\ s^2 & 1 & 10\\ s^1 & \dfrac{1(-3)-2(10)}{1}=-23 & 0\\ s^0 & 10 & \end{array}$$ The first column is $2,\;1,\;-23,\;10$: it changes sign twice (positive to negative, then negative to positive). $$\boxed{\;\text{Two closed-loop poles lie in the right half plane}\;\Rightarrow\;\text{the system is UNSTABLE}\;}$$ Solving the cubic numerically confirms the count: the roots are approximately $s=-1.7297$ and $s=+0.6149\pm j1.6180$, i.e. one stable real root and an unstable oscillatory pair.

It is worth being explicit about why part (b) has no rescue. The gain is fixed at 10, so there is no free parameter to adjust; and even if the numerator gain were treated as variable, the negative $s$-coefficient of $2s^3+s^2-3s+\text{(gain)}$ is contributed entirely by the plant, so no pure gain can remove it. Stabilising this plant requires dynamic compensation — a lead network or a state-feedback design that reshapes the coefficient itself.

Quantity	Result
(a) Characteristic equation	$s^3+3s^2+2s+K=0$
(a) Stability range	$0<K<6$
(a) Imaginary-axis crossing	$s=\pm j\sqrt{2}=\pm j1.4142$ at $K=6$
(b) Characteristic equation	$2s^3+s^2-3s+10=0$
(b) Routh first column	$2,\;1,\;-23,\;10$ — two sign changes
(b) Verdict	Unstable; 2 poles in the RHP ($+0.615\pm j1.618$)

Topic. The Routh–Hurwitz criterion: determining absolute stability and the admissible gain range from the characteristic polynomial, without computing the roots.

Key relations. Closed-loop characteristic equation $1+G(s)H(s)=0$; Routh array construction $b_i=(a_1a_{2i}-a_0a_{2i+1})/a_1$; stability $\iff$ all first-column entries positive; number of RHP roots $=$ number of first-column sign changes; auxiliary polynomial from the row above a vanishing row gives the imaginary-axis crossing frequency.

Why this works. The Routh array is a systematic Sturm-like test that counts sign changes of a sequence built from the polynomial coefficients, and that count equals the number of roots with positive real part. This is exactly the information a designer needs, and it is obtained with a few divisions rather than a root-finding routine — which matters here, since part (a)'s polynomial contains the unknown $K$ symbolically and could not be factored at all. The vanishing-row condition has a direct physical meaning: it is the gain at which a pole pair sits precisely on the imaginary axis, so the loop is a sustained oscillator, and the auxiliary polynomial delivers that oscillation frequency for free.

Common pitfall. The most common error is checking only that all coefficients are positive and declaring the system stable. That condition is necessary but not sufficient for order three and above — in part (a) the polynomial $s^3+3s^2+2s+K$ has all-positive coefficients for every $K>0$, yet the loop is unstable for $K>6$. The mirror-image error is to see the negative coefficient in part (b) and stop without the array, thereby losing the marks for the number of unstable roots. A third trap is expanding $s(s-1)(2s+3)$ carelessly: the leading coefficient is 2, not 1, and dividing through by it at the wrong moment corrupts the array.

Source. Ogata, Modern Control Engineering, 5th ed., §5–7 (Routh's stability criterion) and Example 5–12; Nise, Control Systems Engineering, 7th ed., Ch. 6.

← Question 1 Next: Question 3 →

Quantity	Result
(a) Characteristic equation	\(s^3+3s^2+2s+K=0\)
(a) Stability range	\(0<K<6\)
(a) Imaginary-axis crossing	\(s=\pm j\sqrt{2}=\pm j1.4142\) at \(K=6\)
(b) Characteristic equation	\(2s^3+s^2-3s+10=0\)
(b) Routh first column	\(2,\;1,\;-23,\;10\) — two sign changes
(b) Verdict	Unstable; 2 poles in the RHP (\(+0.615\pm j1.618\))