Routh Stability of a Quartic Characteristic Equation: 23-Chem-A6 Process Dynamics and Control, May 2016, Q3

Problem 3: Routh Stability of a Quartic Characteristic Equation (20%)

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. A fourth-order characteristic polynomial with the gain in the constant term:

Power	$s^4$	$s^3$	$s^2$	$s^1$	$s^0$
Coefficient	$1$	$4$	$6$	$4$	$1+K$

Find. (a) the stabilising range of $K$; (b) the marginal gain(s) and the frequency of sustained oscillation.

Approach. All coefficients are positive only if $1+K>0$; the decisive test is the Routh array—require every first-column entry positive, and locate the stability limit where an entry vanishes (an auxiliary polynomial then gives the imaginary-axis crossing).

Build the Routh array. Lay out the first two rows from the coefficients and reduce: $$\begin{array}{c|ccc} s^4 & 1 & 6 & 1+K\\ s^3 & 4 & 4 & 0\\ s^2 & b_1 & 1+K & \\ s^1 & c_1 & & \\ s^0 & 1+K & & \end{array}$$ with $b_1=\dfrac{4\cdot6-1\cdot4}{4}=5$.
Compute the $s^1$ entry. $$c_1=\frac{b_1\cdot4-4\,(1+K)}{b_1}=\frac{5\cdot4-4(1+K)}{5}=\frac{16-4K}{5}.$$ The first column is therefore $\{1,\,4,\,5,\,\tfrac{16-4K}{5},\,1+K\}$.
(a) Stability conditions. Every first-column entry must be positive. The two binding inequalities are $$\frac{16-4K}{5}>0\ \Rightarrow\ K<4,\qquad 1+K>0\ \Rightarrow\ K>-1.$$ Hence $$\boxed{-1<K<4.}$$
(b) Upper limit $K=4$. At $K=4$ the $s^1$ entry vanishes, signalling a pair of poles on the imaginary axis. Form the auxiliary polynomial from the $s^2$ row: $5s^2+(1+K)=5s^2+5=0\Rightarrow s^2=-1\Rightarrow s=\pm j$. The loop sustains an undamped oscillation at $$\boxed{K=4,\qquad \omega_u=1\ \text{rad/min}.}$$
(b) Lower limit $K=-1$. At $K=-1$ the constant term $1+K=0$, so $s=0$ is a root—a pole migrates through the origin (monotonic, not oscillatory, instability). Physically only positive gains are used, so the operative marginal gain is $K=4$.

Problem 3: the Routh test bounds the gain to $-1<K<4$. The upper edge $K=4$ is an oscillatory (imaginary-axis) limit at $\omega_u=1$ rad/min; the lower edge $K=-1$ is a root crossing through the origin.

Quantity	Value
(a) Stable range	$-1<K<4$
(b) Marginal (oscillatory) gain	$K=4$
(b) Ultimate frequency	$\omega_u=1$ rad/min
(b) Lower marginal gain	$K=-1$ (root at origin)

Concept & Theory

Topic. Routh–Hurwitz stability test; marginal stability and the auxiliary polynomial; ultimate gain and frequency.

Key relations. A polynomial is Hurwitz iff every Routh first-column entry is positive; a zero entry with a non-trivial auxiliary polynomial $A(s)$ gives imaginary-axis roots at $A(s)=0$.

Why this works. The Routh construction counts sign changes in the first column, which equals the number of right-half-plane roots. When an entry passes through zero, a root pair is exactly on the imaginary axis—the boundary between stable and unstable—and the row above supplies the auxiliary equation whose roots are that crossing frequency.

Common pitfall. Quoting only the upper bound; a negative gain can also destabilise (here $K=-1$). Also, a positive-coefficient check alone is necessary but not sufficient for order $>2$—the full array is required.

Source. Stephanopoulos, Ch. 15 (Routh test); Seborg et al., Ch. 11 (stability of closed-loop systems).

Question 3 of 8: Routh Stability of a Quartic Characteristic Equation

Problem 3: Routh Stability of a Quartic Characteristic Equation (20%)