Question 4 of 6: Routh–Hurwitz Root Counts

Question 4: Routh–Hurwitz Root Counts (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 quartic characteristic polynomials, each to be classified by root location without factorising.

Find. For each equation, the number of roots in the open left half-plane (LHP), the open right half-plane (RHP), and on the imaginary axis.

Approach. Construct the Routh array for each polynomial. The number of sign changes down the first column equals the number of RHP roots. Two special cases arise here and each is handled by its standard device: an entire row of zeros signals $j\omega$-axis roots and is repaired by differentiating the auxiliary polynomial, while a single zero leading an otherwise non-zero row is repaired by the $\epsilon$ substitution.

(a) $s^{4}+2s^{2}+1=0$ — the row-of-zeros case

1. Start the array and hit the zero row. With the odd-power coefficients all absent, the $s^{3}$ row is entirely zero: $$\begin{array}{c|ccc} s^{4} & 1 & 2 & 1\\ s^{3} & 0 & 0 & \\ \end{array}$$ An entire zero row means the polynomial contains an even factor whose roots are symmetric about the origin — the classic signature of poles on the imaginary axis.
2. Differentiate the auxiliary polynomial. The auxiliary polynomial is the row above the zero row, $A(s)=s^{4}+2s^{2}+1$, and its derivative is $dA/ds=4s^{3}+4s$. Replacing the zero row by its coefficients and continuing: $$\begin{array}{c|ccc} s^{4} & 1 & 2 & 1\\ s^{3} & 4 & 4 & \\ s^{2} & 1 & 1 & \\ s^{1} & 0\to2 & &\\ s^{0} & 1 & & \end{array}$$ The $s^{1}$ row is zero again (auxiliary $s^{2}+1$, derivative $2s$), which confirms a repeated imaginary pair.
3. Read the result. The first column is $1,\,4,\,1,\,2,\,1$ — all positive, so there are no sign changes and hence no RHP roots. Since the auxiliary polynomial $s^{4}+2s^{2}+1=(s^{2}+1)^{2}$ accounts for all four roots, $$\boxed{\;\text{(a)}\quad 0\ \text{LHP},\qquad 0\ \text{RHP},\qquad 4\ \text{on the } j\omega \text{ axis } (s=\pm j1\ \text{twice}). \;}$$ The system is unstable in the strict sense: a repeated pair on the imaginary axis produces a response growing as $t\sin t$, not a bounded oscillation.

(b) $s^{4}+s^{3}+5s^{2}+5s+2=0$ — the $\epsilon$ case

1. Build the array until the first column fails. $$\begin{array}{c|ccc} s^{4} & 1 & 5 & 2\\ s^{3} & 1 & 5 & \\ s^{2} & 0 & 2 & \\ \end{array}$$ because $b_{1}=\dfrac{(1)(5)-(1)(5)}{1}=0$ while $b_{2}=\dfrac{(1)(2)-(1)(0)}{1}=2$. Only the leading entry vanished, so this is not a zero row and the auxiliary polynomial device does not apply.
2. Substitute $\epsilon$ and continue. Replace the zero by a small positive $\epsilon$ and complete the array: $$c_{1} = \frac{\epsilon(5)-(1)(2)}{\epsilon} = 5-\frac{2}{\epsilon} \;\xrightarrow[\epsilon\to0^{+}]{}\; -\infty .$$ The first column reads $1,\;1,\;\epsilon\,(>0),\;-2/\epsilon\,(<0),\;2$.
3. Count the sign changes. Going down the column the signs are $+,\,+,\,+,\,-,\,+$: one change from $+$ to $-$ and one from $-$ to $+$, giving two sign changes. Therefore $$\boxed{\;\text{(b)}\quad 2\ \text{LHP},\qquad 2\ \text{RHP},\qquad 0\ \text{on the } j\omega \text{ axis.} \;}$$ No row of zeros appeared, so no roots lie on the imaginary axis and the remaining two roots must be in the left half-plane. The system is unstable.

(c) $s^{4}+2s^{3}+3s^{2}+2s+5=0$ — the routine case

1. Complete the array without incident. $$b_{1}=\frac{(2)(3)-(1)(2)}{2}=2,\qquad b_{2}=\frac{(2)(5)-(1)(0)}{2}=5,\qquad c_{1}=\frac{(2)(2)-(2)(5)}{2}=-3,$$ so the array is $$\begin{array}{c|ccc} s^{4} & 1 & 3 & 5\\ s^{3} & 2 & 2 & \\ s^{2} & 2 & 5 & \\ s^{1} & -3 & & \\ s^{0} & 5 & & \end{array}$$
2. Count the sign changes. The first column is $1,\,2,\,2,\,-3,\,5$: one change into the negative entry and one back out, again two sign changes. Hence $$\boxed{\;\text{(c)}\quad 2\ \text{LHP},\qquad 2\ \text{RHP},\qquad 0\ \text{on the } j\omega \text{ axis.} \;}$$ Note that all five coefficients of (c) are positive — a reminder that positive coefficients are necessary but nowhere near sufficient for stability once the order exceeds two.