Question 5 of 6: Routh–Hurwitz Stability Determination

Question 5: Routh–Hurwitz Stability Determination (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 characteristic polynomials, listed in the table below. As printed, part (c) contains a typographical error — two $s^3$ terms and no $s^2$ term — and is read as $s^3+2s^2+3s+6=0$, the standard third-order form the question is testing (see the callout).

Find. For each polynomial, whether all roots lie in the left half plane, and if not, how many are in the right half plane or on the imaginary axis.

Approach. Build the Routh array for each polynomial and count sign changes in the first column; the number of sign changes equals the number of right-half- plane roots. Where an entire row vanishes, replace it with the coefficients of the derivative of the auxiliary polynomial formed from the row above, and solve the auxiliary polynomial for the imaginary-axis roots.

1. Part (a): construct the array. Writing the alternate coefficients into the first two rows and reducing, $$\begin{array}{c|ccc} s^4 & 1 & 33 & 30\\ s^3 & 10 & 46 & 0\\ s^2 & \frac{10(33)-1(46)}{10}=28.4 & 30 &\\ s^1 & \frac{28.4(46)-10(30)}{28.4}=35.44 & &\\ s^0 & 30 & & \end{array}$$
2. Part (a): interpret. The first column reads $1,\ 10,\ 28.4,\ 35.44,\ 30$ — all strictly positive, so there are no sign changes and no right-half-plane roots. The system is $$\boxed{\text{stable: all four roots in the LHP}}$$ Factoring confirms it: the roots are $s=-5,\ -3,\ -1\pm j1$.
3. Part (b): construct the array. Although every coefficient is positive — a necessary but by no means sufficient condition above second order — the array tells a different story: $$\begin{array}{c|ccc} s^4 & 1 & 3 & 5\\ s^3 & 1 & 2 & 0\\ s^2 & \frac{1(3)-1(2)}{1}=1 & 5 &\\ s^1 & \frac{1(2)-1(5)}{1}=-3 & &\\ s^0 & 5 & & \end{array}$$
4. Part (b): interpret. The first column is $1,\ 1,\ 1,\ -3,\ 5$: it changes sign twice (positive to negative, then negative back to positive), so $$\boxed{\text{unstable: 2 roots in the RHP, 2 in the LHP}}$$ Numerically the roots are $s = 0.4068 \pm j1.4078$ (right half plane) and $s = -0.9068 \pm j1.2272$. This part exists precisely to show that all-positive coefficients guarantee nothing beyond second order.
5. Part (c): the array degenerates. For $s^3+2s^2+3s+6=0$, $$\begin{array}{c|cc} s^3 & 1 & 3\\ s^2 & 2 & 6\\ s^1 & \frac{2(3)-1(6)}{2}=0 & \\ \end{array}$$ The entire $s^1$ row vanishes. That signals a factor whose roots are symmetric about the origin — here an imaginary pair.
6. Part (c): apply the auxiliary-polynomial device. Form the auxiliary polynomial from the row above the zero row and differentiate it: $$A(s)=2s^2+6,\qquad \frac{dA}{ds}=4s$$ Replacing the zero row by the coefficient 4 completes the array with first column $1,\ 2,\ 4,\ 6$ — no sign changes, hence no right-half-plane roots. Solving the auxiliary polynomial gives the symmetric pair directly: $$2s^2+6=0 \;\Longrightarrow\; s = \pm j\sqrt3 = \pm j1.732$$ and dividing it out leaves the remaining root at $s=-2$. The verdict is $$\boxed{\text{marginally stable: a simple pair at } s=\pm j\sqrt3,\ \text{one LHP root at } -2}$$ Because the imaginary pair is simple (not repeated), the natural response contains an undamped sinusoid of constant amplitude at $1.732\ \text{rad/s}$ rather than a growing one — the boundary case, not an exponentially unstable one.