Routh–Hurwitz test with a vanishing row: 22-Mec-A3 System Analysis and Control, May 2018, Q3

Question 3: Routh–Hurwitz test with a vanishing row (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. The sixth-order characteristic polynomial

$$\Delta(s)=s^{6}+5s^{5}+14s^{4}+40s^{3}+64s^{2}+80s+96 .$$

Find. Whether the closed loop is stable, and all six roots.

Approach. Build the Routh array; when a row vanishes, form the auxiliary polynomial from the row above, differentiate it to continue the array, and factor the auxiliary polynomial to recover the roots that lie symmetrically about the origin.

Start the array. The first two rows are the alternate coefficients:$$\begin{array}{c|cccc} s^{6} & 1 & 14 & 64 & 96\\ s^{5} & 5 & 40 & 80 & \end{array}$$The $s^{4}$ row follows from the usual $2\times2$ determinants divided by the pivot: $(5\cdot14-1\cdot40)/5=6$, $(5\cdot64-1\cdot80)/5=48$, and the last entry carries down as 96.
The $s^{3}$ row vanishes. Continuing with the $s^{4}$ row as the pivot, $(6\cdot40-5\cdot48)/6=(240-240)/6=0$ and $(6\cdot80-5\cdot96)/6=(480-480)/6=0$. An entire row of zeros signals roots placed symmetrically about the origin — on the imaginary axis, or in real or complex quadruples. Form the auxiliary polynomial from the row above the zero row:$$A(s)=6s^{4}+48s^{2}+96=6\left(s^{4}+8s^{2}+16\right)=6\left(s^{2}+4\right)^{2}.$$Its roots are $s=\pm j2$, each repeated twice — the decisive fact of this question.
Continue the array with $dA/ds$. Replace the zero row by the coefficients of $dA/ds=24s^{3}+96s$ and carry on:$$\begin{array}{c|ccc} s^{4} & 6 & 48 & 96\\ s^{3} & 24 & 96 & \\ s^{2} & 24 & 96 & \\ s^{1} & 48 & & \\ s^{0} & 96 & & \end{array}$$The $s^{1}$ row vanishes as well ($24\cdot96-24\cdot96=0$); its auxiliary polynomial $24s^{2}+96$ has roots $\pm j2$ again, confirming the repetition, and $d/ds\left(24s^{2}+96\right)=48s$ restores the row.
Read the first column. The first column is $1,\ 5,\ 6,\ 24,\ 24,\ 48,\ 96$ — all strictly positive, so there are$$\boxed{\,\text{no roots in the right half-plane}\,}$$That alone does not establish stability: the vanishing rows have already told us that roots sit on the stability boundary, and the Routh test cannot distinguish simple from repeated imaginary roots.
Extract every root. The auxiliary polynomial divides the characteristic polynomial exactly. Dividing $\Delta(s)$ by $\left(s^{2}+4\right)^{2}=s^{4}+8s^{2}+16$ leaves $s^{2}+5s+6=(s+2)(s+3)$, so$$\Delta(s)=(s+2)(s+3)\left(s^{2}+4\right)^{2}$$$$\boxed{\,s=-2,\quad s=-3,\quad s=\pm j2\ \text{(each twice)}\,}$$Multiplying the factors back out reproduces $1,5,14,40,64,80,96$ exactly, which verifies the factorisation.
Judge the stability. Two roots are in the left half-plane and four are on the imaginary axis as a repeated pair. A simple pair $\pm j\omega$ would give a bounded, non-decaying oscillation (marginal stability); a repeated pair contributes a term of the form $t\sin 2t$, whose amplitude grows without bound. Therefore$$\boxed{\,\text{the system is UNSTABLE}\,}$$even though the Routh first column shows no sign changes.

Physically the loop contains an undamped resonance at 2 rad/s that the feedback reinforces rather than damps: excite the system anywhere near that frequency and the response ramps up linearly in time. Any real design would need either damping added at that frequency or a notch/lead network to pull those roots off the axis.

Quantity	Result
Routh first column	$1,\ 5,\ 6,\ 24,\ 24,\ 48,\ 96$ (no sign changes)
Right-half-plane roots	0
Auxiliary polynomial	$6\left(s^{2}+4\right)^{2}$
Factorisation	$(s+2)(s+3)\left(s^{2}+4\right)^{2}$
Roots	$-2$, $-3$, $+j2$ (double), $-j2$ (double) — six roots in all
Verdict	Unstable — repeated imaginary roots give a $t\sin 2t$ term

Topic: Routh–Hurwitz with a premature zero row, and the difference between marginal stability and instability on the imaginary axis.

Key relations: Routh array construction; auxiliary polynomial $A(s)$ taken from the row above a zero row; continuation with $dA/ds$; number of RHP roots = number of sign changes in the first column; a repeated $j\omega$ pair inverts to $t\sin\omega t$.

Why this works: A zero row appears exactly when the polynomial contains a factor whose roots are symmetric about the origin, and that factor is the auxiliary polynomial. Because those roots are already known once $A(s)$ is factored, the rest of the array only has to account for the remaining factor — here $(s+2)(s+3)$. The sign-change count is a statement about the right half-plane only, so a first column that is entirely positive proves nothing about roots sitting exactly on the boundary; their multiplicity has to be inspected separately, and multiplicity two turns bounded oscillation into unbounded growth.

Common pitfall: Declaring the system stable, or 'marginally stable', because the first column has no sign changes. Both readings miss the repetition: $\left(s^{2}+4\right)^{2}$, not $\left(s^{2}+4\right)$, is what the auxiliary polynomial delivers, and the squared factor is what makes the response diverge.

Source: Dorf & Bishop, Modern Control Systems, 14th ed. §6.2 (Routh–Hurwitz, special cases); Nise, Control Systems Engineering, 8th ed. §6.3–6.4 (zero row, stability from the auxiliary polynomial).

Question 3 of 6: Routh–Hurwitz test with a vanishing row

Question 3: Routh–Hurwitz test with a vanishing row (25 marks)