A control system has the characteristic polynomial s³ + 6s² + 11s + 6. All coefficients are positive. A student concludes the system must be stable. Is the student correct, and why or why not?
AYes — positive coefficients guarantee all roots are in the left half-plane for any degree polynomial
BNo — positive coefficients are necessary but not sufficient for degree ≥ 3; the Routh array must be constructed to confirm stability
CNo — a third-order system with all positive coefficients always has at least one right-half-plane root
DYes for this specific polynomial, but not in general — degree-3 systems are a special exception where positive coefficients are sufficient
All positive coefficients is a necessary condition for stability (any negative or zero coefficient immediately implies an unstable root), but it is not sufficient for polynomials of degree 3 or higher. A degree-4 polynomial like s⁴ + s³ + 2s² + s + 1 has all positive coefficients yet is unstable. The Routh array must be completed and the first column checked for sign changes. For this specific polynomial (s³ + 6s² + 11s + 6 = (s+1)(s+2)(s+3)), it happens to be stable, but the student cannot know that from coefficients alone without building the array.
Question 2 Multiple Choice
While constructing a Routh array, you find that the first-column entry in row 3 is zero, but the row contains other nonzero entries. What is the correct next step?
ADeclare the system unstable immediately — any zero in the first column means there is a right-half-plane pole
BSubstitute a small positive number ε for the zero, complete the array symbolically, and count sign changes as ε → 0⁺
CReplace the zero row with the derivative of the auxiliary polynomial formed from the row above
DDeclare the system marginally stable — a zero in the first column (with remaining nonzero entries) always corresponds to a purely imaginary root
A zero in the first column (with other nonzero entries in the same row) requires the ε substitution. You cannot directly divide by zero in the next row calculation, so replace the first-column zero with ε > 0, complete the array, then examine the sign of the resulting expressions as ε → 0⁺. If sign changes occur in the limit, there are right-half-plane roots. The auxiliary polynomial method (option C) is used when an ENTIRE row is zero, which indicates a different special case — symmetric root distribution. Options A and D are wrong: a first-column zero is ambiguous without completing the array.
Question 3 True / False
A polynomial with most positive coefficients is very likely to be a stable characteristic polynomial — meaning most its roots have negative real parts.
TTrue
FFalse
Answer: False
This is the most common misconception when first learning Routh-Hurwitz. Positive coefficients are necessary for stability (a negative or missing coefficient immediately implies instability) but are not sufficient for degree 3 or higher. The classic counterexample is s⁴ + s³ + 2s² + s + 1, which has all positive coefficients but contains right-half-plane roots (verifiable by the Routh array). The Routh-Hurwitz criterion exists precisely because the coefficient sign test is insufficient — you need the full array to determine whether the more subtle stability conditions are satisfied.
Question 4 True / False
An all-zero row appearing in the Routh array during a gain-sweep problem typically means the system has become marginally stable at that gain value — meaning the closed-loop poles are on the imaginary axis.
TTrue
FFalse
Answer: True
An all-zero row means the characteristic polynomial has a symmetric factor — roots that are symmetric about the origin. During a gain sweep, this most commonly means the roots that started in the left half-plane have migrated to the imaginary axis at the critical gain value. The auxiliary polynomial (formed from the row above the zero row) can be solved to find the exact imaginary-axis root locations and the corresponding critical gain. This is, in fact, a useful feature: the all-zero row condition gives you the stability boundary directly, telling you the exact gain at which the system transitions from stable to unstable.
Question 5 Short Answer
What is the Routh-Hurwitz criterion actually counting, and why is this especially useful when designing systems with a free gain parameter K?
Think about your answer, then reveal below.
Model answer: The number of sign changes in the first column of the completed Routh array equals the number of closed-loop poles in the right half of the s-plane (unstable poles). For stability, you need zero sign changes — all first-column entries must share the same sign. When the characteristic polynomial contains a symbolic gain K, the first-column entries become algebraic expressions in K. Setting the condition 'all entries positive' yields a system of inequalities whose solution is the stability range for K — a closed-form answer. This avoids numerically solving for poles at every candidate K value.
This is the practical payoff that makes Routh-Hurwitz a design tool, not just an analysis tool. A designer asked 'for what values of K is this closed-loop system stable?' can write the characteristic polynomial with K symbolic, build the Routh array, express the first-column entries as functions of K, and solve the resulting inequalities. The answer is exact and algebraic. Root locus gives similar information graphically, but Routh-Hurwitz gives it analytically — particularly useful for verifying stability bounds when the polynomial degree is high or when exact pole locations are not needed.