Questions: Routh-Hurwitz Stability Test: Algorithm and Application

Question 1

After constructing the Routh tableau for a 5th-order system, the first column reads: 2, 3, −1, 4, 1. How many poles does the system have in the right half-plane?

• A. Zero — the first column contains no zeros
• B. One — there is one negative entry in the first column
• C. Two — there are two sign changes in the first column (2→3→−1 and −1→4)
• D. Three — there are three positive entries and two sign-change boundaries

Answer: (C)

The Routh-Hurwitz rule counts sign changes in the first column, not negative entries. The sequence 2, 3, −1, 4, 1 has two sign changes: 3→−1 (positive to negative) and −1→4 (negative to positive). Each sign change corresponds to exactly one root in the right half-plane. So the system has two RHP poles and is unstable. A common mistake is counting the single negative entry (option B) — but the theorem is about transitions, not counts of negative values.

Question 2

While constructing the Routh tableau for a closed-loop system with variable gain K, you find that an entire row becomes identically zero for K = 5. What does this indicate, and how do you proceed?

• A. The system is stable at K = 5; an all-zero row means no remaining poles need to be checked
• B. The characteristic polynomial has roots on the imaginary axis at K = 5 — marginal stability; form the auxiliary polynomial from the row above, differentiate it, substitute its coefficients for the zero row, and continue
• C. The tableau calculation is incorrect; no valid characteristic polynomial produces an all-zero row
• D. The gain K = 5 stabilizes the system completely and the all-zero row confirms all remaining poles are in the left half-plane

Answer: (B)

An all-zero row means the characteristic polynomial has a symmetric factor — poles that come in pairs symmetric about the origin, such as ±jω pairs on the imaginary axis. This is marginal stability (poles on the stability boundary), not stability. The procedure is to form the auxiliary polynomial from the row just above the all-zero row, differentiate it with respect to s, use the derivative's coefficients to replace the zero row, and complete the tableau. The marginal gain K = 5 is exactly the answer to 'at what gain does the system first become marginally stable?' — a standard root-locus design question.

Question 3

A system is stable if and only if all entries in the first column of its Routh tableau are positive.

• T. True
• F. False

Answer: True

This is a restatement of the Routh-Hurwitz stability criterion for real-coefficient polynomials. Zero sign changes in the first column means all poles are in the left half-plane, which is exactly the stability condition for a continuous-time system. The test is both necessary and sufficient (absent the special zero-row cases, which indicate marginal stability rather than strict stability). All positive first-column entries implies no RHP poles.

Question 4

The Routh-Hurwitz test determines stability by finding the roots of the characteristic polynomial and checking whether they lie in the left half-plane.

• T. True
• F. False

Answer: False

This is precisely what the Routh-Hurwitz test avoids. Its main practical value is that it determines stability — specifically, the number of right-half-plane poles — using only arithmetic operations on the polynomial coefficients, without computing any roots at all. Finding roots of high-degree polynomials by hand is tedious and error-prone; the Routh tableau reduces the problem to a structured table calculation. The number of sign changes in the first column gives the number of RHP roots directly. Root-finding and Routh analysis answer the same question by completely different means.

Question 5

What does a sign change in the first column of the Routh tableau represent, and why does counting sign changes give the exact number of right-half-plane poles?

Model answer: Each sign change in the first column corresponds to exactly one root crossing from the left half-plane to the right half-plane in the complex s-plane. The Routh-Hurwitz criterion is derived from Cauchy's argument principle and Sturm sequences: the number of times the first column changes sign equals the number of roots of the characteristic polynomial with positive real part. This is a theorem, not a heuristic — the test is both necessary and sufficient, so zero sign changes guarantees all poles are in the LHP (stable) and n sign changes means exactly n unstable poles.

The deep reason is algebraic: the Routh algorithm constructs a Sturm sequence for the characteristic polynomial, and the sign changes in the leading coefficients of a Sturm sequence count real roots in an interval — here adapted to count complex roots with positive real part. The engineering payoff is enormous: stability analysis of an arbitrary-order system reduces to filling in a table by hand. No eigenvalue solver, no factoring, no numerical root-finding — just arithmetic and sign inspection.