What Do You Do When a Routh Array Has a Zero in the First Column?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

If only the first entry of a row is zero, replace it with a small positive number ε, finish the array, and count sign changes in the first column as ε shrinks to zero. If the whole row is zero, the polynomial has roots placed symmetrically about the origin, so form an auxiliary polynomial from the row above and replace the zero row with the coefficients of its derivative. For s⁴ + s³ + 2s² + 2s + 3 the ε method gives two sign changes, and the roots 0.406 ± j1.293 confirm two right-half-plane roots.

What you'll learn

  • Why a zero in the first column stops the ordinary Routh calculation
  • The ε method, worked through in full, plus the multiply-by-(s + 1) cross-check
  • What a whole row of zeros means and how the auxiliary polynomial handles it
  • Two row-of-zeros examples: one marginal, one with a root in the right half-plane
  • How each answer was confirmed against the polynomial's actual roots

Conventions and the basic rule

Write the characteristic polynomial with its highest power first. The first two rows of the array take alternate coefficients; every later entry is formed from the two rows above it, dividing by the first entry of the row immediately above. The number of sign changes down the first column equals the number of roots in the right half-plane. A missing or negative coefficient already guarantees instability, but the array is still needed to count the right-half-plane roots.

Why does a zero in the first column cause trouble?

Every entry in the next row is divided by the first entry of the row above. If that entry is zero, the next row is undefined, and the sign of the zero itself is ambiguous, so the sign changes cannot be counted. Two different situations produce a zero there, and they need different fixes: a single zero with non-zero entries after it, and a row that is entirely zero.

Case 1, a single zero: s⁴ + s³ + 2s² + 2s + 3

Routh array for s to the 4 plus s cubed plus 2 s squared plus 2 s plus 3Rows s to the 4: 1, 2, 3. s cubed: 1, 2. s squared: epsilon, 3, where epsilon replaces the zero that appeared. s to the 1: 2 epsilon minus 3 over epsilon. s to the 0: 3. As epsilon tends to zero from above, the first-column signs are plus, plus, plus, minus, plus: two sign changes, so two roots in the right half-plane.rowcol 1col 2col 3signs⁴123+s³120+s²ε3+s¹(2ε−3)/ε−s⁰3+2 sign changes → 2 right-half-plane roots
The Routh array with ε in place of the zero that appears at the start of the s² row. As ε → 0⁺ the s¹ entry (2ε − 3)/ε becomes a large negative number.
  • Rows s⁴ and s³: 1, 2, 3 and 1, 2.
  • s² row: first entry (1 × 2 − 1 × 2) ÷ 1 = 0, second entry (1 × 3 − 1 × 0) ÷ 1 = 3. The first entry is zero but the row is not, so replace the 0 with ε.
  • s¹ row: (ε × 2 − 1 × 3) ÷ ε = (2ε − 3)/ε. As ε → 0⁺ this tends to −∞, so its sign is negative.
  • s⁰ row: 3.
  • First column signs: +, +, +, −, +. That is two sign changes (+ to − and back to +), so two roots lie in the right half-plane.
  • Check: the roots of the polynomial are 0.406 ± j1.293 and −0.906 ± j0.902. Exactly two have positive real parts.
Computed roots of the example polynomialThe four roots of s to the 4 plus s cubed plus 2 s squared plus 2 s plus 3 on the complex plane: 0.406 plus or minus j 1.293, which are in the right half-plane, and minus 0.906 plus or minus j 0.902, in the left half-plane. This confirms the two sign changes in the Routh array.-1.5-1-0.500.51-1.5-1-0.500.511.5real part σimaginary part ω0.406 ± j1.293−0.906 ± j0.902
The four roots computed numerically. The pair at 0.406 ± j1.293 is in the right half-plane, matching the two sign changes.

A cross-check without ε

Multiplying the polynomial by (s + 1) adds a root at −1, which cannot change the count of right-half-plane roots, and usually moves the zero out of the way. Here (s + 1)(s⁴ + s³ + 2s² + 2s + 3) = s⁵ + 2s⁴ + 3s³ + 4s² + 5s + 3, whose Routh first column is 1, 2, 1, −3, 4.5, 3: again two sign changes, with no ε needed.

Case 2, a whole row of zeros

A row in which every entry is zero means the polynomial has a factor with roots symmetric about the origin: pairs on the imaginary axis (±jω), pairs on the real axis (±σ), or quadruples (±σ ± jω). The row above the zero row gives that factor, the auxiliary polynomial A(s), using only every other power. Replace the zero row with the coefficients of dA/ds and carry on. Sign changes still count right-half-plane roots, and the roots of A(s) itself are found by solving it directly.

Example: s³ + 2s² + s + 2

Rows s³: 1, 1 and s²: 2, 2. The s¹ row is (2 × 1 − 1 × 2) ÷ 2 = 0, all zero. The auxiliary polynomial from the s² row is A(s) = 2s² + 2, so dA/ds = 4s and the s¹ row becomes 4. The s⁰ row is 2. The first column 1, 2, 4, 2 has no sign change, so there are no right-half-plane roots, but A(s) = 0 gives s = ±j: two roots on the imaginary axis. Indeed the polynomial is (s² + 1)(s + 2). A system with this characteristic equation is marginally stable and oscillates at ω = 1 rad/s.

Example: s⁴ + s² − 2

Here the s³ row is zero straight away, because the s³ and s terms are missing. The auxiliary polynomial is the s⁴ row itself, A(s) = s⁴ + s² − 2, so dA/ds = 4s³ + 2s and the s³ row becomes 4, 2. Continuing gives s²: 0.5, −2; s¹: 18; s⁰: −2. The first column 1, 4, 0.5, 18, −2 has one sign change: one right-half-plane root. The factorisation (s² − 1)(s² + 2) confirms it: roots at +1, −1 and ±j1.414.

Common mistakes

  • Using ε for a whole row of zeros. A zero row needs the auxiliary polynomial; ε only fixes a single zero.
  • Building the auxiliary polynomial from the zero row instead of the row above it.
  • Forgetting that A(s) skips powers: from the s² row it is as² + b, not as + b.
  • Taking 'no sign changes' to mean stable when a zero row was found. The roots of A(s) may sit on the jω axis.
  • Letting ε be negative partway through. Keep it small and positive and take the limit only at the end.

Check your understanding

  • Q1. How many right-half-plane roots does s⁴ + 2s³ + 2s² + 4s + 3 have? Answer: two. The s² row starts with 0, ε gives an s¹ entry of (4ε − 6)/ε, which is negative, so the column has two sign changes.
  • Q2. Routh's array for s³ + s² + 4s + 4 has a zero s¹ row. What does that tell you? Answer: A(s) = s² + 4 is a factor, so there are roots at ±j2 and the system is marginally stable; the first column has no sign changes.
  • Q3. How many right-half-plane roots does s³ + 3s + 2 have? Answer: two. The s² coefficient is missing; with ε the s¹ entry is (3ε − 2)/ε, negative, giving two sign changes (roots 0.298 ± j1.807).

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the Routh-Hurwitz and Nyquist stability criteria under Section 6, Control Systems (checked against the official PDF from IIT Madras on 10 October 2026). Every array here was built in code and its sign-change count compared with the polynomial's numerically computed roots.

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