How it works
Construct the Routh array from the characteristic polynomial aₙsⁿ + aₙ₋₁sⁿ⁻¹ + ... + a₀ = 0: first row contains even-indexed coefficients, second row odd-indexed, remaining rows computed as 2×2 determinants divided by the leading element of the previous row. The number of sign changes in the first column equals the number of roots in the right half-plane. For the system to be stable, all elements of the first column must be positive and non-zero. A zero appearing in the first column (not the entire row) is replaced by a small positive ε to continue the array.
Key points to remember
Special case 1: a zero in the first column (other elements non-zero) — replace with ε > 0 and proceed; if the column changes sign around ε, those count as sign changes for instability. Special case 2: an entire row of zeros — indicates roots symmetrically placed about the origin (on jω-axis or real axis pairs). Form the auxiliary polynomial from the row above the zero row, differentiate it, and replace the zero row with the derivative coefficients. For marginal stability (roots exactly on jω-axis), there are no sign changes but a zero row occurs — the system oscillates at ω found from the auxiliary polynomial.
Exam tip
The examiner always asks you to find the range of K for stability in a given closed-loop characteristic equation — set up the Routh array with K as a variable, write the conditions for all first-column elements to be positive, and solve the inequality for K.