Stability and the Routh Test

Judge stability without solving for poles

As with the feedback of the D4 unit, put unity feedback with gain K around K/(s(s + 1)(s + 2)) and the closed-loop characteristic equation is s³ + 3s² + 2s + K = 0. The upper plot is the s-plane: the gold × marks are its three poles, and the green-shaded left half is the stable region. The display below lists the first column of the Routh table built from these coefficients, 1, 3, (6 − K)/3, K, and any entry that turns negative turns red. At the starting K = 8, (6 − K)/3 is negative, so the sign changes twice, and a complex pole pair sits to the right of the imaginary axis. Lower K below 6 and that entry turns positive again, and at the same moment the two poles cross the imaginary axis into the left half, so the Routh table alone reveals the boundary without solving for the poles.

Stability is all poles in the left half

As the D2 unit showed, for a system to be stable every root of the characteristic equation, that is, every pole, must be in the left half-plane (negative real part), and if even one is in the right half, its mode diverges. Yet solving for the roots by hand is a chore once you go past third order, and at higher orders it is practically impossible. Fortunately, what you want to know is not the exact values of the roots but a single yes or no: 'are they all in the left half?' The Routh–Hurwitz test gives that answer by combining only the coefficients of the characteristic equation, without solving for any roots. First there is a necessary condition: every coefficient must be present and all must have the same sign (usually positive). If even one coefficient is missing or has a different sign, that alone is enough to conclude the system is not stable.

The Routh array and its first-column signs

A Routh table starts by writing the coefficients of the characteristic equation alternately in two rows. For s³ + 3s² + 2s + K, you put 1 and 2 in the s³ row and 3 and K in the s² row. Each row below is filled by combining the two rows above it like a 2×2 determinant: the first value of the s¹ row is (3·2 − 1·K)/3 = (6 − K)/3, and the s⁰ row is K. The resulting first column, 1, 3, (6 − K)/3, K, is all you need for the verdict. There is one rule. The number of sign changes going down the first column is exactly the number of right-half-plane poles: with no sign changes there are no right-half poles and the system is stable, with one change there is 1 such pole, and with two there are 2. In this example, if (6 − K)/3 is positive, that is, if K < 6, there are no sign changes and the system is stable.

The boundary gain and design

The point where one entry of the first column becomes exactly 0 is the boundary between stable and unstable. In this example, (6 − K)/3 = 0, that is, K = 6, is the critical gain, and at that gain, setting up the auxiliary equation from the s² row, 3s² + 6 = 0, and solving it gives s² = −2, so the poles sit at s = ±j√2. This is the angular frequency √2 rad/s at which the system oscillates with constant amplitude. In this way, the Routh test goes beyond pass or fail: by hand you can find the boundary gain and the oscillation frequency at that gain. That is why it is used directly in control design to set a safe upper limit on the gain, or to check how far a parameter can vary while the system stays stable, and it leads naturally into the root locus and the Bode plot, which move poles to shape the response you want.

Back to the first screen

On the first screen, you judged the characteristic equation s³ + 3s² + 2s + K of the closed loop formed by putting feedback with gain K around K/(s(s + 1)(s + 2)), using only its coefficients and without solving for the poles. In the Routh table's first column, 1, 3, (6 − K)/3, K, the number of sign changes was the number of right-half-plane poles. Raising K turned (6 − K)/3 negative and produced two sign changes, and at exactly that moment the complex pole pair on the s-plane crossed the imaginary axis into the right half. The boundary was K = 6, where the poles sat at s = ±j√2. Without solving the roots of a high-order equation one by one, a single combination of coefficients put stability and its boundary in your hands.

The Routh test told you whether the system is stable at a given gain, but you have not yet seen what path the poles trace across the s-plane as the gain rises from 0. The next unit, The Root Locus, draws exactly that path. Follow how the closed-loop poles start from the open-loop poles and head toward the open-loop zeros (or to infinity when there are no zeros) as the gain K changes, and you can see at a glance where they cross the imaginary axis and become unstable, and at what K you get the damping ratio you want. The boundary K = 6 you just found with the Routh test is exactly where the root locus meets the imaginary axis.