SC · Control

Stability and the Routh Test

Whether a system is stable — all its poles in the left half-plane — can be told from the coefficients of its characteristic polynomial alone, without ever solving for the poles. Build the Routh table, and the number of sign changes in its first column is the number of right-half-plane poles. Move the gain and watch: the instant a first-column entry passes through zero, a pole crosses the imaginary axis and the system goes unstable.

Judge stability without solving for poles

As in the feedback of D4, putting unit feedback with gain K around K/(s(s+1)(s+2)) gives the closed-loop characteristic polynomial s³ + 3s² + 2s + K = 0. The × marks on the left s-plane are its three poles, and the green left half is the stable region. On the right, the indicator shows the first column of the Routh table built from those coefficients. Move the gain K. As K grows, the (6−K)/3 entry of the first column turns negative, and exactly then a complex pole pair crosses the imaginary axis into the right half. From the Routh table alone, without solving for the poles, you can find that boundary.

s-plane (× = closed-loop pole)

gain K8.0

Routh first column and poles

Routh first column: 1 · 3 · -0.67 · 8.0

sign changes = 2 · RHP poles = 2

The (6−K)/3 entry in the first column of the Routh table has turned negative, giving two sign changes in the first column. Those two are exactly two right-half-plane poles. The complex pole pair has crossed the imaginary axis into the right half, an unstable system whose response diverges (K > 6).

Unstable (RHP pole)

Stability is all poles in the left half

As we saw in D2, for a system to be stable every root of its characteristic polynomial — every pole — must lie in the left half-plane (negative real part). If even one is in the right half, that mode diverges. But past third order, solving for the roots by hand is painful, and at high order practically impossible. Fortunately what we want is not the exact roots but a single yes or no: are they all in the left half? The Routh–Hurwitz test answers that by combining the coefficients of the characteristic polynomial alone, never solving for a root. First, one necessary condition: every coefficient must be present and of the same sign (usually positive). If any coefficient is missing or of a different sign, that alone already means unstable.

Observestable ⟺ all poles in left half

Stable means all poles in the left half.

Choosenecessary: every coefficient ?

A missing or negative coefficient already means unstable.

Fill inRHP poles = first-column ?

First-column sign changes = number of RHP poles.

On your owns³+3s²+2s+K stable ⟺ 0 < K < ?

Stability needs (6−K)/3 > 0.

The Routh array and its first-column signs

The Routh array starts by writing the coefficients in two alternating rows. For s³ + 3s² + 2s + K, the s³ row gets 1 and 2, the s² row gets 3 and K. From the next row down, you fill in by weaving the two rows above 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 first column of this array, 1, 3, (6−K)/3, K, is the whole verdict. The key rule: the number of sign changes going down the first column equals exactly the number of right-half-plane poles. Zero sign changes means no right-half poles, so stable; one means one, two means two are in the right half. In this example, when (6−K)/3 is positive, that is K < 6, there are no sign changes and it is stable.

The boundary gain and design

The point where one first-column entry becomes exactly zero is the boundary between stable and unstable. In this example (6−K)/3 = 0, that is K = 6, is the critical gain, and there the auxiliary equation formed from the s² and s⁰ rows, 3s² + 6 = 0, solves to s² = −2, placing poles at s = ±j√2. That is the frequency at which the system oscillates at constant amplitude. Routh goes beyond a plain pass or fail: it hands you, by hand, the boundary gain and even the oscillation frequency there. So in control design you use it at once to set a safe upper limit on gain, or to ask how far one parameter can vary while staying stable. It leads naturally into the next stage of tools, the root locus and Bode plots, for moving poles to shape the response you want.

Back to the first screen

For the closed loop formed by feeding K/(s(s+1)(s+2)) back with gain K, the characteristic polynomial s³ + 3s² + 2s + K was judged from coefficients alone, without solving for poles. In the Routh 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, creating two sign changes, and exactly then the complex pole pair on the s-plane crossed the imaginary axis into the right half. The boundary was K = 6, where the poles were s = ±j√2. Without solving the roots of a high-order equation one by one, a single combination of coefficients put the stability verdict and its boundary in your hand.

Stability = all poles in the left half-plane (D2). The Routh–Hurwitz test decides this from the coefficients of the characteristic polynomial alone, without solving for poles. Necessary condition: every coefficient must be present and of the same sign. Build the Routh array from the coefficients, and the number of sign changes in the first column = the number of right-half-plane poles (zero means stable). The point where a first-column entry becomes zero is the boundary, and an auxiliary equation gives the oscillation frequency there. Example: s³+3s²+2s+K is stable only for 0 < K < 6 from the (6−K)/3 entry, with s = ±j√2 at K=6. It is the tool for setting a safe gain limit by hand in control design.

On to the next unit

Routh told us whether the system is stable at one gain, but not the path the poles trace across the s-plane as the gain rises from zero. The next unit, the root locus, draws exactly that path. Following the trail of the closed-loop poles as the gain K varies — starting at the open-loop poles and heading toward the zeros — shows at a glance where they cross the imaginary axis into instability and at what K you get the damping ratio you want. The boundary K = 6 we just found with Routh is precisely the point where the root locus meets the imaginary axis.