Training Control Systems Stability & Routh–Hurwitz
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Stability & Routh–Hurwitz

30 min Control Systems

Stability: Poles & Routh–Hurwitz

A linear time-invariant system is stable if and only if every pole of its closed-loop transfer function lies in the open left-half of the $s$-plane (real part $<$ 0). Stability is the first property any feedback design must guarantee — without it, nothing else matters.

For simple systems you can factor the characteristic polynomial and read the pole real parts directly. For higher-order systems, the Routh–Hurwitz criterion gives stability without having to compute roots: all coefficients must have the same sign, and the Routh array's first column must have no sign changes. The number of sign changes equals the number of unstable poles.

We will do stability checks on 2nd–4th order polynomials and find the gain range that keeps a control loop stable.

Stability Condition

All poles $p_i$ of $T(s)$ satisfy $\operatorname{Re}(p_i) < 0$.

Routh Array (order 3: $a_3 s^3 + a_2 s^2 + a_1 s + a_0$)

Build rows for $s^3, s^2, s^1, s^0$:

$s^3: a_3 \; a_1$

$s^2: a_2 \; a_0$

$s^1: (a_2 a_1 - a_3 a_0)/a_2$

$s^0: a_0$

Stable ⇔ first column (leftmost entries) has no sign change.

Example 1 — Second-order always stable if coefficients positive

$s^2 + 3s + 2 = 0$: is it stable?

All coefficients positive. Roots: $s^2 + 3s + 2 = (s+1)(s+2)$ — both negative. Stable.

Example 2 — Third-order Routh

$s^3 + 2s^2 + 3s + K = 0$. For what $K$ is the system stable?

Routh: $s^3: 1, 3$; $s^2: 2, K$; $s^1: (2\cdot 3 - 1\cdot K)/2 = (6-K)/2$; $s^0: K$.

First column: $1, 2, (6-K)/2, K$. Need all positive: $K > 0$ and $(6-K)/2 > 0 \Rightarrow K < 6$.

Stable range: $0 < K < 6$.

Example 3 — Marginal stability

At the boundary $K = 6$ of Example 2, what happens?

The $s^1$ row becomes zero. Substitute $K=6$: $s^3 + 2s^2 + 3s + 6 = (s+2)(s^2+3)$. Roots $s = -2$ and $s = \pm j\sqrt{3}$. Pure-imaginary pair ⇒ system oscillates sustainedly at $\omega = \sqrt{3} \approx 1.73\,\text{rad/s}$.

Interactive Demo: Routh–Hurwitz (3rd Order)
s¹ row0.0
s⁰ row6.0
Stable?Marginal
# unstable poles =0

Practice Problems

1. State the stability criterion in one sentence.
2. Is $s^2 + 4 = 0$ stable, unstable, or marginal?
3. For $s^3 + 4s^2 + 5s + K$, find the stable range of $K$.
4. Why does a pair of pure-imaginary poles give undamped oscillation?
5. What does a sign change in the Routh first column indicate?
6. Why is verifying stability always the first design task?
Show Answer Key

1. All closed-loop poles must lie in the open left-half plane.

2. Marginal — roots at $\pm 2j$ oscillate without decay.

3. $K > 0$ and $K < 20$. So $0 < K < 20$.

4. The impulse response is $\sin(\omega t)$ with constant amplitude — no decay, no growth.

5. Each sign change equals one pole in the right-half plane (unstable).

6. If the closed loop is unstable every other specification (speed, accuracy) is meaningless.

Step Response of First & Second Order PID Controllers