When is a System Stable?
A system is stable if its output remains bounded and well-behaved over time, especially when subjected to bounded inputs.
1. BIBO Stability (Bounded-Input Bounded-Output Stability)
A system is BIBO stable if:
Every bounded input produces a bounded output.
If |x(t)| < Mx < ∞
then |y(t)| < My < ∞
For some finite constant My.
For LTI Systems:
A continuous-time LTI system is stable if and only if:
∫ |h(t)| dt < ∞
That is, the impulse response must be absolutely integrable.
For discrete-time systems:
Σ |h[n]| < ∞
2. Stability in Terms of Poles (Control Systems)
Continuous-Time System:
The system is stable if all poles satisfy:
Re(s) < 0
That means all poles lie in the left-half of the s-plane.
Discrete-Time System:
The system is stable if:
|z| < 1
That means all poles lie inside the unit circle in the z-plane.
3. Physical Interpretation
- Its natural response decays to zero with time.
- It does not oscillate with growing amplitude.
- It does not blow up to infinity.
4. Simple Examples
Stable:
h(t) = e^{-3t}u(t)
Poles at s = -2, -5
Unstable:
h(t) = e^{2t}u(t)
Pole at s = +2
Marginally Stable:
Pure sinusoid (poles on imaginary axis, not repeated). Output oscillates but does not grow.
Conclusion
A system is stable when its response does not grow unbounded with time, and bounded inputs always produce bounded outputs.
Routh–Hurwitz Criterion – Numeric Examples
Case 1: Normal Case (Stable System)
Characteristic Equation: s³ + 6s² + 11s + 6 = 0
| s³ | 1 | 11 |
|---|---|---|
| s² | 6 | 6 |
| s¹ | 10 | 0 |
| s⁰ | 6 |
All first-column elements are positive → Stable
Case 2: Sign Change (Unstable System)
Characteristic Equation: s³ + 2s² − s − 2 = 0
One sign change → 1 RHP root → Unstable
Case 3: Zero in First Column (Epsilon Method)
Characteristic Equation: s³ + 2s² + s + 2 = 0
Replace 0 with small positive ε.
No sign change as ε → 0⁺ → Stable
Case 4: Entire Row Becomes Zero (Auxiliary Equation)
Characteristic Equation: s⁴ + 2s³ + 2s² + 2s + 1 = 0
Auxiliary Equation: A(s) = s² + 1
A′(s) = 2s
Roots of s² + 1 = 0 → s = ±j
Pure imaginary roots → Marginally Stable
Case 5: Repeated Imaginary Roots (Unstable)
Characteristic Equation: s⁴ + 2s² + 1 = 0
(s² + 1)² = 0
Repeated Roots: s = ±j
Repeated imaginary roots → Unstable
Final Summary
| Case | Condition | Example | Stability |
|---|---|---|---|
| 1 | No sign change | s³ + 6s² + 11s + 6 | Stable |
| 2 | Sign change | s³ + 2s² − s − 2 | Unstable |
| 3 | Zero in first column | s³ + 2s² + s + 2 | Stable |
| 4 | Entire row zero | s⁴ + 2s³ + 2s² + 2s + 1 | Marginal |
| 5 | Repeated imaginary roots | (s² + 1)² | Unstable |