Skip to main content

Relative and absolute stability; R-H of s⁵+s⁴+2s³+2s²+3s+5

Short answer

Absolute stability checks if all poles are in the left half-plane; relative stability measures how far they are from instability. The polynomial has two sign changes, so two roots are in the right half-plane and the system is unstable.

What is relative and absolute stability? Check the stability of system with characteristic equation: s⁵ + s⁴ + 2s³ + 2s² + 3s + 5 = 0 using R-H criteria.

Absolute and relative stability

  • Absolute stability answers yes or no: is the system stable or not? A system is absolutely stable if all closed-loop poles lie in the left half of the s-plane, so every bounded input gives a bounded output and the natural response decays to zero. The Routh–Hurwitz criterion checks absolute stability.
  • Relative stability measures how stable a stable system is, i.e. how far its poles are from the jωj\omega axis (instability boundary). Poles further left decay faster and give less oscillation. It is measured by the real part of the dominant poles (settling time), damping ratio, or in the frequency domain by gain margin and phase margin. In Routh's method, it is found by shifting the axis (s=z−σs = z - \sigma) and testing again.
Absolute stabilityRelative stability
Stable or unstable (qualitative)Degree of stability (quantitative)
Checks sign of real parts of polesChecks distance of poles from jωj\omega axis
Routh–Hurwitz, pole locationShifted Routh, GM, PM, damping ratio

Stability of s5+s4+2s3+2s2+3s+5=0s^5 + s^4 + 2s^3 + 2s^2 + 3s + 5 = 0

RowCol 1Col 2Col 3
s5s^5123
s4s^4125
s3s^31(2)−1(2)1=0→ε\frac{1(2)-1(2)}{1}=0 \to \varepsilon1(3)−1(5)1=−2\frac{1(3)-1(5)}{1}=-2
s2s^22ε+2ε\frac{2\varepsilon+2}{\varepsilon}5
s1s^1−2−5ε22ε+2-2-\frac{5\varepsilon^2}{2\varepsilon+2}0
s0s^05

Only the first element of the s3s^3 row is zero (the row is not all zero), so it is replaced by a small positive number ε\varepsilon. Let ε→0+\varepsilon \to 0^+:

RowFirst-column sign
s5s^5+1+1
s4s^4+1+1
s3s^3+ε+\varepsilon
s2s^22ε→+∞\frac{2}{\varepsilon} \to +\infty
s1s^1→−2\to -2
s0s^0+5+5

There are 2 sign changes (+∞→−2+\infty \to -2 and −2→+5-2 \to +5).

Answer: two roots lie in the right half s-plane, so the system is unstable. The other three roots are in the LHP. The actual roots are ≈0.72±j1.17\approx 0.72\pm j1.17, −0.60±j1.34-0.60\pm j1.34 and −1.24-1.24.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Control Systems chapter 4 questions →