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Relative and absolute stability; R-H of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16

Short answer

Absolute stability means all poles in the left half-plane; relative stability quantifies distance from jω-axis. Four simple poles lie on the jω-axis and two in the left half-plane, so the system is marginally stable.

What is relative and absolute stability? Check the stability of system with characteristic equation: s⁶ + 2s⁵ + 8s⁴ + 12s³ + 20s² + 16s + 16 = 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 s6+2s5+8s4+12s3+20s2+16s+16=0s^6 + 2s^5 + 8s^4 + 12s^3 + 20s^2 + 16s + 16 = 0

Routh array

RowCol 1Col 2Col 3Col 4
s6s^6182016
s5s^5212160
s4s^42(8)−1(12)2=2\frac{2(8)-1(12)}{2}=22(20)−1(16)2=12\frac{2(20)-1(16)}{2}=1216
s3s^32(12)−2(12)2=0\frac{2(12)-2(12)}{2}=02(16)−2(16)2=0\frac{2(16)-2(16)}{2}=0

The s3s^3 row is all zeros, so there are roots placed symmetrically about the origin. Auxiliary equation from the s4s^4 row:

A(s)=2s4+12s2+16,dAds=8s3+24sA(s) = 2s^4 + 12s^2 + 16, \qquad \frac{dA}{ds} = 8s^3 + 24s
RowCol 1Col 2
s3s^3824
s2s^28(12)−2(24)8=6\frac{8(12)-2(24)}{8}=68(16)−08=16\frac{8(16)-0}{8}=16
s1s^16(24)−8(16)6=2.67\frac{6(24)-8(16)}{6}=2.670
s0s^016

First column: 1, 2, 2, 8, 6, 2.67, 161,\ 2,\ 2,\ 8,\ 6,\ 2.67,\ 16. No sign change, so no root is in the RHP.

Roots of the auxiliary equation

2s4+12s2+16=2(s2+2)(s2+4)=0s=±j1.414, ±j2\begin{aligned} 2s^4+12s^2+16 &= 2(s^2+2)(s^2+4) = 0 \\ s &= \pm j1.414,\ \pm j2 \end{aligned}

These four roots lie on the jωj\omega axis and are not repeated. The other factor is

s6+2s5+8s4+12s3+20s2+16s+16(s2+2)(s2+4)=s2+2s+2  ⇒  s=−1±j1\frac{s^6+2s^5+8s^4+12s^3+20s^2+16s+16}{(s^2+2)(s^2+4)} = s^2+2s+2 \;\Rightarrow\; s = -1\pm j1
RegionNumberRoots
Right half plane0none
jωj\omega axis4±j1.414\pm j1.414, ±j2\pm j2
Left half plane2−1±j1-1\pm j1

Conclusion

There is no root in the RHP, and the four roots on the jωj\omega axis are simple (non-repeated). The system is marginally stable (not absolutely stable): its response contains sustained oscillations at ω=1.414\omega = 1.414 rad/s and ω=2\omega = 2 rad/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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