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R-H stability of s⁴+8s³+18s²+16s+50

Short answer

Routh array first column entries: 1, 8, 16, −9, 50. Two sign changes occur, so there are 2 roots in the right half-plane and 2 in the left half-plane. The system is unstable.

Check the stability of the system represented by the following characteristic equation using R-H criteria: s⁴ + 8s³ + 18s² + 16s + 50 = 0

Result: the system is unstable, with 2 roots in the right half plane.

The Routh–Hurwitz criterion says a system is stable only if all elements in the first column of the Routh array have the same sign. The number of sign changes equals the number of roots in the right half s-plane.

s4+8s3+18s2+16s+50=0s^4 + 8s^3 + 18s^2 + 16s + 50 = 0

Necessary condition: all coefficients are present and positive, so it is satisfied. This alone does not prove stability, so the array is needed.

Routh array

RowCol 1Col 2Col 3
s4s^411850
s3s^38160
s2s^28(18)−1(16)8=16\frac{8(18)-1(16)}{8}=168(50)−1(0)8=50\frac{8(50)-1(0)}{8}=50
s1s^116(16)−8(50)16=−9\frac{16(16)-8(50)}{16}=-90
s0s^050

Examination of the first column

First column: 1, 8, 16, −9, 501,\ 8,\ 16,\ -9,\ 50

  • +16→−9+16 \to -9: first sign change
  • −9→+50-9 \to +50: second sign change

2 sign changes, so 2 roots lie in the right half of the s-plane.

RegionNumber of roots
Right half plane2
Left half plane2
jωj\omega axis0

The actual roots are s≈0.185±j1.64s \approx 0.185 \pm j1.64 and s≈−4.18±j0.93s \approx -4.18 \pm j0.93, which agrees with the Routh result.

Answer: the system is unstable. It has two roots in the RHP, so its response grows as an oscillation with an increasing envelope.

✓ Answer checked by Claude Opus 5.5 (high effort)

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