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R-H range of K for s⁴+2s³+(4+K)s²+9s+25

Short answer

Form the Routh array and enforce all first-column entries positive. The s² row gives K > 0.5, and the s¹ row requires 18K > 109, so K > 109/18 ≈ 6.056. Thus, the system is stable for K > 6.056 (no upper limit).

Using R-H criteria find the range of K for system having characteristic equation shown below, to be stable. s⁴ + 2s³ + (4+K)s² + 9s + 25 = 0

Form the Routh array with KK and make every first-column element positive.

s4+2s3+(4+K)s2+9s+25=0s^4 + 2s^3 + (4+K)s^2 + 9s + 25 = 0

Routh array

RowCol 1Col 2Col 3
s4s^414+K4+K25
s3s^3290
s2s^22(4+K)−92=2K−12\frac{2(4+K)-9}{2}=\frac{2K-1}{2}25
s1s^12K−12(9)−2(25)2K−12=18K−1092K−1\frac{\frac{2K-1}{2}(9)-2(25)}{\frac{2K-1}{2}}=\frac{18K-109}{2K-1}0
s0s^025

Conditions

  1. s2s^2 row: 2K−12>0⇒K>0.5\frac{2K-1}{2}>0 \Rightarrow K>0.5
  2. s1s^1 row: with 2K−1>02K-1>0, we need 18K−109>0⇒K>10918=6.05618K-109>0 \Rightarrow K > \frac{109}{18} = 6.056
  3. s0s^0 row: 25>025>0 always true.

Condition 2 is stricter, so:

K>6.056K > 6.056

Marginal value

At K=109/18=6.056K = 109/18 = 6.056, the s1s^1 row is zero. The auxiliary equation 509s2+25=0\frac{50}{9}s^2 + 25 = 0 gives s=±j2.12s = \pm j2.12, so the system would oscillate at 2.12 rad/s.

Answer: the system is stable for K>6.056K > 6.056 (i.e. K>109/18K>109/18). There is no upper limit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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