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R-H range of K for s⁴+20s³+15s²+2s+K

Short answer

Construct the Routh array: stability requires all first-column entries positive, so 0 < K < 1.49. The system becomes marginally stable at K = 1.49 where the s¹ row vanishes. The auxiliary equation 14.9s² + 1.49 = 0 gives sustained oscillation at ω = 0.316 rad/s.

The characteristic equation of a feedback control system is s⁴ + 20s³ + 15s² + 2s + K = 0. (i) Determine the range of K for the system to be stable. (ii) Can the system be marginally stable? If so, find the required value of K and the frequency of sustained oscillation.

Apply the Routh array with KK in it. For stability, every element of the first column must be positive. Marginal stability occurs when the s1s^1 row becomes zero.

s4+20s3+15s2+2s+K=0s^4+20s^3+15s^2+2s+K = 0

Routh array

RowCol 1Col 2Col 3
s4s^4115KK
s3s^32020
s2s^220(15)−1(2)20=14.9\frac{20(15)-1(2)}{20}=14.9KK
s1s^114.9(2)−20K14.9=29.8−20K14.9\frac{14.9(2)-20K}{14.9}=\frac{29.8-20K}{14.9}0
s0s^0KK

(i) Range of K for stability

  • From s1s^1 row: 29.8−20K>0⇒K<1.4929.8 - 20K > 0 \Rightarrow K < 1.49
  • From s0s^0 row: K>0K > 0
0<K<1.490 < K < 1.49

(ii) Marginal stability

Yes. At K=1.49K = 1.49 the s1s^1 row becomes all zero, and the system has a pair of roots on the jωj\omega axis (marginally stable, sustained oscillation).

Auxiliary equation from the s2s^2 row:

14.9s2+K=014.9s2+1.49=0s2=−0.1⇒s=±j0.316\begin{aligned} 14.9s^2 + K &= 0 \\ 14.9s^2 + 1.49 &= 0 \\ s^2 &= -0.1 \Rightarrow s = \pm j0.316 \end{aligned}

So the frequency of sustained oscillation is ω=0.1=0.316\omega = \sqrt{0.1} = 0.316 rad/s.

Answer: stable for 0<K<1.490 < K < 1.49; marginally stable at Kmar=1.49K_{mar} = 1.49, oscillating at ω=0.316\omega = 0.316 rad/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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