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R-H range of K for K(s+1)/(s³+bs²+3s+1)

Short answer

The characteristic equation is s³ + bs² + (3+K)s + (1+K)=0. Routh conditions: K > −1 and b(3+K) > 1+K. For K>0, stable range is 0 < K < (3b−1)/(1−b) when 1/3 < b < 1. Marginal stability at K = (3b−1)/(1−b) with oscillation frequency ω = √(3+K) = √(2/(1−b)) rad/s.

A unity feedback system has the feedforward transfer function G(s) = K(s+1)/(s³+bs²+3s+1). Using R-H Criterion, find the range of K for the system to be stable. When does the system just oscillate and what would be that frequency during sustained oscillation?

Result: stable for K>−1K>-1 and b(3+K)>1+Kb(3+K)>1+K. For K>0K>0 this means any KK when b≥1b\ge1, and 0<K<3b−11−b0<K<\frac{3b-1}{1-b} when 13<b<1\frac13<b<1. It just oscillates at K=3b−11−bK=\frac{3b-1}{1-b}, with ω=3+K=21−b\omega=\sqrt{3+K}=\sqrt{\frac{2}{1-b}} rad/s.

The characteristic equation is 1+G(s)=01+G(s)=0. The Routh array gives the stability conditions in terms of KK and bb. Sustained oscillation occurs when the s1s^1 row becomes zero.

Characteristic equation

s3+bs2+3s+1+K(s+1)=0s3+bs2+(3+K)s+(1+K)=0\begin{aligned} s^3+bs^2+3s+1 + K(s+1) &= 0 \\ s^3 + bs^2 + (3+K)s + (1+K) &= 0 \end{aligned}

Routh array

RowCol 1Col 2
s3s^313+K3+K
s2s^2bb1+K1+K
s1s^1b(3+K)−(1+K)b\frac{b(3+K)-(1+K)}{b}0
s0s^01+K1+K

Conditions for stability

  1. b>0b > 0
  2. 1+K>0⇒K>−11+K>0 \Rightarrow K>-1
  3. b(3+K)−(1+K)>0⇒K(b−1)>1−3bb(3+K) - (1+K) > 0 \Rightarrow K(b-1) > 1-3b

Interpreting condition 3 (for K>0K>0, b>0b>0):

  • If b≥1b \ge 1: condition 3 holds for every K>0K>0, so the system is stable for all K>0K>0.
  • If b<1b < 1: K<3b−11−bK < \dfrac{3b-1}{1-b}. A positive range exists only if b>13b > \frac{1}{3}.
Range: 0<K<3b−11−b(13<b<1)\text{Range: } 0 < K < \frac{3b-1}{1-b}\quad\left(\tfrac13<b<1\right)

Sustained oscillation

The system just oscillates when the s1s^1 row is zero:

b(3+K)=1+K⇒Kmar=3b−11−bb(3+K) = 1+K \quad\Rightarrow\quad K_{mar} = \frac{3b-1}{1-b}

Auxiliary equation from the s2s^2 row:

bs2+(1+K)=0ω2=1+Kb=3+K=21−bω=21−b rad/s\begin{aligned} bs^2 + (1+K) &= 0 \\ \omega^2 &= \frac{1+K}{b} = 3+K = \frac{2}{1-b} \\ \omega &= \sqrt{\frac{2}{1-b}}\ \text{rad/s} \end{aligned}

Example: for b=0.5b = 0.5: Kmar=0.50.5=1K_{mar} = \frac{0.5}{0.5}=1 and ω=4=2\omega = \sqrt{4} = 2 rad/s. Check: s3+0.5s2+4s+2=(s2+4)(s+0.5)s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5), which has roots ±j2\pm j2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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