Let's solve the problem step by step: --- ## **Given:** The open-loop transfer function is: \[ G(s) = \frac{Ks(s+2)}{(s^2 - 4s + 8)(s+3)} \] Unity feedback is assumed. --- ## **a. Find the range of \( K \) for stability** ### **Step 1: Closed-Loop Characteristic Equation** For unity feedback, the characteristic equation is: \[ 1 + G(s) = \] So, \[ 1 + \frac{Ks(s+2)}{(s^2 - 4s + 8)(s+3)} = \] \[ (s^2 - 4s + 8)(s+3) + Ks(s+2) = \] Expand \( (s^2 - 4s + 8)(s+3) \): \[ = (s^2 - 4s + 8)s + (s^2 - 4s + 8)3 = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - 4s^2 + 3s^2 + 8s - 12s + 24 = s^3 - s^2 - 4s + 24 \] So the characteristic equation is: \[ s^3 - s^2 - 4s + 24 + Ks(s+2) = \] Expand \( Ks(s+2) \): \[ Ks(s+2) = Ks^2 + 2Ks \] So, \[ s^3 - s^2 - 4s + 24 + Ks^2 + 2Ks = \] \[ s^3 + (K-1)s^2 + (2K-4)s + 24 = \] --- ### **Step 2: Routh-Hurwitz Criterion** Write the characteristic equation: \[ s^3 + (K-1)s^2 + (2K-4)s + 24 = \] Coefficients are: - \( s^3 \): 1 - \( s^2 \): \( K-1 \) - \( s^1 \): \( 2K-4 \) - \( s^ \): 24 **Routh Array:** \[ \begin{array}{c|cc} s^3 & 1 & 2K-4 \\ s^2 & K-1 & 24 \\ s^1 & \frac{(K-1)(2K-4)-24}{K-1} & \\ s^ & 24 & \\ \end{array} \] For stability, all elements in the first column must be positive. #### 1. \( K-1 > \implies K > 1 \) #### 2. \( 2K-4 > \implies K > 2 \) #### 3. Third row, first element: \[ a = \frac{(K-1)(2K-4) - 24}{K-1} = 2K-4 - \frac{24}{K-1} \] But let's expand numerator: \[ (K-1)(2K-4) - 24 = 2K^2 - 4K - 2K + 4 - 24 = 2K^2 - 6K - 20 \] So, \[ \frac{2K^2 - 6K - 20}{K-1} > \] **Find roots:** Set numerator \(2K^2 - 6K - 20 = \): \[ K = \frac{6 \pm \sqrt{36 + 160}}{4} = \frac{6 \pm 14}{4} \] \[ K_1 = \frac{20}{4} = 5,\quad K_2 = \frac{-8}{4} = -2 \] So the numerator is positive for \(K < -2\) or \(K > 5\). Since denominator \(K-1 > \) for \(K > 1\), combine with the other conditions: - \(K > 2\) (from earlier) - \(K > 5\) (from above for positivity of third row) **Therefore, the range for stability is:** \[ \boxed{K > 5} \] --- ## **b. Frequency of Oscillation when System is Marginally Stable** Marginal stability occurs when a row in the Routh table becomes zero, i.e., a pair of purely imaginary roots. Set the third row, first element to zero: \[ 2K^2 - 6K - 20 = \] Already solved above: \[ K = 5 \quad \text{or} \quad K = -2 \] But \(K > 1\), so \(K = 5\). Now substitute \(K=5\) into the characteristic equation: \[ s^3 + (5-1)s^2 + (2 \times 5 - 4)s + 24 = \] \[ s^3 + 4s^2 + 6s + 24 = \] Find the frequency of oscillation (roots on the **imaginary axis**): Try factoring or let \(s = j\omega\): \[ (j\omega)^3 + 4(j\omega)^2 + 6(j\omega) + 24 = \] \[ j\omega^3 + 4(j^2)\omega^2 + 6j\omega + 24 = \] \[ j\omega^3 - 4\omega^2 + 6j\omega + 24 = \] \[ (j\omega^3 + 6j\omega) + (-4\omega^2 + 24) = \] \[ j(\omega^3 + 6\omega) + (-4\omega^2 + 24) = \] Equate real and imaginary parts to zero: - Real: \( -4\omega^2 + 24 = \implies \omega^2 = 6 \implies \omega = \sqrt{6} \) - Imaginary: \( \omega^3 + 6\omega = \implies \omega(\omega^2 + 6) = \implies \omega = \) or \( \omega^2 = -6 \) (not possible for real frequency) So, the non-zero frequency is: \[ \boxed{\omega = \sqrt{6}} \] --- ## **Final Answers** ### **a. Range of \( K \) for stability:** \[ \boxed{K > 5} \] ### **b. Frequency of Oscillation when Marginally Stable:** \[ \boxed{\omega = \sqrt{6}} \]

Let's solve the problem step by step: --- ## **Given:** The open-loop transfer function is: \[ G(s) = \frac{Ks(s+2)}{(s^2 - 4s + 8)(s+3)} \] Unity feedback is assumed. --- ## **a. Find the range of \( K \) for stability** ### **Step 1: Closed-Loop Characteristic Equation** For unity feedback, the characteristic equation is: \[ 1 + G(s) = 0 \] This leads to: \[ (s^2 - 4s + 8)(s+3) + Ks(s+2) = 0 \] ### **Step 2: Expand \( (s^2 - 4s + 8)(s + 3) \)** Calculating the expansion: \[ (s^2 - 4s + 8)(s + 3) = s^3 - 4s^2 + 8s + 3s^2 - 12s + 24 = s^3 - s^2 - 4s + 24 \] ### **Step 3: Formulate the Characteristic Equation** The full characteristic equation becomes: \[ s^3 + (K-1)s^2 + (2K-4)s + 24 = 0 \] ### **Step 4: Apply Routh-Hurwitz Criterion** Using the coefficients: - \( s^3 \): 1 - \( s^2 \): \( K-1 \) - \( s^1 \): \( 2K-4 \) - \( s^0 \): 24 #### **Routh Array:** \[ \begin{array}{c|cc} s^3 & 1 & 2K-4 \\ s^2 & K-1 & 24 \\ s^1 & \frac{(K-1)(2K-4) - 24}{K-1} & \\ s^0 & 24 & \\ \end{array} \] **Conditions for stability:** 1. \( K-1 > 0 \implies K > 1 \) 2. \( 2K-4 > 0 \implies K > 2 \) #### Third Row Analysis: \[ (K-1)(2K-4) - 24 > 0 \] Calculating: \[ = 2K^2 - 6K - 20 \] **Finding Roots:** Set the numerator: \[ 2K^2 - 6K - 20 = 0 \] Using the quadratic formula: \[ K = \frac{6 \pm \sqrt{36 + 160}}{4} = \frac{6 \pm 14}{4} \] This gives: \[ K_1 = 5, \quad K_2 = -2 \] Combining conditions, we find: **Thus, the range for stability is:** \[ \boxed{K > 5} \] --- ## **b. Frequency of Oscillation when System is Marginally Stable** Marginal stability occurs when a row in the Routh table becomes zero. Setting: \[ 2K^2 - 6K - 20 = 0 \] For \( K = 5 \): Substituting into the characteristic equation: \[ s^3 + 4s^2 + 6s + 24 = 0 \] ### **Finding Frequency:** Substituting \( s = j\omega \): \[ (j\omega)^3 + 4(j\omega)^2 + 6(j\omega) + 24 = 0 \] This simplifies to: \[ j\omega^3 - 4\omega^2 + 6j\omega + 24 = 0 \] Separating real and imaginary parts: 1. **Real:** \( -4\omega^2 + 24 = 0 \implies \omega^2 = 6 \implies \omega = \sqrt{6} \) 2. **Imaginary:** \( \omega^3 + 6\omega = 0 \) gives \( \omega(\omega^2 + 6) = 0 \) Thus, we have non-zero frequency: \[ \boxed{\omega = \sqrt{6}} \] --- ## **Final Answers** ### **a. Range of \( K \) for stability:** \[ \boxed{K > 5} \] ### **b. Frequency of Oscillation when Marginally Stable:** \[ \boxed{\omega = \sqrt{6}} \]