# Root Locus and Stability Analysis Given the open-loop transfer function: \[ G(s) =frac{K(s+2)}s^2 (s+3)} \] We are to: 1. **Plot the root locus** 2. **Determine the range of \( K \) for closed-loop stability** --- ## 1. **Rewrite the Characteristic Equation** For unity feedback, the closed-loop characteristic equation is: \[ 1 + G(s) = \implies 1 + \frac{K(s+2)}{s^2 (s+3)} = \] \[ s^2 (s+3) + K(s+2) = \] \[ s^3 + 3s^2 + K(s+2) = \] \[ s^3 + 3s^2 + K s + 2K = \] --- ## 2. **Open-Loop Poles and Zeros** - **Poles:** Set denominator to zero: \( s^2 (s+3) = \) - \( s = \) (double pole) - \( s = -3 \) (single pole) - **Zero:** Set numerator to zero: \( s+2 = \implies s = -2 \) **Summary:** - Poles at: \( \) (multiplicity 2), \( -3 \) - Zero at: \( -2 \) --- ## 3. **Root Locus Diagram (Qualitative)** ### **Step-by-Step Construction** #### **A. Mark Poles and Zeros** - Poles: X at \( \), X at \( \), X at \( -3 \) - Zero: O at \( -2 \) #### **B. Number of Branches** - Number of poles = 3 - Number of zeros = 1 - Number of branches = 3 (each starts at a pole) #### **C. Real Axis Segments** - Root locus exists on the real axis **to the left of an odd number of real poles and zeros.** - Real axis locations: \(-\infty\) to \(-3\), \(-3\) to \(-2\), \(-2\) to \(\), \(\) to \(+\infty\) **Counting to the right:** - \((-\infty, -3)\): 3 poles/zeros to right — odd ⇒ locus exists - \((-3, -2)\): 2 to the right — even ⇒ no locus - \((-2, )\): 1 to the right — odd ⇒ locus exists - \((, \infty)\): to the right — even ⇒ no locus #### **D. Asymptotes** - Number of asymptotes = (number of poles) - (number of zeros) = \(3-1=2\) - Asymptote angles: \[ \theta = \frac{(2q+1)180^\circ}{n-m}, \quad q=,1 \] For \(q=,1\): - \(q=: \theta_1 = 90^\circ\) - \(q=1: \theta_2 = 270^\circ\) (which is \(-90^\circ\)) - **Centroid:** \[ \text{Centroid} = \frac{\sum \text{poles} - \sum \text{zeros}}{n-m} \] \[ \sum \text{poles} = + + (-3) = -3 \] \[ \sum \text{zeros} = -2 \] \[ \text{Centroid} = \frac{(-3) - (-2)}{2} = \frac{-1}{2} = -.5 \] - **So, two asymptotes (at \( \pm 90^\circ \)), intersecting at \( s = -.5 \)** #### **E. Breakaway/Break-in Point** On the real axis between \( -2 \) and \( \) (since two poles at , zero at \(-2\)), there may be a breakaway. --- ## 4. **Stability Range for \( K \)** ### **Routh-Hurwitz Criterion** Given: \[ s^3 + 3s^2 + Ks + 2K = \] Routh array: | | \(s^3\) | \(s^2\) | \(s^1\) | \(s^\) | |----|:-------:|:-------:|:-------:|:-------:| | Row 1 | 1 | \(K\) | | Row 2 | 3 | \(2K\) | | Row 3 | \(a\) | | | Row 4 | \(2K\)| | Where: - Row 3, first entry: \[ a = \frac{3K - 1 \times 2K}{3} = \frac{3K - 2K}{3} = \frac{K}{3} \] So, the full array: | | \(s^3\) | \(s^2\) | \(s^1\) | \(s^\) | |----|:-------:|:-------:|:-------:|:-------:| | | 1 | \(K\) | | | | | 3 | \(2K\) | | | | | \(\frac{K}{3}\) | | | | | | \(2K\) | | | | **First column:** - 1 - 3 - \( \frac{K}{3} \) - \( 2K \) For stability, all first-column entries must be positive. - \( 1 > \): always true - \( 3 > \): always true - \( \frac{K}{3} > \implies K > \) - \( 2K > \implies K > \) **Therefore, the system is stable for:** \[ \boxed{K > } \] --- ## 5. **Summary and Root Locus Sketch** ### **Root Locus Sketch** (ASCII illustration) ``` <---|----|----O----|----X====X----X----> -∞ -3 -2 (real axis) X X (poles at -3, , ) O (zero at -2) ``` - Root locus branches: - From \( s= \) poles, one branch moves toward the zero at \( s=-2 \) - The other \( s= \) pole and \( s=-3 \) pole branches go toward infinity along asymptotes at \( \pm 90^\circ \) from centroid \( s=-.5 \) --- ## 6. **Final Answers** ### **A. Root Locus Diagram** - Poles at \( \) (double), \( -3 \) - Zero at \( -2 \) - 2 asymptotes at \( \pm 90^\circ \) from centroid \( -.5 \) - Real axis segments of the locus: \( (-\infty, -3) \) and \( (-2, ) \) - Sketch as above ### **B. Stability Range** \[ \boxed{K > } \] For all positive values of \( K \), the closed-loop system is stable (all roots in the left half-plane). --- **[If you need a MATLAB or Python code for plotting, ask!]**