Question Q2. Sketch the root loci for the system shown in Figure (2) C(s) R(s) K(+3) ²+2x+3 Figure (2) Q2. Sketch the root loci for the system shown in Figure (2) C(s) R(s) K(+3) ²+2x+3 Figure (2)

Explanation:The open loop transfer function \mathbf{G}(\mathbf{s}) is defined as,\begin{array}{l}\mathbf{G}(\mathbf{s})=\begin{array}{c}\mathbf{k}(\mathbf{s}+\mathbf{3}) \\\mathbf{s}^{2}+2 \mathbf{s}+\mathbf{3}\end{array} \\\mathbf{G}(\mathbf{s})=\frac{\mathbf{k}(\mathrm{s}+\mathbf{3})}{\mathbf{s}^{2}+2 \mathbf{s}+\mathbf{1}+\mathbf{2}} \\\mathbf{G}(\mathbf{s})=\frac{\mathbf{k}(\mathbf{s}+\mathbf{3})}{(\mathbf{s}+\mathbf{1})^{2}+\mathbf{2}} \\\mathbf{G}(\mathbf{s})=\begin{array}{c}\mathbf{k}(\mathbf{s}+\mathbf{3}) \\(\mathbf{s}+\mathbf{1})^{2}+(\sqrt{\mathbf{2}})^{2}\end{array} \\\mathbf{G}(\mathbf{s})=\frac{\mathbf{k}(\mathbf{s}+\mathbf{3})}{(\mathbf{s}+\mathbf{1}+\mathbf{j} \mathbf{2})(\mathbf{s}+\mathbf{1}-\mathbf{j} \mathbf{2})} \\\end{array}Objective: To sketch the root locus diagram for given system.Step-1).Number of zeros: N(z)=1Number of poles: N(p)=2Location of zeros are at: s=-3Location of poles are at: s=-1 \pm j \sqrt{ } 2Step-2) Number of asymptote =|N(p)-N(z)|=|2-1|=1Step-3) Angle of asymptote will be,\theta=\underset{\mathbf{N}(\mathbf{p})-\mathbf{N}(\mathbf{z})}{(\mathbf{2}+1)}Where, q \Rightarrow 0,1,2,3, \ldots, N(p)-N(z)-1=0\begin{array}{l}\theta=\frac{(2 \mathrm{q}+1) 180^{\circ}}{2-1} \\\theta=\frac{(2 * 0+1) 180^{\circ}}{1} \\\theta=180^{\circ} \\\end{array}Step-4). The location of centroid will be,\begin{array}{l}\sigma=\sum \operatorname{Real}(\text { poles })-\sum \operatorname{Real}(\mathbf{z e r o s})-\mathbf{N}(\mathbf{z}) \\\sigma=\begin{array}{c}-\mathbf{1}-\mathbf{1}-(-\mathbf{3}) \\\mathbf{2}-\mathbf{1}\end{array} \\\sigma=\frac{-\mathbf{1}-\mathbf{1}+\mathbf{3}}{1} \\\sigma=\begin{array}{c}-\mathbf{2}+\mathbf{3} \\\mathbf{1}\end{array} \\\sigma=1 \\\mathbf{1} \\\sigma=\mathbf{1}\end{array}Step-5). now to get the location breakaway/break-in point, we need \mathrm{dk} / \mathrm{ds}=0.Characteristics equation will be 1+\mathrm{G}(\mathrm{s})=\begin{array}{l}\mathbf{1}+\mathbf{G}(\mathbf{s})=\mathbf{0} \\\mathbf{1}+\begin{array}{c}\mathbf{k}(\mathbf{s}+\mathbf{3}) \\\mathbf{s}^{2}+\mathbf{2 s}+\mathbf{3}\end{array}=\mathbf{0} \\\frac{\mathbf{k}(\mathbf{s}+\mathbf{3})}{\mathbf{s}^{2}+\mathbf{2 s}+\mathbf{3}}=-\mathbf{1} \\\mathbf{k}=-\mathbf{s}^{2}+\mathbf{2 s}+\mathbf{3} \\\mathbf{s}+\mathbf{3}\end{array}Taking derivative of &quot;k&quot; with respect to \mathbf{s}, using quotient rule of differentiation, we get\begin{array}{l}\underset{\mathbf{d s}}{\mathbf{d k}}=-\left\{\begin{array}{c}(\mathrm{s}+\mathbf{3}) * \frac{\mathrm{d}^{2}+2 \mathrm{~s}+3}{\mathrm{ds}}-\left(\mathrm{s}^{2}+2 \mathrm{~s}+3\right) * \frac{\mathrm{d}[\mathrm{s}+3]}{\mathrm{ds}} \\(\mathrm{s}+3)^{2}\end{array}\right\} \\\frac{\mathrm{dk}}{\mathrm{ds}}=-\left\{\frac{(\mathrm{s}+3) *(2 \mathrm{~s}+2)-\left(\mathrm{s}^{2}+2 \mathrm{~s}+3\right) * 1}{(\mathrm{~s}+3)^{2}}\right\} \\\frac{\mathrm{dk}}{\mathrm{ds}}=-\left\{\begin{array}{c}2 \mathrm{~s}^{2}+8 \mathrm{~s}+6-\mathrm{s}^{2}-2 \mathrm{~s}-3 \\(\mathrm{~s}+3)^{2}\end{array}\right\} \\\frac{\mathrm{dk}}{\mathrm{ds}}=-\left\{\frac{\mathrm{s}^{2}+6 \mathrm{~s}+\mathbf{3}}{(\mathrm{s}+\mathbf{3})^{2}}\right\}=\mathbf{0} \\-\left\{\frac{\mathbf{s}^{2}+6 \mathrm{~s}+3}{(\mathrm{~s}+3)^{2}}\right\}=\mathbf{0} \\\frac{\mathbf{s}^{2}+6 \mathbf{s}+\mathbf{3}}{(\mathrm{s}+\mathbf{3})^{2}}=\mathbf{0} \\\mathbf{s}^{2}+\mathbf{6 s}+\mathbf{3}=\mathbf{0} \\\end{array}Solving above quadratic equation, we will get two value of &quot;s&quot;,\begin{array}{l}\mathbf{s}=\frac{-6 \pm \sqrt{6^{2}-4 * 1 * 3}}{2 * 1} \\\mathbf{s}=\frac{-6 \pm \sqrt{36-12}}{2} \\\mathbf{s}=\frac{-6 \pm 4.89898}{2} \\\mathbf{s}=-0.55,-5.45\end{array}Only s=-5.45 will be valid breakaway point it lies on root locus plane, while the point s=-0.55 doesn't lies on root locus plane.Step-6) The location of zeros, poles and breakaway point is shown belowStep-7). Now the angle of departure for complex poles will be,\phi_{\mathbf{D}}=\mathbf{1 8 0}^{\circ}-\phiWhere, \Phi=< poles -< zeros\begin{array}{l}\phi_{\mathbf{D}}=\mathbf{1 8 0}^{\circ}-\left\{\mathbf{9 0}^{\circ}-\mathbf{3 5 . 2 6 4 ^ { \circ } \}}\right. \\\phi_{\mathrm{D}}=\mathbf{1 8 0}^{\circ}-\mathbf{5 4 . 7 3 6}^{\circ} \\\phi_{\mathrm{D}}=\mathbf{1 2 5 . 2 6 ^ { \circ }} \\\end{array}Step-8). Finally sketching the root locus diagram, ...