Root Locus – GATE ECE Control Systems Practice Questions & PYQs | BattleExam | BattleExam

Q7GATE 2015NAT

The open-loop transfer function of a plant in a unity feedback configuration is given as $G(s)=\frac{K(s+4)}{(s+8)(s^{2}-9)}$ . The value of the gain K(> 0) for which -1+ j2 lies on the root locus is_______.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} G(s) &=\frac{K(s+4)}{(s+8)\left(s^{2}-9\right)} \\ &=\frac{K(s+4)}{(s+8)(s+3)(s-3)} \end{aligned}

For the point (-1+2 j) to lie on the root locus the angle condition must be satisfy i.e. $\angle G(s) H(s)=\pm(2 Q+1) 180^{\circ}$ Taking LHS:

\begin{array}{l} \left.\angle G(s) H(s)\right|_{s=-1+2 j} \\ =\frac{\angle K+\angle(s+4)}{\angle(s+8)+\angle(s+3)+\angle(s-3)} \\ =\frac{\angle K+\angle(-1+2 j+4)}{\angle(-1+2 j+8)+\angle(-1+2 j+3)+\angle(-1+2 j-3)} \\ =\frac{0+\tan ^{-1}\left(\frac{2}{3}\right)}{\tan ^{-1}\left(\frac{2}{7}\right)+\tan ^{-1}(1)+180^{\circ}-\tan ^{-1}\left(\frac{1}{2}\right)} \\ =33.69^{\circ}-15.945^{\circ}-45^{\circ}-180^{\circ}+26.565^{\circ} \\ \simeq 180^{\circ} \end{array}

As angle condition is satisfied the value of system gain K can be obtained by using magnitude condition

\begin{array}{l} i.e. |G(s) H(s)|_{s=-1+2 j}=1 \\ =\frac{K \sqrt{(-1+4)^{2}+2^{2}}}{\sqrt{(-1+8)^{2}+2^{2}} \cdot \sqrt{(-1+3)^{2}+2^{2}} \cdot \sqrt{(-1-3)^{2}+2^{2}}} \\ =1 \\ =\frac{K \sqrt{9+4}}{\sqrt{49+4} \sqrt{4+4} \sqrt{16+4}}=1 \\ =\frac{K \sqrt{13}}{\sqrt{53} \sqrt{8} \sqrt{20}}=1 \\ K=\frac{\sqrt{53 \times 8 \times 20}}{\sqrt{13}}=25.54 \end{array}