Root Locus Techniques Practice Questions

Given: The Nyquist plot of a strictly stable transfer function $G(s)$ has the numerator polynomial $(s - 3)$ , and it encircles the critical point $-1$ once in the anti-clockwise direction. Step 1: Understand the Nyquist criterion The Nyquist criterion provides a relationship between the open-loop transfer function's Nyquist plot and the stability of the closed-loop system. According to the Nyquist criterion, the number of encirclements of the critical point $-1$ determines the stability of the closed-loop system. Step 2: Analyze the information given The Nyquist plot encircles the critical point $-1$ once in the anti-clockwise direction. The open-loop transfer function is strictly stable, implying that all poles are in the left half of the complex plane. The system's Nyquist plot encircles the critical point $-1$ once in the anti-clockwise direction. According to the Nyquist criterion, this indicates that the number of right-half-plane poles of the open-loop transfer function is 1. Step 3: Apply the Nyquist criterion for stability The Nyquist criterion states that for the closed-loop system to be stable, the number of encirclements of the point $-1$ must be equal to the number of right-half-plane poles of the open-loop transfer function. Since the Nyquist plot encircles $-1$ once in the anti-clockwise direction, the number of right-half-plane poles is 1. This suggests that the closed-loop system is unstable because the number of encirclements of $-1$ is non-zero, and the system has a pole in the right half of the complex plane. Final Answer: The system is unstable.

Characteristic equation:

R-H criteria:

Hence, for stable system, $k-5 \gt 0 \;\; \Rightarrow \; k \gt 5$

$G(s)H(s)=\frac{6Ks}{s^{2+}2s+5}$ The point at which root locus enters real axis (break away point) is given by $\frac{dK}{ds}=0$ $K=-\frac{(s^{2+}2s+5)}{6s}$ $\frac{dK}{ds}=0\;\Rightarrow \;s=\pm \sqrt{5}$ $\therefore \;\;s=-\sqrt{5}$

$OLTS\Rightarrow G(s)=\frac{Ks}{(s-1)(s-4)}$ Now, characteristic equation, $1+G(s)H(s)=0$ $\frac{Ks}{(s-1)(s-4)}+1=0$ $\Rightarrow \;\; Ks+(s^{2-}5s+4)=0$ $\;\;K=-\frac{(s^{2-}5s+4)}{s}=\left[ s-5+\frac{4}{s} \right]$ For break away point $: \frac{dK}{ds}=0$ \frac{dK}{ds}=-\left[ 1-0-\frac{4}{s^2} \right]\=0$We get,$s=\pm 2$Therefore valid break away point is s=2 Now gain at s=2 is K=(Product of distances from all the poles to break away point)/(Product of distances from all the zero to break away point) Gain,$K=\frac{1 × 2}{2}=1$

$\frac{K(s+3)}{s(s+1)(s+2)}$ $P=3, \; Z=1$ $P-Z=3-1=2$ asymptotes angle $=90^{\circ}, 180^{\circ}$ Centroid $=\frac{(-3)-(-3)}{2}=0$ as root locus never cross asymptotes so, will remain in left sideof x-axis.

$G_1(s)=\frac{k}{s(s+1)(s+2)}$ $\Rightarrow \;\; 1+\frac{k}{s(s+1)(s+2)}=0$ $\Rightarrow \;\; s(s^{2+}3s+2)+k=0$ $\Rightarrow \;\; s^{3+}3s^{2+}2s+k=0$ $\Rightarrow \;\; k=-s^{3-}3s^{2-}2s\;\;\;...(i)$ $\Rightarrow \;\;\frac{dk}{ds}=0$ $\;\;\;\;\frac{dk}{ds}=-3s^{2-}6s-2=0$ $\;\;\;\;s=-0.422,-1.577$ Only s=-0.422 lie on root locus. Therefore brakaway point is s=-0.42.

This is converse root locus having no zero. As, its $G(s)H(s)=\frac{K}{(s+1)(s+2)}$ Its converse root locus only valid when $K \lt 0$ So, C.L.T.F. $=\frac{G(s)H(s)}{1-G(s)H(s)} =\frac{\frac{K}{(s+1)(s+2)}}{1-\frac{K}{(s+1)(s+2)}} =\frac{K}{(s+1)(s+2)-K}$

$G(s)=\frac{k\left ( s+\frac{2}{3} \right )}{s^2(s+2)}$ and $H(s)=1$ Characteristic equation, $1+G(s)H(s)=0$ $\Rightarrow \;\; 1+\frac{k\left ( s+\frac{2}{3} \right )}{s^2(s+2)}=0$ $\Rightarrow \;\; s^{3+}2s^{2+}k\left ( s+\frac{2}{3} \right )=0$ $\Rightarrow \;\; s^{3+}2s^{2+}ks+\frac{2k}{3}=0$ Routh array: As $k \gt 0$ , there is no sign change in the 1st column of routh array. So the system is stable and all the three roots lie on LHS of s-plane. For $k \gt 0(k\neq 0)$ , none of the row of Routh array becomes zero. So root loci does not cross the $j\omega$ -axis. Number of Zero = Z = 1 Number of poles= P =3 Number of branches terminating at infinity = P-Z = 3-1 =2 Angle of asymptotes $=\frac{(2k+1) \times 180^{\circ}}{P-Z}$ $\;\;\;=\frac{(2k+1) \times 180^{\circ}}{2}$ $\;\;\;=(2k+1) \times 90^{\circ}$ $\;\;\;=90^{\circ} \; and \; 270^{\circ}$ $Centroid=\frac{\Sigma poles-\Sigma zero}{P-Z}$ $\;\;=\frac{0+0-2-\left ( -\frac{2}{3} \right )}{2}=-\frac{2}{3}$ Since, all the three branches terminates at $R_e(s)=-\frac{2}{3}$ . So, all the three roots have nearly equal real parts.

Characteristic equation, $s(s+1)(s+3)+k(s+2)=0$ $1+\frac{k(s+2)}{s(s+1)(s+3)}=0$ Comparing with $1+G(s)H(s)=0$ $G(s)H(s)=$ open-loop transfer function (OLTF) $=\frac{k(s+2)}{s(s+1)(s+2)}$ Number of zeros=Z=1 zero at -2 Number of poles = P= 3 poles at 0, -1 and -3 Number of branches terminating at infinity $= P-Z = 3-1 =2$ Angle of asymptotes $=\frac{(2k+1)\times 180^{\circ}}{P-Z}$ $=\frac{(2k+1)\times 180^{\circ}}{2}$ $=(2k+1) \times 90^{\circ}$ $=90^{\circ} \; and \; 270^{\circ}$ $Centroid=\frac{\Sigma poles -\Sigma zeros}{P-Z}$ $=\frac{0-1-3-(-2)}{2}=1$ Breakaway point lies in the range $-1 \lt Re[s] \lt 0$ and two branches terminates at infinityalong the asymptotes $Re(s)=-1$ .

Characteristic function $\Rightarrow \;\;(s^{2-}4)(s+1)+k(s-1)$ $\Rightarrow \;\;1+\frac{k(s-1)}{(s^{2-}4)(s+1)}\equiv 1+G(s)H(s)$ Open loop transfer function $=G(s)H(s)$ $=\frac{k(s-1)}{(s^{2-}4)(s+1)}$ Zero of OLTF s=1; z=1 Poles of OLTS s=-1,-2,+2,P=3 The root locus starts from open-loop poles and terminates either on open-loop zero or infinity. Root locus exist on section of real axis it the sum of the open-loop poles and zeros to the right of the section is odd. Number of branches teminating on infinity. =P-Z=3-1=2 Angle of asymptotes $=\frac{(2k+1)\times 180^{\circ}}{P-Z}$ $=\frac{(2k+1)\times 180^{\circ}}{2}$ $=90^{\circ} \; and \; 270^{\circ}$ Intersection of asymptotes on real axis (centroid) $=\frac{\Sigma poles-\Sigma zeros}{P-Z}$ $=\frac{(-1-2+2)-(1)}{2}=-1$ Option (B) is correct on the basic of above analysis.

These are three asymptotes with angle $60^{\circ},180^{\circ}$ and $300^{\circ}$ Angle of asymptotes $=\frac{(2k+1)\times 180^{\circ}}{P-Z}$ Where, k=0,1,2 up to (P-Z)-1 as angle are $60^{\circ},180^{\circ}$ and $300^{\circ}$ it means P-Z=3 Intersection of asymptotes on real axis $x=\frac{\Sigma poles -\Sigma zeros}{P-Z}$ Since, system does not have zeroes $x=\frac{\Sigma poles}{P}$ As asymptotes intersect at origin, it means all the three poles are on right Hence, Option (A) is correct.

$G(s)H(s)=\frac{K}{s^2}$ $\therefore \;\;P_Z=2$ $Centroid=\frac{0}{2}=0$ Angle of asymptotes $=90^{\circ},270^{\circ}$ $\therefore \;\;$ Option (B) is correct.

Closed loop system limits the peak value.