# Root Locus Techniques

 Question 1
The open loop transfer function of a unity gain negative feedback system is given by $G(s)=\frac{k}{s^2+4s-5}$.
The range of $k$ for which the system is stable, is
 A $k \gt 3$ B $k \lt 3$ C $k \gt 5$ D $k \lt 5$
GATE EE 2022   Control Systems
Question 1 Explanation:
Characteristic equation:
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
$\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}$
Hence, for stable system,
$k-5 \gt 0 \;\; \Rightarrow \; k \gt 5$
 Question 2
The root locus of the feedback control system having the characteristic equation
$s^{2}+ 6Ks +2s +5 =0 \; where \; K \gt 0$, enters into the real axis at
 A s=-1 B s=$-\sqrt{5}$ C s=-5 D s=$\sqrt{5}$
GATE EE 2017-SET-2   Control Systems
Question 2 Explanation:
$G(s)H(s)=\frac{6Ks}{s^2+2s+5}$
The point at which root locus enters real axis (break away point) is given by $\frac{dK}{ds}=0$
$K=-\frac{(s^2+2s+5)}{6s}$
$\frac{dK}{ds}=0\;\Rightarrow \;s=\pm \sqrt{5}$
$\therefore \;\;s=-\sqrt{5}$

 Question 3
The gain at the breakaway point of the root locus of a unity feedback system with open loop transfer function $G(s)=\frac{Ks}{(s-1)(s-4)}$ is
 A 1 B 2 C 5 D 9
GATE EE 2016-SET-2   Control Systems
Question 3 Explanation:
$OLTS\Rightarrow G(s)=\frac{Ks}{(s-1)(s-4)}$
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, $K=\frac{1 \times 2}{2}=1$
 Question 4
An open loop transfer function G(s) of a system is
$G(s)=\frac{K}{s(s+1)(s+2)}$
For a unity feedback system, the breakaway point of the root loci on the real axis occurs at,
 A -0.42 B -1.58 C -0.42 and -1.58 D none of the above
GATE EE 2015-SET-2   Control Systems
Question 4 Explanation:
$G_1(s)=\frac{k}{s(s+1)(s+2)}$
$\Rightarrow \;\; 1+\frac{k}{s(s+1)(s+2)}=0$
$\Rightarrow \;\; s(s^2+3s+2)+k=0$
$\Rightarrow \;\; s^3+3s^2+2s+k=0$
$\Rightarrow \;\; k=-s^3-3s^2-2s\;\;\;...(i)$
$\Rightarrow \;\;\frac{dk}{ds}=0$
$\;\;\;\;\frac{dk}{ds}=-3s^2-6s-2=0$
$\;\;\;\;s=-0.422,-1.577$ Only s=-0.422 lie on root locus. Therefore brakaway point is s=-0.42.
 Question 5
The open loop poles of a third order unity feedback system are at 0, -1, -2. Let the frequency corresponding to the point where the root locus of the system transits to unstable region be K. Now suppose we introduce a zero in the open loop transfer function at -3, while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits to unstable region?
 A It corresponds to a frequency greater than K B It corresponds to a frequency less than K C It corresponds to a frequency K D Root locus of modified system never transits to unstable region
GATE EE 2015-SET-1   Control Systems
Question 5 Explanation:
$\frac{K(s+3)}{s(s+1)(s+2)}$
$P=3, \; Z=1$
$P-Z=3-1=2$
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.

There are 5 questions to complete.