# Root Locus

 Question 1
The root-locus plot of a closed-loop system with unity negative feedback and transfer function $KG(s)$ in the forward path is shown in the figure. Note that $K$ is varied from 0 to $\infty$.
Select the transfer function $G(s)$ that results in the root-locus plot of the closed-loop system as shown in the figure. A $G(s)=\frac{1}{(s+1)^5}$ B $G(s)=\frac{1}{(s^5+1)}$ C $G(s)=\frac{s-1}{(s+1)^6}$ D $G(s)=\frac{s+1}{(s^6+1)}$
GATE EC 2022   Control Systems
Question 1 Explanation:
Here 5 Root Locus branches are diverging from same point, this can possible only when if we have 5 poles in the system at the same point because Root Locus branch departs from open loop pole and
Number of Root Locus branches = Number of open loop poles or Number of zero (Whichever is greater).
Here, there are 5 multiple Real poles, and matching with option (A).
 Question 2
The characteristic equation of a system is

$s^3 + 3s^2 + (K + 2)s + 3K = 0$

In the root locus plot for the given system, as K varies from 0 to $\infty$, the break-away or break-in point(s) lie within
 A (-1,0) B (-2,-1) C (-3,-2) D (-$\infty$ ,-3)
GATE EC 2020   Control Systems
Question 2 Explanation:
$Q(s)=1+G(s)H(s)=0$
$s^{3}+3s^{2}+2s+ks+3k=0$
$-k=\frac{s^{3}+3s^{2}+2s}{s+3}$
$-\frac{\mathrm{d} k}{\mathrm{d} s}=\frac{(s+3)(3s^{2}+6s+2)-(s^{3}+3s^{2}+2s)}{(s+3)^{2}}=0$
$3s^{3}+6s^{2}+2s+9s^{2}+18s+6-s^{3}-3s^{2}-2s=0$
$2s^{3}+12s^{2}+18s+6=0$
$s=-0.46,-3.87,-1.65$ $\therefore$ Break-away point lies between (0, -1), i.e. (-1, 0)

 Question 3
A linear time invariant (LTI) system with the transfer function

$G(s)=\frac{K(s^{2}+2s+2)}{(s^{2}-3s+2)}$

is connected in unity feedback configuration as shown in the figure. For the closed loop system shown, the root locus for $0 \lt K \lt \infty$ intersects the imaginary axis for K = 1.5. The closed loop system is stable for
 A $K>1.5$ B $1 \lt K \lt 1.5$ C $0 \lt K \lt 1$ D no positive value of K
GATE EC 2017-SET-1   Control Systems
Question 3 Explanation:
Given that,
$G(s)=\frac{K\left(s^{2}+2 s+2\right)}{s^{2}-3 s+2}$
The root locus plot of the given system is follows: $\therefore$System is stable for $K \gt 1.5$.
 Question 4
The forward-path transfer function and the feedback-path transfer function of a single loop negative feedback control system are given as

$G(s)=\frac{K(s+2)}{s^{2}+2s+2}$ and H(s)=1,

respectively. If the variable parameter K is real positive, then the location of the breakaway point on the root locus diagram of the system is __________
 A -2 B -3.4 C -1.5 D -6.5
GATE EC 2016-SET-3   Control Systems
Question 4 Explanation:
\begin{aligned} G(s) &=\frac{K(s+2)}{s^{2}+2 s+2} \\ H(s)&=1 \\ 1+G(s) H(s) &=0 \\ 1+\frac{K}{\left(s^{2}\right)+2 s+2}(s+2) &=0\\ K&=-\frac{\left(s^{2}+2 s+2\right)}{(s+2)} \end{aligned} $\frac{d K}{d s}=-\left(\frac{(2 s+2)(s+2)-\left(s^{2}+2 s+2\right)}{(s+2)^{2}}\right)$
For break away points, $\frac{d K}{d s}=0$
$\begin{array}{l} (2 s+2)(s+2)-\left(s^{2}+2 s+2\right)=0 \\ 2 s^{2}+6 s+4-s^{2}-2 s-2=0 \\ s^{2}+4 s+2=0 \end{array}$
\begin{aligned} s&=\frac{-4 \pm \sqrt{16-8}}{2} \\ &=\frac{-4 \pm 2 \sqrt{2}}{2}=-2 \pm \sqrt{2} \\ s&=-0.58 ; s=-3.41 \\ \text{But,}\quad s&=-3.41 \text{ lies on root locus }\\ \text{Hence}\quad s&=-3.41 \end{aligned}
 Question 5
The open-loop transfer function of a unity-feedback control system is
$G(s)=\frac{K}{s^{2}+5s+5}$
The value of K at the breakaway point of the feedback control system's root-locus plot is ________
 A 0.65 B 1.25 C 2.25 D 3.15
GATE EC 2016-SET-1   Control Systems
Question 5 Explanation:
Characteristic equation $1+G(s)H(s)=0$
$1+\frac{K}{s^2+5s+5}=0$
$K=-s^2-5s-5$
For break away point $\frac{dK}{ds}=0$
$\frac{dK}{ds}=-2s+5=0\;\Rightarrow \; s=-2.5$
According to magnitude condition,
\begin{aligned} |G(s)H(s)|_{s=-2.5} &=1 \\ |G(s)H(s)|_{s=-2.5} &=\frac{|K|}{|(2.5)^2 +5 \times -2.5+5|}=1 \\ |K|&=|(6.25+5-12.5)|=1.25 \\ K&=\pm 1.25 \end{aligned}

There are 5 questions to complete.