# Root Locus Techniques

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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.
 Question 5
The root locus of a unity feedback system is shown in the figure. The closed loop transfer function of the system is
 A $\frac{C(s)}{R(s)}=\frac{K}{(s+1)(s+2)}$ B $\frac{C(s)}{R(s)}=\frac{-K}{(s+1)(s+2)}+K$ C $\frac{C(s)}{R(s)}=\frac{K}{(s+1)(s+2)-K}$ D $\frac{C(s)}{R(s)}=\frac{K}{(s+1)(s+2)+K}$
GATE EE 2014-SET-1   Control Systems
Question 5 Explanation:
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.$=\frac{G(s)H(s)}{1-G(s)H(s)} =\frac{\frac{K}{(s+1)(s+2)}}{1-\frac{K}{(s+1)(s+2)}} =\frac{K}{(s+1)(s+2)-K}$
 Question 6
The open loop transfer function G(s) of a unity feedback control system is given as

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

From the root locus, at can be inferred that when k tends to positive infinity,
 A Three roots with nearly equal real parts exist on the left half of the s-plane B One real root is found on the right half of the s-plane C The root loci cross the $j\omega$ axis for a finite value of k; k$\neq$0 D Three real roots are found on the right half of the s-plane
GATE EE 2011   Control Systems
Question 6 Explanation:
$G(s)=\frac{k\left ( s+\frac{2}{3} \right )}{s^2(s+2)}$
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.
 Question 7
The characteristic equation of a closed-loop system is s(s+1)(s+3)k(s+2)=0, $k \gt 0$. Which of the following statements is true ?
 A Its root are always real B It cannot have a breakaway point in the range$-1 \lt Re[s] \lt 0$ C Two of its roots tend to infinity along the asymptotes Re[s] =-1 D It may have complex roots in the right half plane.
GATE EE 2010   Control Systems
Question 7 Explanation:
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 $Re(s)=-1$.
 Question 8
A closed-loop system has the characteristic function $(s^{2}- 4) (s + 1) + K(s - 1) = 0$. Its root locus plot against K is A a B b C c D d
GATE EE 2006   Control Systems
Question 8 Explanation:
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.
 Question 9
Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is A $\frac{K}{s^{3}}$ B $\frac{K}{s^{2}(s+1)}$ C $\frac{K}{s(s^{2}+1)}$ D $\frac{K}{s(s^{2}-1)}$
GATE EE 2005   Control Systems
Question 9 Explanation: 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.
 Question 10
A unity feedback system has an open loop transfer function, $G(s)=\frac{K}{s^{2}}$. The root locus plot is A a B b C c D d
GATE EE 2002   Control Systems
Question 10 Explanation:
$G(s)H(s)=\frac{K}{s^2}$
$\therefore \;\;P_Z=2$
$Centroid=\frac{0}{2}=0$
Angle of asymptotes$=90^{\circ},270^{\circ}$
$\therefore \;\;$ Option (B) is correct.
There are 10 questions to complete. 