Root Locus Technique.pptx

Knowing the locations of the closed loop poles is important because,
• It affects the transient response.
• It indicates whether the system is stable or not.
• An important property of the control system is the variation of these locations as
the controller gain changes, therefore we need to quantify these variations.
• In this session we introduce the idea “the root locus” which as a graphical means
of quantifying the variations in pole locations (but not the zeros).
Introduction

Introduction
Consider a closed loop system with unity feedback that uses a simple controller,
Figure 1: Closed-loop control system
with a variable K

What is Root Locus??
• For the simple single loop system shown in Figure 1, we have characteristic
equation,
1 + 𝐾𝐺 𝑠 = 0
Where ‘K’ is the variable parameter.
• A root locus is a plot in s-plane of all possible locations of closed-loop poles with
some system parameter , usually controller gain varied from 0 to ∞.
• All the points on the root locus satisfies Equ(1) and Equ(1) can be rewritten in
polar form as,
• Therefore it is necessary that
(1)

Angle and Magnitude Condition
• The Angle Condition: ∠𝐾𝐺 𝑠 = ± 2𝑞 + 1 180
• The Magnitude Condition : 𝐾𝐺(𝑠) = 1
• In practice the angle condition is used to determine whether a point s lies on the
root locus, and if it does, the magnitude condition is used to determine the gain K
associated with that point, since 𝐾 = 1
𝐺(𝑠)
Example :
• Consider system with 𝐺 𝑠 =
𝑲
𝒔(𝒔+𝟒)
, test a point 𝑠 = −2 + 𝑗5 for its existence on
root locus. Also find corresponding K.

Definition
• The root locus is a plot of the roots of the characteristic equation of the closed-
loop system as a function of one system parameter varies, such as the gain of the
open-loop transfer function.
• It is a method that determines how the poles move around the s-plane as we
change one control parameter.
• This plot was introduced by Evans in 1948 and has been developed and used
extensively in control engineering.

Difference between Root Locus & Routh-
Hurwitz method
• The Root Locus Method ----- tells us the position of the poles in the s-plane for
each value of a control parameter.
• The Routh -Hurwitz Method ---- could only tell us for could only tell us for
which values of the control parameter the poles would be to the left of a given
vertical axis in the s - plane.

Plotting roots of a characteristic equation-
Example 1
• Let us consider a position control
servomechanism system.
• The plant consists of servomotor and
load, controlled by controller power
amplifier.
• The open loop transfer function
𝐺 𝑠 =
𝑲
𝒔(𝒔 + 𝟐)
• Open loop poles, 𝑠 = 0 𝑎𝑛𝑑 𝑠 = −2
K
𝟏
𝒔(𝒔 + 𝟐)
R(s) C(s)
Power
Amplifier
Servomotor
+ Load

Continued…
• The closed loop transfer function is given by,
𝑇 𝑠 =
𝑲
𝒔𝟐 + 𝟐𝒔 + 𝑲
• The characteristic equation is,
𝒔𝟐 + 𝟐𝒔 + 𝑲 = 0
• The second order system under consideration is stable for positive values of ‘K’.
• The relative stability of the system depends upon the location of poles
𝑠1,2 = −1 ± 1 − 𝐾
• Values of the poles depends upon the system parameter ‘K’.
Gain of the open loop
transfer function

• At K=0, 𝑠1 = 0 , 𝑠2 = −2 (which are the open loop poles of the system)
• As ‘K’ increases , the roots move towards each other.
• 𝟎 < 𝑲 < 𝟏  the roots 𝑠1,2 are real and lie on the negative real axis of the s-plane
between -2 & -1 and 0 to -1 , respectively
• At K=1, 𝑠1,2 = −1.
• As ‘K’ is increased further , the roots breaks away from real axis and becomes
complex conjugate and real part remains fixed at s=-1.
• 𝑲 > 𝟏, the roots complex and are given by, 𝑠1,2 = −𝜁𝜔𝑛 ± 𝑗𝜔𝑛 1 − 𝜁2 = −1 ±
𝑗 1 − 𝐾
Continued…

Continued…
• There are two branches A-C-E and
B-C-D in the plot.
• Number root locus branches is equal to
number of closed loop poles.
• Each root loci starts at open loop pole
value at K=0 and terminates at zero as
‘K’ tends to infinity.
• For each value of ‘K’ gives one closed
loop pole value.
Note: Each point on the locus
is closed loop root, so the
locus is called root locus

Example 2
Consider 𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠+1)
𝑠(𝑠+5)
. Obtain the root locus using direct method.
The characteristic equation is ,
1 + 𝐺 𝑠 𝐻 𝑠 = 1 +
𝐾(𝑠 + 1)
𝑠(𝑠 + 5)
= 0
𝑠2 + 𝑠 𝐾 + 5 + 𝐾 = 0
Roots are 𝑠1,2 =
−(𝐾+5)
2
±
𝐾2+6𝐾+25
2
K
𝑠1 =
−(𝐾 + 5)
2
+
𝐾2 + 6𝐾 + 25
2
𝑠2 =
−(𝐾 + 5)
2
−
𝐾2 + 6𝐾 + 25
2
0 0 -5
1 -0.1715 -5.828
5 -0.527 -9.472
.
.
.
.
.
.
∞ -1 -∞

• Consider a second order multi-loop system with CLTF 𝑇 𝑠 =
𝐾
𝑠2+ 0.2𝐾+2 𝑠+𝐾
,
construct root locus.
• The characteristic equation , 𝑠2 + 2𝑠 + 𝐾 0.2𝑠 + 1 = 0
• Which can be rewritten as 1 +
0.2𝐾(𝑠+5)
𝑠(𝑠+2)
= 0
• Open loop poles 𝑠1 = 0 and 𝑠1 = −2 and open loop zero z = -5
• Closed loop roots , 𝑠1,2 = −(0.1𝐾 + 1) ± (0.1𝐾 + 1)2−𝐾
Example 3

Drawbacks of drawing root locus using direct
method of substitution
• It is very difficult to plot the root locus for higher order systems by the method of
substituting different values of ‘K’ in the roots of the characteristic equation.
• To simplify the construction of root locus for higher order systems certain rules
are specified.

• To draw a rough draft of root locus , some construction rules are devised.
• Let us discuss the rules, first stating the rule, then giving justification for its
validity and citing an example.
Rules for construction of root locus

Rule 1
Locate open loop poles and zeros on ‘s’ plane.
• The values of closed loop poles & zeros will be equal to open loop poles &
zeros at K=0.
• Rule 1: Root locus are always symmetrical about the real axis .
• The root of the characteristic equation are either real or complex or
combination of both.

• Rule 2: The number of branches in the root locus
• Let G(s)H(s) = open loop transfer function of a given closed loop system,
P = no of open loop poles & N= no of open loop zeros
Note: The root locus branch always starts from open loop pole and ends on open
loop zero.
• The number of root locus branches N is equal to the number of finite open
loop poles P or the number of finite open loop zeros Z, whichever is greater.
• If P > Z , N = P , no of branches that terminate at infinity is P-Z.
• If Z > P , N = Z , no of branches that originates from infinity is Z-P.
Rule 2

• Rule 3: A point on the real axis lies on the root locus if the sum of number of open
loop poles and open loop zeros , on the real axis , to the right hand side of that
point is ODD.
Example1 : 𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠+1)(𝑠+4)
𝑠(𝑠+3)(𝑠+5)
, find on which sections of the real axis root
locus exists
Rule 3

Example2: 𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠+2)
𝑠2(𝑠2+2𝑠+2)(𝑠+3)
. Find the sections of real axis which
belongs to the root locus.
Rule 4

• Rule 4:
• It is a normal occurrence to have less finite zeros than finite poles, so system will
have infinite zeros.
• For each infinite zero, the root locus will have a branch that travels to the infinite
zero along a line called Asymptote.
• So there will be one asymptote for one infinite zero.
• But the question is where will these asymptotes located and what will be there
angle of orientation??
Rule 4

Rule 4
• Each asymptote is oriented at an angle from the positive real axis.
• The asymptote angles are designated θ,
𝜃 =
(2𝑘 + 1)𝜋
𝑃 − 𝑍
; 𝑘 = 0,1, … (𝑃 − 𝑍 − 1)
• So if there is one infinite zero, there is one asymptote and its asymptote angle
is180° 𝑜𝑟 𝜋.
• If there are two infinite zeros, there will be two asymptotes with angles
𝜋
2
and
3𝜋
2
.

Rule 4
• Next question is where these asymptotes are located??
• All the asymptotes intersect at a point on the real axis called centroid 𝜎𝐴.
• The coordinates of this centroid can be calculated as,
𝜎𝐴 =
𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺 𝑠 𝐻 𝑠 − 𝑧𝑒𝑟𝑜𝑠 𝑜𝑓 𝐺 𝑠 𝐻 𝑠
𝑃 − 𝑍
• Centroid is always real , it may be located on negative or positive real axis.
• Note: Centroid may or may not be the part of the root locus.

Rule 4
Example1: 𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠+2)
(𝑠+1+𝑗4)(𝑠+1−𝑗4)(𝑠+3)(𝑠+4)
, calculate the angles of
asymptotes and the centroid.
Solution: The root locus of 1+G(s)H(s)=0, will consist of four root loci (branches)
starting from open loop pole with K=0.
One root will terminate on open loop zero s = -2 as 𝐾 ⟶ ∞.
Other three roots will terminate at infinity as 𝐾 ⟶ ∞ along the asymptotes
radiating from the centroid 𝜎𝐴.

Rule 4
• Angles of the asymptotes ,
𝜃 =
(2𝑘 + 1)𝜋
𝑃 − 𝑍
; 𝑘 = 0,1, … (𝑃 − 𝑍 − 1)
Here P-Z = 3, k= 0,1,2.
Angles of asymptotes, 𝜃 =
(2𝑘+1)𝜋
𝑃−𝑍
are 60° , 180° and 300° .
Centroid,𝜎𝐴 =
𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺 𝑠 𝐻 𝑠 − 𝑧𝑒𝑟𝑜𝑠 𝑜𝑓 𝐺 𝑠 𝐻 𝑠
𝑃−𝑍
=
−1−1−3−4 −2
3
= −2.33

Rule 4
• Example1:𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠+10)
𝑠(𝑠+1)(𝑠+2)
, calculate the angles of asymptotes and the
centroid.

Rule 4
Conclusion from Rule 4:
• The angles of the asymptotes are fixed for fixed values of P – Z.
• The values of P & Z may be different but for particular values of P – Z, angles of
asymptotes are fixed.
P-Z Number of asymptotes
required
Angles of asymptotes
0 0 -
1 1 180
2 2 90, 180
3 3 60,180,300
4 4 45,135,225,315

Rule 5
• Rule 5: Breakaway point where multiple roots occur at same point , for a
particular value of K
• Example 1: 𝐺 𝑠 =
𝑲
𝒔(𝒔+𝟐)
, in this case both roots 𝑠1& 𝑠2 starts from open loop
values and approach to common value -1 for K=1.
• From this breakaway point at s=-1, the roots becomes complex conjugate pairs as
K increases towards infinity from 1.
• Such a point where two or more roots occur for a particular value of K is called
breakaway point.
• The root locus branches will leave this breakaway point at an angle of ±
𝟏𝟖𝟎
𝒏
,
where ‘n’ is number of branches approaching breakaway point.

Rule 5
• Foer this example breakaway point is
at s=-1.
• Two root locus branches approaches
the breakaway point.
• Branch B-C-D depart away from
breakaway point at 90 degree.
• Branch A-C-E depart away from
breakaway point at -90 degree
• Note: the breakaway point will always
be on root locus.

Rule 5
• General predictions about existence of breakaway points
1. If there are adjacently placed poles on the real axis and the real axis between
them is part of root locus then minimum one breakaway point exists between
the adjacent poles
Example1 : 𝐺 𝑠 =
𝑲
𝒔(𝒔+𝟐)

Rule 5
2. If there are adjacently placed zeros on the real axis and the real axis between them
is part of root locus then minimum one breakaway point exists between the adjacent
zeros
Example 2: 𝐺 𝑠 =
𝑲(𝒔+𝟐)(𝒔+𝟒)
𝒔𝟐(𝒔+𝟔)

Rule 5
3. If there is a zero on the real axis and to the left of that zero there is no zero or pole
existing and complete real axis to the left of that zero is part of root locus then there
exist breakaway point to the left of that zero
Example 3: 𝐺 𝑠 =
𝑲(𝒔+𝟐)(𝒔+𝟒)
𝒔(𝒔𝟐+𝟐𝒔+𝟐𝟎)

Rule 5
Determination of breakaway point:
1. Construct the characteristic equation 1+G(s)H(s)=0.
2. Write the characteristic equation as, K = f(s).
3. Differentiate K = f(s) equation and equate it to zero
𝑑𝐾
𝑑𝑠
= 0
4. Roots of the equation
𝑑𝐾
𝑑𝑠
= 0 gives us breakaway points for range of K from -∞
to + ∞
5. So the valid breakaway point is obtained by substituting the point in equation of
K = f(s), if value of K is positive that breakaway point is valid & if value of K is
negative that breakaway point is invalid

• Example 1: For 𝐺 𝑠 𝐻 𝑠 =
𝑲
𝒔(𝒔+𝟏)(𝒔+𝟒)
, determine the valid breakaway points.

Rule 6
• Rule 6: Intersection of root loci with imaginary axis can be determined using
Routh’s criteria.
• Segment of root loci can exist in right half ,this indicates the stability.
• The point at which root loci crosses the imaginary axis give stability limits.
• Basic Routh’s array gives this stability limit.
• Substituting the value of K margin obtained in auxiliary equation we get
intersection point of root loci with imaginary axis.

Rule 6
• Example 1: 𝐺 𝑠 =
𝑲
𝒔(𝒔+𝟏)(𝒔+𝟒)

Rule 6
• We obtain,

• Rule 7: Angle of departure at complex conjugate poles and angle of arrival at
complex conjugate zeros.
• A branch always leaves from an open loop pole.
• If the open loop pole is complex, we calculate angle of departure 𝜙𝑑
• The angle of departure is given as,
𝜙𝑑 = 180∘ − 𝜙 , 𝑤ℎ𝑒𝑟𝑒 𝜙 = 𝜙𝑃 − 𝜙𝑍
𝜙𝑃 = sum of angles made by other open loop poles at the pole under consideration
𝜙𝑍 = sum of angles made by open loop zeros at the pole under consideration
Rule 7

Rule 7
• Example: 𝐺 𝑠 =
𝑲(𝒔+𝟐)
𝒔(𝒔+𝟒)(𝒔𝟐+𝟐𝒔+𝟐)
, calculate angles of departure at complex
conjugate poles

• Rule 7: Angle of departure at complex conjugate poles and angle of arrival at
complex conjugate zeros.
• If the open loop zero is complex, we calculate angle of arrival 𝜙𝑎
• The angle of arrival is given as,
𝜙𝑎 = 180∘ + 𝜙 , 𝑤ℎ𝑒𝑟𝑒 𝜙 = 𝜙𝑃 − 𝜙𝑍
𝜙𝑃 = sum of angles made by other open loop poles at the pole under consideration
𝜙𝑍 = sum of angles made by open loop zeros at the pole under consideration
Rule 7

Rule 7
• Example: 𝐺 𝑠 =
𝑲(𝒔𝟐+𝟑𝒔+𝟏𝟎)
𝒔(𝒔+𝟐)(𝒔𝟐+𝟐𝒔+𝟏𝟎𝟏)
, calculate arrival angles at complex conjugate
zeros

Example : Construction Root Locus without
complex poles
• Draw the approximate root locus diagram for a closed loop system whose loop
transfer function is given by,
𝐺 𝑠 =
𝑲
𝒔(𝒔 + 𝟓)(𝒔 + 𝟏𝟎)
Comment on stability.

Example : Construction Root Locus without
complex poles
• Consider the feedback system with the characteristic equation,
1 +
𝑲
𝒔(𝒔 + 𝟏)(𝒔 + 𝟐)
= 𝟎 ; 𝐊 ≥ 𝟎
Sketch the root locus.

Example : Construction Root Locus with
complex poles
• Sketch the root locus for the system having
𝐺 𝑠 =
𝑲
𝒔(𝒔𝟐+𝟐𝒔+𝟐)

Root Locus Technique.pptx

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