Control Systems I
ME344
Chapter 8: Root Locus Technique
Dr.Mariam Wajdi Ibrahim, IEEE member
Department of Mechatronics Engineering,
German Jordanian University
Introduction
 The root locus is a graphical technique that can be used to describe qualitatively the
performance of a system as various parameters are changed. For example, the effect of
varying gain upon percent overshoot, settling time, and peak time can be vividly
displayed.
 The root locus can provide solutions for systems of order higher than 2. (e.g., gains and
other system parameters can be designed to yield a desired transient response )
 Besides transient response, the root locus also gives a graphical representation of a system's
stability. We can clearly see ranges of stability, ranges of instability, and the conditions that
cause a system to break into oscillation.
The Control System Problem

A typical closed-loop feedback control system is shown in Figure 8.1(a).
𝑇 𝑠 =

𝐾𝐺 𝑠
1 + 𝐾𝐺 𝑠 𝐻 𝑠
Letting
,
,
⇒
𝑲𝑮 𝒔 𝑯(𝒔) ≡open-loop
transfer function
where N and D are factored polynomials and signify numerator and denominator terms,
respectively.



The zeros of 𝑇(𝑠) consist of the zeros of 𝐺(𝑠) and the poles of 𝐻(𝑠).
The poles of 𝑇(𝑠) are not immediately known and in fact can change with 𝐾.
Since the system's transient response and stability are dependent upon the poles of 𝑇(𝑠), we
have no knowledge of the system's performance unless we factor the denominator for
specific values of 𝐾. The root locus will be used to give us a picture of the poles of 𝑇(𝑠) as
𝐾 varies.
Defining the Root Locus
(𝒂)
One closed-loop pole moves
upward while the other moves
downward. We cannot tell
which pole moves up or which
moves down.
The closed-loop
poles of the system
change location as the
gain, 𝐾, is varied.
the individual closedloop pole locations
are removed and their
paths are represented
with solid lines
the representation
of the paths of the
closed-loop poles as
the gain is varied is
called a root locus.
Sketching the Root Locus
Ex : Sketch the root locus for the system shown in the Figure.
𝑲(𝒔 + 𝟑)(𝒔 + 𝟒)
𝑪. 𝑳. 𝑻. 𝑭: 𝑻 𝒔 =
𝒔 + 𝟏 (𝒔 + 𝟐) + 𝒔 + 𝟑 𝒔 + 𝟒 𝑲
•
Starting points of the root locus when 𝐾 = 0:
s = −1, −2 = 𝐩𝐨𝐥𝐞𝐬 𝐨𝐟 𝐾𝐺(𝑠)𝐻(𝑠)
•
Ending points of the root locus when 𝐾 = ∞:
s = −𝟑, −𝟒 = zeros 𝐨𝐟 𝐾𝐺(𝑠)𝐻(𝑠)
• Number of poles = Number of zeros = 2
• Real-axis segments. On the real axis, for K > 0 the
pole-zero plot of 𝐺(𝑠)
root locus exists to the left of an odd number of real
axis, finite open-loop poles and/or finite open-loop
zeros.
• The root locus begins at the finite and infinite poles
(𝟑)
(𝟏)
of 𝐾𝐺(𝑠)𝐻(𝑠) and ends at the finite and infinite zeros
of 𝐾𝐺(𝑠)𝐻(𝑠) these poles and zeros are the openloop poles and zeros.
• Number of branches. The number of branches of the
root locus equals the number of closed-loop poles, where
a branch is the path that one pole traverses.
• Symmetry. The root locus is symmetrical about the real
axis.
Sketching the Root Locus\cont.
5. Behavior at infinity
(∗)
 There are three finite poles in (∗), at 𝑠 = 0, −1, and − 2,
and no finite zeros.
 A function can also have infinite poles and zeros:
1. If the function approaches infinity as 𝑠 approaches infinity, then the function has a pole at infinity.
2. If the function approaches zero as 𝑠 approaches infinity, then the function has a zero at infinity.
 For example, the function 𝐺(𝑠) = 𝑠 has a pole at infinity, since 𝐺(𝑠) approaches
infinity as 𝑠 approaches infinity.
 On the other hand, 𝐺(𝑠) = 1/𝑠 has a zero at infinity, since 𝐺(𝑠) approaches zero as
𝑠 approaches infinity.
 Every function of s has an equal number of poles and zeros if we include the infinite poles
and zeros as well as the finite poles and zeros.
There are three finite poles in (∗), and three infinite zeros. (let 𝑠 approach infinity. The openloop transfer function becomes
)
Sketching the Root Locus\cont.
Behavior at infinity. The root locus approaches straight lines as asymptotes as the locus
approaches infinity.
where k = 0, ±1, ±2, ±3 and the angle is given in radians with respect to the positive
extension of the real axis.
 Note that the running index, 𝑘, yields a multiplicity of lines that account for the
many branches of a root locus that approach infinity.
Sketching the Root Locus\cont.
Ex 8.2: Sketch the root locus for the system shown in the Figure.
 The open-loop finite poles of 𝐾𝐺 𝑠 𝐻 𝑠 , 𝐻 𝑠 = 1, at 𝑠 = 0, −1, − 2, and −4, and a finite
zero at 𝑠 = −3 ⇒ since number of poles≠number of zeros, we have three zeros at infinity. (the
asymptotes tell us how we get to these zeros at infinity.)
 To calculate the asymptotes for the infinite zeroes, the real axis intercept is evaluated as
⇒
4
 The angles of the lines that intersect at − 3 are:
⇒
 The number of lines obtained equals
the difference between the number
of finite poles and the number of
 If the value for 𝑘 continued to increase, the angles
finite zeros= 4 − 1 = 3 lines.
would begin to repeat.
Sketching the Root Locus\cont.
Ex 8.2: Sketch the root locus for the system shown in the Figure.
 The real-axis segments (1), (3), and (5) lie to
the left of an odd number of poles and/or
zeros.
 The locus starts at the open-loop poles and
ends at the open-loop zeros. Since there is
only one open-loop finite zero and three
infinite zeros, the three zeros at infinity are at
the ends of the asymptotes.
𝜽=𝝅
(𝟓)
𝝅
𝜽=
𝟑
(𝟑)
𝟓𝝅
The root locus 𝜽 = 𝟑
(𝟏)
Sketching the Root Locus\cont.
Ex : Sketch the root locus for the system shown in the Figure.
𝟏
R(s) +
𝑲
−
(𝒔 + 𝟐)(𝒔 + 𝟒)
 The open-loop finite poles of 𝐾𝐺 𝑠 𝐻 𝑠 , 𝐻 𝑠 = 1, at 𝑠 = −2, −4, and no finite zero since
number of poles≠number of zeros, we have two zeros at infinity. (the asymptotes tell us how
we get to these zeros at infinity.)
 To calculate the asymptotes for the infinite zeroes, the real axis intercept is evaluated as
 The number of lines obtained equals
−𝟔
⇒ 𝝈𝒂 =
= −𝟑 the difference between the number
𝟐
of finite poles and the number of
finite zeros= 2 − 0 = 2 lines.
 The angles of the lines that intersect at −3 are:
𝜋
2
3𝜋
𝒌 = 𝟏 ⇒ 𝜽𝒂 =
2
𝒌 = 𝟎 ⇒ 𝜽𝒂 =
Sketching the Root Locus\cont.
Ex 8.7: Sketch the root locus for the system shown in the Figure.
R(s)
+
−
𝑲(𝒔𝟐 − 𝟒𝒔 + 𝟐𝟎)
(𝒔 + 𝟐)(𝒔 + 𝟒)
 The open-loop finite poles of 𝐾𝐺 𝑠 𝐻 𝑠 , 𝐻 𝑠 = 1, at 𝑠 = −2, −4, and zeros of
𝐾𝐺 𝑠 𝐻 𝑠 = 2 + 4𝑗 , 2 − 4𝑗
 number of poles = number of zeros = 2
\angle
\angle
Properties of the Root Locus

The closed-loop transfer function for the system is

A pole, 𝑠, exists on the root locus when the characteristic
polynomial in the denominator becomes zero, or
𝑲𝑮 𝒔 𝑯(𝒔)
≡open-loop transfer
function

−1 can be represented in polar form as 1∠(𝟐𝑘 + 𝟏)180°.
𝑘 = 0, ±1, ±2, ± 3 , . . . (the angle of the complex number is odd multiple of 180°)

Alternately, a value of 𝑠 is a closed-loop pole if
(∗) -magnitude criterionand
(∗∗) -angle criterion-
⇒

We conclude that a pole of the closed-loop system causes the angle of 𝐾𝐺(𝑠)𝐻(𝑠), or simply
𝐺(𝑠)𝐻(𝑠) since 𝐾 is a scalar, to be an odd multiple of 180°.

Furthermore, the magnitude of 𝐾𝐺(𝑠)𝐻(𝑠) must be unity, implying that the value of 𝐾 is the
reciprocal of the magnitude of 𝐺(𝑠)𝐻(𝑠) when the pole value is substituted for 𝑠.
Properties of the Root Locus\cont.
 For the following system the open-loop transfer
function is
𝑲𝑳𝟏 𝑳𝟐
=
∠(𝜽𝟏 + 𝜽𝟐 ) − (𝜽𝟑 + 𝜽𝟒 )
𝑳𝟑 𝑳𝟒
 The closed-loop transfer function, 𝑇(𝑠), is
 If point 𝑠 is a closed-loop system pole for some value
of gain, 𝐾, then 𝑠 must satisfy (∗) and (∗∗)
pole-zero plot of 𝐺(𝑠)
 EX) is the point −2 + 𝑗3 on the root locus of the system
If this point is a closed-loop pole for some value of gain,
then the angles of the zeros minus the angles of the poles
must equal an odd multiple of 180°.
(−2 , 𝑗3)
≠ 𝟏𝟖𝟎
⇒ −2 + 𝑗3 is not a point on the root locus, (it is not a
closed-loop pole for any gain).
𝑰𝒎
Note that 𝜽 = 𝒕𝒂𝒏−𝟏 𝑹𝒆
Vector representation of 𝐺(𝑠) at −2 + 𝑗3
Properties of the Root Locus\cont.
2
 EX) is the point −2 + 𝑗 2 on the root locus of the system
If this point is a closed-loop pole for some value of gain, then the angles of the zeros minus
the angles of the poles must equal an odd multiple of 180°.
𝜽𝟏 + 𝜽𝟐 − 𝜽𝟑 − 𝜽𝟒 = 𝟏𝟗. 𝟒𝟕 + 𝟑𝟓. 𝟐𝟔 − 𝟗𝟎 − 𝟏𝟒𝟒. 𝟕𝟒 = −𝟏𝟖𝟎°
2
(−2 + 𝑗 2 is a point on the root locus, (it is a closed-loop pole for some gain 𝐾))
 To evaluate that value of gain.
2
(−2, 𝑗 )
2
2
𝑗
2
𝑲𝑳𝟏 𝑳𝟐
=𝟏
𝑳𝟑 𝑳𝟒
Check EX.8.1
Refining the RL Sketch
•
Real-Axis Breakaway and Break-in Points:
σ1 : Breakaway point where the locus leaves the real axis
σ2 : Break-in point where the locus returns to the real axis


Breakaway point: at maximum gain on the real axis
between -2 and -1.
Break-in point: at minimum gain on real axis between +3
and +5.
 For all points on the RL:


Root locus example showing real- axis
breakaway (-1) and break-in points (2)
For points along the real-axis segment of the RL where
breakaway and break-in points could exist, 𝑠 = 𝜎 .
Differentiating with respect to 𝜎 and setting the
derivative equal to zero, results in points of maximum
and minimum gain and hence the breakaway and breakin points.
©2000, John Wiley & Sons, Inc.
Nise/Control Systems Engineering, 3/e
Example:
https://ocw.mit.edu/courses/mechanical-engineering/2-004-systems-modeling-and-control-ii-fall-2007/lecturenotes/
Refining the RL Sketch
• Imaginary axis J 𝝎 –crossings:
https://ocw.mit.edu/courses/mechanical-engineering/-004-2systems-modeling-and-control-ii-fall--erutcel/2007
fdp.18erutcel/seton
Angles of Departure and Angles of Arrival
 Angle of departure is used to determine the slope of a root locus at its
starting point (pole) while the angle of arrival are used to determine
the slope of a root locus at its ending point (zero).
𝑘 = 0, ±1, ±2, ± 3 , . . .
sites.google.com/site/ziyadmasoud
Angles of Departure and Angles of Arrival
 EX) Find the angle of departure 𝜽𝟏
𝜽𝟏 = 𝜽𝟑 − 𝜽𝟐 − 𝜽𝟒 + 𝟏𝟖𝟎
𝜽𝟏 = 𝟒𝟓 − 𝟗𝟎 − 𝟐𝟔. 𝟓𝟔 + 𝟏𝟖𝟎 = 𝟏𝟎𝟖. 𝟒𝟒°