Automatic Control 2

Lecture-3
• Examples of root locus

Dr. Mountasser Mohamed Ramadan

email: mountasser.m.r@gmail.com
 Lecture Outline

• Root Locus of 1st order systems

• Root Locus of 2nd order systems

• Root Locus of Higher order systems

 Root Locus of 1st Order System
• 1st order systems (without zero) are represented by following
transfer function.
K
G( s) H ( s) 
s 
• Root locus of such systems is a horizontal line starting from -α
and moves towards -∞ as K reaches infinity.

jω

-∞ σ
-α
 Home Work
• Draw the Root Locus of the following systems.

K
1) G( s) H ( s) 
s2

K
2) G( s) H ( s) 
s 1

K
3) G( s) H ( s) 
s
 Root Locus of 1st Order System
• 1st order systems with zero are represented by following
transfer function.
K (s   )
G( s) H ( s) 
s 
• Root locus of such systems is a horizontal line starting from -α
and moves towards -β as K reaches infinity.

jω

σ
-β -α
 Home Work
• Draw the Root Locus of the following systems.

Ks
1) G( s) H ( s) 
s2

2) G( s) H ( s)  K ( s  5)
s 1

3) G( s) H ( s)  K ( s  3)
s
 Root Locus of 2nd Order System
• Second order systems (without zeros) have two poles and the
transfer function is given
K
G( s) H ( s) 
( s  1 )( s   2 )

• Root loci of such systems are vertical lines.

jω

σ
-α2 -α1
 Home Work
• Draw the Root Locus of the following systems.

K
1) G( s) H ( s) 
K 4) G( s) H ( s)  2
s( s  2) s  3s  10

K
2) G( s) H ( s)  2
s

K
3) G( s) H ( s) 
( s  1)(s  3)
 Root Locus of 2nd Order System
• Second order systems (with one zero) have two poles and the
transfer function is given
K (s   )
G( s) H ( s) 
( s  1 )( s   2 )
• Root loci of such systems are either horizontal lines or circular
depending upon pole-zero configuration.

jω jω jω

σ σ σ
-α2 -β -α1 -β -α2 -α1 -α2 -α1 -β
 Home Work
• Draw the Root Locus of the following systems.

K ( s  1)
1) G ( s) H ( s) 
s( s  2)

2) G( s) H ( s)  K ( s  2)
2
s

K ( s  5)
3) G( s) H ( s) 
( s  1)(s  3)
 Example
• Sketch the root-locus plot of following system
with complex-conjugate open loop poles.
 Example
• Step-1: Pole-Zero Mao

• Step-2: Determine the root loci on real axis

• Step-3: Asymptotes
 Example
• Step-4: Determine the angle of departure from the
complex-conjugate open-loop poles.

– The presence of a pair of complex-conjugate open-loop

poles requires the determination of the angle of
departure from these poles.

– Knowledge of this angle is important, since the root

locus near a complex pole yields information as to
whether the locus originating from the complex pole
migrates toward the real axis or extends toward the
asymptote.
 Example
• Step-4: Determine the angle
of departure from the
complex-conjugate open-loop
poles.
 Example
• Step-5: Break-in point
 Root Locus of Higher Order System
• Third order System without zero

K
G( s) H ( s) 
( s  1 )(s   2 )(s   3 )
 Root Locus of Higher Order System
• Sketch the Root Loci of following unity feedback system

K ( s  3)
G( s) H ( s) 
s( s  1)(s  2)(s  4)
• Let us begin by calculating the asymptotes. The real-axis intercept is
evaluated as;

• The angles of the lines that intersect at - 4/3, given by

• The Figure shows the complete root locus as well as the asymptotes
that were just calculated.
 Example: Sketch the root locus for the system with the characteristic equation
of;

a) Number of finite poles = n = 4.

b) Number of finite zeros = m = 1.
c) Number of asymptotes = n - m = 3.
d) Number of branches or loci equals to the number of finite poles (n) = 4.
e) The portion of the real-axis between, 0 and -2, and between, -4 and -∞, lie
on the root locus for K > 0.
• Using Eq. (v), the real-axis asymptotes intercept is evaluated as;

−2 + 2 −4 − (−1) −10 + 1
σ𝑎 = = = −3
𝑛 −𝑚 4 −1

• The angles of the asymptotes that intersect at - 3, given by Eq. (vi), are;

(2𝑘 + 1)π (2𝑘 + 1)π For K = 0, θa = 60o

θ𝑎 = = For K = 1, θa = 180o
𝑛 −𝑚 4−1 For K = 2, θa = 300o
• The root-locus plot of the system is shown in the figure
below.
• It is noted that there are three asymptotes. Since n – m = 3.
• The root loci must begin at the poles; two loci (or branches)
must leave the double pole at s = -4.
• Using Eq. (vii), the breakaway point, σ, can be determine as;
𝑠(𝑠 + 2)(𝑠 + 4)2
𝐾=
(𝑠 + 1)
𝑑𝐾 𝑑 𝑠(𝑠 + 2)(𝑠 + 4)2
= =0
𝑑𝑠 𝑑𝑠 (𝑠 + 1)

• The solution of the above equation is 𝜎 = −2.59.

 Example: Sketch the root loci for the system.

• A root locus exists on the real axis between points s = –1 and s = –3.6.
• The intersection of the asymptotes and the real axis is determined as,
0 + 0 − 3.6 − (−1) −2.6
σ𝑎 = = = −1.3
𝑛 −𝑚 3 −1
• The angles of the asymptotes that intersect at – 1.3, given by Eq. (vi),
are;
(2𝑘 + 1)π (2𝑘 + 1)π For K = 0, θa = 90o
θ𝑎 = = For K = 1, θa = -90o or 270o
𝑛 −𝑚 3−1

• Since the characteristic equation is

• We have (a)
• The breakaway and break-in points are found from Eq. (a)
as,

From which we get,

• Point s = 0 corresponds to the actual breakaway point. But

points are neither breakaway nor break-in points, because
the corresponding gain values K become complex
quantities.
• To check the points where root-locus branches may cross the imaginary
axis, substitute 𝑠 = 𝑗𝜔 into the characteristic equation, yielding.

• Notice that this equation can be

satisfied only if 𝜔 = 0, 𝐾 = 0.

• Because of the presence of a double

pole at the origin, the root locus is
tangent to the 𝑗𝜔axis at 𝑘 = 0.

• The root-locus branches do not cross

the 𝑗𝜔axis.

• The root loci of this system is shown

in the Figure.