Chapter 4

Control Systems Analysis

and Design by the
Root-Locus Method

1
 4-1 INTRODUCTION
• The transient response is related to the location of the
closed-loop poles.
• If a system has a variable loop gain, then the location of
the closed-loop poles depends on the value of the loop
gain chosen.
• How the closed-loop poles move in s plane as the loop
gain is varied.
• Roots of the characteristic equation of degree higher 3
will be hard to find.
• Changes in the location of roots.

2
 4-1 INTRODUCTION
• Root-locus method for finding the roots of the
characteristic equation has been developed by Evans
and used extensively in control engineering.
• The roots of the characteristic equation are plotted
for all values of a system parameter. The roots
corresponding to a particular value of this parameter
can then be located on the resulting graph.
• The gain of the open-loop transfer function is the
parameter to be varied through all values, from zero
to infinity.

3
 4-2 ROOT-LOCUS PLOTS
• The closed-loop transfer function is

• The characteristic equation is

• Magnitude condition:

• Angle condition:

• The values of s that fulfill both the angle and magnitude

conditions are the roots of the characteristic equation, or the
closed-loop poles.
4
-p1, -p2, -p3, -p4 are poles
-z1 is zero
• EXAMPLE: Consider the negative
feedback system shown in Figure. Let
us sketch the root-locus plot and then
determine the value of K such that the
damping ratio of a pair of dominant
complex-conjugate closed-loop poles is
0.5.

The magnitude condition is

The angle condition becomes

6
 • A typical procedure for
sketching the root-locus plot.
1. Determine the root loci on the
real axis.
• locate the open-loop poles:
s =0, s = -1, s=-2, in the complex
plane. The locations of the open-loop
poles are indicated by crosses.
• If the test point is on the positive real axis, then

• A test point on the negative real axis between 0 and –1. Then

• A test point is selected between –1 and –2, then

7
• A test ranges from –2 to –infinity, all angles are 180 degree.
2. Determine the asymptotes of the root loci
• A test point s is selected very
far from the origin.
K K
lim  lim
s  s ( s+1)( s+ 2) s  ( s+1)3

3( s+1)  180ο(2 k +1)

Substituting s=σ+jω into the angle equation:

( +1  j  )  60ο(2 k +1)

• Three equations represent three straight lines for the

asymptotes.
8
3. Determine the breakaway point

The root-locus branches originating

from the poles at 0 and –1 break away
(as K is increased) from the real axis
and move into the complex plane.
K
The characteristic equation is 1 0
s ( s+1)( s+ 2)
K  ( s 3 + 3 s 2  2 s )

dK
By setting dK/ds=0, we obtain  (3 s 2 + 6 s  2) = 0
ds
s = -0.4226, s = -1.5774

Since the breakaway point must lie on a root locus between 0 and –1,
it is clear that s=–0.4226 corresponds to the actual breakaway point.

9
4. Determine the points where the root
loci cross the imaginary axis.

The characteristic equation is

s3+ 3 s 2  2 s  K  0

the Routh array becomes

Stability range 0<K<6

K 6 s  j 2
• 5. Choose a test point in the broad
neighborhood of the jω axis and the
origin
If a test point is on the root loci, then
the sum of the three angles must be 180°.

6. Draw the root loci.

11
• 7. Determine a pair of dominant
complex-conjugate closed-loop
poles such that the damping ratio
is 0.5.

Closed-loop poles with ζ=0.5 lie on

lines passing through the origin and
making the angles cos-1 0.5=±60
degree with the negative real axis.

s1 = -0.3337 + j0.5780
s2 = -0.3337 - j0.5780

K=1.0383

12
• Example: Sketch the root-
locus plot of a system with
complex-conjugate open-loop
poles. Consider the negative
feedback system shown in
Figure.
• For this system, K>=0,
open loop transfer function has
a pair of complex-conjugate
poles at

13
1. Determine the root loci on
the real axis.
• For any test point s on the
real axis, the sum of the
angular contributions of the
complex-conjugate poles is
360°.
• The location of the root locus
on the real axis is determined
from the open-loop zero on
the negative real axis.
Between –2 and –infinity.

14
2. Determine the angle of departure from the
complex-conjugate open-loop poles.
• If we choose a test point and move it in
the very vicinity at –p1 , we find that the
sum of the angular contributions from
the pole -p2 and zero –z1 to the test
point can be considered remaining the
same.

15
3. Determine the break-in point.

s = -3.7320 K=5.4641
s = -0.2680 K=–1.4641

4. Sketch a root-locus plot, based

on the information obtained in the
foregoing steps.
General Rules for Constructing Root Loci

First, obtain the characteristic equation:

1 + G(s)H(s) = 0
and re-arrange into the factor form
1. Locate the poles and zeros of G(s)H(s) on the s plane. The root-
locus branches start from open-loop poles and terminate at zeros.
2. Determine the root loci on the real axis.
17
General Rules for Constructing Root Loci

3. Determine the asymptotes of root loci. Angles of asymptotes,

asymptotic to straight lines. Intersection point is determined by
dividing the denominator by the numerator, and setting the new
denominator equal to zero.
4. Find the breakaway and break-in points. Either lie on the real
axis or occur in complex-conjugate pairs.
B(s)+ KA(s)=0, dK/ds=0.
5. Determine the angle of departure (angle of arrival) of the root
locus from a complex pole (at a complex zero).
6. Find the points where the root loci may cross the imaginary axis.
Letting s=jω in the characteristic equation.

18
General Rules for Constructing Root Loci

7. Take a series of test points in the broad neighborhood of the

origin of the s plane, sketch the root loci. The most important part
of the root loci is on neither the real axis nor the asymptotes, but is
in the broad neighborhood of the jω axis and the origin.

8. Determine the closed-loop poles. A particular point on each

root-locus branch will be a closed-loop pole if the K satisfies the
magnitude condition.

19
 4–3 ROOT-LOCUS APPROACH TO
CONTROL SYSTEMS DESIGN
• Design by Root-Locus Method.
– In designing a control system, if a gain adjustment is
required, we must modify the original root loci by
inserting a suitable controller/compensator.
– Once the effects on the root locus of the addition of poles
and zeros are fully understood, we can readily determine
the locations of the pole(s) and zero(s) of the controller
that will reshape the root locus as desired.
– In essence, in the design by the root locus method, the
root loci of the system are reshaped through the use of a
controller so that a pair of dominant closed-loop poles
can be placed at the desired location. 20
 4–3 ROOT-LOCUS APPROACH TO
CONTROL SYSTEMS DESIGN
• Series Compensation.
– The compensator Gc(s) is placed in series with the plant.

• Parallel (or Feedback) Compensation.

– The compensator is placed in the resulting inner feedback path.

– Series compensation may be simpler than parallel compensation,

however, series compensation frequently requires additional
amplifies to increase the gain

21
 4–3 ROOT-LOCUS APPROACH TO
CONTROL SYSTEMS DESIGN
• Commonly Used Compensators
Lead compensators, Lag compensators, lag-lead
compensators, velocity-feedback compensators.
• Effects of the Addition of Poles
pull the root locus to the right, tend to lower the system’s
relative stability and to slow down the settling.
• Effects of the Addition of Zeros
Pull the root locus to the left, tend to make the system
more stable and to speed up the settling of the response.
22
• Sketch the root loci for the system shown in Figure. The gain
K is assumed to be positive. Observe that for small or large
values of K the system is overdamped and for medium values
of K it is underdamped.

• Note that this system is stable for any

positive value of K since all the root
loci lie in the left half s plane.
• Small values of K (0<K<0.0718)
correspond to an overdamped system.
• Medium values of K (0.0718<K<14)
correspond to an underdamped system.
• Finally, large values of K (14<K)
correspond to an overdamped system.
• With a large value of K, the steady state
can be reached in much shorter time
than with a small value of K.

23
• Example: Sketch the root loci of
the control system shown in Figure.
Determine the range of gain K for
stability.

• The root-locus branches cross

the imaginary axis at (where
K=16) and (where K=7).
• Since the value of gain K at the
origin is 7, the range of gain
value K for stability is 7<K<16.

24
 4–4 LEAD COMPENSATION
• Transfer function of a lead compensator Gc(s)
is

• where a, T, and Kc are gains.

25
 Procedure for designing a lead compensator for
the system shown by the root-locus method

• From the performance specifications, determine the

desired location for the dominant closed-loop poles.
• By drawing the root-locus plot of the uncompensated
system (original system), ascertain whether or not the
gain adjustment alone can yield the desired closed loop
poles. If not, calculate the angle deficiency. This angle
must be contributed by the lead compensator if the new
root locus is to pass through the desired locations for
the dominant closed-loop poles. 26
• If static error constants are not specified, determine
the location of the pole and zero of the lead
compensator so that the lead compensator will
contribute the necessary angle ϕ. If no other
requirements are imposed on the system, try to make
the value of α as large as possible. A larger value of α
generally results in a larger value of Kv, which is
desirable. Note that
K v  lim  sGc G   K c lim  sG 
s 0 s 0

• Determine the value of Kc of the lead compensator

from the magnitude condition.

27
• Example:
– The feedforward transfer function is

– The root-locus plot for this system is

28
• The closed-loop transfer function for the system is

• The closed-loop poles are located at

• The damping ratio of the closed-loop poles is

0.5/sqrt(10).The undamped natural frequency of
the closed-loop poles is sqrt(10)=3.1623 rad/sec.
Because the damping ratio is small, this system
will have a large overshoot in the step response
and is not desirable.
29
• It is desired to design a lead compensator Gc(s) so
that the dominant closed-loop poles have the
damping ratio of 0.5 and the undamped natural
frequency 3 rad/sec. The desired location of the
dominant closed-loop poles can be determined
from

30
• lead compensator is

• Desired characteristic equation is

s2 + 3s + 9=0
• Kc=0.9
• The static velocity error constant for the
present case is obtained :

31
4–5 LAG COMPENSATION

Lag compensation: for the case where the system

exhibits satisfactory transient-response
characteristics but unsatisfactory steady-state
characteristics.
32
4-6 LAG-LEAD COMPENSATION

• For improvements in both transient response and

steady-state response.
33
 Example: Design a controller for Planar
Crane by the Root-Locus Method

•Consider a proportional-plus-derivative (PD) controller for vibration reduction

in a industrial crane.
•The ratio of proportional gain Kp to the derivate gain Kd is 10.
•Let us sketch the root-locus plot and then determine the derivative gain Kd such
that the damping ratio of a pair of dominant complex-conjugate closed-loop
poles is 0.707.
•Note that Cable length was fixed at 1 m.

34
• Transfer function of the open-loop system 10 kd + kd s
Gc  G 
s 2  9. 8

Gc  G 10 kd + kd s
• 
Transfer function of the closed-loop system 1  G  G s 2 + k s  10 k  9.8
c d d

• Characteristic equation s 2 + kd s  (10 kd  9.8)  0

• Draw root-locus plot

Kd = 0 s   j 9 .8 pole
cos 1   0.707

K d = 40.9571 s  -20.4817 break-in

K d =  s  10 zero

• Design the derivative gain Kd

K d = 20.9
so that the damping ratio of the closed-loop systems is 0.707.
 Example: Design a controller for mass-spring
systems by the Root-Locus Method
• Mass 2.6 Kg
• Spring stiffness 200 N/m
• Zero Damper
• Input: force, u
• Output: mass displacement, y
• Design a PD controller, Kp+ Kd*s, for vibration
reduction by root-locus method so that the
damping ratio of the closed-loop systems is 0.1.
• 1) Draw root locus plot
• 2) Design gains
• 3) Transfer Function of closed-loop system
• 4) State space equation
• 5) Stability
36
• Transfer function of the open-loop system
k p + kd s s  10
Gc  G   kd 
2.6 s 2  200 2.6 s 2  200

• Draw root-locus plot

Kd = 0 s   j 8.7706 pole

K d = 121.1665 s  -23.3145 break-in

K d =  s  10 zero
• Design the derivative gain Kd

K d = 5.14

so that the damping ratio of the closed-loop systems is 0.1

• Transfer Function of closed-loop system

Gc  G kd s  10 kd 5.14 s  51.4
 
1  Gc  G 2.6 s 2  kd s  200  0.1 kd 2.6 s 2  5.14 s  251.4
• Closed-loop differential equation
  
y+1.9769 y+ 96.6923 y = 1.9769 r+19.769 r

• Input is r, while the output is y

• State space equation

 x1 = y  
 x1 = y
    
 x2 = y- 1.9769 r  x 2 = y- 1.9769 r

 
 x1    0 1   x1  1.9769 
   r
    96.6923  
-1.9769   x2  -1.9312 
 x2 

38
• Characteristic equation

s 2 +1.9769 s+ 96.6923 = 0

• Stability

s2 1 96.6923
s 1.9769 0

• Natural frequency is 9.8332 rad/s, damping ratio is 0.1

• 1  2
1
• Rise time π  tan ( )

tr 
n 1   2

3
• Settling time, 2% ts 
n
 / 1 2
• Overshoot Mp e
39
• Steady-state error to a step input

1 1 5.14 s  51.4 51.4

ess  lime(t )  lim E ( s )  lim s  
t  s 0 s 0 1  Gc  G s 2.6 s 2  200 200

• Steady-state is very large.

40
• In order to overcome the steady-state error, the controller will add the
integral control, then to be a proportional-integral-derivative (PID)
controller

• Then the open-loop transfer function is:

1 kp k
kd s  k p + ki ( s 2
 s+ i )
s k  kd kd
Gc  G  d
2.6 s 2  200 2.6 s 3  200 s
• Letting
kp ki
 10  0 .1
kd kd

• Draw root-locus plot

• Zeros are -0.1
• Poles are 0, j8.7706, -j8.7706

• K=5.2 so that the damping ratio of the closed-loop systems is 0.1

41
• Closed-loop transfer function
kp ki
kd  s 2  kd  s+ kd 
Gc  G kd kd

1  Gc  G kp ki
2.6 s 3  kd  s 2  (200  kd  ) s+ kd 
kd kd
• Steady-state error to a step input

1 1 2.6 s 3  200 s
ess  lime(t )  lim E ( s )  lim s  0
t  s 0 s 0 1  G  G s kp ki
c
2.6 s 3  200 s  kd  ( s 2  s+ )
kd kd

42
 4–7 PARALLEL COMPENSATION

closed-loop transfer function for the system with series compensation is

43
closed-loop transfer function for the system with parallel compensation is

44
 EXAMPLE 6–10

Characteristic equation for the system is

where

45
damping ratio of 0.4

𝑘 1.4130 𝑘 0.4490

When k=0.4490, the three closed-

loop poles are
s = -1.0490 + j2.4065,
s = -1.0490 - j2.4065,
s = -2.9021

When k=1.4130, three closed-loop

poles
s = -2.1589 + j4.9652,
s = -2.1589 - j4.9652,
s = -0.6823
The system with k=0.4490 (which exhibits a faster response with relatively small overshoot)
has a much better response characteristic than the system with k=1.4130 (which exhibits a
slow overdamped response).

47