Modern Control - Lec 04 - Analysis and Design of Control Systems using Root Locus

1. ‫ر‬َ‫ـد‬ْ‫ق‬‫ِـ‬‫ن‬،،،‫لما‬‫اننا‬ ‫نصدق‬ْْ‫ق‬ِ‫ن‬‫ر‬َ‫د‬
LECTURE (4)
Analysis and Design of Control
Systems using Root Locus
Assist. Prof. Amr E. Mohamed

2. Lecture Outline
4.1 Introduction
4.2 Root Locus Construction
4.3 Root Locus Examples
4.4 Stability Analysis
4.5 Compensation and Controller Design using Root Locus
2

3. 4.1 Introduction
 Motivation
 To satisfy transient performance requirements, it may be necessary to know
how to choose certain controller parameters so that the resulting closed-loop
poles are in the performance regions, which can be solved with Root Locus
technique.
 Definition
 A graph displaying the roots of a polynomial equation when one of the
parameters in the coefficients of the equation changes from 0 to .
3

4. 4.1 Introduction
 Consider the feedback control system shown in Fig. 4.1, the closed loop
transfer function is given by:
 Where k is a constant gain parameter. The poles of the transfer
function are the roots of the characteristic equation given by:
)()(1
)(
)(
)(
)(
sHsGk
sGk
sR
sC
sT


Fig. 4.1
)(sR )(sC
)(sG
)(sH
k
0)()(1  sHsGk
4

5. Definition of Root Locus
 The root locus is the locus of the characteristic equation of the closed-
loop system as a specific parameter (usually, gain K) is varied from zero
to infinity. For 0 ≤ 𝑘 < ∞ , if
 then
 and for positive k, this means that a point s which is a point on locus
must satisfy the magnitude and angle and conditions:
 The magnitude condition:
 The angle condition:
0)()(1  sHskG
k
sHsG
1
)()( 
k
sHsG
1
|)()(| 
,....3,2,1,0360180)()(  iwhereisHsG oo
5

6. 4.2.1 Root Locus Rules
 Rule 1: The number of branches of the root locus is equal to the
number of closed-loop poles (or roots of the characteristic equation).
 Rule 2: Root locus starts at open-loop poles (when K= 0) and ends at
open-loop zeros (when K=). If the number of open-loop poles is
greater than the number of open-loop zeros, some branches starting
from finite open-loop poles will terminate at zeros at infinity.
 Rule 3: Root locus is symmetric about the real axis, which reflects the
fact that closed-loop poles appear in complex conjugate pairs.
6

7. 4.2.1 Root Locus Rules
 Rule 4: Along the real axis, the root locus includes all segments that
are to the left of an odd number of finite real open-loop poles and
zeros.
 Rule 5: If number of poles (n) exceeds the number of zeros (m), then as
K, (n - m) branches will become asymptotic to straight lines. These
straight lines intersect the real axis with angles A at A .
zerosloopopenofpolesloopopenof
zerosloopopensumpolesloopopensum
mn
zp ii
A
## 





 
,....3,2,1,0
360180



 iwhere
mn
i
A
7

8. 4.2.1 Root Locus Rules
 Rule 6: “Breakaway and/or break-in points” where the locus between
two poles on the real axis leaves the real axis is called the breakaway
point and the point where the locus between two zeros on the real axis
returns to the real axis is called the break-in point. The loci leave or
return to the real axis at the maximum gain k of the following equation:
 Rule 7: The departure angle for a pole pi ( the arrival angle for a zero
zi) can be calculated by slightly modifying the following equation:
k
sHsG oo
1
|)()(| 
8

9. 4.2.1 Root Locus Rules
 The angle of departure p of a locus from a complex pole is given by:
p = 180 – [sum of the other GH pole angles to the pole under consideration] +
[sum of the GH zero angles to the pole]
 The angle of approach z of a locus to a complex zero is given by:
z = 180 + [sum of the other GH pole angles to the zero under consideration] –
[sum of the GH zero angles to the zero]
 Rule 8: If the root locus passes through the imaginary axis (the
stability boundary), the crossing point j and the corresponding gain
K can be found using Routh-Hurwitz criterion.
9

10. Steps to Sketch Root Locus (1/2)
 Step#1: Transform the closed-loop characteristic equation into the
standard form for sketching root locus:
 Step#2: Find the open-loop zeros and the open-loop poles. Mark the
open-loop poles and zeros on the complex plane. Use ‘x’ to represent
open-loop poles and ‘o’ to represent the open-loop zeros.
 Step#3: Determine the real axis segments that are on the root locus by
applying Rule 4.
 Step#4: (If necessary) Determine the number of asymptotes and the
corresponding intersection and angles by applying Rules 2 and 5.
0)()(1  sHsGk
10

11. Steps to Sketch Root Locus (2/2)
 Step#5: (If necessary) Determine the break-away and break-in points using
Rule 6.
 Step#6: (If necessary) Determine the departure and arrival angles using
Rule 7.
 Step#7: (If necessary) Determine the imaginary axis crossings using Rule 8.
 Step#8: Use the information from Steps 1-7, sketch the root locus.
11

12. Fig. 4.2
4.2.2 Root Locus Procedure – Example 4.1
 Example 4.1: Sketch the root locus for the system shown in Fig.4.2
 Solution:
 Step 1: “Write the open loop transfer function of the system and find the
open loop Poles & zeros” as follows:
 Open loop Poles = {0, -1, -2, -4}  # of poles = n =4
 Open loop Zeros= {-3}  # of zeros = m =1
)4)(2)(1(
)3(
)()(



ssss
sk
sHskG
12

13.  Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments exists
after the poles at 0, -2 and -4
 Asymptotic
 Beakaway
 Intersection with imaginary??
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
13

14.  Step 3. Asymptotic Angles: “As k approaches +∞, the branches of the locus become
asymptotic to straight lines with angles:
• The number of asymptotes are three [n-m = 3]
• for i=0  1 = 60, for i=1  2=180, and for i= -1  3=-60.
 Step 4. Center of Asymptotes: The starting point of the asymptotes. These linear
asymptotes are centered at a point on the real axis at:
,....3,2,1,0
360180



 iwhere
mn
i
A
33.1
3
3421




 

mn
GHofvalueszeroGHofvaluespole
A
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
14

15. 4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
15

16.  Step 5. Breakawayand break-in or entry points:
|
)3(
)4)(2)(1(
|
|)()(|
11
|)()(|



s
ssss
sHsG
kand
k
sHsG oo
 Therefore, the break away point lies between 0 and -1 at
maximum value for k which is calculated as follows:
s -0.3 -0.4 -0.45 -0.5
k 0.489 0.532 0.534 0.525
 It is clear that the breakaway point is at s = -0.45.
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
16

17.  Step 7. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The Routh-table for the characteristic equation is shown
• The coefficient of s1 = 0
• Then k = 9.65
• The intersection points are
ksksss
sk
sHsG
sG
sT
3)8(147
)3(
)()(1
)(
)( 234





kskssssHsG 3)8(147)()(1 234

0720652
 kk
021)90( 2
 ksk
07.20235.80 2
s
59.12,1 jsthus 
 We also conclude that the system is stable for 0 < k < 9.65.
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
17

18. 4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
18

19. 4.2.2 Root Locus Procedure – Example 4.2
 Example 4.2: Sketch the root locus for the system shown in Fig.4.4
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation
of the system as follows:
Fig. 4.4
)(sR )(sC
)2)(1(
)5)(3(


ss
ssk
)2)(1(
)5)(3(
)()(



ss
ssk
sHskG
19

20.  Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments
exists after the zero at 5
and after the pole at -1.
 Beakaway & Break-in
 Intersection with imaginary
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
20

21.  Step 3: Breakaway and break-in or entry points
 Therefore, the breakaway point lies between -1 and -2 and the break in point lies
between 3 and 5.
)5)(3(
)2)(1(
|)()(|
11
|)()(|



ss
ss
sHsG
kand
k
sHsG oo
at s=-1.45 breakaway
&
at s=-3.82 break-in points
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
21

22.  Step 7. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The coefficient of s1 = 0
• Then k = 3/8=0.375
• The intersection points are
)152()83()1(
)158(
)()(1
)(
)( 2
2
ksksk
ssk
sHsG
sG
sT





)152()83()1()()(1 2
ksksksHsG 
083  k
0)152()1( 2
 ksk
0625.7375.1 2
s
35.22,1 jsthus 
 We also conclude that the system is stable for 0 < k < 0.375.
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
22

23. 4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
23

24.  Example 4.3: Sketch the root locus for the system shown in Fig.4.6
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation
of the system as follows:
Fig. 4.6
)(sR )(sC
)22)(3(
)2(
2


sss
sk
)22)(3(
)2(
)()( 2



sss
sk
sHskG
4.2.2 Root Locus Procedure – Example 4.3
24

25.  Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments
exists after the zero at -2.
 Asymptotic
 Departure Angles
 Intersection with imaginary????
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
25

26.  Step 3. Asymptotic Angles:
• The number of asymptotes are Two [n-m = 2]
• for i=0  1 = 90, for i=, 2=-90 .
 Step 4. Center of Asymptotes:
,....3,2,1,0
13
360180



 iwhere
i
A
1
3
211113




 

jj
mn
GHofvalueszeroGHofvaluespole
A
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
26

27.  Step 5. Angles of departure: The angle of
departure p of a locus from a complex
pole is given by:
p = 180 – [sum of the other GH pole
angles to the pole under consideration]
+ [sum of the GH zero angles to the
pole]
p = 180 – [2+4] + [3]
= 180 – [90+30] + [45]
= 105 degree
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
27

28. 4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
28

29. 4.2.2 Root Locus Procedure – Example 4.4
 Example 4.2: Sketch the root locus for the system shown in Fig.4.8 (a) and
find the following:
a) The point and the gain on the root locus where the damping ration (ξ=0.45).
b) The range of k within which the system is stable.
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation of the
system as follows:
Fig.4.8 (a)
)4)(2(
)204(
)()(
2



ss
ssk
sHskG
29

30.  Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments
exists after the pole at -2.
 Beakaway
 Arrival Angles
 Intersection with imaginary
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
30

31.  Step 3. Angles of Arrival: The angle of
Arrival z of a locus from a complex pole
is given by:
z = 180 + [sum of the other GH pole
angles to the zero under consideration]
– [sum of the GH zero angles to the
zero]
z = 180 + [1+2] - [3]
= 180 + [45+30] – [90]
= 165 degree
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
31

32.  Step 4: Breakaway points
 Therefore, the breakaway point lies between -2 and -4.
 The breakaway point lies at s=-2.88 at maximum value for k = 0.0248.
)204(
)4)(2(
|)()(|
11
|)()(| 2



ss
ss
sHsG
kand
k
sHsG oo
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
32

33.  Step 5. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The coefficient of s1 = 0
• Then k = 1.5
• The intersection points are
)208()46()1(
)204(
)()(1
)(
)( 2
2
ksksk
ssk
sHsG
sG
sT





)208()46()1()()(1 2
ksksksHsG 
046  k
0)208()1( 2
 ksk
0385.2 2
s
89.32,1 jsthus 
 We also conclude that the system is stable for 0 < k < 1.5
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
33

34. a) The angle of the desired poles
is at = cos-1(0.45) =63.3 °. A
line is drawn from the origin to
intersect the root locus at the
point 3.4 116.7° with a gain k
= 0.417.
b) Therefore, this system is stable
for 0 < k < 1.5.
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
34

35. CONTROL SYSTEMS DESIGN
USING ROOT-LOCUS APPROACH
35

36. Control Design Using Root Locus
 Case study: double integrator, transfer function 𝐺 𝑠 =
1
𝑠2
 Control objective: ensure stability; meet time response specifications.
 First, let’s try a simple P-gain:
 Closed-loop transfer function:
𝐾
𝑠2
1 +
𝐾
𝑠2
=
𝐾
𝑠2 + 𝐾
36

37. Double Integrator with P-Gain
 Closed-loop transfer function:
𝐾
𝑠2
1 +
𝐾
𝑠2
=
𝐾
𝑠2 + 𝐾
 Characteristic equation: 𝑠2 + 𝐾 = 0
 Closed-loop poles: 𝑠 = ±𝑗 𝐾
37
 This confirms what we already knew: P-gain alone does not deliver
stability.

38. Double Integrator with PD-Control
 Characteristic equation:
1 + 𝐾 𝑃 + 𝐾 𝐷 𝑠 .
1
𝑠2
= 0
𝑠2 + 𝐾 𝐷 𝑠 + 𝐾 𝑃 = 0
 To use the Root-Locus method, we need to convert it into the Evans form
1 + 𝐾𝐿(𝑠) = 0, where 𝐿 𝑠 =
𝑏(𝑠)
𝑎(𝑠)
1 + 𝐾 𝑃 + 𝐾 𝐷 𝑠 .
1
𝑠2 = 1 + 𝐾 𝐷
𝑠 +
𝐾 𝑃
𝐾 𝐷
𝑠2 = 0
⇒ 𝐾 = 𝐾 𝐷, 𝐿 𝑠 =
𝑠 +
𝐾 𝑃
𝐾 𝐷
𝑠2 (𝑎𝑠𝑠𝑢𝑚𝑒
𝐾 𝑃
𝐾 𝐷
fi𝑥𝑒𝑑, = 1)
38

39. Double Integrator with PD-Control
 So, the effect of D-gain was to introduce an open-loop zero into LHP, and
this zero “pulled” the root locus into LHP, thus stabilizing the system.
39
 But let’s actually draw the Root-
Locus using the rules:
 What can we conclude from this root
locus about stabilization?
 all closed-loop poles are in LHP (we
already knew this from Routh, but now
can visualize)
 nice damping, so can meet reasonable
specifications.

40. Dynamic Compensation
 Objectives: stabilize the system and satisfy given time response
specifications using a stable, causal controller.
 Characteristic equation:
1 + 𝐾
𝑠 + 𝑧
𝑠 + 𝑝
.
1
𝑠2
= 1 + 𝐾 𝐿(𝑠) = 0
40

41. Approximate PD Using Dynamic Compensation
 Reminder: we can approximate the D-controller 𝐾 𝐷 𝑠 by
𝐾 𝐷
𝑝𝑠
𝑠 + 𝑝
⟹ 𝐾 𝐷 𝑠 𝑎𝑠 𝑝 ⟶ ∞
 where, p is the pole of the controller.
 So, we replace the PD controller 𝐾 𝑃 + 𝐾 𝐷 𝑠 by
𝐾 𝑠 = 𝐾 𝑝 + 𝐾 𝐷
𝑝𝑠
𝑠 + 𝑝
 Closed-loop poles: 1 + 𝐾 𝑝 + 𝐾 𝐷
𝑝𝑠
𝑠+𝑝
𝐺 𝑠 = 0
41

42. Lead & Lag Compensators
 Consider a general controller of the form
𝐾
𝑝𝑠
𝑠 + 𝑝
⟹ 𝐾, 𝑧, 𝑝 > 0 𝑎𝑟𝑒 𝑑𝑒𝑠𝑖𝑔𝑛 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠
 Depending on the relative values of z and p, we call it:
 a lead compensator when z < p
 a lag compensator when z > p
 Why the name “lead/lag?”
 think frequency response
 if 𝑧 < 𝑝, then 𝜓 − 𝜑 > 0 (phase lead)
 if 𝑧 > 𝑝, then 𝜓 − 𝜑 < 0 (phase lag)
42

43. Back to Double Integrator
 Controller transfer function is 𝐾
𝑠
𝑠+𝑝
, where:
 so, as 𝑝 → ∞ , 𝑧 tends to a constant, so we get a lead controller.
43
 We use lead controllers as dynamic compensators for approximate PD
control.

44. Double Integrator & Lead Compensator
 To keep things simple, let’s set 𝐾 𝑃= 𝐾 𝐷. Then:
𝐾 = 𝐾 𝑃+𝑝𝐾 𝐷 = (1 + 𝑝)𝐾 𝐷
𝑧 =
𝑝𝐾 𝑝
𝐾 𝑝 + 𝑝𝐾 𝐷
=
𝑝𝐾 𝐷
(1 + 𝑝)𝐾 𝐷
=
𝑝
(1 + 𝑝)
≈ 1 𝑎𝑠 𝑝 → ∞
 Since we can choose p and z directly, let’s take
𝑧 = 1 𝑎𝑛𝑑 𝑝 = 𝑙𝑎𝑟𝑔𝑒.
44 We expect to get behavior similar to PD control.

45. Double Integrator & Lead Compensator
 Let’s try a few values of p. Here’s p = 10:
 Close to jω-axis, this root locus looks similar to the PD root locus. However,
the pole at 𝑠 = − 10 makes the locus look different for s far into LHP.
45

46. Double Integrator & Lead Compensator
 The design seems to look good: nice damping, can meet reasonable
specs.
 Any concerns with large values of p?
 When p is large, we are very close to PD control, so we run into the
same issue: noise amplification.
 (This is just intuition for now — we will confirm it later using
frequency-domain methods.)
46

47. Control Design Using Root Locus
 Case study: double integrator, transfer function 𝐺 𝑝 𝑠 =
1
𝑠−1
 Control objective: ensure stability and constant reference tracking.
 In earlier lectures, we saw that for perfect steady-state tracking we
need PI control :
 Closed-loop poles are determined by the Characteristic equation:
1 + 𝐾 𝑃 +
𝐾𝐼
𝑠
1
𝑠 − 1
= 0
47

48. Control Design Using Root Locus
 To use the Root-Locus method, we need to convert it into the Evans
form 1 + 𝐾𝐿(𝑠) = 0, where 𝐿 𝑠 =
𝑏(𝑠)
𝑎(𝑠)
1 + 𝐾 𝑃 +
𝐾𝐼
𝑠
1
𝑠 − 1
= 1 +
𝐾 𝑃 𝑠 + 𝐾𝐼
𝑠
1
𝑠 − 1
= 1 + 𝐾 𝑃
𝑠 +
𝐾𝐼
𝐾 𝑃
𝑠 𝑠 − 1
= 0
⇒ 𝐾 = 𝐾 𝑃, 𝐿 𝑠 =
𝑠 +
𝐾𝐼
𝐾 𝑃
𝑠 𝑠 − 1
(𝑎𝑠𝑠𝑢𝑚𝑒
𝐾𝐼
𝐾 𝑃
fi𝑥𝑒𝑑, = 1)
48

49. Root Locus for PI Compensation
 The system is stable for K > 1
(from Routh-Hurwitz)
 For very large K, we get a
completely damped system,
with negative real poles
 Perfect steady-state tracking of
constant references.
49

50. Approximate PI via Dynamic Compensation
 PI control achieves the objective of stabilization and perfect steady-
state tracking of constant references; however, just as with PD earlier,
we want a stable controller.
 Here’s an idea:
 More generally, if 𝑧 =
𝐾 𝐼
𝐾 𝑃
, then
 This is lag compensation (or lag control)!
50
 We use lag controllers as dynamic compensators for approximate PI control.

51. Compensation
51

52. Design by Root-Locus Method
 The root-locus approach to design is very powerful when the
specifications are given in terms of time-domain quantities, such as the
damping ratio and undamped natural frequency of the desired dominant
closed-loop poles, maximum overshoot, rise time, and settling time.
 The design by the root-locus method is based on reshaping the root
locus of the system by adding poles and zeros to the system’s open-loop
transfer function and forcing the root loci to pass through desired
closed-loop poles in the s plane.
 Consider a design problem in which the original system either is
unstable for all values of gain or is stable but has undesirable transient-
response characteristics. In such a case, the reshaping of the root locus
is necessary in the broad neighborhood of the 𝑗𝜔 axis and the origin in
order that the dominant closed-loop poles be at desired locations in the
complex plane. This problem may be solved by inserting an appropriate
lead compensator in cascade with the feedforward transfer function.
52

53. Series Compensation and Parallel Compensation
53
(a) Series compensation;
(b) parallel or feedback
compensation.

54. Commonly Used Compensators
 Lead Compensator
 If a sinusoidal input is applied to the input of a network, and the steady-
state output (which is also sinusoidal) has a phase lead.
 Lag Compensator
 If a sinusoidal input is applied to the input of a network, and the steady-
state output (which is also sinusoidal) has a phase lag.
 Lead-Lag Compensator
 If a sinusoidal input is applied to the input of a network, and the steady-
state output (which is also sinusoidal) has both phase lag and phase lead
occur in the output but in different frequency regions; phase lag occurs in
the low-frequency region and phase lead occurs in the high-frequency
region.
54

55. Effects of the Addition of Poles
 The addition of a pole to the open-loop transfer function has the effect
of pulling the root locus to the right, tending to lower the system’s
relative stability and to slow down the settling of the response.
55
(a) Root-locus plot of
a single-pole system;
(b) root-locus plot of
a two-pole system;
(c) root-locus plot of
a three-pole system.

56. Effects of the Addition of Zeros
 The addition of a zero to the open-loop
transfer function has the effect of
pulling the root locus to the left,
tending to make the system more
stable and to speed up the settling of
the response. (Physically, the addition
of a zero in the feedforward transfer
function means the addition of
derivative control to the system. The
effect of such control is to introduce a
degree of anticipation into the system
and speed up the transient response.)
56
(a) Root-locus plot of a three-pole system;
(b), (c), and (d) root-locus plots showing effects of
addition of a zero to the three-pole system.

57. Lead or Lag Compensation
 The transfer function is
 where
57

58. Lead or Lag Compensation (Cont.)
 This network has a dc gain of
 This network is
 a lead network if
 a lag network if
58

59. Lead Compensation Techniques Based on the Root-Locus Approach
 The designing a lead compensator for the system shown by the root-locus
method may be stated as follows:
59
1) From the performance specifications, determine the desired location for the
dominant closed-loop poles.
2) 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 𝜑. 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.

60. Lead Compensation Techniques Based on the Root-Locus
Approach (Cont.)
3) Assume the lead compensator 𝐺𝑐(𝑠) to be
where 𝛼 and 𝑇 are determined from the angle deficiency. 𝐾𝑐 is determined
from the requirement of the open-loop gain.
4) 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 a generally
results in a larger value of 𝐾𝑣 , which is desirable. Note that
5) Determine the value of 𝐾𝑐 of the lead compensator from the magnitude
condition. 60

61. EXAMPLE
 Consider the position control system with the feedforward transfer
function is
 It is desired to design a lead compensator 𝐺𝑐 𝑠 , so that the dominant
closed-loop poles have the damping ratio 𝜂 = 0.5 and the undamped
natural frequency 𝜔 𝑛 = 3 𝑟𝑎𝑑/𝑠𝑒𝑐.
61

62. Solution
 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 𝜂 = 0.5
 The undamped natural frequency of the closed-loop poles
is 𝜔 𝑛 = 3 𝑟𝑎𝑑/𝑠𝑒𝑐
 Because the damping ratio is small, this system will have
a large overshoot in the step response and is not
desirable.
62

63. Solution (Cont.)
 The desired location of the dominant closed-loop poles can be
determined from
 as follows:
63

64. Solution (Method -1)
 First, determine the necessary angle 𝜑 to be added
so that the total sum of the angles is equal to
± 180(2𝑘 + 1).
 The angle from the pole at the origin to the desired
dominant closed-loop pole at 𝑠 = – 1.5 + 𝑗2.5981 is
120°.The angle from the pole at 𝑠 =– 1 to the desired
closed-loop pole is 100.894°. Hence, the angle deficiency
is
 Thus, if we need to force the root locus to go
through the desired closed-loop pole, the lead
compensator must contribute 𝜑 = 40.894° at this
point.
64

65. Solution (Method -1)
 Then we bisect angle APO and take 40.894°/2 each side, then the
locations of the zero and pole are found as follows:
65

66. Solution (Method -1)
 Thus, 𝐺𝑐 𝑠 can be given as
 For this compensator the value of a is α = 1.9432/4.6458 = 0.418.
 The value of 𝐾𝑐 can be determined by use of the magnitude condition.
 Hence, the lead compensator 𝐺𝑐 𝑠 just designed is given by
 the closed-loop transfer function becomes
66

67. Solution (Method -2)
 If we choose the zero of the lead compensator at 𝑠 = −1 so that it will cancel
the plant pole at 𝑠 = −1, then the compensator pole must be located at 𝑠 = −3.
 The value of 𝐾𝑐 can be determined by use
of the magnitude condition.
 Hence, the lead compensator 𝐺𝑐 𝑠 just designed is given by
.
67

68. Lead Compensation Techniques Based on the Root-Locus Approach
 The designing a lag compensator
for the system shown by the root-
locus method may be stated as
follows:
68
1) Draw the root-locus plot for the uncompensated system whose open-loop transfer
function is 𝐺(𝑠) . Based on the transient-response specifications, locate the
dominant closed-loop poles on the root locus.
2) Assume the lag compensator 𝐺𝑐(𝑠) to be
Then the open-loop transfer function of the compensated system becomes
𝐺𝑐 𝑠 𝐺(𝑠).

69. Lead Compensation Techniques Based on the Root-Locus
Approach (Cont.)
3) Evaluate the particular static error constant specified in the problem.
4) Determine the amount of increase in the static error constant necessary to
satisfy the specifications.
5) Determine the pole and zero of the lag compensator that produce the
necessary increase in the particular static error constant without appreciably
altering the original root loci. (Note that the ratio of the value of gain required
in the specifications and the gain found in the uncompensated system is the
required ratio between the distance of the zero from the origin and that of the
pole from the origin.)
6) Adjust gain 𝐾𝑐 of the compensator from the magnitude condition so that the
dominant closed-loop poles lie at the desired location.
69

70. Example
 Consider the control system shown. Design a lag compensator 𝐺𝑐(𝑠) such
that the static velocity error constant 𝑘 𝑣 is 50 sec−1
without appreciably
changing the location of the original closed-loop poles, which are at
𝑠 = − 2 ± 𝑗16.
70

71. Solution
 Assume that the transfer function of the lag compensator is
 Since 𝑘 𝑣 is specified as 50 sec−1, we have
 Thus
 Now choose then
71

72. Solution
 Choose 𝑇 = 10. Then the lag compensator can be given by
 The angle contribution of the lag compensator at the closed - loop pole
𝑠 = − 2 ± 𝑗16.
 which is small. The magnitude of 𝐺𝑐(𝑠) at is 0.981. Hence the change
in the location of the dominant closed-loop poles is very small.
72

73. Solution
 The open-loop transfer function of the system becomes
 The closed-loop transfer function is
73

74. 74