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EE3331CPART 2 ~PAGE 1EE3331CPART 2 ~PAGE 1 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
EE3331C: Feedback Control SystemsPart 2: Frequency Domain Methods
EE3331C: Feedback Control SystemsPart 2: Frequency Domain Methods
Ben M. Chen
Professor & Provost’s ChairDepartment of Electrical & Computer Engineering
Office: E4‐06‐08, Phone: 6516‐2289Email: [email protected] http://www.bmchen.net
EE3331CPART 2 ~PAGE 2EE3331CPART 2 ~PAGE 2 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Introduction & Background Materials
EE3331CPART 2 ~PAGE 3EE3331CPART 2 ~PAGE 3 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
o G. F. Franklin, J. D. Powell and A. Emami‐Naeini, Feedback Control of Dynamic Systems, Pearson Prentice Hall
o R. C. Dorf and R. H. Bishop, Modern Control Systems, Pearson Prentice Hall
Text and References
o No fixed time for consultation
o Appointment through email is recommended
Consultation
EE3331CPART 2 ~PAGE 4EE3331CPART 2 ~PAGE 4 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Motivation
Learning Objectives
o Ability to find the gain and the phase of a system when the sinusoidal waveforms at its input and output are given
o Ability to compute the gain and the phase of a transfer function at a given frequency
o Ability to find the steady‐state output of a transfer function for a given sinusoidal input
o Ability to compute the crossover frequencies
The design of feedback control systems in industry is probably accomplished using frequency‐response methods more often than any other. This approach provides good designs in the face of uncertainty in the plant model. The so‐called frequency response of a system lies at the core of these methods.
An introduction to frequency response is covered in this unit.
EE3331CPART 2 ~PAGE 5EE3331CPART 2 ~PAGE 5 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Final Grades for Part 2
Final Grade = 80% Final exam marks for Part 2 (max = 50) +
20% Part 2 Test marks (max = 50)
There will be a in class (10–11:30am in LT4) test for Part 2 on Tuesday,
24 October 2017.
EE3331CPART 2 ~PAGE 6EE3331CPART 2 ~PAGE 6 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Controller
What is a control system?
System to be controlled
Desired Performance
REFERENCE
INPUTto the system
Information about the system:
OUTPUT
+–
DifferenceERROR
Objective: To make the system OUTPUT and the desired REFERENCE as close
as possible, i.e., to make the ERROR as small as possible.
Key Issues: (1) How to describe the system to be controlled? (Modeling)
(2) How to design the controller? (Control)
aircraft, missiles, economic systems,
cars, etc
EE3331CPART 2 ~PAGE 7EE3331CPART 2 ~PAGE 7 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Some Control Systems Examples
System to be controlledController+
–
OUTPUTINPUTREFERENCE
Economic SystemDesired
Performance Government
Policies
EE3331CPART 2 ~PAGE 8EE3331CPART 2 ~PAGE 8 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
A Control System Design Example – Unmanned Helicopter
Bare Helicopter
Positioning/SLAM
Flight Control System
Motion Planning
Mission Management
Drone
Autonomous UAS
Model Aircraft
EE3331CPART 2 ~PAGE 9EE3331CPART 2 ~PAGE 9 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Modeling – Data Collection
Data Collection Procedure:
1). Chirp‐like signal issued
in single channel;
2). Chirp‐like signal issued
in multi‐channels;
3). Step‐like and random
signals issued for validation.
Example of chirp‐like signal
Chirp‐like signal and corresponding responses
EE3331CPART 2 ~PAGE 10EE3331CPART 2 ~PAGE 10 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Modeling – Test Flights
Flight testing for modeling purpose
EE3331CPART 2 ~PAGE 11EE3331CPART 2 ~PAGE 11 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Domain Responses
EE3331CPART 2 ~PAGE 12EE3331CPART 2 ~PAGE 12 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Unmanned Helicopter Control System
EE3331CPART 2 ~PAGE 13EE3331CPART 2 ~PAGE 13 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
A Hybrid UAV Control System
Auto‐landing on moving platform…
03.201303.2013 09.201309.2013 03.201403.2014
20172017
20162016
EE3331CPART 2 ~PAGE 14EE3331CPART 2 ~PAGE 14 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
International Micro Air Vehicle Competition, Toulouse, France
Positioning/SLAM
Flight Control System
Motion Planning
Mission Management
NUS Team Instinct Cougar
EE3331CPART 2 ~PAGE 15EE3331CPART 2 ~PAGE 15 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Linear Systems
Let y1(t) be the output produced by an input signal u1(t) and y2(t) be the output
produced by another input signal u2(t). Then, the system is said to be linear if
a) the input is u1(t), the output is y1(t), where is a scalar; and
b) the input is u1(t) + u2(t), the output is y1(t) + y2(t).
Or equivalently, the input is u1(t) + u2(t), the output is y1(t) + y2(t). Such a
property is called superposition. For the circuit example on the previous page,
It is a linear system! We will mainly focus on linear systems in this course.
Systemu(t) y(t)
)()()()()()()( 21221
21
21
221
21
2 tytytuRR
RtuRR
RtutuRR
Rty
EE3331CPART 2 ~PAGE 16EE3331CPART 2 ~PAGE 16 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Summary of Laplace transform properties
Property f (t) F (s)
Linearity
Scaling
Time shift
Frequency shift
Time derivative
Time integration
Time periodicity
Initial value
Final value
Convolution
t
df0
sTesF1)(1
)0( f
EE3331CPART 2 ~PAGE 17EE3331CPART 2 ~PAGE 17 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Some commonly used Laplace transform pairs
2
1
2
1
1
!
1
1)(1
1
)(
)(
)(
aste
ase
snt
st
st
sF
t
tf
at
at
nn
2
1
2
1
1
!
1
1)(1
1
)(
)(
)(
aste
ase
snt
st
st
sF
t
tf
at
at
nn
22
22
22
22
22
22
cos
sin
sincos)cos(
cossin)sin(
cos
sin
)()(
asaste
aste
sst
sst
sst
st
sFtf
at
at
22
22
22
22
22
22
cos
sin
sincos)cos(
cossin)sin(
cos
sin
)()(
asaste
aste
sst
sst
sst
st
sFtf
at
at
EE3331CPART 2 ~PAGE 18EE3331CPART 2 ~PAGE 18 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Operations of complex numbers
Coordinates: Cartesian Coordinate and Polar Coordinate
125tan
2239.01
51213512j
j eejreal part imaginary part magnitude argument
Euler’s Formula:
Additions: It is easy to do additions (subtractions) in Cartesian coordinate.
)()()()( wbjvajwvjba
Multiplication's: It is easy to do multiplication's (divisions) in Polar coordinate.
)()( jjj eruuere )(
j
j
j
eur
uere
sincos je j
EE3331CPART 2 ~PAGE 19EE3331CPART 2 ~PAGE 19 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Responses
EE3331CPART 2 ~PAGE 20EE3331CPART 2 ~PAGE 20 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The frequency response of a system is defined as the steady‐state responseof the system to a sinusoidal input signal.
Ref: Modern Control Systems, Richard C. Dorf & Robert H. Bishop
When a stable linear system is subject to sinusoidal input of frequency i rad/s
the output and all intermediate signals are sinusoidal of frequency i rad/s in the steady state
)sin()( tAtu i
)sin()( tBty iss
Linear system G(s)
)()()( tytyty sstr
the output and the intermediate signals differ from the input in amplitude and phase.
Frequency Response
EE3331CPART 2 ~PAGE 21EE3331CPART 2 ~PAGE 21 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Response Example
Example: Response of a 1st‐order system to sinusoidal input
10010)()10sin()(,
11)( 2
s
sUttus
sG
)100()1(
)100(10
)1(1)(
2
2
sCBs
sA
sssYSolution:
)100()()()1()1()100(10
2
2
CAsCBsBAsCsBssA
10110,
10110,
10110
10100,0,0
CBA
CACBBA
)10sin(101
1)10cos(10110
10110)(
)10(10
1011
)10(10110
)1(101/10)( 2222
ttety
sss
ssY
t
)8410sin(1011
10110)( tety t )8410sin(
1011
10110
2
1
ty
ey t
EE3331CPART 2 ~PAGE 22EE3331CPART 2 ~PAGE 22 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Response
)sin()( tAtu i
)sin()( tBty iss
Linear system G(s)
)()()( tytyty sstr
Gain of the system:ABM
Phase of the system:
Gain & phase vary with the frequency of the input sinusoid
M() and () define the frequency response of the linear system
[amplitude scaling]
[shift in time]
EE3331CPART 2 ~PAGE 23EE3331CPART 2 ~PAGE 23 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Response
)sin()( tAtu i
)sin()( tBty iss
Linear system G(s)
)()()( tytyty sstr
Measurement of frequency response:a) Apply a sinusoidal input to the system under testb) Allow sufficient time for transient to decayc) Measure amplitudes of the input and the output; ratio of amplitudes
is the gaind) Measure time‐shift of the output with reference to the input and
determine the phasee) Repeat steps (a) to (d) for different frequencies min < i < max
1 2 3 4 5 6 7 8 …..
Gain M1 M2 M3 M4 M5 M6 M7 M8 ….Phase 1 2 3 4 5 6 7 8 ….
In modern equipment, the measurement steps are automated
EE3331CPART 2 ~PAGE 24EE3331CPART 2 ~PAGE 24 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Frequency Response (cont.)
G (s): Transfer function of a linear systemM (i): Gain of the system at frequency i rad/s (i): Phase of the system at frequency i rad/s
Then,
)()(,)()( iiii jGjGM
9310)( 2
ss
sGExample: Find the gain and the phase at = 4 rad/s for
25.12072.0
712tan127
10127
109)4(3)4(
10)4(122
2 jjjjG
EE3331CPART 2 ~PAGE 25EE3331CPART 2 ~PAGE 25 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
57.0)(,2.0)(01.0 jGjG
71.5)(,199.0)(1.0 jGjG
31.11)(,196.0)(2.0 jGjG
57.26)(,179.0)(5.0 jGjG
45)(,141.0)(1 jGjG
43.63)(,089.0)(2 jGjG
69.78)(,039.0)(5 jGjG
29.84)(,020.0)(10 jGjG
43.89)(,002.0)(100 jGjG
94.89)(,0002.0)(1000 jGjG
10-2 10-1 100 101 102 1030
0.05
0.1
0.15
0.2
Frequency (rad/sec)
Mag
nitu
de
10-2 10-1 100 101 102 103-100
-50
0
Frequency (rad/sec)
Pha
se (d
egre
es)
Example:
Let us a series RL circuit with the following transfer function:
Thus, it is straightforward, but very tedious, to compute its amplitude and phase.
1
2.0)(s
sG
12
tan1
2.0)()(1
2.0)(
jGjGj
jG
EE3331CPART 2 ~PAGE 26EE3331CPART 2 ~PAGE 26 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Note that in the plots of the magnitude and phase responses, we use a log scale for the
frequency axis. If we draw in a normal scale, the responses will look awful.
There is another way to draw the frequency response. i.e., directly draw both magnitude
and phase on a complex plane, which is called the polar plot. For the example
considered, its polar plot is given as follows:
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
read axis
imag
axi
s
10-2
10-1
100
101
102
103
0
0.05
0.1
0.15
0.2
Frequency (rad/sec)
Mag
nitu
de
10-2 10-1 100 101 102 103-100
-80
-60
-40
-20
0
Frequency (rad/sec)
Pha
se (d
egre
es)
The red‐line curves are more accurate plots using MATLAB.
EE3331CPART 2 ~PAGE 27EE3331CPART 2 ~PAGE 27 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
What can we observe from the frequency response?
10-2 10-1 100 101 102 1030
0.05
0.1
0.15
0.2
Frequency (rad/sec)
Mag
nitu
de
10-2 10-1 100 101 102 103-100
-80
-60
-40
-20
0
Frequency (rad/sec)
Pha
se (d
egre
es)
At low frequencies, magnitude response is relatively a large. Thus, the corresponding output will be large.
For large frequencies, the magnitude response is small and thus signals with large frequencies is attenuated or blocked.
EE3331CPART 2 ~PAGE 28EE3331CPART 2 ~PAGE 28 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Lowpass System
0 10 20 30 40 50 60 70 80 90 100-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Time (second)0 10 20 30 40 50 60 70 80 90 100
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
Time (second)
Lowpass System
Lowpass Systems
0 1 2 3 4 5 6 7 8 9 10-0.02
-0.01
0
0.01
0.02
0.03
0.04
Time (second)0 1 2 3 4 5 6 7 8 9 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Time (second)
rad/sec1.0
rad/sec10
EE3331CPART 2 ~PAGE 29EE3331CPART 2 ~PAGE 29 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Time Delay
Delay(td)
Element causing delay of td seconds
)()( dttuty )(tu )(tu
)()(
)()(
..
..
sUettu
sUtu
dst
TLd
TL
dd
tsts
esU
sUesUsYsG
)(
)()()()(delay
)(ty
t
t
dt
dtjejG )(delay
d
tj
t
eM d
1
EE3331CPART 2 ~PAGE 30EE3331CPART 2 ~PAGE 30 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Time Delay
No difference between amplitudes of two signals
)sin()(
tAtu
i )sin()(
tAty
i
dtT
degree180rad
didi tt
Tttftt d
ddii
d
22
The phase of the delay unit is rad
1M
Phase of delay unit is
)sin()sin()(
iii tAtAty
EE3331CPART 2 ~PAGE 31EE3331CPART 2 ~PAGE 31 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 1‐1
A sine wave of unity amplitude and frequency 5 Hz is applied to a linear system described by the transfer function
)50)(1()10()(
ssssG
What is the steady‐state output?
Solution:
22 10010)()10sin()2sin()(,
)50)(1(10)(
s
sUttftuss
ssG
0.00735473 1002500
104940
10010
110
0.00584059100110
499
10010
5010
10050110010
)50)(1(10)()()(
250
22
21
22
2222
s
s
sssB
sssA
sDCs
sB
sA
sssssUsGsY
EE3331CPART 2 ~PAGE 32EE3331CPART 2 ~PAGE 32 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
2222 100500.00735473
10.00584059
10010
)50)(1(10)()()(
sDCs
sssssssUsGsY
)100)(50)(1())(50)(1()100)(1(0.00735473)100)(50(0.00584059
22
2222
sss
DCsssssss
)100)(50)(1()501000.0073547350000.00584059()0.007354739(0.0058405
22
223
sss
DsC
0.01319532 00.007354730.00584059 CC
0.37357712 100501000.0073547350000.00584059 22 DD
2222 10037357712.0
10001319532.0
500.00735473
10.00584059)(
sss
sssY
)4810sin(0.01776289 0.007354730.00584059)10sin(0.01189133)10cos(0.013195320.007354730.00584059)(
50
50
teetteety
tt
tt
EE3331CPART 2 ~PAGE 33EE3331CPART 2 ~PAGE 33 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Alternatively, we calculate
480.01776289
32.1201856.06534.729691.32
1051)10(501010)10( 2
j
jjG
)4810sin(0.01776289 )( ttyss
-60
-55
-50
-45
-40
-35
-30
-25
-20
-15
-10
System: sysFrequency (rad/sec): 31.2Magnitude (dB): -35
Mag
nitu
de (
dB)
Bode Diagram
Frequency (rad/sec)10-2 10-1 100 101 102 103
-90
-60
-30
0
System: sysFrequency (rad/sec): 31.5Phase (deg): -48
Pha
se (d
eg)
It is even easier
to use
MATLAB…
EE3331CPART 2 ~PAGE 34EE3331CPART 2 ~PAGE 34 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
For any system, there may exist one or more frequencies at which the gain of the system is unity (1). Such a frequency is called the gain‐crossover frequency. We’ll use the symbol cg to represent it, i.e.,
9310)( 2
ss
sGExample: Find the gain‐crossover frequency cg of
Solution:
Gain‐crossover frequency
1)( cgjG
1
39
1039
1093)(
10)(22222
cgcgcgcgcgcgcg jjj
jG
100391039 222222 cgcgcgcg
rad281.3765.100199 2222 cgcgcgcg
EE3331CPART 2 ~PAGE 35EE3331CPART 2 ~PAGE 35 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Phase‐crossover frequency
For any system, there may exist one or more frequencies at which the phase of the system is 180. Such a frequency is called the phase‐crossover frequency. We’ll use the symbol cp to represent it, i.e.,
)2)(1(10)(
ssssGExercise 1‐3: Find the phase‐crossover frequency cp of
180)( cpjG
-120
-100
-80
-60
-40
-20
0
20
40
60
Mag
nitu
de (
dB)
10-2 10-1 100 101 102-270
-225
-180
-135
-90
System: sysFrequency (rad/sec): 1.41Phase (deg): -180
Pha
se (d
eg)
Bode Diagram
Frequency (rad/sec)
Frankly, it is really troublesome to calculate the crossover frequencies (gain and frequency) using pen and paper…
EE3331CPART 2 ~PAGE 36EE3331CPART 2 ~PAGE 36 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 1‐2
Find the gain‐crossover frequency of the following transfer function
)50)(1()10(100)(
ssssG
Use MATLAB to verify your result.
Solution:
1)50()1(
10100
501
10100)(
222
222
22
222
jj
jjG
rad/s 35.877629.7384
09975007499)(2
222
10-2
10-1
100
101
102
103
-90
-45
0
Phas
e (d
eg)
Bode Diagram
Frequency (rad/s)
-20
-10
0
10
20
30
System: sysFrequency (rad/s): 87.2Magnitude (dB): -0.0184
Mag
nitu
de (d
B)
EE3331CPART 2 ~PAGE 37EE3331CPART 2 ~PAGE 37 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Key Concepts Learnt
o Frequency response is defined for linear systems only
o If input to a stable linear system is sinusoidal of frequency i, then its output in the steady‐state is also sinusoidal of frequency i Input an output may differ in amplitude and phase
o Gain of the linear system is the ratio between the output (steady‐state) amplitude and the input amplitude Gain varies with frequency
o Phase defines the delay between the input sinusoid and the output sinusoid Phase varies with frequency
)()(,)()( iiii jGjGM
o If a system’s transfer function is G(s), then its gain and phase at i are
o For a transfer function G(s), the crossover frequencies are defined as
180)(,1)( cpcg jGjG
EE3331CPART 2 ~PAGE 38EE3331CPART 2 ~PAGE 38 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot
The plots of the magnitude and phase responses of a transfer function are called the Bode
plot. The easiest way to draw Bode plot is to use MATLAB (i.e., bode function). However, there
are some tricks that can help us to sketch Bode plots (approximation) without computing
detailed values. To do this, we need to introduce a scale called dB (decibel). Given a positive
scalar a, its decibel is defined as . For example,)(log20 10 a
dB 00)(log201 10 aaa
dB 2020)(log2010 10 aaa
dB 4040)(log20100 10 aaa
dBin dBin )(log20)(log20)(log20 101010 aa
dBin dBin )(log20)(log20)(log20 101010
aa
In the dB scale, the product of two scalars becomes an addition and the division of two
scalars becomes a subtraction.
EE3331CPART 2 ~PAGE 39EE3331CPART 2 ~PAGE 39 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – an integrator
We start with finding the Bode plot asymptotes for a simple system characterized by
Examining the amplitude in dB scale, i.e.,
it is simple to see that
901)()(1)(1)(
jGjGj
jGs
sG
dBlog201log20)(log20 101010
jG
dB01log20)1(log201 1010 jG
dB2010log20)10(log2010 1010 jG
dBlog2020log2010log20
10log20log20)(log2010
11011010
11021021012
jG
Thus, the above expressions clearly indicate that the magnitude is reduced by 20 dB
when the frequency is increased by 10 times. It is equivalent to say that the magnitude is
rolling off 20 dB per decade.
EE3331CPART 2 ~PAGE 40EE3331CPART 2 ~PAGE 40 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The phase response of an integrator is 90 degrees, a constant. The Bode plot of an
integrator is given by
-60
-40
-20
0
20
Mag
nitu
de (d
B)
10-1
100
101
102
103
-91
-90.5
-90
-89.5
-89
Phas
e (d
eg)
Bode Diagram
Frequency (rad/sec)
rolling off 20
dB per
decade
constant
phase
response
EE3331CPART 2 ~PAGE 41EE3331CPART 2 ~PAGE 41 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – an differentiator
The Bode plot of G(s) = s can be done similarly…
-20
0
20
40
60
Mag
nitu
de (d
B)
10-1
100
101
102
103
89
89.5
90
90.5
91
Phas
e (d
eg)
Bode Diagram
Frequency (rad/sec)
constant
phase
response
rolling up 20
dB per
decade
EE3331CPART 2 ~PAGE 42EE3331CPART 2 ~PAGE 42 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – a general first order system
The Bode plot of a first order system characterized by a simple pole, i.e.,
Let us examine the following situations.
1
12
111
1
1 tan1
1
1
1)(1
1)(
jjGss
sG
dB) 3707.0(45707.0tan1
1)(1
112
1
1
11
jG
01tan1
1)(1
12
1
1
jG
90tan1
1)( 1
1
12
1
1
jG
These give us the approximation (asymptotes) of the Bode curves…
EE3331CPART 2 ~PAGE 43EE3331CPART 2 ~PAGE 43 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Example
Consider a 1st order system where 1 = 10 rad/sec is called the corner
frequency. 1011)( ssG
10-1 100 101 102 103-40
-30
-20
-10
0
10
Mag
nitu
de (d
B)
Frequency (rad/sec)
10-1 100 101 102 103-100
-80
-60
-40
-20
0
Frequency (rad/sec)
Pha
se (d
egre
es)
101
rolling off 20
dB per
decade
101
110red: asymptotes
blue: actual curves
EE3331CPART 2 ~PAGE 44EE3331CPART 2 ~PAGE 44 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – a general first order unstable system
The Bode plot of a first order system characterized by a simple pole, i.e.,
Let us examine the following situations.
1
12
111
1
1 tan1801
1
1
1)(1
1)(
jjGss
sG
dB) 3707.0(135707.0tan1
1)(1
112
1
1
11
jG
1801tan1801
1)(1
1
2
1
1
jG
90tan1801
1)( 1
1
1
2
1
1
jG
These give us the approximation (asymptotes) of the Bode curves…
EE3331CPART 2 ~PAGE 45EE3331CPART 2 ~PAGE 45 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
10-1
100
101
102
103
-50
-40
-30
-20
-10
0
10
Mag
nitu
de (
dB)
10-1
100
101
102
103
-200
-180
-160
-140
-120
-100
-80
Pha
se (
degr
ee)
Frequency (rad/sec)
Example
Consider a 1st order system where 1 = 10 rad/sec is called the corner
frequency. 1011)( ssG
101
rolling off 20
dB per
decade
101
110
red: asymptotes
blue: actual curves
EE3331CPART 2 ~PAGE 46EE3331CPART 2 ~PAGE 46 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – a simple zero factor
The Bode plot of can be done similarly…1
1)( ssG
10-1 100 101 102 103-10
0
10
20
30
40
Frequency (rad/sec)
Mag
nitu
de (d
B)
10-1 100 101 102 103
0
20
40
60
80
100
Frequency (rad/sec)
Pha
se (d
egre
es)
1 rolling up 20
dB per
decade
101
110
red: asymptotes
blue: actual curves
EE3331CPART 2 ~PAGE 47EE3331CPART 2 ~PAGE 47 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
10-1 100 101 102 103-10
0
10
20
30
40
Mag
nitu
de (d
B)
10-1 100 101 102 103-100
-80
-60
-40
-20
0
Pha
se (d
eg)
Frequency (rad/s)
Bode plot – a simple unstable zero factor
The Bode plot of can be done similarly…1
1)( ssG
1 rolling up 20
dB per
decade
101
110red: asymptotes
blue: actual curves
EE3331CPART 2 ~PAGE 48EE3331CPART 2 ~PAGE 48 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot – putting all together
Assume a given system has only simple poles and zeros, i.e.,
or
In the dB scale, we have
Similarly,
Thus, the Bode plot of a complex system can be broken down to the additions and subtractions of some simple systems…
)()()()()(
1
1
011
1
011
1
m
mmn
nn
mm
mm
pspszszsb
asasasbsbsbsbsG
)1()1(
)1()1(
)()()()()(
1
1
1
1
1
1
n
qm
m
mm
ps
pss
ps
zsk
pspszszsbsG
dB in1dB in 1 dB in dB in 1dB in 1dB in dB )(11 11 nm p
jp
jqzj
zjkjG
11
11 9011)(11 nm p
jp
jqzj
zjkjG
EE3331CPART 2 ~PAGE 49EE3331CPART 2 ~PAGE 49 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
10-3 10-2 10-1 100 101 102 103-60
-40
-20
0
20
40
60
Frequency (rad/sec)
Mag
nitu
de (d
B)
10-3 10-2 10-1 100 101 102 103-100
-50
0
50
100
Frequency (rad/sec)
Pha
se (d
egre
es)
Example
We consider a system characterized by)101(
)1.01(1.0
)10()1.0(10)( ss
s
ssssG
s1
1.01 s
1011s
1.0k
EE3331CPART 2 ~PAGE 50EE3331CPART 2 ~PAGE 50 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The actual Bode plot
-50
0
50
Mag
nitu
de (d
B)
10-3
10-2
10-1
100
101
102
103
-90
-45
0
Phas
e (d
eg)
Bode Diagram
Frequency (rad/sec)
EE3331CPART 2 ~PAGE 51EE3331CPART 2 ~PAGE 51 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot of a typical 2nd order system with complex poles
So far, we haven’t touched the case when the system has complex poles. Consider
When < 1, it has two complex conjugated poles at
2222
2
22
2
21
1)1()(2
)(
nn
nn
n
nn
n
ssssssG
21 nn
is called the damping ratio of the system
n is called the natural frequency
This 2nd order prototype is the
most important system for classical
control.
EE3331CPART 2 ~PAGE 52EE3331CPART 2 ~PAGE 52 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Examining
we have
nnnnjjj
jH
21
1
21
1)(22
01121
1)( 2
nn
n
jjH
18021
1)(22
2
nn
nn
n
jjH
n
10 10 1 1001 1 1( ) in dB 20log 20log 20log
2 26 dB 20logH j
9021
21
21
1)( 2
jj
jH
nn
0 dB in magnitude and 0 degree phase at low
frequencies
roll off 40 dB per decade at
high frequencies
G
G
G
G
|G
EE3331CPART 2 ~PAGE 53EE3331CPART 2 ~PAGE 53 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot of the 2nd order prototype
rolling off 40
dB per
decade
221
bygiven isfrequency (resonant)peak
then the,707.0 If
nr
2121
is magnitudepeak The
rM
EE3331CPART 2 ~PAGE 54EE3331CPART 2 ~PAGE 54 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
10-1 100 101-40
-30
-20
-10
0
10
20
30
Frequency (rad/s)
Mag
nitu
de (d
B)
05.0
1.0
2.0
3.0
4.0
5.0
nrr MMw
dB 20dB 20rad/s 1
dB 14dB 1.14rad/s 99.0
dB 8dB 3.8rad/s 96.0
dB 5.4dB 3.5rad/s 91.0
dB 0.2dB 6.3rad/s 82.0
dB 02.0dB 3rad/s 71.0
Resonant frequency r is considerably lower than natural frequency n when > 0.2.
EE3331CPART 2 ~PAGE 55EE3331CPART 2 ~PAGE 55 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Modeling and System Identification
EE3331CPART 2 ~PAGE 56EE3331CPART 2 ~PAGE 56 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Modeling from the Bode Plot
Motivation
Learning Objectives
Ability to estimate the transfer function model of a system from its Bode plot
Frequency response data can be generated by performing experiments. It is a common practice to find a transfer function model of the dynamic system from the experimentally obtained frequency response. In this unit, we learn how to estimate a transfer function model from the Bode plot.
EE3331CPART 2 ~PAGE 57EE3331CPART 2 ~PAGE 57 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Features of Bode (Magnitude) Plot
a) Low frequency gradient: dB/decade20)(N(gradient)low_freq
N: Number of integrators in G(s)negative value of N implies differentiator
o By measuring the low‐frequency gradient of Bode (magnitude) plot, we can determine the value of N
N = ?N = ?
EE3331CPART 2 ~PAGE 58EE3331CPART 2 ~PAGE 58 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
(n‐m) = ? (n‐m) = ?
b) High frequency gradient:dB/decade20)(m)(n
dB/decade20)(m20)(n(gradient)high_freq
n: Number of poles in G(s)m: Number of zeros in G(s)
By measuring the high frequency gradient of Bode (magnitude) plot, we can determine the value of (n‐m) Pole Excess
EE3331CPART 2 ~PAGE 59EE3331CPART 2 ~PAGE 59 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
c) At the corner frequency of a zero, the slope is changed by +20 dB/decade
d) At the corner frequency of a pole, the slope is changed by ‐20 dB/decade
o Change in the gradient is an indicator of the presence of corner frequency or natural frequency
o Bending upward indicates presence of zeroo Bending downward means pole
EE3331CPART 2 ~PAGE 60EE3331CPART 2 ~PAGE 60 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Two 1st‐order factors shown below have identical magnitude plots but different phase plots
Phase of System with RHS Pole/ Zero
A transfer function with RHS zeros is called non‐minimum phase system
)tan180()()(
tan)()(
122
122
aaajas
aaajas
js
js
o Two transfer functions can have identical magnitude plots though their phase plots are different)10(
)1(10
ss
)10()1(10
ss
EE3331CPART 2 ~PAGE 61EE3331CPART 2 ~PAGE 61 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Two functions may have identical magnitude plots but different phase plots if transmission delays are different
Phase of System with Delay
180tan)()(
tan)()(
122
122
dtj
jsst
js
ta
aeajeas
aaajas
dd
Magnitude plot alone is not enough for identification of model
EE3331CPART 2 ~PAGE 62EE3331CPART 2 ~PAGE 62 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 2‐1: Find the transfer function model from the Bode plot given
-60
-40
-20
0
20
Mag
nitu
de (
dB)
10-1 100 101 102 103 104-90
-60
-30
0
Pha
se (d
eg)
Bode Diagram
Frequency (rad/s)
10011
41)(
s
ssksG
4.010
dB 820/8
k
)100(
410)(
ss
ssG
EE3331CPART 2 ~PAGE 63EE3331CPART 2 ~PAGE 63 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 2‐2: Find the transfer function model from the Bode plot given
10-1
100
101
102
103
104
-150
-100
-50
0
50
10-1
100
101
102
103
104
-300
-250
-200
-150
-100
-50
Rolling off 20 dB per decade
Rolling off 40 dB per decade
– 300
20 dB
dste
s
ssG
8011
10)(
180180300
10000
dd tt
ms 0.21 s00021.0
dt
80 rad/s
EE3331CPART 2 ~PAGE 64EE3331CPART 2 ~PAGE 64 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Steady State Responses
EE3331CPART 2 ~PAGE 65EE3331CPART 2 ~PAGE 65 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Motivation
Learning Objectives
Putting a controller in cascade with the plant effectively modifies the properties of the loop transfer function and hence its frequency response. What modification is desirable and what is not? It is important to know the properties of the loop transfer function that would give the desired performance.
Understanding Design Specifications (Steady-State Error)
1) Ability to determine the target properties of the loop transfer function, given the desired specifications on steady‐state error
2) Ability to design the feedback loop so that it meets the steady‐state error specifications (only)
EE3331CPART 2 ~PAGE 66EE3331CPART 2 ~PAGE 66 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Design of feedback: start with the simplest feedback loop possible
o Only one parameter (K) to choose
o Needs
o Minimum number of components (in analog implementation) or
o Minimum computational delay (in digital implementation)
K Gp(s)er
+
+++ y+
d nu (t)
EE3331CPART 2 ~PAGE 67EE3331CPART 2 ~PAGE 67 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
A simple control system: Toilet water tank…
EE3331CPART 2 ~PAGE 68EE3331CPART 2 ~PAGE 68 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Can we meet the design requirements with the simplest solution?
1. Tracking Performance: How closely does the output follow the reference?
If the desired speed of a motor is 100 RPM, does it spin at 100 RPM or 99.5 RPM?
How do we specify the design requirements?
During transient, the output is still changing towards the final steady‐state value. Tracking performance is defined by the difference between output and reference in the steady‐state.
ssss yre
Reference is a step function )(tr
)(ty )()(lim)(lim tytrtett
EE3331CPART 2 ~PAGE 69EE3331CPART 2 ~PAGE 69 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
2. Transient Performance:
How fast does the output reaches the steady‐state?
What is the nature of the transient response? Does the response reach steady‐state exponentially? Does it oscillate?
How do we specify the design requirements?
In this unit, we consider only the tracking performance. We’ll also consider stability issue. Transient specifications are discussed later.
EE3331CPART 2 ~PAGE 70EE3331CPART 2 ~PAGE 70 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The output signal may or may not follow the reference in a feedback controlsystem. If the output is not following the reference, there is an error.
How do we find the steady‐state error?
)()()( tytrte
According to the Final Value Theorem, for any signal x(t), )(lim)(0
ssXtxst
K Gp(s)er
+
+++ y+
d nu (t)
EE3331CPART 2 ~PAGE 71EE3331CPART 2 ~PAGE 71 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
sAsX )(
A
t
x(t)
x(t)t
Aas
AssAsssX
sss
000
limlim)(lim
asA
sAsX
eAtx at
)(
1)(
1t
x(t)
x(t)t
t
ss
txA
sAsssX
)(
lim)(lim00
A A
eAtx at
1)(
Final Value Theorem (Illustrations)
EE3331CPART 2 ~PAGE 72EE3331CPART 2 ~PAGE 72 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
K Gp(s)er +
+++ y
+
d n
u (t)
How do we find the steady‐state error?
)(lim)(0
ssEtest
NUDGYEKU
p
)(
)()(1
1)()(1
)()(
)(11
)()(1
1)()(1
)()(
)(11)(
)1(
sNsL
sDsL
sGsR
sL
sNsKG
sDsKG
sGsR
sKGsE
NDGREKH
p
pp
p
p
pp
NEKGDGRNUGDGR
YRE
pp
pp
function transfer loop open called is )()( sKGsL p
EE3331CPART 2 ~PAGE 73EE3331CPART 2 ~PAGE 73 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
K Gp(s)er
+
+++ y+
d nu
How do we find the steady‐state error?
It is easily concluded from the previous slide that, the steady‐state error depends on
a) the Open Loop Transfer Function L(s) and b) the type of reference signal
)()(1
1lim)(0
sRsL
stest
If r(t) is a step function,
)(lim1)(1lim)(
00 sL
AsL
Ates
st
psksL
)(lim
0
(Position Error Constant)
sAsRAttr )(,)( 0
EE3331CPART 2 ~PAGE 74EE3331CPART 2 ~PAGE 74 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
)()(1
1lim)(0
sRsL
stest
If r(t) is a ramp function,
)(lim
)(lim
)(11lim)(
0
0
20
ssLA
ssLsA
sA
sLste
s
s
st
vskssL
)(lim
0
(Velocity Error Constant)
If r(t) is a parabolic function,
)(lim2
)(2lim
2)(1
1lim)(
2
0
220
30
sLsA
sLssA
sA
sLste
s
s
st
asksLs
)(lim 2
0
(Acceleration Error Constant)
21 )(,)(
sAsRAttr
32 2)(,)(
sAsRAttr
EE3331CPART 2 ~PAGE 75EE3331CPART 2 ~PAGE 75 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 3‐1: Find the steady‐state error for ramp‐input in the following system
10er
+
+++ y
+
d n
u (t))12(
1ss
)12(10)(
ss
sL
10)12(
10lim)(lim00
ss
sssLss
1.01011)(
vt k
te
EE3331CPART 2 ~PAGE 76EE3331CPART 2 ~PAGE 76 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Ker
+
+
y+
nu (t)
Exercise 3‐2: Find the value of K so that steady‐state error for step‐input is < 10%
10310
2 ss
10310)( 2
ssKsL
KssKsLk
ssp
103
10lim)(lim 200
91.0
1.011.01
1)(
pp
t kKk
te
EE3331CPART 2 ~PAGE 77EE3331CPART 2 ~PAGE 77 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Exercise 3‐3: Find the closed loop poles of the simple proportional control feedback system from the previous slide.
o What is the damping coefficient?o Sketch the step response of this feedback system? o Is this a good design?
)()(1
)()(
)(111)()()()(
)(11)( sR
sKGsKG
sRsKG
sYsYsRsRsKG
sEp
p
pp
222
2
2
101015.0290
100390
103901
10390
)(1)(
)()(
ssss
ss
sssKG
sKGsRsY
p
p
The closed loop poles of the control feedback system are
9.95.12,1 js
10,15.0 n
0 0.5 1 1.5 2 2.5 3 3.50
0.5
1
1.5Step Response
Time (sec)
Am
plitu
de
EE3331CPART 2 ~PAGE 78EE3331CPART 2 ~PAGE 78 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Steady‐state error for different input signals:
Step Ramp Parabolic
Steady‐state error ps
kA
sLA
1)(lim10 vs
kA
ssLA
)(lim0 as
kA
sLsA 2
)(lim2
2
0
If zero‐error is desired for step‐input, kp must be infinity which requires the loop transfer function L(s) to have at least one integrator
1),(1)( NsPs
sL NP(s) has no integrator in it
Number of integrators present in the loop transfer function is called System Type. A Type 0 has no integrator, Type 1 has one integrator, and so on
To achieve zero steady‐state error,
o Step input: the loop transfer function L(s) must be Type 1 or highero Ramp input: L(s) must be Type 2 or highero Parabolic input: L(s) must be Type 3 or higher
EE3331CPART 2 ~PAGE 79EE3331CPART 2 ~PAGE 79 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Going back to the slide where we said, “start with the simplest feedback loop possible”, such design may not satisfy many other design specifications. We may even end up with an unstable system.
o How to test stability of the closed loop system?o How to predict the transient response of the closed loop?
K Gp(s)er
+
+++ y+
d n
u (t)
So, we need to answer two more questions:
EE3331CPART 2 ~PAGE 80EE3331CPART 2 ~PAGE 80 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Stability Margins
EE3331CPART 2 ~PAGE 81EE3331CPART 2 ~PAGE 81 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
A given system is stable if the system does not have poles on the right‐half plane (RHP). It is
unstable if it has at least one pole on the RHP. In particular,
marginally stable
unstable
stable
The above diagram also shows the relationship the locations of poles and natural responses.
EE3331CPART 2 ~PAGE 82EE3331CPART 2 ~PAGE 82 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Motivation
Learning Objectives
o While designing a feedback system, we start with a known model of plant and then add a controller such that target specifications (for example, steady‐state error specification) are met.
o We didn’t consider the stability issue in earlier. o How to test stability?
We can derive the closed loop transfer function and find its poles to test stability.
But it is more convenient if we can assess closed loop stability from the open loop transfer function.
o We also need to determine the range of stable operation as model parameters are not accurate for any practical system.
o Ability to assess the closed loop stability from the knowledge of the open loop transfer function L(s)
o Ability to find the gain margin and phase margin of a given L(s)o Ability to explain the significance of the stability margins
EE3331CPART 2 ~PAGE 83EE3331CPART 2 ~PAGE 83 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Recap:
o Gain‐crossover frequency (cg) – frequency at which gain is 1 or unity (0 dB)
o Phase‐crossover frequency (cp) – frequency at which phase is 180
o If cg and cp are the same frequency for a loop transfer function L(s) then the resulting closed loop has poles on the imaginary axis
Closed loop will be marginally stable if no closed loop pole lies in the right hand side of the complex s‐plane
)(1)(
)(1)(
)()(
sLsL
sKGsKG
sRsY
p
p
Recall the transfer function of the closed‐loop system is given by
The closed‐loop poles can be found by solving
18011)(0)(1 sLsL
If we have an such that such that L(j) = 1 180, which implies j is a pole.
EE3331CPART 2 ~PAGE 84EE3331CPART 2 ~PAGE 84 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
K=10er+
+
y+
nu (t)
Exercise 4‐1: Find the crossover frequencies of the system of Exercise 3‐2
10310
2 ss
EE3331CPART 2 ~PAGE 85EE3331CPART 2 ~PAGE 85 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode plot of L(s) for different values of K
10310)( 2
ssKsL
K=10K=20K=50
What is the effect of varying K on cg?
What is the relation between K and cp?
EE3331CPART 2 ~PAGE 86EE3331CPART 2 ~PAGE 86 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
In this example, gain can be increased before cg and cp become equal
cg
180)( cg
180
cp
0)( cpdBM
)56(15)( 2
sss
sL
The resulting closed loop will not have poles on the j‐axis.
cpcg
For this case:
Gain Margin (GM): Variation in gain that can be allowed before the condition for marginal stability is met
EE3331CPART 2 ~PAGE 87EE3331CPART 2 ~PAGE 87 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Next slide shows numerical example of computing gain margin
180
cp
0)( cpdBM
)56(15)( 2
sss
sL
GM
)(log20
)(0
cp
cpdB
jL
MGM
EE3331CPART 2 ~PAGE 88EE3331CPART 2 ~PAGE 88 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
In this example, cg & cp can become equal due to delay (negative phase).
cg
180)( cg
180
cp
0)( cpdBM
)56(15)( 2
sss
sL
Phase Margin (PM): Variation in phase that can be allowed before the condition for marginal stability is met
EE3331CPART 2 ~PAGE 89EE3331CPART 2 ~PAGE 89 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
180
cg
180)( cp
)56(15)( 2
sss
sL
PM )180()(PM cgjL
GM expressed in dB
EE3331CPART 2 ~PAGE 90EE3331CPART 2 ~PAGE 90 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
gaincrossoverfrequency
phasecrossoverfrequency
gainmargin
phasemargin
Gain and phase margins (recap)
EE3331CPART 2 ~PAGE 91EE3331CPART 2 ~PAGE 91 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The stability margins (GM and PM) can be used as indicators of closed loop stability
For most practical systems (second order or higher), increasing loop gain makes the system more oscillatory and eventually may lead to instability.We use this assumption to find a relation between GM/PM and closed loop stability.
dBjGK )(1
dBjGK )(2
dBm jGK )(
12 KKK m
)()( jGKjL mFor
cpcg
Closed loop will have poles on the j‐axis
)(1 jGK
)(2 jGK
Based on the assumption made above,
Unstable CL
Stable CL
EE3331CPART 2 ~PAGE 92EE3331CPART 2 ~PAGE 92 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
)180()(PM cgjL
Stable UnstablePM > 0 PM < 0
0 dB 0 dB
0PM 0PM
EE3331CPART 2 ~PAGE 93EE3331CPART 2 ~PAGE 93 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/sec)
-100
-80
-60
-40
-20
0
20
40
60
80
Mag
nitu
de (
dB)
10-2 10-1 100 101 102-270
-225
-180
-135
-90
Pha
se (d
eg)
*
*
*
*
*
*
K = 10, PM > 0
K =100, PM < 0K = 30, PM = 0
)56()( 2
sssKsL
Closed‐loop Poles
K = 30
– 6; 0 j 2.24
K = 10
– 5.42; – 0.29 j 1.33
K = 100
– 7.22; 0.61 j 3.67
Unstable!
Example
EE3331CPART 2 ~PAGE 94EE3331CPART 2 ~PAGE 94 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
)(log20)(0GM cpdBcp jLjL
0 dB 0 dB0GM
0GM
Stable UnstableGM > 0 GM < 0
EE3331CPART 2 ~PAGE 95EE3331CPART 2 ~PAGE 95 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Nyquist Plot
Instead of separating into magnitude and phase diagrams as in Bode plots, Nyquist plot
maps the open‐loop transfer function L(s) directly onto a complex plane, e.g.,
Harry Nyquist 1889–1976
EE3331CPART 2 ~PAGE 96EE3331CPART 2 ~PAGE 96 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
–1
PM
1GM
Gain and phase margins
The gain margin and phase margin can also be found from the Nyquist plot by zooming in
the region in the neighbourhood of the origin.
Remark: Gain margin is the maximum
additional gain you can apply to the closed‐
loop system such that it will still remain
stable. Similarly, phase margin is the
maximum phase you can tolerate to the
closed‐loop system such that it will still
remain stable.
EE3331CPART 2 ~PAGE 97EE3331CPART 2 ~PAGE 97 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Summary
Stable CL poles on the j-axis
Unstable
K < Km K = Km K > Km
cg< cp cg = cp cg > cp
GM > 0 GM = 0 GM < 0|L(jcp)| < 1 |L(jcp)| = 1 |L(jcp)| > 1
PM > 0 PM = 0 PM < 0L(jcg)| > 180 L(jcg)| = 180 L(jcg)| < 180
o These conditions are based on the assumption that increasing gain makes a marginally stable system unstable and decreasing gain makes it stable
o This assumption holds for most practical systemso However, there are cases when it is violated
EE3331CPART 2 ~PAGE 98EE3331CPART 2 ~PAGE 98 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Transient Responses
EE3331CPART 2 ~PAGE 99EE3331CPART 2 ~PAGE 99 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Motivation
Learning Objectives
If we know how the frequency domain properties of the open loop transfer function L(s) are related to the transient response of the closed loop, we can modify the loop by adding controller in a way to meet the transient specifications.
Ability to determine the target properties of the loop transfer function, given the desired specifications on transient response
EE3331CPART 2 ~PAGE 100EE3331CPART 2 ~PAGE 100 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Taking the process output from one level to another is not the only need. How it reaches the final level is just as important. Transient profile is used to define the trajectory of the output in transition.
Transient specifications are typically described in terms of step response.
We have learnt how to modify the loop transfer function to meet specifications on steady‐state error with polynomial input signal.
If the closed loop is dominant 1st‐order, then its transient response is characterized by the time constant.
Initial slope =
2 3 4 5
1
EE3331CPART 2 ~PAGE 101EE3331CPART 2 ~PAGE 101 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
If the closed loop transfer function is dominant 2nd‐order, then its transient is characterized by the rise time, the settling time and the overshoot.
In this note, closed loop represented by dominant 2nd‐order transfer function (two complex poles) is considered for quantifying the frequency‐domain specifications.
22
2
2)(
nn
nCL ss
KsG
Step response of a typical 2nd order system
EE3331CPART 2 ~PAGE 102EE3331CPART 2 ~PAGE 102 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Unit step response for the 2nd order prototype
This is very important for the 2nd part of this course in designing a meaningful control system.
We consider the 2nd order prototype
)1()(2)( 222
2
22
2
nn
n
nn
n
ssssH
21 nn
is the damping ratio of the system
n is the natural frequency
It can be shown that its unit step response is
given as
( ) 1 cos sintd d
d
y t e t t
EE3331CPART 2 ~PAGE 103EE3331CPART 2 ~PAGE 103 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Unit step response for the 2nd order prototype (cont.)
Graphically,
It can be observed that the smaller damping ratio yields larger overshoot.
EE3331CPART 2 ~PAGE 104EE3331CPART 2 ~PAGE 104 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
104
Unit step response for the 2nd order prototype (cont.)
The typical step response can be depicted as
1 or 2%pMpt
rt
stt
19.0
1.0
rise time
1% settling time
peak time
nrt
8.1
4.6s
n
t
21
n
pt
ttety d
dd
t sincos1)(
overshoot21 eM p
all in sec
2% settling time 4s
n
t
EE3331CPART 2 ~PAGE 105EE3331CPART 2 ~PAGE 105 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
A loop transfer function with two poles gives a closed loop transfer functionof this type
22
2
2)(
nn
nCL ss
KsG
L(s)+
-
Y(s)
R(s)
)(1)()(sL
sLKsGCL
22
2
2
2)(
)2()(
nn
nCL
n
n
ssKsG
sssL
1)(1)(
))((1)(
2
absbasKsG
bsassL
CL
K
Examples
EE3331CPART 2 ~PAGE 106EE3331CPART 2 ~PAGE 106 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Let’s find the gain‐crossover frequency and the phase margin of the loop transfer function
)2()(
2
n
n
sssL
)2)(()(
2
n
n
jjjL
2
12
1)(
nnn
n
n
jjjjjL
21)(
jj
jLn
22 41)(
jL
Gain‐crossover frequency can be obtained by solving
14
1)(22
jL
EE3331CPART 2 ~PAGE 107EE3331CPART 2 ~PAGE 107 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
1414
1 2222
cgcgcgcg
014 224 cgcg 142 422 cg
142 42 cg
142 42 ncgn
cgcg
Conclusion:
Higher the gain‐crossover frequency of the loop transfer function, higher is the natural frequency (n) of the dominant 2nd‐order CL poles shorter rise time
This can be used as a design guideline
EE3331CPART 2 ~PAGE 108EE3331CPART 2 ~PAGE 108 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Conclusion:PM is related to the damping factor of the dominant closed loop pole
njjjL
,
21)(
2
tan90)( 1 jL
2
tan90)( 1 cgcgjL
cg
cg
cgjL
2tan
2tan90
)(180PM
1
1
142
2tanPM42
1
2
cg
cg2tan 1
2
tan 1 cg
EE3331CPART 2 ~PAGE 109EE3331CPART 2 ~PAGE 109 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The relation between PM and the damping factor of the closed loop poles can
be approximated by the following for PM < 60
)2()(
2
n
n
sssL
142
2tanPM42
1
100PM
(This is commonly used as the design formula)
EE3331CPART 2 ~PAGE 110EE3331CPART 2 ~PAGE 110 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
110
This example shows that if we try to meet the time‐domain specifications, the frequency domain parameters (cg and PM) are also fixed…
1 or 2%pMpt
rt
stt
19.0
1.0
rise time
peak time
nrt
8.1
21
n
pt
overshoot21 eM p
2% settling time 4s
n
t
Time‐domain specs
142 42 ncg
142
2tanPM42
1
Frequency‐domain parameters
EE3331CPART 2 ~PAGE 111EE3331CPART 2 ~PAGE 111 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Lead and Lag Compensators
EE3331CPART 2 ~PAGE 112EE3331CPART 2 ~PAGE 112 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Motivation
Learning Objectives
o Feedback design is usually carried out in stages starting with the attempt to meet the steady‐state error which can be done by adjusting DC‐gain (gain‐adjustment phase). However, it doesn’t take into consideration the transient response or stability.
o A compensator is a circuit/system that compensates for the discrepancy remaining after the gain‐adjustment stage.
1) Ability to explain the role of compensator in a feedback system
2) Ability to explain the effects of 1st‐order compensator on the frequency response of the loop transfer function
3) Ability to recognize the difference between two types of 1st‐order compensator
EE3331CPART 2 ~PAGE 113EE3331CPART 2 ~PAGE 113 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Recap:
The steady state error with polynomial signals depends on different error constants
)(lim
,1
1
0sLk
ke
sp
pss
)(lim
,1
0ssLk
ke
sv
vss
)(lim
,1
2
0sLsk
ke
sa
ass
For step input For ramp input For parabolic input
o These error constants are limiting values of certain transfer function as s0
In frequency domain, they represent the DC gain of the corresponding transfer function
)(lim)(lim00
jLksLk p
js
sp
EE3331CPART 2 ~PAGE 114EE3331CPART 2 ~PAGE 114 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Recap:
Parameters that characterize the transient response (for example, and n)are related to the gain‐crossover frequency (cg) and phase‐margin (PM) ofthe loop transfer function.
60PM,100PM
142 42 ncg
142
2tanPM42
1
o Gain adjustment to meet the steady‐state error specificationsEffect in frequencies around cg are not considered at this stage
o Design compensator to modify the loop around crossover frequencycg and PM can also be changed by adjusting the gain but such adjustment affects the DC‐gain as well and result of gain‐adjustment stage is therefore undone.
Phase margin is an indicator of stability and stability margin.
Design in stages:
EE3331CPART 2 ~PAGE 115EE3331CPART 2 ~PAGE 115 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Ker
+
–
+
y+
n
ym
u (t)
Exercise 3‐2 revisited:
10310
2 ss
Desired steady‐state error is less the 10% for step input:
9,1.01
11.0,
pp
stepss
kk
e
Position error constant: KssKk
sp
103
10lim 20
10let,99 KKkp
103100)( 2
ss
sL
EE3331CPART 2 ~PAGE 116EE3331CPART 2 ~PAGE 116 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Plot of L(s) of this example:
PM 20
Can you find PM numerically?What kind of transient response will the closed loop give?What is the GM for this case?
EE3331CPART 2 ~PAGE 117EE3331CPART 2 ~PAGE 117 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
What can be done to achieve higher PM?
K=10PM 20
Use lower value of K so that cgis shifted to a lower frequency
K=2PM 45
K=10K=2
Change in K has no effect on phase
What will happen to steady‐state error performance if lower value of K is used?
EE3331CPART 2 ~PAGE 118EE3331CPART 2 ~PAGE 118 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Ideally what we want?
Increase the phase at this frequency without changing the gain
These solutions are not feasible by varying K. Compensator gives a realizable approximation of these solutions by
modifying gain/phase in a selected range of frequency.
K=10 (meets steady-state specs)
Reduce the gain at this frequency without changing DC gain
Soln#1
Soln#2
EE3331CPART 2 ~PAGE 119EE3331CPART 2 ~PAGE 119 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
C(s)er
+
yu (t)
G(s)
Solution?
Add a compensator…
0,1
1)(
TTs
TssC
o If 0 < < 1, C(s) is called a lead compensator as it will add a positive phase to the resulting open loop transfer function L(s).
o If > 1, C(s) is called a lag compensator as it will add a negative phase to the resulting open loop transfer function L(s).
EE3331CPART 2 ~PAGE 120EE3331CPART 2 ~PAGE 120 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Compensators: Sketch the Bode plot of the following two transfer functions:
112)(1
sssC 15
1)(2
sssC
1
15.0
s
s
12.0
1
s
s
10-2
10-1
100
101
102
-2
0
2
4
6
8
Mag
nitu
de (
dB)
10-2
10-1
100
101
102
0
5
10
15
20
Pha
se (
deg)
Frequency (rad/sec)
10-2
10-1
100
101
102
-14
-12
-10
-8
-6
-4
-2
0
2
Mag
nitu
de (
dB)
10-2
10-1
100
101
102
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
Pha
se (
deg)
Frequency (rad/sec)
EE3331CPART 2 ~PAGE 121EE3331CPART 2 ~PAGE 121 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Transfer functions shown on the previous slide can be generalized in this form
10,0,1
1)(
TTs
TssC
Such a transfer function is a 1st‐order compensator.
o The value of T determines the corner frequency of the zero (let’s name it z)
o The values of and T determine the corner frequency of the pole (let’s name it p)
T1
z
o The value of determines whether z > p or p > z and also the separation between these two frequencies
T 1
p
pz
EE3331CPART 2 ~PAGE 122EE3331CPART 2 ~PAGE 122 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Two examples:
o < 1
125.012
112)(1
s
ssssC
151
151)(2
s
ssssC
rad/s5.021,2 zT
rad/s0.125.0
1,5.0
pzp
sT z rad/0.111,1
rad/s2.015
1,5
ppz
o > 1
EE3331CPART 2 ~PAGE 123EE3331CPART 2 ~PAGE 123 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Compensator with 0 < < 1
10,1
1)(
TsTssC
pz
m
mIt adds positive phase (phase lead). So it is called Lead Compensator.
It can be used to improve phase margin by choosing m of the compensator at a frequency near gain‐crossover frequency of the loop before compensation. Positive phase of the lead compensator is the feature that we want to use.
Soln#1 can be approximated by this!
EE3331CPART 2 ~PAGE 124EE3331CPART 2 ~PAGE 124 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Soln#1: By a lead compensation…
Increase the phase at this frequency without changing much in the gain
EE3331CPART 2 ~PAGE 125EE3331CPART 2 ~PAGE 125 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Compensator with > 1
1,1
1)(
TsTssC
It adds negative phase (phase lag). So it is called Lag Compensator.
Soln#2 can be approximated by this!
p
z
m
m
o Gain is negative dB and fairly constant for >> z.
o Phase is negligibly small for >> z. o DC gain is 0 dB.
If we need to reduce the gain at some frequency x without changing the phase significantly and without affecting DC gain, we can use a lag compensator such that z << x.
EE3331CPART 2 ~PAGE 126EE3331CPART 2 ~PAGE 126 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Soln#2: By a lag compensation…
Reduce the gain at this frequency without changing DC gain
EE3331CPART 2 ~PAGE 127EE3331CPART 2 ~PAGE 127 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Undesirable (but Unavoidable) Consequence:
pz
m
m
Phase lead is the desirable effect. But the gain is also altered in frequencies greater than z. Attention must be paid to this when we design lead compensator.
p
z
m
m
To achieve gain reduction without significant change in phase, z must be chosen much lower than the frequency at which gain reduction is desired.
This affect low‐frequency gain of the loop though the DC‐gain is unaffected.
EE3331CPART 2 ~PAGE 128EE3331CPART 2 ~PAGE 128 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Issues to Consider while Designing a Lead Compensator
Phase of lead compensator
Phase after compensation
Gain of lead compensator
cg gets shifted after compensation
EE3331CPART 2 ~PAGE 129EE3331CPART 2 ~PAGE 129 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Issues to Consider while Designing a Lag Compensator
Bode plot of Lu(s)
Desired PMPhase after compensation
cg shifted to lower frequency after compensation
Gain of lag compensator
Phase of lag compensator
EE3331CPART 2 ~PAGE 130EE3331CPART 2 ~PAGE 130 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
How to design a lag compensator?
Example: Consider the 1st‐order process
15.01)(p
s
sG
Design a controller to meet the following specifications ‐
a) Zero steady‐state error for step inputb) 5% error for ramp inputc) Phase margin of 45
First use gain‐adjustment so that specifications (a) and (b) are met.
C(s)er
+
yu (t) G(s)
EE3331CPART 2 ~PAGE 131EE3331CPART 2 ~PAGE 131 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Loop transfer function that meets specs (a) and (b):
)15.0(20)()( pu
ss
sGsKsL
The subscript u is used to imply that the loop is uncompensated.
Does it meet the 3rd specification of PM (related to transient response)?
rad/s 2.6ωcg
162)ω( cgu jL
18PMu
)15.0(20)(u
jj
jL
To find PM, we need to know the gain‐crossover frequency.
Phase at the gain‐crossover frequency:
Phase margin: 18)180(162 Let’s use the symbol PMu for this.
EE3331CPART 2 ~PAGE 132EE3331CPART 2 ~PAGE 132 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Lag compensator can be used to reduce gain in a range of frequencies
-130
(1) Find the frequency where phase is ‐180+target PM.
(2) Find how much gain reduction is required at that frequency.
Caution: Lag compensator introduces a small phase lag. Choose target PM slightly greater (5) than the one given in the specifications. (50 and not 45)
EE3331CPART 2 ~PAGE 133EE3331CPART 2 ~PAGE 133 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Find the frequency at which phase of Lu is 130
)5.0(tan90 1
rad/s7.1x
)15.0(20)(u
jj
jL
Find the uncompensated gain at this frequency
We want to make the compensated gain at this frequency to be 1 (equivalent to 0 dB).
130)5.0(tan90 1x
40)5.0(tan 1x
125.020)(
2u
xx
xjL
91)7.1(25.07.1
20)7.1(2u
jL
91
)7.1(1)7.1(1)7.1()7.1(
uu
jLjCjLjC
EE3331CPART 2 ~PAGE 134EE3331CPART 2 ~PAGE 134 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
If we choose z of the compensator << 1.7 rad/s,
11)(
11)(
Tj
TjjCTs
TssC
Tz1
7.1
17.117.1)7.1(
Tj
TjjC
z7.1
Under this condition,
1,17.17.1
17.117.1)7.1(
Tj
TjTj
TjjC
Required reduction in gain at this frequency,
91
)7.1(1)7.1( jL
jCu
9
T17.1 17.1 T
EE3331CPART 2 ~PAGE 135EE3331CPART 2 ~PAGE 135 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
How to find the 2nd parameter T ?
11)(
11)(
Tj
TjjCTs
TssC
Tz1
7.1
17.117.1)7.1(
Tj
TjjC
15416
11)(6
7.110107.1
ss
TsTssCTT
A factor of 10 is usually good enough,
Homework: Find gain and phase of C(s) at = 1.7 rad/s and verify if they satisfy the requirements, i.e., gain equal to 1/9 and small negative phase.
Tz17.1 17.1 T
EE3331CPART 2 ~PAGE 136EE3331CPART 2 ~PAGE 136 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/s)10
-310
-210
-110
010
1-60
-30
0
System: sysFrequency (rad/s): 1.7Phase (deg): -4.99
Phas
e (d
eg)
-20
-15
-10
-5
0
System: sysFrequency (rad/s): 1.7Magnitude (dB): -19
Mag
nitu
de (d
B)
15416)(
sssC 19 dB = 8.9
EE3331CPART 2 ~PAGE 137EE3331CPART 2 ~PAGE 137 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/s)
-100
-50
0
50
100
150
System: sysFrequency (rad/s): 1.71Magnitude (dB): -0.0475
Mag
nitu
de (d
B)
10-3
10-2
10-1
100
101
102
-180
-135
-90
System: sysFrequency (rad/s): 1.7Phase (deg): -135
Phas
e (d
eg)
Green Curve:
Uncompensated open loop transfer function
cg = 6.2 rad/sPM = 18
Blue Curve:
Compensated open loop transfer function
cg,new = 1.7 rad/sPM = 45
EE3331CPART 2 ~PAGE 138EE3331CPART 2 ~PAGE 138 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Procedure to design a lag compensator…
i. Find the phase margin of the uncompensated system and compare it to the
design specs on phase margin (say, req). To be safe, we usually choose a
value m > req to start with the design.
ii. Find the frequency x such that the uncompensated loop at which has a
phase response of –180 + m.
iii. Find the corresponding gain value (say, Gx) of the uncompensated loop at x.
iv. Compute = Gx and .
v. The required lag compensator is given as .
vi. Verify your result. Repeat the above steps if necessary.
xω10
T
11)(
Ts
TssC
EE3331CPART 2 ~PAGE 139EE3331CPART 2 ~PAGE 139 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
How to design a lead compensator?
Example: Consider the same problem but repeat the design using lead compensator
15.01)(p
s
sG
Specifications:a) Zero steady‐state error for step inputb) 5% error for ramp inputc) Phase margin of 45
Gain‐adjustment stage is same as in the previous example.
)15.0(20)()( pu
ss
sGsKsL
First two specifications are met but phase margin is only 18. We now use a lead compensator to improve phase margin to 45.
EE3331CPART 2 ~PAGE 140EE3331CPART 2 ~PAGE 140 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Lead compensator can be used to increase phase lead
-130
Try a compensator with m = cg that gives m = 27.
m
PMu = 18
rad/s2.6cg
m
27PMPM uspecreq,app
EE3331CPART 2 ~PAGE 141EE3331CPART 2 ~PAGE 141 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
How m and m of compensator (lead or lag) are related to the compensator parameters T and ?
11)(
11)(
Tj
TjjCTs
TssC
TT 11 tantan
One way to find m is to express as function of and solve , where0
dd
Tz1 Tz
1Tp 1
Tp 1
Alternatively,
)log(21
2loglog
log pzpz
m
pzm
T1
m
TT 1,1
pz
EE3331CPART 2 ~PAGE 142EE3331CPART 2 ~PAGE 142 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Find the relation between m and compensator parameters
11)(
11)(
Tj
TjjCTs
TssC
TT 11 tantan
TT m1
m1
m tantan
11m tan1tan
11
1
tan m
21tan m
BABABA
BABABA
tantan1tantan)tan(
tantan1tantan)tan(
)1(
m
)1(
2
11sin m
11
mm TT
EE3331CPART 2 ~PAGE 143EE3331CPART 2 ~PAGE 143 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Let’s choose a compensator that gives m = 27 at m = 6.2 rad/s
11sin m
T1
m
1127sin
1145.0 4.0
4.012.6
T
25.0T
11.0125.0)(
sssC
)15.0(20)(u
ss
sL
)11.0)(15.0()125.0(20)(
sssssL
Lead Compensator
Uncompensated open loop
Compensated open loop
EE3331CPART 2 ~PAGE 144EE3331CPART 2 ~PAGE 144 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/sec)
0
1
2
3
4
5
6
7
8
System: sysFrequency (rad/sec): 6.22Magnitude (dB): 3.91
Mag
nitu
de (
dB)
10-1 100 101 102 1030
5
10
15
20
25
30 System: sysFrequency (rad/sec): 6.22Phase (deg): 25.4
Pha
se (d
eg)
11.0125.0)(
sssC
EE3331CPART 2 ~PAGE 145EE3331CPART 2 ~PAGE 145 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/sec)
-100
-50
0
50
System: sysFrequency (rad/sec): 8.32Magnitude (dB): 0.037
Mag
nitu
de (
dB)
10-1 100 101 102 103-180
-150
-120
-90
System: sysFrequency (rad/sec): 8.32Phase (deg): -142
Pha
se (d
eg)
The new gain cross‐over frequency is about
cg,new = 8.3 rad/s
PM = 38
EE3331CPART 2 ~PAGE 146EE3331CPART 2 ~PAGE 146 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
rad/s2.6cg
We get the desired phase at the original cg. However, the gain‐crossover is shifted to higher frequency as the lead compensator adds positive dB gain.
newcg,
Do we know this?
Phase of Luat cg,new is slightly more negative than phase at cg.
)( jL
EE3331CPART 2 ~PAGE 147EE3331CPART 2 ~PAGE 147 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
30)1845(1.1
)PMPM(1.1 uspecreq
req,app
req,appreq,appreq
1.11.0
Rule of Thumb:
We need a compensator that would add maximum phase
30m
11sin m 3.0
1)()( mum jLjC
Then we choose m in a way so that the compensated gain becomes 1 at the frequency where compensator phase is maximum.
Do we know any of these two terms?
EE3331CPART 2 ~PAGE 148EE3331CPART 2 ~PAGE 148 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
11)(
11)(
m
mm
Tj
TjjCTj
TjjC
What is |C(jm)| of a compensator?
11
mm TT
11)(
2m
jC
jj
jj
jC
11
)(1
1
m
As we have already chosen , we know what |C(jm)| is.
3.011)( m
jC
3.0)(
1)(m
mu
jC
jL1)()( mum jLjC
As the uncompensated transfer function Lu is known, we can find the frequency at which this condition is met.
EE3331CPART 2 ~PAGE 149EE3331CPART 2 ~PAGE 149 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
rad/s4.871m
13333.0
20)125.0(2
2m
2m
75or712m
1065.01217.0)(
sssC
)15.0(20)(u
ss
sL)1065.0)(15.0(
)1217.0(20)(
sssssL
125.020)(
)15.0(20)(
2uu
jLjj
jL
Find m by solving 3.0)( mu jL
3.0125.0
202mm
53324 2m
4m
By solving this equation:
217.03.04.8
11m
T
T
EE3331CPART 2 ~PAGE 150EE3331CPART 2 ~PAGE 150 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
1065.01217.0)(
sssC
Bode Diagram
Frequency (rad/s)
0
2
4
6
8
10
12
System: sysFrequency (rad/s): 8.39Magnitude (dB): 5.22
Mag
nitu
de (d
B)
10-1
100
101
102
103
0
5
10
15
20
25
30
35
System: sysFrequency (rad/s): 8.4Phase (deg): 32.6
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 151EE3331CPART 2 ~PAGE 151 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The new gain cross‐over frequency is
about
cg,new = 8.4 rad/s
PM = 46
Bode Diagram
Frequency (rad/s)
-100
-50
0
50
System: sysFrequency (rad/s): 8.41Magnitude (dB): 0.0367
Mag
nitu
de (d
B)
10-1
100
101
102
103
-180
-135
-90
System: sysFrequency (rad/s): 8.4Phase (deg): -134
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 152EE3331CPART 2 ~PAGE 152 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Procedure to design a lead compensator…
i. Find the phase margin of the uncompensated system and compare it to the
design specs to obtain the necessary phase needed (say, req) to be added
to the loop. To be safe, we usually choose m > req to start with the design.
ii. From , we find .
iii. We look for a frequency (m) that has an uncompensated gain of .
iv. Computer .
v. The required lead compensator is given as .
vi. Verify your result. Repeat the above steps if necessary.
11sin m
m
1T
11)(
Ts
TssC
EE3331CPART 2 ~PAGE 153EE3331CPART 2 ~PAGE 153 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Step Response of the Closed Loop
)15.0(20)(u
ss
sL
205.020)( 2CL
ss
sG
15416)(lag
sssC
201215.5427)16(20)( 23CL
sssssG
106.012.0)(lead
sssC
6654.1667.18)5(133)( 23CL
sssssG
Without compensator
With lag compensator
With lead compensator
Without compensator
Which one is with lead compensator?
Which one is with lag compensator?
EE3331CPART 2 ~PAGE 154EE3331CPART 2 ~PAGE 154 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Example A.1: For the system given below
C(s)er
+
yu (t) G(s)
)1000)(2()(
sssKsG
1) Design a lead compensator C(s) and find K such that
PM 45ess (ramp) 0.02
2) Obtain the step response of the uncompensated and compensated systems.
EE3331CPART 2 ~PAGE 155EE3331CPART 2 ~PAGE 155 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Solution: (1) Since the compensator we studied in this module has a unity DC gain, i.e.,
The velocity error constant is then given by
000,10002.01
2000)1000)(2(lim)(lim)()(lim
000v
KKssKsCsGsCsk
sss
11
1lim)(lim00
Ts
TssCss
i. Find the phase margin of the uncompensated system and compare it to the
design specs to obtain the necessary phase needed (say, req) to be added
to the loop. To be safe, we usually choose m > req to start with the design.
From the Bode plot of the uncompensated open loop transfer function given on the next page, we have
PM = 180 – 169 = 11 (at frequency around 10 rad/s)
EE3331CPART 2 ~PAGE 156EE3331CPART 2 ~PAGE 156 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
To be safe, we choose to add an additional phase of
req = (45–11)+5 = 39
Bode Diagram
Frequency (rad/s)
-200
-150
-100
-50
0
50
100
System: sysFrequency (rad/s): 9.9Magnitude (dB): -0.00191
Mag
nitu
de (d
B)
10-1
100
101
102
103
104
105
-270
-225
-180
-135
-90
System: sysFrequency (rad/s): 9.97Phase (deg): -169
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 157EE3331CPART 2 ~PAGE 157 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
ii. From , we find .
11sin m
227.063.039sin11
iii. We look for a frequency (m) that has an uncompensated gain of .
rad/s 6.14dB 44.6227.0 m
iv. Computer
iv. The required lead compensator is given as
iv. Verify your result. Repeat the above steps if necessary.
m
1T
144.0227.06.14
11
m
T
1033.01144.0
1144.0227.01144.0
11)(
ss
ss
TsTssC
EE3331CPART 2 ~PAGE 158EE3331CPART 2 ~PAGE 158 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/s)
0
5
10
15
System: sysFrequency (rad/s): 14.6Magnitude (dB): 6.45
Mag
nitu
de (d
B)
10-1
100
101
102
103
0
10
20
30
40
System: sysFrequency (rad/s): 14.6Phase (deg): 38.8
Phas
e (d
eg)
1033.01144.0)(
sssC
EE3331CPART 2 ~PAGE 159EE3331CPART 2 ~PAGE 159 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The new gain cross‐over frequency is
about
cg,new = 14.3 rad/s
PM = 46
Bode Diagram
Frequency (rad/s)
-200
-150
-100
-50
0
50
100
System: sysFrequency (rad/s): 9.9Magnitude (dB): -0.00191
System: sysFrequency (rad/s): 14.2Magnitude (dB): 0.125
Mag
nitu
de (d
B)
10-1
100
101
102
103
104
105
-270
-225
-180
-135
-90
System: sysFrequency (rad/s): 9.97Phase (deg): -169
System: sysFrequency (rad/s): 14.2Phase (deg): -134
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 160EE3331CPART 2 ~PAGE 160 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
(2) Obtain the step response of the uncompensated and compensated systems.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8Step Response
Time (sec)
Am
plitu
de
Blue: UncompensatedGreen: Compensated
Exercise A.1: Solve the problem in Example A.1 using a lag compensator.
EE3331CPART 2 ~PAGE 161EE3331CPART 2 ~PAGE 161 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Example A.2: For the system given below
C(s)er
+
yu (t) G(s)
)5)(4()(
sssKsG
1) Design a lag compensator C(s) and find K such that
PM 45Kv 5
2) Obtain the step response of the uncompensated and compensated systems.
EE3331CPART 2 ~PAGE 162EE3331CPART 2 ~PAGE 162 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Solution: (1) The velocity error constant is then given by
100520)5)(4(
lim)(lim)()(lim000v
KK
ssKsCsGsCsk
sss
To be safe, we choose
m = 45 + 5 = 50
i. Find the phase margin of the uncompensated system and compare it to the
design specs on phase margin (say, req). To be safe, we usually choose a
value m > req to start with the design.
ii. Find the frequency x such that the uncompensated loop at which has a
phase response of –180 + m.
x = 1.63 rad/s
EE3331CPART 2 ~PAGE 163EE3331CPART 2 ~PAGE 163 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Looking for a frequency that gives us the required m.
Bode Diagram
Frequency (rad/s)
-140
-120
-100
-80
-60
-40
-20
0
20
40
System: sysFrequency (rad/s): 1.63Magnitude (dB): 8.59
Mag
nitu
de (d
B)
10-1
100
101
102
103
-270
-225
-180
-135
-90
System: sysFrequency (rad/s): 1.63Phase (deg): -130
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 164EE3331CPART 2 ~PAGE 164 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
iii. Find the corresponding gain value (say, Gx) of the uncompensated loop at x.
Gx = 8.59 dB Gx = 2.7
iv. Compute = Gx and .
= 2.7 and T = 6.135
v. The required lag compensator is given as
vi. Verify your result. Repeat the above steps if necessary.
xω10
T
156.161135.6
1135.67.21135.6
11)(
ss
ss
TsTssC
EE3331CPART 2 ~PAGE 165EE3331CPART 2 ~PAGE 165 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
Bode Diagram
Frequency (rad/s)
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
System: sysFrequency (rad/s): 1.62Magnitude (dB): -8.59
Mag
nitu
de (d
B)
10-3
10-2
10-1
100
101
-30
-25
-20
-15
-10
-5
0
System: sysFrequency (rad/s): 1.62Phase (deg): -3.62
Phas
e (d
eg)
156.161135.6)(
sssC
EE3331CPART 2 ~PAGE 166EE3331CPART 2 ~PAGE 166 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
The new gain cross‐over frequency is
about
cg,new = 1.64 rad/s
PM = 46
Bode Diagram
Frequency (rad/s)
-150
-100
-50
0
50
100
System: sysFrequency (rad/s): 1.63Magnitude (dB): 0.0389
Mag
nitu
de (d
B)
10-3
10-2
10-1
100
101
102
103
-270
-225
-180
-135
-90
System: sysFrequency (rad/s): 1.63Phase (deg): -134
Phas
e (d
eg)
EE3331CPART 2 ~PAGE 167EE3331CPART 2 ~PAGE 167 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
(2) Obtain the step response of the uncompensated and compensated systems.
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6Step Response
Time (sec)
Am
plitu
de
Blue: UncompensatedGreen: Compensated
Exercise A.2: Solve the problem in Example A.2 using a lead compensator.
EE3331CPART 2 ~PAGE 168EE3331CPART 2 ~PAGE 168 BEN M.CHEN,NUSECEBEN M.CHEN,NUSECE
That’sall,folks!
ThankYou!
That’sall,folks!
ThankYou! BugsBunny1940–
