ControlEngineeringI

See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/282503995
Notes on Control Engineering I
Book · October 2015
CITATIONS
READS
0
1,151
4 authors, including:
Luis E Moreno
Santiago Garrido
University Carlos III de Madrid
University Carlos III de Madrid
233 PUBLICATIONS 1,647 CITATIONS
124 PUBLICATIONS 858 CITATIONS
SEE PROFILE
SEE PROFILE
Miguel Angel Salichs
University Carlos III de Madrid
171 PUBLICATIONS 1,694 CITATIONS
SEE PROFILE
Some of the authors of this publication are also working on these related projects:
UAV path planning View project
SMA based exoskeletons for hand, elbow and shoulder rehabilitation View project
All content following this page was uploaded by Santiago Garrido on 12 November 2015.
The user has requested enhancement of the downloaded file.
UNIVERSIDAD CARLOS III DE MADRID
Departamento de Ingeniería de Sistemas y Automática
NOTES ON CONTROL SYSTEMS I
L. Moreno, S. Garrido, C. Balaguer and
M. A. Salichs
ii
CONTENTS
1. Introduction . . . . . . . . . . . . . . . . . . .
1.1 Introduction . . . . . . . . . . . . . . . . . .
1.2 Transform concept . . . . . . . . . . . . . .
1.3 Fourier Transform . . . . . . . . . . . . . .
1.3.1 Inverse Fourier Transform . . . . . .
1.4 Laplace Transform . . . . . . . . . . . . . .
1.4.1 Properties of the Laplace transform .
1.4.2 Laplace transforms of some functions
1.4.3 Inverse Laplace transform . . . . . .
1.4.4 Resolution of differential equations .
1.4.5 Derivative operator . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
1
1
1
1
3
3
4
6
7
10
11
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
13
13
14
14
14
17
20
21
24
26
26
27
3. Time Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Introduction. Time Analysis of Systems . . . . . . . . . . . . . .
3.2 Temporal Response . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Impulse response . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Response of First Order systems . . . . . . . . . . . . . .
3.2.3 Response of the second order systems . . . . . . . . . . .
3.3 Parameters characterizing the response of a second order system or
specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
45
45
46
47
50
2. Classical techniques for system modeling
2.1 Introduction . . . . . . . . . . . . . . . . .
2.2 Input/Output models . . . . . . . . . . . .
2.3 Time models . . . . . . . . . . . . . . . .
2.3.1 Basic concepts . . . . . . . . . . .
2.3.2 Models in continuous time . . . . .
2.3.3 Rotational motion . . . . . . . . .
2.3.4 Electrical systems . . . . . . . . .
2.4 Linearization . . . . . . . . . . . . . . . .
2.5 Block diagrams . . . . . . . . . . . . . . .
2.5.1 Diagrams reduction . . . . . . . .
2.5.2 Transposition of sum points . . . .
.
.
.
.
.
.
.
.
.
.
.
60
4. Stability of dynamical systems . . . . . . . . . . . . . . . . . . .
4.1 Classical stability analysis methods . . . . . . . . . . . . . . . . .
4.1.1 Routh’s method . . . . . . . . . . . . . . . . . . . . . .
73
74
74
5. Errors in steady state . . . . . . . . . . . . . . . . . . . . . . . .
5.1 Introduction to errors . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Gain position (static) . . . . . . . . . . . . . . . . . . . .
5.1.2 Gain speed . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.3 Gain acceleration . . . . . . . . . . . . . . . . . . . . . .
5.1.4 Type system . . . . . . . . . . . . . . . . . . . . . . . .
5.1.5 Relationship between type and gain of a system . . . . . .
5.1.6 Relationship between type and gain a system . . . . . . .
5.1.7 Steady state error of a unit feedback system . . . . . . . .
5.1.8 Errors to standard input . . . . . . . . . . . . . . . . . . .
5.1.9 The relationship between errors and system type . . . . .
5.1.10 Error input for polynomial in t . . . . . . . . . . . . . . .
81
81
82
83
83
83
84
85
86
86
88
88
6. Root Locus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1 Concept of Root Locus . . . . . . . . . . . . . . . . . . . . . . .
6.2 Magnitude and Angle Condition (or Module and Argument Criteria)
6.3 Tracing Rules of the Root Locus. . . . . . . . . . . . . . . . . . .
91
91
92
93
7. Classical Control Techniques . . . . . . . . . . . . . . . . . . .
7.1 Problem approach . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Design of classical controllers . . . . . . . . . . . . . . . . . . .
7.3 Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 PID controllers . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4.1 Basic structure of a PID controller . . . . . . . . . . . . .
7.4.2 Ziegler-Nichols methods . . . . . . . . . . . . . . . . . .
7.5 Analytical methods for PID controllers design . . . . . . . . . . .
7.5.1 Pole placement method for adjusting PID . . . . . . . . .
7.5.2 Method based on the dominant poles for adjusting PID . .
99
99
100
101
102
103
104
107
107
110
8. FreQuency models. Bode diagrams . . . . . . . . . . . . . . . .
8.1 Asymptotic Bode diagram . . . . . . . . . . . . . . . . . . . . .
8.1.1 Asymptotic Bode diagram. Constants K . . . . . . . . . .
8.1.2 Asymptotic Bode diagram. Poles / zeros in the origin s1n .
8.1.3 Asymptotic Bode plot. Negative real Polo/ Zeros (1+T1 s)n
8.1.4 Asymptotic Bode plot. Negative complex Pole/Zeros . . .
8.1.5 Asymptotic Bode plot. Positive real Poles/Zeros. . . . . .
8.1.6 Asymptotic Bode plot. Positive complex Poles/Zeros. . . .
123
125
126
126
127
128
130
132
9. Nyquist Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
vi
10. Design of regulators by frecuency methods . . . . . . . . . . .
10.1 Relationship between parameters of relative stability and transient
response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Lead Networks . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Lag networks . . . . . . . . . . . . . . . . . . . . . . . . . . . .
147
149
149
153
Some historical data . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
vii
viii
LIST OF TABLES
7.1
7.2
Parameter values given by Ziegler-Nichols method for the step response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Parameter values given by Ziegler-Nichols frequency method . . 107
x
LIST OF FIGURES
Impulse response and a sum of delayed pulses . . . . . . . . . .
Mass-spring-damper system . . . . . . . . . . . . . . . . . . . .
Mechanical translation system with two inputs and two outputs . .
Example of mechanical system: 1/4 Car Model . . . . . . . . . .
Elements of mechanical systems of translation and rotation . . . .
Example of mechanical rotation system . . . . . . . . . . . . . .
RLC circuit with voltage as input . . . . . . . . . . . . . . . . .
System example . . . . . . . . . . . . . . . . . . . . . . . . . .
Example of RLC network with two meshes . . . . . . . . . . . .
Linearization of a function on two points (x0 , y0 ) y (x1 , y1 ) . . .
Diagrams reduction . . . . . . . . . . . . . . . . . . . . . . . . .
Transposition of sum points . . . . . . . . . . . . . . . . . . . .
a) Starting system of the example, and b) first simplification: feedback loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14 a) Transposition of the bifurcation simplifying and streamlining the
feedback blocks in parallel. b) Final transfer function. . . . . . . .
2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
3.1
3.2
3.3
3.4
3.5
3.6
Response of a system . . . . . . . . . . . . . . . . . . . . . . . .
Impulse response of a first order system . . . . . . . . . . . . . .
Impulse response of a first order system . . . . . . . . . . . . . .
Unit step response of a first order system . . . . . . . . . . . . . .
Unit step response of a first order system . . . . . . . . . . . . . .
Determination of the transfer function from the unit step response
of a first order system . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Response to a ramp unit of a first order system . . . . . . . . . . .
3.8 Unit ramp response of a first order system . . . . . . . . . . . . .
3.9 Relation among the parameters of a second order system. . . . .
3.10 Classification of second-order systems . . . . . . . . . . . . . . .
3.11 Impulse response of the second order underdamped systems. . . .
3.12 The real exponential function eσt is increased when the real coefficient σ is positive and decreasing when it is negative. . . . . . .
16
19
19
20
21
21
23
23
24
24
27
28
28
28
32
34
46
47
48
48
49
49
50
50
52
53
54
56
3.13 The function eαt sin(βt + φ) is a growing sinusoid if α is positive
and decreasing if negative. . . . . . . . . . . . . . . . . . . . . .
3.14 The unit step responses of a second order system according to the
coefficient of friction: 1) ζ = 0, oscillator (poles on the imaginary
axis). 2) 0 < ζ < 1, underdamped system (complex conjugate
poles in the left half. 3) ζ = 1, critically damped system (real
and equal poles). 4) ζ > 1, overdamped system (real and different
poles). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.15 Parameters of the temporal response . . . . . . . . . . . . . . . .
57
60
4.1
4.2
Feedback loop . . . . . . . . . . . . . . . . . . . . . . . . . . . .
System of example 5.1.4 . . . . . . . . . . . . . . . . . . . . . .
73
77
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
Relationship between type and gain a system . .
Relationship between type and gain a system . .
Unit feedback system. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
The relationship between errors and system type.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
81
82
84
85
86
87
87
88
88
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
Meaning of the magnitudes and angles.
. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .
Tracing rules for Root Locus. . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
93
94
94
95
95
95
96
98
7.1
7.2
7.3
7.4
7.5
Block diagram of a general control scheme . . . . . . . . . . . .
Controller in the principal loop . . . . . . . . . . . . . . . . . . .
Parameters of the time response . . . . . . . . . . . . . . . . . .
a −sL
e
. . . . . . . . . . . . .
Response to step input of G(s) = sL
Obtaining the critical gain and critical period for the second method
of Ziegler-Nichols . . . . . . . . . . . . . . . . . . . . . . . . . .
Response system of Example 7.5.1: (a) open loop and (b) closed
loop with the designed controller . . . . . . . . . . . . . . . . .
System of the example 7.5.2 . . . . . . . . . . . . . . . . . . . .
Response system of Example 7.5.2: (a) open loop and (b) closed
loop with the controller designed . . . . . . . . . . . . . . . . . .
99
100
102
105
7.6
7.7
7.8
xii
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
56
106
110
113
114
Effect of the addition of a pole on the Root Locus . . . . . . . . . 115
Effect of the addition of a zero on the Root Locus . . . . . . . . . 116
Root locus for the example system 7.5.3 . . . . . . . . . . . . . . 116
Application of the magnitude and angle conditions for Example 7.5.3118
Response system of Example 7.5.3: (a) open loop and (b) closed
loop with the controller designed . . . . . . . . . . . . . . . . . . 119
7.14 Example 7.5.5 System . . . . . . . . . . . . . . . . . . . . . . . 119
7.15 Distribution of poles and zeros of example 7.5.5 . . . . . . . . . . 120
7.9
7.10
7.11
7.12
7.13
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
Bode plot of a second order system . . . . . . . . . . . . . . . . .
Asymptotic Bode Plot of Gain Term . . . . . . . . . . . . . . . .
Asymptotic Bode plot. Poles in origin s1n . . . . . . . . . . . . .
Asymptotic Bode plot. Negative real Poles/Zeros (1+T1 s)n . . . .
Asymptotic Bode plot. Negative complex Poles/Zeros. . . . . . .
Bode plot. Negative complex Poles/Zeros . . . . . . . . . . . . .
Asymptotic Bode diagram of 1/s − 1. (Real positive poles). . . .
Asymptotic Bode diagram ofs − 1. (Real positive zero). . . . . .
Bode plot with complex poles with: a) negative real part, b) positive real part. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Closed path ρ and its image F (s) . . . . . . . . . . . . . . . . . .
Sistema en bucle cerrado . . . . . . . . . . . . . . . . . . . . . .
Metodo de Nyquist . . . . . . . . . . . . . . . . . . . . . . . . .
Modified Nyquist method . . . . . . . . . . . . . . . . . . . . . .
System of the example 5.1.6 . . . . . . . . . . . . . . . . . . . .
Nyquist path for the example 5.1.8 . . . . . . . . . . . . . . . . .
Nyquist diagram of example 5.1.6 . . . . . . . . . . . . . . . . .
Nyquist path for a system with a pole at the origin . . . . . . . . .
Nyquist path for a system with poles on the imaginary axis . . . .
Stable system . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Unstable system . . . . . . . . . . . . . . . . . . . . . . . . . . .
The system a) is more stable than the b) and c systems) . . . . . .
Calculation of gain margin . . . . . . . . . . . . . . . . . . . . .
Nyquist plot, Root locus and closed loop step response according
1
system gain increases G(s) = s(s+a)(s+b)
con a, b > 0. . . . . . .
9.15 The two systems have the same profit margin, but a) is more stable
than the b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.16 Phase Margin . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.17 Gain margin and phase margin in Bode diagram . . . . . . . . . .
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
10.1
10.2
125
126
127
128
129
130
131
131
132
133
134
135
135
136
136
137
137
138
139
139
140
141
142
143
144
144
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
xiii
10.3 RLC lead network and its corresponding diagram of poles and zeros. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Bode and Nyquist plot for a lead network according to different
parameter values α. . . . . . . . . . . . . . . . . . . . . . . . . .
10.5 Bode diagram of the system and lead network. . . . . . . . . . . .
10.6 Comparison of Bode plots and unit step responses. . . . . . . . .
10.7 Adding a lead network to the system changes the magnitude and
phase, so it is difficult to predict the new crossing point. . . . . . .
10.8 Designing a lead network. . . . . . . . . . . . . . . . . . . . . .
10.9 Lag network and pole-zero diagram. . . . . . . . . . . . . . . . .
10.10Frequency response of lag networks. . . . . . . . . . . . . . . . .
10.11Comparison of the response of the systems. . . . . . . . . . . . .
10.12Designing a lag network. . . . . . . . . . . . . . . . . . . . . . .
10.13The effect of the change of T. . . . . . . . . . . . . . . . . . . . .
xiv
150
150
150
151
151
152
153
153
153
154
154
1. INTRODUCTION
1.1
Introduction
One of the problems that have to face those working in the control or identification of processes lies in the existence of an variety of mathematical tools to study,
analyze and design such systems.
For example, if we analyze and control a continuous system, we can model
the system in terms of its temporal response by using the transfer function. The
system can be controlled in continuous time which will require the use of Laplace
transform, in the case of continuous time. Sometimes, we are interested in the study
of the frequency response of the system for this reason we will study the response
in steady state (based on the transfer function).
In the study and analysis of a system we will find different mathematical tools
that allow us to characterize, analyze and design a control system so that we consider the different effects in the system.
The basic objective of this chapter is to introduce the tools mathematics that
are typically used to analyze and synthesize systems of deterministic control in the
time domain, in continuous time as well as in frequency.
1.2
Transform concept
When analyzing and designing control systems for a wide variety of processes,
we find the need to solve differential or integro-differential equations in order to
know the response the process will give us to certain inputs or control actions.
These differential equations can be solved directly or can be converted to solve
other equations of easier resolution. There are a number of operational methods
that allow the resolution of problems using transform equations in an easier way.
Among the most frequently used transformations for solution of ordinary differential equations of linear transformations are an integral, and among these a
widely used in control is called Laplace transform.
1.3
Fourier Transform
Given a real function f (t), the Fourier transform of the function f (t) is defined
as:
2
1. Introduction
F (ω) = F [f (t)] =
Z
+∞
f (t)e−jωt dt
(1.3.1)
−∞
where ω a variable frequency, −∞ < ω < +∞, expressed in (rad/s). The function
f (t) is said to have Fourier transform if the integral 1.3.1 exists for all values of
ω. Sufficient conditions for the integral existence is that the function f (t) has only
a finite number of discontinuities, maxima and minima over a finite time interval,
and that function f (t) is absolutely integrable, ie
Z
+∞
−∞
|f (t)|dt < ∞
(1.3.2)
The complex exponential term of the transformed 1.3.1 can be expressed by applying Euler’s formula, as
e−jωt = cos(ωt) − j sin(ωt)
(1.3.3)
The complex exponential term of the transformed 1.3.1 can be expressed by
applying Euler’s formula, as
F (ω) =
=
Z
+∞
−∞
Z +∞
−∞
f (t)[cos(ωt) − j sin(ωt)]dt
f (t) cos(ωt)dt − j
Z
+∞
f (t) sin(ωt)dt
(1.3.4)
−∞
In this expression 1.3.4 of the Fourier transform you can see the existence of
two terms, one real and one imaginary, which we call R(ω) and I(ω), respectively,
so F (ω) is given by
F (ω) = R(ω) + jI(ω)
(1.3.5)
In the equation 1.3.5 can be seen that the Fourier transform F (ω) is a complex
function of the variable ω. For every value of the variable ω, will have a value of
F (ω) which has a module, |F (ω)|, and an argument, ∠F (ω).
|F (ω)| =
p
R2 (ω) + I 2 (ω)
I(ω)
−1
∠F (ω) = tan
R(ω)
(1.3.6)
If we represent the values on a plot |F (ω)| for every ω would obtain what is
called the spectrum signal magnitudes f (t), and if we did the same for argument
∠F (ω) would obtain what is known as phase spectrum signal f (t).
1.4. Laplace Transform
1.3.1
3
Inverse Fourier Transform
The inverse transform is an operation that allows us to obtain the original function f (t) from the Fourier transform F (ω). This operation is given by the following
expression
f (t) = F −1 [F (ω)] =
1.4
1
2π
Z
+∞
F (ω)ejωt dω
(1.3.7)
−∞
Laplace Transform
Given a real function f (t) such that f (t) = 0 < t < ∞, the unilateral Laplace
transform of the function f (t) is defined as:
F (s) = L[f (t)] =
Z
∞
f (t)e−st dt
(1.4.1)
0
There is also the bilateral Laplace transform, in which integration limits range from
−∞ to +∞, but it is not typically used in control engineering, and we are not going
to study it. Unilateral Laplace transform, which in and for simplicity we will call
hereinafter as Laplace transform, exists always if the integral 1.4.1 converges, and
it is a necessary and sufficient condition that:
Z
∞
0
|f (t)e−σt |dt < ∞
(1.4.2)
for σ finite and real.
Given the Laplace transform of a function f (t), the set of all complex numbers
s for which the 1.4.1 exists is called region of convergence.
Given a function f (t), if the Laplace transform L(f (s)) is particularized for
those complex numbers in which s = jω we obtain that
F (s)|s=jω = F (jω) =
Z
∞
f (t)e−jωt dt
(1.4.3)
0
if in addition the function f (t) = 0 for t < 0, the equation 1.4.3 is equal to the
Fourier transform F (ω) of the function f (t) (compare with the expression that we
saw in 1.3.1). Namely
F (ω) = F (s)|s=jω
(1.4.4)
This tells us that we can obtain the Fourier transform, F (ω) of a function f (t)
(where the value of this function for t < 0 is zero) from its transform Laplace,
F (s), replacing in the expression of the Laplace transform the variable s for jω.
4
1. Introduction
1.4.1
Properties of the Laplace transform
The Laplace transform has a number of very useful properties which means it
can be considered as an operator that converts a temporal function f (t) in the time
domain in a function F (s) of complex variable s. These properties are:
1. Linearity
A very important property of the Laplace transform is that it is a linear operator. Given two functions f (t) and g(t) and a constant k is verified that:
L[f (t) + g(t)] = L[f (t)] + L[g(t)]
(1.4.5)
L[kf (t)] = kL[f (t)]
(1.4.6)
and
2. Differentiation
The Laplace transform of the differential of a function f (t) can be expressed
as follows:
d
L
f (t) = sF (s) − lim f (t) = sF (s) − f (0+ )
t→0
dt
(1.4.7)
where F (s) the transform of the function f (t) and f (0+ ) is the limit of the
function f (t) when t tends to zero from the right. In general, for higher
derivatives we have the following expression:
dn−1
dn
n
n−1
+
n−2 d
+
f
(0
L
+
.
.
.
+
f
(t)
=
s
F
(s)
−
s
f
(0
)
−
s
)
f (0+ )
dt
dtn
dtn−1
(1.4.8)
3. Integration
The Laplace transform of the integral of a function f (t) with respect to time
is the Laplace transform of the function divided by s, ie:
L
Z
t
0
f (τ )dτ =
F (s)
s
(1.4.9)
and if generalized to higher orders integrations
L
Z
t1
0
Z
t2
...
0
Z
tn
0
f (τ )dτ dτ1 . . . dτn =
F (s)
sn
(1.4.10)
1.4. Laplace Transform
5
4. Time shift
A time shift of a units of the function f (t), in the complex domain is equivalent to multiply the Laplace transform of the function F (s) for e−s , that
is:
(1.4.11)
L[f (t − a)u0 (t − a)] = e−sa F (s)
where u0 (t − a) represents the unit step function shifted in time a units.
5. Initial value theorem
If the Laplace transform of the function f (t) is F (s), then
lim f (t) = lim sF (s)
t→0
s→∞
(1.4.12)
if the limit when time t → 0 exists.
6. Final value theorem
If the Laplace transform of the function f (t) is F (s), and if sF (s) is analytic
on the right side of the plane-s and on the imaginary axis, then
lim f (t) = lim sF (s)
t→∞
s→0
(1.4.13)
This theorem is useful in the analysis and design of control systems because
it tells us the final value of a function in the time domain knowing the value
of its Laplace transform for s = 0.
7. Complex shift
The Laplace transform of a function f (t) multiplied by e∓αt , where α is
a constant, is equivalent to replace the Laplace transform of the function,
F (s), s with s ± a, ie
L[e∓αt f (t]] = F (s ± a)
(1.4.14)
8. Real convolution
Let F1 (s) and F2 (s) the Laplace transforms of the functions f1 (t) and f2 (t),
respectively, and f1 (t) = 0, and f2 (t) = 0, for t < 0 then
F1 (s)F2 (s) = L[f1 (t) ∗ f2 (t)]
Z t
= L[ f1 (τ )f2 (t − τ )dτ ]
0
Z t
= L[ f2 (τ )f1 (t − τ )dτ ]
0
(1.4.15)
(1.4.16)
(1.4.17)
6
1. Introduction
in this expression the symbol ∗ denotes the convolution operation in the time
domain.
1.4.2
Laplace transforms of some functions
Example 1.4.1: The Laplace transform of the unit step function u0 (t) defined as
f (t) = u0 (t) =
1, t ≥ 0
0, t < 0
(1.4.18)
is obtained as
F (s) = L[u0 (t)] =
Z
∞
0
1
u0 (t)e−st dt = − e−st
s
∞
0
=
1
s
(1.4.19)
Example 1.4.2: The Laplace transform of the exponential function f (t) = e−αt
for t ≥ 0 is obtained as
F (s) = L[e
−αt
]=
Z
∞
0
e−αt e−st dt = −
1 −(s+α)t
e
s+α
∞
0
=
1
s+α
(1.4.20)
In the table you can see the transforms of some of the more usual functions.
f (t)
δ(t)
u0 (t)
t
e−at
te−at
tn , t ≥ 0
tn e−at , t ≥ 0
sin(ωt)
cos(ωt)
e−at sin ωt
e−at cos ωt
L(s)
1
1
s
1
s2
1
s+a
1
(s+a)2
n!
, n = 1, 2, . . .
sn+1
n!
, n = 1, 2, . . .
(s+a)n+1
ω
(s2 +ω 2 )
s
(s2 +ω 2 )
ω
(s+a)2 +ω 2
s+a
(s+a)2 +ω 2
1.4. Laplace Transform
1.4.3
7
Inverse Laplace transform
The operation to obtain the function f (t) from the Laplace transform F (s) is
called the inverse Laplace transform, and it is written as
f (t) = L−1 [F (s)]
The inverse of the Laplace transformation can be calculated as
Z c−j∞
1
F (s)est ds
f (t) =
2πj c+j∞
(1.4.21)
(1.4.22)
where c is a real constant greater than the real part of all singularities of F (s).
This integral is evaluated in the complex s-plane. From the standpoint of process
control, to obtain the inverse transformation is extremely costly and impractical.
So, tables of Laplace transforms are used on the partial fraction expansion.
Partial fraction expansion
In most control applications, it is not necessary to use the integral inversion
given by 1.4.22 for the inverse Laplace transform. Usually both the differential
equations and its solutions in control problems take the form of rational functions
in s,
F (s) =
N (s)
D(s)
(1.4.23)
where N (s) and D(s) are polynomials in s. It is assumed that the degree of the
polynomial N (s) in the numerator s is less or equal than the degree of the polynomial D(s) in the denominator.
This expression can be written as follows
F (s) =
sn
+ a1
sn−1
N (s)
+ . . . + an−1 s + an
(1.4.24)
where the coefficients ai are real coefficients. The roots of the polynomial N (s) are
called zeros and the roots of the denominator D(s) are called poles of the function
F (s).
The basic idea of the method of partial fraction expansion is to decompose the
polynomial F (s) into a sum of simple fractions such that these are transformed
into simple and known functions such that for each fraction becomes very easy
to calculate the inverse transform. As the Laplace transform is a linear operator,
after calculating the inverse transforms of the simple fractions, to obtain the inverse
transform of the function F (s) simply we need to add the inverse transforms of the
simple fractions. We consider the different cases.
8
1. Introduction
Case of simple and real poles
the equation 1.4.24 takes the form
F (s) =
If all poles of D(s) are simple and real, then
N (s)
(s + s1 )(s + s2 ) . . . (s + sn )
(1.4.25)
and it ∀i 6= j, Si 6= sj . Decomposing the expression 1.4.25 we obtain the partial
fraction
F (s) =
A1
An
A2
+
+ ... +
(s + s1 ) (s + s2 )
(s + sn )
(1.4.26)
where the coefficients A1 , . . . , An are obtained as
N (s)
Ai = (s + si )
D(s) s=−si
(1.4.27)
Note that each of these simple fractions are Laplace transforms of exponential
functions with exponent −si and coefficient Ai .
Case of multiple real poles If some of the poles of D(s) are multiples, then
the equation 1.4.24 takes the form
F (s) =
N (s)
(s + s1 )(s + s2 ) . . . (s + sk−1 )(s + sk )n−k+1
(1.4.28)
and it has a −sk root with multiplicity n − k + 1. Decomposing the expression in
fractions 1.4.28 obtain the partial expansion
F (s) =
Ak−1
A2
A1
+
+ ... +
+
(s + s1 ) (s + s2 )
(s + sk−1 )
Akn−k
Ak2
Ak1
+ ... +
+
+
2
(s + sk ) (s + sk )
(s + sk )n−k+1
(1.4.29)
where the coefficients A1 , . . . , Ak−1 corresponding to the simple poles are obtained as in the previous case, and for the multiple pole the coefficients are obtained
as
Akn−k
Akn−k+1
=
=
...
Ak 1
=
(s + sk )
n−k+1 N (s)
D(s)
(1.4.30)
s=−sk
d
n−k+1 N (s)
(s + sk )
ds
D(s) s=−sk
1
dn−k
n−k+1 N (s)
(s + sk )
(n − k)! dsn−k
D(s)
(1.4.31)
(1.4.32)
s=−sk
1.4. Laplace Transform
9
Case of complex conjugate poles The partial fraction expansion seen in
the case of poles simple and real, it is also valid for the case of poles complex
conjugates of the form s = −σ +jw and s = −σ −jw. In this case the coefficients
A−σ+jw and A−σ−jw are obtained as follows
N (s) D(s)
N (s) = (s + σ + jw)
D(s)
A−σ+jw =
A−σ−jw
(s + σ − jw)
s=−σ+jw
s=−σ−jw
(1.4.33)
(1.4.34)
Example 1.4.3: Suppose the following function
2s + 3
s(s + 1)(s + 3).
F (s) =
(1.4.35)
If the partial fraction is expanded as follows
A1
A3
A2
+
+
s
(s + 1) (s + 3)
F (s) =
(1.4.36)
the coefficients will be determined by
A1 =
A2 =
A3 =
thus
sF (s)
s=0
=
(s + 1)F (s)
(s + 3)F (s)
F (s) =
2s + 3
=1
(s + 1)(s + 3) s=0
2s + 3
1
=
=−
s(s + 3) s=−1
2
s=−1
2s + 3
1
=
=−
s(s + 1) s=−3
2
s=−3
− 12
− 12
1
+
+
s (s + 1) (s + 3)
(1.4.37)
and its inverse transform will be the function
1
1
f (t) = 1 − e−t − e−3t
2
2
(1.4.38)
Example 1.4.4: Suppose the following function
F (s) =
s2 + 3s
(s + 1)3 .
If the partial fraction will be expanded as follows
(1.4.39)
10
1. Introduction
F (s) =
A3
A1
A2
+
+
2
(s + 1) (s + 1)
(s + 1)3
(1.4.40)
The coefficients are determined by
A3 =
A2 =
A1 =
thus
(s + 1)3 F (s)
s=−1
= s2 + 3s
s=−1
= −2
d
(s + 1)3 F (s)
= 2s + 3
=1
ds
s=−1
s=−1
1 d2 2
=
=1
(s + 1)3 F (s)
2
2! ds
2! s=−1
s=−1
F (s) =
−2
1
1
+
+
2
(s + 1) (s + 1)
(s + 1)3
(1.4.41)
and its inverse transform will be the function
f (t) = e−t + te−t − t2 e−t
1.4.4
(1.4.42)
Resolution of differential equations
One of the most useful properties of the Laplace transform is the simplicity
with which you can solve linear differential equations. Here’s an example.
Example 1.4.5: Suppose a system whose differential equation is as follows
dx
d2 x
+δ =0
+2
2
dt
dt
(1.4.43)
besides, we know that for this system the values of x and ẋ for t = 0, are, respectively, x(0) = 1 and ẋ(0) = 0, and we want to obtain the output x(t) that give the
system 1.4.43.
Applying the property of the Laplace transform of the derivative will have the
following
[s2 X(s) − sx(0) − ẋ(0)] + 2[sX(s) − x(0)] + 1 = 0
Now if we replace x(0) and ẋ(0) by its value, we it will be
[s2 X(s) − s] + 2[sX(s) − 1] + 1 = 0
s2 X(s) + 2sX(s) = s + 1
s+1
X(s) =
s(s + 2)
1.4. Laplace Transform
11
Expanding this expression into partial fractions
X(s) =
A1
A2
+
s
(s + 2)
(1.4.44)
The coefficients are determined by
(s + 1)
1
=
=
A1 = sX(s)
(s + 2) s=0 2
s=0
1
s+1
A2 = (s + 2)X(s)
=
=
s s=−2 2
s=−2
thus
X(s) =
1
2
+
1
2
s
(s + 2)
and its inverse transform is the solution of the differential equation 1.4.43 for the
initial conditions are assumed, i.e.
x(t) =
1.4.5
1
(1 + e−2t ).
2
(1.4.45)
Derivative operator
It is frequently used the expression derivative operator, which is usually represented by the symbol p or D, although here the first one will be used,
d
y(t)
dt
Using this symbol, a differential equation of order n
py(t) =
dn
dn−1
d
y(t)+a
y(t) + . . . + an−1 y(t) + an y(t) =
1
n
n−1
dt
dt
dt
dn−1
d
= b1 n−1 u(t) + . . . + bn−1 u(t) + bn u(t)
dt
dt
can be expressed as a polynomial
(1.4.46)
(1.4.47)
pn y(t)+a1 pn−1 y(t)+. . .+an−1 py(t)+an y(t) = b1 pn−1 u(t)+. . .+bn−1 pu(t)+bn u(t)
(1.4.48)
thus
[pn + a1 pn−1 + . . . + an−1 p + an ]y(t) = [b1 pn−1 + . . . + bn−1 p + bn ]u(t)
A(p)y(t) = B(p)u(t)
B(p)
y(t) =
u(t)
A(p)
(1.4.49)
This operator has the advantage of working directly in the time domain, which
is sometimes useful. This operator is frequently used in the classical methods of
resolution of ordinary differential equations.
12
1. Introduction
Bibliografia
[1] L. Ljung, System Identification: Theory for the User, Prentice Hall, 1987.
[2] L. Ljung and T. Söderstrom, Theory and Practice of Recursive Identification,
The MIT Press, 1983.
[3] T. Söderstrom and P. Stoica, System Identification, Prentice Hall, 1989.
[4] J. P. Norton, An Introduction to Identification, Academic Press, 1986.
[5] G.R. Cooper and C.D McGuillem, Probabilistic Methods of Signal and System Analysis, Oxford University Press, 1999.
[6] A. Papoulis, Probabilidad, variables aleatorias y procesos estocasticos, Ed.
Universitaria de Barcelona, 1980.
[7] B.C. Kuo, Automatic Control Systems, Ed. Prentice Hall, 1991.
[8] K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edition, 1995.
[9] K. Ogata, Ingenieria de control moderna, Prentice Hall, 1998.
[10] E. Andres Puente, Regulacion automatica, Servicio de publicaciones ETSIIMUPM, 1993.
2. CLASSICAL TECHNIQUES FOR SYSTEM
MODELING
2.1
Introduction
A basic problem in control engineering and in general in many other sciences
is to be able to predict what effect it will have some action on a physical system.
Because what you want is to predict future response of the system, it is necessary
to use some kind of model that allows us to make this prediction.
There are several types of models, some of them such as scale models of what is
sought is to reproduce the actual physical system by some model or more manageable system on which to experiment and extract results. These, though interesting,
are not the models more interested in control, but those where predictions are made
on a mathematical model of the system.
Within the mathematical models that can be used to analyze the effects that
different actions will have on the system can distinguish two groups of models:
• Input/Output models.
In this class of models seeks a mathematical description that expresses the
relationship between the system input and the output thereof. These models
do not describe the inner workings of the system, but merely the relation
between input and output. We may find different systems but having the
same ratio input/output so lead to the same mathematical model. These types
of models are what we might call classic models.
• State Space model based.
Such models seek a deeper description of the system, since not only the relationship between the input and output features, but besides the behavior of
a number of variables or internal variables of the system features. There are
many systems in which the internal variables that can be used to characterize
it can be chosen in different ways, so that the same system may be described
by different mathematical models. These types of techniques have given rise
to what is called modern control theory.
This chapter will address the study of what are the techniques classical modeling systems
14
2.2
2. Classical techniques for system modeling
Input/Output models
These types of models are what we might call classic models , and within them
can distinguish several types : temporal patterns , and stochastic frecuenciales .
The temporal patterns of input / output feature allow the study of the temporal
response of the system to a any input . Knowing the temporal response of the system allow us to analyze transient effects of short duration in the system, such as
response peaks , rise and settling times . These models can be temporal models
based on differential continuous-time or discrete-time models described by differential equations equations .
The frecuenciales called models are based on the characterization of the relationship / out of a steady-state system to sinusoidal inputs .
Finally the most common models are stochastic or deterministic models combine temporal and frequential noise or disturbance signals that add a stochastic
behavior to the system and that needs to be modeled when possible .
2.3
2.3.1
Time models
Basic concepts
Linear systems invariant in time
The model input/output linear systems corresponding to the following general
form,
dn
dn−1
y(t)
+
a
n−1 n−1 y(t) + . . . + a0 y(t) =
dtn
dt
dm
dm−1
bm m u(t) + bm−1 n−1 u(t) + . . . + b0 u(t)
dt
dt
(2.3.1)
(2.3.2)
where the coefficients ai , i = 0, 1, . . . , n − 1 and bj , j = 0, 1, . . . , m are real
numbers, u(t) is the input to system and y(t) is the output. If these coefficients ai
and bj do not depend on time, the system is said to be time invariant.
These linear models have several advantages, including and perhaps the most
important is the linearity in the response. That is, if α and β are two arbitrary
constants and u1 (t) and u2 (t) are two inputs to the system, it is verified that if
u(t) = αu1 (t) + βu2 (t)
(2.3.3)
then the output of the system is given by
y(t) = αy1 (t) + βy2 (t)
(2.3.4)
2.3. Time models
15
Impulse response
Since an invariant linear system in time, which for a input u(t) gives an answer
y(t), is defined as bf impulse response of the system output g(t) that would give
the entry system at the same is a unit impulse δ(t).
This impulse response g(t) contains all the information needed on the system,
and you can get the output of same, at any input u(t), no more to perform convolution in the time domain with the signal, ie
y(t) = g(t) ∗ u(t)
Z t
=
g(τ )u(t − τ )dτ
0
Z t
u(τ )g(t − τ )dτ
=
(2.3.5)
(2.3.6)
(2.3.7)
0
In general terms the impulse response g(t) is not very practical when modeling
a system and that it is much easier to work in the field of Laplace, to solve integral
expressions can be complex.
Figure 2.1 you can see that impulse δ(t), if an input the output of the system is
g(t), then, by linearity, to an input δ(t), the output shall be subject of ag(t). Since
the system is time invariant, momentum shifted to a δ(t − t0 ) shall be subject of a
displaced g(t − t0 ) response. To a composite input a superposition of scaled pulse
corresponds to one output will be the sum of the responses.
Transfer function
It is defined as transfer function of a linear and time invariant system the
Laplace transform of the impulse response of the system, assumed zero initial conditions.
G(s) = L[g(t)]
(2.3.8)
The concept of transfer function is very useful when determining the output y
of the system to a given input u, since this is given by the expression
Y (s) = G(s)U (s)
(2.3.9)
from whose values ??in the complex field, developing in partial fractions and
making the inverse transforms of each of these partial fractions was easy to obtain the analytical response that give the system in the time domain as seen in the
previous chapter.
This concept of transfer function is widely used in the future and it is convenient to make some clarifications on the matter:
16
2. Classical techniques for system modeling
y(t)
u(t)
g(t)
δ(t)
t
t
u(t)
g(t)
y(t)
y(t)
u(t)
ag(t)
Escalado
aδ(t)
t
t
y(t)
u(t)
Desplazamiento
δ(t-to)
t
t
to
u(t)
2δ(t)
δ(t-to)
to
g(t-to)
y(t)
1.5δ(t-t1)
t1
2g(t)+g(t-to)+1.5g(t-t1)
Superposición
t
to
t1
t
Fig. 2.1: Impulse response and a sum of delayed pulses
• The transfer function is defined for linear time-invariant system over time,
so it is not defined in the case of nonlinear systems.
• For the transfer function are assumed zero initial conditions.
• The transfer function expresses the relationship between an input signal and
an output signal, so one can express completely the dynamics of a system
with one input and one output. If the system had more of an input or output,
various transfer functions would be required to express all the possible relationships between each input and each output. In this case we would have
multivariable systems: MIMO (Multiple Input - Multiple Output).
2.3. Time models
17
• The transfer function is independent of the input to the system.
• The transfer function of a system is only rational function of s or z, and
also with a denominator greater than or equal to the numerator degree, to be
physically realizable.
2.3.2
Models in continuous time
A very important part of the systems we are interested in corresponds to control
physical systems whose dynamics is described by the following type of ordinary
differential equations
dn
dn−1
y(t)
y(t) + . . . + a0 y(t) =
+
a
n−1
dtn
dtn−1
(2.3.10)
dm
dm−1
bm m u(t) + bm−1 n−1 u(t) + . . . + b0 u(t)
dt
dt
where u(t) is the input to the system and y(t) the output. Assuming zero initial
conditions and take Laplace transforms on both sides of the equation 2.3.10, the
result is
(sn + an−1 sn−1 + . . . + a0 )Y (s) = (bm sm + bm−1 sm−1 + . . . + b0 )U (s) (2.3.11)
Solving Y (s) in Eq 2.3.11 we obtain that
Y (s) =
sn + an−1 sn−1 + . . . + a0
U (s)
bm sm + bm−1 sm−1 + . . . + b0
(2.3.12)
and here the general form of the transfer function G(s) for a physical system described by an ordinary differential equation of the form 2.3.10 is a rational function
of the type s
G(s) =
sn + an−1 sn−1 + . . . + a0
bm sm + bm−1 sm−1 + . . . + b0
(2.3.13)
Example 2.3.1: A car responds to an applied force f (t) from the engine power
and a force due to friction ρv(t), proportional to the velocity v(t) of the car. The
differential equation of the system is
mv̇(t) = f (t) − ρv(t)
(2.3.14)
Taking Laplace transforms is obtained
msV (s) = F (s) − ρV (s)
(2.3.15)
18
2. Classical techniques for system modeling
i.e.
(ms + ρ)V (s) = F (s)
(2.3.16)
and the transfer function of the relative speed of the force is
V (s)
1
=
F (s)
ms + ρ
(2.3.17)
Example 2.3.2: Consider the mass-spring-damper system of Fig. 2.2. Newton’s
second law states that the sum of all forces acting on a rigid body is equal to its
mass times its acceleration.
X
(2.3.18)
Fi = m ∙ ÿ
where the minus sign represents the force Fm opposes the change of position.
A spring is an element that stores potential energy and when stretched or compressed exert a force that is expressed by Hooke’s law:
Fm = −K ∙ y
(2.3.19)
where the minus sign represents the force Fm opposes the change of position.
A damper is a device that converts energy into heat and represents a viscous
friction. The force exerted on their ends can be expressed as
Fa = −b ∙ ẏ
(2.3.20)
That is, the damper exerts a force that opposes the speed changing.
The mass-spring-damper system shown in Fig. 2.2 can be modelled using these
three expressions in the form of differential equation
m ∙ ÿ = u − K ∙ y − b ∙ ẏ
where u it is an external force, which is the system input.
Rearranging
m ∙ ÿ + b ∙ ẏ + K ∙ y = u
(2.3.21)
(2.3.22)
By taking Laplace transforms
ms2 Y (s) + bsY (s) + KY (s) = U (s)
(2.3.23)
The transfer function is the output input ratio (in the domain s)
Y (s)
1
=
U (s)
ms2 + bs + K
(2.3.24)
2.3. Time models
19
Fig. 2.2: Mass-spring-damper system
Example 2.3.3: In the multivariate case of Figure 2.3, you must write two differential equations, one for each mass. It is useful to draw the corresponding diagram
of forces (free body diagram).
In this case the differential equations are
m1 ÿ1 = u1 + b1(ẏ2 − ẏ1 ) − k1 y1
m2 ÿ2 = u2 − b1 (ẏ2 − ẏ1 ) − k2 y2
(2.3.25)
(2.3.26)
Fig. 2.3: Mechanical translation system with two inputs and two outputs
Example 2.3.4:
20
2. Classical techniques for system modeling
Fig. 2.4: Example of mechanical system: 1/4 Car Model
2.3.3
Rotational motion
In this case considered movements about a fixed axis. It is rather like translation
movements, but the variables of interest are angular displacement, angular velocity and angular acceleration and instead of mass and forces, consider moments of
inertia and torques.
Newton’s second law for rotational motion says that the sum of all torques is
equal to the moment of inertia by the angular acceleration.
X
Ti (t) = J ∙ θ̈(t)
(2.3.27)
In addition to Newton’s second law can be considered the torsion springs, which
Hooke’s law
Tm (t) = −K ∙ θ(t)
(2.3.28)
and the corresponding elements to the viscous dampers are friction rotation,
whose expression is
Ta (t) = −b ∙ θ̇(t)
(2.3.29)
Example 2.3.5: Consider the mechanical system of rotation of the figure ??.
If it starts from rest and a small torque is applied to the flywheel J1 , in the early
stages of the movement, θ1 > θ2 > θ3 . This means that the torsion spring K is
θ1 − θ2 and D2 is the damper θ2 − θ3 . Therefore equations are
J1 θ̈1 = T (t) − D1 θ1 − K(θ1 − θ2 )
J2 θ̈2 = K(θ1 − θ2 ) − D2(θ̇1 − θ̇2 )
J3 θ̈3 = D2(θ̇1 − θ̇2 ) − D3 θ̇3
(2.3.30)
(2.3.31)
(2.3.32)
2.3. Time models
21
Fig. 2.5: Elements of mechanical systems of translation and rotation
Fig. 2.6: Example of mechanical rotation system
2.3.4
Electrical systems
The three basic elements of electrical circuits are resistors, capacitors and inductors.
Resistance is an element that dissipates energy by converting it into heat and
its equation is given by Ohm’s law
v(t) = R ∙ i(t)
A capacitor is an element that stores energy and its equation is
Z
1
i(t)dt
v(t) =
C
(2.3.33)
(2.3.34)
Finally, the coil or inductance has the equation
u(t) = L
di(t)
dt
In addition, the basic laws are the laws of Kirchoff:
(2.3.35)
22
2. Classical techniques for system modeling
• At any node (junction) in an electrical circuit, the sum of currents flowing
into that node is equal to the sum of currents flowing out of that node
or equivalently
The algebraic sum of currents in a network of conductors meeting at a point
is zero.
• The directed sum of the electrical potential differences (voltage) around any
closed network is zero, or:
More simply, the sum of the emfs in any closed loop is equivalent to the sum
of the potential drops in that loop, or:
The algebraic sum of the products of the resistances of the conductors and
the currents in them in a closed loop is equal to the total emf available in that
loop.
In addition, to calculate the transfer function is preferable working in the frequency
domain, ie, with the Laplace transforms of the equations:
• For resistors, the expression v(t) = Ri(t) becomes V (s) = RI(s).
You can also say that its complex impedance is ZR (s) =
• For capacitors, the term v(t) =
1
C
R
V (s)
I(s)
=R
i(t)dt becomes V (s) =
You can also say that its complex impedance is ZC (s) =
1
Cs I(s).
V (s)
1
I(s) = Cs
• For the coils or inductors, the expression v(t) = L di(t)
dt becomes V (s) =
LSI(s).
In this case the complex impedance is ZL (s) =
V (s)
I(s)
= Ls.
Example 2.3.6: Consider the RLC network in Fig. 2.7. At the right appears the
result when using complex impedances. The input loop, according to the 2nd Law
of Kirchoff is
1
V (s) = LSI(s) + RI(s) +
I(s)
Cs
and the output voltage is
1
VC (s) =
I(s)
Cs
. The transfer function is the ratio
1
VC (s)
Cs I(s)
=
V (s)
LsI(s) + RI(s) +
1
Cs I(s)
=
1
Cs
Ls + R +
1
Cs
=
LCs2
1
+ RCs + 1
2.3. Time models
23
Fig. 2.7: RLC circuit with voltage as input
Fig. 2.8: System example
Example 2.3.7:
24
2. Classical techniques for system modeling
Fig. 2.9: Example of RLC network with two meshes
2.4
Linearization
Fig. 2.10: Linearization of a function on two points (x0 , y0 ) y (x1 , y1 )
The actual physical systems are nonlinear. In the above examples, the nonlinearities are small and can be considered negligible. But in many cases the system
is clearly nonlinear. In these cases, you can get a simplified model, which is a
linear approximation in a working point, for small increases in the variable around
that point (x0 , y0 ). As you can see in Figure ref Picture9, linearization elsewhere
(x1 , y1 ), will generally be different.
If we consider a small variation around the working or equilibrium point , ex-
2.4. Linearization
25
panding in Taylor series
y = y0 +
dy
dx
0
(x − x0 ) +
1 d2 y
2! dx2
0
(x − x0 )2 + . . . +
1 dm y
m! dxm
0
(x − x0 )m + . . .
(2.4.1)
The best linear approximation is the tangent line, corresponding to the expansion
for m = 1
dy
y = y0 +
(x − x0 )
(2.4.2)
dx 0
For multivariate functions, the linear approximation is
y = y0 +
∂y
∂x1
0
(x1 − x10 ) +
∂y
∂x2
0
(x2 − x20 ) + ∙ ∙ ∙ +
∂y
∂xn
0
(xn − xn0 ) (2.4.3)
and corresponds to a plane or a hyperplane.
Example 2.4.1: Obtain the transfer function VL (s)/V (s) of the nonlinear electrical network figure for small signs v(t), containing a non-linear resistor whose
voltage-current relationship is defined byir = 2e0.1vr , where ir y vr are the current
and voltage of the resistor, respectively.
If we consider the system at rest or in equilibrium, the small signal v(t) = 0
and all derivatives are zero, i.e. VL (t) = L di(t)
dt = 0.
Then, at the equilibrium point, I0 = 2e0.1∙20 = 14.8.
The linearized equation of resistance is
ir = ir0 +
∂2e0.1vr
∂vr
0
(vr − vr0 ) = 14.8 + 2 ∙ 0.1e0.1vr0 (vr − vr0 )
= 14.8 + 0.2e2 (v − 20) = 14.8 + 1.48(vr − 20)
Considering increments around the equilibrium point
Δ(ir (t)) = 1.48Δ(vr (t))
26
2. Classical techniques for system modeling
and the Laplace transform is ir (s) = 1.48Vr (s).
Once linearized the nonlinear element, we can write the equations of the circuit
Δv(t) = LΔ
di(t)
+ RΔi(t)
dt
where R = 1.48 is the value obtained in the linearization.
The Laplace transform is
V (s) = VL (s) + Vr (s) = LsI(s) + RI(s)
The output is VL (s) = LsI(s). Dividing the output from the input, we obtain the
transfer function
LsI(s)
VL (s)
Ls
s
=
=
=
V (s)
LsI(s) + RI(s)
Ls + R
s + 1.48
2.5
Block diagrams
The block diagrams are a way to represent systems that are composed of several
parts which are represented by blocks and which are linked together by arrows
depicting the flow of signals.
The aim is to represent as blocks the transfer functions of each part, because
it is easier, and then simplify the diagram to obtain the transfer function of the
complete system.
2.5.1
Diagrams reduction
a) The output of a system is the product of the transfer function from the input
(this is true only in the frequency domain).
b) The transfer function of two blocks in series is the product of the transfer
functions of the blocks.
c) The transfer function of two blocks in parallel is the sum of the transfer
functions of the blocks.
d) The transfer function of a negative feedback loop is the transfer function of
the direct line divided by one plus the transfer function of the direct line and the
feedback.
2.5. Block diagrams
27
Fig. 2.11: Diagrams reduction
2.5.2
Transposition of sum points
Example 2.5.1: Consider the block diagram of Figure ref BL3a. To simplify the
diagram the steps include:
1. Simplification of the negative feedback loop consisting of H1 (s) and H2 (s)
for the diagram in Figure ref BL3b.
2. Transposition of the bifurcation of the above line , you will H6 (s), for the
diagram in Figure ref BL4a.
3. Replacing the negative feedback loop for the TF and the corresponding sum
of the signals in parallel on the right side as seen in the diagram of Fig ref
BL4a.
4. Finally, the product is both TF and it is simplified as shown in Figure ref
BL4b.
28
2. Classical techniques for system modeling
Fig. 2.12: Transposition of sum points
Fig. 2.13: a) Starting system of the example, and b) first simplification: feedback
loop
2.5. Block diagrams
29
Example 2.5.2: For each of the systems given in the sketches below, find the
transfer functions.
Solution:
a) Using the 2nd Newton’s Law
J θ̈ = τ − mg ∙ sin θ ∙ l
where J = ml2 and τ is the external applied torque.
The linearization is trivial because sin θ ≈ θ, when θ ≈ 0.
Applying Laplace transform
ml2 s2 Θ(s) + mglΘ(s) = τ (s)
The transfer function is
1
Θ(s)
=
2
2
τ (s)
ml s + mgl
b) Aplying the 2nd Newton’s Law
J θ̈ = τ (t) − kθ − bθ̇
30
2. Classical techniques for system modeling
Applying Laplace transform
Js2 Θ(s) = τ (s) − KΘ(s) − bsΘ(s)
The transfer function is
1
Θ(s)
=
2
τ (s)
Js + bs + K
c) The equations of the system are
I(s)=i1 (s) + i2 (s)
V (s)=Ri1 (s)
1
V (s)= i2 (s)
Cs
considering i1 the intensity trough the resistor and i2 the the intensity trough the
capacitor.
Solving for the intensities
I(s)=i1 (s) + i2 (s)
R
i1 (s)=
V (s)
i2 (s)=CsV (s)
Substituting
I(s) =
1
V (s) + CsV (s)
R
and the transfer function is
V (s)
=
I(s)
1
R
1
R
=
RCs
+1
+ Cs
d) Considering the position u(t) as input and y(t) as output, when the suspension system is in steady state and suddenly a small positive input is applied,
the spring and the damper are compressed and it appear in the mass two forces:
Fs = K(u − y) and Fb = b(u̇ − ẏ). Using the 2nd Newton’s Law
mÿ = K(u − y) + b(u̇ − ẏ)
The Laplace transform is
ms2 Y (s) + bsY (s) + KY (s) = bsU (s) + KU (s)
and the transfer function
bs + K
Y (s)
=
2
U (s)
ms + bs + K
2.5. Block diagrams
31
Example 2.5.3: For raising and lowering a mass m = 1 kg, it is used a pulley
with a radius r = 0.5 m and moment of inertia J = 0.1 kg Δm2 powered by a DC
motor whose characteristics are:
• Internal resistance Ri = 70 Ω
• Inductance Li = 10 H
• Torque constant Ki = 2 N ∙ m/A
• Constant of electromotive force Ke = 6 V ∙ s
The vertical displacement of the mass is proportional to the position of the
pulley and is measured using the dynamic behavior which is given by the position
sensor expression:
r
2θ (t)
cos(θ(t))
us (t) = −5
π
Measuring voltage us (t) is compared with the reference voltage uref (t) with
which the system is governed, being the gain of the amplifier Ku = 1.
Find:
• 1. Linearize the system around the equilibrium point defined by x0 = π4 .
• 2. Draw the block diagram of the above system having as signal input
Uref (t) and as output the weight displacement x(t).
• 3. Calculate the transfer function of the system
X(s)
Uref (s)
• 4. The displacement changes of the weight if the voltage value becomes
uref = 180.5 V .
• 5. If you want to convert the weight that drives the motor to a variable value
(Such as what occurs in an elevator, when you do not know the accurate
number people entering), where the disturbance signal has to be introduced
the block diagram?
32
2. Classical techniques for system modeling
Fig. 2.15:
Example 2.5.4: In the figure it is shown a system that controls the height of
liquid in a tank h(t) depending on the reference signal, r(t).
The float sensor is a negligible dimensions weight attached to a negligible
weight rod whose end is the cursor of a potentiometer of 1/4 in circumference.
Tension rod end, b(t), is compared to the reference, r(t), and the difference
is amplified with a gain A = 20. This tension regulates the inlet valve whose
equation is:
Q1 (t) = Kv ∙ [50v(t)]
with 0.5Kv =
m3
s∙V
2.5. Block diagrams
33
p
√ 5
Data: K = 5 2 ms2 (coefficient of Torricelli ’s equation q2 (t) = K h(t),
drain of the tank).
a) Write the equations that represent the physical model of the system.
b) Linearize around the equilibrium point given by H0 = 0.5m.
c) Draw a representative block diagram of the system.
d) Calculate the transfer function H(s)/R(s).
e) Calculate the final value of h(t) when the input undergoes a change abrupt
value of 1.
f) Does the liquid overflow the tank in this case?
Resolution:
a) Write the equations that represent the physical model of the system.
• The amplifier output is:
V (t) = A[r(t) − b(t)]
• The valve regulates the inflow according to the relationship:
Q1 (t) = Kv [50 − v(t)]
• The equations of deposit in the figure with section S = 1 m3 are:
q1 (t) − q2 (t) = S
q2 (t) = K
√ 5
with K = 5 2 ms2 .
dh(t)
dt
p
h(t)
• The potentiometer equation is:
b(t) = Vmax
θ(t)
θmax
with Vmax = 20 V y θmax = π/2.
• The angle is related to the height of liquid:
hmax − h(t) = l cos θ(t)
with hmax = 1m y l = 1m.
34
2. Classical techniques for system modeling
Fig. 2.16:
• Finally we write the system equations:
v(t) = 20[r(t) − b(t)]
1
q1 (t) = [50 − v(t)]
2
dh(t)
q1 (t) − q2 (t) =
√ p dt
q2 (t) = 5 2 h(t)
cos θ(t) = 1 − h(t)
θ(t)
b(t) = 40
π
b) Linearize around the given equilibrium point H0 = 0.5 m.
The equilibrium condition is:
h0 = 1 rad
θ0 = arccos(1 − h0 ) =
π
rad
3
40
θ0 = 13.3 V
πp
= 5 2h0 = 5 m3 /s
b0 =
q2,0
q1,0 = q2,0
v0 = 50 − 2q1,0 = 40 V
1
r0 =
v0 + b0 = 15.3 V
20
The linearized system around the equilibrium condition is:
2.5. Block diagrams
35
Δv = A(Δr − Δb) = 20(Δr − Δb)
1
Δq1 = −Kv Δv = − Δv
2
Δq1 − Δq2 = SΔḣ = Δḣ
K
Δq2 = √ Δh = 5Δh
2 h0
√
3
Δθ
Δh = l sin θ0 Δθ =
2
Δθ
40
Δb = Vmax
= Δθ
θmax
π
c) Draw a representative block diagram of the system.
To draw the block diagram, we must apply the Laplace Transform to the linearized equations:
V (s) = A(R(s) − B(s))
Q1 (s) = −Kv V (s)
Q1 (s) − Q2 (s) = SsH(s)
K
Q2 (s) = √ H(s)
2 h0
H(s) = l sin θ0 Θ(s)
Vm ax
Θ(s)
B(s) =
θm ax(s)
The block diagram is:
d) Calculate the transfer function H(s)/R(s).
C ∙ G(s)
H(s)
=
R(s)
1 + C ∙ G(s) ∙ Z
36
2. Classical techniques for system modeling
where
C = −Kv A = −10
G(s) =
Z=
m3
s∙V
H(s)
=
Q1 (s)
1+
1
sS
1 √
K
sS 2 h0
=
1
s
1+
5
s
=
1
s+5
B(s)
B(s) Θ(s)
Vmax 1
80
= √
=
=
θmax sin θ0
H(s)
Θ(s) H(s)
π 3
Finally:
10
− s+5
−10
H(s)
=
80
10 =
√
√
R(s)
1 − π 3 s+5
s + 5 − π800
3
e) Calculate the final value of h (t) when the input suffers a sudden change of
value 1.
Applying the final value theorem, it is obtained:
Δh∞ = lim Δh(t) = lim sH(s) = lim s
t→∞
s→0
= lim s
s→0
s→0
−10
R(s) =
√
s + 5 − π800
3
1
−10
= −0.0704 m
800
s + 5 − π √3 s
f) It is overflowing the fluid from the reservoir in this case? The height reached
by the liquid is:
h∞ = h0 + Δh∞ = 0.5 − 0.0704 m = 0.5704 m < hmax = 1 m
So the liquid does not overflow.
Example 2.5.5: It a furnace in which the heat, generated by an internal resistance qg (t), is proportional (K1 = 10) to the square of the current flowing through
it (q(t) = K1 i2 (t)), while the lost heat qp (t) is proportional (K2 = 2.5) to the difference between the temperature inside the oven θi (t) and the external temperature
θe (t).
Knowing that the variation of the internal furnace temperature is proportional
(K3 = 1) to the difference between the heat generated and wasted heat, and the
system control is performed by an amplifier which generates a current i(t) =
θref (t) − θi (t) + 5.
Find:
a) Equations of the system.
b) Linearize the system around the equilibrium condition θi0 = θref 0 = 100o C
y θe0 = 0o C
2.5. Block diagrams
37
c) Draw the block diagram of the device. Inputs: Δθref (s), Δθe (s). Output:
Δθi (s).
Resolution:
a) System equations.
q(t) = K1 i2 (t)
qp (t) = K2 [θi (t) − θe (t)]
dθi (t)
= K3 [qg (t) − qp (t)]
dt
i(t) = θref (t) − θi (t) + 5
with K1 = 10, K2 = 2.5, and K3 = 1.
b) Linearize the system around the equilibrium condition θi0 = θref 0 = 1000 C
y θe0 = 00 C.
Initial conditions:
qg0 = K1 i20
qp0 = K2 [θi0 − θe0 ]
0 = K3 [qg0 − qp0 ]
i0 = θref 0 − θi0 + 5
namely
i0 = 5
qg0 = 250
qp0 = 250
38
2. Classical techniques for system modeling
The linearized system around the equilibrium point is
Δqg (t) = 2i0 K1 Δi(t)
Δqp (t) = K2 [Δθi (t) − Δθe (t)]
Δθ̇i (t) = K3 [Δqg (t) − Δqp (t)]
Δi(t) = Δθref (t) − Δθi (t)
c) Draw the block diagram of the device.
Inputs: Δθref (s), Δθe (s).
Output: Δθi (s)
Laplace transforms are:
ΔQg (s) = 2i0 K1 ΔI(s)
ΔQp (s) = K2 [ΔΘi (s) − ΔΘe (s)]
sΔΘi (s) = K3 [ΔQg (s) − ΔQp (s)]
ΔI(s) = ΔΘref (s) − ΔΘi (s)
The block diagram is:
Example 2.5.6: Is a physical system consisting of a power amplifier, a motor, a
gear train and an encoder as shown in the figure:
2.5. Block diagrams
39
The voltage that reaches the power section is proportional to the difference
between the reference voltage Ur and the output voltage of the encoder Ur , where
the constant of proportionality Kp = 2.
The power section produces a current i from the input voltage i, according to
the following differential equation:
p
d i(t)
+ i(t) − 2 e(t) + 1.75 = 0
dt
In the engine, the torque τ1 is proportional to the intensity i , being the proportionality constant Km = 4 N m/A.
The motor shaft is connected to a reducer, being N1 = 16 teeth, N2 = 32
teeth. On the second axis we have coupled a flywheel with a moment of inertia
I = 10 Kg m2 , a friction B = 5 N ms/rad, showing the axis an elastic behavior
with K = 2N m/rad.
Finally, the angle rotated by the second axis, θ2 , is measured with an encoder,
that measures the output voltage Us iand it is proportional to the rotated angle θ2 ,
being the proportionality constant Ke = 0.5 V /rad.
Find:
a) System equations.
b) Linearized around equilibrium point given by τ20 = 2N m.
c) Blocks diagram.
Us (s)
d) Find the relation U
.
r (s)
e) Find the output voltage of the encoder Us when the reference voltage experiences a sharp jump of two units.
I(t)
Solution: a) System equations. The equations of the system are as follows
Amplifier:
e(t) = Kp [Ur (t) − Us (t)], con Kp = 2
Power section:
i(t)
Motor:
p
di(t)
+ i(t) − 2 e(t) + 1.75 = 0
dt
τ1 (t) = Km i(t),
con Km = 4N m/A
Reductor:
τ1 (t)
N1
=
,
τ2 (t)
N2
con
N1
= 0.5
N2
Flywheel
I
dθ2 (t)
d2 θ2 (t)
= τ2 (t) − Kθ2 (t) − B
2
dt
dt
40
2. Classical techniques for system modeling
con I = 10Kgm2 , B = 5N ms/rad y K = 2N m/rad
Encoder
Us (t) = Ke θ2 (t),
con Ke = 0.5V /rad
b) Linearize around equilibrium point given by τ = 220N m.
We consider the equilibrium point:
τ20 = 2N m
N1
= 1 Nm
τ10 = τ20 N
2
τ10
i0 = Km = 0.25 A
2
e0 = i0 +1.75
= 1V
2
The only nonlinear equation is the one that characterizes the power section. So:
1
i0 Δi̇ + Δi − √ Δe = 0
e0
The equations of changes are:
Δe = Kp (ΔUr − ΔUs )
i0 Δi̇ + Δi − √1e0 Δe = 0
Δτ1 = Km Δi
2
Δτ2 = N
N1 Δτ
IΔθ̈2 = Δτ2 − KΔθ2 − BΔθ̇2
ΔUs = Ke Δθ2
Applying the Laplace transform
E(s) = Kp [Ur (s) − Us (s)]
i0 sI(s) + I(s) − √1e0 E(s) = 0
τ1 (s) = Km I(s)
2
τ2 (s) = N
N 1 τ1 (s)
2
Is Θ(s) = τ2 (s) − KΘ(s) − BsΘ(s)
Us (s) = Ke Θ2 (s)
The blocks diagram is:
2.5. Block diagrams
41
Us (s)
Ur (s)
c) Find the relation
G(s)
Us (s)
=
Ur (s)
1 + G(s)
where
G(s) =
1
1
N2
1
16
Us (s)
= Kp √
Km
Ke =
U (s)
N 1 Is2 + Bs + K
2.5s3 + 11.25s2 + 5.5s + 2
e0 i0 s + 1
So
Us (s)
1
=
Ur (s)
2.5s3 + 11.25s2 + 5.5s + 18
d) Find the output voltage of the encoder Us when the reference voltage experiences a sharp jump of two units.
Applying the final value theorem is obtained:
ΔUs,∞ = lim ΔUs (t) = lim sU s(s) = lim s
t→∞
s→0
= lim s
s→0
2.5s3
s→0
2.5s3
16
Ur (s) =
+ 11.25s2 + 5.5s + 18
16
2
= 1.778 V
2
+ 11.25s + 5.5s + 18 s
Us,∞ = Us,0 + ΔUs,∞
Us0 = Ke θ20 =
Ke
τ20 = 0.25V
K
So
Us,∞ = Us,0 + ΔUs,∞ = 2.028 V
Example 2.5.7: A thermistor has a response to the temperature represented by
R = R0 e−0.1T , where R = resistance and T = temperature in Celsius. Find the
linear model for the thermistor operating around T = 20o C with a small range of
temperature variation.
Solution:
In the balance point or point of operation T = 20o C
Req = R0 e−0.1∙20 = 0.135R0
Linearizing around this point of equilibrium
ΔR =
∂R0 e−0.1T
∂T
ΔT = R0 (−0.1)e−0.1T
T =20
T =20
ΔT = −0.0135ΔT
42
2. Classical techniques for system modeling
By taking Laplace transforms
R(s) = −0.0135T (s)
Example 2.5.8: Determine the transfer function H(s)/Qe (s) of the system consisting of a water reservoir, linearizing around the point of equilibrium given by
qe0 = 1m3 /s. if the section of the outlet pipe is A = 1m and the area of the base
is B = 1m2 .
√
Data: Consider the Torricelli’s equation qs (t) = K ∙ A h.
Solution:
The inflow minus outflow is the variation in volume
qe (t) − qs (t) = B
dh(t)
dh(t)
=
dt
dt
According to Torricelli’s equation
qs (t) =
√
h
In the equilibrium point, qe0 = 1m3 /s
qe0 =qs0 ⇒ qs0 = 1m3 /s
√
1
qs0 =K h0 ⇒ h0 = m
K
(2.5.1)
Linearizing
Δqe − Δqs =Δḣ
K 3/2
K
Δqs = √ Δh =
Δh
2
2 h0
(2.5.2)
2.5. Block diagrams
43
By taking Laplace transforms
Qe (s) − Qs (s)=sH(s)
Qs (s)=
K 3/2
H(s)
2
Sustituting
Qe (s) −
The transfer function is
K 3/2
H(s) = sH(s)
2
1
H(s)
=
3/2
Qe (s)
s + K2
(2.5.3)
44
2. Classical techniques for system modeling
Bibliografia
[1] E.A. Puente, Regulacion Automatica I, Seccion de Publicaciones ETSIIM,
1980.
[2] K. Ogata, Modern Control Engineering, Prentice Hall International Edition,
1995.
[3] B. Kuo, Automatic Control Systems, Prentice Hall, 1996.
3. TIME ANALYSIS
3.1
Introduction. Time Analysis of Systems
The previous chapters have studied the mathematical modeling on the basis of
their relationship between input and output of the systems. This leads us to the
concept of transfer function. In this chapter we analyze how the systems behave
using their transfer functions.
3.2
Temporal Response
When the input signal u(t) is introduced into a system transfer function G(s),
its output is
Y (s) = G(s)U (s) ⇒ y(t) = g(t) ∗ u(t) = L−1 (G(s)U (s))
This time response has two parts: the transient response and steady state response. Transient response is part of the temporal response that tends to zero as
time goes to infinity and the steady-state response is that which remains over time.
y(t) = yrt (t) + yrp (t)
lim yrt (t) = 0
t→∞
yrp (t) = lim y(t)
t→∞
The dynamic behavior of the system is primarily related to transient response
and the speed and stability can be studied. In the steady-state response the accuracy
of the response can be studied, ie the stationary system error.
To analyze the response, different test signals are introduced. The most important are:
The signal impulse or Dirac delta
u(t) = δ(t)⇒U (s) = 1
The unit step signal
0,t < 0
⇒U (s) =
u(t) = u0 (t) =
1,t > 0
1
s
46
3. Time Analysis
Fig. 3.1: Response of a system
and the ramp signal unit
u(t) = tu0 (t)⇒U (s) =
3.2.1
1
s2
Impulse response
The main reason for the classical control theory uses the frequency domain
rather than the time domain is given by the convolution theorem.
In the time domain, the output is the convolution of the input and the impulse
response of the system
y(t) = g(t) ∗ u(t) =
Z
t
0
g(t − τ )u(τ )dτ =
Z
t
0
g(τ )u(t − τ )dτ
while in the frequency domain, the output is a product
Y (s) = G(s)U (s).
In the event that the input is an impulse unit u(t) = δ(t), its Laplace transform
is the corresponding signal in the frequency domain, U (s) = 1.
y(t) = g(t) ∗ δ(t) =
Z
t
0
g(t − τ )δ(τ )dτ = g(t)
That is, when the input signal is an impulse, the output is
y(t) = g(t), where g(t) = L−1 (G(s))
The function g(t) is called the impulse response of the system and is a mathematical model of the system, as can be the differential equations or transfer function
G(s).
3.2. Temporal Response
3.2.2
47
Response of First Order systems
The First Order systems are the simplest and are given by the differential equation
T ẏ(t) + y(t) = Ku(t)
(3.2.1)
its Laplace transform with zero initial conditions is
T ∙ sY (s) + Y (s) = KU (s)
and its transfer function can be written as
K
Y (s)
=
U (s)
1 + Ts
The constant K is the static gain, T is the time constant and σ =
of attenuation.
It is interesting to note that the pole is located at −1/T .
1
T
is the constant
Impulse response
Consider a first order system transfer function G(s) =
unit impulse u(t) = δ(t) ⇒ U (s) = 1, the output is
Y (s) = G(s) ∙ 1 =
K
1+T s .
If the input is a
K
1 + Ts
and from the time domain is
y(t) = g(t) = L−1
K
1 + Ts
=
k −t/
e T
T
t≥0
Fig. 3.2: Impulse response of a first order system
As shown in Fig. 3.3, the constant T is the time it takes the system to reach the
37% of the initial value, g(0) = K/T .
48
3. Time Analysis
Fig. 3.3: Impulse response of a first order system
Step response
If the input signal to the system G(s) is a unit step u(t) = u0 (t) ⇒ U (s) = 1/s
Fig. 3.4: Unit step response of a first order system
then output in the time domain is
k
−1
−1 1 −
y(t) = L−1 G(s)
=
L
=
kL
s
s
s(1+T s)
t
= k(1 − e− /T ) ,
when t ≥ 0.
T
1+T s
=
In figure Fig. 3.5 the output of a first-order system to a step input unit is shown.
The time constant T is the time it takes for the system to reach the 63% of the final
value K and measures the speed of system response. The time it takes the system
to reach the 95% of the final value is called settling time and can be calculated as
ts ≈ 3T ≈ πσ .
Example 3.2.1: It is also interesting to note that from the response unit step is
possible to determine the parameters K (final value) and T (time it takes for the
system to reach the 63% of the final value) and therefore , you can determine the
transfer function.
For example if the unit step response is shown in Fig. 3.6, the final value
K = 0.72 (which is also the gain because the input is a unit step) and the time
3.2. Temporal Response
49
Fig. 3.5: Unit step response of a first order system
Fig. 3.6: Determination of the transfer function from the unit step response of a
first order system
constant is corresponding to the time at which 0.63 ∙ 0.72 , which is T = 0.13, and
the transfer function is
0.72
K
=
G(s) =
1 + Ts
1 + 0.13s
Response to ramp
If the system input G(s) is the unit ramp
u(t) = t ∙ u0 (t) ⇒ U (s) = 1/s2
the output y(t) can be expressed as
50
3. Time Analysis
Fig. 3.7: Response to a ramp unit of a first order system
y(t) = L−1
G(s)
s2
= L−1
k
s2 (1+T s)
= KL−1 − Ts +
t
t
= K(−T + t + T e− /T ) = K(t − T ) + kT e− /T ,
1
s2
+
T2
1+T s
para t ≥ 0
=
In this expression, T is the delay between the ramp input and the response.
Fig. 3.8: Unit ramp response of a first order system
3.2.3
Response of the second order systems
The standard equation of a second order system is
G(s) =
where
Kωn 2
s2 + 2ζωn s + ωn 2
(3.2.2)
3.2. Temporal Response
51
K = Static gain.
ωn = Undamped natural frequency.
ζ = Damping ratio.
σ=ζωn =
p Attenuation.
ωd = ωn 1 − ζ 2 = Damped frequency.
The system behavior is determined by K, ζ y ωn .
The poles can be determined from the characteristic equation (denominator of
G(s) equated to zero).
s2 + 2ζωn s + ωn 2 = 0
(3.2.3)
p
p
s = −ζωn ±p ζ 2 ωn 2 − ωn2 = −ζωn ± ωn ζ 2 − 1 =
−ζωn ± jωn 1 − ζ 2
(3.2.4)
The poles are
s = −σ ± jωd
(3.2.5)
52
3. Time Analysis
Fig. 3.9: Relation among the parameters of a second order system.
3.2. Temporal Response
Fig. 3.10: Classification of second-order systems
53
54
3. Time Analysis
Impulse response
Consider a second order system
G(s) =
kKωn 2
s2 + 2ζωn s + ωn 2
(3.2.6)
and assume that your input is a unit impulse u(t) = δ(t) ⇒ U (s) = 1
The most interesting case is when the system is underdamped 0 < ζ < 1. In
this case the poles are complex conjugates with negative real part.
p
−ζωn ± jωn 1 − ζ 2 = −σ ± jωd
The response is
Kωn −σt
y(t) = p
e
sin(ωd t + ϑ),
1 − ζ2
para t ≥ 0
That is a damped sine with envelope e−σt . The decay factor σ = ζωn indicates
the rate at which decreases the amplitude of the oscillation.
Fig. 3.11: Impulse response of the second order underdamped systems.
Step response
Consider a system with transfer function G(s) and step input unit,
Y (s) = U (s)G(s) =
y(t) = L
−1
G(s)
s
=
Z
G(s)
s
t
0
g(τ )dτ
3.2. Temporal Response
55
If the denominator is decomposed into simple factors, which may correspond
to simple roots or pairs of complex conjugate roots. The output to a unit step is
G(s) = Q Y (s) =
N (s)
(s − αi )2 + βi 2
Q
(s − σi )
N (s)
G(s)
=
= Q
Q
2
2
s
s (s − σi )
(s − αi ) + βi
X
Ci s + D i
A X Bi
+
+
s − σi
s
(s − αi )2 + βi 2
X
X
y(t) = L−1 (Y (s)) = A +
Ei eαi t sen(βi t + ϕi )
B i e σi t +
=
y(t) = G(0) +
X
B i e σi t +
X
Ei eαi t sen(βi t + ϕi )
t≥0
t≥0
(3.2.7)
The output consists of three parts: the constant A = G(0) is due to the input
and it is the forced response and the stationary system response gain and the second
part
X
X
Ei eαi t sin(βi + tϕi )
Bi eσti +
which is the natural response of the system, or the part that it is due to the transfer
function. In it, the expression
X
Bi eσti
corresponds to the real roots of the denominator (real poles) and
X
Ei eαt
i sin(βi + tϕi )
corresponds to pairs of complex conjugate roots of the denominator (complex conjugate poles).
It is interesting to note that if any of the coefficients σi or αi of the real exponential is positive, the corresponding exponential is growing and the response y(t)
tends to infinity, ie, the system is unstable.
A dynamic system is stable if and only if all poles are located in the left half
semi plane.
If the poles are real, ie the coefficients Ei are zero, and the response
X
y(t) = A +
B i e σi t
consists of two parts: a constant and an exponential that is increasing if σi is positive, and it is decreasing if σi is negative, and one if σi = 0.
56
3. Time Analysis
Fig. 3.12: The real exponential function eσt is increased when the real coefficient
σ is positive and decreasing when it is negative.
Fig. 3.13: The function eαt sin(βt + φ) is a growing sinusoid if α is positive and
decreasing if negative.
Adding the constant A and the part of the exponential, we have the responses
shown in Fig. 3.14.
The constant A is the steady-state response
yrp (t) = lim y(t) = A
t→∞
and according to the final value theorem coincides with G(0):
lim y(t) = lim s
t→∞
s→0
G(s)
= G(0)
s
3.2. Temporal Response
57
Fig. 3.14: The unit step responses of a second order system according to the coefficient of friction: 1) ζ = 0, oscillator (poles on the imaginary axis). 2)
0 < ζ < 1, underdamped system (complex conjugate poles in the left
half. 3) ζ = 1, critically damped system (real and equal poles). 4) ζ >
1, overdamped system (real and different poles).
Example 3.2.2: Consider the system of Fig.3.2.3, but with the parameters: m=1,
b=3, K=2.
a) Find the transfer function G(s) = X(s)/F (s).
b) Find the response to unit step of the system.
58
3. Time Analysis
The differential equation system is
mẍ + bẋ + kx = f (t)
By taking Laplace transforms
(ms2 + bs + k)X(s) = F (s)
The transfer function is
1
1
X(s)
=
= 2
2
F (s)
ms + bs + k
s + 3s + 2
b) If the input is a unit step F (s) =
X(s) =
1
s
and output is
1
1
2
s s + 3s + 2
Decomposing into partial fractions
X(s) =
1
1
A
B
C
= +
+
2
s s + 3s + 2
s
s+1 s+2
where
A = lim
s→0 s2
1
= 0.5
+ 3s + 2
1 1
= −1
s→−1 s s + 2
B = lim
1 1
= 0.5
s→−2 s s + 1
C = lim
namely
−1
0.5
0.5
+
+
s
s+1 s+2
Taking inverse Laplace transforms, we have
X(s) =
x(t) = 0.5 − e−t + 0.5e−2t para t > 0
Example 3.2.3: 1.- For the system of Fig. 3.2.3
a) Find the transfer function G(s) = X(s)/F (s).
b) Find the unit step response of the system.
3.2. Temporal Response
59
a) The differential equation of the system is
mẍ + bẋ + kx = f (t)
By taking Laplace transforms
(ms2 + bs + k)X(s) = F (s)
The transfer function is
X(s)
1
1
=
= 2
F (s)
ms2 + bs + k
3s + 15s + 33
b) If the input is a unit step F (s) =
X(s) =
1
s
and output is
1
1
1
1
=
s 3s2 + 15s + 33
3 s(s2 + 5s + 11)
Decomposing into partial fractions
X(s) =
1
=
3
X(s) =
1 A Ms + N
=
3 s s2 + 5s + 11
−(1/11)s − 5/11
1/11
+
s
s2 + 5s + 11
1
33
s+5
1
−
s (s2 + 2.5s)2 + 2.182
This last expression can be written as the sum of the transformed damped sine
and cosine
s + 4.75
0.25
1 1
X(s) =
−
−
33 s (s2 + 2.5s)2 + 2.182 (s2 + 2.5s)2 + 2.182
1
X(s) =
33
0.25
s + 2.182
1
− 2
− 2
2
2
s (s + 2.5s) + 2.18
(s + 2.5s)2 + 2.182
The temporal response is the inverse transform
x(t) =
1
1 − e−2.5t cos(2.18t) − e−2.5t sin(2.18t) for t > 0
33
60
3. Time Analysis
Parameters characterizing the response of a second order
system or specifications
3.3
The specifications express the characteristics to be achieved in the system when
a controller is introduced into it. They can be expressed in different ways by using
the frequency domain or the time domain. In our case we will refer to specifications
in the time domain.
Specifications regarding the answer in the time domain of a system are usually
defined with respect to the temporary response to a step input of a second order
system. To do so a reference system is taken, with a transfer function:
G(s) =
Kωn2
s2 + 2ζωn s + ωn2
(3.3.1)
whose unit step response is
with
1
y(t) = 1 − p
1 − ζ2
e−ζωn t sin(ωn t
φ = arctan
p
1 − ζ 2 + φ)
p
(3.3.2)
1 − ζ2
ζ
(3.3.3)
p
(3.3.4)
This system has its poles located in
p1,2 = −ζωn ± jωn
1 − ζ2
where ζ is called damping ratio damping ratio and ωn is the undamped natural
frequency of the system. For a system with damping rate 0 < ζ ≤ 1 (underdamped
system) the system response curve has the form shown in Fig.7.3
y
To
Mp
d·Mp
ep
Tp
t
Ts
Fig. 3.15: Parameters of the temporal response
3.3. Parameters characterizing the response of a second order system or specifications 61
Among the parameters that can be used to specify the response, we have the
following:
• Peak time, Tp .
Tp =
π
ωd
(3.3.5)
where ωd is the damped frequency ωd = ωn
• Overshoot, Mp .
Mp = e
• Settling time, Ts .
−π √
Ts '
ζ
1−ζ 2
p
1 − ζ 2.
=
√
d
π
ζωn
(3.3.6)
(3.3.7)
Finally, the steady state position error, Ep , is defined as the difference between
the desired reference value and the output value steady state of the controlled variable of the system (we are assuming for simplicity that the feedback signal is unitary).
Example 3.3.1: Given the following circuit, find the maximum voltage on the
capacitor when closing the switch:
Solution:
The system equations are calculated, considering as input the battery-switch
voltage (0 to 10 V, depending on the switch is open or closed) and as output the
voltage across the capacitor:
Z
di(t)
1
+
i(t)dt
vi = vR + vL + vC = Ri(t) + L
dt
C
Z
1
i(t)dt
vo =
C
Applying the Laplace transform
Vi (s) = RI(s) + LsI(s) +
1
I(s)
Cs
62
3. Time Analysis
Vo (s) =
1
I(s)
Cs
Dividing the transfer function is obtained
1
Vo (s)
Cs I(s)
=
Vi (s)
RI(s) + LsI(s) +
=
1
Cs I(s)
1
Cs
R + Ls +
1
Cs
=
LCs2
1
+ RCs + 1
With the values of the statement you have
108
Vo (s)
= 2
s + 12 ∙ 103 + 108
Vi (s)
As seen is a second order system. Compared with the standard expression
Kωn2
s2 + 2ζωn s + ωn2
it is obtained
ωn2 =108 ⇒ ωn = 104
Kωn2 =108 ⇒ K = 1
2ζωn =12 ∙ 103 ⇒ ζ =
12 ∙ 103
= 0.6 ⇒ Underdampened.
2 ∙ 10
(3.3.8)
The maximum voltage is reached when the overshoot is maximized
Mp = e
− √ πζ
1−ζ 2
= 0.0948 = 9.48%
When the input is a step of 10V , then
Vo,max = 10 + 10 ∗ 0.0948 = 10.948V
Example 3.3.2: Compare these systems in terms of overshoot and settling time,
drawing responses from the closest possible way:
Ga =
Gc =
10
s2 +2s+5
5
(s2 +2s+5)(s+0.5)
Solution:
a) First system
Ga =
Gb =
Gd =
20
(s2 +2s+5)(s+2)
5(s+2)
(s2 +2s+5)
10
s2 + 2s + 5
3.3. Parameters characterizing the response of a second order system or specifications 63
b) Comparison among the systems:
64
3. Time Analysis
• The Gb (s) system equals Ga (s) with a further (farther from the imaginary
axis) pole.
• The Gc (s) system equals Ga (s) with an additional influential pole on the
case of Gb (s): the system will be much slower.
• The Gd (s) equals Ga (s) system with an extra zero: the system will be faster.
Looking at the graph can be concluded that:
• Adding a zero to a system makes the peak occurs earlier and is greater.
• The closer the zero will be to the imaginary axis, the faster the system is.
• Adding a real pole system makes the peak occurs later and is minor.
3.3. Parameters characterizing the response of a second order system or specifications 65
• The closer the pole to the imaginary axis is, the slower the system is.
Example 3.3.3: A system consists of the following block diagram figure:
The response of each subsystem is shown in the following figure:
Calculate the final value when a 2 units step is applied.
Solution: The subsystem response to a unit step input is:
66
3. Time Analysis
It can be seen that:
• That’s the typical response of an underdamped second order system.
• It stabilizes around the value 1, thus the static gain of the system is one.
• Reaches a maximum at 1.08 therefore Mp =
1.08−1
1
= 0.08.
• The stabilization time is ts = 3.14
With these data we will calculate the transfer function of the system.
Compared with the standard expression
Kωn2
s2 + 2ζωn s + ωn2
It can be obtained:
• Static gain: K = 1.
• Stabilization time: ts =
π
σ
→σ=
π
ts
= 1 → ζωn = 1.
• Damping ratio:
Mp = e
π
− tan
θ
−π
→ θ = atan
= 0.8936rad.
ln(Mp )
To calculate the damping ratio: ζ = cos(θ) = 0.6266 < 1
To calculate the undamped natural frequency:
ζωn = 1 → ωn =
1
= 1.596rad/s
ζ
The system is:
GA =
You can verify in this plot:
s2
2.5
+ 2s + 2.5
3.3. Parameters characterizing the response of a second order system or specifications 67
The impulse response of the system B is:
It can be seen that this is the typical response to a system of first order.
GB (s) =
K
1 + Ts
The generic response to a first order system facing an input impulse is:
68
3. Time Analysis
We note that
y(0) = 2 →
K
=2
T
T =1
With the data T = 1 and K = 2 we will calculate the transfer function of the
system:
GB =
2
1+s
Finally, the response of the system C to a ramp with slope 0.5 is:
3.3. Parameters characterizing the response of a second order system or specifications 69
That is the answer to a ramp of a system of first order:
Gc (s) =
K
1 + Ts
The response to a ramp with slope 0.5 has a value:
y(t) =
K
K
(t − T ) + T e−t/T
2
2
The generic response is
We note that
T = 1, the output delay with respect to the line r(t) =
output signal.
The steady slope of the output signal is 1:
K
=1→K=2
2
We can also verify that using the final value theorem:
K
2 (t − T )
once stabilized
70
3. Time Analysis
The transfer function is:
Gc (s) =
2
1+s
The complete system will be:
Example 3.3.4: If Vi (t) is a voltage step in the network shown in the figure, find
the resistance value so that the overshoot is 20% in the voltage across the capacitor,
if C = 10−6 F y L = 1H.
Solution:
The system equations in the frequency domain are
Vi (s)=LsI(s) + RI(s) +
Vo (s)=
1
I(s)
Cs
1
I(s)
Cs
Dividing the transfer function is obtained
1
Vo
Cs
=
Vi
Ls + R +
1
Cs
=
s2 +
1
LC
R
1
L s + LC
Compared to the standard form of a second order system
Kωn2
s2 + 2ζωn s + ωn2
q
q
1
Se obtiene que K = 1, ωn = LC
, y ζ = R2 C
L.
Furthermore, the overshoot is required to be the 20%
π
Mp = e− tanθ = 0.20 ⇒ θ = 62.370 ⇒ ζ = cosθ = 0.456
3.3. Parameters characterizing the response of a second order system or specifications 71
Considering C = 10−6 F y L = 1H,
r
ωn =
1
= 103
LC
R
R
=
= 2ζωn = 2 ∙ 0.456 ∙ 103 = 912Ω ⇒ R = 912Ω
L
1
Example 3.3.5: Consider the following RC-circuit implementation of a lowpass filter.
(a) The transfer function for this circuit is given below. Employing this model,
find the output response of the circuit e(∞) for a step input of magnitude E:
1
Eo (s)
=
Ei (s)
R1 C 1 + 1
(b) If R1 = 5Ω and C1 = 2F , determine the settling time ts = σ4 for the filter,
that is, the time it takes for the circuit to reach and stay within 2% of its final value.
(c) An alternative implementation of a low-pass filter is given below. Find the
transfer function for this circuit for an input of ei (t) and an output of eo (t) .
(d) You wish to design the filter in (c) to achieve similar performance to the filter
in (a). Specifically, if L2 = 5H, then find the resistance R2 and capacitance C2 so
that:
i. The two filters have the same settling time ts (corresponds to the filter’s pass
band).
ii. The two filters reach the same steady-state value for a constant input.
iii. The second filter has a maximum percent overshoot equal to 5%.
72
3. Time Analysis
Solution:
a) e(∞) = E.
4
= 40s.
b) ts = σ4 = 0.1
c)
1
Eo (s)
C2 s
=
Ei (s)
L2 s + R 2 +
1
C2 s
=
s2 +
1
LC
R
1
L s + LC
=
Kωn2
s2 + 2ζωn s + ωn2
d)
π π
= 0.05 ⇒ tan θ = −
= 1.0487
Mp = exp −
tan θ
ln(0.05)
θ = 46.36o = 0.81rad ⇒ ζ = 0.7
ωn =
0.1
σ
=
= 0.14
ζ
0.7
Using the value L = 5
R
= 2ζωn ⇒ R = 2 ∙ 0.7 ∙ 0.14 ∙ 5 = 0.98Ω
L
1
1
= 10F
= ωn2 ⇒ C =
LC
0.144 ∙ 5
4. STABILITY OF DYNAMICAL SYSTEMS
The first property they must have control systems is that they are stable. Moreover, in these systems, the closed loop operation is the risk that they are unstable
in open loop systems were stable.
The simplest concept of stability is the stability of input-output that is bounded
input bounded output will correspond, i.e. the gain is always finite.
If we consider the system of Figure 4.1, where G(s) and H(s) are linear or
nonlinear, the error e is e = r + H(s)y = r + H(s)G(s)e. The norm error is given
r(t)
+
e(t)
G(s)
y(t)
-
H(s)
Fig. 4.1: Feedback loop
by the expression
kek =
kuk
kuk
≤
k1 − G(s)H(s)k
1 − kG(s)kkH(s)k
for any norm we take.
If kG(s)kkH(s)k < 1 the system gain is finite and therefore the system is
input-output stable.
Theorem 4.0.1: Small gain theorem Consider the system of Figure 4.1. A sufficient condition for the system is input-output stable is that
kG(s)kkH(s)k < 1.
(4.0.1)
In the case that the system is linear, it can require a little less demanding condition,
it is sufficient that kG(s)H(s)k < 1.
Of the numerous methods to study the stability of dynamical systems, they
are going to be seen in this chapter some of the best known, both for linear and
nonlinear systems.
74
4. Stability of dynamical systems
Classical stability analysis methods
4.1
In linear systems, the stability of the system is determined by the position of the
closed-loop poles. Consider a system with an input and an output whose transfer
function in a closed loop
Qm
(s + zi )
Qi=1
M (s) = K Qn
p
2
2
(s
+
p
)
j
j=1
k=1 (s + 2ζk ωk s + ωk )
The impulse response is
y(t) =
k
X
Aj e
−pj t
+
j=1
p
X
Bk
k=1
1
ωk
e−ζk t sin(ωk t)
Therefore, for a bounded input bounded output to a need for both the real poles pj ,
as the real parts of the complex poles ζk are negative. The sufficient condition for
a linear SISO system (Single Input-Single Output) in continuous time, is stable is
that All of the poles are located in the left half-plane (negative real part).
The problem with this approach is that it is very difficult to apply when parameters. In this case we must apply other criteria such as Routh.
4.1.1
Routh’s method
Consider a continuous-time system with characteristic equation
p(s) = an sn + an−1 sn−1 + an−2 sn−2 + . . . + a1 s + a0 = 0
(4.1.1)
As prerequisites we require that:
1. All coefficients have the same sign
2. No coefficient is zero.
The Routh criterion is based on the construction of a table called Routh table and
which is constructed in the following way: The first row is formed by the coefficients an , an−2 , an−4 . . . and the second row of the remaining an−1 , an−3 , an−5 , . . . .
Namely
sn
sn−1
an
an−1
an−2
an−3
The remaining rows are constructed as
an−4
an−5
...
...
4.1. Classical stability analysis methods
sn
sn−1
sn−2
sn−3
...
s1
s0
an
an−1
bn−1
cn−1
...
u1
v1
75
an−2
an−3
bn−3
cn−3
...
u2
an−4
an−5
bn−5
cn−5
...
...
...
...
...
...
(4.1.2)
wherein the third row is
bn−1
bn−3
an an−2
an−1 an−3
1
=−
=
(an−1 an−2 − an an−3 )
an−1
an−1
an an−4
an−1 an−5
1
=−
=
(an−1 an−4 − an an−5 )
an−1
an−1
Analogously, the fourth row is
cn−1
cn−3
an−1 an−3
bn−1 bn−3
1
=−
=
(bn−1 an−3 − an−1 bn−3 )
bn−1
bn−1
an−1 an−5
bn−1 bn−5
1
=−
=
(bn−1 an−5 − an−1 bn−5 )
bn−1
bn−1
and so on down to zero degree.
Note 4.1.1: In developing the table, we can multiply or divide a row by a positive
constant without the result is changed (this is useful to simplify the table).
The Routh criterion states that the number of roots in the right half-plane
coincides with the number of sign changes of the coefficients of the first column.
That is, the system will be stable if all the signs in the first column are equal.
There are a number of remarkable special cases:
There is a zero in the first column. The value 0 is replaced by a very small
positive number ε and the development of the table continues. To count the
number of sign changes in the first column is necessary to consider that ε
goes to zero.
76
4. Stability of dynamical systems
There is a row of zeros. ( Poles symmetrically placed about the origin of the
complex plane).
The coefficients of the row above the row of zeros are considered. With
these coefficients the auxiliary equation is constructed, which will be the
grade that indicates the power of s respectively. The row of zeros is replaced
by the coefficients of the derivative and the method continues.
Example 4.1.1: Normal case. It is considered the system with characteristic
equation
a 0 s 3 + a 1 s2 + a 2 s + a3 = 0
with a0 , a1 , a2 y a3 positives.
The corresponding Routh table is
s3
s2
s1
s0
a0
a1
a1 a2 −a0 a3
a1
a2
a3
a3
For all coefficients in the first column be positive, shall be verified that
a 1 a 2 − a 0 a3
>0
a1
Therefore, the system will be stable when a1 a2 > a0 a3 .
Example 4.1.2: Case with a zero in the first column. Consider the system with
characteristic equation
p(s) = s4 + 2s3 + 4s2 + 8s + 5
The corresponding Routh’s table is
s4
s3
s2
s1
s0
1
2
0≈ε
4
8
5
5
8ε−10
ε
5
where the zero value of the first column by the small positive number is replaced ε.
= sign −10
= −1. Since there are two sign
When ε → 0, sign 8ε−10
ε
ε
changes in the first column, there are two poles in the right half, and therefore the
system is unstable.
4.1. Classical stability analysis methods
77
Example 4.1.3: Case with a zeros row. Consider a system with characteristic
equation
p(s) = s4 + 2s3 + 11s2 + 18s + 18 = 0.
Solution:
This system has its roots located symmetrically respect to the origin of the plane s:
±σ ± jω.
Routh’s table is
s4
s3
s2
s1
s0
1
2
2
0→ 4
18
11
18
18
0→ 0
18
The auxiliary equation associated with the third row is A(s) = 2s2 + 18 and
its derivative is A(s)
ds = 4s + 0 to replace the old values third row. Since all the
coefficients of the first column are positive, the system is stable.
Example 4.1.4: Consider the system of Fig. 4.2 Is desired to find the stability of
r(t) +
-
e(t)
1
s2+2s+5
K
y(t)
1
s+3
Fig. 4.2: System of example 5.1.4
the system as a function of the parameter K.
Solution:
The transfer function in closed loop is
K
Gc (s) =
KG(s)
K(s + 3)
s2 +2s+5
=
=
K
1
1 + KG(s)H(s)
(s + 3)(s2 + 2s + 5) + K
1 + s2 +2s+5 s+3
78
4. Stability of dynamical systems
Equating the denominator to zero the characteristic equation is obtained, whose
roots are the poles of the closed loop system:
p(s) = (s + 3)(s2 + 2s + 5) + K = s3 + 5s2 + 11s + 15 + K = 0
The precondition requires that all the coefficients are of the same sign:
15 + K > 0 ⇒ K > −15
Routh’s table is
s3
s2
s1
s0
1
5
11
15 + K
40−k
5
15 + K
For the system to be stable it is necessary that all the coefficients of the first
column be greater than zero:
40 − K
> 0 ⇒ 40 − K > 0 ⇒ K < 40
5
Putting together the two conditions, we find that the system is stable for
−15 < K < 40.
4.1. Classical stability analysis methods
79
Bibliografia
[1] E. A. Puente, Regulacion Automatica Seccion de Publicaciones de la ETSIIM, 1993.
[2] E. Umez-Eronini, Dinamica de Sistemas y Control, Thomson Learning, 2001.
[3] S. Barnett y R.G. Cameron, Introduction to Mathematical Control Theory,
Oxford University Press, 1992.
[4] T. Glad y L. Ljung, Control Theory, Taylor and Francis, 2000.
[5] JJ.E. Slotine y W.Li, Applied Nonlinear Control, Prentice Hall, 1991.
80
4. Stability of dynamical systems
5. ERRORS IN STEADY STATE
5.1
Introduction to errors
Consider the system errors to the standard inputs. The behavior of different
systems with the same input signal can be very different. As an example, consider
the behavior of an engine when you want to control it in position and speed:
Control of the axis position of a DC motor
Fig. 5.1:
By varying the reference signal:
• varies the output of the adder (error)
• is given to the motor voltage
• The motor rotates
• When the motor position coincides with the reference output the adder is
zero
• The engine is off power
• (possible oscillations)
• The motor is stopped at the position indicated
The steady system output matches the reference signal
82
5. Errors in steady state
Speed control of a DC motor
Fig. 5.2:
By varying the reference signal:
• varies the output of the sum block (error)
• voltage to the motor is given
• The motor rotates
• When the engine speed matches the reference output the sum is zero
• The engine power is off
• The engine brakes
• varies the output of the sum
• ...
• The system will stabilize at a speed different from the reference (If the error
were zero, it would stop)
The steady system output does not match the reference signal
These two examples show the different behavior of the two systems to the step
input. While in the case of motor position control, the error in the steady may be
zero, in the case of speed control, the error must be nonzero.
5.1.1
Gain position (static)
Constant position error is the value of the steady output when the input is a unit
step, ie Kp = lim y(t). It is normally calculated as
t→∞
Kp = lim G(s)
s→0
Is only meaningful in stable systems.
Interpretation
Kp = lim y(t) = lim s
t→∞
s→0
G(s)
= lim G(s)
s→0
s
5.1. Introduction to errors
5.1.2
83
Gain speed
Constant speed error as the value of the steady output when the input is a unit
ramp, ie Kv = lim ẏ(t). It is normally calculated as
t→∞
Kv = lim sG(s)
s→0
Interpretation
Kv = lim y 0 (t) = lim s s
t→∞
5.1.3
s→0
G(s)
= lim sG(s)
s→0
s
Gain acceleration
Error constant acceleration steady as the value of the output is set when the
input is a unit parabola, namely Kv = lim ÿ(t).
t→∞
It is normally calculated as
Ka = lim s2 G(s)
s→0
Ka = lim y 00 (t) = lim s s2
t→∞
5.1.4
s→0
G(s)
= lim s2 G(s)
s→0
s
Type system
Is named type of a system the number of integrators of the system (poles at the
origin). That is, if the transfer function is factored as
Q
(s − zi )
G(s) = K r Q
s
(s − pi )
84
5. Errors in steady state
the system type is r.
Examples
G(s) =
G(s) =
G(s) =
G(s) =
5.1.5
3
Type 0 (order 1)
s+2
s+1
Type 1 (order 2)
s(s+3)
4
Type 2 (order
s2 (s+2)(s+1)
5
Type 3 (order 3)
s3
4)
Relationship between type and gain of a system
The relationship between gain and type of system is:
Kp Type = 0
Kp = lim G(s) =
∞ Type > 0
s→0

0 Type = 0
Kv = lim sG(s) = Kv Type = 1
s→0

∞ Type > 1

0 Type < 2
2
Ka = lim s G(s) = Ka Type = 2
s→0

∞ Type > 2
Fig. 5.3: Relationship between type and gain a system
5.1. Introduction to errors
5.1.6
Relationship between type and gain a system
Fig. 5.4: Relationship between type and gain a system
85
86
5. Errors in steady state
5.1.7
Steady state error of a unit feedback system
Consider a unitary feedback system as shown on Fig. 5.5
Fig. 5.5: Unit feedback system.
The error signal is
e(t) = r(t) − y(t)
in the frequency domain is
E(s) = R(s) − Y (s) = R(s) −
E(s) =
G(s)
R(s)
R(s) =
1 + G(s)
1 + G(s)
R(s)
1 + G(s)
and the steady-state error
erp = lim e(t) = lim sE(s) = lim s
t→∞
s→0
erp = lim s
s→0
5.1.8
s→0
R(s)
1 + G(s)
R(s)
1 + G(s)
(5.1.1)
Errors to standard input
To find the steady-state response of the systems the value of the steady-state
error to the standard input step, ramp and parabola is calculated.
• Unit step input
The Laplace transform of the unit step is
R(s) =
1
s
Substituting in the expression of the steady-state error 5.1.1
erp = ep = lim s
s→0
R(s)
1
1
1
=
= lim
=
1 + G(s) s→0 1 + G(s)
1 + lim G(s)
1 + Kp
s→0
(5.1.2)
This error is called position error.
5.1. Introduction to errors
87
Fig. 5.6:
• Unit ramp input
The Laplace transform of the unit ramp is
1
s2
Substituting in the expression of steady-state error 5.1.1
R(s) =
erp = ev = lim s
s→0
1
1
1
R(s)
=
= lim
=
1 + G(s) s→0 s + sG(s)
lim sG(s)
Kv
s→0
(5.1.3)
This error is called velocity error.
Fig. 5.7:
• Unit parabola input
The Laplace transform of 1/2 unit parabola is
1
2
r(t) = t2 ⇒ R(s) = 3
R(s) = 3
s
s
Substituting in the expression of steady-state error 5.1.1
erp = ea = lim s
s→0
1
1
1
R(s)
=
=
= lim 2
2
2
s→0
1 + G(s)
s + s G(s)
lim s G(s)
Ka
s→0
(5.1.4)
This error is called acceleration error.
88
5. Errors in steady state
Fig. 5.8:
5.1.9
The relationship between errors and system type
The relationship between the constant error and system type is represented in
the table 5.9
Fig. 5.9: The relationship between errors and system type.
and the relationship between errors and the types in the following table
ep
Type 0
Type 1
Type 2
5.1.10
1
1+kp
0
0
e v ea
∞ ∞
1
∞
kv
0 k1a
Error input for polynomial in t
A lot of input signals can be approximated by a polynomial entry. For example,
if the input is
r(t) = 1 + 3t + 5t2
5.1. Introduction to errors
89
then the error in steady state is a composition of the errors of position, speed and
acceleration
1
3
10
+
+
erp = ep + 3ev + 10ea =
1 + k p kv
ka
90
5. Errors in steady state
6. ROOT LOCUS
In control theory and stability theory, root locus analysis is a graphical method
for examining how the roots of a system change with variation of a certain system
parameter, commonly a gain within a feedback system. This is a technique used
as a stability criterion in the field of control systems developed by Walter R. Evans
which can determine stability of the system. The root locus plots the poles of
the closed loop transfer function in the complex S plane as a function of a gain
parameter.
The dynamical behavior of a system is determined by the position of its poles.
The place of the roots to determine the position of the poles in closed loop varies
according constant K the system.
6.1
Concept of Root Locus
Consider the closed loop system of the figure
Its transfer function is closed loop
M (s) =
G(s)
1 + G(s)H(s)
Factoring the transfer function in open loop
Q
(s − zi )
G(s)H(s) = K Q
(s − pi )
If we change the value of K, the poles of M (s) vary. (zeros remain unchanged).
The Root Locus of M (s) is the place followed by its poles when K vary
from zero to infinity.
92
6. Root Locus
The poles of F (s) are the roots of 1 + G(s)H(s) = 0, nad this is called
characteristic equation.
Q
(s − zi )
1+KQ
=0
(s − pi )
(6.1.1)
where K is proportional to the static gain.
6.2
Magnitude and Angle Condition (or Module and Argument
Criteria)
From factoring in open chain
Q
(s − zi )
G(s)H(s) = K Q
(s − pi )
The characteristic equation is
Q
(s − zi )
Q
1+K
=0
(s − pi )
namely,
Q
(s − zi )
Q
K
= −1
(s − pi )
As this equation is complex, it can be expressed as two real equations, first
magnitudes (modules) and secondly the angles (arguments).
The equation of the magnitudes
Q
|s − zi |
Q
=1
K
|s − pi |
It is usually written as
Q
|s − pi |
product of the distances to the poles
K= Q
⇒K≡
|s − zi |
product of the distances to the zeros
(6.2.1)
This equation is the Magnitude Condition.
Furthermore, the equation of the angles is
X
X
arg(s − pi ) = (2q + 1)π
arg(s − zi ) −
that it is the Angle Condition and is usually written in the form
X
arg(s − pi ) −
X
arg(s − zi ) = (2q + 1)π
(6.2.2)
The meaning of the magnitudes and the angles of these formulas can be in Fig.
6.1.
6.3. Tracing Rules of the Root Locus.
93
Fig. 6.1: Meaning of the magnitudes and angles.
6.3
Tracing Rules of the Root Locus.
• Number of branches
If
n = number of poles of G(s)H(s)
m = number of zeroes of G(s)H(s)
then
Number of branches = max number of poles and number of zeroes of M(s)
M (s) = max(m, n)
This is because the characteristic equation
Q
(s − zi )
=0
1+KQ
(s − pi )
This is because the characteristic equation
Y
Y
(s − zi ) = 0
(s − pi )+K
is a real polynomial coefficients whose degree is equal to the number of
roots (Fundamental Theorem of Algebra). Each of these roots, to vary K,
produces a branch.
• Rule 2. Initial and Final points
For K = 0, the characteristic equation is
Y
(s − pi ) = 0
94
6. Root Locus
That is the Root Locus, starts in the closed-loop poles.
On the other hand, if K = ∞, the characteristic equation can be written as
and as
1
K
=0
Y
1 Y
(s − zi ) = 0
(s − pi )+
K
K=∞⇒
Y
(s − zi ) = 0
meaning that the place of roots ends in open loop zeros.
The difference between poles and zeros corresponds to points in the infinite.
• Rule 3. Real axis.
If we apply the angle condition to a point of the real axis we get:
X
X
arg(s − zi ) = (2q + 1)π
arg(s − pi ) −
The contribution of any of these terms if the real root is on the right is π.
Fig. 6.2:
The contribution of any real root on the right is 0.
Fig. 6.3:
Finally, the contribution of any pair of complex conjugate roots is es π −π =
0.
6.3. Tracing Rules of the Root Locus.
95
Fig. 6.4:
As a consequence, to satisfy the angle condition we must have an odd number of roots on the right side of the point s.
Fig. 6.5:
A point on the real axis will belong to the root locus if it has an odd number
of real roots on the right.
• Rule 4. Symmetry of the Root Locus
The root locus is symmetric w.r.t. to the real axis.
• Rule 5. Asymptotes
Fig. 6.6:
96
6. Root Locus
If we consider a point that is far, as is shown in Figure Fig. 6.6 point at
infinity, then all angles of view of the argument
X
X
arg(s − pi ) −
arg(s − zi ) = (2q + 1)π
They can be considered equal, and therefore
X
arg(s − pi ) −
namely,
X
arg(s − zi ) = nθa − mθa = (2q + 1)π
θa =
2q + 1
π
n−m
(6.3.1)
• Rule 6. Centroid
The cutting point of the asymptotes. This point is called ‘centroid’.
σ0 =
P
P
pi − z i
n−m
(6.3.2)
• Rule 7. Angles of departure and arrival
Departure angles:
If we take a point infinitely close to a pole in a departure branch.
Fig. 6.7:
X
arg(s − pi ) −
X
arg(s − zi ) =
X
αi + θ −
X
βi = (2q + 1)π
6.3. Tracing Rules of the Root Locus.
97
θ = (2q + 1)π −
Arrival angles:
X
αi +
X
βi
(6.3.3)
The calculation is the same with a point in an arrival branch.
• Rule 8. Points of dispersion and confluence
They coincide with local maxima (dispersions) and minima (confluence) of
K on the real axis
dK(s)
=0
ds
or also
X
(6.3.4)
X 1
1
=
σ − pi
σ − zi
• Rule 9. Intersection with the imaginary axis
It is calculated by using the Routh method.
• Rule 10. Value of K in a point of the Root Locus
It is calculated by using the magnitude condition.
• Rule 11. Sum of the roots
Q
|s − pi |
K= Q
|s − zi |
If the characteristic eqn. is
sn + an−1 sn−1 + an−2 sn−2 + ... + a1 s + a0 = 0
We have
X
raices = − an−1
(6.3.5)
98
6. Root Locus
Fig. 6.8: Tracing rules for Root Locus.
7. CLASSICAL CONTROL TECHNIQUES
7.1
Problem approach
The objective of any control system is that given a certain system, which provides a certain amount of information based on a series of measures of the input /
output signals, try to determine a certain control inputs feasible so that the system
variable to be control continues to track as accurate as possible a certain reference
signal, although the influence of the possible disturbances, measurement errors and
variations in load system.
A very general process control scheme Gp is displayed on the figure7.1.
r
Gf
u
+
y
Gp
-
Gc
Fig. 7.1: Block diagram of a general control scheme
In this scheme, the control signal is
U (s) = Gf (s)R(s) − Gc (s)Y (s)
(7.1.1)
We can see a part formed by a controller Gc located in the feedback and a filter
Gf at the input of system. This control scheme has as joint transfer function of the
entire system the following expression
Gp (s)
Y (s)
.
= Gf (s)
1 + Gp (s)Gc (s)
R(s)
(7.1.2)
In this expression it can be seen that the term filtering Gf directly affects the
steady-state gain of the feedback system as well as to the poles and zeros of the
same to be located in the main chain loop before the feedback. However, does
not change the positions of the poles and zeros of the feedback loop , except that
pole-zero cancellations occur between Gf and the closed loop transfer function.
100
7. Classical Control Techniques
Moreover the term Gc to be located in the feedback loop significantly affect
the positions of the poles of the closed loop system, altering system dynamics
. Simplifying a bit , we may think that while the design Gf adjust or improve
steady-state response of the system, is sought Gc adjust the transient response.
A particular case of this control structure is when Gf = Gc . In this case,
the freedom in design is lost as it only has a transfer function . Thus, the control
scheme is simplified as can seen in the diagram (a) of Figure ?? , being able to be
expressed by the equivalent circuit diagram ( b ) of this Figure.
r
Gc
u
+
y
Gp
(a)
Gc
r
+
-
u
e
Gc
y
Gp
(b)
Fig. 7.2: Controller in the principal loop
This equivalent circuit (b) is the we could call classic control scheme, at least
academically. Within this group we can include controllers PID control, which is
what we will study in this chapter.
7.2
Design of classical controllers
Among the drivers that match the control scheme Classic can distinguish two
basic approaches when address controller design Gc . These two approaches are
fundamentally different ways of defining the function transfer controller .
begin enumerate item controller with fixed structure. This group of techniques are characterized by using a transfer function defined a priori. This group
includes the so-called PID control. In these controllers the transfer function of the
controller is fixed and in them are present: proportional control to error ( P action
), proportional control to the integral of error ( action I) and proportional control to
the derivative of the error ( action D ).
The weight given to each of the control actions can be determined from theoretical form ( in the case of the known transfer function of the process) or the
empirical form if such transfer function is not known. Clearly it can happen that
7.3. Specifications
101
one or more of the parameters of control override some actions if they do not will
be necessary.
item controller with variable structure. This second controllers group has
not a predefined structure of its transfer function, but this result is obtained using
the desired specifications for the system and the transfer function of the process
to be controlled. This is the case of the design of controllers by direct synthesis
technique. end enumerate
7.3
Specifications
The specifications express the characteristics that are desired to achieve in the
system when a controller is introduced therein. Can be expressed in different ways
using for this the frequency domain or the time domain. In our case we will refer
to domain specifications time.
Specifications regarding the domain response of time of a system are often defined with respect to the response time to a step input of a second system order. For
this purpose, taking as reference system a system with theoretical transfer function:
G(s) =
Kωn2
s2 + 2ζωn s + ωn2
(7.3.1)
cuya respuesta a un escalon es
1
y(t) = 1 − p
1 − ζ2
con
e−ζωn t sin(ωn t
φ = arctan
p
p
1 − ζ 2 + φ)
(7.3.2)
1 − ζ2
ζ
(7.3.3)
p
(7.3.4)
π
ωd
(7.3.5)
This system has its poles located at
p1,2 = −ζωn ± jωn
1 − ζ2
where ζ is called damping ratio and damping ratio ωn is the undamped natural
frequency index undamped natural frequency of the system. For a system with
damping ratio 0 < ζ ≤ 1 (underdamped system) the system response curve has the
form shown in Figure7.3
Among the parameters that can be used to specify the response can be included
the following:
• Peak time,
Tp =
where ωd is the damped frequency ωd = ωn
p
1 − ζ 2.
102
7. Classical Control Techniques
y
To
d·Mp
Mp
ep
Tp
t
Ts
Fig. 7.3: Parameters of the time response
• Overshoot, Mp .
• Settling time, Ts .
Mp = e
−π √
Ts '
ζ
1−ζ 2
π
ζωn
=
√
d
(7.3.6)
(7.3.7)
Finally, the position error in steady-state Ep is defined as the difference between the desired reference value and the output value of the controlled variable of
the system (for simplicity we are assuming that the feedback signal is unitary).
It is relatively easy to obtain, from a certain specifications, the positions which
must have the poles in the closed loop system that meet these specifications.
7.4
PID controllers
PID controllers are by far the most frequently used for industrial process control , where more than 95% of the control loops used are PID controllers. Their
simplicity, versatility and ability to solve the basic problems that arise in the process favorable dynamics and modest performance requirements make them an essential tool in controlling processes.
Over the past decades the PID controllers have survived various technological
changes ranging from first controllers developed from pneumatic elements to the
microprocessors developed going by electronic valves, the analog devices, transistors and integrated circuits . The incorporation of microprocessors has had a big
impact because it has allowed the PID controllers to be enriched in other functions
without losing any of their properties, so they have added possibility of automatic
7.4. PID controllers
103
adjustment of the parameters , possibilities of execution of logical rules or sequential automatisms.
Therefore, and although there are more sophisticated control techniques, in the
lowest level are still present or even is the majority of control processes. In the next
few pages we will refer to them.
7.4.1
Basic structure of a PID controller
The academic version of this driver is characterized by a control law of the
form:
Z
1 t
de(t)
(7.4.1)
u(t) = K e(t) +
e(τ )dτ + Td
Ti 0
dt
where u(t) is the control variable and e(t) is the error expressed as e(t) = r(t) −
y(t) (Figure 7.2 b). The expression 7.4.1 shows the three types of control actions:
the proportional action with weight K, the integral action with weight TKi and the
derivative action with weight KTd .
To set the behavior of the controller has three parameters: the proportional gain
K, the integral time Ti and the derivative time Td . It can be seen that if the time
integrated becomes infinite, the integral action disappears and if the time derivative
becomes zero the derivative action disappears.
The transfer function of the PID controller is as follows
U (s)
K(1 + sTi + s2 Ti Td )
1
+ Td s] =
= K[1 +
E(s)
Ti s
sTi
(7.4.2)
In this controller, the setting seeks to determine the positions of the two zeros
of the transfer function and the static gain of the same, so that the desired design
specifications are met the best possible in the system .
Equation 7.4.2 is not physically realizable , so it is called theoretical PID controller. For a PID to be physically realizable there will be added to the derivative
action a pole with limited influence. A real PID controller will have the form
Gc (s) = K
1 +2 Ti Td sTi + s
(1 + sTi )(1 + sTd )
(7.4.3)
The adjustment of the parameters of the controller can be done in two ways :
1. Empirically . For what it is adjusted experimental values ??to the desired
response. This system may be too slow in many systems. Especially if the
response time of these is great.
To avoid this problem, the Ziegler-Nichols methods are used. These methods
are based on some very simple measures observed in the response of the system and give some theoretical values ??of the parameters of the controller.
104
7. Classical Control Techniques
Values ??are taken as reference and then they require a finer adjustment.
These methods and others derived from them are considerably used in industry, both for purely manual adjustment and industrial PID controller with
auto-tuning systems.
2. Theoretically. To determine what are the analytical values ??of the regulator. The explicit knowledge of the transfer function of the process is required.
7.4.2
Ziegler-Nichols methods
The Ziegler -Nichols methods are two classic methods for the empirical adjustment of the parameters of a PID controller. They were presented by these authors
in 1942. These methods are still widely used, either in its original form or with
improved versions.
Both methods are based on the determination of some characteristics of the
process response, (time or frequency), to determine from these features and by
means of simple mathematical relationships the controller parameters that givie a
response rate of decay of a fourth between the values ??of the first and the second
overshoot .
Step response Tuning PID method
Step response Tuning PID method
This first method is based on the observation of the open loop response of the
system to a step input. Noting this response, two parameters that are obtained are
determined as follows :
1. The point of maximum slope is determined by the response curve to a step,
and at that point the tangent is drawn to the response curve at that point.
2. The intersection of the tangent obtained above with the coordinate axes is
determined and the distances a and L. These two parameters correspond to
the theoretical response of a mathematical model of the form
G(s) =
a −sL
e
sL
(7.4.4)
corresponding to a time delay integrator. This system can be characterized
by two parameters a and L as can be seen in the figure 7.4.
3. Once you have determined the parameters a and L in response, Ziegler and
Nichols proposed as the PID controller parameters indicated in Table 7.1
obtained directly as a function of the parameters a and L measured on the
system response.
7.4. PID controllers
105
y
L
t
a
Fig. 7.4: Response to step input of G(s) =
Controller
P
PI
PID
K
1/a
0.9/a
1.2/a
Ti
Td
3L
2L
L/2
a −sL
sL e
Tab. 7.1: Parameter values given by Ziegler-Nichols method for the step response
This controller is designed to give a decay rate of a quarter ( d = 0.25) between
the magnitudes of the first and second overshoot, so it usually has a high overshoot.
It has the advantage that from these values ??is easy to make a finer adjustment to
fit the response you want without the need of a long process of trial and error .
subsubsection Method of the frequency response index method to adjust
PID’s using the frequency response
This second method developed by Ziegler and Nichols is based on determining
the point on the Nyquist curve for the transfer function G(s), in which that curve
intersects the negative real axis ( point of relative stability ) . Once obtained this
point is characterized by two parameters named : Gain critical Ku and critical
period Tu .
This method is based on the observation that many systems become unstable
when proportional control has high gain. The method of determination of the parameters of the PID controller has the following steps:
1. the feedback loop is closed by a proportional regulator (Figure 7.5a). It
necessary to increase the gain of the proportional controller until you get the
critical gain Ku that correspond to an oscillatory response as indicated in
the Figure 7.5. For this gain Ku , is determined the Tu critical period of the
oscillatory output signal.
2. As in the first method, are determined the values of the parameters of the con-
106
7. Classical Control Techniques
r
+
-
e
u
K
y
G(s)
a)
y
K<Ku
b)
K=Ku, Tu
c)
K>Ku
d)
t
y
t
y
t
Fig. 7.5: Obtaining the critical gain and critical period for the second method of
Ziegler-Nichols
troller from from Ku and Tu . The values proposed by Ziegler and Nichols is
shown in Table 7.2.
As in the previous method the design criterion is to regulators decay rate of
a quarter. These controllers give good response to changes in system load, but
produce frequently a high overshoot, as already discussed above. In any case, it’s
easy to make a further adjustment of the regulatory on the basis of these parameters.
7.5. Analytical methods for PID controllers design
Controlador
P
PI
PID
K
0.5Ku
0.4Ku
0.6Ku
107
Ti
Td
0.8Tu
0.5Tu
0.125Tu
Tab. 7.2: Parameter values given by Ziegler-Nichols frequency method
7.5
Analytical methods for PID controllers design
These methods seek to determine the most appropriate form analytical values ??of the parameters of a PID controller so that certain specifications are met.
See two analytical techniques for calculating these parameters: the first of these
techniques (pole assignment) is applicable to systems of second order, and the second (design based on the dominant poles) is applicable to complex systems. Both
techniques require prior knowledge of the transfer function of the process to be
controlled.
7.5.1
Pole placement method for adjusting PID
Pole placement method for adjusting PID
The objective of the technique of assignment of poles index pole assignment is
to find the parameters of a controller such that the transfer function in closed loop
has its poles in the predetermined positions. Closed loop poles are determined so
that a certain specifications are obtained in the system response. In general terms a
regulator of type P to have a single parameter alone would allow us to assign only
a pole, a PI controller or a PD having two parameters would allow us to assign two
poles and a PID controller has three parameters that would allow us to assign three
poles. The use of either type of controller depends on the transfer function of the
system.
Suppose first a process whose transfer function is first order, with the following
parameterization
Kp
Gp (s) =
(7.5.1)
1 + sT
and suppose we want to control the system through a PI controller type, which
responds to the following expression
1
(7.5.2)
Gc (s) = K 1 +
1 + sTi
the closed loop system has the following transfer function
G(s) =
Gp (s)Gc (s)
1 + Gp (s)Gc (s)
(7.5.3)
108
7. Classical Control Techniques
The characteristic equation of the closed loop system defined by the equation
7.5.3 is the following
1 + Gp (s)Gc (s) = 0
(7.5.4)
and substituting in 7.5.4 Gp (s) y Gc (s) by the expessions 7.5.1 y 7.5.2 obtain that
Kp
1
1+K 1+
=0
sTi (1 + sT )
1 + Kp K
Kp K
=0
+
s2 + s
T
T Ti
(7.5.5)
Suppose we want the response of closed loop system is that of a second order
system with a damping rate ζ and frequency ωn . That is, you want
p to place the
poles in the closed chain system at positions p1,2 = −ζωn ± ωn 1 − ζ 2 . So the
closed loop characteristic equation will have the expression
s2 + 2ζωn s + ωn2 = 0
(7.5.6)
Igualando las expresiones 7.5.5 y 7.5.6 obtenemos que
1 + Kp K
T
K
K
p
ωn2 =
T Ti
2ζωn =
(7.5.7)
(7.5.8)
y despejando en 7.5.7 y 7.5.8 tenemos que
2ζωn T − 1
Kp
2ζωn T − 1
Ti =
ωn2 T
K=
(7.5.9)
(7.5.10)
This technique requires the introduction of a controller which have two possible
adjustment parameters, a PI controller, when the open-chain system is first order
since it is necessary to fix arbitrarily two pole positions in a closed loop. Similarly,
when the system is second order, it is usually convenient to use a PID controller
that has three degrees of freedom to set arbitrarily the three poles that will have the
closed-loop system.
Let the transfer function of the second order system of the form
Kp
(7.5.11)
(1 + sT1 )(1 + sT2 )
and suppose we want to control the system via a PID controller, which responds to
the following expression
Gp (s) =
Gc (s) = K
(1 + sTi + s2 Ti Td )
sTi
(7.5.12)
7.5. Analytical methods for PID controllers design
109
so that the characteristic equation of the closed loop system 1 + Gp (s)Gc (s) = 0,
will be
Kp KTd
Kp K
Kp K
1
1
1
3
2
+
+
+
= 0 (7.5.13)
s +s
+s
+
T1 T 2
T1 T 2
T1 T2
T i T2
T 1 T2
If you want the system to have the following closed-loop characteristic equation,
(s + α)(s2 + 2ζωn s + ωn2 ) = 0
(7.5.14)
equating coefficients of the polynomial expressions 7.5.13 y 7.5.14 we have the
following expressions
Kp K
T T
1 2
Kp K
1
ωn2 + 2ζωn α=
+
T 1 T2
T 1 T2
Kp KTd
1
1
+
+
2ζωn + α=
T i T2
T1 T 2
ωn2 α=
(7.5.15)
(7.5.16)
(7.5.17)
and solving this system of equations we obtain
T1 T2 (2ζωn + α) + 1
Kp
Kp K
Ti =
T1 T2 ωn2 α
T1 T2 ωn3 α(ωn + 2ζα)
Td =
T1 T2 (2ζωn + α) − 1 + T1 ωn2 α(ωn2 α + 1)
K=
(7.5.18)
(7.5.19)
(7.5.20)
Example 7.5.1: Suppose you want to control a system whose transfer function
1
is GP (s) = s(s+2)
so that has a closed loop damping rate ζ = 0.5 and a frequency
ωn = 3.14 rad
s .
To control this system we need is a PD type controller with transfer function of
the form
Gc (s) = K(1 + Td s)
(7.5.21)
because the system is second order but has a pole at the origin in the transfer function, so that the system does not require adding an integral effect on the controller
to override the position error in steady-state. By not including regulatory effect on
the whole system will have a closed loop characteristic equation of order 2. The
closed loop system has a characteristic equation of the form
110
7. Classical Control Techniques
1 + K(1 + Td s)
1
=0
s(s + 2)
(7.5.22)
and you want to have the characteristic equation ?? closed loop. To do this we
equate 7.5.6 to 7.5.22,
s2 + s(2 + KTd ) + K = s2 + 2ζωn s + ωn2 = 0
(7.5.23)
and equating coefficients of two polynomials have to
K=ωn2
2ζωn − 2
Td =
ωn2
(7.5.24)
(7.5.25)
and substituting the values ??of the specifications we have indicated that K = 9.86
and Td = 0.115. So that the the resulting controller has the expression:
Gc (s) = 9.86(1 + 0.115s)
(7.5.26)
2.5
y(t)
2
(a)
1.5
1
(b)
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
t
5
Fig. 7.6: Response system of Example 7.5.1: (a) open loop and (b) closed loop
with the designed controller
7.5.2
Method based on the dominant poles for adjusting PID
The pole assignment technique is to locate the positions of all the poles in
the closed loop system. This complicates considerably the method for the case of
7.5. Analytical methods for PID controllers design
111
complex systems, the problem may even have no feasible solution. Thus, if you
want to control with a PID controller a system through this technique, the system
should be restricted to first and second order to assures the existence of a solution.
If the system is complex, or we are looking for an approximation of the model
to a first or second order, this is the technique commonly used design based on the
dominant poles . This technique assigns only a few poles of the system.
This allows to design simple controllers for controlling complex systems, although the controller behaviour do not assure the given specifications since the
design ignores the effect of some poles. However, if the choice of positions is
appropriate, the behaviour will be very close to what is desired .
This technique also requires that the transfer function of the system to be
known, and the idea of assigning only a few poles of the closed loop system has
been widely used in the early design of PID control, especially when the power of
computers did not allow other possibilities.
Even today , this technique has considerable interest since many industrial controllers implement a fixed PID structure and are used to be applied to processes of
variable complexity.
This allocation based on the dominant poles can be performed in two ways:
either by determining the values of the controller parameters algebraically which
place the dominant poles in the desired positions or by using the Root Locus. Both
techniques allow us to obtain analytically the parameters of the PID controllers.
Method based on algebraic determination for adjusting PID
The problem is similar to the previous section. If you enter P controller, we
need to put a pole in order to determine the controller parameter. If you want to
introduce a PI or PD controller is necessary to assign two poles to calculate the
parameters controller and a PID controller if it is necessary to set three poles.
Suppose, for illustrating the operation of this technique, it is desired to design
a PI controller for controlling a system. This controller has two parameters by
which to determine the value of them need to locate the position of both poles of
the closed loop system.
Assume the controller equation is of the form
1
Gc (s) = K 1 +
1 + sTi
(7.5.27)
the closed loop system has the following transfer function
G(s) =
Gp (s)Gc (s)
1 + Gp (s)Gc (s)
(7.5.28)
112
7. Classical Control Techniques
so that the characteristic equation of the closed loop system is defined by
1
1 + Gp (s)K 1 +
1 + sTi
=0
(7.5.29)
We need to set two pole positions p1,2 , so we can assume that these are complex
conjugates so that they have a certain damping ζ and a certain frequency ωn , ie
p
(7.5.30)
p1,2 = ωn (−ζ ± j 1 − ζ 2 )
In the event that the driver you need is type of PD are in a similar situation,
because we have to determine two parameters of the controller, whose transfer
function is
Gc (s) = K(1 + sTd )
(7.5.31)
If what you want is to design a PID controller to control such a system. This
controller has three parameters by which to determine the value of them need to
locate the position of three poles of the closed loop system.
Suppose that the controller equation is of the form
1
+ sTd
Gc (s) = K 1 +
1 + sTi
(7.5.32)
the closed loop system has the following transfer function
G(s) =
Gp (s)Gc (s)
1 + Gp (s)Gc (s)
(7.5.33)
so that the characteristic equation of the closed loop system is defined by
1
+ sTd
1 + Gp (s)K 1 +
1 + sTi
=0
(7.5.34)
We need to set three pole positions p1,2,3 , so we can assume that two of these are
complex conjugates so that they have a certain damping ζ and a certain frequency
ωn , and the other is a pole located at −α ie
p
p1,2 =ωn (−ζ ± j 1 − ζ 2 )
(7.5.35)
p3 =−α
(7.5.36)
The determination of the values of the parameters of the PID controller would
be analogous to that seen for controllers with two parameters.
7.5. Analytical methods for PID controllers design
113
Fig. 7.7: System of the example 7.5.2
Example 7.5.2: Consider the system shown in Figure 7.7, whose transfer function is
1
Gp (s) =
(7.5.37)
s(s + 2)(s + 8)
and suppose we want to introduce a controller in the system so that the system
has a damping rate ζ = 0.5 and ωn = 3.14. If we fix only the dominant poles,
these specifications would have to said poles should be
p1,2 = ωn (−ζ ± j
p
1 − ζ 2 ) = −1.57 ± j2.72
(7.5.38)
Since the system is of type 1, which includes an integrator in open chain, introduce a PD type controller in the K(s + a) system. First obtain the characteristic
equation of the closed loop system
1 + K(s + a)
1
= s3 + 10s2 + (16 + K)s + Ka = 0 (7.5.39)
s(s + 2)(s + 8)
Moreover, we fix the positions of two of the poles at positions 7.5.38. As the thirdorder system is the third pole will be in a position (s + b) so that the closed-loop
characteristic equation will have the following appearance
(s+b)(s2 +2ζωn s+ωn2 ) = s3 +(2ζωn +b)s2 +(2ζωn b+ωn2 )s+ωn2 b = 0 (7.5.40)
By matching the characteristics equations 7.5.39,7.5.40 we obtain the following equations
10=2ζωn + b
16 + K=2ζωn b + ωn2
(7.5.41)
Ka=ωn2 b
and solving for in 7.5.41 we have:
b=10 − 2ζωn = 6.86
K=2ζωn b + ωn2 − 16 = 15.4
ω2 b
a= n = 4.392
K
(7.5.42)
114
7. Classical Control Techniques
so that the PID controller that puts the dominant poles at desired positions in closed
chain is
Gc (s) = 15.4(s + 4.392)
(7.5.43)
Note that the position of the third pole in b = 6.86 is far away from the dominant
poles, so their influence is very small.
1.4
y(t)
1.2
(b)
1
0.8
0.6
0.4
(a)
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
t (s)
5
Fig. 7.8: Response system of Example 7.5.2: (a) open loop and (b) closed loop
with the controller designed
Method based on the Root Locus for adjusting PID Method
The method of Root Locus , introduced by Evans in 1954, is a graphical
method to determine the positions of the poles in closed loop from the positions of
the poles and zeros of the system in open open as any of the parameters (typically
the gain) varies from zero to infinity .
In practice, the drawing of the Root Locus of some system can tell us if it will
be sufficient to vary gain system to meet the specifications that have given us. That
is, in case that the location of the roots pass through the desired locations for the
poles closed chain with enough to adjust to the proper value the system gain to
meet the required transient response specifications. If this not, it will be required
to the system controller to alter the place of the required roots in the right way.
If we remember the effect that had the addition of poles and zeros system is
relatively simple to determine what type of structure must have the driver that we
introduce in the system.
7.5. Analytical methods for PID controllers design
115
• Effect of addition of poles. The addition of a pole to the transfer function
of an open-loop system has the effect of adding a branch more to the Root
Locus and branches as a result tend to approach or even enter the positive
real half-plane. This tends to decrease the relative stability of the system and
that the settling time of the response is longer.
ω
ω
ω
Fig. 7.9: Effect of the addition of a pole on the Root Locus
This effect can be seen in Figure 7.9, where poles are added as the number
of branches and these are increasingly closer to the half-plane with positive
real part. Coming in case (c), the introduction of the branches on the positive
real part, for high values ??of gain and to become the system unstable.
• Effect of the addition of zeros. The addition of a zero to the transfer function in open loop decreases the number of branches that go to infinity so
increases the relative stability of the system to move the branches to the negative real part of the complex plane. The system as a faster response by
introducing a derivative effect that anticipates the evolution of the system.
This effect can be seen in Figure 7.10, where adding a zero to a system with
three poles as shown in 7.9c only two branches go to infinity and none of
them comes in part positive real, so that the system becomes unstable for
any value of the gain.
Example 7.5.3: For the system of Figure 7.7 can be seen that if a proportional
controller is introduced and the value of gain K is varied in
1 + KGp (s) = 0
(7.5.44)
the poles of the characteristic equation of the system move through the branches of
the Root Locus, which for this system would have an aspect as indicated in figure
116
7. Classical Control Techniques
ω
ω
ω
Fig. 7.10: Effect of the addition of a zero on the Root Locus
7.11. Looking at the positions of the dominant poles of the system PD can be
seen that the Root Locus does not pass through them, so it is not enough with a
proportional controller. If you want the Root Locus pass through these points and
that these become the dominant poles of the closed chain system, we just need to
introduce a PD type controller. This regulator introduce a zero and if the position of
it is adequately chosen, the Root Locus can pass trough the location of the desired
positions for the roots.
Pd
ω
P'd
Fig. 7.11: Root locus for the example system 7.5.3
7.5. Analytical methods for PID controllers design
117
In this case, it is necessary to determine the positionof the zero and the value
of the gain necessary for the poles to be located exactly at the desired points. Recalling that the Root Locus was the locus of the points that verify the characteristic
equation of the system in closed loop 1+G(s) = 0, this equation led to the module
and angle conditions.
|G(s)|=1
(7.5.45)
∠G(s)=(2q + 1)π
q = 0, ±1, ±2, . . . (K > 0)
(7.5.46)
if the transfer function G(s) has the form
G(s) = K
(s + z1 )(s + z2 ) . . . (s + zm )
(s + p1 )(s + p2 ) . . . (s + pn )
where n ≥ m the expressions 7.5.45 and 7.5.46 will be as follows
Qm
|s + zi |
Q
=1
|G(s)| = |K| ni=1
|s
+ pj |
j=1
or what is the same
Qn
|K| =
and
|s + pj |
Qj=1
m
i=1 |s + zi |
∠G(s) = ∠K +
m
X
i=1
Qn
j=1
= Qm
distances from pj to Pd
i=1 distances
∠(s + zi ) −
n
X
desde zi to Pd
∠(s + pj ) = (2q + 1)π
(7.5.47)
(7.5.48)
(7.5.49)
(7.5.50)
j=1
or what is the same
∠K +
m
n
X
−−→ X −−→
∠ zi Pd −
∠pj Pd = (2q + 1)π
i=1
(7.5.51)
j=1
In our case, given the positions of the poles of the process and controller, we
have to if we want the Root Locus pass through the positions
p
(7.5.52)
p1,2 = ωn (−ζ ± j 1 − ζ 2 ) = −1.57 ± j2.72
must meet the magnitude 7.5.48 and angle conditions 7.5.50. The arguments for
the poles whose positions are known (fig 7.12):
β1 =120o
2.72
= 81.01o
β2 =arctan
2 − 1.57
2.72
β3 =arctan
= 22.93o
8 − 1.57
(7.5.53)
(7.5.54)
(7.5.55)
118
7. Classical Control Techniques
as the angle condition must be met
180o = α1 − (β1 + β2 + β3 )
(7.5.56)
and solving for α1 = 43.93◦ . Knowing this angle, we can determine the position
where it should be located the zero to form that angle with the point where the
dominant pole is located −1.57 + 2.72j. The result is that the zero is located at
a = 4.39. To determine the gain K we apply the magnitude condition
K=
n1 n2 n3
m1
(7.5.57)
and substituting the values of n1 , n2 , n3 , we obtain that m1 = 15.4K. So that the
PD controller that puts the dominant poles at desired positions in closed chain is
Gc (s) = 15.4(s + 4.39)
(7.5.58)
Fig. 7.12: Application of the magnitude and angle conditions for Example 7.5.3
As can be seen the values obtained by both methods are similar (Fig.7.13).
Example 7.5.4: Find a controller for an continuous system with transfer function
1
G(s) = s(s+2)
, with temporal specifications damping coefficient ζ = 0.5 and
settling time ts = 2 sec.
The position of the dominant pole of the system is obtained from the temporal
specifications
ζ = cos(θ) → θ = 60o
7.5. Analytical methods for PID controllers design
119
1.4
y(t)
1.2
(b)
1
0.8
0.6
0.4
(a)
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
t (s)
5
Fig. 7.13: Response system of Example 7.5.3: (a) open loop and (b) closed loop
with the controller designed
r +
-
e
Gc(s)
u
1
s(s+2)
y
Fig. 7.14: Example 7.5.5 System
ts ≈
π
→ σ = 1.57
σ
p1,2 = −1.57 ± j2.72
As shown in Fig. 7.15, to place the dominant poles in the desired positions, it is
necessary to add a regulator P D to the the system with transfer function Gc (s) =
K s+a
s+b . To set the pole position in −b, use the method of cancellation of the zero in
−a with a regulator with a pole in −2.
Using the angle condition we have
X
α−
X
β = (2q + 1)π
α1 = 81o , β1 = 120o , β2 = 81o
120
7. Classical Control Techniques
ω
Fig. 7.15: Distribution of poles and zeros of example 7.5.5
α1 − β1 − β2 − β3 = (2q + 1)π
β3 = −(2q + 1)π − 120o = 60o
Thus the position of pole of the controller b is −3.14. For the determination of
the gain we use the magnitude condition
K=
n1 n 2 n 3
3.14 × 2.75 × 3.14
= 9.86
=
2.75
m1
Thus the continuous controller that meets specifications is shaped as
Gc (s) = 9.86
s+2
s + 3.14
7.5. Analytical methods for PID controllers design
121
Bibliografia
[1] R. Aracil y A. Jimenez, Sistemas discretos de control, Servicio de publicaciones ETSIIM-UPM, 1980.
[2] K. Aström and T. Hägglund, PID Controllers:Theory, Design, and Tuning,
ISA - The Instrumentation, Systems, and Automation Society, 1994.
[3] G.F. Franklin, J.D. Powell and M. Workman, Digital Control of Dynamics
Systems, Addison Wesley, 1997.
[4] R. Isermann, Digital Control Systems: Volume I Fundamentals, Deterministic Control , Springer-Verlag, 1989.
[5] K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edition, 1995.
[6] K. Ogata, Ingenieria de control moderna, Prentice Hall, 1998.
[7] E. Andres Puente, Regulacion automatica, Servicio de publicaciones ETSIIMUPM, 1993.
122
7. Classical Control Techniques
8. FREQUENCY MODELS. BODE
DIAGRAMS
When a sinusoidal signal of unit amplitude and variable frequency as u(t) =
1 sin omegat is introduced into a stable linear system, the system responds by
outputting a signal whose Laplace transform is given by
w
Y (s) = G(s) 2
.
(8.0.1)
s + ω2
If this signal is expanded in partial fractions, it appears a set of terms corresponding to the poles of the transfer function G(s) and the terms corresponding to the
sinusoidal input. Namely
c∗
c
+
.
(8.0.2)
s + jω s − jω
If all poles of the system give a stable response, your response will decrese
over time so the the steady-state response will retain only the last two terms of
expansion 8.0.2. So the steady-state response in the time domain will take the
form
Y (s) = . . . +
y(t) = 2|c| sin(ωt + φ) = A sin(ωt + φ)
(8.0.3)
A = |G(jω)|
(8.0.4)
where
Im[G(jω)]
= ∠G(jω)
(8.0.5)
Re[G(jω)]
The frequency response of a system is usually represented by two curves, of
which, the first is the magnitude |G(jω)| and in the other is the angle ∠G(jω) for
each value of ω.
The the stability of the closed loop system can be determined using the frequency response open loop system.
Besides these response curves may be obtained experimentally without have
knowledge of the mathematical model of the system, no more than excite the system with a variable frequency sinusoidal signal and measuring the amplitude and
the phase of the sinusoidal output signal.
φ = arctan
124
8. FreQuency models. Bode diagrams
When in the diagrams represent the amplitude in decibels 20 log|G(jω)| and
the frequency represent logarithmically, the resulting diagram is called Bode diagram.
This diagram is formed by two response curves: one it is the magnitude response of the system for different frequencies of the input signal and another one
in which the phase delay of the response is represented for different frequencies of
the input signal.
This way of characterizing a system has some properties that make it very
interesting, the first is the ease with which they can be obtained experimentally
these curves for a considerable number of (electrical, electronic, mechanical, etc.)
systems. The second is interpretative in that it allows us to see or treat any physical system as a filter it were (normally low pass) that tend to amplify or dampen
sinusoidal signals of certain frequencies to be presented at the entrance system
(remember that any periodic signal could be decomposed into series of Fourier).
Example 8.0.5: Consider a system whose transfer function is
5
(8.0.6)
+s+5
In order to represent its frequency response, we use the command bf bode () MATLAB, as shown below
G(s) =
s2
num=[5];
den=[1 1 5];
sys=tf(num,den);
bode(sys);
and the result can be seen in Fig. 8.1.
In the reasoning process that was followed to determine the frequency response
of a system we started the mathematical model in s = σ+jω, and has been replaced
by jω considering that the value is studied steady state.
It is clear therefore that this model G(jω) mathematically derived from the
temporal model.
8.1. Asymptotic Bode diagram
125
Bode Diagrams
From: U(1)
10
5
0
-5
Phase (deg); Magnitude (dB)
-10
-15
-20
-25
-30
0
To: Y(1)
-50
-100
-150
-200
-1
10
0
1
10
10
Frequency (rad/sec)
Fig. 8.1: Bode plot of a second order system
8.1
Asymptotic Bode diagram
The Asymptotic Bode diagram can be obtained as sum of the Asymptotic Bode
diagram of each of the factors of the transfer function.
Consider a system with transfer function factored as
Q
(s − zi )
G(s) = K Q
(s − pi )
The frequency response is obtained by replacing s with jω.
Q
(jω − zi )
⇒
G(jω) = K Q
(jω − pi )
The magnitude and the angle are
Q
|jω − zi |
|G(jω)| = |k| Q
|jω − pi |
arg (G(jω)) = arg(k) +
X
arg(jω − zi ) −
By taking logarithms products become sums
X
arg(jω − pi )
126
8. FreQuency models. Bode diagrams
⇒
P
P
log |jω − pi |
log |G(jω)| = log |k| + Plog |jω − zi | − P
arg (G(jω)) = arg(k) + arg(jω − zi ) − arg(jω − pi )
(8.1.1)
For this reason, the modules are taken in decibels (dB), which are 20 log10
of the given quantity. Also on the horizontal axis, the frequency is taken on a
logarithmic scale.
The asymptotic Bode diagram is an approximate diagram of the transfer function from the sum of the representation of each term of the expression 8.1.1.
8.1.1
Asymptotic Bode diagram. Constants K
The constant K is a real number. Therefore, the magnitude is 20log10 |K| dB
and the phase is 0 if K > 0 and −1800 if K < 0.

|K| = 20 log |K| dB
| | = 00 k > 0
K=
ϕ =
1800 k < 0
Fig. 8.2: Asymptotic Bode Plot of Gain Term
8.1.2
Asymptotic Bode diagram. Poles / zeros in the origin
1
sn
The poles at the origin s1n of order n have as module −20n log(ω) as shown in
the semilog plot of magnitude as a straight line with a negative slope of −20n and
the phase as a horizontal line at the position −900 n.
1
=
(jω)n
1
=
(jω)n
(
20 log
ϕ=
1
(jω)n
−n900
| | = ω1n = −20n log(ω) dB
ϕ = −n 900
= 20 log ω1n = 20 log ω −n = −20n log(ω) dB
8.1. Asymptotic Bode diagram
127
Fig. 8.3: Asymptotic Bode plot. Poles in origin
1
sn
The magnitude diagram is a straight line that intersects the horizontal line of 0
dB in ω = 1, because −20n log ω = 0 =⇒ ω = 1 .
For the zeros at the origin of order n module is 20n log(ω) as it is shown in the
semi-log plot as a straight line with slope 20n positive argument and as a horizontal
line in the 900 n position.
8.1.3
Asymptotic Bode plot. Negative real Polo/ Zeros
1
(1+T s)n
The negative real poles /zeros can be written as (1+T1 s)n , with n > 0 for the
poles y n < 0 for the zeros.
For negative real poles, the amplitude is a horizontal line 0 dB for low frequencies and a line of slope −20N dB/dec to high frequencies.
The phase is a horizontal line 00 for low frequencies and −900 n for high frequencies.
1
=
(1 + T jω)n

| | = 0dB


ω → 0 1 = ϕ = 0 0
=
1
| | = (T ω)
n = −20n log(T ω) dB

1

ω → ∞ (T jω)n =
ϕ = −n900
For negative real zeros, the amplitude is a horizontal line 0 dB/dec for low
frequencies and a line of slope 20n dB/dec for high frequencies.
The phase is a horizontal line of 00 for low frequencies and for high frequencies
0
90 n.
128
8. FreQuency models. Bode diagrams
(1 
+ T jω)n = | | = 0dB


ω → 0 1 =
ϕ =
00
=
| | = (T ω)n = 20n log(T ω) dB


ω → ∞(T jω)n =
ϕ = +n900
Cutoff Frequency
Cut of the two asymptotes
0 db = −20 log(ωc T )
log(ωc T ) = 0 ⇒ ωc T = 1
ωc = T1
Fig. 8.4: Asymptotic Bode plot. Negative real Poles/Zeros
8.1.4
1
(1+T s)n
Asymptotic Bode plot. Negative complex Pole/Zeros
The complex negative Poles/Zeros can be represented as
ωn 2
=
s2 + 2ζωn s + ωn 2
1
1+
2ζ
ωn s
+
s
ωn
2 n
with n > 0 for the poles and n < 0 for the zeros.
If the equation s2 + 2ζωn s + ωn2 = 0 has complex roots with negative real
part, then 0 < ζ < 1.
The asymptotic approximation is
8.1. Asymptotic Bode diagram
1
1+ ω2ζ jω+
n
jω
ωn
2 n
129
=

| | = 0dB


ω → 0 1 = ϕ = 00

(
−2n
=
−2n

= −40n log( ωωn ) dB
| | = ωωn
jω

ω
→
=
∞


ωn
ϕ = −2n900 = −n1800
Cutoff frequency
Cut of the two asymptotes
0 db = −40n log( ωωn ) dB
log( ωωn ) = 0 ⇒ ωωn = 1
ω c = ωn
Fig. 8.5: Asymptotic Bode plot. Negative complex Poles/Zeros.
Resonance Frequency ωr
ω r = ωn
p
1 − 2ξ 2
(0 < ξ < 0.707)
The Resonance Frequency ωr exists when:
1 − 2ξ 2 > 0
⇒
1
ξ < √ = 0.707
2
Resonance peak according to ζ:
|G(jω)|max = |G(jωr )| = Mr =
2ξ
1
p
1 − ξ2
(0 < ξ < 0, 707)
130
8. FreQuency models. Bode diagrams
Fig. 8.6: Bode plot. Negative complex Poles/Zeros
8.1.5
Asymptotic Bode plot. Positive real Poles/Zeros.
1
Positive real Poles/Zeros can be expressed as (−1+T
, with n > 0 for the
s)n
poles and n < 0 for the zeros. For positive real poles, the amplitude is a horizontal
line 0 dB for low frequencies and a line of slope −20n dB / dec to high frequencies.
The phase is a horizontal line of −1800 n for low frequencies and −900 n for
high frequencies.
1
n
(−1+T
 jω)
=
=

1

ω → ∞ (T jω)
n
| | = 0dB
ϕ = −1800 n
1
| | = (T ω)
n = −20n log(T ω) dB
=
0
ϕ = −90 n

n

ω → 0 (−1) =
For positive real zeros, the amplitude is a horizontal line 0 dB for low frequencies and a line of slope 20n dB/dec to high frequencies.
The phase is a horizontal line of −1800 n for low frequencies and 900 n for high
8.1. Asymptotic Bode diagram
131
frequencies.
n
(−1
+ T jω) =
| | = 0dB

n

ω → 0 (−1) =
ϕ = 1800 n
=
| | = (T ω)n = 20n log(T ω) dB


ω → ∞(T jω)n =
ϕ = +900 n
Fig. 8.7: Asymptotic Bode diagram of 1/s − 1. (Real positive poles).
Fig. 8.8: Asymptotic Bode diagram ofs − 1. (Real positive zero).
In short, in the Bode plots of positive real poles and zeros, the magnitude is the
same as in the negatives and the phase is reversed.
132
8. FreQuency models. Bode diagrams
8.1.6
Asymptotic Bode plot. Positive complex Poles/Zeros.
Negative complex Poles/Zeros can be expressed as
ωn 2
=
s2 + 2ζωn s + ωn 2
1
1+
2ζ
ωn s
with n > 0 for the poles and n < 0 for the zeros.
The asymptotic approximation for the poles is
1
1+ ω2ζ jω+
n
jω
ωn
2 n
+
s
ωn
2 n
=

| | = 0dB



ω → 0 1 = ϕ = 00
(
−2n
=
−2n
ω

|
= −40n log( ωωn ) dB
|
=
jω

ωn
ω
→
=
∞


ωn
ϕ = 2n900 = n1800
Fig. 8.9: Bode plot with complex poles with: a) negative real part, b) positive real
part.
9. NYQUIST METHOD
Nyquist stability criterion
The method determines Nyquist stability through a graphic-analytical method,
the stability of a closed-loop system based on the frequency response G(jω)H(jω)
in open loop.
In addition to determining the absolute stability of a system, we also used to
determine the degree of stability-instability of a system of regulation. Provides
information about how to improve the system performance in both steady-state
and transient.
It is based on the argument principle Cauchy for functions of a complex variable.
Concepts and principles of the complex variable theory
If we choose a closed path ρ in the complex plane, being F (s) analytic at all
points of the road. Then the curve image of ρ in the plane F (s) will be also closed
(function analytical means that they exist all its derivatives).
ρ
F(s)
F(ρ)
Fig. 9.1: Closed path ρ and its image F (s)
Theorem 9.0.1: Cauchy argument principle
Let F (s) a complex function of a complex variable, and ρ a closed path in
the complex plane which does not pass through singular points of F (s). Then F
134
9. Nyquist Method
(ρ) gives a number of turns around the origin equal to the difference between the
number of zeros and poles of F (s) within ρ (Fig. 9.1).
That is, if
Z =number of zeros of F (s) within ρ
P = number of poles of F (s) within ρ
N = number of turns of F (ρ) around the origin.
Then N = Z − P.
Introduction to Nyquist method
The Nyquist method allows us to study the stability of a feedback system from
open loop frequency response.
If we consider the system of Figure 9.2, the transfer function in closed loop
will be:
G(s)
M (s) =
1 + G(s)H(s)
r(t)
+
G(s)
y(t)
H(s)
Fig. 9.2: Sistema en bucle cerrado
The system will be stable if 1 + G(s)H(S) has no zeros in the positive halfplane (the zeros 1 + G(s)H(s) are the poles of M (s)).
Q
(s + zi )
Q
G(s)H(s) = k
(s + pi )
Q
Q
Q
(s + pi ) + k (s + zi )
(s + zi )
Q
F (s) = 1 + G(s)H(s) = 1 + k Q
=
(s + pi )
(s + pi )
Applying the Cauchy argument principle to the function F (s), and taking a path
that surrounds the positive half-plane (Nyquist path).
If P is known (positive poles in the path ρ, in the unstable half plane) and N
(turns to the origin of F (ρ)) it is possible to know Z (positive zeros of F (s), ie,
unstable) making Z = N + P
Since the zeros Z of F (s) are the closed loop poles system, they define whether
the system is stable. If Z is nonzero, the system will be unstable.
135
ρ
F(s) = 1+G(s)H(s)
F(ρ)
∞
Fig. 9.3: Metodo de Nyquist
Modified method.
One way to simplify the calculations in complex variable, maintaining the
principle of the argument is to take the function F (s) = G(s)H(s), instead of
1 + G(s)H(s).
Thus, the curve γ corresponding to F (ρ), remain displaced a unit on the real
axis, and must now count the number N of turns around the point −1, instead from
around the origin (Figure 9.4).
ρ
F(s) = G(s)H(s)
F(ρ)
∞
-1
Fig. 9.4: Modified Nyquist method
Example 9.0.1: Study the system stability of the Figure9.5 using the method of
Nyquist.
Solution:
The transfer function in open loop is
G(s)H(s) =
s+2
(s + 1)(s + 3)
The Nyquist path and the location of the zeros and poles of open loop is:
136
9. Nyquist Method
Fig. 9.5: System of the example 5.1.6
II
I
III
Fig. 9.6: Nyquist path for the example 5.1.8
The image of section I is the polar diagram. Taking the limit for ω → 0 y
ω → ∞ we have
jω + 2
G(jω)H(jω) =
(jω + 1)(jω + 3)
lim |G(jω)| =
ω→0
2
3
lim arg(G(jω)) =
ω→0
0◦
lim |G(jω)| = 0
ω→∞
lim arg(G(jω)) = −90◦
ω→∞
The image of the section II will be the origin, since Num poles> Num zeroes.
The image of the section III will be the symmetric section of I’ to the real axis.
As the number of poles inside the Nyquist path is P = 0, and the image of the
Nyquist path does not encircle the point -1 (N = 0), then Z = N + P = 0. That is,
there are no zeros of the characteristic equation (poles of the closed loop transfer
function) in the right half plane and therefore the system is stable (see Figure 9.6).
137
0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Fig. 9.7: Nyquist diagram of example 5.1.6
For systems with poles on the imaginary axis. If there are singular points
in the Nyquist path, ie, if GH has poles at the origin or on the imaginary axis, it
will be necessary to select a path that avoids these poles Nyquist.
• Open loop system with pole at the origin. In this case you should add a new
section (IV) around this pole.
The section IV is defined by a semicircle of infinitesimal radius and centered
at the origin:
jθ ε → 0
s = εe
π
θ ∈ −π
2 , 2
I
II
IV
III
Fig. 9.8: Nyquist path for a system with a pole at the origin
138
9. Nyquist Method
• Open loop system with pure imaginary poles
The new sections will now be:
I s = jω
ω ∈ (0, a)
π
jθ
ε → 0 θ ∈ ( −π
II s = aj + εe
4 , 4)
III s = jω
ω ∈ (a, ∞)
R → ∞ θ ∈ ( π4 , −π
IV s = Rejθ
4 )
V s = jω
ω ∈ (−∞, −a)
π
V I s = −aj + εejθ ε → 0 θ ∈ ( −π
4 , 4)
ω ∈ (−a, 0)
V IIs = jω
Fig. 9.9: Nyquist path for a system with poles on the imaginary axis
Open-loop stable systems (minimum phase systems)
In the open loop stable systems poles are located in the left half of the plane s
therefore P = 0. In this case, the Nyquist criterion is reduced to
Z = N.
Therefore, the system is stable if N = 0.
To see if the system is stable just draw the image of the section I (polar diagram)
and check that cuts on the right of the point -1 of the real axis.
Most real systems have open loop transfer function of G(s)H(s) with all its
poles and zeros in the left half plane. This kind of systems are called Minimum
phase.
139
Then, since G(s)H(s) has no poles in the right half-plane, we have that P = 0,
which will be fulfilled
Z=N
and therefore for the system to be stable must apply:
N =0
In minimum phase systems the condition of stability means that the polar diagram
of G(s)H(s) does not enclose the point -1
-1
Fig. 9.10: Stable system
-1
Fig. 9.11: Unstable system
Relative stability
The Nyquist method allows the study of the relative stability of a closed loop
system from the frequency response of the open loop system. This relative stability
140
9. Nyquist Method
is given by the proximity of the polar diagram at point s = -1 (see Figure ref
estabrel1). In this figure the polar diagrams of three systems are shown. From
the observation of them can be said for the system a) will be more stable than
the b) and c) as its polar diagram is farthest from -1 than those of the other two
systems (assuming all three are of minimum phase). To have a more quantitative
and accurate picture of the relative stability, the concepts of gain margin and phase
are used.
ω
ω
ω
ω
ω
ω
Fig. 9.12: The system a) is more stable than the b) and c systems)
Gain Margin M G. Provides a measure of the distance in magnitude between
point -1 and the polar diagram of G(jω)H(jω).
It is the inverse of the value of the polar diagram where this intersects the
negative real axis (Figure 9.13).
Kg =
1
A
The frequency at which this occurs is called emph phase crossover frequency,
and is the frequency at which the phase is 180◦ .
arg(GH(ωϕ )) = π
|GH(ωϕ )| = A
141
ω
ω=ωϕ
ω
Fig. 9.13: Calculation of gain margin
For a system to be stable is needed that Kg > 1.
Kg > 1⇒sistema estable
Kg = 1⇒sistema marginalmente estable
Kg < 1⇒sistema inestable
Expressed in dB; the gain margin M G is:
M G = 20 log
1
= −20 log |G(ωϕ )H(ωϕ )|
|G(ωϕ )H(ωϕ )|
For a stable system is verified |G(ωφ )H(ωφ )| < 1. Therefore, M G > 0. This
value indicates how much gain can be increased without causing instability.
For an unstable system is verified |G(ωφ )H(ωϕ )| > 1. Therefore, M G < 0,
indicating there is to reduce the gain so that the system becomes stable.
When the system is critically stable, |G(ωφ )H(ωφ )| = 1. Therefore, M G = 0,
no gain can be increased without causing open loop instability.
If the curve of the polar diagram does not intersect the negative real axis, as
occurs in systems of first and second order, M G = ∞. Theoretically gain can
be increased indefinitely without cause instability (FigureFor an unstable system is
verified |G(ωφ )H(ωϕ )| > 1. Therefore, M G < 0, indicating there is to reduce the
gain so that the system becomes stable.
When the system is critically stable, |G(ωφ )H(ωφ )| = 1. Therefore, M G = 0,
no gain can be increased without causing open loop instability.
If the curve of the polar diagram does not intersect the negative real axis, as
occurs in systems of first and second order, M G = inf ty. Theoretically gain can
be increased indefinitely without cause instability (Figure9.14).
142
9. Nyquist Method
ω
ω
ω ∞
σ
ω
ω
ω
ω ∞
σ
ω
ω
ω
ω ∞
σ
ω
ω
ω
ω ∞
ω
σ
Fig. 9.14: Nyquist plot, Root locus and closed loop step response according system
1
gain increases G(s) = s(s+a)(s+b)
con a, b > 0.
Phase margin γ. Figure 9.15 shows polar diagrams of two systems whose gain
margin is the same. However, the System a) is more stable than the b). To quantify
143
ω
ω
ω
ω
Fig. 9.15: The two systems have the same profit margin, but a) is more stable than
the b)
this circumstance the phase margin is used.
Gain crossover frequency (ωg ): the one for which
|G(jω)H(jω)| = 1 (0 dB)
The phase margin is the difference between the phase angle to ωg and 180◦ .
M F = γ = arg(GH(ωg )) + π
Graphically, γ is reflected in Figure 9.16.
In stable and minimum phase open loop systems, the condition of stability of
the closed loop system is that
M G|dB > 0
MF = γ > 0
Calculation of gain margin (Kg ) and phase margin (γ) Given the definition of gain margin, to calculate its value we will have to find the point of intersection with the real axis and find the value of phase margin as the constant by which
multiply |G(jω)H(jω)| for the value 1, which is the module |(−1.0)|. In decibels
(20log1 0 of the corresponding value) this product becomes sum. therefore
arg(G(jωϕ )H(jωϕ )) = −180
|G(jωϕ )H(jωϕ )|dB + Kg dB = 0
Kg dB = − |G(jωϕ )H(jωϕ )|dB
1
Kg =
|G(jωϕ )H(jωϕ )|
144
9. Nyquist Method
ω
ω
ω=∞
ω=∞
ω
ω
ϕ
ϕ
ωγ
ω=0
ω=0
∞
Fig. 9.16: Phase Margin
Fig. 9.17: Gain margin and phase margin in Bode diagram
To find the phase margin must be calculated as the cut of the trace lef t|G(j omega)H(j omega) right|
145
with the unit circle and, then the frequency corresponding to this cut point.
|G(jωg )H(jωg )| = 1 = 0 dB
− arg(G(jωg )H(jωg )) − γ = −180
γ = 180 − arg(G(jωg )H(jωg ))
The phase and gain margins are measured on the loop system opened.
The gain/phase margin represents how you can increase the gain/phase shift of
the open-loop system without the system closed loop becomes unstable.
146
9. Nyquist Method
10. DESIGN OF REGULATORS BY
FRECUENCY METHODS
The design of regulators for frecuency methods are based on the concepts of
phase margin and gain margin in relation to Nyquist and Bode diagrams.
The margins of phase and gain are measured in the open loop system.
Phase margin and gain represent how much you can increase the gain or phase
of the open loop system without turning the closed loop system unstable.
The gain margin is the ratio of the distance from the origin to the point −1 + 0j
and the distance to the cutoff of G(jω)H(jω) with the real axis.
Fig. 10.1:
In this cut point of G(jω)H(jω) with the real axis, the argument is zero:
arg(G(jωϕ )H(jωϕ )) = −180◦
and its magnitude
148
10. Design of regulators by frecuency methods
Fig. 10.2:
|G(jωϕ )H(jωϕ )|dB + Kg dB = 0
Therefore,
Kg dB = −|G(jωϕ )H(jωϕ )|dB
and natural scale (not decibels)
Kg =
1
|G(jωϕ )H(jωϕ )|
(10.0.1)
On the other hand, if we consider cutting G(jω)H(jω) with the unit circle, at
this point the module is one.
|G(jωg )H(jωg )| = 1 = 0 dB
The difference to the argument of −1 is the missing phase to reach the instability.
arg(G(jωg )H(jωg )) − γ = −180◦
Therefore, the phase margin can be calculated as
γ = 180◦ + arg(G(jωg )H(jωg ))
(10.0.2)
10.1. Relationship between parameters of relative stability and transient response
10.1
149
Relationship between parameters of relative stability and
transient response
The approximate relationship between the parameters of relative stability and
transient response
γ ≈ 100ζ para 0 < ζ < 0, 7
(10.1.1)
ts ≈
2π
1
=
ωg tan(γ)
fg tan(γ)
(10.1.2)
where
• ζ It is the damping ratio
• ts It is the settling time
Conclusions: In the underdamped systems, the phase margin (γ) is directly
related to the damping ratio (ζ).
When the phase margin decreases, overshoot increases.
When the cutoff frequency gain (ωg ) and the phase margin (γ) increase the
system becomes faster (the time of settlement (ts ) decreases).
10.2
Lead Networks
The RC network shown in Fig. is equivalent to a phase lead controller
We denote the impedances
Z1 = R1 / (1 + R1 Cs)
y
Z 2 = R2
The transfer function of the phase-lead network can be written as
150
10. Design of regulators by frecuency methods
Fig. 10.3: RLC lead network and its corresponding diagram of poles and zeros.
Fig. 10.4: Bode and Nyquist plot for a lead network according to different parameter values α.
Fig. 10.5: Bode diagram of the system and lead network.
10.2. Lead Networks
151
Fig. 10.6: Comparison of Bode plots and unit step responses.
Fig. 10.7: Adding a lead network to the system changes the magnitude and phase,
so it is difficult to predict the new crossing point.
152
10. Design of regulators by frecuency methods
Fig. 10.8: Designing a lead network.
10.3. Lag networks
10.3
Lag networks
Fig. 10.9: Lag network and pole-zero diagram.
Fig. 10.10: Frequency response of lag networks.
Fig. 10.11: Comparison of the response of the systems.
153
154
10. Design of regulators by frecuency methods
Fig. 10.12: Designing a lag network.
Fig. 10.13: The effect of the change of T.
SOME HISTORICAL DATA
156
10. Design of regulators by frecuency methods
Jacopo Francesco Riccati
(1676 Venice (Italia) –
1754 Treviso (Italia))
Taylor, Brooks
(1685 Edmonton (Reino
Unido) –
1731 Londres (Reino
Unido))
Leonhard Euler
(1707 Basilea (Suiza) –
1783 San Petersburgo (Rusia))
Watt, James
(1736 Greenock,
land) –
1736)
Laplace, Pierre Simon
(1749 Normandia (Francia)
–
1827 Paris (Francia))
Jean Baptiste
Fourier,
Joseph
(1768 Borgoña (Francia) –
1830 Paris (Francia))
(Scot-
10.3. Lag networks
157
Cauchy, Agustin Louis
(1789 Paris (Francia) –
1857 Paris (Francia))
Sir William Rowan Hamilton
(1805 Dublin (Irlanda) 1865 Dublin (Irlanda))
Routh, Edward John
(1831 Quebec (Canada) –
1907 Cambridge (Reino
unido))
Jordan, Marie Ennemond
Camille
(1838 Lyon (Francia) –
1922 Paris (Francia))
Liapunov, Aleksandrer Mijailovich
(1857 Yaroslavl (Rusia) –
1918 Odessa (Rusia))
Hurwitz, Adolf
(1859 Hanover (Germany)1919 Zurich (Switzerland))
158
10. Design of regulators by frecuency methods
Evans, Griffith Conrad
(1887 Boston (USA) –
1973 Berkeley (USA))
Nyquist, Harry
(1889, Nilsby (Suecia) –
1976 (USA))
Norbert Wiener
(1894, Columbia (USA) –
1964 Estocolmo (Suecia))
Bode, Henrick W.
(1905 Madison (USA) –
1982 Maryland (USA))
Pontriagin,
Lev
Semionovich
(1908 Moscú (Rusia) –
1988 Moscú (Rusia))
Shannon, Claude E.
(1916 Michigan(USA) –
2001
Massachusetts
(USA))
10.3. Lag networks
159
Yakov Zalmanovitch Tsypkin
( 1919 (Rusia) –
1997 (Rusia))
Richard Bellman
(1920 Brooklyn USA) –
actualidad)
Kalman, Rudolf E.
(1930 Budapest (Hungria)
– actualidad)
Truxal, John G.
Eliahu I. Jury
Karl J. Åström
(1934 Östersund (Sweden)
- actualidad)
INDEX
, 60, 101, 104, 107
Small gain theorem, 73
Bode plot, 124
Time models, 14
Transfer function, 15
Cauchy argument principle, 133
undamped natural frequency, 60
Final value theorem, 5
Fourier transform, 1
Ziegler-Nichols, 104
Ziegler-Nichols methods, 104
Gain Margin, 140
impulse response, 15
Initial value theorem, 5
Input/Output models, 14
Inverse Fourier Transform , 3
Inverse Laplace transform, 7
Laplace Transform, 3
method based on algebraic determination for adjusting PID, 111
Method based on the dominant poles
for adjusting PID, 110
Method based on the Root Locus for
adjusting PID Method, 114
Nyquist, 133
Nyquist stability criterion, 133
Overshoot, 61, 102
Peak, 61
Peak time, 101
Phase margin, 142
PID controllers, 102
Routh’s method, 74
Settling time, 61, 102
160
View publication stats