Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
Chapter 1 Overview
1.1 Concept of Automatic Control
1.2 A Brief History of Automatic Control
1.3 Stages of Automatic Control Theory
1.4 Categories of Classical Control Systems
1.5 A First Analysis of Feedback
1.6 Components of Feedback Control Systems
1.1 Concept of Automatic Control
What is automatic control ?
Definition: Use a control devices (or controller) without
direct intervention of human to keep a physical quantity
(temperature, pressure, PH value, etc.) of the object running
with expected rules.
Another Definition: Machine or equipment achieves the
desired goals by auto-detection, information processing,
analyzing, manipulating without direct intervention of
human.
1.1 Concept of Automatic Control
The core of automation : Automatic Control
Two aspects of automation technology:
① extension of human limbs - dynamic control
domain of automation technology: industrialization
② extension of the human brain - information
processing domain of automation technology:
informationization
1.2 A Brief History of Automatic Control
James Watt’s steam engine speed control system
1.2 A Brief History of Automatic Control
James Watt’s steam engine speed control system
1.2 A Brief History of Automatic Control
First numerical control
systems, developed in 1950
Henry Ford’s mechanized
assembly machine introduced for
automobile production, 1913.
1.2 A Brief History of Automatic Control
US space shuttle "Columbia"
successfully launched in 1981
First industrial robot ,
made in 1954
1.2 A Brief History of Automatic Control
Automatic Production Line for
Automobile Welding
Assembly Robot
1.2 A Brief History of Automatic Control
Robot Soccer
Satellite
1.2 A Brief History of Automatic Control
James Webb Space Telescope
1.2 A Brief History of Automatic Control
Telescope deployment
1.2 A Brief History of Automatic Control
Aerospace vehicles
Lunar rover
1.2 A Brief History of Automatic Control
Theseus
RQ-4 Global Hawk
Proteus
Helios
2005-11-29
33
U2
Zephyr
Modern aircrafts
1.2 A Brief History of Automatic Control
Lunar Exploration Project of CHINA
circling
Chang’e-1
2007.10
falling
Chang’e-2
2010.10
Chang’e-3
2013.12
returning
Chang’e-4
2014.10
Chang’e-5
2020.12
The three-step idea of "circling, falling and returning".
Now, the United States, Russia, China, Germany, India, Japan and other
countries have new lunar exploration plans. The design of the control system is
the key technology.
1.2 A Brief History of Automatic Control
Commercial Airplane Project of CHINA
➢ In January 2006, the C919 project was
initiated
➢ In November 2015, the first C919 rolled
off the production line
➢ The C919 is scheduled to obtain
certificate in 2021 and start commercial
operations
➢ The C919 perform its first flight on Jan.
1st, 2023
Aviation Industry Corporation of China (AVIC) is the main manufacturer of major
components of the C919.
However, the supply of engine, avionics, flight control, power supply and other airborne
systems have been completed in cooperation with 16 multinational companies.
1.3 Stages of Automatic Control Theory
1.3 Stages of Automatic Control Theory
1.3 Stages of Automatic Control Theory
1. Classical Control Theory
2. Modern Control Theory
3. Large-Scale System Theory
4. Intelligent Control Theory
1.3 Stages of Automatic Control Theory
Classical control theory and model control theory
Subject
Classical control theory
Modern control theory
Description/Mode Transfer function
ling Methods
(Input-output description)
Vector space
(State-space
representation)
Research
Methods
Root locus method and
frequency method
State-space method
Objective
Systems analysis and
system integration by the
given input and output
Reveal the inherent law
of the system and realize
controller design
1.3 Stages of Automatic Control Theory
Large-Scale System Control Theory
◼
Large-scale system control theory is a dynamic system engineering
theory combining process control and information processing.
◼
The object of study has the characteristics of large-scale, complex
structure, Integrated Functions, variety of objectives, plenty of factors.
◼
Example: The human body could be taken as a large scale system.
There are temperature control, emotional control, blood components
control in this system.
1.3 Stages of Automatic Control Theory
Intelligent Control
◼
Intelligent Control is a new control technology used artificial
intelligence and neural network. The main idea is to solve complex
control problems that require human intelligence methods.
◼
It is a new subject, many previously intractable control problems
(such as control without model, uncertain environment …) could be
solved by using data and algorithms. This technology is still in
developing stage.
1.4 Categories of Classical Control Systems
1. According to the transmission paths of signals
◼
Open-loop control system
◼
Closed-loop control system
Open Loop or closed-loop system?
Diagram of automobile cruise system
1.4 Categories of Classical Control Systems
◼
Open-loop control system
◼
Features: There is no feedback loop between system's
output and input side. The output has no influence on
control effect of the system.
1.4 Categories of Classical Control Systems
◼
Closed-loop control system (feedback control system)
◼
Features: There is a feedback loop between the measuring
component and the output signal.
◼
"Closed loop" means that the output signal is feedback to the
input side by measuring component, and errors could be
reduced by comparison and control.
1.4 Categories of Classical Control Systems
Open Loop or closed-loop system?
1.4 Categories of Classical Control Systems
Questions:
◼
What are the minimal components of an open-loop
system?
◼
What are the minimal components of a closed-loop
system (feedback control system)?
◼
What is the main difference between open-loop
systems and closed-loop systems?
1.4 Categories of Classical Control Systems
2. According to signal properties
Continuous system
Feature: All signals in the system are analog and continuous
functions.
◼
Discrete-time system
Feature: In a system, there is one place or several signals that
are of pulse sequence or digital form in control system.
◼
1.4 Categories of Classical Control Systems
Discrete control system
Input +
e (t )
e(t )
Output
Hold
_
Process
Sampler
Input
Output
A/D
Computer
D/A
Amplifier
Feedback
Component
Actuator
Object
1.4 Categories of Classical Control Systems
3. According to mathematical model
◼
Linear system
𝑑 𝑛 𝑐(𝑡)
𝑑 𝑛−1 𝑐(𝑡)
𝑑𝑐(𝑡)
𝑎𝑛
+
𝑎
+
⋯
+
𝑎
+ 𝑎0 𝑐(𝑡)
𝑛−1
1
𝑑𝑡 𝑛
𝑑𝑡 𝑛−1
𝑑𝑡
𝑑 𝑚 𝑟(𝑡)
𝑑 𝑚−1 𝑟(𝑡)
𝑑𝑟(𝑡)
= 𝑏𝑚
+ 𝑏𝑚−1
+ ⋯ + 𝑏1
+ 𝑏0 𝑟(𝑡)
𝑑𝑡 𝑚
𝑑𝑡 𝑚−1
𝑑𝑡
𝑟(𝑡) : Input signal； 𝑐(𝑡) : Output signal
Feature: The system could be described using ordinary
differential equations (ODEs) and transfer function.
1.4 Categories of Classical Control Systems
◼
Nonlinear system
Feature: There are one or more than one nonlinear elements in
the system.
1
𝑥ሶ = 2𝑥
2
𝑥ሶ = 𝑥 2
3
𝑥ሶ = sin( 𝑥)
4
𝑥ሶ = 𝑥 + 1
Which is (are) linear system?
1.4 Categories of Classical Control Systems
◼
Categories of Control System (Summary)
Open-loop system
1. According to transmission path
𝟐. According to signal properties
Closed-loop system
Continuous system
Discrete-time system
𝟑. According to mathematical model
Linear system
Nonlinear system
1.5 A First Analysis of Feedback
The speed control of an aircraft
Wind
1.5 A First Analysis of Feedback
The speed control of an aircraft
Control variable – the throttle angle of aircraft engine
Output variable – the aircraft speed
The throttle angle changes 1 degree
speed changes 20km/h （without wind）
Assumptions:
1. Ignore the dynamic characteristics
2. Assume the system is linear
1.5 A First Analysis of Feedback
wind
Desired speed
controller
process
actuator
??
Engine
Actual speed
Aircraft
Throttle
sensor
Gyro
Measured speed
When throttle angle changes 1°
speed changes 20km/h;（without wind）
When wind (disturbance) changes 100km/h
speed changes 100km/h.
1.5 A First Analysis of Feedback
Problem:
Suppose in nominal state, the aircraft speed is 1000km/h;
Around the nominal state, the throttle angle changes 1 degree, aircraft speed
changes 20km/h;
wind (disturbance) changes from -100km/h to 100km/h;
We wish the actual aircraft speed is 1000km/h.
Design an open-loop system and a closed-loop (feedback control) system to
control aircraft speed.
Analyze output (actual) speeds of open-loop and closed- loop (feedback) systems.
1.5 A First Analysis of Feedback
Question:
We cannot control the wind (disturbance) ;
The wind (disturbance) do affect the speed (output);
We wish to minimize the influence of the wind (disturbance) and keep the
speed as we wish.
Which scheme we should choose? (open-loop control or closed-loop control)
WHY?
1.5 A First Analysis of Feedback
Open-loop control and closed-loop control
𝑤 -100km/h
𝑤 -100km/h
1000km/h
Disturbance
Desired
speed
1
+
𝑢
20
Disturbance
Output
speed
+
𝑢 = 𝑟/20
𝑟 =1000km/h
Open-loop control
𝑦
𝑟
Speed
error
+
10
1
+
𝑢
20
Measured
speed
+
1
Closed-loop control
(feedback control)
𝑢 = 10*Speed Error
Gain (增益)
Output
speed
𝑦
Page . 40
1.5 A First Analysis of Feedback
Open Loop Control
◼
𝑤 -100km/h
Disturbance
1
+
𝑢
20
Without disturbance
(𝑟 = 1000km/h, 𝑤 = 0)
𝑦 = 20 ∗ 𝑢 + 𝑤 = 𝑟 = 1000km/h
𝑒ol = 𝑟 − 𝑦 = 0
No error
Output
speed
+
𝑢 = 𝑟/20
𝑟 =1000km/h
Open-loop control
𝑦
◼
With disturbance
(𝑟 = 1000km/h, 𝑤 = −100)
𝑦 = 20 ∗ 𝑢 + 𝑤 = 1000 − 100
= 900km/h
𝑒ol = 𝑟 − 𝑦 = 𝑤 = 100km/h
Big difference
Open loop control system is sensitive to disturbance
1.5 A First Analysis of Feedback
Question: Which relationship is correct?
𝑤
Disturbance
Desired
speed
𝑟
Speed
error
+
𝐴
𝐷
𝑥 +
𝑢
𝐵
Measured
speed
Output
speed
𝑦
+
𝑦 = 𝐷𝑤 + 𝑥
𝑥 = 𝐴𝐵(−𝐶𝑦 + 𝑟)
𝑦 =?
𝐶
B. y =
AB
D
r+
w
1 + ABC
1 + ABC
1.5 A First Analysis of Feedback
Closed-Loop Control
◼
𝑤 -100km/h
Disturbance
1000km/h
Desired
speed
𝑟
Speed
error
+
10
1
+
𝑢
20
Measured
speed
+
1
𝑢 = 10*Speed Error
Output
speed
𝑦
Without disturbance
𝑟 = 1000km/h, 𝑤 = 0
𝑦 = 995km/h
𝑒cl = 𝑟 − 𝑦 = 5km/h
Small error
◼
With disturbance
(𝑟 = 1000km/h, 𝑤 = −100)
𝑦 = 994.5km/h
𝑒𝑐l = 𝑟 − 𝑦 = 5.5km/h
Small error
Feedback reduces output sensitivity to disturbances
1.5 A First Analysis of Feedback
Question:
Why Feedback could reduce output sensitivity to disturbances?
- High loop gain is the reason.
- Unfortunately, loop gain is limited by dynamics of systems.
The key work of feedback control design is to find appropriate
loop gain to make system stable, fast enough as well as reduce
sensitivity to disturbances.
1.6 Components of Feedback Control Systems
General form
◼
◼
◼
◼
Process – the physical system to be controlled
Controller – the component that computes the desired control signal
Actuator – the device that can influence the controlled variable
Sensor – the device can measure the output signal
1.6 Components of Feedback Control Systems
Example: Room Temperature Control
Control Objective – the room temperature holds at the expected
constant value
◼
◼
Open-loop or closed-loop system?
Could you identify components of this system?
1.6 Components of Feedback Control Systems
Summary of Chapter 1
1.1 Concept of Automatic Control
1.2 A Brief History of Automatic Control
1.3 Stages of Automatic Control Theory
1.4 Categories of Classical Control Systems
1.5 A First Analysis of Feedback
1.6 Components of Feedback Control Systems
KEY POINTS IN THIS COURSE
1. One Concept: Feedback
2. Three Problems:
- Stability
- Steady (error)
- Dynamic Performance (speed & oscillation)
3. Three Design Methods:
- Root Locus
- Frequency Response
- State Space
Homework
◼
Textbook, Page 40-42, Problem 1.6
◼
“Control is dead” -- Some researchers supported this view in a few
international conference and public speech/blog. What do you think of
“Control is dead”？
Some references for you:
- https://blog.sciencenet.cn/blog-1565-344686.html
- https://www.zhihu.com/question/34560411
- ChatGPT reply
Homework
Notes
◼
For students in classroom, please submit your last week’s homework on
every Monday to TA and I will return them to you ASAP.
◼
For students online, please submit your last week’s homework on Canvas
before every Monday’s class.
◼
I will upload brief answers of some of the last week’s homework for your
reference in canvas.
Notes
For more background on the history of control, see the
survey papers appearing in the IEEE Control Systems
Magazine of November 1984 and June 1996.
Thanks！
Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
2
Chapter 2 Dynamic Models
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
2.4 Model of Eletromechanical System
2.5 Nonlinear Model of Aircraft System & Linearization
2.6 Typical Elements
2.1 Concept of Dynamic Models
◼
Mathematical model - mathematical description of a
physical system
◼
Dynamic Model - mathematical description for a dynamic
process to be controlled
Control system model reveals the relationship among system
variables (state, input, output) using analytical formula
2.1 Concept of Dynamic Models
Question: What kind of model we need?
- Relative accurate
- Describing main system behaviors
- Simple enough
Ordinary Differential Equation Model
(ODE model) – Linear Model
2.1 Concept of Dynamic Models
Typical Dynamic Model Examples
◼
Model of Mechanical System
◼
Model of Electric Circuit
◼
Model of Electromechanical System
◼
Model of Aircraft
5
2.1 Concept of Dynamic Models
The step of analyzing and studying dynamic system
1. Define the system and its components;
2. Formulate the mathematic model and list the necessary
assumptions;
3. Write the differential equations describing the model;
4. Solve the equations for the desired output/input variables;
5. Examine the solutions and the assumptions;
6. If necessary, reanalyze or redesign the system.
2.1 Concept of Dynamic Models
“Tri-domain” models and mutual relations
◼
◼
◼
Ordinary Differential equation
(time domain)
Transfer function
(complex frequency domain)
Frequency response
(frequency domain)
2.1 Concept of Dynamic Models
Example: Establish RC circuit dynamic equation
R
𝑟(𝑡): Input; 𝑐(𝑡): Output
r (t )
◼
i (t )
Time domain
d𝑐(𝑡)
Differential equation: 𝑅𝐶
+ 𝑐(𝑡) = 𝑟(𝑡)
dt
◼
Complex frequency domain
Transfer function:
◼
𝐶(𝑠)
1
𝐺(𝑠) =
=
𝑅(𝑠) 𝑅𝐶𝑠 + 1
Frequency domain
Frequency response: 𝐺 𝑗𝜔 =
𝐶 𝑗𝜔
1
=
𝑅 𝑗𝜔
𝑅𝐶𝑗𝜔 + 1
C
c (t )
2.1 Concept of Dynamic Models
◼
Set up different mathematical model based on different needs
▪ Engineering analysis:
Transfer function, frequency characteristics
▪ Mathematical analysis:
Ordinary differential equation(ODE)
◼
Engineering analysis method is more intuitive and convenient
than mathematical analysis one, and is easier to be understand
and used.
◼
Complex domain (transfer function) and frequency domain
model to be the main mathematical model in classical control.
2.1 Concept of Dynamic Models
◼
Mathematical analysis method is more accurate and complex
than engineering analysis one.
◼
Matrix analysis and state space are introduced into modern
control theory.
2.2 Models of Mechanical Systems
Newton’s law: (linear motion)
𝐹 =𝑚⋅a
Force = Mass × Acceleration
Theory of Moment of Momentum: (rotation)
𝜏 =𝐽⋅α
Torque = Inertia × Angular Acceleration
2.2 Models of Mechanical Systems
Example 1: Spring-Mass-Damper System
𝑘𝑦(𝑡)
𝑏𝑦(𝑡)
ሶ
Elastic force:
𝑦(𝑡)
𝑘𝑦(𝑡)
Damping force:
𝑏
𝑑𝑦(𝑡)
𝑑𝑡
𝑦(𝑡)
𝑟(𝑡)
𝑟(𝑡)
Input: Pulling force
Output: Displacement
2.2 Models of Mechanical Systems
Example 1: Spring-Mass-Damper System
𝑘𝑦(𝑡)
𝑏𝑦(𝑡)
ሶ
𝑦(𝑡)
𝑑𝑦(𝑡)
𝑑 2 𝑦(𝑡)
𝑟(𝑡) − 𝑘𝑦(𝑡) − 𝑏
=𝑚
𝑑𝑡
𝑑𝑡 2
𝑑 2 𝑦(𝑡)
𝑑𝑦(𝑡)
𝑚
+𝑏
+ 𝑘𝑦(𝑡) = 𝑟(𝑡)
𝑑𝑡 2
𝑑𝑡
𝑟(𝑡)
Input: 𝑟(𝑡); Output: 𝑦(𝑡)
𝑚𝑦ሷ + 𝑏𝑦ሶ + 𝑘𝑦 = 𝑟
2.2 Models of Mechanical Systems
𝑑 2 𝑦(𝑡)
𝑑𝑦(𝑡)
𝑚
+𝑏
+ 𝑘𝑦(𝑡) = 𝑟(𝑡)
2
𝑑𝑡
𝑑𝑡
◼
◼
The solution of the equation will be studied in next chapter.
(Laplace Transform)
For a given input signal of the form 𝑟 𝑡 = 𝑅𝑒 𝑠𝑡 , and assume
the solution of output signal is of the form 𝑦 𝑡 = 𝑌𝑒 𝑠𝑡
𝑟(𝑡)
ሶ
= s𝑅𝑒 𝑠𝑡
y(𝑡)
ሶ
= s𝑌𝑒 𝑠𝑡
y(𝑡)
ሷ
= s2 𝑌𝑒 𝑠𝑡
2.2 Models of Mechanical Systems
◼
The differential equation can be written as
𝑚s2 𝑌𝑒 𝑠𝑡 + 𝑏𝑠𝑌𝑒 𝑠𝑡 + 𝑘𝑌𝑒 𝑠𝑡 = 𝑅e𝑠𝑡
◼
The transfer function (will be clear in next chapter)
𝑌(𝑠)
1
=
𝑅(𝑠) 𝑚𝑠 2 + 𝑏𝑠 + 𝑘
𝑅(𝑠)
Input
1
𝑚𝑠 2 + 𝑏𝑠 + 𝑘
𝑌(𝑠)
Output
𝑑
𝑑𝑡
The transfer function is obtained by substituting 𝑠 for in
the differential equation for zero initial condition.
2.2 Models of Mechanical Systems
Example 2: Mechanical Rotational System with Damper
Write Out Transfer Function
Output: Angle
𝜃
𝜔
角动量守恒：
合外力矩 = 角动量改变
𝜏 =𝐽⋅α
𝐽
𝑇
𝑐⋅𝜔
Input: Torque
Damper
2.2 Models of Mechanical Systems
Example 2: Mechanical Rotational System with Damper
Output: Angle
𝜃
𝜔
𝐽
𝑇
𝑐⋅𝜔
Input: Torque
Damper
𝑑𝜔
𝑇 − 𝑐𝜔 = 𝐽
,
𝑑𝑡
𝑑𝜃
𝜔=
𝑑𝑡
𝐽: rotary inertia
𝜔: angular velocity
2.2 Models of Mechanical Systems
Example 2: Mechanical Rotational System with Damper
Output: Angle
𝜃
𝜔
Damper
𝑑2 𝜃
𝑑𝜃
𝐽 2 +𝑐
=𝑇
𝑑𝑡
𝑑𝑡
𝐽
𝑇(𝑠)
𝑇
𝑐⋅𝜔
Input: Torque
1
𝐽𝑠 2 + 𝑐𝑠
𝜃(𝑠)
2.2 Models of Mechanical Systems
Example 3: Mechanical Accelerometer
Write Out Transfer Function
𝑦(𝑡)
𝑟 𝑡 = 𝑥ሷ 𝑡
Input: Force
Output: deformation of
the spring
2.2 Models of Mechanical Systems
Example 3: Mechanical Accelerometer
Input: 𝑟 𝑡 = 𝑥ሷ 𝑡
Output: deformation of
the spring, 𝑦 𝑡
𝑑𝑦(𝑡)
𝑚 𝑥(𝑡)
ሷ
− 𝑦(𝑡)
ሷ
= 𝑘𝑦(𝑡) + 𝑐
𝑑𝑡
𝑚𝑥(𝑡)
ሷ
=𝑚
𝑑 2 𝑦(𝑡)
𝑑𝑡 2
𝑑𝑦(𝑡)
+𝑐
+ 𝑘𝑦(𝑡)
𝑑𝑡
𝑚𝑅(𝑠) = 𝑚𝑠 2 𝑌(𝑠) + 𝑐𝑠𝑌(𝑠) + 𝑘𝑌(𝑠)
𝑅(𝑠)
1
𝑚𝑠 2 + 𝑐 ′ 𝑠 + 𝑘 ′
𝑌(𝑠)
2.3 Models of Electric Circuits
Kirchhoff’s current law(KCL):
The algebraic sum of currents leaving a node equals the
sum of currents entering that node.
Kirchhoff’s voltage law(KVL):
The algebraic sum of voltages taking around a closed
path in a circuit is zero.
2.3 Models of Electric Circuits
Electric Elements
Resistor
𝑅
𝑣(𝑡) = 𝑅𝑖(𝑡)
Inductor
𝐿
𝑣(𝑡) = 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
Capacitor
𝐶
𝑖(𝑡) = 𝐶
𝑑𝑣(𝑡)
𝑑𝑡
2.3 Models of Electric Circuits
Example 1: Simple Electric Circuit System
𝑅
𝑢𝑖
(KVL):
𝐿
𝑖
𝐶
Input: 𝑢𝑖 ; Output: 𝑢𝑜
𝑢𝑜
𝑑𝑖 𝑡
𝑢𝑖 𝑡 = 𝑅𝑖 𝑡 + 𝐿
+ 𝑢𝑜 𝑡
𝑑𝑡
1
𝑢𝑜 (𝑡) = න𝑖(𝑡)𝑑𝑡
𝐶
𝑑 2 𝑢𝑜 (𝑡)
𝑑𝑢𝑜 (𝑡)
𝐿𝐶
+ 𝑅𝐶
+ 𝑢𝑜 (𝑡) = 𝑢𝑖 (𝑡)
𝑑𝑡 2
𝑑𝑡
𝑈𝑖 (𝑠)
1
𝐿𝐶𝑠 2 + 𝑅𝐶𝑠 + 1
𝑈𝑂 (𝑠)
2.3 Models of Electric Circuits
Example 2: Bridged Tee Circuit System
𝐶2
𝐶𝑅22
𝑅1
𝑉𝑖
input
𝐶1
𝑉𝑜
output
Question:
Please find the transfer function of this system.
2.3 Models of Electric Circuits
Example 2: Bridged Tee Circuit System
𝐶2
1
2
𝐶1
𝑉𝑜
4
𝑉𝑖 = 𝑉1
𝑉𝑜 = 𝑉3
3
𝐶𝑅22
𝑅1
𝑉𝑖
Node 2 (KCL):
𝑉2 − 𝑉1 𝑉2 − 𝑉3
𝑑𝑉2
+
+ 𝐶1
=0
𝑅1
𝑅2
𝑑𝑡
Node 3 (KCL):
𝑉3 − 𝑉2
𝑑(𝑉3 − 𝑉1 )
+ 𝐶2
=0
𝑅2
𝑑𝑡
𝑉𝑜 (𝑠)
𝑅1 𝐶1 𝑅2 𝐶2 s 2 + (𝑅1 𝐶2 + 𝑅2 𝐶2 )𝑠 + 1
=
𝑉𝑖 (𝑠) 𝑅1 𝐶1 𝑅2 𝐶2 s2 + (𝑅1 𝐶1 + 𝑅1 𝐶2 + 𝑅2 𝐶2 )𝑠 + 1
2.3 Models of Electric Circuits
Example 3: OP Amp (Operational Amplifier 运算放大器) Circuit
◼
Suppose the OP Amp is ideal:
𝑣0 = 𝐴(𝑣+ − 𝑣− )
𝐴 = 𝑉0 /(𝑣+ −𝑣− ) = ∞
𝑣+ − 𝑣− = 0
𝑖− = 𝑖+ = 0
◼
Suppose the OP Amp is non-ideal:
107
𝑉0 =
(𝑣+ −𝑣− )
𝑠+1
2.3 Models of Electric Circuits
Example 3: OP Amp (Operational Amplifier 运算放大器) Circuit
Question: Please find the transfer function of the following system
2.3 Models of Electric Circuits
Example 3: OP Amp (Operational Amplifier 运算放大器) Circuit
◼
Ideal case:
(1 − N )Vin = ( P − N )Vout
𝑣+ − 𝑣− = 0
Vout 1 − N
=
Vin P − N
2.3 Models of Electric Circuits
Example 3: OP Amp (Operational Amplifier 运算放大器) Circuit
◼
107
𝑉0 =
(𝑣+ −𝑣− )
𝑠+1
Non-ideal case:
Summary
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
30
Homework
◼
Textbook, Page 92-97, Problem 2.1, 2.3, 2.11, 2.15(b)(c)
Thanks！
Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
2
Chapter 2 Dynamic Models
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
2.4 Model of Eletromechanical System
2.5 Nonlinear Model of Aircraft System & Linearization
2.6 Typical Elements
2.4 Models of Electromechanical Systems
A common actuator used in
electromechanical system (机电系
统) is the direct current (DC) motor
to provide rotary motion.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
𝑇𝑑
𝐽 : moment of inertia
𝑏: damping coefficient
𝑇𝑑 : torque from load (disturbance)
𝑇𝑚: torque from DC motor
𝐸: back electromotive force (emf)
𝜃: rotating angle
𝜔: rotating angular velocity
𝑉𝑎: control voltage
𝐼𝑎: control current
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
The Motor Equations:
𝑇𝑑
𝑇m = 𝐾m 𝐼𝑎
𝐸 = 𝐾𝑒 𝜔 = 𝐾𝑒 𝜃ሶ
armature current
back emf
The generated electromotive force (emf) works against the
applied armature voltage, so it is called back emf.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
𝑇𝑑
Output:
𝜽: rotating angle
Input:
𝑽𝒂: control voltage
armature current
back emf
Question:
Please Establish dynamic model of this system.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑉𝑎 (𝑠) = 𝐸(𝑠) + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝑇𝑚
𝑇𝑑
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝐸(𝑠) = 𝐾e 𝜔(𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝜔(𝑠) = 𝑠𝜃(𝑠)
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑉𝑎 (𝑠) = 𝐸 + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝑉𝑎 (𝑠)
1
𝑅𝑎 + 𝑠𝐿𝑎
𝐸
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝐼𝑎 (𝑠)
𝐾𝑚
𝜔(𝑠)
𝐾e
𝐸 = 𝐾e 𝜔
𝜔(𝑠) = 𝑠𝜃(𝑠)
𝜔(𝑠)
1
𝑠
𝐼𝑎 (𝑠)
𝑇𝑚 (𝑠)
𝐸(𝑠)
𝜃(𝑠)
𝑇𝑑 (𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝑇𝑚 (𝑠)
1
𝐽𝑠 + 𝑏
𝜔(𝑠)
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝑉𝑎 (𝑠) = 𝐸 + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝐸 = 𝐾e 𝜔
𝜔(𝑠) = 𝑠𝜃(𝑠)
𝐾e
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
TG: Tachogenerator （测速发电机）
SM: Servomotor （伺服电机）
𝐾𝑝
𝑢𝜔 = 𝐾𝑇 𝜔
𝜃𝑖𝑛
𝑢𝑖𝑛 = 𝐾𝑝 𝜃𝑖𝑛
𝐾𝑇
𝐾𝑝
𝑢𝜃 = 𝐾𝑝 𝜃𝑜𝑢𝑡
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
𝐾𝑝
𝑢𝜔 = 𝐾𝑇 𝜔
𝜃𝑖𝑛
𝑣𝑎
𝑢1
𝑢2
𝑢𝑖𝑛 = 𝐾𝑝 𝜃𝑖𝑛
𝐾𝑇
𝐾𝑝
𝑢𝜃 = 𝐾𝑝 𝜃𝑜𝑢𝑡
𝑢𝑖𝑛 −𝑢𝜃 𝑢1
+
+
=0
10
10
30
𝑢1 −𝑢𝜔 𝑢2
+
+
=0
10
10
20
𝑣𝑎 = 𝐾3 𝑢2
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
DC Servo Motor
𝑣𝑎
𝜔
𝑇𝑑
𝑣𝑎
𝐾𝑚
𝑅𝑎
𝜔
1
𝐽𝑠 + 𝑏
𝐾𝑏
1
𝑠
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
𝑇𝑑
𝜃𝑖𝑛
𝐾𝑝
3
2
𝑣𝑎
𝐾3
𝐾𝑚
𝑅𝑎
1
𝐽𝑠 + 𝑏
𝐾𝑏
𝐾𝑇
Speed feedback
𝐾𝑝
Position feedback
Block Diagram of Position Tracking System
𝜔
1
𝑠
𝜃𝑜𝑢𝑡
14
2.5 Nonlinear Model of Aircraft & Linearization
◼
Many complex processes are nonlinear in physics,
for example the aircraft system.
◼
However, the analysis and design of linear system
is much easier than that of nonlinear system.
◼
So, we need Linearization to find a linear model to
approximate a nonlinear one.
15
2.5 Nonlinear Model of Aircraft & Linearization
6 DOF nonlinear aircraft dynamic model
16
2.5 Nonlinear Model of Aircraft & Linearization
Force equations
𝑋𝑎𝑒𝑟𝑜 + 𝑇𝑥
− 𝑔𝑠𝑖𝑛𝜃
𝑚
𝑌𝑎𝑒𝑟𝑜 + 𝑇𝑦
𝑣ሶ + 𝑟𝑢 − 𝑝𝑤 =
+ 𝑔𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝛷
𝑚
𝑍𝑎𝑒𝑟𝑜 + 𝑇𝑧
𝑤ሶ + 𝑝𝑣 − 𝑞𝑢 =
+ 𝑔𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛷
𝑚
𝑢ሶ + 𝑞𝑤 − 𝑟𝑣 =
Moment equations
𝑢, 𝑣, 𝑤 – velocity of X,Y,Z
𝑝, 𝑞, 𝑟 – angular velocity of
roll, pitch, yaw
𝜃, 𝛷, 𝜓 – angles of
roll, pitch, yaw
𝑔 – gravitational force
per unit mass
𝑙, 𝑚, 𝑛 – aerodynamic torques
𝐼𝑖𝑖 – the inertias in body axes
𝐼𝑥𝑥 𝑝ሶ + 𝐼𝑧𝑧 − 𝐼𝑦𝑦 𝑞𝑟 − 𝐼𝑥𝑧 𝑟ሶ + 𝑝𝑞 = 𝑙
𝐼𝑦𝑦 𝑞ሶ + 𝐼𝑥𝑥 − 𝐼𝑧𝑧 𝑝𝑟 − 𝐼𝑥𝑧 𝑟 2 − 𝑝2 = 𝑚
𝐼𝑧𝑧 𝑟ሶ + 𝐼𝑦𝑦 − 𝐼𝑥𝑥 𝑝𝑞 − 𝐼𝑥𝑧 𝑝ሶ − 𝑞𝑟 = 𝑛
𝑋𝑎𝑒𝑟𝑜, 𝑌𝑎𝑒𝑟𝑜, 𝑍𝑎𝑒𝑟𝑜 – aerodynamic
force in X,Y, Z
𝑇𝑥 , 𝑇𝑦, 𝑇𝑧 – thrust in X, Y, Z
17
2.5 Nonlinear Model of Aircraft & Linearization
Kinematic equations
𝛷ሶ = 𝑝 + 𝑞𝑠𝑖𝑛𝛷 + 𝑟𝑐𝑜𝑠𝛷 𝑡𝑎𝑛𝜃
𝜃ሶ = 𝑞𝑐𝑜𝑠𝛷 − 𝑟𝑠𝑖𝑛𝛷
𝛹ሶ = 𝑞𝑠𝑖𝑛𝛷 + 𝑟𝑐𝑜𝑠𝛷 𝑠𝑒𝑐𝜃
The related force and moment components
𝑇𝑥, 𝑇𝑦, 𝑇𝑧 are thrust related force
𝑋𝑎𝑒𝑟𝑜, 𝑌𝑎𝑒𝑟𝑜, 𝑍𝑎𝑒𝑟𝑜 are aerodynamic related force
𝑙, 𝑚, 𝑛 are aerodynamic related moment
They are function of elevator deflection 𝞭𝑒 (升降舵偏转)，aileron deflection
𝞭𝑎 (副翼偏转), rudder deflection 𝞭𝑟 (方向舵偏转), and throttle setting
𝞭𝜋 (油门调节).
2.5 Nonlinear Model of Aircraft & Linearization
19
2.5 Nonlinear Model of Aircraft & Linearization
So, 6 DOF flight dynamics is a set of 9 ODEs
𝑥ሶ = 𝑓(𝑥, 𝑢)
states:
controls:
𝑥 = {𝑢, 𝑣, 𝑤, 𝑝, 𝑞, 𝑟, 𝛷, 𝜃, 𝛹}𝑇
or {𝛼, 𝛽, 𝑉𝑇 , 𝑝, 𝑞, 𝑟, 𝛷, 𝜃, 𝛹}𝑇
𝑢 = {𝛿𝑎, 𝛿𝑒, 𝛿𝑟, 𝛿π}𝑇
𝛼: angle of attack
𝛽: sideslip angle
20
2.5 Nonlinear Model of Aircraft & Linearization
Linear model of aircraft
Basic trim condition:
– steady, symmetric, level flight
旋转
侧滑
倾斜
21
2.5 Nonlinear Model of Aircraft & Linearization
For a moderate rate in states, the behavior of aircraft
dynamics can be decoupled into 2 sets
纵向
横向
22
2.5 Nonlinear Model of Aircraft & Linearization
Linearization of nonlinear aircraft model
23
2.5 Nonlinear Model of Aircraft & Linearization
Linearization of nonlinear aircraft model
24
2.5 Nonlinear Model of Aircraft & Linearization
With linearization, we have 2 sets of linear ODEs
𝑢
{𝛼,
, 𝑞, 𝜃}𝑇
𝑉𝑇0
𝑥ሶ = 𝐴 𝑥 + 𝐵 𝑢
State: 𝑥 =
∆ is omitted
Input: 𝑢 = {𝛿𝑒}
𝐴 =
𝑍𝛼
𝑋𝛼
𝑚𝛼
0
𝑍𝑢 1 + 𝑍𝑞
𝑋𝑢
𝑋𝑞
𝑚 𝑢 𝑚𝑞
0
1
0
−𝑔/𝑉𝑇0
0
0
𝐵 =
𝑍𝛿
𝑋𝛿
𝑚𝛿
0
25
2.5 Nonlinear Model of Aircraft & Linearization
𝑥ሶ = 𝐴 𝑥 + 𝐵 𝑢
Input: 𝑢 = {𝛿𝑎, 𝛿𝑟}
∆ is omitted
𝐴 =
𝑌𝛽
𝑙𝛽
𝑛𝛽
0
State: 𝑥 = {𝛽, 𝑝, 𝑟, 𝛷}𝑇
𝑌𝑝 𝑌𝑟 − 1 𝑔/𝑉𝑇0
𝑙𝑝
𝑙𝑟
0
𝑛𝑝
𝑛𝑟
0
0
0
1
𝐵 =
𝑍𝛿
𝑋𝛿
𝑚𝛿
0
26
2.6 Typical Elements of Linear Dynamic Systems
1. Proportion Element（or Amplifying Element）
2. Differential Element
3. Integral Element
4. Inertial Element (First-order Integral )
5. Oscillation Element (Second-order Integral )
6. First-order Differential Element
7. Second-order Differential Element
27
1. Proportion Element（or Amplifying Element）
R (s )
C (s )
K
Features: Output depends on input according to a certain proportion.
Dynamic equation :
c(t)=Kr(t)
K——Amplification coefficient, are usually have dimension
Transfer function：
Frequency response：
H(s) =
C(s)
=K
R(s)
H(j  ) =
C(j )
R(j )
=K
28
1. Proportion Element（or Amplifying Element）
电位器：将线位移或角位移变换为电压量的装置
Input：(t)——angle
E——constant voltage
Output：u(t)——voltage
+
E
-
u(t)
 (s )
 (t )
U (s )
K
•
+
Dynamic equation ：𝑢 𝑡 = 𝐾θ(𝑡)
Transfer function ：𝐻 𝑠 =
𝑈(𝑠)
=𝐾
𝜃(𝑠)
Frequency response: 𝐻 𝑗ω = 𝐾
29
1. Proportion Element（or Amplifying Element）
Gear 传动装置
Ex 2：Input：n1(t)——rotate speed, Z1—— No. of teeth of active wheel
Output：n2(t)——rotate speed, Z2—— No. of teeth of passive wheel
passive wheel
active wheel
Z1
n1 (t )
n2 ( t )
N1 ( s )
Z2
𝑧1
𝑛 (𝑡)
𝑧2 1
𝑁2 (𝑠)
𝑧1
=
= =𝐾
𝑁1 (𝑠)
𝑧2
𝑁2 (𝑗𝑤)
𝑧1
𝑗𝑤 =
=
𝑁1 (𝑗𝑤)
𝑧2
z1
z2
Dynamic equation ：𝑛2 𝑡 =
Transfer function：𝐻 𝑠
Frequency response: 𝐻
=𝐾
N2 ( s)
30
1. Proportion Element（or Amplifying Element）
Other Proportion Elements
voltage divider（分压器）
proportional circuit
R(s)
-
r1
r2
r2
+
r (t )
c (t )
C (s)
r1 + r2
𝐶(𝑠)
𝑟2
𝐻(𝑠) =
=
𝑅(𝑠) 𝑟1 + 𝑟2
+ Ec
R2
R1
r (t )
Triode(三极管)
K
c (t )
R3
R(s)
−
ic (t )
R
R2
C (s)
ib (t )
Ib (s)
R1
𝐶(𝑠)
𝑅2
𝐻(𝑠) =
=−
𝑅(𝑠)
𝑅1
𝐻(𝑠) =

Ic (s)
𝐶(𝑠)
=𝛽
𝑅(𝑠)
31
2. Differential Element
Features：Dynamic process, output is proportional to the rate
(differentiation) of input.
R (s )
S
Equation of motion：𝐶 𝑡 = 𝐾
C (s )
d𝑟 (𝑡)
𝑑𝑡
𝐶(𝑠)
𝑅(𝑠)
= 𝐾𝑠
Frequency response：𝐻 𝑗𝑤 =
𝐶(𝑗𝑤)
𝑅 𝑗𝑤
Transfer function：𝐻 𝑠 =
= 𝑗𝐾𝑤
32
2. Differential Element
Ex1. RC circuit
i (t )
Input——ur(t)
C
u r (t )
𝐶
R
𝑑𝑢(𝑡)
= 𝑖(𝑡)
𝑑𝑡
u c (t )
Output——uc(t)
1
𝑢 𝑡 = න 𝑖(𝑡) 𝑑𝑡
𝐶
Try to write transfer function.
33
2. Differential Element
EX1 RC circuit
C
i (t )
Input——ur(t)
1
𝑢𝑟 𝑡 = න 𝑖 𝑡 𝑑𝑡 + 𝑖 𝑡 R
𝐶
u r (t )
R
u c (t )
𝑖 𝑡 =
Output——uc(t)
𝑢𝑐 (𝑡)
𝑅
1
Equation of motion：𝑢𝑟 𝑡 = 𝑅𝐶 ‫ 𝑡𝑑)𝑡( 𝑐𝑢 ׬‬+ 𝑢𝑐 (𝑡)
Transfer function：𝐻 𝑠 =
𝑈𝑐 (𝑠)
𝑈𝑟 (𝑠)
=
𝑇𝑐 𝑠
𝑇𝑐 𝑠+1
(Tc=RC – product of R and C)
𝑈 (𝑠)
When Tc<<1，the up formula is change to：𝐻 𝑠 = 𝑈𝑐 (𝑠) = 𝑇𝑐 𝑠
𝑟
Frequency response：𝐺 𝑗𝑤 = 𝑗𝑇𝑐 𝑤
34
2. Differential Element
Ex2：Mathematical Model of Generator
 (t )
u d (t )
F
D
u f (t )
Input： (𝑡)——angle of the rotor of motor D
Output： 𝑢𝑓(𝑡)——armature voltage of generator F
Dynamic equation：𝑢𝑓 𝑡 = 𝐾
𝑑𝜑(𝑡)
𝑑𝑡
Transfer function：𝐻 𝑠 = 𝐾𝑠
Frequency response ：
𝐻 𝑗𝑤 = 𝑗𝐾𝑤
35
2. Differential Element
Other Differential Elements
C
i (t )
C
i (t )
I (s)
Cs
U (s)
eL (t )
L
u (t )
uc (t )
U c (s)
i (t )
R
Cs
1
R
+
+
I (s)
I (s)
Ls
EL ( s )
Try to write transfer function
I (s)
H ( s) =
= Cs H (s) = I (s) = Cs + 1
U ( s)
U ( s)
R
E ( s)
H ( s) =
= Ls
I ( s)
36
3. Integral Element
R (s )
1
s
C (s )
Feature： The change rate of output is proportional
to the input. (or the output is proportional to the
integration of input).
Dynamic
𝑑𝑐(𝑡)
equation：
𝑑𝑡
= 𝐾𝑟(𝑡)
Transfer function：𝐻 𝑠 =
𝐾
𝑠
Frequency Response： 𝐻 𝑗𝑤 =
𝐾
𝑗𝑤
37
3. Integral Element
Ex1：Integral circuit
Input : r(t); Output: c(t)
Transfer function?
38
3. Integral Element
Ex1：Integral circuit
ic (t )
i1 (t ) R1
+
C
K
r (t )
c (t )
R3
Input : 𝑟(𝑡)，Output: 𝑐(𝑡)
𝑖𝑐 𝑡 = 𝑖1 𝑡 =
R (s )
𝑟(𝑡)
𝑅1
1
1
‫𝑟׬‬
1𝐶
Dynamic equation：𝑐 𝑡 = − ‫ = 𝑡𝑑 𝑡 𝑐𝑖 ׬‬−
𝐶
𝑅
𝐶(𝑠)
𝑠
Transfer function:（𝑇 = 𝑅1𝐶）𝐺 𝑠 = 𝑅
Frequency function: 𝐺 𝑗𝑤 =
𝐶(𝑗𝑤)
𝑅(𝑗𝑤)
=−
1
−
R1Cs
1
𝑡 𝑑𝑡 = − ‫𝑡𝑑 𝑡 𝑟 ׬‬
𝑇
𝐾
= − 𝑇𝑠 = − 𝑠
𝐾
−𝑗𝑇𝑤
C (s )
1
39
3. Integral Element
Other Integral Elements
i (t )
u (t )
Current
I (s)
Voltage
1
Cs
U (s)
𝑈(𝑠)
1
𝐻(𝑠) =
=
𝐼(𝑠) 𝐶𝑠
40
4. Inertia Element
Features： This link has an independent energy storage components,
so that the input cannot be transferred to output immediately,
existing delay in time.
R (s )
𝑑𝑐(𝑡)
+𝑐
𝑑𝑡
𝐾
𝑠 =
𝑇𝑠+1
Dynamic equation: 𝑇
Transfer function:𝐺
1
Ts + 1
𝑡 = 𝐾𝑟(𝑡)
Frequency response：𝐺 𝑗𝑤 =
𝐾
𝑗𝑇𝑤+1
C (s )
41
4. Inertia Element
id
Ex1：DC Motor
+
Input: 𝑢𝑑 ——armature voltage
Output: 𝑖𝑑 ——armature current
Dynamic equation： 𝐿𝑑
𝜏𝑑
ud
𝑑𝑖𝑑
+ 𝑅𝑑 𝑖𝑑 = 𝑢𝑑
𝑑𝑡
𝑑𝑖𝑑
𝑢𝑑
+ 𝑖𝑑 =
𝑑𝑡
𝑅𝑑
1
𝐼 (𝑠)
𝑅𝑑
Transfer function: 𝐻(s) = 𝑑
=
𝑈𝑑 (𝑠) 𝜏𝑑 𝑠 + 1
𝐿𝑑 ——armature (电枢) inductance
𝑅𝑑 ——armature resistance
𝜏𝑑 ——armature time constant
𝜏𝑑 =
𝐿𝑑
𝑅𝑑
D
42
4. Inertia Element
Other inertial elements
L
r(t)
R
c(t)
v(t )
f (t )
M
T (t )
 (t )
B
J
B
R( s)
1
L
s +1
R
C (s)
F (s)
1
𝐵
𝑀
𝑠+1
𝐵
V (s)
T ( s)
1
B
J
s +1
B
(s)
43
5. Oscillation Element
Features：Contains two independent energy storage components,
when the input of the change, two energy storage components
exchange the energy, to make the output with the nature of the
oscillation.
2 c(t)
d
d𝑐(t)
Dynamic equation： 𝑇
+ 2𝜍𝑇
+ 𝑐(t) = Kr(t)
2
dt
dt
2
Transfer function：
——Damping ratio， 𝑇——Time constant
Frequency response：𝐺(𝑗𝜔) = 𝐶(𝑗𝜔) =
𝑅(𝑗𝜔)
1
(1 − 𝑇 2 𝜔 2 ) + 𝑗2𝜍𝑇𝜔
44
5. Oscillation Element
L
R
+
EX1：RLC circuit
+
r(t)
i (t )
_
𝑟(𝑡) = 𝐿
𝑑𝑖(𝑡)
1
+ 𝑅𝑖(𝑡) + න𝑖(𝑡)𝑑𝑡
𝑑𝑡
𝐶
𝑐(𝑡) =
1
න𝑖(𝑡)𝑑𝑡
𝐶
Transfer
Elimination
Variables
𝑖(𝑡)
c(t)
C
_
d2 c(t)
dc(t)
LC
+
RC
+ c(t) = r(t)
dt 2
dt
ODE model
1
function：G(s) = LCs2 + RCs + 1
Frequency
response：G(jω) =
1
1
=
LC(jω)2 + RC(jω) + 1 (1−LCω2 ) + jRCω
_
45
5. Oscillation Element
Ex2: Armature-controlled DC Motor
i (t )
+
R
L
 (t )
eb (t )
ea (t )
+
_
D
_
𝑒𝑎(𝑡) --- Input voltage applied to the armature
(𝑡) --- Output : angular displacement of the shaft
𝑅 --- Armature resistance；
𝐿 --- Armature (电枢绕组) inductance ；
𝑖(𝑡)--- Armature current；
𝑒𝑏(𝑡) --- Generator back-EMF(反电势)；
𝑇(𝑡) --- Generator torque ；
𝐽 --- moment of inertia ；
𝐵 --- Viscous friction coefficient
J
B
46
5. Oscillation Element
1）𝑇 𝑡 = 𝐾𝑖 𝑡 , 𝑇(𝑡)——Torque 𝐾——Torque coefficient
𝑑𝜃(𝑡)
2）𝑒𝑏 𝑡 = 𝐾𝑏 𝑑𝑡 , 𝑒𝑏(𝑡)——Back-EMF, 𝐾𝑏——Back-EMF coefficient
𝑒𝑎(𝑡)——Armature Voltage
3）𝐿
𝑑𝑖(𝑡)
𝑑𝑡
+ 𝑅𝑖 𝑡 + 𝑒𝑏 𝑡 = 𝑒𝑎 (𝑡)
i a (t )
𝑑 2 𝜃(𝑡)
4）𝐽
𝑑𝑡 2
+
𝑑𝜃(𝑡)
𝐵
𝑑𝑡
L
+
= 𝑇(𝑡)
Laplace transform:
1）
2）
3）
4）
R
𝑇(𝑠) = 𝐾𝐼(𝑠)
𝐸𝑏(𝑠) = 𝐾𝑏𝑠(𝑠)
𝐸𝑎(𝑠) = (𝐿𝑠 + 𝑅)𝐼(𝑠) + 𝐸𝑏(𝑠)
𝑇(𝑠) = (𝐽𝑠2 + 𝐵𝑠) (𝑠)
 (t )
eb (t )
ea ( t )
_
+
_
D
J
B
47
5. Oscillation Element
Elimination Variables 𝐸𝑏 (𝑠), 𝑇(𝑠) and 𝐼(𝑠)
𝜃(𝑠)
𝐾
=
𝐸𝑎 (𝑠) 𝑠[𝐿𝐽𝑠 2 + 𝐿𝐵 + 𝑅𝐽 𝑠 + (𝑅𝐵 + 𝐾𝐾𝑏 )]
Input 𝐸𝑎(𝑠), output speed (𝑠), we can get：
Ω(𝑠)
𝐾
=
𝐸𝑎 (𝑠) 𝐿𝐽𝑠 2 + 𝐿𝐵 + 𝑅𝐽 𝑠 + (𝑅𝐵 + 𝐾𝐾𝑏 )
This is a typical oscillation element.
48
5. Oscillation Element
If we Ignore the effect of 𝐿, i.e., 𝐿=0:
𝜃(𝑠)
𝐾𝑚
=
𝐸𝑎 (𝑠) 𝑠(𝑇𝑚 𝑠 + 1)
Ω(𝑠)
𝐾𝑚
=
𝐸𝑎 (𝑠) 𝑇𝑚 𝑠 + 1
where:𝐾𝑚 =
𝐾
,
𝑅𝐵+𝐾𝐾𝑏
Generator gain constant
𝑇𝑚 =
𝑅𝐽
,
𝑅𝐵+𝐾𝐾𝑏
Generator time constant
When 𝑇𝑚 approximates to zero (𝑅 → 0), we can get：
𝜃(𝑠)
1/𝐾𝑏 𝐾 ′
1
′
=
= , (𝐾 = )
𝐸𝑎 (𝑠)
𝑠
𝑠
𝐾𝑏
Ω(𝑠)
= 𝐾′
𝐸𝑎 (𝑠)
49
5. Oscillation Element
EX3：Mechanical System
K
f (t )
Input---Force 𝑓(𝑡)，
Output---Displacement 𝑥(𝑡)
𝐾---Elastic coefficient
𝑀---Mass of the object
𝐵---Viscous friction coefficient
Transfer function?
M
x (t )
B
图2-16 机械振荡
Mechanic
Movement
50
5. Oscillation Element
K
f (t )
EX3：Mechanical System
M
x (t )
Input-----Force 𝑓(𝑡)
Output----Displacement 𝑥(𝑡)
B
Differential equation:
𝑑 2 𝑥(𝑡)
𝑑𝑥(𝑡)
𝑓 𝑡 =𝑀
+𝐵
+ 𝐾𝑥(𝑡)
2
𝑑𝑡
𝑑𝑡
where：𝐾 — Elastic coefficient
𝑀 — Mass of the object，
𝐵 — Viscous friction coefficient
Transfer function：
𝑋(𝑠)
1/𝐾
𝐻 𝑠 =
=
𝐹(𝑠) 𝑀 𝑠 2 + 𝐵 𝑠 + 1
𝐾
𝐾
图2-16 机械振荡
Mechanic
Movement
51
6. First-order Differential Element
Features：the output not only is related to input itself, but also
related to the change rate of input
Dynamic equation：𝑐 𝑡 =
𝑑𝑟(𝑡)
𝑇
𝑑𝑡
+ 𝑟(𝑡)
Transfer function： 𝐻 𝑠 = 𝑇𝑠 +1
Frequency response： 𝐻 𝑗𝑤 = 𝑗𝑇𝑤 + 1
52
6. First-order Differential Element
RC Circuit
Input：𝑢(𝑡)，Output：𝑖(𝑡) ，hence
𝑖 𝑡 = 𝑖1 𝑡 + 𝑖2 𝑡 = 𝐶
=
Transfer
1
𝑅
𝑅𝐶
𝑑𝑢 𝑡
𝑑𝑡
+𝑢 𝑡
𝐼(𝑠)
function：
𝑈(𝑠)
（assume 𝑅 = 1, 𝑅𝐶 =
𝑑𝑢(𝑡)
𝑑𝑡
=𝜏
+
𝑢(𝑡)
𝑅
𝑑𝑢(𝑡)
𝑑𝑡
+ 𝑢(𝑡)
= 𝜏𝑠 + 1
）
Frequency response ：𝐻 𝑗𝑤 = 1 + 𝑗𝜏𝑤
u (t )
i1 (t )
C
i2 (t )
R
i (t )
53
7. Second-order Differential Element
Dynamic equation：𝑐 𝑡 =
Transfer function：𝐺 𝑠 =
2
2 𝑑 𝑟(𝑡)
𝑇
𝑑𝑡 2
𝐶(𝑠)
𝑅(𝑠)
+ 2𝜉𝑇
𝑑𝑟(𝑡)
𝑑𝑡
+ 𝑟(𝑡)
= 𝑇 2 𝑠 2 + 2𝜉𝑇𝑠 + 1
Frequency function：𝐺 𝑗𝑤 = 𝑇 2 (𝑗𝑤) 2 +2𝜉𝑇 𝑗𝑤 + 1
= (1 + 𝑇 2 𝑤 2 ) + 𝑗2𝜉𝑇𝑤)
2.6 Typical Elements of Linear Dynamic Systems
Example
?
How many typical elements the transfer function have?
2.6 Typical Elements of Linear Dynamic Systems
Solution:
56
Summary of Chapter 2
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
2.4 Model of Eletromechanical System
2.5 Nonlinear Model of Aircraft System & Linearization
2.6 Typical Elements
57
Homework
Page 99-100，Problem 2.20
Thanks！
Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
Chapter 3 Dynamic Response
3.1 Laplace Transforms
3.2 Transfer Function
3.3 Modeling Diagrams
3.4 Time-Domain Specifications
3.5 Effect of Pole & Zero Locations
3.6 Amplitude & Time Scaling
3.7 Stability
3
3.1 Laplace Transforms
4
3.1 Laplace Transforms
• 赫维赛德最伟大的贡献是简化了麦克斯韦
的原始方程组，挖掘出了蕴含在麦克斯韦
方程内部的深刻意义，从而使简化后麦克
斯韦方程组呈现出无与伦比的对称性，成
为历史上是最漂亮的方程式（没有唯一）
• 靠“直觉”，引入微积分算子，将常微分
方程转换为普通代数方程（民科？）
奥列弗. 赫维赛德（1985-1925）
5
3.1 Laplace Transforms
• 微积分算子
常微分方程：𝑚𝑥ሷ + 𝑏𝑥ሶ = 𝑢
代数方程：𝑚𝑝2 + 𝑏𝑝 = 𝑢
奥列弗. 赫维赛德（1985-1925）
赫维赛德的微积分算子，就是拉普拉斯变换的前身!
6
3.1 Laplace Transforms
• 傅里叶变换
傅里叶《热的解析》：任何连续周期信号可
以由一组适当的正弦曲线组合而成
拉格朗日：无法表示有棱角的信号
15年后正式发表，狄利赫里完善傅里叶变换
3.1 Laplace Transforms
• 只要一个函数满足如狄利赫里条件，都能分解为复指数函数之和。
包括拉格朗日提到的带有棱角的方波函数。
方波函数傅里叶变换
𝑓 𝑡 = 𝑡 2 ???
3.1 Laplace Transforms
𝑓 𝑡 = 𝑡 2 ???
Solution: 把不满足绝对的可积的函数乘以一个快速衰减的函数，
在趋于 ∞ 时原函数也衰减到零，从而满足绝对可积。
select proper 𝜎 so that the limit exists
为保证 𝑒 −𝜎𝑥 一直为衰减函数，把𝑥定义域缩减到正半轴，傅里叶变换就变成
(拉普拉斯变换)
3.1 Laplace Transforms
𝑒 −𝑖𝜔𝑡 , (𝑒 −𝑖𝜔𝑡 = cos −𝜔𝑡 + 𝑖sin(−𝜔𝑡))
螺旋曲线
𝑒 −𝑠𝑡 , (𝑠 = 𝜎 + 𝑖𝜔𝑡, 𝜎 > 0)
半径衰减的螺旋曲线
3.1 Laplace Transforms
总结：
• 傅里叶变换是将函数分解到频率不同、幅值恒为1的单位圆上
• 拉普拉斯变换是将函数分解到频率幅值都在变化的圆上
• 因为拉普拉斯变换的基有两个变量，因此更灵活，适用范围更广
11
3.1 Laplace Transforms
ℒ
(双边拉普拉斯变换)
（因果性)
ℒ
12
3.1 Laplace Transforms
Example 1
e−𝑡 𝑢(𝑡)
（α < 0）
（𝑅e(𝑠) > α）
13
3.1 Laplace Transforms
Example 1
（𝑅e 𝑠 > ⍺）
lim 𝑒 −
𝑡→∞
𝑠−⍺ 𝑡
− 𝑒 0 = −1
14
3.1 Laplace Transforms
Unit Step Signal
15
3.1 Laplace Transforms
Example 2
Volunteer？
16
3.1 Laplace Transforms
Example 2
17
3.1 Laplace Transforms
Impulse Signal
1 𝑎 −𝑠𝑡
lim න 𝑒 𝑑𝑡
𝑎→0+ 𝑎 0−
1
1
= lim+ − 𝑒 −𝑠𝑡 |𝑎0−
𝑎
𝑠
−1
= lim+ 𝑎𝑠 𝑒 −𝑎𝑠 − 1
𝑎→0
−1
= lim+ 𝑠 −s𝑒 −𝑎𝑠
𝑎→0
𝑎→0
=1
18
3.1 Laplace Transforms
19
3.1 Laplace Transforms
Example 3
ℒ
ℒ
20
3.1 Laplace Transforms
21
3.1 Laplace Transforms
Example 4
Volunteer？
22
3.1 Laplace Transforms
Example 4
ℒ
ℒ
23
3.1 Laplace Transforms
24
3.1 Laplace Transforms
Proof of
Integration Property？
25
3.1 Laplace Transforms
More Examples
ℒ
ℒ
Volunteer？
ℒ
ℒ
26
3.1 Laplace Transforms
More Examples
ℒ
ℒ
ℒ
ℒ
27
3.1 Laplace Transforms
Properties for Laplace Transform
ℒ
ℒ
ℒ
28
3.1 Laplace Transforms
Properties for Laplace Transform
ℒ
(实卷积)
ℒ
29
3.1 Laplace Transforms
Properties for Laplace Transform
30
3.1 Laplace Transforms
31
3.1 Laplace Transforms
Properties for Laplace Transform
32
Page . 32
3.1 Laplace Transforms
33
Homework
Page 179-180，Problems
⚫
3.3 (a) (b) (c)
⚫
3.4 (a) (b) (e)
Thanks！
Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
Chap3 Dynamic Response
3.1 Laplace Transform
3.2 Transfer Function
3.3 Modeling Diagrams
3.4 Time-Domain Specifications
3.5 Effect of Pole & Zero Locations
3.6 Amplitude & Time Scaling
3.7 Stability
Page . 33
3.1 Laplace Transform
Inverse Laplace Transform
𝑥 𝑡
𝐿−1 𝑋 𝑠
𝐿
𝑋 𝑠
𝐿−1
𝑥(𝑡)
1 𝜎1+𝑗∞
=
න
𝑋(𝑠)𝑒 𝑠𝑡 𝑑𝑠
𝑗2𝜋 𝜎1−𝑗∞
• 𝜎1 is such that the integral is taken over a line in the region of
convergence
• Very difficult to apply directly
• Instead, we convert 𝑋(𝑠) to a form such that we can easily find the
inverse
4
Page . 4
3.1 Laplace Transform
Example 1
𝑠+8
4
3
𝑋 𝑠 =
= −
𝑠(𝑠 + 2) 𝑠 𝑠 + 2
Since we know
1
𝑠
𝑢 𝑡 ↔ ,
𝑒 −𝑎𝑡 𝑢 𝑡 ↔
1
𝑠+𝑎
𝑎1 𝑥1 𝑡 𝑢 𝑡 + 𝑎2 𝑥2 𝑡 𝑢 𝑡 ↔ 𝑎1 𝑋1 𝑠 + 𝑎2 𝑋2 𝑠
so
𝐿−1 𝑋 𝑠
= 4𝑢 𝑡 − 3𝑒 −2𝑡 𝑢(𝑡)
5
3.1 Laplace Transform
Solving Differential Equations
The differentiation theorem
𝑑
𝐿
𝑓 𝑡
𝑑𝑡
= 𝑠𝐿 𝑓 𝑡
− 𝑓(0− )
Higher order derivatives
𝑑2
𝐿
𝑓 𝑡
𝑑𝑡 2
𝑑
= 𝑠𝐿
𝑓 𝑡
𝑑𝑡
𝑑𝑛
𝐿
𝑓 𝑡
𝑑𝑡 𝑛
= 𝑠 𝑛 𝐹 𝑠 − 𝑠 𝑛−1 𝑓 0 − ⋯ − 𝑠𝑓
𝑑𝑓 −
𝑑𝑓 −
2
−
−
0 = 𝑠 𝐹 𝑠 − 𝑠𝑓 0 −
(0 )
𝑑𝑡
𝑑𝑡
𝑛−2
0 −𝑓
𝑛−1
0
6
3.1 Laplace Transform
𝑛−1
𝑑𝑛
𝑑
𝐿 𝑛
𝑛−1 𝑓 0− − 𝑠 𝑛−2 𝑓 0− − ⋯ −
−
𝑓
𝑡
↔
𝑠
𝐹
𝑠
−
𝑠
𝑓
0
𝑑𝑡 𝑛
𝑑𝑡 𝑛−1
Differentiation Theorem when initial conditions are zero
𝐿
𝑑
𝑓 𝑡
𝑑𝑡
= 𝑠𝐹(𝑠)
𝑑2
𝐿
𝑓 𝑡
𝑑𝑡 2
= 𝑠 2 𝐹(𝑠)
𝑑𝑛
𝐿
𝑓 𝑡
𝑑𝑡 𝑛
= 𝑠 𝑛 𝐹(𝑠)
7
3.1 Laplace Transform
Example 2
Consider
𝑑𝑥(𝑡)
= 1, 𝑡 ≥ 0
𝑑𝑡
What is 𝑥(𝑡)?
Since the functions are equal on the left and right, the Laplace Transforms will
be equal.
1
−
𝑠𝑋 𝑠 − 𝑥 0 =
𝑠
Solve for 𝑋(𝑠):
1 𝑥(0− )
𝑋 𝑠 = 2+
𝑠
𝑠
8
3.1 Laplace Transform
Example 2
1 𝑥(0− )
𝑋 𝑠 = 2+
𝑠
𝑠
We take the inverse Laplace Transforms on the right hand side
𝑥 𝑡 = 𝑡 + 𝑥 0− , 𝑡 ≥ 0
Solution Summary
• Use differentiation theorem to take Laplace Transform of the differential
equation
• Solve for the unknown Laplace Transform Function
• Find the inverse Laplace Transform
9
3.1 Laplace Transform
Example 2
Find the Laplace Transform for the solution to
𝑥ሷ + 3𝑥ሶ + 2𝑥 = 1, 𝑡 ≥ 0, 𝑥 0 = 1, 𝑥ሶ 0 = −3
𝑑𝑛
𝐿
𝑓 𝑡
𝑑𝑡 𝑛
= 𝑠 𝑛 𝐹 𝑠 − 𝑠 𝑛−1 𝑓 0 − ⋯ − 𝑠𝑓
𝑛−2
0 −𝑓
𝑛−1
Take Laplace Transform of both sides:
𝑠 2 𝑋 𝑠 − 𝑠𝑥 0 − 𝑥ሶ 0 + 3 𝑠𝑋 𝑠 − 𝑥 0 + 2𝑋 𝑠 =
1
𝑠 𝑋 𝑠 − 𝑠 + 3 + 3 𝑠𝑋 𝑠 − 1 + 2𝑋 𝑠 =
𝑠
𝑠2 + 1
𝑋 𝑠 =
𝑠(𝑠 2 + 3𝑠 + 2)
2
1
𝑠
0
3.1 Laplace Transform
Partial Fraction Expansions
In general, ODEs can be transformed into a function that is expressed as a ratio of
polynomials.
In a partial fraction expansion we try to break it into its parts, so we can use
Laplace Transform Table to go back to the time domain:
𝑠2 + 1
𝐴
𝐵
𝐶
𝑋 𝑠 =
= +
+
𝑠(𝑠 + 1)(𝑠 + 2) 𝑠 𝑠 + 1 𝑠 + 2
Three ways of finding coefficients
• Put partial fraction expansion over common denominator and equate
coefficients of s
• Residue formula
• Equate both sides for several values of s
3.1 Laplace Transform
Partial Fraction Expansions
Have to consider that in general we can encounter:
-
Real, distinct roots
-
Real repeated roots
-
Complex conjugate pair roots (2𝑛𝑑 order terms)
-
Repeated complex conjugate roots
𝑁(𝑠)
𝐴
𝐵
𝐶𝑠 + 𝐷
𝐸𝑠 2 + 𝐹𝑠 + 𝐺
𝑋 𝑠 =
=𝐾+
+
+
+
+⋯
𝐷(𝑠)
𝑠 + 𝑝1 (𝑠 + 𝑝1 )2 (𝑠 + 𝑎)2 +𝑏2 ( 𝑠 + 𝑎 2 + 𝑏2 )2
3.1 Laplace Transform
Example 3
Given 𝑋 𝑠 , find 𝑥(𝑡)
1+𝑠 2
𝑋 𝑠 =
𝑠(𝑠 2 + 3𝑠 + 2)
Step 1:
Factorize the denominator, then use partial fraction expansion:
𝑋 𝑠 =
𝑠 2 +1
𝑠(𝑠+1)(𝑠+2)
𝐴
𝑠
= +
𝐵
𝐶
+
𝑠+1 𝑠+2
3.1 Laplace Transform
Example 3
Step2: Finding A, B, and C (“Equate coefficients” method)
reduction of fraction to
a common denominator
3.1 Laplace Transform
Example 3
Step 3: Finally,
1/2
−2
5/2
𝑋 𝑠 =
+
+
𝑠
𝑠+1 𝑠+2
Since
𝐾𝑒 −𝑎𝑡 , 𝑡 ≥ 0 ↔
𝐾
𝑠+𝛼
Take Inverse Laplace Transformation
1
5 −𝑡
−𝑡
𝑥 𝑡 = − 2𝑒 + 𝑒 , 𝑡 ≥ 0
2
2
1
5
3.1 Laplace Transform
The residue formula allows us to find one coefficient at a time by multiplying
both sides of the equation by the appropriate factor.
𝑠+1
𝑠 2 +1
𝑠 𝑠+1 𝑠+2
𝑠 2 +1
𝑠 𝑠+2
=
=
𝑠+1 𝐴
𝑠+1 𝐵
𝑠+1 𝐶
+
+
𝑠
𝑠+1
𝑠+2
𝑠+1 𝐴
+
𝑠
𝐵+
𝑠+1 𝐶
𝑠+2
Evaulate both sides at 𝑠 = −1
(−1)2 +1
=0+𝐵+0
(−1)(−1 + 2)
−2 = 𝐵
1
6
3.1 Laplace Transform
For Laplace Transform with non-repeating roots,
𝐹 𝑠 =
𝐴
𝑠+𝑎
+
𝐵
𝑠+𝑏
+
𝐶
𝑠+𝑐
+⋯
The general residue formula is:
𝐴 = 𝑠 + 𝑎 𝐹(𝑠)|𝑠=−𝑎
1
7
3.1 Laplace Transform
Example 4
𝑥ሷ + 3𝑥ሶ + 2𝑥 = 0,
𝑥 0 = 1, 𝑥(0)
ሶ
=0
Solution:
𝑠 2 𝑋 𝑠 − 𝑠𝑥 0 − 𝑥ሶ 0 + 3 𝑠𝑋 𝑠 − 𝑥 0 + 2𝑋 𝑠 = 0
𝑠 2 𝑋 𝑠 − 𝑠 + 3 𝑠𝑋 𝑠 − 1 + 2𝑋 𝑠 = 0
𝑠 2 + 3𝑠 + 2 𝑋 𝑠 = 𝑠 + 3
𝑋 𝑠 =
𝐴 = 𝑠 + 1 𝑋 𝑠 |𝑠=−1
𝑋 𝑠 =
𝑠+3
𝐴
𝐵
=
+
(𝑠 + 1)(𝑠 + 2) 𝑠 + 1 𝑠 + 2
−1 + 3
=
=2
−1 + 2
2
−1
+
𝑠+1 𝑠+2
𝐵 = 𝑠 + 2 𝑋 𝑠 |𝑠=−2
−2 + 3
=
= −1
−2 + 2
𝑥 𝑡 = 2𝑒 −𝑡 − 𝑒 −2𝑡 , 𝑡 ≥ 0
1
8
3.1 Laplace Transform
Now we will consider partial fraction expansion rules for functions with
repeated (real)roots:
Number of constants = Order of repeated roots
Example 5:
(𝑠 + 1)4
𝐴
𝐵
𝐶
𝐷 𝐸
=
+
+
+ +
(𝑠 + 3)3 𝑠 2 𝑠 + 3 (𝑠 + 3)2 (𝑠 + 3)3 𝑠 𝑠 2
(𝑠 + 1)4
𝐴
𝐵
𝐶
𝐷 𝐸
𝐹
=
+
+
+
+
+
𝑠 + 3 3 𝑠 2 (𝑠 − 1) 𝑠 + 3 (𝑠 + 3)2 (𝑠 + 3)3 𝑠 𝑠 2 𝑠 − 1
1
9
3.1 Laplace Transform
2
0
3.1 Laplace Transform
Example 8
2
1
3.1 Laplace Transform
Example 6
Terms with repeated roots:
𝑠+3
𝐴
𝐵
𝐶
=
+
+
(𝑠 2 + 2𝑠 + 1)(𝑠 + 2) 𝑠 + 1 (𝑠 + 1)2 𝑠 + 2
The highest order terms (of each distinct root) can be found using residue
formula with appropriate factor:
2
(𝑠
+
1)
𝑠 + 1 2𝑋 𝑠 = 𝐴 𝑠 + 1 + 𝐵 +
𝐶
𝑠+2
𝑖𝑓 𝑠 = −1, 𝑇ℎ𝑒𝑛, 𝑠 + 1 2 𝑋 𝑠 |𝑠=−1 = 𝐵
𝐵 = 𝑠+1
2𝑋
𝑠 |𝑠=−1
𝑠+3
=
|
=2
𝑠 + 2 𝑠=−1
𝑠+3
𝐶=
|
=1
(𝑠 + 1)2 𝑠=−2
2
2
3.1 Laplace Transform
𝑠+3
𝐴 𝑠 2 + 3𝑠 + 2 + 2 𝑠 + 2 + 𝑠 2 + 2𝑠 + 1
=
2
𝑠 + 2𝑠 + 1 𝑠 + 2
𝑠 2 + 2𝑠 + 1 (𝑠 + 2)
Equate the 𝑠 2 coefficient on both sides
0=𝐴+1
𝐴 = −1
Thus
−1
2
1
𝑋 𝑠 =
+
+
2
𝑠 + 1 (𝑠 + 1)
𝑠+2
𝑥 𝑡 = −𝑒 −𝑡 + 2𝑡𝑒 −𝑡 + 𝑒 −2𝑡 , 𝑡 ≥ 0
2
3
3.1 Laplace Transform
Example 7
Find the solution to the following differential equation
𝑥ሷ + 4𝑥ሶ + 4𝑥 = 0, 𝑥 0 = 1, 𝑥ሶ 0 = 0
Volunteer？
2
4
3.1 Laplace Transform
Example 7
Find the solution to the following differential equation
𝑥ሷ + 4𝑥ሶ + 4𝑥 = 0, 𝑥 0 = 1, 𝑥ሶ 0 = 0
𝑠 2 𝑋 𝑠 − 𝑠𝑥 0 − 𝑥ሶ 0 + 4 𝑠𝑋 𝑠 − 𝑥 0 + 4𝑋 𝑠 = 0
𝑠 2 𝑋 𝑠 − 𝑠 − 0 + 4𝑠𝑋 𝑠 − 4 + 4𝑋 𝑠 = 0
𝑋 𝑠 𝑠 2 + 4𝑠 + 4 = 𝑠 + 4
𝑠 + 2 2𝑋 𝑠
= 𝑠+2 𝐴+𝐵 =𝑠+4
𝑠𝑜, 𝐴 = 1, 𝐵 = 2
𝑠+4
𝐴
𝐵
=
+
(𝑠 + 2)2 𝑠 + 2 (𝑠 + 2)2
𝐵 = 𝑠 + 2 2 𝑋 𝑠 |𝑠=−2 = 𝑠 + 4 |𝑠=−2 = 2
𝑋 𝑠 =
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝐴 = 1
𝑥 𝑡 = 𝑒 −2𝑡 + 2𝑡𝑒 −2𝑡 ,
𝑡≥0
2
5
3.1 Laplace Transform
Inverse Laplace Transform with
Complex Roots
To simplify your algebra, don’t use first-order denominators such as
1
𝐴
𝐾1
𝐾2
= +
+
𝑠(𝑠 2 + 2𝑠 + 2) 𝑠 𝑠 + 1 − 𝑗 𝑠 + 1 + 𝑗
Instead, rename variables
𝐵 = 𝐾1 + 𝐾2
𝐶 = ( 1 + 𝑗 𝐾1 + (1 − 𝑗)𝐾2 )
So that
1
𝐴
𝐵𝑠 + 𝐶
=
+
𝑠(𝑠 2 + 2𝑠 + 2) 𝑠 𝑠 2 + 2𝑠 + 2
2
6
3.1 Laplace Transform
Inverse Laplace Transform with Complex Roots
2
7
3.1 Laplace Transform
Example 8
Find the solution to the following differential equation
𝑥ሷ + 2𝑥ሶ + 2𝑥 = 1, 𝑥 0 = 1, 𝑥ሶ 0 = 0
Volunteer？
2
8
3.1 Laplace Transform
Example 8
𝐿 𝑥ሷ 𝑡
= 𝑠 2 𝑋 𝑠 − 𝑠𝑥 0 − 𝑥 ′ (0)
2
9
3.1 Laplace Transform
Example 8
3
0
3.1 Laplace Transform
Example 8
3
1
3.1 Laplace Transform
Example 8
3
2
3.1 Laplace Transform
Example 9
Find the solution to the following differential equation
𝑥ሷ + 4𝑥ሶ + 8𝑥 = 0, 𝑥 0 = 1, 𝑥ሶ 0 = 0
𝑠 2 𝑋 𝑠 − 𝑠𝑥 0 − 𝑥ሶ 0 + 4 𝑠𝑋 𝑠 − 𝑥 0 + 8𝑋 𝑠 = 0
𝑠 2 𝑋 𝑠 − 𝑠 + 4𝑠𝑋 𝑠 − 4 + 8𝑋 𝑠 = 0
𝑋 𝑠 𝑠 2 + 4𝑠 + 8 = 𝑠 + 4
𝑠+4
𝑠+2
2
𝑋 𝑠 = 2
=
+
𝑠 + 4𝑠 + 8 (𝑠 + 2)2 +22 (𝑠 + 2)2 +22
𝑥 𝑡 = 𝑒 −2𝑡 cos 2𝑡 + 𝑒 −2𝑡 sin 2𝑡 ,
𝑡≥0
3.2 Transfer Function
The transfer function 𝐻 𝑠 of a linear time-invariant (LTI)
system is defined as the radio of the Laplace transform of
the output of the system to its input assuming that all zero
initial conditions.
3.2 Transfer Function
If Input--𝑟(𝑡)，output--𝑦(𝑡). Transfer function was define as：
𝐿[𝑦(𝑡)] 𝑌(𝑠)
𝐻(𝑠) =
=
𝐿[𝑟(𝑡)] 𝑅(𝑠)
𝑌(𝑠) = 𝐿[𝑦(𝑡)]——Laplace transform for output
𝑅(𝑠) = 𝐿[𝑟(𝑡)]——Laplace transform for input
The time response of control system 𝑦(𝑡) equals the inverse
Laplace transform of 𝑌(𝑠) ：
𝑦(𝑡) = 𝐿−1 [𝑦(𝑠)] = 𝐿−1 [𝐻(𝑠)𝑅(𝑠)]
3.2 Transfer Function
• Response by convolution
（叠加原理）
For linear time-invariant system, the principle of superposition holds.
The principle of superposition states that if the system has an input that
can be expressed as a sum of signals, then the response of the system can
be expressed as the sum of the individual responses to the respective
signals.
We are able to solve the response of a linear system to a general signal
simply by decomposing the given signal into a sum of the elementary
components.
3.2 Transfer Function
Paul Direc（狄拉克）finds the idea of using “Impulse” signal.
For linear time-invariant system
+∞
Input signal: 𝑟 𝑡 = ‫׬‬−∞ 𝑟 𝜏 𝛿 𝑡 − 𝜏 𝑑𝜏
The response of 𝛿 𝑡 − 𝜏 : ℎ(𝑡 − 𝜏)
+∞
*代表卷积
Output signal: 𝑦 𝑡 = ‫׬‬−∞ 𝑟 𝜏 ℎ 𝑡 − 𝜏 𝑑𝜏 = 𝑟(𝑡) ∗ ℎ(𝑡)
The response of a linear time-invariant system is the
integration (sum) of impulse response.
3.2 Transfer Function
• Explanation of Transfer Function
R(s) is Laplace Transform of input signal
+
R( s ) =  r ( )e − s d
−
H(s) is Laplace Transform of impulse input response
+
H ( s) =  h( )e − s d
−
Y(s) is Laplace Transform of output signal
Y( s) = 
+
−
y( )e− s d
Can you find the relation of H(s), Y(s) and R(s) ?
3.2 Transfer Function
For linear time-invariant system, we have
+∞
𝑦(𝑡) = න
+∞
𝑟(𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏 = න
−∞
ℎ(𝜏)𝑟(𝑡 − 𝜏)𝑑𝜏
−∞
+∞
+∞
𝑌(𝑠) = න
ℎ(𝜏)𝑟(𝑡 − 𝜏)𝑑𝜏 𝑒 −𝑠𝑡 𝑑𝑡
න
−∞
−∞
+∞
=න
+∞
𝑟(𝑡 − 𝜏) 𝑒 −𝑠𝑡 𝑑𝑡 ℎ(𝜏)𝑑𝜏
න
−∞
−∞
+∞
+∞
=න
න
−∞
𝑟(𝜂) 𝑒 −𝑠(𝜂+𝜏) 𝑑𝜂 ℎ(𝜏)𝑑𝜏
−∞
+∞
=න
−∞
𝜂 = 𝑡 − 𝜏 with 𝑑𝜂 = 𝑑τ
𝑟(𝜂) 𝑒
+∞
−𝑠𝜂
𝑑𝜂 න
−∞
ℎ(𝜏)𝑒 −𝑠𝜏 𝑑𝜏
= 𝐻(𝑠)𝑅(𝑠)
𝑌(𝑠)
𝐻(𝑠) =
𝑅(𝑠)
𝐻 𝑠 is the Transfer Function, which is also the Laplace Transform
of impulse input response.
3.2 Transfer Function
3.2 Transfer Function
In general, the time domain mathematical model of the system Differential
Equation is：
𝑑𝑛 𝑦(𝑡)
𝑑 𝑛−1 𝑦(𝑡)
𝑑𝑦 𝑡
𝑑𝑚 𝑟 𝑡
𝑑𝑟 𝑡
𝑎𝑛
+
𝑎
+
⋯
+
𝑎
+
𝑎
𝑦
𝑡
=
𝑏
+
⋯
+
𝑏
+ 𝑏0 𝑟(𝑡)
𝑛−1
1
0
𝑚
1
𝑑𝑡 𝑛
𝑑𝑡 𝑛−1
𝑑𝑡
𝑑𝑡 𝑚
𝑑𝑡
where，𝑎𝑖 ,b_j (𝑖 = 0,1,2, … , 𝑛; 𝑗 = 0,1,2 … , 𝑚) are real numbers，which
are determined by the system structure parameters. The Laplace
transformation are used to both sides :
𝑎𝑛 𝑠 𝑛 𝑌 𝑠 + 𝑎𝑛−1 𝑠 𝑛−1 𝑌 𝑠 + ⋯ + 𝑎1 𝑠𝑌 𝑠 + 𝑎0 𝑌 𝑠
= 𝑏𝑚 𝑠 𝑚 𝑅 𝑠 + 𝑏𝑚−1 𝑠 𝑚−1 𝑅 𝑠 + ⋯ + 𝑏1 𝑠𝑅 𝑠 + 𝑏0 𝑅 𝑠
So, the general expression of control system transfer function:
𝑌(𝑠) 𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 + ⋯ + 𝑏1 𝑠 + 𝑏0
𝐺 𝑠 =
=
𝑅(𝑠)
𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎1 𝑠 + 𝑎0
41
3.2 Transfer Function
1. Open loop transfer function
R(s) +
𝑌(𝑠)
C
( s)
E ( s)
_
G (s)
B(s)
H (s)
Open loop transfer function is equal to the transfer function 𝐺(𝑠) in forward
path multiply the transfer function 𝐻(𝑠) in feedback path.
𝐵(𝑠)
𝐿(𝑠) =
= 𝐺(𝑠)𝐻(𝑠)
𝐸(𝑠)
3.2 Transfer Function
The general situation ：
𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 + ⋯ + 𝑏1 𝑠 + 𝑏0
𝐿 𝑠 =𝐺 𝑠 𝐻 𝑠 =
𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎1 𝑠 + 𝑎0
𝜂
𝜇
2
𝐾 ς𝑖=1(𝜏𝑖 𝑠 + 1) ς𝑖=1(𝜏𝑑𝑖
+ 2𝜉𝑑𝑖 𝜏𝑑𝑖 𝑠 + 1)
= 𝑣 𝜌
𝑖 2
𝑠 ς𝑖=1(𝑇𝑖 𝑠 + 1) ς𝜎𝑖=1(𝑇𝑛𝑖
𝑠 + 2𝜉𝑛𝑖 𝑇𝑛𝑖 𝑠 + 1)
where 𝐾 is the open loop amplification coefficient (also called
the open loop magnification or open-loop gain), is an important
parameter to influence the system performance.
When transfer function in feedback loop 𝐻(𝑠) = 1 (unit
feedback system)，the open loop transfer function is equal to
forward transfer function, that is 𝐺( 𝑠 ).
3.2 Transfer Function
2. Closed-loop Transfer Function
The transfer function of the output and input when main
feedback loop is connected, usually represented by Φ(𝑠).
𝑌(𝑠)
𝐺(𝑠)
Φ(𝑠) =
=
𝑅(𝑠) 1 + 𝐺(𝑠)𝐻(𝑠)
R(s) +
C𝑌(𝑠)
( s)
E ( s)
_
G (s)
𝐵 𝑠 =𝑌 𝑠 𝐻 𝑠
𝐸 𝑠 =𝑅 𝑠 −𝐵 𝑠
𝑌 𝑠 = 𝐺 𝑠 𝐸(𝑠)
𝑌 𝑠 =𝐺 𝑠 𝑅 𝑠 −𝑌 𝑠 𝐻 𝑠
𝑌 𝑠 =𝐺 𝑠 𝑅 𝑠 −𝐺 𝑠 𝑌 𝑠 𝐻 𝑠
1 + 𝐺 𝑠 𝐻 𝑠 𝑌 𝑠 = 𝐺 𝑠 𝑅(𝑠)
B(s)
H (s)
𝑌(𝑠)
𝐺(𝑠)
=
𝑅(𝑠) 1 + 𝐺 𝑠 𝐻(𝑠)
3.2 Transfer Function
3. Disturbance Transfer Function
Disturbance
signal
R( s) +
E (s)
G1 ( s )
+
N (s)
+
𝑌(𝑠)
C
(s)
G2 ( s )
_
B( s)
H (s)
Consider both input and disturbance signals
𝑌 𝑠 = 𝑌𝑅 𝑠 + 𝑌𝑁 𝑠
=
𝐺1 𝑠 𝐺2 𝑠
𝐺2 𝑠
𝑅 𝑠 +
𝑁 𝑠
1 + 𝐺1 𝑠 𝐺2 𝑠 𝐻 𝑠
1 + 𝐺1 𝑠 𝐺2 𝑠 𝐻 𝑠
=
𝐺2 𝑠
[𝐺1 𝑠 𝑅 s + 𝑁(𝑠)]
1 + 𝐺1 𝑠 𝐺2 𝑠 𝐻 𝑠
When input 𝑹(𝒔) = 𝟎
𝛷𝑁
𝑌𝑁 (𝑠)
𝐺2 (𝑠)
𝑠 =
=
𝑁(𝑠) 1 + 𝐺1 𝑠 𝐺2 𝑠 𝐻(𝑠)
we hypothesis :
𝑮𝟏 𝒔 𝑮𝟐 𝒔 𝐇 𝐬
≫ 𝟏, 𝑮𝟏 𝒔 𝐇 𝐬
𝑌𝑁 (𝑠)
→0
𝑁(𝑠)
≫𝟏
The disturbance signals can be restrained.
If disturbance signal 𝑁(𝑠) = 0
we hypothesis :
𝑌𝑅 (𝑠)
𝐺1 (𝑠)𝐺2 (𝑠)
𝛷𝑅 (𝑠) =
=
𝑅(𝑠) 1 + 𝐺1 (𝑠)𝐺2 (𝑠)𝐻(𝑠)
|𝑮𝟏 𝒔 𝑮𝟏 𝒔 𝐇(𝐬)| ≫ 𝟏
𝒀𝑹 (𝒔)
𝟏
≈
𝑹(𝒔)
𝑯(𝒔)
It shows that the transfer function of closed-loop system only has
relationship with 𝐻(𝑆) , which is not depend on 𝐺1 𝑠 , 𝐺2 𝑠 ,
3.2 Transfer Function
4. Error Transfer Function
a) Error transfer function with reference signal
If 𝑁(𝑠)＝0，then
𝐸(𝑠) 𝑅(𝑠) − 𝑌(𝑠)𝐻(𝑠)
𝑌(𝑠)𝐻(𝑠)
=
=1−
𝑅(𝑠)
𝑅(𝑠)
𝑅(𝑠)
=1−
𝐺1 (𝑠)𝐺2 (𝑠)
1
𝐻(𝑠) =
1 + 𝐺1 (𝑠)𝐺2 (𝑠)𝐻(𝑠)
1 + 𝐺1 (𝑠)𝐺2 (𝑠)𝐻(𝑠)
Disturbance
signal
R( s) +
E (s)
G1 ( s )
+
N (s)
+
𝑌(𝑠)
C (s)
G2 ( s )
_
B( s)
47
H (s)
3.2 Transfer Function
Disturbance
signal
R( s) +
E (s)
G1 ( s )
+
N (s)
+
𝑌(𝑠)
C (s)
G2 ( s )
_
B( s)
H (s)
b) Error transfer function with disturbance signal ：
−G 2 ( s ) H ( s )
E ( s)
=
N ( s ) 1 + G1 ( s )G 2 ( s ) H ( s )
Volunteer？
48
3.2 Transfer Function
Disturbance
signal
R( s) +
E (s)
G1 ( s )
+
N (s)
+
C𝑌(𝑠)
(s)
G2 ( s )
_
B( s)
H (s)
c) Total Error with 𝑅(𝑠) and 𝑁(𝑠)
G2 (s) H (s)
1
E ( s) =
R( s ) −
N ( s)
1 + G1 ( s)G 2 ( s) H ( s)
1 + G1 ( s)G2 ( s) H ( s)
3.2 Transfer Function
Summary for Transfer Function ：
1) Transfer function is the system mathematical model (or link)
in complex domain, which is the system characteristics
description, reflecting the relation between input and output
for linear time-invariant system.
2) Transfer function depends only on the structural
parameters of the system itself, and have no relationship
with system input
50
3.2 Transfer Function
3) Transfer function is the rational real fractional function
with respect to the complex variable 𝒔, that is 𝒎𝒏 ( m, n
are the highest order times for numerator and denominator
respectively)
4) If the input is the unit impulse function，that is 𝒓(𝒕) =
𝜹(𝒕)，𝑹(𝒔) = 𝑳[𝒓(𝒕)] = 𝟏，we have
y 𝒕 = 𝑳−𝟏 𝑹 𝒔 𝑮 𝒔
= 𝑳−𝟏 𝑮 𝒔
51
5
2
Homework
▪
Textbook, Page 180-181, Problems
3.7 (a) (h)
3.9 (a) (e)
3.12(a)(b)(c)
Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
Chap3 Dynamic Response
3.1 Laplace Transforms
3.2 Transfer Function
3.3 Modeling Diagrams
3.4 Time-Domain Specifications
3.5 Effect of Pole & Zero Locations
3.6 Amplitude & Time Scaling
3.7 Stability
3.3 Block Diagram
The Block Diagram Model
which consists of block, arrow, differencing junction and pickoff point.
Pickoff point
3.3 Block Diagram
A block diagram（方框图）represents the flow of
information and the function performed by each
component in the system.
 Arrows(箭头)are used to show the direction of the flow
of information.
The block（方框）represents the function or dynamic
characteristics of the component and is represented by a
transfer function.
The complete block diagram shows how the functional
components are connected and the mathematic equations
that determine the response of each component.
3.3 Block Diagram
1. Basic Element of Block Diagram
r (t )
c (t )
C (s)
R(s)
C (s)
(a)
a)
b)
c)
d)
R( s ) − C ( s )
R(s) +
(b)
R(s)
G (s)
C ( s)
C (s)
(c )
(d )
Signal lines (信号线) ;
Branch point (also called pickoff point, 引出点)；
Comparison point (also called summing/differencing junction，比较点) ;
Block （方框或环节）
The system block diagram combines diagrams and mathematical equations
together, it describes the comprehensive characteristics of a system.
3.3 Block Diagram
Example 1
𝑟 𝑡 − 𝑢1 (𝑡)
= 𝑖1 (𝑡)
𝑅1
𝑢1 𝑡 =
1
න[𝑖1 𝑡 − 𝑖2 (𝑡)]𝑑𝑡
𝐶1
𝑢1 𝑡 − 𝑐(𝑡)
= 𝑖2 (𝑡)
𝑅2
𝑐 𝑡 =
1
න 𝑖2 𝑡 𝑑𝑡
𝐶2
3.3 Block Diagram
Assemble all block diagrams above：
R(s) +
_
1
R1
+
1
C1s
+
_
1
R2
How can simplify this diagrams?
1
C2 s
C ( s)
3.3 Block Diagram
2. Block Diagram Simplification
(1) Series principle
X1 ( s )
X 2 ( s)
G1 ( s )
𝑋2 (𝑠)
𝐺1 (𝑠) =
𝑋1 (𝑠)
𝐺(𝑠) =
X 3 (s)
G2 ( s )
𝐺2 (𝑠) =
𝑋3 (𝑠)
𝑋2 (𝑠)
𝑋3 (𝑠) 𝑋2 (𝑠) 𝑋3 (𝑠)
=
•
= 𝐺1 (𝑠)𝐺2 (𝑠)
𝑋1 (𝑠) 𝑋1 (𝑠) 𝑋2 (𝑠)
𝐺(𝑠) is equal to the product of the total series links of the transfer
function 𝐺(𝑠) = 𝐺1(𝑠) 𝐺2(𝑠) ⋯ 𝐺𝑛(𝑠)
3.3 Block Diagram
(2) Parallel principle
G1 ( s )
R( s)
X1 ( s )
+
C (s)
-
G2 ( s )
X 2 (s)
G1 (s) =
G(s) =
X1 (s)
R(s)
G 2 (s) =
X 2 (s)
R(s)
X1 (s) + X 2 (s) = C(s)
C(s) X 1 (s) + X 2 (s) X 1 (s) X 2 (s)
=
=
+
R(s)
R(s)
R(s)
R(s)
= G 1 (s) + G 2 (s)
The transfer function of parallel link is equal to the sum of all transfer
function 𝐺(𝑠) = 𝐺1(𝑠) + 𝐺2(𝑠) + ⋯ + 𝐺𝑛(𝑠)
3.3 Block Diagram
(3) Feedback principle
R( s ) +
E (s)
_
C (s)
G( s)
B(s)
H ( s)
Here the two blocks are connected in a feedback arrangement, so that
each feeds into each other. When the feedback B(s) is subtracted, we
call it Negative feedback.
Note: negative feedback is usually required for system stability
𝐶(𝑠) = 𝐺(𝑠)𝐸(𝑠)
= 𝐺(𝑠)[𝑅(𝑠) − 𝐵(𝑠)]
= 𝐺(𝑠)[𝑅(𝑠) − 𝐻(𝑠)𝐶(𝑠)]
𝐶(𝑠)
𝐺(𝑠)
=
𝑅(𝑠) 1 + 𝐺(𝑠)𝐻(𝑠)
The transfer function for negative feedback
The gain of a single-loop negative feedback system is given by the
forward gain divided by the sum of 1 plus the loop gain.
3.3 Block Diagram
(3) Feedback principle
When the feedback is added instead of
subtracted, we call it Positive feedback. In this
case, the gain is given by the forward gain
divided by the sum of 1 minus the loop gain.
R( s) +
B(s)
E (s)
+_
C (s)
G ( s)
H ( s)
C(s) = G(s)E(s) = G(s)[R(s) + B(s)]
= G(s)[R(s) + H(s)C(s)]
C(s)−G(s)H(s)C(s) = G(s)R(s)
C(s)
G(s)
=
R(s) 1 − G(s)H(s)
C(s) = G(s)[R(s) + H(s)C(s)]
C(s)[1−G(s)H(s)] = G(s)R(s)
The transfer function for positive feedback
序号
变换方式
原方块图
A +
1
比较点分解
+
A
比较点前移
A
C
+
A− B+C
G
4
+
比较点后移
comparison point move
backward
5
分支点前移
A
B
G
+
_
C
B
C
+
A− B+C
+
B
AG − B
+
A +
-
AG − BG
A
G
AG − BG
+
-
B
AG
A
G
G
AG
Pickoff point moved forward
B
1
G
B
G
AG − B
G
-
comparison point move forward
A
A− B+C
+
A +
B
comparison point decomposition
3
+
A
+
C
B
comparison point exchange
2
A− B+C
+
比较点交换
等效方块图
G
AG
AG
6
A
分支点后移
A
AG
G
AG
G
A
Pickoff point moved backward
B
+
A− B
7
比较点与分支点
交换
+
A
8
化成单位并联
A
G1
9
+
化成单位反馈
-
10
分支点交换
Exchange Pickoff points
AG1
-
B
+
AG1 + AG2
1
G2
A G
2
+
G1
B
A
1
G2
G1
+
-
G2
Transform into units feedback
A
A− B
AG1 + AG2
+
+
G2
Transform into units Parallel
A
A−B
-
+
A
A− B
B
Exchange comparison and
Pickoff point (不推荐)
A
1
G
G1
G2
B
AG1
A
AG1
G1
B
G1
G2
G2
B
AG1
3.3 Block Diagram
Example 1
_
R(s) +
_
1
R1
+
A
1
C1s
B+ C
_
1
R2
Please simplify the Block Diagram.
Volunteer？
D
1
C2s
C (s)
3.3 Block Diagram
Example 1
_
R(s)
+
1
R1
_
+
1
C1s
A
B+ C
_
1
R2
D
1
C2s
C (s)
Solution：Employ Block Diagram Algorithms
(a) move comparison point A forwards, pickoff point D afterwards
C2s
R1
_
R(s) +
1
R1
1
C1s
B + C
_
(a)
1
R2
1
C2s
C(s)
3.3 Block Diagram
(b) Eliminate local feedback loop
R(s)
+
_
1
R1C1s + 1
1
R2C2s + 1
C (s)
R1C2 s
(b)
Note: There are three local feedback loops in this examples
3.3 Block Diagram
（c）Eliminate main feedback loop and get the result
R(s)
1
C (s)
R1C1R2C2 s 2 + ( R1C1 + R2C2 + R1C2 ) s + 1
Conclusion: The simplification method for block diagram is
not unique, we should make full use of all kinds of
transformation skills, and choose the simple path to
achieve the simplest purposes.
3.3 Block Diagram
Exercise 1
3.3 Block Diagram
Exercise 1
3.3 Block Diagram
Signal-Flow Graph & Mason’s Rule
Signal-flow graph is a graphical representations of a set of linear algebraic
equations.
Consider the following set of algebraic equations







x1 = x1
x2 = ax1 + dx2 + ex3
x3 =
x4 =
x5 =
x5
+ fx5
bx2
cx3
x5
f
x1
a
x2
c
b
Signal-flow graph representation
x3
d
e
x4
3.3 Block Diagram
1. Definitions
▪ Input node (or source node，输入节点) : nodes have output branchs only, such as x1 , x5.
▪ Output node (or sink node，输出节点): nodes have input branches only, such as x4 .
▪ Mixed Node （混合节点）: nodes have both output and input branchs, another branch of
the node type, such as x2 , x3 .
▪ Transmission （传递）: the gain between two nodes. For example: the gain between x1 → x2
is a, then the transmission is a.
▪ Forward path（前向通路）: the path that pass each node only once, when a signal is
transmitted from a input node to a output node. Such as: x1 → x2 → x3 → x4.
x5
f
x1
a
x2
c
b
x3
d
e
x4
3.3 Block Diagram
▪ Overall Gain of forward path （前向通路总增益）: the gain product of each branch on
forward path, Example: overall gain of x1→x2→x3→x4 is 𝑎𝑏𝑐.
▪ Loop （回路）：a closed path that originates and terminates on the same nodes, and no
node is met twice along the path.
▪ Loop Gain （回路增益）: the gain product of each branch of the loop. There are two loop in
the graph, one is x2→x3→x2 and the loop gain is 𝑏𝑒, the other is x2→x2 , also known as selfloop, whose gain is 𝑑.
▪ Non-touching Loops （不接触回路）: loops that have no common node with each other.
x5
There is no such loops in the following graph.
f
x1
a
x2
c
b
x3
d
e
x4
3.3 Block Diagram
2. Properties and Algorithms
1) Properties:
• 1. Each node represents a variable, and transmit the accumulation of
all input signals to each output branch.
• 2. A branch represents the functional relationship between one signal
and another. The direction of the arrow on the branch represents the
flow direction of the signal.
• 3. Mixed nodes will generate output nodes by increasing a branch with
the gain of 1, and two ends of this branch represent the same variables.
3.3 Block Diagram
2) Algorithms
x1
(a)
a
x2
a
x2
x1
(b)
x1
a+b
x2
b
a
x1
(c )
x1
x2
b
x2
a
b
x3
x3
(d )
x1
x1
a
(e)
x2
b
x1
x3
x2
x1
ab x3
c
x4
ab
1 − bc
x3
正反馈
bc
c
x1
ab
ac
x4
x2
bc
24
3.3 Block Diagram
Example 2
Simplify the signal-flow in graph (d):
since: x2 = ax1 + cx3
x3 = bx2
eliminate intermediate variable x2 , we have:
x3 =
ab
x
1 − bc 1
序号
方块图
R( s)
1
信号流程图
C (s)
G (s)
E(s)
R(s) +
C(s)
G(s)
2
G (s) C (s)
R( s)
R( s)
1
E (s)
G (s)
C (s)
_
H(s)
− H (s)
N(s)
N(s)
R(s) +
E(s)
G1(s)
_
3
+
+
1
C(s)
G2(s)
R(s) 1 E ( s ) G1(s)
H(s)
− H (s)
N(s)
R(s) +
E(s)
_
4
+
+
G(s)
G2(s) C ( s )
N(s)
C(s)
R( s)
1
G (s)
E (s)
1
1 C (s)
C (s)
H(s)
R1 ( s )
G11 ( s )
− H ( s)
C1 ( s )
+
+
R1 ( s )
G11 ( s )
C1 ( s )
G21 ( s )
5
G12 ( s )
R2 ( s )
G12 ( s )
R2 ( s )
+
G22 ( s )
+
C2 ( s )
G21 ( s )
C2 ( s )
G22 ( s )
3.3 Block Diagram
3. Mason’s rule (Mason Formula)
The overall transmission (or overall gain) between input node and output node
could be determined by Mason formula:
𝑁
1
𝐺 = ෍ 𝑝𝑘 ∆𝑘
∆
𝑘=1
Where:
Δ = the system determinant；
Δ = 1 - (sum of all individual loop gains)
+ (sum of the gain products of all combinations of two non-touching loops)
- (sum of the gain products of all combinations of three nont-ouching loops)
+ ......
= gain of the kth forward path ;
N = the total number of forward path;
Δk = cofactor（余因数）of the kth path; （和第k个前向通路不接触的回路增益）
3.3 Block Diagram
Example 3
Determine C(s)/R(s) by using Mason formula.
G7
G6
R(s)
G1
+
G2
G3
+
+
G4
G5
+
+ C ( s)
-
-
H1
H2
Solution：Plot the signal-flow graph of the system
G7
G6
R ( s ) G1
G4
G2
G3
− H1
−H2
G5
1 C (s)
3.3 Block Diagram
There are 4 independent loops in this graph：
L1 = -G4H1
L2 = -G2G7H2
L3 = -G6G4G5H2
L4 = -G2G3G4G5H2
The only nontouching loops are L1 L2
hence, the determinant is
Δ=1-（L1 + L2 + L3 + L4）+ L1 L2
The 3 forward paths are：
R ( s ) G1
P1= G1G2G3G4G5
Δ1=1
P2= G1G6G4G5
Δ2=1
P3= G1G2G7
Δ3=1-L1
G7
G6
G4
G2
G3
− H1
−H2
G5
1 C (s)
3.3 Block Diagram
hence，the closed-loop transfer function C(s) / R(s) is
C(s)
1
= G = (p1Δ 1 + p 2Δ 2 + p 3Δ 3 )
R(s)
Δ
G 1G 2 G 3 G 4 G 5 + G 1G 6 G 4 G 3 + G 1G 2 G 7 (1 + G 4 H1 )
=
1 + G 4 H 1 + G 2 G 7 H 2 + G 6 G 4 G 5 H 2 + G 2 G 3 G 4 G 5 H 2 + G 4 H 1G 2 G 7 H 2
G7
G6
R ( s ) G1
G4
G2
G3
− H1
−H2
G5
1 C (s)
3.3 Block Diagram
Exercise 2
Determine C(s) / R(s) by using Mason formula.
R(s) +
A
_
1
R1
+
-
B
1
C1s
C +
D
_
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1
R2
E
1
C2s
C ( s)
3.3 Block Diagram
Exercise 2
Determine C(s) / R(s) by using Mason formula.
R(s) +
A
_
1
R1
+
-
B
C +
1
C1s
1
R2
D
_
E
1
C2s
C ( s)
Solution：Plot the signal-flow graph of the system
R( s )
1
A
1
R1
1
C1s
B
−1
−1
C
1
D
1
C2 s
1
R2
E
−1
1
C ( s)
3.3 Block Diagram
NOTES：node C is in front of comparison node D, in order to obtain output signal
of node C, we need a branch with gain 1 to separate signals of C and D.
There are 3 independent loops L1,L2 and L3. Non-touching loops are L1L2 ：
• 𝐿1 =
−1
𝑅1 𝐶1 𝑠
, 𝐿2 =
−1
𝑅2 𝐶2 𝑠
, 𝐿3 =
−1
𝑅2 𝐶1 𝑠
𝐿1 𝐿2 =
1
𝑅1 𝐶1 𝑠𝑅2 𝐶2 𝑠 2
• ∆= 1 − (𝐿1 + 𝐿2 + 𝐿3 )+ 𝐿1 𝐿2
=1+
1
𝑅1 𝐶1 𝑠
+
1
𝑅2 𝐶2 𝑠
+
1
𝑅2 𝐶1 𝑠
+
1
𝑅1 𝐶1 𝑠𝑅2 𝐶2 𝑠 2
There is only one forward path 𝑃1 =
hence,
𝐶(𝑠)
𝑅(𝑠)
=𝐺=
𝑃1 ∆1
∆
=
1
, ∆1 =
𝑅1 𝑅2 𝐶1 𝐶2 𝑠 2
1
1
𝑅1 𝑅2 𝐶1 𝐶2 𝑠 2 +𝑅1 𝐶1 𝑠+𝑅2 𝐶2 𝑠+𝑅2 𝐶1 𝑠+1
3.3 Block Diagram
Exercise 3
3.3 Block Diagram
Exercise 2
，
Homework
▪ Textbook, Page 185, Problems
3.20 (a) (b) (c)
3.21 (a) (b) (c)