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Control Systems
Lect. 1 Introduction
Basil Hamed
Introduction
1. What is a control system.
2. Why control systems are important.
3.What are the basic components of a control system.
4. Some examples of control-system applications.
5.Why feedback is incorporated into most control systems.
6. Types of control systems.
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What is a control system
A control system is considered to be any
system which exists for the purpose of
regulating or controlling the flow of
energy, information, money, or other
quantities in some desired fashion.
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What is a control system
• An interconnection of components forming a
system configuration that will provide a
desired system response
• The study of control provides us with a
process for analyzing and understanding the
behavior of a system given some input
• It also introduces methods for achieving the
desired system response
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Systems and Control
A System is a device or process that takes a
given input and produces some output:


A DC motor takes as input a voltage and
produces as output rotary motion
A chemical plant takes in raw chemicals and
produces a required chemical product
Input
System
Output
Why control systems are important
In recent years, control systems have assumed an
increasingly important role in the
development and
advancement of modern civilization and technology.
Practically every aspect of our day-to-day activities is
affected by some type of control system.
Control systems are found in abundance in all sectors of
industry, such as quality control of manufactured products,
automatic assembly lines, machine-tool control, space
technology and weapon systems, computer control,
transportation systems, power systems, robotics, MicroElectro-Mechanical Systems, nanotechnology, and many
others.
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Basic Components of a Control System
1. Objectives of control.
2. Control-system components.
3. Results or outputs.
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Examples of Control-System Applications
Potential applications of control of these systems may benefit the
following areas:
• Machine tools. Improve precision and increase productivity by
controlling chatter.
• Flexible robotics. Enable faster motion with greater accuracy.
•Photolithography. Enable the manufacture of smaller
microelectronic circuits by controlling vibration in the
photolithography circuit-printing process.
• Biomechanical and biomedical. Artificial muscles, drug
delivery systems, and other assistive technologies.
• Process control. For example, on/off shape control of solar
reflectors or aerodynamic surfaces.
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Examples of Control Applications
Aerospace Applications:
Aircraft or missile guidance and control
Space vehicles and structures
Why feedback is incorporated into most
control systems
Control Systems can be classified as :
open loop system (Nonfeedback System)
closed loop system (Feedback System).
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Open-Loop Control Systems
(Nonfeedback Systems)
The elements of an open-loop control system can
usually be divided into two parts: the controller and the
controlled process, as shown by the block diagram
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Open Loop Control Systems
A system in which the output has no effect on
the control action is known as an open loop
control system. For a given input the system
produces a certain output. If there are any
disturbances, the out put changes and there is no
adjustment of the input to bring back the output
to the original value.
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Open-Loop Control Systems
• The controlled ‘output’ is the
resulting toast
• System does not reject
changes
in
component
characteristics
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Closed-Loop Control Systems
(Feedback Control Systems)
What is missing in the open-loop control system for more
accurate and more adaptive control is a link or feedback
from the output to the input of the system.
To obtain more accurate control, the controlled signal y
should be fed back and compared with the reference
input.
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Open-Closed Loop Control
Open-loop control is ‘blind’ to actual output
Closed-loop control takes account of actual
output and compares this to desired output
Desired
Output
Input
+
Controller/
Amplifier
-
Measurement
Process
Dynamics
Output
TYPES OF FEEDBACK CONTROL SYSTEMS
Feedback control systems may be classified in a number
of ways, depending upon the purpose of the
classification. For instance, according to the method of
analysis and design, control systems are classified as:
Linear or Nonlinear
Time-varying or Time-invariant
Continuous-data or Discrete-data
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Control
Many control systems can be characterised
by these components
Disturbance
Plant
Reference
r(t)
Error
e(t)
+
Control
Control
Signal
u(t)
Actuator
Feedback
Sensor
Sensor Noise
Process
Outpu
t
y(t)
Actuation
A device for acting on the environment
Sensing
A device for measuring some aspect of the
environment
The Control Problem
Generally a controller is required to filter the error
signal in order that certain control criteria or
specifications, be satisfied. These criteria may involve,
but not be limited to:
1. Disturbance rejection
2. Steady state errors
3. Transient response characteristics
4. Sensitivity to parameter changes in the plant
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The Control Problem
Solving in control problem generally involves;
1. Choosing sensors to measure the plant output
2. Choosing actuators to drive the plant
3. Developing the plant, actuator, and sensors equations
4. Designing the controller
5. Evaluating the design analytically by simulation, and
finally by testing the physical system.
6. If the physical tests are unsatisfactory, iterating these
steps.
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The Control Problem
Problem Formulation
Mathematical
model
system
Physical
System
Solution Translation
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Mathematical
solution of
mathematical
problem
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Modeling in The Frequency Domain
Mathematical Modelling
To understand system
performance,
a
mathematical model of
the plant is required
This will eventually allow
us to design control
systems to achieve a
particular specification
Modeling Physical Systems - Overview
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Modeling – Remarks
• Modeling is the most important and difficult task
in control system design.
• No mathematical model exactly represents a
physical system.
Math Model  Physical System
Math Model  Physical System
• Do not confuse models with physical systems!
• In this course, we may use the term “system” or
“plant” to mean a mathematical model.
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2.2 Laplace Transform Review
The defining equation above is also known as the onesided Laplace transform, as the integration is
evaluated from t = 0 to ∞.
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Laplace Transform Review
Example 2.3 P.39
PROBLEM: Given the following differential equation, solve for
y(t) if all initial conditions are zero. Use the Laplace transform.
Solution
Solving for the response, Y(s), yields
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Laplace Transform Review
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2.3 Transfer Function
T.F of LTI system is defined as the Laplace
transform of the impulse response, with all the
initial condition set to zero
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Transfer Functions
Transfer Function
component
G(s)
describes
Described as a Laplace transform because
Y ( s)  G ( s)U ( s )
X ( s)
y (t )  g (t )u (t )
G (s)
Y (s)
system
T.F
Example 2.5 P. 46
PROBLEM: Use the result of Example 2.4 to find the response,
c(t) to an input, r(t) = u(t), a unit step, assuming zero initial
conditions.
SOLUTION: To solve the problem, we use G(s) = l/(s + 2) as
found in Example 2.4. Since r(t) = u(t), R(s) = 1/s, from Table
2.1. Since the initial conditions are zero,
Expanding by partial fractions, we get
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2.4 Electric Network Transfer Function
• In this section, we formally apply the transfer
function to the mathematical modeling of electric
circuits including passive networks
• Equivalent circuits for the electric networks that we
work with first consist of three passive linear
components: resistors, capacitors, and inductors.“
• We now combine electrical components into circuits,
decide on the input and output, and find the transfer
function. Our guiding principles are Kirchhoff s laws.
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2.4 Electric Network Transfer Function
Table 2.3 Voltage-current, voltage-charge, and
impedance
relationships
for
capacitors,
resistors, and inductors
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Modeling – Electrical Elements
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Example 2.6 P. 48
Problem: Find the transfer function relating
the 𝑣𝑐 (t) to the input voltage v(t).
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Example 2.6 P. 48
SOLUTION: In any problem, the designer must first
decide what the input and output should be. In this
network, several variables could have been chosen to be
the output.
Summing the voltages around the loop, assuming zero
initial conditions, yields the integro-differential equation
for this network as
𝑖 𝑡 =
𝑑𝑣𝑐
𝑐
Taking Laplace 𝐼 𝑠 = 𝑐𝑠𝑣𝑐 (𝑠)
𝑑𝑡
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substitute in above eq.
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Example 2.9 P. 51
PROBLEM: Repeat Example 2.6
using the transformed circuit.
Solution
using voltage division
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Modeling – Summary (Electrical System)
• Modeling
– Modeling is an important task!
– Mathematical model
– Transfer function
– Modeling of electrical systems
• Next, modeling of mechanical systems
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2.5 Translational Mechanical System T.F
• The motion of Mechanical elements can be described in
various dimensions as translational, rotational, or
combinations of both.
• Mechanical systems, like electrical systems have three
passive linear components.
• Two of them, the spring and the mass, are energystorage elements; one of them, the viscous damper,
dissipate energy.
• The motion of translation is defined as a motion that takes
place along a straight or curved path. The variables that are
used to describe translational motion are acceleration,
velocity, and displacement.
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2.5 Translational Mechanical System T.F
Newton's law of motion states that the algebraic sum of
external forces acting on a rigid body in a given
direction is equal to the product of the mass of the
body and its acceleration in the same direction. The
law can be expressed as
𝐹𝑜𝑟𝑐𝑒𝑠 = 𝑀𝑎
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2.5 Translational Mechanical System T.F
Table 2.4 Forcevelocity,
forcedisplacement, and
impedance
translational
relationships
for
springs,
viscous
dampers, and mass
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Modeling – Mechanical Elements
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Modeling – Spring-Mass-Damper Systems
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Modeling – Free Body Diagram
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Modeling – Spring-Mass-Damper System
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2.6 Rotational Mechanical System T.F
• Rotational mechanical systems are handled the
same way as translational mechanical systems,
except that torque replaces force and angular
displacement replaces translational displacement.
• The mechanical components for rotational systems
are the same as those for translational systems,
except that the components undergo rotation
instead of translation
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2.6 Rotational Mechanical System T.F
• The rotational motion of a body can be defined as
motion about a fixed axis.
• The extension of Newton's law of motion for
rotational motion :
𝑇𝑜𝑟𝑞𝑢𝑒𝑠 = 𝐽𝛼
where J denotes the inertia and α is the angular acceleration.
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Modeling – Rotational Mechanism
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Example
Problem: The rotational system shown
in Fig below consists of a disk mounted
on a shaft that is fixed at one end.
Assume that a torque is applied to the
disk, as shown.
Solution:
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2.8 Electromechanical System Transfer
Functions
• Now, we move to systems that are hybrids of electrical and
mechanical variables, the electromechanical systems.
• A motor is an electromechanical component that yields a
displacement output for a voltage input, that is, a mechanical
output generated by an electrical input.
• We will derive the transfer function for one particular kind of
electromechanical system, the armature-controlled
servomotor.
• Dc motors are extensively used in control systems
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dc
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DC Motor
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Modeling – Model of DC Motor
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The Mathematical Model Of Dc Motor
The relationship between the armature current, ia(t), the applied
armature voltage, ea(t), and the back emf, vb(t), is found by
writing a loop equation around the Laplace transformed
armature circuit
The torque developed by the motor is proportional to the
armature current; thus
where Tm is the torque developed by the motor, and Kt is a constant of
proportionality, called the motor torque constant, which depends on the
motor and magnetic field characteristics.
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The Mathematical Model Of Dc Motor
Mechanical System
Since the current-carrying armature is rotating in a magnetic
field, its voltage is proportional to speed. Thus,
Taking Laplace Transform
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The Mathematical Model Of Dc Motor
We have
Electrical System
GIVEN
Mechanical System
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The Mathematical Model Of Dc Motor
To find T.F
If we assume that the armature inductance, La, is small compared to
the armature resistance, Ra, which is usual for a dc motor, above Eq.
Becomes
the desired transfer function of DC Motor:
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Modeling – Why Linear System?
• Easier to understand and obtain solutions
• Linear ordinary differential equations (ODEs),
– Homogeneous solution and particular solution
– Transient solution and steady state solution
– Solution caused by initial values, and forced solution
• Easy to check the Stability of stationary states (Laplace
Transform)
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Modeling in The Time Domain
Modeling
Derive mathematical models for
• Electrical systems
• Mechanical systems
• Electromechanical system
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Electrical Systems:
• Kirchhoff’s voltage & current laws
Mechanical systems:
• Newton’s laws
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3.1 Introduction
• Two approaches are available for the analysis and design of
feedback control systems. The first, which we began to study
in Chapter 2, is known as the classical, or frequency-domain,
technique.
• The 1st approach is based on converting a system's
differential equation to a transfer function, thus generating a
mathematical model of the system that algebraically relates a
representation of the output to a representation of the input.
• The primary disadvantage of the classical approach is its
limited applicability: It can be applied only to linear, timeinvariant systems or systems that can be approximated as
such.
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3.1 Introduction
• The 2nd approach is state-space approach (also referred to as
the modern, or time-domain, approach) is a unified method
for modeling, analyzing, and designing a wide range of
systems.
• For example, the state-space approach can be used to
represent nonlinear systems, Time-varying systems, Multipleinput, multiple-output systems.
• The time-domain approach can also be used for the same class
of systems modeled by the classical approach.
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3.2 Some Observations
• We proceed now to establish the state-space approach
as an alternate method for representing physical
systems.
• In general, an nth-order differential equation can be
decomposed into n first-order differential equations.
• Because, in principle, first-order differential equations
are simpler to solve than higher-order ones, first-order
differential equations are used in the analytical studies
of control systems.
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3.3The General State-Space Representation
State space model composed of 2 equations;
1. State equation
State
Space Model
2. Output equation
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3.3The General State-Space Representation
x = state vector
𝑋= derivative of the state vector with respect to time
y = output vector
u = input or control vector
A = system matrix
B = input matrix
C = output matrix
D = feedforward matrix
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3.3The General State-Space Representation
Where
The state variables of a system are defined as a minimal set of variables,
x1(t),x2(t), ... ,xn(t), such that knowledge of these variables at any time
to and information on the applied input at time t0 are sufficient to
determine the state of the system at any time t > to
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Example
Given 2nd order Diff Eq.
1
Above eq. can be transform into state eq;
Let
then Eq. (1) is decomposed into the following two first-order differential equations:
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Example
𝑥1
0
𝑥=
=
−1/𝐿𝐶
𝑥2
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0
𝑥1
1
1 𝑒(𝑡)
+
−𝑅/𝐿 𝑥2
𝐿
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3.4 Applying the State-Space Representation
In this section, we apply the state-space formulation to the
representation of more complicated physical systems. The first
step in representing a system is to select the state vector, which
must be chosen according to the following considerations:
1. A minimum number of state variables must be selected as
components of the state vector. This minimum number of state
variables is sufficient to describe completely the state of the
system.
2. The components of the state vector (that is, this minimum
number of state variables) must be linearly independent.
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Minimum Number of State Variables
• Typically, the minimum number required equals the order
of the differential equation describing the system. For
example, if a third-order differential equation describes the
system, then three simultaneous, first-order differential
equations are required along with three state variables.
• From the perspective of the transfer function, the order of
the differential equation is the order of the denominator of
the transfer function after canceling common factors in the
numerator and denominator.
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Minimum Number of State Variables
• In most cases, another way to determine the number of state
variables is to count the number of independent energystorage elements in the system.
• The number of these energy-storage elements equals the
order of the differential equation and the number of state
variables.
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Example
Find state model of
System shown in the Fig.
Solution
• A practical approach is to assign the current in the inductor L, i(t), and
the voltage across the capacitor C, ec(t), as the state variables.
• The reason for this choice is because the state variables are directly
related to the energy-storage element of a system. The inductor stores
kinetic energy, and the capacitor stores electric potential energy.
• By assigning i(t) and ec(t) as state variables, we have a complete
description of the past history (via the initial states) and the present and
future states of the network.
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Example
The state equation:
This format is also known as the state form if we set
OR
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3.5 Converting a Transfer Function to State
Space
In the last section, we applied the state-space representation to
electrical and mechanical systems. We learn how to convert a
transfer function representation to a state-space representation in
this section.
One advantage of the state-space representation is that it can be
used for the simulation of physical systems on the digital
computer. Thus, if we want to simulate a system that is
represented by a transfer function, we must first convert the
transfer function representation to state space.
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Converting T.F to S.S
• System modeling in state space can take on many
representations
• Although each of these models yields the same output for a
given input, an engineer may prefer a particular one for
several reasons.
• Another motive for choosing a particular set of state
variables and state-space model is ease of solution.
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Converting T.F to S.S
There are many ways of converting T.F into S.S but the
most useful and famous are:
1. Direct Decomposition
2. Cascade Decomposition
3. Parallel Decomposition
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Direct Decomposition
Direct Decomposition is applied to T.F that is not
factored form.
Example
𝐶(𝑆)
24
=
𝑅(𝑆) 𝑆 3 +9𝑆 2 +26𝑆+24
Solution:
Step1: Express T.F in negative powers of S
𝐶(𝑆)
24𝑆 −3
=
𝑅(𝑆) 1+9𝑆 −1 +26𝑆 −2 +24𝑆 −3
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Direct Decomposition
Step 2: Multiply the numerator & denominator of T.F by
a dummy variables X(S)
𝐶(𝑆)
24𝑆 −3
𝑋(𝑆)
=
𝑅(𝑆) 1+9𝑆 −1 +26𝑆 −2 +24𝑆 −3 𝑋(𝑆)
Step 3: 𝐶 𝑆 = (24𝑆 −3 )
𝑅 𝑆 = (1 + 9𝑆 −1 + 26𝑆 −2 + 24𝑆 −3 ) X(S)
Step 4: Construct state diagram using above equation
𝑋 𝑆 = 𝑅 𝑆 − 9𝑆 −1 X(S) − 26𝑆 −2 𝑋(𝑆) − 24𝑆 −3 X(S)
𝐶 𝑆 = 24𝑆 −3 X(S)
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Direct Decomposition
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Direct Decomposition
From State diagram
In vector-matrix form,
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Direct Decomposition
𝐶(𝑆)
𝑅(𝑆)
=
𝑏𝑛−1 𝑆 𝑛−1 +𝑏𝑛−2 𝑆 𝑛−2 +⋯+𝑏0
𝑆 𝑛 +𝑎𝑛−1 𝑆 𝑛−1 +⋯+𝑎0
General form of Direct Decomposition
0
𝑋1
0
⋮
⋮ =
0
𝑋𝑛
−𝑎0
𝑦 = 𝑏0
𝑏1
1
0
⋮
0
−𝑎1
…
0
0
1 …
⋮
0 1
1
…
−𝑎𝑛−1
…
… 𝑏𝑛−1
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0
𝑋1
⋮ + ⋮ r(t)
0
𝑋𝑛
1
𝑋1
⋮ + 0 r(t)
𝑋𝑛
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Cascade (Series) Decomposition
May applied to T.F that are written as product of simple
first or 2nd Order components (factored form)
Example
𝐶(𝑆)
24
= 3 2
𝑅(𝑆) 𝑆 +9𝑆 +26𝑆+24
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=
24
(𝑆+2)(𝑆+3)(𝑆+4)
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Cascade (Series) Decomposition
𝑊
1
=
24𝑅 𝑆+2
𝑆 −1
=
1+2𝑆 −1
𝑉1
𝑉1
𝑊 𝑆 = 𝑆 −1 𝑉1
𝑉1 = 24𝑅 − 2𝑆 −1 𝑉1
𝑍
𝑊
=
1
𝑆+3
=
(1)
𝑆 −1
𝑉2
1+3𝑆 −1 𝑉2
𝑍 𝑆 = 𝑆 −1 𝑉2
𝑉2 = 𝑊 𝑆 − 3𝑆 −1 𝑉2
(2)
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Cascade (Series) Decomposition
𝐶(𝑆)
𝑍(𝑆)
=
1
𝑆+4
=
𝑆 −1
𝑉3
1+4𝑆 −1 𝑉3
𝐶 𝑆 = 𝑆 −1 𝑉3
𝑉3 = 𝑍 𝑆 − 4𝑆 −1 𝑉3
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(3)
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Cascade (Series) Decomposition
Now write the state equations for the new representation of the system.
The state-space representation is completed by rewriting above
Eqs in vector-matrix form:
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Parallel Decomposition
Parallel subsystems have a common input and an output formed
by the algebraic sum of the outputs from all of the subsystems.
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Parallel Decomposition
Example
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Parallel Decomposition
𝑋1
𝑅
𝑋2
𝑅
𝑋3
𝑅
=
1
𝑆+2
𝑋1 = 𝑟 − 2𝑋1
=
1
𝑆+3
𝑋2 = 𝑟 − 3𝑋2
=
1
𝑆+4
𝑋3 = 𝑟 − 4𝑋3
𝐶 = 12𝑋1 − 24𝑋2 + 12𝑋3
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Parallel Decomposition
𝑋1
−2
𝑋2 = 0
0
𝑋3
0
0 𝑋1
1
−3 0 𝑋2 + 1 𝑟(𝑡)
0 −4 𝑋3
1
𝑋1
𝐶 = 12 −24 12 𝑋2 + 0𝑟(𝑡)
𝑋3
Thus, our third representation of the system yields a diagonal
system matrix. What is the advantage of this representation?
Each equation is a first-order differential equation in only one
variable. Thus, we would solve these equations independently.
The equations are said to be decoupled.
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3.6 Converting from State Space to a
Transfer Function
In Chapters 2 and 3, we have explored two methods of
representing systems: the transfer function representation and
the state-space representation. In the last section, we united the
two representations by converting transfer functions into statespace representations. Now we move in the opposite direction
and convert the state-space representation into a transfer
function.
Given the state and output equations
𝑋 = Ax + Bu
y = Cx + Du
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Converting From S.S to T.F
Take the Laplace transform assuming zero initial conditions:
SX(s) = AX(s) + BU(s)
Y(s) = CX(s) + DU(s)
Solving for X(s) ,
(SI- A)X(s) = BU(s)
X(s) = (SI-A)−1 BU(s)
where I is the identity matrix.
𝑌 𝑠 = 𝐶(𝑆𝐼 − 𝐴)−1 𝐵𝑈 𝑠 + 𝐷𝑈 𝑠 = 𝐶(𝑆𝐼 − 𝐴)−1 𝐵 +
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Example
𝑋1
0
1
0 𝑋1
0
𝑋2 = 0
0
1 𝑋2 + 0 𝑟(𝑡)
−24 −26 −9 𝑋3
1
𝑋3
𝐶 = 24 0
𝑋1
0 𝑋2 + 0𝑟(𝑡)
𝑋3
Find T.F
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Example
𝑆 −1
0
= 0
𝑆
−1
24 26 𝑆 + 9
(𝑆𝐼 − 𝐴)−1
−1
𝐴
𝐶(𝑆𝐼
=
−1
𝑎𝑑𝑖 𝐴
𝐴
− 𝐴)−1 𝐵
24
= 3 2
𝑆 +9𝑆 +26𝑆+24
Basil Hamed
93
H.W
1.18, 2.18, 2.26, 2.30, 3.3, 3.4. 3.14
Due Next Class
Basil Hamed
94