Post on 18-Nov-2021
● What is Control System?
● Control systems types
● Examples
● Control systems objectives
● Design Process
1 6/10/2012 Dr. Ahmad Al-Jarrah
A control system is an interconnection of components forming a system configuration that will provide a desired system response.
2
Input Output
linear system theory assumes a cause–effect
relationship for the components of a system
and is the basis of system analysis
The variable which may be
adjusted to bring about the
required control action
(also know on the actuating
Signal).
The physical
system which is to
be controlled.
The variable to
be controlled.
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Open Loop System
An open-loop control system utilizes an
actuating device to control the process
directly without using feedback
Actuating signal
Disadvantages:
Open-loop control system cannot compensate for disturbance
Inputs to the process or for process parameter variations.
6/10/2012 Dr. Ahmad Al-Jarrah
5
Closed Loop System
A closed-loop control system uses a
measurement of the output and feedback of this
signal to compare it with the desired output
(reference or command)
The error can be compensated for by the
controller which generates a correcting
signal u(t)
6/10/2012 Dr. Ahmad Al-Jarrah
1. System/Plant to be controlled
2. Actuators Converts u(t) to power signal
3. Sensors gives measurement of system output
4. Reference input desired output
5. Error detector
6. Controller operates on the error signal to form the required control signal u(t)
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Control system components…!!
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• Stabilizing closed loop system • Accuracy • Achieving proper transient and steady-state response • Reduction of sensitivity to process parameters • Disturbance rejection • Performance and robustness
6/10/2012 Dr. Ahmad Al-Jarrah
1. Establishment of goals, variables to be controlled, and specifications.
2. System definition and modeling.
3. Control system design, simulation, and analysis.
4. If the performance meets the specifications, then finalize the design.
5. Otherwise iterate.
14 6/10/2012 Dr. Ahmad Al-Jarrah
Mathematical models of physical systems are key elements in the design and analysis of control systems.
We will consider electrical and mechanical systems
Obtain the input-output relationship for components and subsystems of the system in the form of transfer functions using Laplace transforms.
Forming Different graphical representations of the system model (Block diagram and signal flow).
1 6/10/2012 Dr. Ahmad Al-Jarrah
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Recognize that differential equations can describe the dynamic behavior of physical systems.
Be able to utilize linearization approximations through the use of Taylor series expansions.
Understand the application of Laplace transforms and their role in obtaining transfer functions.
Be aware of block diagrams (and signal-flow graphs) and their role in analyzing control systems.
Understand the important role of modeling in the control system design process
6/10/2012 Dr. Ahmad Al-Jarrah
3
Electrical Components
The differential
equations describing
the dynamic
performance of a
physical system
are obtained by
utilizing the physical
laws of the process
like Kirchhoff's laws
(KCL, KVL) and
Newton's second law
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4
Mechanical Components: Translational motion
s
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Mechanical Components: Rotational motion
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Element Type Physical Element Describing Equation
Inductive storage Electrical Inductance v = L di/dt
Translational spring F= kx
Rotational spring T=kθ
Capacitive storage Electrical capacitance i = C dv/dt
Translational mass F = M d 2x/dt 2
Rotational mass T = J dω /dt
Energy dissipators Electrical resistance v = iR
Translational damper F = b dx/dt
Rotational damper T = bω
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Examples
8
The Laplace transform is a mathematical tool for solving
linear time invariant differential equation.
It allows a time domain differential equation model of a
system to be transformed in to algebraic model
Definition
Therefore simplifying the
analysis and design of a control
system
0),(for )()}({)(0
ttfdttfetfLsF st
)}({)( 1 sFLtf
x(t) can be found by applying the
inverse Laplace transform of X(s) 6/10/2012 Dr. Ahmad Al-Jarrah
In general, a necessary condition for a linear system can be
determined in terms of an excitation x(t) and a response y(t)
When the system at rest is subjected to an excitation
, it provides a response and when the
system is subjected to an excitation , it provides
a corresponding response ;
For a linear system, it is necessary that the excitation
result in a response
This is usually called the principle of superposition
If the system is nonlinear a linear one can be obtained using
Taylor series expansion around a known operating conditions
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)(1 tx )(1 ty)(2 tx
)(2 ty
)()( 21 tt xx )()( 21 tt yy
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( , )x y
Consider a system whose input variable is x(t) and output
variable is y(t) where the relationship between them is
nonlinear given by y=f(x); If the operation conditions
corresponds to , then a linear relationship around this
point can be found using the Taylor series as follows:
y f x
y f xdf
dxx x
d f
dxx x
( )
( ) ( )!
( ) .......1
2
2
2
2
Where the derivatives are evaluated . df
dx
d f
dx, ,.......
2
2 x x
Where, y f x kdf
dx x x
( ) ,
Eliminating higher order terms from Taylor series gives: ( ) ( )y y k x x
y k x
Linear relationship
11
Time domain Differential
equation
Laplace domain Algebraic
equation
Solution of Differential
Equation
Solution of Algebraic
Equation
Laplace Transform
Inverse Laplace
Transform
F(s)F(t)
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Laplace transform of
time Differentiation
)0()(])(
[ fssFdt
tdfL
We can extend the time
differentiation to be:
)0(...
)0(')0()()(
)0('')0(')0()()(
)0(')0()()(
)1(
21
23
3
3
2
2
2
n
nnn
n
n
f
fsfssFsdt
tdfL
casegeneral
fsffssFsdt
tdfL
fsfsFsdt
tdfL
Laplace transform of
the unit step (u(t)=1) se
sdtetuL stst 11
1)]([ |0
0
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f(t) L(f) f(t) L(f)
1 Unit-impulse 1 7 cos t
2 Unit-step 1 1/s 8 sin t
3 Unit-ramp t 1/s2 9 cosh at
4 t2 2!/s3 10 t eat
5 tn (n is +ve integer)
11 eat cos t
6 eat 12 eat sin t
1
!ns
n
as
1
22 s
s
22
s
22 as
s
2)(
1
as
22)(
as
as
22)(
as
)(t
For most engineering purposes the
inverse Laplace transformation can be
accomplished simply by referring to
Laplace transform tables
6/10/2012 13 Dr. Ahmad Al-Jarrah
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The Laplace Transform….cont
Initial Value Theorem
0
)(lim)(lim)0(
ts
tfssFf
Final Value Theorem
ts
tfssFf
0
)(lim)(lim)(
For function
f(t)
The final value theorem is very useful for
analysis and design of control systems,
since it gives the final value of a time
function f(t)
The Laplace variable s can be considered
to be the differential operator so that dt
ds
And we also have the integral operator
0
1dt
s
s-operator is a complex quantity has a real and imaginary parts
jbas
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Examples
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The Transfer Function (T.F) of Linear Systems
The transfer function of a linear system is defined as the
ratio of the Laplace transform of the output variable to the
Laplace transform of the input variable, with all initial
conditions assumed to be zero.
The transfer function of a system (or element) represents the
relationship describing the dynamics of the system under
consideration
Represents system dynamics in s-domain
)(. sGinput
outputFT
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)(
)(
)(.
sq
sp
sGinput
outputFT
Where p(s) and q(s ) are polynomials
The roots of p(s) are called the zeros of the
system where the roots of q(s ) are called the
poles of the system
))()((
))((
)(
)()(
321
21
pspsps
zszs
sq
spsG
q(s ) is also known as the characteristic equation of the systems
The location of the roots of q(s ) in s-plane gives a
character to the system performance
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Example: TF’s of Operational Amplifier circuits
The operational amplifier (op-amp) belongs to an important class of analog
integrated circuits commonly used as building blocks in the implementation
of control systems and in many other important applications.
• Compensators
• Control laws
• Filters
• Comparator/summing elements
6/10/2012 Dr. Ahmad Al-Jarrah
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DC motors are widely
used in numerous control
applications such as
robotic manipulators, tape
transport mechanisms,
disk drives, and machine
tools
The DC motor converts direct
current (DC) electrical energy into
rotational mechanical energy
DC motors features:
• High output torque
• Speed controllability over a wide range
• Portability
• Adaptability to various types of controllers
TF’s of DC motors Examples: A DC motor is used to move loads and is
called an actuator.
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This relationship is typically represented by the TF of
the subsystem relating the input and output variables
Block Diagram (BD) Models
Again:
Control systems consists of elements that are represented
mathematically by a set of simultaneous differential equations
Laplace transformation reduces the problem of differential
equations to the solution of a set of linear algebraic equations.
Since control systems are concerned with the control of specific
variables, the controlled variables must relate to the controlling
variables
The importance of the TF is evidenced by the ability
to represent the relationship of system variables by
diagrammatic means called BD
Hence, the control system with all its elements can be
represented by one BD showing all variables relations
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Armature controlled DC motor BD
6/10/2012 Dr. Ahmad Al-Jarrah
In order to find the cause-effect relationship of a system
BD, we simplify the BD (reduction) by applying the rules
of BD algebra.
23
Block Diagram (BD) Algebra
Original Diagram Equivalent Diagram
(1)
(2)
(3)
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(5)
(4)
(6)
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Example
For the following control system, find the input-output relationship
(i.e. TF) relation the output variable Y(s) to the input variable R(s).
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Signal-Flow (SF) Graph Models
Block diagrams are adequate for the representation of the system
interrelationships. However, for a system with reasonably complex
interrelationships, the block diagram reduction procedure is often
quite difficult to complete.
An alternative method for determining the relationship
between system variables has been developed by Mason
which is called the signal-flow graph method
A signal-flow graph is a diagram consisting of nodes that are
connected by several directed branches and is a graphical
representation of a set of linear relations.
The reduction procedure (used in the BD method) is not
necessary to determine TF (input-output relationship) of a system
represented by SF graph.
We apply Mason`s Gain Formula to find the TF
6/10/2012 Dr. Ahmad Al-Jarrah 28
General SF Graph
Node: acts like a summing point and also represents a system variable.
Transmittance: real or complex gain between two nodes.
Branch: directed line segment joining two nodes.
Input node (source): only outgoing branches.
Output node (sink): only incoming branches.
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Mixed node: both incoming and outgoing branches
Path: traversal of connected branches in the direction of arrows.
Loop: closed path.
Loop gain: product of branch transmittance at a loop.
Non touching loops: they do not posses any common nodes.
Forward path: path from an input to an output node that does
not cross any node more than once.
Forward path gain: product of transmittances of a
forward path
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SF Graph Algebra
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Examples
(1)
(2) (3)
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Mason`s Gain Formula
The formula is often used to relate the output variable
Y(s) to the input variable R(s) (i.e. finding the TF) and is
given by
K KKP
TF
KP
where,
is the gain of path K from input node to output node in the
direction of the arrows and without passing node than once.
Δ : determinant of the graph
K KP : Cofactor o the path
Δ = 1 – ( sum of all different 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 non
touching loops )
6/10/2012 Dr. Ahmad Al-Jarrah 34
Example
For the following control system, find the input-output relationship
(i.e. TF) relation the output variable Y(s) to the input variable R(s).
6/10/2012 Dr. Ahmad Al-Jarrah 35
Example
For the following control system, find the input-output relationship
(i.e. TF) relation the output variable Y(s) to the input variable R(s).
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Example: Armature Controlled DC Motor (page 94)
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Example: Disk Drive Read System (Page 118)
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1 6/19/2012 Dr. Ahmad Al-Jarrah
So far we have used the Laplace transform to obtain a transfer
function model representing liner time invariant physical
systems described by ordinary differential equations.
An alternative method of system modeling using time domain
methods can be used State-space (SS) model
The state variables (states) of a dynamic system is a set of
variables such that the knowledge of these variables and the
input function will with the equations describing the dynamics
provide the future state and output of the system.
SS can be utilized for nonlinear, time
varying and multivariable systems
6/19/2012 Dr. Ahmad Al-Jarrah 2
For a dynamic system, the state of a system is described in
terms of a set of state variables . The state
variables are those variables that determine the future behavior
of a system when the present state (initial conditions) of the
system and the excitation signals are known.
)(),......,(),( 21 txtxtx n
6/19/2012 Dr. Ahmad Al-Jarrah 3
The response of a system is described by the set of
first-order differential equations written in terms of
the state variables and the inputs.
dt
dxx
ububxaxaxax
ububxaxaxax
ububxaxaxax
mnmnnnnnnn
mmnn
mmnn
where
....................................
...................................
.....................................
112211
212122221212
111112121111
6/19/2012 Dr. Ahmad Al-Jarrah 4
mnmn
m
nnmnn
n
n
n u
u
bb
bb
x
x
x
aaa
aaa
aaa
x
x
x
1
1
111
2
1
21
22221
11211
2
1
In matrix form:
nx
x
x
X2
1Where is known as state vector
And the state differential equation is given by
MatrixInput B
(dynamics)Matrix StateA where
equation) [state
BUAXX
matrix mnB
matrix square
nnA
6/19/2012 Dr. Ahmad Al-Jarrah 5
And the outputs of a linear system can be related to the state
variables and the input signals by the output equation:
Matrixon TransmissiDirect D
iableoutput var theDecides C
formtor column vecin expressed signalsoutput ofSet
Y
DUCXY
Where
The solution of the state differential equation can be obtained in a manner
similar to the method for solving a first-order differential equation
x ax bu
Taking the Laplace transform gives sX s x aX s bU s
X sx
s a
b
s aU s
( ) ( ) ( ) ( )
( )( )
( )
0
0
6/19/2012 Dr. Ahmad Al-Jarrah 6
Taking the inverse Laplace transform gives
x t e x e bu dat a tt
( ) ( ) ( )( )
00
Similarly, the solution of the state differential equation but using matrix
exponential function as follows:
X t e X e BU dAt A tt
( ) ( ) ( )( )
00
The solution of external input
(forced system)
This solution can be verified by taking the Laplace transform of the state
Differential equation : X AX BU
X s sI A X sI A BU s( ) [ ] ( ) [ ] ( ) 1 10
Noting that is the Laplace transform of [ ] ( )sI A s 1 ( )t e At
Where is called the state transition matrix ( )t
6/19/2012 Dr. Ahmad Al-Jarrah 7
The solution of the state differential equation for initial conditions input
(natural or unforced system when U=0) is simply:
x t
x t
x t
t t
t t
t t
x
x
xn
n
n
n nn n
1
2
11 1
21 2
1
1
2
0
0
0
( )
( )
.
.
( )
( ) ............ ( )
( )............. ( )
.
.
( )............. ( )
( )
( )
.
.
( )
How to find the transfer function from the state equation?
The TF of a SISO system can be derived as follows
Taking Laplace
0
assuming ( ) 0
X Ax Bu
SX(s) - X( ) AX(s) Bu(s)
And Y (s) CX(s) Du(s)
X o
6/19/2012 Dr. Ahmad Al-Jarrah 8
sX s AX s Bu s
sI A X s Bu s
X s sI A Bu s
( ) ( ) ( )
( ) ( ) ( )
( ) [ ] ( )
1
y s CX s Du s( ) ( ) ( )
y s C sI A Bu s Du s
y s
u sC sI A B D
( ) [ ] ( ) ( )
( )
( )[ ]
1
1
Now, for the output equation ,
substitute for X(s) to have:
6/19/2012 Dr. Ahmad Al-Jarrah 9
Examples
1 6/22/2012 Dr. Ahmad Al-Jarrah
This chapter will emphasize on the advantages of the feedback
closed-loop control system compared to the open-loop.
Introducing a feedback in a control system is often necessary
to improve the control system.
Generally speaking, the areas of interest in a control system
response are:
Minimizing the error signal.
Reducing the effect of system parameter uncertainties or changes
(i.e. reducing the system sensitivity to parameter
uncertainties/variations).
Reducing the effect of unwanted signals like disturbance or noise.
Improving the transient and steady-state performance of a system.
6/22/2012 Dr. Ahmad Al-Jarrah 2
An open-loop system operates without feedback and directly
generates the output in response to an input signal.
The disturbance directly influence the output; without
feedback, the control system is highly sensitive to disturbance.
T sd ( )
Open-loop control system
6/22/2012 Dr. Ahmad Al-Jarrah 3
A closed-loop system uses a measurement of the output signal
and a comparison with the desired output to generate an error
signal [E(s)=desired response-actual response] that is used by
the controller to adjust the actuator be generating a control signal
u(t).
Closed-loop control system
Error
Advantages of closed-loop system:
Reducing the error.
Reducing the sensitivity to parameter variations.
Improving rejection of disturbances
Improving the transient response.
6/22/2012 Dr. Ahmad Al-Jarrah 4
E s R s Y s
But
Y sG s G s
G s G sR s
G s
G s G sT s
G s G s
G s G sN s
E sG s G s
R sG s
G s G sT s
G s G s
G s G
C
C C
dC
C
C C
dC
C
( ) ( ) ( )
,
( )( ) ( )
( ) ( )( )
( )
( ) ( )( )
( ) ( )
( ) ( )( )
( )( ) ( )
( )( )
( ) ( )( )
( ) ( )
( ) (
1 1 1
1
1 1 1 sN s
)( )
Desired Input Disturbance input Noise input
L s G s G s
F s L s
S sF s L s G s G s
C
C
( ) ( ) ( )
( ) ( )
( )( ) ( ) ( ) ( )
1
1 1
1
1
1
Define
Where,
L(s) is loop gain
S(s) is the sensitivity function
Error signal analysis:
6/22/2012 Dr. Ahmad Al-Jarrah 5
Where C(s) is the complementary sensitivity function
Note that sensitivity can be reduced by increasing the
controller gain and the error can be reduced by reducing
the sensitivity S(s) and C(s), but
E s S s R s S s G s T s C s N sd( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
C sL s
L s( )
( )
( )
1
S s C s( ) ( ) 1
i.e. S(s) and C(s) can not reduced simultaneously; so,
design compromises must be made.
6/22/2012 Dr. Ahmad Al-Jarrah 6
T sY s
R s
G s G s
G s G s
C
C
( )( )
( )
( ) ( )
( ) ( )
1
System sensitivity S(s) is the ratio of the percentage change in the
system transfer function to the percentage change of a process
transfer function (or parameter).
Sensitivity of Control Systems to Parameter Variations
Assume the system TF to be
Where T(s) is the system TF, is the controller, and G(s) is the
process TF.
G sC ( )
S sT s T s
G s G s( )
( ) / ( )
( ) / ( )
6/22/2012 Dr. Ahmad Al-Jarrah 7
S sT T
G G
S sT
G
G
T
SG
G G
G
G G G G
SG s G s
G
T C
C C C
G
T
C
( )/
/
( ) .
( ).
/ ( )
( ) ( )
1 1
1
1
2
For small incremental changes, given the previous T(s), the sensitivity of the
closed-loop system T(s) with respect to a small changes in the process G(s)
becomes
If we seek to determine , where is a parameter within G(s), using the
chain rule gives S T
S S ST
G
T G
6/22/2012 Dr. Ahmad Al-Jarrah 8
Note: Uncertainties in the system model might come from : Aging ,
changing environment, and ignorance of exact values of
the system parameters which all affect the control process.
In open-loop : inaccurate output result from
these effects.
In closed-loop: the system attempts to
compensate and correct for these effects
by use of the controller
Sensitivity is reduced in closed-loop by increasing
system compared to the open-loop case where S=1 in the case
of T(s)=G(s)
G s G sC ( ) ( )
6/22/2012 Dr. Ahmad Al-Jarrah 9
Example: Feedback Amplifier
Study the sensitivity changes for the two cases: open-loop
and closed-loop.
6/22/2012 Dr. Ahmad Al-Jarrah 10
Open-loop speed control system (without tachometer feedback)
Disturbance Signals in a Feedback Control System
Disturbance signal is unwanted undesired signal that
affects the output signal
( )
( )
/ ( )
.
s
T s
Js b
K K
R Js bd m b
a
1
11
Assuming very small
inductance and only
disturbance input
L Va a 0 0
6/22/2012 Dr. Ahmad Al-Jarrah 11
( )
( )
s
T sJs b
K K
Rd m b
a
1
Steady-state speed due to a step disturbance is T sD
sd ( )
ss
m b
a
ss openm b
a
s s
s
s
Js bK K
R
D
s
D
bK K
R
lim ( )
lim
( )
1
s
6/22/2012 Dr. Ahmad Al-Jarrah 12
Closed-loop speed tachometer control system
( )
( )
/ ( )
( )
s
T s
Js b
Js b
K K
RK
K
K
Js bK
RK K K
d m a
a
tb
a
m
a
t a b
1
11
1
controller
6/22/2012 Dr. Ahmad Al-Jarrah 13
( )
( ).
ss closed
ss open
a m b
a m t
R b K K
K K K
002
ss
m
a
t a b
ss closedm
a
t a b
s s
s
s
Js bK
RK K K
D
s
D
bK
RK K K
lim ( )
lim
( )
( )
( )
1
s
For high controller
gain
Ka
6/22/2012 Dr. Ahmad Al-Jarrah 14
Control of the Transient Response
Voltage
divider DC motor
( )
( )
( )
,
s
V s
K
R Js R b K K
G sK
s
where
R J
R b K K
KK
R b K K
a
m
a a m b
a
a b m
m
a b m
1
1
1
1
1
1 is known as the time constant
v t k E
V sk E
s
a
a
( )
( )
2
2
And the input is :
The system TF is
{Open-Loop}
6/22/2012 Dr. Ahmad Al-Jarrah 15
( ) ( ) ( )
( )
s G s V sK
s
k E
s
K k Es
s
a
1
1
2
1 2
1
1
1 1
1
To find the response : ( )t
( ) { ( )}
( )/
t L s
K k et
1
1 2 11
If the speed is too
slow, we reduce the time
Constant of the motor
( )t
i.e. Choose another motor
with different time constant
ss K K k 1 2
6/22/2012 Dr. Ahmad Al-Jarrah 16
{Closed-Loop}
( )
( )
( )
( )
/
( )
/
,
s
R s
K G s
K K G s
K K
s K K K
K K
sK K K
K K
s
where
K K K
a
a t
a
a t
a
a t
a
c
c
a t
1
1 1 1
1
1
1 1
1 1
1
1
1 1
1
1
1 ci.e. the Closed-loop system has a
faster response compared to the
Open-loop system
cAlso, can be reduced by increasing the controller gain Ka
6/23/2012 Dr. Ahmad Al-Jarrah 17
The response is For High
amplifier gain
Noting different steady-
state values for the open-
loop and closed-loop
( ) { ( )}
( )
,
/
t L s
K e
where
KK K k E
K K K
t
a
a t
C
1
1 2
1
1
1
Ka
6/23/2012 Dr. Ahmad Al-Jarrah 18
Steady-State Error
The steady-state error is the error after the transient
response has decayed, leaving only the continuous
steady response.
E s R s Y s
E s
R s
Y s
R s
G s
( ) ( ) ( )
( )
( )
( )
( )
( )
1
1
{Open-Loop}
6/23/2012 Dr. Ahmad Al-Jarrah 19
{Closed-Loop}
E s R s Y s
E s
R s
Y s
R s
T s T sG s G s
G s G s
G s G s
and
e e s E s
c
c
c
ss
( ) ( ) ( )
( )
( )
( )
( )
( ) , ( )( ) ( )
( ) ( )
( ) ( )
( ) lim ( )
1
11
1
1
s
For unity-feedback
control system, i.e.,
H s( ) 1
1 3/27/2013 Dr. Ahmad Al-Jarrah
One of the first steps in the design process is to specify the
measures of performance (performance specifications). In this
chapter we introduce the common time-domain specifications
such as percent overshoot, settling time, peak time, rise time and
steady-state error.
Time domain performance specifications can be found from the
response of the system to a given input.
?? ??
The time response of a control system is usually divided into two
parts. The transient response and the steady-state response.
3/27/2013 Dr. Ahmad Al-Jarrah 2
In control systems transient response is defined as the part
of the time response that goes to zero as time becomes
very large 0)(lim
tytt
The steady state response is simply the part of the total
response that remains after the transient has died out.
response statesteady responsetransient
Where
system data continuous a of response time thedenote )(Let
sst
tss
YY
(t)Y(t)YY(t)
tY
Y t e tt( ) cos 5 5 2
3/27/2013 Dr. Ahmad Al-Jarrah 3
Transient Steady-state
3/27/2013 Dr. Ahmad Al-Jarrah 4
For the purposes of analysis and design it is necessary to
assume some basic types of test input signals so that the
performance of a system can be evaluated
Test Input Signals for the Time Response of
Control Systems
These input signals are close to a real life input signals
Step input , ramp input, parabolic input, and unit-impulse input
0 0
0 )(
t
tAtr
I. Step input r(t) :
s
AsR )(
Where A is the amplitude of the step input
3/27/2013 Dr. Ahmad Al-Jarrah 5
II. Ramp input r(t) :
2)(
s
AsR
0 0
0 )(
t
tAttr
Slope=A
III. Parabolic input r(t) :
r tAt t
t( )
2 0
0 03
2)(
s
AsR
IV. Unit impulse input : ( )t
Unit impulse is a special case from rectangular function f(t)
f tt
otherwise
( )/
12 2
0
As approaches to zero, the function
f(t) approaches the unit-impulse function
0where
( )t dt
1 R s( ) 1
3/27/2013 Dr. Ahmad Al-Jarrah 6
Performance of Control Systems
I. Zero order system (static TF)
II. First order system
III. Second order system
IV. Higher order system
We will study the performance specification of the following
control systems:
Static TF
3/27/2013 Dr. Ahmad Al-Jarrah 7
II. First order system:
s
1E(s) R(s) C(s)
Test signal is a unit-step function, R(s)=1/s
A first-order system without
zeros can be represented by
the following TF:
1
1
)(
)(
ssR
sC
1
11
)1(
1)(
1
1)(
ssss
sRs
sC
To find the response
t
etc
1)(
If t= , , so the step response is
C ( ) = (1− 0.37) = 0.63
3/27/2013 Dr. Ahmad Al-Jarrah 8
Settling time ( ) is defined to be the time taken for the step
response to come to within 2% of the final value of the step
response (i.e. entering the steady-state region).
4st
st
If t= , , so the step response is
C ( ) = (1− 0.018) = 0.982
4
4
3/27/2013 Dr. Ahmad Al-Jarrah 9
a
1
3/27/2013 Dr. Ahmad Al-Jarrah 10
First order system response for different input signals:
3/27/2013 Dr. Ahmad Al-Jarrah 11
10
9
sR(s) Y(s)
Example: The following system, find the unit-step response of the system and
the steady-state error.
Y s G s R s
s s
( ) ( ) ( )
( )
9
10 Again, what is the time constant??
y t L Y s
Ls s
e t
( ) { ( )}
. .. ( )
1
1 100 9 0 9
100 9 1
y y y t
t
e
ss
ss
( ) lim ( ) .
. .
0 9
1 0 9 01
And
3/27/2013 Dr. Ahmad Al-Jarrah 12
Example: The following figure gives the measurements of the step response
of a first-order system, find the transfer function of the system.
3/27/2013 Dr. Ahmad Al-Jarrah 13
III. Second order system:
Second-order systems exhibit a wide range of responses which
must be analyzed and described.
For a first-order system, varying a single parameter changes the
speed of response,
Changes in the parameters of a second order system can
change the form of the response not only the speed of the
response.
For example: a second-order system can display characteristics
much like a first-order system or, depending on the system’s
parameters values, pure oscillations or damped behavior might
result for its transient response.
3/27/2013 Dr. Ahmad Al-Jarrah 14
Assume the following transfer function of a general closed-loop
second-order system:
We can re-write the above transfer function in the following form of
a standard second order system closed-loop transfer function:
is referred to as the un-damped natural frequency
of the system, which is the frequency of oscillation of
the system without damping.
T ss s
n
n n
( )
2
2 22
T sb
s as b( )
2
3/27/2013 Dr. Ahmad Al-Jarrah 15
is referred to as the damping ratio of the second
order system, which is a measure of the degree of
resistance to change in the system output.
And the poles of the closed loop system
(roots of the characteristic equation)are:
According the value of ζ, a second-order system can be set into
one of the following four cases:
1. Overdamped - when the system has two real distinct poles (ζ >1).
2. Underdamped - when the system has two complex conjugate poles (0 <ζ <1)
3. Undamped - when the system has two imaginary poles (ζ = 0).
4. Critically damped - when the system has two real but equal poles (ζ = 1).
3/27/2013 Dr. Ahmad Al-Jarrah 16
Case I:
1s polesdifferent Real
1 (Stable) case dampedOver
2
1,2
nn
jw
Over-damped
n
2s1s
Case II:
1s polesComplex
1 0 (Stable) case dUnderdampe
2
1,2
nn j jw
n
21 nj
21 nj
Under damped
y te
s
e
s
n
s t s t
( )
12 1
1 2
1 2
y te
t
where
nt
d
d
n
n
n
d n
( ) sin( )
tan tan
11
1
1
2
1 1
2
2
Assume a unit-step input for the coming slides,
: is the damped natural frequency
3/27/2013 Dr. Ahmad Al-Jarrah 17
Case III:
s polesImaginary , 0
(Stable) case )( damped Zero
1,2 nj
Undamped
Undamped
nj
nj
jw
Case IV:
- pole double equall Real
1 (Stable) damped Critically
n
jw
Critically damped
n
y t e tnt
n( ) ( ) 1 1
y t tn( ) cos( ) 1
Remark:
3/27/2013 Dr. Ahmad Al-Jarrah 18
0
Poles path (location) in
s-plane as changes
Second order system
responses for a unit-
step input and different
values of
3/27/2013 Dr. Ahmad Al-Jarrah 19
Transient response specifications of an underdamped second order system for
a unit-step input:
1. Rise time( ) : rise time is the time required for the response to change
from a lower prescribed value to a higher one.
2. Peak time( ): the peak time is the time required for the response
to reach the first peak
3. Settling time ( ): the settling time is the time required for the
amplitude of the sinusoid to decay to 2% or 5% of the steady-state
value.
3/27/2013 Dr. Ahmad Al-Jarrah 20
t r
t p
t s
21
nd
pt
y te
tnt
d( ) sin( )
11 2
All performance specifications are derived from:
tr
d n
1 2
3/27/2013 Dr. Ahmad Al-Jarrah 21
2% criterion
n
st
4
4. Maximum overshoot percentage: the percent overshoot is defined as the
amount that the waveform at the peak time overshoots the steady-state
value.
n
st
3 5% criterion
%100)(
)()(% x
y
ytyMP
p
%100_
_max_% x
finaly
finalyyOS
OR
Maximum Peak Percentage MP%
Overshoot Percentage OS%
MP e%
100
1 2
3/27/2013 Dr. Ahmad Al-Jarrah 22
For given OS%, the damping ratio can
be solved from the OS% equation;
100/%ln
100/%ln
22 MP
MP
3/27/2013 Dr. Ahmad Al-Jarrah 23
Examples:
3/27/2013 Dr. Ahmad Al-Jarrah 24
C sa
s
A
s
B s D
s s
i
ii
q
k k
k kk
r
( )
1
2 21 2
IV. Higher order system:
Assume the following TF:
For step input and using partial fraction,
3/27/2013 Dr. Ahmad Al-Jarrah 25
The response of a higher order system is a combination of responses of
first and second order systems.
Observations:
The dominant pole is the one nearer to the j-axis.
Poles and zeros that are several order (usually five order )of magnitude smaller
than the dominate, Poles and zeros can be ignored.
Generally speaking, the order of the denominator is larger than that of
the numerator.
For the system to be stable, all the poles of the transfer function must be
negative, that is to say must lay in the left hand side of the s-plane.
A pole and a zero that are near to each other tend to cancel each other.
A pole and a zero that coincide cancel each other.
The poles far away from the j-axis (Im-axis) can be ignored (fast response).
3/27/2013 Dr. Ahmad Al-Jarrah 26
Steady-state Error Analysis
For the feedback system shown in block
diagram below, the transfer function is given by:
The system error is given by:
This last expression shows that the loop gain G(s)H(s) determine the amount
and nature of the steady state error of a system.
3/27/2013 Dr. Ahmad Al-Jarrah 27
The loop gain G(s)H(s) can be expressed in the general form;
The error in this case would be given by:
3/27/2013 Dr. Ahmad Al-Jarrah 28
The steady state error is calculated as follows:
When the standard test signals of a step (A/s), a ramp (A/s2), and an acceleration
(A/s3) are used, the Laplace operator “s” in the input test signal denominator will
cancel or reduce from the power of “s” in the numerator of the expression above.
The power of “s” (the poles of the G(s)H(s) located on the origin of s-plane), i.e. N,
determines the steady state error response of the system when subjected to standard
test signals, and is called the “type number” of the system.
For N = 0, the system is a type zero, for N = 1, the system is a type one, and so on.
3/27/2013 Dr. Ahmad Al-Jarrah 29
Type Zero System:
The steady state error for a step input; A/s is given by
Position error constant
K G s H s
s
p
lim ( ) ( )
0
3/27/2013 Dr. Ahmad Al-Jarrah 30
The steady state error for a ramp input; A/s2 is given by
Type One System:
The steady state error for a step input; A/s is given by
3/27/2013 Dr. Ahmad Al-Jarrah 31
The steady state error for a ramp input; A/s2 is given by
The steady state error for an acceleration input; A/s3 is given by
Velocity error constant
K sG s H s
s
v
lim ( ) ( )
0
3/27/2013 Dr. Ahmad Al-Jarrah 32
Type Two System:
The steady state error for a step input;
A/s is given by
The steady state error for an
acceleration input; A/s3 is given by
Acceleration error constant
K s G s H s
s
a
lim ( ) ( )2
0
3/27/2013 Dr. Ahmad Al-Jarrah 33
3/27/2013 Dr. Ahmad Al-Jarrah 34
1 7/8/2012 Dr. Ahmad Al-Jarrah
Stability of closed-loop feedback systems is central to control
system design.
A stable system should exhibit a bounded output if the
corresponding input is bounded. This is known as bounded-input-
bounded-output (BIBO) stability.
The stability of a feedback system is directly related to the location
of the roots of the characteristic equation of the system transfer
function.
The Routh-Hurwitz method is introduced as a useful tool for
assessing system stability.
The technique allows us to compute the number of roots of the
characteristic equation in the right half plane without actually
computing the values of the roots.
7/8/2012 Dr. Ahmad Al-Jarrah 2
A stable system is a dynamic system with a bounded
response to a bounded input.
7/8/2012 Dr. Ahmad Al-Jarrah 3
u(t)
N
-N
t
y(t)
M
-M
t
Bounded input signal
Bounded output signal
INPUT – OUTPUT STABLE if and only if every
bounded input produces a bounded output
A system is said to be unstable if it’s not BIBO stable
7/8/2012 Dr. Ahmad Al-Jarrah 4
Input/output stability is characterized by the location of the poles of
the system closed loop transfer function (i.e. roots of the
characteristic equation).
Undammped critically (marginally) stable
System with poles of ZERO real parts
7/8/2012 Dr. Ahmad Al-Jarrah 5
Here we present a method for investigating the stability of high
order systems without having to obtain a complete time response
or determining the precise position of the poles in s-plane.
Assume the following general closed loop transfer function of a
system :
The denominator ( ) can always be expressed as a polynomial of
nth order as follows:
7/8/2012 Dr. Ahmad Al-Jarrah 6
Now, the following conditions are necessary (but not
sufficient) for the system to be stable:
1- All the coefficients have the same sign.
2- No coefficient is zero.
o n, , , ....,1 2 3
If the roots of are positive, that means that the system is
unstable. In this case the polynomial will have
alternating signs
If all the coefficients of the polynomial are positive, this does not
mean that the system is stable
If the roots of are negative, that means that the system is stable
and this case the polynomial will have all positive signs.
Build the Routh-Hurwitz array from
7/8/2012 Dr. Ahmad Al-Jarrah 7
7/8/2012 Dr. Ahmad Al-Jarrah 8
Routh-Hurwitz stability criterion states that the number of
roots of the characteristic equation that have positive real
roots is equal to the number of changes in sign of the first
column of the ordered array.
Now it is sufficient condition for the system to be stable if
there is no changes in sign of the first column of the Routh-
Hurwitz array.
1 7/11/2012 Dr. Ahmad Al-Jarrah
So far in the studies of control systems the role of the characteristic
equation polynomial in determining the behavior of the system has
been highlighted.
The roots of that polynomial are the poles of the control system, and
their locations in the complex s-plane reveal information about the
stability and the performance of the system.
Consider the system shown below:
7/11/2012 Dr. Ahmad Al-Jarrah 2
The transfer function of the system is given by:
The loop gain of the system can be expressed as a numerator
polynomial over a denominator polynomial as follows:
The characteristic equation is then given by:
7/11/2012 Dr. Ahmad Al-Jarrah 3
As an example consider that the system in the previous figure
have the following loop gain:
The characteristic equation for the system is given by:
The numerator is in fact the denominator of the closed loop system TF.
The roots of this polynomial are the closed loop poles of the system.
7/11/2012 Dr. Ahmad Al-Jarrah 4
As K varies between zero and infinity, the closed loop poles of
the system are changing and can be calculated as follows:
s s K2 3 2 0( )
7/11/2012 Dr. Ahmad Al-Jarrah 5
The plot of system poles as they move throughout the s-domain
when K varies between zero and infinity are as shown below:
The root locus is nothing but the path
of the system closed loop poles as
the gain of the system K varies
between zero and infinity.
Since the root locus represent the path of the roots of the characteristic
equation as the gain varies from zero to infinity, it follows that every
point on the root locus must satisfy the characteristic equation, namely;
1 0G s H s( ) ( )
7/11/2012 Dr. Ahmad Al-Jarrah 6
Rearranging the characteristic equation gives;
G s H s( ) ( ) 1
Both G(s) and H(s) are complex quantities. The above equation
can hence be cast in polar or vector form as follows:
This last relationship specifies the conditions that must prevail for
any point on the root locus.
I. Magnitude Condition
II. Angle Condition
7/11/2012 Dr. Ahmad Al-Jarrah 7
I. Magnitude Condition:
II. Angle Condition:
This equation can be rewritten as:
For a given point on Root locii,
s=a+jb
s=a+jb
For a given point on Root locii,
s=a+jb
7/11/2012 Dr. Ahmad Al-Jarrah 8
For the system analyzed before and shown below at the point given
by -1.5 ± j 1.3229, the angles of the vectors from that point to the
two poles are calculated as follows:
G s H s zeros poles
s s
( ) ( )
. .
0 1 2
110 7 69 3
180
1 2
Angle Condition,
7/11/2012 Dr. Ahmad Al-Jarrah 9
Magnitude Condition,
G s H s
K s z s z
s p s p
K s s
j
j
( ) ( )
( )( ).......
( )( )........
( )( )
. .
. .
1 5 1 3229
1 2
1 2
1 5 1 3229
1
1
1 2
2
The point -1.5 ± j 1.3229 satisfied the both conditions since it is on root locus
Before starting the steps of sketching Root locus, we have to know
1. The Start and End Points of a Root Locus
2. Number of segments (branches) of root locus
3. Location of root locus segments on real axis
The characteristic equation can be rearranged as follows:
7/11/2012 Dr. Ahmad Al-Jarrah 10
When K = 0, the last equation becomes;
This indicates that the root locii starts at the poles of the system when K = 0.
The characteristic equation can also be arranged as follows:
When K is at infinity the above equation becomes;
This indicates that the root locii ends at the zeros of the system when K
Root locii starts at the system poles (when ) and ends at the
system zeros (when ). K K 0
7/11/2012 Dr. Ahmad Al-Jarrah 11
Since root locii must start at poles and end at zeros, then it is fair to
assume that the number of segments of root locii is equal to the
number of poles.
If the number of poles in the loop gain equation is larger than the
number of zeros, this means that a number of root locii segments
equal to the number of poles and zeros will be ending at infinity.
Since poles and zeros that occur off the real axis must be in complex
conjugate pairs, it follows that complex portions of the root locii
always occur as complex conjugate portions. These portions appear
as mirror images of one another with the parting line being the real
axis.
Root locii segments on the real axis can be found by applying the angle
criteria as follows:
7/11/2012 Dr. Ahmad Al-Jarrah 12
Test point
This indicates that the assumed
point violates the angle criteria and
hence cannot be on a segment of
the root locus.
Now, let us move the test point as follows:
7/11/2012 Dr. Ahmad Al-Jarrah 13
Applying the angle criteria gives:
This indicates that the assumed
point conforms to the angle criteria
and hence is on a segment of the
root locus.
Now, let us move the test point as follows:
7/11/2012 Dr. Ahmad Al-Jarrah 14
Applying the angle criteria gives:
This indicates that the assumed
point violates the angle criteria and
hence cannot be on a segment of
the root locus.
Now, let us move the test point as follows:
7/11/2012 Dr. Ahmad Al-Jarrah 15
Applying the angle criteria gives:
This indicates that the assumed
point conforms to the angle criteria
and hence is on a segment of the
root locus.
7/11/2012 Dr. Ahmad Al-Jarrah 16
Root locii on the real axis will occur to the left of odd
number of poles and zeros .
Steps of Constructing Root Locus of a System: 1- Write the characteristic equation of the system in the following standard form
1 01 2
1 2
Ks z s z s z
s p s p s p
m
n
( )( ).....( )
( )( ).....( )
Where K might be a controller gain (or system gain) and is
the parameter of interest
7/11/2012 Dr. Ahmad Al-Jarrah 17
2- Locate all poles and zeros in s-plane. p p pn1 2, , ..... z z zm1 2, , .....
3- Determine the root locus segments on the real axis
4- Determine the asymptotes of the root locus:
number of asymptotes = n-m
intersection of asymptotes with real axis
Angles of asymptotes
a
poles zeros
n m
a
k
n mk n m
180 2 11
( ),
5- Find the break away / in points if any
dK
ds 0
Rearrange the characteristic equation then find the
break away/in points that should result from
Why ???!!
7/11/2012 Dr. Ahmad Al-Jarrah 18
6- Find the points of intersection with Im-axis by applying Routh-Hurwitz criteria.
7- Determine the departure angle (arrival angle) of there is complex poles
(complex zeros) by applying the angle condition
8- Calculate the desired gain K that corresponds to a particular desired closed loop
poles by applying the magnitude condition.
-1 -2
-1.5
dK
ds 0
Back to our example with
11 2
k
s s( )(