DEPARTMENT OF ELECTRICAL AND ELECTRONICS ...Control system components can be mechanical, electrical,...
Transcript of DEPARTMENT OF ELECTRICAL AND ELECTRONICS ...Control system components can be mechanical, electrical,...
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 1
M.I.E.T. ENGINEERING COLLEGE
(Approved by AICTE and Affiliated to Anna University Chennai)
TRICHY – PUDUKKOTTAI ROAD, TIRUCHIRAPPALLI – 620 007
DEPARTMENT OF ELECTRICAL AND
ELECTRONICS ENGINEERING
COURSE MATERIAL
IC6501 CONTROL SYSTEMS ENGINEERING
III YEAR - V SEMESTER
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IV– CONTROL SYSTEMS 2
SYLLABUS IC6501 CONTROL SYSTEMS ENGINEERING
L T P C 3 0 0 3 UNIT I: SYSTEMS AND THEIR REPRESENTATION [9 Hours] Basic Elements In Control Systems – Open And Closed Loop Systems – Electrical Analogy Of Mechanical And Thermal Systems – Transfer Function – Synchros – AC And DC Servomotors – Block Diagram Reduction Techniques – Signal Flow Graphs. UNIT II: TIME RESPONSE [9 Hours] Time Response – Time Domain Specifications – Types Of Test Input – I And II Order System Response – Error Coefficients – Generalized Error Series – Steady State Error – Root Locus Construction- Effects Of P, PI, PID Modes Of Feedback Control –Time Response Analysis. UNIT III : FREQUENCY RESPONSE [9 Hours] Frequency Response – Bode Plot – Polar Plot – Determination Of Closed Loop Response From Open Loop Response – Correlation Between Frequency Domain And Time Domain Specifications- Effect Of Lag, Lead And Lag-Lead Compensation On Frequency Response- Analysis. UNIT IV : STABILITY AND COMPENSATOR DESIGN [9 Hours] Characteristics Equation – Routh Hurwitz Criterion – Nyquist Stability Criterion- Performance Criteria – Lag, Lead And Lag-Lead Networks – Lag/Lead Compensator Design Using Bode Plots. UNIT V : STATE VARIABLE ANALYSIS [9 Hours] Concept Of State Variables – State Models For Linear And Time Invariant Systems – Solution Of State And Output Equation In Controllable Canonical Form – Concepts Of Controllability And Observability – Effect Of State Feedback. [TOTAL (L:45+T:15): 60 PERIODS]
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UNIT-I
SYSTEMS AND THEIR REPRESENTATION
Introduction to Control Systems The goal of control engineering is to design and build real physical systems to perform given
tasks. An engineer is asked to design an install a heat exchanger to control the temperature and
humidity of a large building. He determines the required capacity of the exchanger and then
proceed to install the system. After installation the exchanger is found to be insufficiently
powerful to control the buildings environment. He has to replace the unit and place a powerful
one. This is an example of empirical method.
Consider again the task of sending astronauts in moon and bringing them back safely. It cannot
be carried out by empirical method. In this case analytical method is indispensable. This method
consists of the following steps:
- modeling
- setting up mathematical equations
- analysis and design
Empirical methods may be expensive and dangerous. Analytical methods are simulated in
computers to see the result. If the design is satisfactory, the system is implemented using
physical devices.
A control system is an interconnection of components or devices so that the output of the overall
system will follow as closely as possible a desired signal. The reasons of designing control
systems include:
- Automatic control (e.g., control of room temperature)
- Remote control (e.g., antenna position control)
- Power amplification(e.g., control system will generate sufficient power to turn the
heavy antennas)
Control system components can be mechanical, electrical, hydraulic, pneumatic, thermal, or it
may be a computer program. It plays an important role in the development of modern
civilization. We use heating and air-conditioning in domestic domain for comfortable living. It
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 4
has found application in quality control of manufacturing products, machine-tool control,
weapon systems, power systems, robotics and many other places.
Classification of Control Systems
Control systems are basically classified as –
Open-loop control system
Closed-loop control system
In open-loop system the control action is independent of output. In closed-loop system control
action is somehow dependent on output. Each system has at least two things in common, a
controller and an actuator (final control element). The input to the controller is called reference
input. This signal represents the desired system output.
Open-loop control system is used for very simple applications where inputs are known ahed of
time and there is no disturbance. Here the output is sensitive to the changes in disturbance
inputs. Disturbance inputs are undesirable inputs that tend to deflect the plant outputs from their
desired values. They must be calibrated and adjusted at regular intervals to ensure proper
operation.
Closed-loop systems are also called feedback control systems. Feedback is the property of the
closed-loop systems which permits the output to be compared with the input of the system so that
appropriate control action may be formed as a function of inputs and outputs. Feedback systems
has the following features:
- reduced effect of nonlinearities and distortion
- Increased accuracy
- Increased bandwidth
- Less sensitivity to variation of system parameters
- Tendency towards oscillations
- Reduced effects of external disturbances
The general block diagram of a control system is shown below.
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Some Definitions
Reference input – It is the actual signal input to the control system.
Output (Controlled variable) – It is the actual response obtained from a control system.
Actuating error signal – It is the difference between the reference input and feedback signal.
Controller – It is a component required to generate control signal to drive the actuator.
Control signal – The signal obtained at the output of a controller is called control signal.
Actuator – It is a power device that produces input to the plant according to the control signal, so
that output signal approaches the reference input signal.
Plant – The combination of object to be controlled and the actuator is called the plant.
Feedback Element – It is the element that provides a mean for feeding back the output quantity
in order to compare it with the reference input.
Servomechanism – It is a feedback control system in which the output is mechanical position,
velocity, or acceleration.
Example of Control Systems Toilet tank filling system:
Figure: Toilet tank filling system Position control system: [antenna]
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Figure: Position control system Velocity control system: [audio/ video recorder]
Figure: Velocity control system
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Clothes Dryer:
Figure: Automatic dryer
Temperature control system: [oven, refrigerator, house]
Figure: Temperature control system Computer numerically controlled (CNC) machine tool:
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(a)
(b)
Figure: CNC machine tool control system
Control System Design In order to design and implement a control system, we need knowledge about the following
things:
Knowledge of desired value, (performance specification)
Knowledge of the output value, (feedback sensor, its resolution and dynamic
response)
Knowledge of controlling device,
Knowledge of actuating device,
Knowledge of the plant.
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With all of this knowledge and information available for the control system designer, he can start
the design steps shown below in the flow diagram.
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Mathematical Models of Physical Systems
We use mathematical models of physical systems to design and analyze control systems.
Mathematical models are described by ordinary differential equations. If the coefficients of the
describing differential equations are function of time, then the mathematical model is linear time-
varying. On the other hand, if the coefficients describing differential equations are constants, the
model is linear time-invariant.
The differential equations describing a LTI system can be reshaped into different forms
for the convenience of analysis. For transient response or frequency response analysis of single-
input-single-output linear systems, the transfer function representation is convenient. On the
other hand, when the system has multiple inputs and outputs, the vector-matrix notation may be
more convenient.
Powerful mathematical tools like Fourier and Laplace transforms are available for linear
systems. Unfortunately no physical system in nature is perfectly linear. Certain assumptions must
always be made to get a linear model. In the presence of strong nonlinearity or in presence of
distributive effects it is not possible to obtain linear models.
A commonly adopted approach is to build a simplified linear model by ignoring certain
nonlinearities and other physical properties that may be present in a system and thereby get an
approximate idea of the dynamic response of the system. A more complete model is then built
for more complete analysis.
Transfer function
The transfer function of an LTI system is the ratio of Laplace transform of the output variable to
the Laplace transform of the input variable assuming zero initial conditions. Following are some
examples of how transfer functions can be determined for some dynamic system elements.
[See chapter 3 of Ogata for details]
2
( ) 1
( ) 1o
i
E s
E s LCs RCs
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The insertion of an isolation amplifier between the two RC-circuits will produce no loading
effect.
A. Transfer Function of Armature Controlled DC Motor
In Laplace domain,
2
0
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
b b
a a a b
T a
E s K s s
L s R I s E s E s
Js f s s K I s
; 0
( ) ( )
( ) ( )( )T
a a T b
KsG s
E s s R sL Js f K K
Neglecting aL , 2
0
/ /( )
( / ) ( ) ( 1)T a T a m
T b a m
K R K R KG s
Js s f K K R s Js f s s ;
where, 0 /T b af f K K R and / ; / .m T a mK K R f J f
mK and m are called the motor gain and time constant respectively. These two parameters are
usually supplied by the manufacturer.The block diagram model is,
21 1 2 2 1 1 2 2 1 2
( ) 1
( ) ( ) 1o
i
E s
E s RC R C s RC R C RC s
1;f f M f f a T aK i T K K i i K i
; ab b a a a b
dide K L R i e e
dt dt
2
02 M T a
d dJ f T K i
dt dt
M.I.E.T ENGG COLLEGE DEPT OF EEE
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B. Transfer Function of a Field-controlled DC Motor
'1 1
2
2
TM a f f a f
ff f f
M T f
T K i K K i i K i
diL R i e
dt
d dJ f T K i
dt dt
; 2
( ) ( ) ( )
( ) ( ) ( ) ( )
f f f
M T f
L s R I s E s
Js fs s T s K I s
We obtain,
Mechanical System
Translational system and rotational system
( )( )
( ) ( )( )T
f f
KsG s
E s s L s R Js f
2
2
1 2 1 2 2 1( ); ( )
m
d s
du d xu ma m m
dt dtu k u u u k x x
2
2
1 1 2 2 1 2
;
( ); ( )
J
d s
d dT J J
dt dtT k T k
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 13
Mechanical Accelerometer
For a constant input acceleration y becomes constant.
Then, x ky . Taking Laplace transform of the previous equation,
2( ) ( ) ( )X s s c s k Y s ;
( ) ( )2
1X s Y s
s c s k
Rotary Potentiometer
2
1 22
( ) ( )( )
d y t dy tm k k y t u
dt dt
maxmax
max max
( )( ) ( )
or, ( ) ( )
o
o p
VtV t V t
V t k t
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IV– CONTROL SYSTEMS 14
Example: Draw the block diagram of the following system.
Example: Control of flaps in airplane
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Example: Figure below shows a reduction gearbox being driven by a motor that develops a torque( )mT t . It
has a gear reduction ratio of ‘n = b / a’. Find a differential equation relating the motor
torque ( )mT t and the output angular position( )o t .
2
2( ) ( ) m m
a m m m
d dT t T t I C
dt dt
; 2
02( ) o o
b o
d dT t I C
dt dt
; ( ) 1
( )a
b
T t a
T t b n ; m
o
bn
a
From above, 2 2
02 2( ) o o o o
m m m o
d d d dn T t nI nC I C
dt dt dt dt
; N.B. 2 2
2 2;m o m od d d d
n ndt dt dt dt
Gearbox parameters:
, , From above, 2 2( ) ( ) ( )o m o o m o mI n I C n C nT t
Inputting parameters, 0.0225 0.3 50 ( )o o mT t [Ans]
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 16
Example: Gear train and its equivalence
max
( ) ( ) ( )
/p r o
p batt
V s k s s
k V
1 1 1 1motor eq eqT J f
Tachometers
Error detector
( ) ( ) ( )t tV s k s k s s
max
( ) ( ) ( )
/p r o
p batt
V s k s s
k V
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IV– CONTROL SYSTEMS 17
Example
Example
Sketch the analogous electrical circuit of the following mechanical system.
Example
Draw the analogous electric circuit of the system below using f-i analogy.
Error detector using op amp e = r - vw
Obtain the transfer function X(s)/E(s) for the electromechanical system shown left assuming
that the coil has a back emf 1b
dxe K
dt and the
coil current i2 produces a force 2 2cF k i on the
mass M.
24 3
21 2 1 2
( )
( ) ( ) (2
) ( 2 ) 2
KX s
E s RLCMs L M RCB s RC LK
k k s RB LK k k s RK
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[Ans] Mechanical Systems: There are two Types
1. Mechanical Translation system. 2. Mechanical rotational system.
The basic elements used in Mechanical translation system and Mechanical Rotational Systems are
Mechanical Translational system
Mechanical Rotational system
1. The weight of mechanical translational system is represented by the element Mass(M)
1. The weight of the mechanical rotational system is represented by the moment of inertia of the mass (J)
2. The elastic deformation of the body can be represented by a Spring ( K)
2. The elastic deformation of the body can be represnted by a Torsional spring with stiffness (K).
3. The frictional existing in rotating mechanical system can be represented by Dash pot (B)
3. The friction existing in the mechanical rotational system can be represented by the Dash pot (B)
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Symbols used in the mechanical system are:
Mechanical Translational Mechanical rotational
1. x = Displacement(m) 1. θ – Angular displacement (rad)
2. V = dx / dt (m/sec) 2. dθ / dt – Angular velocity.
3. a = dv / dt (m/s) 3. dω / dt – Angular acceleration
4. f = Applied force ( N) 4. T - Applied Torque (N- m)
5. fm = Opposing force offered by mass of the body (N)
5. Tj – Opposing torque due to moment of inertia.
6. fk = Opposing force offered by the elasticity of the body (N)
6. Tk – Opposing Torque due to elasticity.
7. fb = Opposing force offered by friction of the body (N)
7. Tb – Opposing torque due to friction.
8. M= Mass (Kg) 8. J = Moment of inertia (kg-m/rad)
9. K = Stiffness of Spring( N /m) 9. K - Stiffness of the spring (N-m /
rad)
10. B= Viscous friction coefficient 10. B - rotational frictional
coefficient(N-m/(rad/sec))
Balance equation of idealized elements:
Force Balance Equation Torque Balance Equation 1. An ideal mass element has
negligible friction and elasticity. A force applied on it. The opposing force offered by the mass is proportional to the acceleration.
2
2
dt
tdfm X
2
2
dt
XdMfm f
By Newton’s 2nd Law
1. An ideal mass element has negligible friction and elasticity. A force applied on it. The opposing force offered by the mass is proportional to the angular acceleration.
2
2
dt
dTj
2
2
dt
dJTj
T θ
By Newton’s 2nd Law 2. An ideal frictional element dashpot
which has negligible mass and elasticity. A force applied on it. The dash pot will offered the opposing
2. Negligible moment of inertia and elasticity.
dt
dTb
B
M J
M.I.E.T ENGG COLLEGE DEPT OF EEE
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force which is proportional to the velocity.
X f B
dt
dxfb
dt
dxBfb
By Newton’s 2nd Law
dt
dBTb
T θ
21 dt
dTb
21 dt
dBTb T1 θ1 θ2
3. Consider an Ideal elastic spring which has negligible mass and friction
xfk f
X1 X2
kxf k f
If the displacement at both the ends means 2121 XXKfXXf kk
3.
T θ k kT
kTk k
T θ1 θ2
21 kT
TkTk 21
Problems In Mechanical Translational System
1. Write the differential equation governing the mechanical system shown in the fig. And determine the transfer function.
X1 X k1 B
f(t) B1 k B2 Solution:
Transfer function = )(
)(
sF
sX
Step 1: Free Body diagram for M1
M1
M2
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The opposing force acting on Mass M1 are fm1, fb, fk1, fb and fk
w.k.t. 2
12
11 dt
Xdmfm
dt
dXBfb
111
111 Xkf k
dt
XXdBfb
1
XXkf k 1
By Newton’s 2nd Law:
Applied force = Opposing Force
0111 kkbbm fffff - - - - - (1)
01
111
112
12
1 XXKdt
XXdBXK
dt
dXB
dt
XdM - - - - - (2)
Taking Laplace transform for equation (2)
0))()(()())()(()()( 11111112
1 sXsXKsXKsXsXBSsSXBsXM
0))()()())()()()( 11111112
1 sKXsKXsXKsBsXsBsXssXBsXsM
])[()]()()[( 112
11 KBssXKKsBBsMsX
)()(
)()()(
112
11 KKsBBsM
KBssXsX
- - - - - (3)
Step 2:
Free Body Diagram for M2
The opposing force acting on Mass M2 are fm2, fb, fk, fb2
w.k.t. 2
2
22 dt
XdMfm
)( 1XXdt
dBfb
)( 1XXKfk
dt
dXBfb 22
By Newton’s 2nd Law:
M.I.E.T ENGG COLLEGE DEPT OF EEE
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)(22 tfffff bkbm
)(2112
2
2 tfdt
dxBXXKXX
dt
dB
dt
XdM - - - - - (4)
Taking Laplace transform:
)()(())()(())()(()( 2112
2 sFsXsBsXsXKsXsXBssXsM
Subs X1(s) value in the above equation:
)()()(
)()(])()[(
112
1
2
22
2 sFKKsBBsM
KBssXKsBBsMsX
)()()(
)()()(][)([)(
112
1
112
122
2 sFKKsBBsM
KBsKKsBBsMKsBBsMsX
2
22
2112
1
112
1
)(])()][()([
)()(
)(
)(
kBsKsBBsMKKsBBsM
KKsBBsM
sF
sX
Result:
The differential equations governing the system are
1. 01
111
112
12
1 XXKdt
XXdBXK
dt
dXB
dt
XdM
2. )(2112
2
2 tfdt
dxBXXKXX
dt
dB
dt
XdM
Transfer function of the system is
1. 2
22
2112
1
112
1
)(])()][()([
)()(
)(
)(
kBsKsBBsMKKsBBsM
KKsBBsM
sF
sX
Block diagram reduction technique
Block diagram:
A block diagram is a short hand, pictorial representation of cause and effect
relationship between the input and the output of a physical system. It characterizes the functional
relationship amongst the components of a control system.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 23
Lower case letters usually represent functions of time. Upper case letters denote
Laplace transformed quantities as a function of complex variable sor Fourier transformed
quantity i.e. frequency function as function of imaginary variablejωs .
Summing point: It represents an operation of addition and / or subtraction.
Negative feedback: Summing point is a subtractor.
Positive feedback: Summing point is an adder.
Stimulus: It is an externally introduced input signal affecting the controlled output.
Take off point: In order to employ the same signal or variable as an input to more than block or
summing point, take off point is used. This permits the signal to proceed unaltered along several
different paths to several destinations.
.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 24
1.Find the transfer function of the closed-loop system below.
To simplify the inner feedback loop to obtain the following block diagram.
To combine the two series blocks into one.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 25
To obtain the transfer function for the standard feedback system.
2.Find the transfer function of the closed-loop system below.
To simplify the inner feedback blocks.
M.I.E.T ENGG COLLEGE DEPT OF EEE
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To get the following block diagram.
To combine the two sets of series blocks.
Calculate the overall transfer function of the system
SIGNAL FLOW GRAPHS
An alternate to block diagram is the signal flow graph due to S. J. Mason. A signal flow graph is
a diagram that represents a set of simultaneous linear algebraic equations. Each signal flow graph
consists of a network in which nodes are connected by directed branches. Each node represents a
system variable, and each branch acts as a signal multiplier. The signal flows in the direction
indicated by the arrow.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 27
Definitions:
1) Node: A node is a point representing a variable or signal.
2) Branch: A branch is a directed line segment joining two nodes.
3) Transmittance: It is the gain between two nodes.
4) Input node: A node that has only outgoing branche(s). It is also, called as source and
corresponds to independent variable.
5) Output node: A node that has only incoming branches. This is also called as sink and
corresponds to dependent variable.
6) Mixed node: A node that has incoming and out going branches.
7) Path: A path is a traversal of connected branches in the direction of branch arrow.
8) Loop: A loop is a closed path.
9) Self loop: It is a feedback loop consisting of single branch.
10) Loop gain: The loop gain is the product of branch transmittances of the loop.
11) Nontouching loops: Loops that do not posses a common node.
12) Forward path: A path from source to sink without traversing an node more than once.
13) Feedback path: A path which originates and terminates at the same node.
14) Forward path gain: Product of branch transmittances of a forward path.
M ASON ’S GAIN FORMULA: The relation between the i/p variable & the o/p variable of a signal flow graphs
is given by the net gain between the i/p & the o/p nodes and is known as Overall gain of the
system. Mason’s gain form ula for the determ ination of overall system gain is given by,
Properties of Signal Flow Graphs:
1) Signal flow applies only to linear systems.
2) The equations based on which a signal flow graph is drawn must be algebraic equations
in the form of effects as a function of causes.
Nodes are used to represent variables. Normally the nodes are arranged left to right,
following a succession of causes and effects through the system.
3) Signals travel along the branches only in the direction described by the arrows of the
branches.
M.I.E.T ENGG COLLEGE DEPT OF EEE
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4) The branch directing from node Xk to Xj represents dependence of the variable Xj on Xk
but not the reverse.
5) The signal traveling along the branch Xk and Xj is multiplied by branch gain akj and
signal akjXk is delivered at node Xj.
Guidelines to Construct the Signal Flow Graphs: The signal flow graph of a system is constructed from its describing equations, or by direct
reference to block diagram of the system. Each variable of the block diagram becomes a node
and each block becomes a branch. The general procedure is
1) Arrange the input to output nodes from left to right.
2) Connect the nodes by appropriate branches.
3) If the desired output node has outgoing branches, add a dummy node and a unity gain
branch.
4) Rearrange the nodes and/or loops in the graph to achieve pictorial clarity.
Signal Flow Graph Algebra
Addtion rule
The value of the variable designated by a node is equal to the sum of all signals entering the
node.
Transmission rule
The value of the variable designated by a node is transmitted on every branch leaving the node.
Multiplication rule
A cascaded connection of n-1 branches with transmission functions can be replaced by a single
branch with new transmission function equal to the product of the old ones.
Masons Gain Formula
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The relationship between an input variable and an output variable of a signal flow graph is given
by the net gain between input and output nodes and is known as overall gain of the system.
Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs.
Gain P is given by
k
kkPP1
Where, Pk is gain of kth forward path,
∆ is determinant of graph
∆=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations of
two nontouching loops – sum of gain products of all possible combination of three
nontouching loops) + ∙∙∙
∆k is cofactor of kth forward path determinant of graph with loops touching kth forward path. It is
obtained from ∆ by removing the loops touching the path Pk.
Example 1 Obtain the transfer function of C/R of the system whose signal flow graph is shown in Fig.1
Figure 1 Signal flow graph of example 1
There are two forward paths:
Gain of path 1 : P1=G1
R
G4
G2
-G3
G1
C 1 1
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Gain of path 2 : P2=G2
There are four loops with loop gains:
L1=-G1G3, L2=G1G4, L3= -G2G3, L4= G2G4
There are no non-touching loops.
∆ = 1+G1G3-G1G4+G2G3-G2G4
Forward paths 1 and 2 touch all the loops. Therefore, ∆1= 1, ∆2= 1
The transfer function T = 42324131
212211
1 GGGGGGGG
GGPP
sR
sC
Example 2 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.2.
Figure 2 Signal flow graph of example 2
There is one forward path, whose gain is: P1=G1G2G3
There are three loops with loop gains:
L1=-G1G2H1, L2=G2G3H2, L3= -G1G2G3
There are no non-touching loops.
∆ = 1-G1G2H1+G2G3H2+G1G2G3
Forward path 1 touches all the loops. Therefore, ∆1= 1.
R(s) C(s) 1 1 1 G1 G2 G3
H1
-1
-H2
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 31
The transfer function T = 321231121
32111
1 GGGHGGHGG
GGGP
sR
sC
Example 3 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.3.
Figure 3 Signal flow graph of example 3
There are three forward paths.
The gain of the forward path are: P1=G1G2G3G4G5
P2=G1G6G4G5
P3= G1G2G7
There are four loops with loop gains:
L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2
There is one combination of Loops L1 and L2 which are nontouching with loop gain product
L1L2=G2G7H2G4H1
∆ = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1
Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.
Forward path 3 is not in touch with loop1. Hence, ∆3= 1+G4H1.
The transfer function T =
2174225432254627214
14721654154321332211
1
1
HHGGGHGGGGHGGGHGGHG
HGGGGGGGGGGGGGPPP
sR
sC
G1 C(s) R(s)
G7 G6
-H1
G2 G3 G4 G5
-H2
X1 X2 X3 X4 X5
1
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 32
Example 4
Find the gains 1
3
2
5
1
6 ,,X
X
X
X
X
X for the signal flow graph shown in Fig.4.
Figure 4 Signal flow graph of MIMO system
Case 1: 1
6
X
X
There are two forward paths.
The gain of the forward path are: P1=acdef
P2=abef
There are four loops with loop gains:
L1=-cg, L2=-eh, L3= -cdei, L4=-bei
There is one combination of Loops L1 and L2 which are nontouching with loop gain product
L1L2=cgeh
∆ = 1+cg+eh+cdei+bei+cgeh
Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.
The transfer function T = cgehbeicdeiehcg
abefcdefPP
X
X
1
2211
1
6
b
a d c f e
-h
-g
-i
X1 X6 X5
X4 X3 X2
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 33
Case 2: 2
5
X
X
The modified signal flow graph for case 2 is shown in Fig.5.
Figure 5 Signal flow graph of example 4 case 2
The transfer function can directly manipulated from case 1 as branches a and f are removed
which do not form the loops. Hence,
The transfer function T=cgehbeicdeiehcg
becdePP
X
X
1
2211
2
5
Case 3: 1
3
X
X
The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in
Fig.6.
b
1 d c 1 e
-h
-g
-i
X2 X5 X5
X4 X3 X2
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 34
Figure 6 Signal flow graph of example 4 case 3
There are two forward paths.
The gain of the forward path are: P1=abcd
P2=ac
There are five loops with loop gains:
L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg
There is one combination of Loops L1 and L2 which are nontouching with loop gain product
L1L2=ehcg
∆ = 1+eh+cg+bei+efd+befg+ehcg
Forward path 1 touches all the five loops. Therefore ∆1= 1.
Forward path 2 does not touch loop L1. Hence, ∆2= 1+ eh
The transfer function T =
ehcgbefgefdbeicgeh
ehacabefPP
X
X
1
12211
1
3
Example1 Draw the signal flow graph of the block diagram shown in Fig.1
G2 G1 G3
H2
H1
− −
R X1 X2 X3 X4 X5 X6 C
a e b f
-h
-g
-i
X1 X5
X4 X3
X2
c
d
X3 1
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 35
Figure 7 Multiple loop system Choose the nodes to represent the variables say X1 .. X6 as shown in the block diagram..
Connect the nodes with appropriate gain along the branch. The signal flow graph is shown in
Fig. 2
Figure 8 Signal flow graph of the system shown in Fig. 1
Example 2 Draw the signal flow graph of the block diagram shown in Fig.3.
R X1 X2 X3
X4 X5 X6
C G1
H1
-H2
G2 G3
-1
1 1 1 1
G1
G2
G3
G4
R
−
C
−
X1 X2
X3
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 36
Figure 9 Block diagram feedback system The nodal variables are X1, X2, X3. The signal flow graph is shown in Fig. 4.
Figure 10 Signal flow graph of example 2
Example 3 Draw the signal flow graph of the system of equations.
3332321313
223232221212
113132121111
XaXaXaX
ubXaXaXaX
ubXaXaXaX
The variables are X1, X2, X3, u1 and u2 choose five nodes representing the variables. Connect the various nodes choosing appropriate branch gain in accordance with the equations. The signal flow graph is shown in Fig. 5.
Figure 11 Signal flow graph of example 2
R
G4
G2
-G3
G1
C 1 1
X1
X2 X3
u1 b1
b2
a21
a12
a33
a31
X1
X2 a11
a13
a32
X3
u2
a23
a22
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 37
Synchros
Figure: Synchro error detector
( ) sinr r cv t V t ; 1
2
3
( ) sin cos( 120 )
( ) sin cos
( ) sin cos( 240 )
s r c
s r c
s r c
v t KV t
v t KV t
v t KV t
;
1 2
2 3
3 1
( ) 3 sin sin( 240 )
( ) 3 sin sin( 120 )
( ) 3 sin sin
s s r c
s s r c
s s r c
v t KV t
v t KV t
v t KV t
When 0 , 3 1( ) 0s sv t and maximum voltage is induced on S2 coil. This position of the rotor is
defined as the “electrical zero” of the transmitter and used as reference position of the rotor.
The output of the synchro transmitter is applied to the stator winding of a “synchro control
transformer”. Circulating current of the same phase but of different magnitude flows through the
two sets of stator coil. The pair acts as an error detector.
The voltage induced in the control transformer rotor is proportional to the cosine of the angle
between the two rotors and is given by,
( ) sin cos ; where, 90r ce t K V t .
( ) sin cos(90 ) sin sin( ) ( )sinr c r c r ce t K V t K V t K V t …(01)
The equation above holds for small angular displacement.
A synchro is an electromagnetic transducer that is used to convert angular shaft position into an electric signal. The basic element of a synchro is a synchro transmitter whose construction is very similar to that of the 3- alternator. An ac voltage is applied to the rotor winding through slip-rings. The schematic diagram of synchro synchro transmitter- control transformer pair is shown above.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 38
Thus the synchro transmitter-control transformer pair acts as an error detector which gives a
voltage signal at the rotor terminal of the control transformer proportional to the angular
difference between the shaft positions.
Equation (01) is represented graphically in figure below for an arbitrary time variation of
( ).
It is seen that the output of the synchro error detector is a modulated signal, where the ac signal
applied to the rotor of synchro transmitter acts as carrier and the modulating signal is,
( ) ( ); m s s re t K K K V
A.C. Servomotor
For low power application a.c. motors are preferred, because of their light weight, ruggedness
and no brush contact. Two phase induction motors are mostly used in control system.
The motor has two stator windings displaced 90 electrical degree apart. The voltages applied to
the windings are not balanced. Under normal operating conditions a fixed voltage from a
constant voltage source is applied to one phase. The other phase, called control phase, is
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 39
energized by a voltage of variable magnitude which is 90 out of phase w.r.t. the voltage of fixed
phase.
The torque speed characteristic of the motor is different from conventional motor. X / R ratio is
low and the curve has negative slope for stabilization.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 40
The torque-speed curve is not linear. But we assume it as linear for the derivation of transfer
function. The troque is afunction of both speed and the r.m.s. control voltage, ie., ( , )MT f E .
Using Tailor series expansion about the normal operating point0 0 0( , , )MT E we get,
0 0
0 0
0 0 0( ) ( )M MM M
E E E E
T TT T E E
E
For position control system, 0 0E , 0 0 , 0MT
Thus, the above equation may be simplified as,
0 ;MT kE m J f where, 0
0
M
E E
Tk
E
and 0
0
M
E E
Tm
.
Performing Laplace transform, 20( ) ( ) ( ) ( )kE s ms s Js s f s
Or, 2
0
( )( )
( ) ( ) ( 1)m
m
Ks kG s
E s Js f m s s s
; where,
0m
kK
f m and
0m
J
f m .
Since ‘m’ is negative the transient part is decaying as m is positive. If ‘m’ would positive and
0m f the transient part will increase with time and the system would be unstable. k and m are
the slope of the torque-voltage and torque-speed curve.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 41
UNIT-II
TIME RESPONSE
Time Response:
The time response of the system is the output of the closed loop system as a function of
time. It is denoted by c(t).
It is obtained by solving the differential equation governing the system
Otherwise, it can be obtained from the transfer function of the system and input to the
system.
The closed loop transfer function,)()(1
)(
)(
)(
sHsG
sG
sR
sC
)()(
)(sM
sR
sC
)()()( sMsRsC
From the above equation the output (or) response in S-domain is obtained by product of
the transfer function and the input R(s).
Response in S – domain = C(s)
Response in t – domain = c(t)
Parts in Time Response:
It consists of two parts. They are
1. Transient Response
2. Steady State Response.
1. Transient Response:
It is the response of the system when the input changes from one state to another
2. Steady State Response:
It is the response as time t approaches infinity.
Test Signals:
Input Signals:
It is required to predict the response of a system.
Characteristics of Input Signals:
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 42
1. A sudden shock
2. A sudden Change
3. A constant velocity and
4. A constant acceleration
Need of Test Signals:
The test signals which resemble these characteristics are used as input signals to predict
the performance of the system.
Standard test signals are
1. i) Step Signal r(t) = 1; t ≥ 0
ii) Unit step Signal = 0; t < 0
r(t)
A
t
2. i) Ramp Signal r(t) = At; t ≥ 0
ii) Unit Ramp Signal = 0; t < 0
r(t)
A
1 2 t
3. i) Parabolic Signal r(t) = At2 / 2; t ≥ 0
ii) Unit Parabolic Signal = 0; t < 0
r(t)
2A
0.5
1 2 t
4. Impulse Signal δ (t) = 0; t ≠ 0
5. Sinusoidal Signal
Impulse Signal:
A signal of very large magnitude which is available for very short duration is called
impulse signal
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 43
It is denoted by δ (t) r(t) = δ (t) r(t) = δ (t)
δ (t) = ∞; t = 0 and Adtt)(
= 0; t ≠ 0
Note: 0 t 0 YΔ t
i) Impulse signal is derivative of a step signal
ii) Laplace transform of the Impulse function is Unity.
Weighting Function:
The response of the system, with input as impulse signal is called weighting function (or
impulse response ) of the system.
Impulse response, m(t) = L-1 {R(s), M(s)}
= L-1 {M(s)} where R(s) = 1
The response for any input can be obtained by convolution of input with impulse
response.
Name of the Signal Time domain r(t) Laplace Transform
Step A A / s
Unit Step 1 1 / s
Ramp At A / s2
Unit Ramp t 1 / s2
Parabolic At2 / 2 A / s3
Unit Parabolic t2 / 2 1 / s3
Impulse δ (t) 1
Note:
The zeroes and poles are critical value of s, at which the function T(s) attains extreme
values 0 or ∞.
1. When s takes the value of a zero, the function T(s) will be zero. [T(s)=0]
2. When s takes the value of a pole, the function T(s) will be ∞. [T(s) = ∞]
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 44
Response of First Order System:
Unit Step Input:
The closed loop order system with unity feedback is shown in below figure.
The closed loop transfer function of first order system, TssR
sC
1
1
)(
)(
If the input is Unit step r(t) = 1; R(s) = 1/s.
Ts
sRsC 1
1)()(
sT
sTTss 1
1
1
11
sT
s
TsC1
1
)(
Ts
B
s
A
Tss
TsC11
1
)(
BsTsAT
11
(i) Put s = 0: 001 T
AT
A = 1
(ii) Put S = -1 / T: TB
T1
B = -1
The response in s – domain is given by
TsssC
111
)(
The response in time domain is
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 45
TssLscLtc
111
)}({)( 11
Tt
etc1)(
The above equations is the response of the closed loop first order system for unit step
Input
1. For Closed Loop first order system,
Unit Step response = T
te1
Step Response = A( Tt
e1 )
When t = 0, c(t) = 1 – e0 = 1
When t = 1T, c(t) = 1 – e-1 = 0.632
When t = 2T, c(t) = 1 – e-2 = 0.865
When t = 3T, c(t) = 1 – e-3 = 0.95
When t = 4T, c(t) = 1 –e-4 = 0.9817
When t = 5T, c(t) = 1 – e-5 = 0.993
When t = ∞, c(t) = 1 – e-∞ = 1
T – is the time constant of the system
In a time of 5T, the system is assumed to have attained the steady state/
C(t)
r(t) 0.95
0.865
0.632
t=0 t 0 T 2T 3T t
Unit step Input
Second Order System:
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 46
ss
ss
ss
ss
ss
sR
sC
n
nn
n
n
n
n
n
n
2
2
2
21
2
)(
)(
2
22
2
2
2
2
2
2
22
2
2)(
)(
nn
n
sssR
sC
Where ωn = Undamped natural frequency in radians per second
ζ = Damping ratio
Damping Ratio:
Definition:
It is defined as the ratio of the actual damping to the critical damping
The response c(t) of second order system depends on the value of damping ratio.
The second order system can be classified into four types depending upon the value of
damping ratio
1. Undamped system ζ = 0
2. Under damped system 0 < ζ < 1
3. Critically damped system ζ = 1
4. Over damped system ζ > 1
The characteristics equation of the second order system is 22 2 nnss
It is a quadratic equation and the roots of this equation is given by
2
442,
222
21nnnss
2
)1(42 22 nn
)1( 2 nn
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 47
i) When ζ = 0, s1, s2 = ±jωn; ׀ roots are purely imaginary and the system
is under damped
ii) When ζ = 1, s1, s2 = -ωn ׀ roots are real and equal and the system is
critically damped
iii) When ζ > 1, s1, s2 = 12 nn roots are real and unequal and the system is over ׀
damped
iv) When 0 < ζ < 1, s1, s2 = 12 nn
)1(1 2 nn
21 nn j
dn j
Here ωd is called damping frequency of Oscillation
Response of Undamped Second Order system:
For Unit Step Input:
22
2
2)(
)(...
nn
n
sssR
sCtkw
For Undamped system ζ = 0
22
2
)(
)(
n
n
ssR
sC
When the input is unit step, r(t) = 1 and R(s) = 1/s
The response in s – domain 22
2
)()(n
n
ssRsC
22
21
n
n
ss
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 48
)( 22
2
n
n
ss
2222
2
)( nn
n
s
B
s
A
ss
BssA nn )( 222
i) Put s = 0
22nn A
A = 1
ii) Put s2 = - 2n s = j n nn Bj 2
Bj n
n 2
B = -jωn
B = -s
2222
)(1)(
nn s
s
ss
B
s
AsC
2222)(cos
1
nn s
stL
s
s
s
22
11 1)}({
ns
s
sLsCL
c(t) = 1 - cosωnt
r(t) c(t)
2
1 1
Input (Unit Step) output (Response) t
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 49
Every practical system has same amount of damping. Hence undamped system does not
exist in practice.
The response of the undamped closed loop second order system is
For Unit step Response = 1 - cosωnt
Step Response = A (1 - cosωnt)
The response is completely oscillatory.
Response of Under Damped Second Order System:
For Unit Step Input:
The standard form of closed loop transfer function of second order system is
22
2
2)(
)(
nn
n
sssR
sC
For under damped system, 0 < ζ < 1 and root of the denominator are complex conjugate.
The roots of the denominator are
12 nns
Since ζ < 1, ζ2 is also less than 1
Therefore, the damped frequency of oscillation, 21 nd
dn js
The response in s – domain, 22
2
2)()(
nn
n
sssRsC
For unit step input, r(t) = 1 and R(s) = 1/s
)2()(
22
2
nn
n
ssssC
)2()2()(
2222
2
nnnn
n
ss
CBs
s
A
ssssC
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 50
scBsssA nnn )()2( 222
1. Put s = 0
22nn A
A = 1
2. Equating the co-efficient of s2.
CsBsssA nnn 2222 )2(
0 = A + B
0 = 1 + B
B = -1
3. Equating co-efficient of s
0 = 2Aωnζ + C
= 2ωnζ + C
C = - 2ωnζ
22 2
21)(
nn
n
ss
s
ssC
Let us add and subtract 22n to denominator of the second term
222222 2
21)(
nnnn
n
ss
s
ssC
222222 2
21
nnnn
n
ss
s
s
2222)(
21
nnn
n
s
s
s
)1()(
21222
nn
n
s
s
s
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 51
22)(
21
dn
n
s
s
s
22 )()(
1
dn
n
dn
n
ss
s
s
Multiply and divide by ωd
2222 )()(
1)(
dn
d
d
n
dn
n
ss
s
ssC
The response in time domain is given by
222211
)()(
1)}({)(
dn
d
d
n
dn
n
ss
s
sLsCLtc
tete dt
d
nd
t nn sincos1
tte dd
nd
tn sincos1
tte d
n
nd
tn sin1
cos12
tte ddtn
sin1
cos12
tte
tc dd
tn
sincos11
1)( 2
2
Note:
On constructing on right angle triangle with ζ and 21 , we get
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 52
1
21
ζ
sin θ = 21
cos θ = ζ
2-1
tan
tt
etc dd
tn
sincoscossin1
1)(2
)sin(
11)(
2
t
etc d
tn
2
1 1tan
For closed loop under damped second order system:
Unit step response =
)sin(1 2
te
d
tn
Step Response =
)sin(1 2
te
A d
tn
r (t) c(t)
1 1
Input t Output Response t
θ
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 53
The response of the under damped second order system for unit step input sketched and
observed that the response oscillates before settling to a final value. The oscillation depends on
the value of damping ratio.
Response of Critically Damped second Order System for Step Input:
The standard form of closed loop transfer function of second order system is,
22
2
2)(
)(
nn
n
sssR
sC
For critically damping ζ = 1
22
2
2)(
)(
nn
n
sssR
sC
2
2
)( n
n
s
When input is unit step, r(t) = 1 and R(s) = 1/s
2
2
)()()(
n
n
ssRsC
2
2
)(
1)(
n
n
sssC
2
2
2 )()()()(
n
n
nn sss
C
s
B
s
AsC
ssCBssA nnn )()( 22
i) Put s = 0
022 nn A
ii) Put s = -ωn
))(()( 22nnnnnnn CBA
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 54
nn B 2
nB
iii) Equating the co-efficient of s2
0 = A + C
= 1 + C
C = -1
)(
1
)(
1
)()()(
22nn
n
nn ssss
C
s
B
s
AsC
The response in time domain,
nn
n
sssLscLtc
1
)(
1)}({)(
211
tt
nnn etetc 1)(
]1[1)( tetc ntn
For closed loop critically damped second order system,
Unit Step Response = ]1[1 te ntn
Step Response = ]1[1 teA ntn
r (t) c(t)
1
1
Input t Response t
Time response (Transient ) Specification (Time domain) Performance :-
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 55
The performance characteristics of a controlled system are specified in terms of the
transient response to a unit step i/p since it is easy to generate & is sufficiently drastic.
The transient response of a practical C.S often exhibits damped oscillations before
reaching steady state. In specifying the transient response characteristic of a C.S to unit step
i/p, it is common to specify the following terms.
1) Delay time (td)
2) Rise time (tr)
3) Peak time (tp)
4) Max over shoot (Mp)
5) Settling time (ts)
Response curve
1) Delay time :- (td)
It is the time required for the response to reach 50% of its final value for the 1st
time.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 56
2) Rise time :- (tr)
It is the time required for the response to rise from 10% and 90% or 0% to 100% of
its final value. For under damped system, second order system the 0 to 100% rise time is
commonly used. For over damped system, the 10 to 90% rise time is commonly used.
3) Peak time :- (tp)
It is the time required for the response to reach the 1st of peak of the overshoot.
4) Maximum over shoot :- (MP)
It is the maximum peak value of the response curve measured from unity. The amount
of max over shoot directly indicates the relative stability of the system.
5) Settling time :- (ts)
It is the time required for the response curve to reach & stay with in a range about the
final value of size specified by absolute percentage of the final value (usually 5% to 2%).
The settling time is related to the largest time const., of C.S.
Transient response specifications of second order system :-
W. K.T. for the second order system,
T.F. = C(S) = Wn2 . ------------------------------(1)
R(S) S2+2 WnS+ Wn2
Assuming the system is to be under damped (< 1)
Rise time tr
W. K.T. C (tr) = 1- e-Wnt Cos Wdtr + . Sin wdtr
1-2
Let C(tr) = 1, i.e., substituting tr for t in the above Eq:
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 57
Then, C(tr) = 1 = 1- e-Wntr Cos wdtr + . sin wdtr
1-2
Cos wdtr + . sin wdtr = tan wdtr = - 1-2 = wd .
1-2
jW
Thus, the rise time tr is , jWd
tr = 1 . tan-1 - w d = - secs Wn 1-2 Wn
Wd wd
When must be in radians. -
Wn
Peak time :- (tp)
Peak time can be obtained by differentiating C(t) W.r.t. t and equating that
derivative to zero.
dc = O = Sin Wdtp Wn . e-Wntp
dt t = t p 1-2
Since the peak time corresponds to the 1st peak over shoot.
Wdtp = = tp = .
Wd
The peak time tp corresponds to one half cycle of the frequency of damped oscillation.
Maximum overshoot :- (MP)
The max over shoot occurs at the peak time.
S- Plane
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 58
i.e. At t = tp = .
Wd
Mp = e –(/ Wd) or e –( / 1-2)
Settling time :- (ts)
An approximate value of ts can be obtained for the system O < <1 by using the
envelope of the damped sinusoidal waveform.
Time constant of a system = T = 1 .
Wn
Setting time ts = 4x Time constant.
= 4x 1 . for a tolerance band of +/- 2% steady state.
Wn
Delay time :- (td)
The easier way to find the delay time is to plot Wn td VS . Then approximate
the curve for the range O<< 1 , then the Eq. becomes,
Wn td = 1+0.7
td = 1+0.7 Wn
PROBLEMS:
(1) Consider the 2nd order control system, where = 0.6 & Wn = 5 rad / sec, obtain the
rise time tr, peak time tp, max overshoot Mp and settling time ts When the system is subject
to a unit step i/p.,
Given :- = 0.6, Wn = 5rad /sec, tr = ?, tp = ?, Mp = ?, ts = ?
Wd = Wn 1- 2 = 5 1-(0.6) 2 = 4
= Wn = 0.6 x 5 = 3.
tr = -, = tan-1 Wd = tan-1 4 = 0.927 rad
Wd 3
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 59
tr = 3.14 – 0.927 = 0.55sec.
4
tp = = 3.14 = 0.785 sec.
Wd 4
. = e / Wd
MP = e 1- 2
MP = e (3/4) x 3.14 = 0.094 x 100 = 9.4%
ts :- For the 2% criteria.,
ts = 4 . = 4 . = 1.33 sec.
Wn 0.6x5
For the 5% criteria.,
ts = 3 = 3 = 1 sec
3
3) The open loop T.F. of a unity feed back system is given by G(S) = K . where,
S(1+ST)
T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so
that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25%
(b) The damping ratio increases from 0.1 to 0.6.
Solution: G(S) = K .
S(1+ST)
Let the value of damping ratio is, when peak overshoot is 75% & when peak
overshoot is 25%
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 60
Mp = .
e 1- 2
ln 0. 75 = . 0.0916 = . 1- 2 1- 2
1 = 0.091 (0.0084) (1-2) = 2
2 = 0.4037 (1.0084 2) = 0.0084
= 0.091
k .
S + S2T .
w.k.t. T.F. = G(S) = 1 + K . = K .
1+ G(S) . H(S) S + S2T S + S2T+K
T.F. = K / T .
S2 + S + K .
T T
Comparing with std Eq :-
Wn = K . , 2 Wn = 1 .
T T
Let the value of K = K1 When = 1 & K = K2 When = 2.
Since 2 Wn = 1 . , = 1 . = 1 .
T 2TWn 2 KT
1 .
1 . = 2 K1T = K2 .
2 1 K1
2 K2T
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 61
0.091 = K2 . K2 . = 0.0508
0.4037 K1 K1
K2 = 0.0508 K1
a) The amplitude K has to be reduced by a factor = 1 . = 20
0.0508
b) Let = 0.1 Where gain is K1 and
= 0.6 Where gain is K2
0.1 = K2 . K2 . = 0.027 K2 = 0.027 K1
0.6 K1 K1
The amplitude gain should be reduced by 1 . = 36
0.027
4) Find all the time domain specification for a unity feed back control system whose open loop
T.F. is given by
G(S) = 25 .
S(S+6)
Solution:
25 .
G(S) = 25 . G(S) . = S(S+6) .
S(S+6) 1 + G(S) .H(S) 1 + 25 .
S(S+6)
= 25 .
S2 + ( 6S+25 )
W2n = 25 , Wn = 5, 2 Wn = 6 = 6 . = 0.6
2 x 5
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 62
Wd = Wn 1- 2 = 5 1- (0.6)2 = 4
tr = - , = tan-1 Wd = Wn = 0.6 x 5 = 3
Wd
= tan-1 ( 4/3 ) = 0.927 rad.
tp = . = 3.14 = 0.785 sec.
Wd 4
MP = . = 0.6 . x3.4 = 9.5%
e 1- 2 e 1- 0.62
ts = 4 . for 2% = 4 . = 1.3 ………3sec.
Wn 0.6 x 5
5) The closed loop T.F. of a unity feed back control system is given by
C(S) = 5 .
R(S) S2 + 4S +5
Solution:
C(S) = 5 . , Wn2 = 5 Wn = 5 = 2.236
R(S) S2 + 4S +5
2Wn = 4 = 4 . = 0.894. Wd = 1.0018
2 x 2.236
MP = . = 0.894 . X 3.14 = 0.19%
e 1- 2 e 1-(0.894)2
Determine (1) Damping ratio (2) Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 63
W. K.T. C(t) = e-Wnt Cos Wdtr + . sin wdtr
1-2
= e-0.894x2.236t Cos 1.0018t + 0.894 . sin 1.0018t
1-(0.894)2
6) A servo mechanism is represent by the Eq:-
d2 + 10 d = 150E , E = R- is the actuating signal calculate the
dt2 dt value of damping ratio, undamped and damped
frequency of ascillation.
Soutions:- d2 + 10 d = 15 ( r - ) , = 150r – 150.
dt2 dt
Taking L.T., [S2 + 10S + 150] (S) = 150 R (S).
(S) = 150 .
R(S) S2 + 10S + 15O
Wn2 = 150 Wn = 12.25. ………………………….rad sec .1
2Wn = 10 = 10 . = 0.408.
2 x 12.25
Wd = Wn 1 - 2 = 12.25 1- (0.408)2 = 11.18. rad 1sec.
6) Fig shows a mechanical system and the response when 10N of force is applied to the
system. Determine the values of M, F, K,.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 64
K x(t)inmt
f(t) 0.00193 The T.F. of the mechanical system is , 0.02 X(S) = 1 .
F(S) MS2 + FS = K
f(t) = Md2X + F dX + KX
F x dt2 dt
F(S) = (MS2 + FS + K) x (S)
1 2 3 4 5
Given :- F(S) = 10
S.
X(S) = 10 .
S(MS2 + FS + K)
SX (S) = 10 .
MS2 + FS + K
The steady state value of X is By applying final value theorem,
lt. SX(S) = 10 . = 10 = 0.02 ( Given from Fig.)
S O M(0) + F (0) + K K. ( K = 500.)
MP = 0.00193 = 0.0965 = 9.62%
0.02
MP = e . ln 0.0965 = . 1 - 2 1 - 2
= . 0.5539 = 2 . 0.744
1 - 2 1 - 2
M
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 65
0.5539 – 0.5539 2 = 2
= 0.597 = 0.6
tp = = .
Wn 1 – 2 Wd
3 = . Wn = 1.31…… rad / Sec.
Wn1 – (0.6)2
Sx(S) = 10/ M .
(S2 + F S + K )
M M
Comparing with the std. 2nd order Eq :-, then,
Wn2 = K Wn = K (1.31)2 = 500 . M = 291.36 kg.
M M M
F = 2Wn F = 2 x 0.6 x 291 x 1.31
M F = 458.7 N/M/ Sec.
8) Measurements conducted on sever me mechanism show the system response to be c(t) =
1+0.2e-60t – 1.2e-10t , When subjected to a unit step i/p. Obtain the expression for closed
loop T.F the damping ratio and undamped natural frequency of oscillation .
Solution:
C(t) = 1+0.2e-60t –1.2e-10t
Taking L.T., C(S) = 1 . + 0.2 . – 1.2 .
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 66
S S+60 S+10
C(S) . = 600 / S .
S2 + 70S + + 600
Given that :- Unit step i/p r(t) = 1 R(S) = 1 .
S
C(S) . = 600 / S .
R(S) S2 + 70S + + 600
Comparing, Wn2 = 600, 24.4 …..rad / Sec
2 Wn = 70, = 70 . = 1.428
2 x 24.4
Steady state Error :-
Steady state errors constitute an extremely important aspect of system
performance. The state error is a measure of system accuracy. These errors arise from the nature
of i/p’s type of system and from non-linearties of the system components. The steady state
performance of a stable control system are generally judged by its steady state error to step, ramp
and parabolic i/p.
Consider the system shown in the fig.
R(S) C(S)
G(S)
H(S)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 67
C(S) = G(S) . …………………………(1)
R(S) 1+G(S) . H(S)
The closed loop T.F is given by (1). The T.F. b/w the actuating error signal e(t) and the
i/p signal r(t) is,
E(S) = R(S) – C(S) H(S) = 1 – C(S) . H(S)
R(S) R(S) R(S)
= 1 – G(S) . H(S) . = 1 + G(S) . H(S) – G(S)H(S)
1 + G(S) . H(S) 1+G(S) . H(S)
= 1 .
1 + G(S) . H(S)
Where e(t) = Difference b/w the i/p signal and the feed back signal
E(S) = 1 . .R(S) ……………………….(1)
1 + G(S) . H(S)
The steady state error ess may be found by the use of final value theorem
and is as follows;
ess = lt e(t) = lt SE(S)
t S O
Substituting (1), ess = lt S.R(S) . ……………….(2)
S O 1+G(S) . H(S)
Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward T.F.
G(S) and loop T.F G(S) . H(S).
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 68
The expression for steady state errors for various types of standard test signals are
derived below;
1) Steady state error due to step i/p or position error constant (Kp):-
The steady state error for the step i/p is
I/P r(t) = u(t). Taking L.T., R(S) = 1/S.
From Eq:- (2), ess = lt S. R(s) . = 1 .
S O 1 +G(S). H.S 1 + lt G(S). H(S)
S O
lt G(S) . H(S) = Kp
(S O )
Where Kp = proportional error constant or position error const.
ess = 1 .
1 + Kp
(1 + Kp) ess = 1 Kp = 1 - ess
ess
Note :- ess = R . for non-unit step i/p
1 + Kp
2) Steady state error due to ramp i/p or static velocity error co-efficient (Kv) :-
The ess of the system with a unit ramp i/p or unit velocity i/p is given by,
r ( t) = t. u(t) , Traking L -T, R(S) = 1/S2
Substituting this to ess Eq:-
ess = lt S . . 1 . = lt 1 .
S O 1 + G(S) . H(s) S2 S O S +S G(S) H(s)S
lt = SG(S) . H(S) = Kv = velocity co-efficient then
S O
ess = lt 1 . ess = 1 .
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 69
S O S + Kv Kv
Velocity error is not an error in velocity, but it is an error in position error due to a ramp
i/p
3) Steady state error due to parabolic i/p or static acceleration co-efficient (Ka) :-
The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p)
which is defined by r(t) + 1 . t2 Taking L.T. R(S)= 1 .
2 S3
ess = lt S . 1 . lt 1 .
S O 1 + G(S) . H(S) S3 S O S2 + S2 G(S) . H(S)
lt S2 G(S) . H(S) = Ka.
S O
ess = lt 1 . = 1 .
S O S2 + Ka Ka
Note :- ess = R . for non unit parabolic.
Ka
Types of feed back control system :-
The open loop T.F. of a unity feed back system can be written in two std, forms;
1) Time constant form and 2) Pole Zero form,
G(S) = K (TaS +1) (TbS +1)…………………..
Sn(T1 S+1) (T2S + 1)……………….
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 70
Where K = open loop gain.
Above Eq:- involves the term Sn in denominator which corresponds to no, of
integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1,
2, ………….. Respectively. The Type no., determines the value of error co-efficient. As
the type no., is increased, accuracy is improved; however increasing the type no.,
aggregates the stability error. A term in the denominator represents the poles at the origin
in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin.
The steady state errors co-efficient for a given type has definite values. This is
illustration as follows.
1) Type – O system: - If, n = 0, the system is called type – 0, system. The
Steady state error is as follows;
Let, G(S) = K. [ .. . H(s) = 1]
S + 1
ess (Position) = 1 . = 1 . = 1 .
1 + G (O). H(O) 1 + K 1 + Kp
.. . Kp = lt G(S) . H(S) = lt K . = K
S O S O S + 1
ess (Velocity) = 1 . = 1 . =
Kv O
Kv = lt G(S) . H(S) = lt S K . = O.
S O S O S + 1
ess (acceleration) = 1 . = 1 . =
Ka O
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 71
Ka = lt S2 G(S) . H(S) = lt S2 K . = O.
S O S O S + 1
2) Type 1 –System :- If, n = 1, the ess to various std, i/p, G(S) = K .
S (S + 1)
ess (Position) = 1 . = O
1 +
Kp = lt G(S) . H(S) = lt K . =
S O S O S( S + 1)
Kv = lt S K . = K
S O S(S+1)
ess (Velocity) = 1 .
K
ess (acceleration) = 1 . = 1 . =
O O
Ka = lt S2 K . = O.
S O S (S + 1)
3) Type 2 –System :- If, n = 2, the ess to various std, i/p, are , G(S) = K .
S2 (S + 1)
Kp = lt K . =
S O S2 (S + 1)
. . . ess (Position) = 1 . = O
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 72
Kv = lt S K . =
S O S2 (S + 1)
. . . ess (Velocity) = 1 . = O
Ka = lt S2 K . = K.
S O S2 (S + 1)
. . . ess (acceleration) = 1 .
K
STEADY STATE ERRORS IN UNITY FEEDBACK CONTROL SYSTEMS
Errors in control systems may be attributed to many factors. Changes in the reference input will
cause unavoidable errors during transient periods, and also cause steady state errors. Any
physical control system inherently suffers from steady state error in response to certain types of
inputs. Whether a given system will exhibit steady state error for a given input depends on the
type of the open loop transfer function of the system.
Classification of Control Systems
Control systems may be classified according to their ability to follow step inputs, ramp inputs,
parabolic inputs etc. The magnitudes of the steady state errors due to these individual inputs are
indicative of the goodness of the system. Consider the unity feedback control system with the
following open loop transfer function G(s):
1s
pT1s
2T1s
1TNs
1sm
T1sb
T1sa
TKsG
It involves the term sN in the denominator, representing the pole of multiplicity N at the origin
which also indicates the number of integrations in the open loop. A system is called Type 0,
Type 1, Type 2 …. If N = 0, 1, 2, … respectively. As the type number is increased accuracy is
improved; however, increasing the type number aggravates the stability problem.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 73
Steady State Errors
A unity feedback closed loop control system is shown in Fig.1.
Figure 12 Unity feedback control system
The closed loop transfer function is
sG1
sG
sR
sC
The transfer function between the error signal e(t) and input signal r(t) is
sG1
1
sR
sC1
sR
sE
, Which can be rearranged as sRsG1
1sE
The final value theorem provides a convenient way to find the steady state error
sG1
ssRlimssElimtelime
0s0stss
Three types of static error constants are
1) Static position error constant Kp due to unit step input.
2) Static velocity error constant Kv due to unit ramp input.
3) Static acceleration error constant Ka due to unit parabolic input
Which indicate the figures of merit of control systems?
Static Position Error Constant
The steady state error of the system for a unit step input is
0G1
1
s
1
sG1
slime
0sss
The static position error constant Kp is defined by
0GsGlimK0sp
−
R(s) G(s)
E(s) C(s)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 74
Thus, the steady state error in terms of the static position error constant Kp is given by
pss K1
1e
For a type 0 system,
K
1sp
T1s2
T1s1
T
1sm
T1sb
T1sa
TK
limK0sp
and
K1
1e
ss
For a type 1 or higher system
1N
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TK
limKN0sp
and ss
e
Response of a feedback control system to a step input involves a steady state error if there
is no integration in the feed forward path. If a zero steady state error for a step input is desired,
the type of the system must be one or higher.
Static Velocity Error Constant
The steady state error of the system with a unit ramp input is given by
ssG
1lim
s
1
sG1
slime
02
0sss
s
The static velocity error constant Kv is defined by
ssGlimK0sv
Thus, the steady state error in terms of the static velocity error constant Kv is given by
vss K
1e
The term velocity error is used here to express the steady state error for a ramp input. The
velocity error is an error in position due to ramp input.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 75
For a type 0 system,
0
1sp
T1s2
T1s1
T
1sm
T1sb
T1sa
TsK
limK0sv
and
vss K
1e
For a type 1 system
K
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TsK
limK0sv
and
K
1
K
1e
vss
For a type 2 or
higher system
2N
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TsK
limKN0sv
and 0K
1e
vss
Type 0 system is incapable of following a ramp input in the steady state. Type 1 system
with unity feedback can follow the ramp input with a finite error. Type 2 and higher order
systems can follow a ramp input with zero steady state error.
Static Acceleration Error Constant
The steady state error of the system with a unit parabolic input is given by
sGs
1lim
s
1
sG1
slime
20
30s
ss
s
The static acceleration error constant Ka is defined by
sGslimK 2
0sa
Thus, the steady state error in terms of the static acceleration error constant Ka is given by
ass K
1e
For a type 0 system,
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 76
0
1sp
T1s2
T1s1
T
1sm
T1sb
T1sa
TKs
limK
2
0sa
and
ass K
1e
For a type 1 system
0
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TKs
limK
2
0sa
and
ass K
1e
For a type 2 system
K
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TKs
limK2
2
0sa
and K
1
K
1e
ass
For a type 3 or higher system
3N
1sp
T1s2
T1s1
Ts
1sm
T1sb
T1sa
TKs
limKN
2
0sa
and 0K
1e
ass
Type 0 and type 1 systems are incapable of following a parabolic in the steady state. Type 2
system with unity feedback can follow the parabolic input with a finite error. Type 3 and higher
order systems can follow a parabolic input with zero steady state error.
Table 1 shows the summary of error constants and steady state errors for unit step, unit
ramp and unit parabolic inputs to a unity feedback loop.
Table 1: Summary of steady state errors for unity feedback systems
Type of
system
Error constants Steady state error ess
Kp Kv Ka Unit step
input
Unit ramp
input
Unit parabolic
input
0 K 0 0 1/(1+K)
1 K 0 0 1/K
2 K 0 0 1/K
3 0 0 0
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 77
Note:
1) The step, ramp, parabolic error constants are significant for the error analysis only
when the input signal is a step, ramp and parabolic function respectively.
2) Since the error constants are defined with respect to forward path transfer function
G(s), the method is applicable to a unity feedback system only.
3) Since error analysis relies on use of final value theorem, it is important first to
check to see if sE(s) has any poles on the j axis or in the right half of s-plane.
4) Principle of superposition can be used if combination of the three basic inputs is
present.
5) If the configuration differs, we can either simplify to unity feedback system or
establish error signal and apply final value theorem.
Examples:
For a negative unity feedback system determine the steady state error due to unit step, unit ramp
and unit parabolic input of the following systems.
102sss
4s12s1KsGiii)
2004sss
KsGii)
2s10.1s1
50sGi)
22
2
Solution
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 78
ass
2
0sv
vss0sv
pss0sp
K
1e G(s)slimK
K
1e sG(s)limK
K1
1e G(s)limK
Table 2 shows the results of error constants and steady state errors.
Table 2 Results of example
Problem Error constants Steady state error ess
Kp Kv Ka Unit step
input
Unit ramp
input
Unit parabolic
input
i 50 0 0 1/51
ii K/200 0 0 200/K
iii K/10 0 0 10/K
CONTROL ACTIONS
An automatic controller compares the actual value of the system output with the reference input
(desired value), determines the deviation, and produces a control signal that will reduce the
deviation to zero or a small value. The manner in which the automatic controller produces the
control signal is called the control action. The controllers may be classified according to their
control actions as
1) Two position or on-off controllers
2) Proportional controllers
3) Integral controllers
4) Proportional-plus- integral controllers
5) Proportional-plus-derivative controllers
6) Proportional-plus-integral-plus-derivative controllers
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 79
A two position controller has two fixed positions usually on or off.
A proportional control system is a feedback control system in which the output forcing function
is directly proportional to error.
A integral control system is a feedback control system in which the output forcing function is
directly proportional to the first time integral of error.
A proportional-plus-integral control system is a feedback control system in which the output
forcing function is a linear combination of the error and its first time integral.
A proportional-plus-derivative control system is a feedback control system in which the output
forcing function is a linear combination of the error and its first time derivative.
A proportional-plus-derivative-plus-integral control system is a feedback control system in
which the output forcing function is a linear combination of the error, its first time derivative
and its first time integral.
Many industrial controllers are electric, hydraulic, pneumatic, electronic or their
combinations. The choice of the controller is based on the nature of plant and operating
conditions. Controllers may also be classified according to the power employed in the operation
as
1) Electric controllers
2) Hydraulic controllers
3) Pneumatic controllers
4) Electronic controllers.
The block diagram of a typical controller is shown in Fig.1. It consists of an automatic
controller, an actuator, a plant and a sensor. The controller detects the actuating error signal and
amplifies it. The output of a controller is fed to the actuator that produces the input to the plant
according to the control signal. The sensor is a device that converts the output variable into
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 80
another suitable variable to compare the output to reference input signal. Sensor is a feedback
element of the closed loop control system.
Figure 13 Block diagram of control system
Two Position Control Actions
In a two position control action system, the actuating element has only two positions which are
generally on and off. Generally these are electric devices. These are widely used are they are
simple and inexpensive. The output of the controller is given by Eqn.1.
0
0
2
1
teU
teUtu ….. (1)
Where, U1 and U2 are constants (U2= -U1 or zero)
The block diagram of on-off controller is shown in Fig. 2
−
Actuating Error signal e(t)
Reference Input r(t)
Error Detector
Feed back signal b(t)
Controller Output u(t)
−
Output c(t)
automatic controller
actuating error signal e(t)
reference input r(t)
error detector
Amplifier Actuator Plant
Sensor feed back signal b(t)
Controller output u(t)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 81
Figure 14 Block diagram of on off controller
Proportional Control Action
The proportional controller is essentially an amplifier with an adjustable gain. For a controller
with proportional control action the relationship between output of the controller u(t) and the
actuating error signal e(t) is
teKtup
..... (2)
Where, Kp is the proportional gain.
Or in Laplace transformed quantities
pK
sE
sU ….. (3)
Whatever the actual mechanism may be the proportional controller is essentially an
amplifier with an adjustable gain. The block diagram of proportional controller is shown in
Fig.3.
Figure 15 Block diagram of a proportional controller
Integral Control Action
The value of the controller output u(t) is changed at a rate proportional to the actuating error
signal e(t) given by Eqn.4
t
0ii
dtteKtuor teKdt
tdu …..(4)
−
Actuating Error signal e(t)
Reference Input r(t)
Error Detector
Feed back signal b(t)
Controller Output u(t)
Kp
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 82
Where, Ki is an adjustable constant.
The transfer function of integral controller is
s
K
sE
sU i ….(5)
If the value of e(t) is doubled, then u(t) varies twice as fast. For zero actuating error, the
value of u(t) remains stationary. The integral control action is also called reset control. Fig.4
shows the block diagram of the integral controller.
Figure 16 Block diagram of an integral controller
Proportional-plus-Integral Control Action
The control action of a proportional-plus-integral controller is defined by
t
0i
p
pdtte
T
KteKtu …..(6)
The transfer function of the controller is
sT
11K
sE
sU
ip
.....(7)
Where, Kp is the proportional gain, Ti is the integral time which are adjustable.
The integral time adjusts the integral control action, while change in proportional gain
affects both the proportional and integral action. The inverse of the integral time is called reset
rate. The reset rate is the number of times per minute that a proportional part of the control action
is duplicated. Fig.5 shows the block diagram of the proportional-plus-integral controller. For an
actuating error of unit step input, the controller output is shown in Fig.6.
−
Actuating Error signal e(t)
Reference Input r(t)
Error Detector
Feed back signal b(t)
Controller Output u(t)
s
Ki
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 83
Figure 17 Block diagram of a proportional-plus-integral control system
Figure 18 Response of PI controller to unit actuating error signal
Proportional-plus-Derivative Control Action
The control action of proportional-plus-derivative controller is defined by
dt
tdeTKteKtu
dpp …..(8)
The transfer function is
sT1KsE
sUdp
…..(9)
Where, Kp is the proportional gain and Td is a derivative time constant.
−
Actuating Error signal e(t)
Reference Input r(t)
Error Detecto
r
Feed back signal b(t)
Controller output u(t)
sT
sT1K
i
ip
t 0
Kp
u(t)
2Kp
proportional only
P-I control action
Ti
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 84
Both, Kp and Td are adjustable. The derivative control action is also called rate control. In
rate controller the output is proportional to the rate of change of actuating error signal. The
derivative time Td is the time interval by which the rate action advances the effect of the
proportional control action. The derivative controller is anticipatory in nature and amplifies the
noise effect. Fig.7 shows the block diagram of the proportional-plus-derivative. For an actuating
error of unit ramp input, the controller output is shown in Fig.8
Figure 19 Block diagram of a proportional-plus-derivative controller
Figure 20 Response of PD controller to unit actuating error signal
Proportional-plus-Integral-plus-Derivative Control Action
It is a combination of proportional control action, integral control action and derivative control
action. The equation of the controller is
dt
tdeTKdtte
T
KteKtu
dp
t
oi
p
p …..(10)
or the transfer function is
t 0
u(t)
proportional only
P-D control action
Td
−
actuating error signal e(t)
reference input r(t)
error detector
feed back signal b(t)
Controller output u(t)
ST1KDP
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 85
sT
sT
11K
sE
sUd
ip
…..(11)
Where, Kp is the proportional gain, Ti is the integral time, and Td is the derivative time. The
block diagram of PID controller is shown in Fig.9. For an actuating error of unit ramp input, the
controller output is shown in Fig.10.
Figure 21 Block diagram of a proportional-plus-integral-plus-derivative controller
Figure 22 Response of PID controller to unit actuating error signal
t 0
u(t)
proportional only
PD control action PID control action
−
actuating error signal e(t)
reference input r(t)
error detector
feed back signal b(t)
Controller output u(t)
sT
sTTsTK
d
diip
21
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 86
UNIT-III
FREQUENCY RESPONSE ANALYSIS
FREQUENCY RESPONSE
The Frequency response is the steady state response (output) of a system when the input
to the system is a sinusoidal signal.The frequency response of a system is normally obtained by
varying the frequency of the input signal by keeping the magnitude of the input signal at a
constant value.
In a system transfer function T(s),if s is replaced by jω then the resulting transfer
function T(jω) is called sinusoidal transfer function. The frequency response of the system can
be directly obtained from the sinusoidal transfer function T(jω) of the system.
Open loop transfer function, G(jω) => | G(jω)| G(jω) G(jω)
Loop transfer function, G(jω)H(jω) =>| G(jω) H(jω) | G(jω) H(jω)
Closed loop transfer function,C(jω)/R(jω)=> M((jω)= | M((jω)| M((jω)|
ADVANTAGES OF FREQUENCY RESPONSE ANALYSIS
1. The absolute and relative stability of the closed loop system can be estimated from the
knowledge of the open loop frequency response
2. The practical testing of system can be easily carried with available sinusoidal signal
generators and precise measurements equipments
3. The transfer function of complicated function can be determined experimentally by
frequency response test
4. The design and parameter adjustment can be carried more easily
5. The corrective measure for noise disturbance and parameter variation can be easily
carried
6. It can be extended to certain non-linear systems.
FREQUENCY DOMAIN SPECIFICATIONS
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 87
The performance and characteristics of a system in frequency domain are measured in
terms of frequency domain specifications. The frequency domain specifications are
1. Resonant peak,Mr
2. Resonant frequency, ωr
3. Bandwidth
4. cut-off rate
5. phase margin,Kp
Peak response Mp :
The peak response Mp is defined as the maximum value of M() that is given in
Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of a feed
back control system. Normally, a large Mp corresponds to a large peak overshoot in the step
response. For most design problems it is generally accepted that an optimum value Mp of should
be somewhere between 1.1 & 1.5.
Resonant frequency p :
The resonant frequency p is defined as the frequency at which the peak resonance Mp occurs.
Bandwidth :
The bandwidth , BW, is defined as the frequency at which the magnitude of M(j), M(),
drops at 70.7 percent of its zero-frequency level, or 3 dB down from the zero-frequency gain. In
general, the bandwidth of a control system indicates the noise-filtering characteristics of the
system. Also, bandwidth gives a measure of the transient response properties, in that a large
bandwidth corresponds to a faster rise time, since higher-frequency signals are passed on to the
outputs. Conversely, if the bandwidth is small, only signals of relatively low frequencies are
passed, & the time response will generally be slow & sluggish.
Cutoff rate:
Often, bandwidth alone is inadequate in the indication of the characteristics of the system in
distinguishing signals from noise. Sometimes it may be necessary to specify the cutoff rate of the
frequency response at the higher frequencies. However, in general, steep cutoff characteristics
may be accompanied by a large Mp, which corresponds to a system with a low stability margin.
The performance criteria defined above for the frequency-domain analysis are illustrated on
the closed-loop frequency response, as shown in Fig. 1.3.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 88
There are other criteria defined that may be used to specify the relative stability &
performance of a feedback control system. These are defined in the ensuring sections of this
chapter.
M() Mp 1 Bandwidth 0 p BW Fig.1.3. Typical magnification curve of a feedback control system. Mp, , p & the bandwidth of a second-order system: For a second-order feedback control system, the peak resonance Mp, the resonant frequency p,
& the bandwidth are all uniquely related to the damping ratio & the natural undamped
frequency n of the system. Consider the second-order sinusoidal steady-state transfer function
of a closed-loop system,
C(j) 2n
M(j) = = R(j) (j)2 + 2 n (j) + 2
n
1 = 1 + j 2 (/n) - (/n)2
We may simplify the last expression by letter u = / n. The Eq. (1.6) becomes 1 M(ju) = 1 + j2 u - u2
The magnitude & phase of M (j) are 1 M(ju) = M(u) = [( 1 – u2 )2 + ( 2 u)2] ½
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 89
And 2 u M(ju) = m(u) = - tan -1 1 – u2
The resonant frequency is determined first by taking the derivative of M(u) with respect to u &
setting it equal to zero. Thus
dM(u) 1 = - [( 1 – u2 )2 + ( 2 u)2] –3/2 ( 4 u3 – 4u + 8u2) = 0 du 2 from which 4u3 – 4u + 8u2 = 0 The root of Eq. (1.11) are up = 0 and up = 1 - 22 The solution Eq. (1.12) merely indicates that the slope of the M() versus curve is zero at = 0; it is not true maximum. The solution of eq. (1.13) gives the resonant frequency, p = n 1 - 22 (1.14) Since frequency is a real quantity, Eq. (1.14) is valid only for 1 22 or 0.707. This means
simply that for all values of greater than 0.707, the solution of p = 0 becomes the valid one, &
Mp = 1.
Substituting Eq. (1.13) into Eq. (1.8) & simplifying, we get 1 Mp = 2 1 - 2 It is important to note that Mp is a function of only, whereas p is a function of & n
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 90
5 4 3 Mp 2 1 0 0.5 0.707 1.0 1.5 2.0 Damping ratio 1 Fig.1.4 Mp versus-damping ratio for a second – order system, Mp = 2 1 - 2 1.0 0.8 up = p /n 0.6 0.4 0.2 0 0.5 0.707 1.0 Damping ratio Fig 1.5. Normalized resonant frequency- versus-damping ratio for a second order system,
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 91
Up = 1 - 2 . Fig.1.4 & 1.5 illustrate the relationship between Mp & , & u = p / n & , respectively. Bandwidth: Bandwidth BW of a system is a frequency at which M() drops to 70.7% of its zero
frequency level or 3 dB down from the zero frequency gain. Equating the Eq.
1 M(u) = = 0.707 [( 1 –u2)2 + ( 2u)2] ½
1 = [( 1 –u2)2 + ( 2u)2] ½ = 0.707 = 2. Squaring both sides ( 1 –u2)2 + 42u2 = 2 1 + u4 – 2u2 + 42u2 = 2
Let u2 = x 1 + x2 – 2x + 42 x = 2 x2 – 2x + 42 x = 1
x2 – x ( 2 - 42 ) = 1 x2 – x ( 2 - 42 ) – 1 = 0 a = 1, b = - ( 2 - 42 ) , c = -1 -b b2 – 4ac x = 2a (2 – 42 ) (2 – 42 )2 + 4 = 2 (2 – 42 ) (4 + 164 – 16 2 + 4 =
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 92
2 (2 – 42 ) 16 4 – 16 2 + 8 = 2 2 (1 – 22 ) 4 + 164 – 16 2 + 4 = 2 2 (1 – 22 ) 22 + 44 – 42 = 2 2 (1 – 22 ) 22 + 44 – 42 = 2 2 [(1 – 22 ) 2 + 44 – 42] = 2 u2 = x = (1 – 22 ) 2 + 44 – 4 2 / n = u = [(1 – 22 ) (2 + 44 – 4 2 ] 1/2 BW = n [ ( 1 – 22 ) + 44 - 42 +2]1/2
For the second order system under consideration, we easily establish some simple relationship
between the time – domain response & the frequency-domain response of the system.
1. The maximum over shoot of the unit step response in the time domain depends upon
only.
2. The response peak of the closed - loop frequency response Mp depends upon only.
3. The rise time increases with , & the bandwidth decreases with the increase of , for a
fixed n, therefore, bandwidth & rise time are inversely proportional to each other.
4. Bandwidth is directly proportional to n.
5. Higher bandwidth corresponds to larger Mp.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 93
Determination of closed loop response from open loop response
• The steady-state response of a linear system to sinusoidal excitation may be determined
from the system transfer function T(s) by replacing s by j
• The resulting frequency response function T(j) has the polar form :
• For any value of , T(j) reduces to a complex number whose amplitude, A(j) gives
, and () gives .
\ Steady-state vs. general response • In control work. the response of the system to general input functions, e.g. step input,
impulse input, are of most interest (rather than steady state sinusoidal response)
• However, with Fourier analysis, any time function can be represented as a series sum of
sinusoids :
Therefore it is possible to predict the nature of a response, y(t) to any given x(t) from a
knowledge of how the gain, A, and phase shift, , vary with angular frequency
• In particular, the closed-loop behaviour of a feedback system may be usefully predicted
from the open-loop gain and phase-shift characteristics
Open loop response: Bode plots
T(s) X(s ) Y(s)
)()()( jeAjT
)(
)(
sX
sY
)(
)(arg
sX
sY
22
2)(
2 ss
sT
......)2()()( 22110 tSintSintx
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 94
• Amplitude is plotted in dB
• Phase is plotted in degrees
• Used to identify system order, corner frequencies, etc
Open loop response: Polar plots • Polar plots are more useful in control system design work
– open-loop polar plot indicates whether the closed-loop system is stable
– also gives a measure of how stable the system is
• Represent T(jw) on an Argand diagram :
• Polar plot is traced out by the trajectory of T(jw) as w increases progressively from zero
Re Im
T(jw)
A(jw)
)
fai(jw)
T(s)-plane
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 95
• A polar plot may be constructed from experimental data or from a system transfer
function
• If values of w are marked along the contour, a polar plot has the same information as a
Bode plot .Usually, the shape of a polar plot is of most interest.
FREQUENCY RESPONSE PLOTS Frequency response analysis of control system can be carried either analytically (or)
graphically. The various graphical techniques available for frequency response analysis are
1. Bode plot
2. Polar plot
3. Nichols plot
4. M & N Circles
5. Nichols chart
Among the above five, Bode plot, Polar plot and Nichols plot are used to get the frequency
response of closed loop system
BODE PLOT It is one of the plot used to obtain the frequency response for the open loop or closed loop
system it consist of
1. Log magnitude plot
2. Phase magnitude plot
The log magnitude plot can be drawn with the help of magnitude in decibel and varying
frequencies (in angular)
The phase angle plot ca be drawn with the help of phase angle and varying frequencies (in
angular)
ADVANTAGE OF BODE PLOT The main advantage of the bode plot is that multiplication of magnitudes can be
converted into addition. also simple method for sketching an approximate log=magnitude curve
is available
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 96
THE FACTORS USED TO DRAW THE BODE PLOT ARE GIVEN BELOW
1. Constant gain factor, K
2. Integral factor, K / jω or K / (jω)n
3. Derivative factor, K(jω) or K (jω) n
4. First order factor found in dr
5. First order factor found in nr
6. Second order factor found in dr
7. Second order factor found in nr
CONSTANT GAIN FACTOR, K Let G(S) = K
G(jω)= K=K 0˚
A=|G(jω)| in db
=20 log K
Φ= G(jω)|= 0˚
When K>1,20 log K is Positive
When 0<K<1,20 log K is Negative
When K=1,20 log K is Zero
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 97
INREGRAL FACTOR Let G(S) = K/s
G(jω)= K/(jω)=K /ω -90˚
A=|G(jω)| in db
=20 log K/ω
Φ= G(jω)|= -90˚
When ω =0.1K, A=20 log K is 20db
When ω =K , A=20 log K is Zero db
When ω =10K , A=log K is -20db
Let G(S) = K/sn
G(jω)= K/(jω) n=K /ω -90n˚
A=|G(jω)| in db
=20 log K/ωn=20 log [K1/n/ω]n
Φ= G(jω)|= -90n˚
Now the magnitude plot of integral factor is a straight line with a slope of -20ndb/dec and
passing through zero db when ω=KI/n.the phase plot is a straight line at -90 n˚
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 98
DERIVATIVE FACTOR Let G(S) = Ks
G(jω)= K(jω) =Kω -90
A=|G(jω)| in db
=20 log Kω=20 log Kω
Φ= G(jω)|= +90˚
When ω =0.1/K, A=20 log K is -20db
When ω =1/K , A=20 log K is Zero db
When ω =10/K , A=log K is +20db
Let G(S) = Ksn
G(jω)= K(jω)n =Kωn -90˚
A=|G(jω)| in db
=20 log Kωn=20 log Kω
Φ= G(jω)|= +90n˚
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 99
FIRST ORDER FACTOR IN DENOMINATOR G(S) =1/(1+ST)
G(jω) =1/(1+jωT)=1/√(1+ω2T2) -tan-1ωT
A=|G(jω)|in db=20log1/√(1+ω2T2)= -20 log√(1+ω2T2)
At very low frequencies ωT<<1
A=-20log√(1+ω2T2)≈-20log1=0
At very high frequencies ωT>>1
A=-20 log√(1+ω2T2)≈-20log√ ω2T2=-20log√ ω2T2=-20logωT
At ω=1/T,A=-20log1=0
At ω=10/T,A=-20log1=-20db
The actual magnitude at corner frequency, ωc=1/T is A=-20log√1+1=-3db
Hence by approximation the loss in db at corner frequency is 3db
The phase angle is obtained by calculating the phase angle of G(jω) for various values of ω
Phase angle, Φ= G(jω)=-tan-1 ωT
At corresponding, ω=ωc=1/T, Φ= -tan-1 ωT=-tan-11=-45˚
At ω0, Φ0 and At ω∞, Φ=-90˚
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 100
G(S)=1/(1+ST)m=1/(√(1+ ω2T2)m m tan-1 ωT
A=|G(j ω)| in db=20log 1/√(1+ ω2T2)m=-20 m log1/√(1+ ω2T2) Φ= G(jω)= -m tan-1 ωT
FIRST ORDER FACTOR IN THE NUMERATOR G(S)=1+ST
G(jω)=1+ jωT=/√(1+ ω2T2) tan-1 Ωt
A= |G(jω)| in db=20 log√(1+ ω2T2)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 101
Φ= G(jω)= -tan-1 ωT
QUADRATIC FACTOR IN DENOMINATOR G(S)=1/(1+2ζs/ωn+(s/ωn)2)
G(jω)=1/(1+j2ζω/ωn+(jω/ωn)2=1/(1-(ω/ωn)2+j2 ζω/ωn)=1/√(1- ω2/ωn2)2+4ζ2 ω2/ωn
2
-tan-1(2ζ ω/ωn)/(1- ω2/ωn2)
A=|G(jω)| in db
A=20log 1//√(1- ω2/ωn2)2+4ζ2 ω2/ωn
2=- -20log√(1- ω2/ωn2(2-4ζ2)+ ω4/ωn
4)
At low frequencies when ω<< ωn
A= -20 log √(1- ω2/ωn2(2-4ζ2)+ ω4/ωn
4)≈-20log 1=0
At very high frequencies, when ω>> ωn
A=-20 log √(1- ω2/ωn2(2-4ζ2)+ ω4/ωn
4≈-20log√ ω4/ωn4=-20log ω2/ωn
2
A =-40 log ω/ωn
At ω= ωn,,A=-40log 1=0db
At ω= 10 ωn=-40log10=-40db
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 102
Φ= G(jω)= tan-1 2 ζ ω/ωn At ω= ωn Φ=-90˚
1-ω2/ωn2 At ω0 Φ0˚
ω ∞ Φ180˚
QUADRATURE FACTOR IN THE NUMERATOR
G(S)=1+2ζs/ωn+(s/ωn)2
G(jω)= √(1- ω2/ωn2)2+4ζ2 ω2/ωn
2 -tan-1(2ζ ω/ωn)/(1- ω2/ωn2)
ζ=0.1
ζ=0.3
ζ=0..5.
ζ=1.0
ζ=0.1 ζ=0.3
ζ=0..5
ζ=1.
0
0˚
-90˚
-180˚
+db
-db
A
Log ω
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 103
PROBLEMS 1. Sketch bode plot for the following transfer function and determine the system gain K for the
gain cross over frequency to be 5rad/sec
G(S) = Ks2/(1+0.2s)(1+0.02s)
Solution:
The sinusoidal transfer function G(j ω) is obtained by replacing S by j ω in the given S-domain
transfer function
G(j ω)=K(j ω)2/(1+0.2 j ω)(1+0.002 j ω)
Let K=1, G(j ω)= (j ω)2/(1+0.2 j ω)(1+0.002 j ω)
MAGNITUDE PLOT The corner frequencies are ωc1=1/0.2=5rad/sec and ωc2=1/0.02=50rad/sec
The various values of G(j ω) are below tabulation , in the increasing order of their corner
frequencies. also slope contributed by each term and the change in slope at the corner frequency
ζ=0.5
ζ=1.0
ζ=0.3
ζ=0.1
ζ=1.0
ζ=0.3
ζ=0.5
ζ=0.1
A in db
Log ω
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 104
TREM CORNER FREQUENCY
SLOPE IN db/dec CHANGE IN SLOPE db/dec
(jω)2 ----- +40 1/(1+j0.2) ωc1=1/0.2=5 -20 40-20=20 ωc2=1/0.02=50 -20 20-20 Choose a low frequency ωl such that ωl< ωc1& Choose a high frequency ωh such that
ωh> ωc2.Let ωl=0.5 rad/sec and ωh=100 rad/sec
Let A=|G(jω)| in db
Let us calculate A at ωl, , ωc1, ωc2, ωh
At ω= ωl, A=20 log|(jω)2| =20 log(ω2)=20 log(0.5)2= -12db
At ω= ωc1, A=20 log|(jω)2|=20log(5)2=28db
At ω= ωc2, A=[slope from ωc1 to ωc2×log ωc2/ωc1]+A (at ω= ωc1)
=20log50/5+28=48db
At ω= ωh,A=[slope from ωc2 to ωh×log ωh/ωc2]+A (at ω= ωc2)
=0×log (100/50)+48=48db
Let points a,b,c,d be the points corresponding to frequencies ωl, , ωc1, ωc2, ωh respectively on the
magnitude plot.
PHASE PLOT The phase angle G(jω) as a function of ω is given by Φ= G(jω)= 180˚-tan-10.2ω-tan-10.02 ω The phase angle of G(jω) are calculated for various values of ω ω rad/sec Tan-1(0.2 ω) deg Tan-1(0.02 ω) deg Φ= G(jω)
0.5 5.7 0.6 173.7≈174 1 11.3 1.1 167.6≈168 5 45 5.7 129.3≈130 10 63.4 11.3 105.3≈106 50 84.3 45 50.7≈50 100 87.1 63.4 29.5≈30 CALCULATION OF K Given that the gain crossover frequency is 5rad/sec
At ω=5,gain= 28db
At ω=5,frequency the db gain should be zero
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 105
20log K=-28db
Log K=-28/20
K=10-28/20=0.0398
NOTE: The frequency ω rad/sec is a corner frequency .hence in the exact plot the db gain at
ω=5rad/sec will be 3db less than the approximate plot.therefore for exact plot the 20logK=-25
LogK=-25/20=>K=10-25//20=0.0562
2. Given that G(S)=Ke-0.2s/s(s+2)(s+8). Find K so that the system is stable with
A) Gain margin equal to 6db
b) Phase margin equal to 45˚
Solution:
G(S)= Ke-0.2s/s(s+2)(s+8)= Ke-0.2s/s×2(1+s/2)×8(1×s/8)
Let K=1; G(j ω)= e-0.2jω/j ω(j ω+2)( j ω+8)
NOTE:
|0.0625 e-0.2jω|=0.0625 and e-0.2jω= -0.2 ω rad/sec
The various values of G(j ω) are below tabulation , in the increasing order of their corner
frequencies.also slope contributed by each term and the change in slope at the corner frequency
TREM CORNER FREQUENCY
SLOPE IN db/dec CHANGE IN SLOPE db/dec
0.0625/jω ----- -20 1/(1+j0.2) ωc1=1/0.2=5 -20 -20-20=-40 ωc2=1/0.02=50 -20 -40-20=-60 Choose a low frequency ωl such that ωl< ωc1& Choose a high frequency ωh such that
ωh> ωc2.Let ωl=0.5 rad/sec and ωh= 50 rad/sec
Let A=|G(jω)| in db
Let us calculate A at ωl, , ωc1, ωc2, ωh
At ω= ωl, A=20 log|0.0625/jω|=20log0.0625/0.5=-18db
At ω= ωc1, A=20 log|0.0625/jω|=20log 0.0625/2=-30db
At ω= ωc2, A=[slope from ωc1 to ωc2 × log ωc2/ ωc1]+A at ω= ωc1
=-40×log8/2+(-30)=-54db
At ω= ωh, A=[slope from ωc2 to ωh × log ωh/ ωc2]+A at ω= ωc2
=-60log50/8+(-54)=-102db
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 106
Example
Problem : Construct the polar plot for the (critically damped system)
defined by :
Solution :
Limiting conditions :
(i)
(ii)
(iii)
0j1)j(T:0 o
2
180j
1
e0)j(T.e.i
)j(T:
2j)j(T:1
Re-0.2 0 0.2 0.4 0.6 0.8 1
-0.6
-0.4
-0.2
Im
2)1s(
1)s(T
2)j1(
1)j(T
w 0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0
T(jw) 1 +j0
0.83 -j0.44
0.48 -j0.64
0.18 -j0.61
0 -j0.5
-0.12 -j0.28
-0.12 -j0.16
-0.08 -j0.06
-0.04 -j0.01
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IV– CONTROL SYSTEMS 107
ssT
1)(
Re
Im
1
22
2
2)(
nn
n
sssT
Re
Im
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UNIT –IV
STABILITY OF CONTROL SYSTEM
Introduction
The stability of a linear closed loop system can be determined from the locations of closed loop poles in the S-plane. If the system has closed loop T.F. C(s) 10 R(s) (S+2) (S+4) 1 Output response for unit step input R(s) = S C(s) 10 A B C S(S+2) (S+4) S S+2 S+4 Find out partial fractions 1 1 1 C(s) = 8 4 8 10 S S+2 S+4
= 1.25 2.5 1.25
S S+2 S+4 C(s) = 1.25 – 2.5e-2t+1.25e-4t =Css + Ct (t) If the closed loop poles are located in left half of s-plane, Output response contains exponential
terms with negative indices will approach zero & output will be the steady state output.
i.e. Ct (t) = 0 t Transient output = 0 Such system is called absolutely stable systems. Now let us have a system with one closed loop pole located in right half of s- plane C(s) 10 R(s) S(S-2)(s+4) A + B + C 10 = S S-2 S + 4 C(t) = - 1.25 + 0.833e 2t + 0.416e – 4 t Here there is one exponential term with positive in transient output
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IV– CONTROL SYSTEMS 109
Therefore Css = - 1.25
t C(t) 0 0 1 + 4.91 2 + 44.23 4 +2481.88
From the above table, it is clear that output response instead of approaching to steady state value
as t due to exponential term with positive index, transients go on increasing in amplitude.
So such system is said to be unstable. In such system output is uncontrollable & unbounded one.
Output response of such system is as shown in fig(4).
C(t) C(t) OR Steady state output t (a) fig(4) t (b) For such unstable systems, if input is removed output may not return to zero. And if the input
power is turned on, output tends to . If no saturation takes place in system & no mechanical
stop is provided then system may get damaged.
If all the closed loop poles or roots of the characteristic equation lie in left of s-plane, then in the
output response contains steady state terms & transient terms. Such transient terms approach to
zero as time advances eventually output reaches to equilibrium & attains steady state value.
Transient terms in such system may give oscillation but the amplitude of such oscillation will be
decreasing with time & finally will vanish. So output response of such system is shown in fig5
(a) & (b).
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IV– CONTROL SYSTEMS 110
C(t) C(t) Damped oscillations Steady state ----------------------- --------------------------- OR Steady state output t t (a) fig 5 (b) BIBO Stability: This is bounded input bounded output stability. Definition of stable system:
A linear time invariant system is said to be stable if following conditions are satisfied.
1. When system is excited by a bounded input, output is also bounded & controllable.
2. In the absence of input, output must tend to zero irrespective of the initial conditions.
Unstable system:
A linear time invariant system is said to be unstable if,
1. for a bounded input it produces unbounded output.
2. In the absence of input, output may not be returning to zero. It shows certain output
without input.
Besides these two cases, if one or more pairs simple non repeated roots are located on the
imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output
response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such
systems are said to be critically or marginally stable systems.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 111
3. Complex
conjugate with negative real part
J x J 1
-a1 x -J 2
C(t) t Damped oscillation
Absolutely stable.
4. Complex conjugate with positive real part
J1 x -J 1 -x
Ct
t
oscillations with increasing amplitude
Unstable.
Sl.No
Nature of closed loop poles.
Location of closed loop poles in s-plane
Step response Stability condition
1. Real negative i.e in LHS of splane
J x x -02 -01
C(t) --------------------- t
Absolutely stable
2.
Real positive in RHS of s-plane
J x a1
C(t) ------------------------ t increasing towards
Unstable
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IV– CONTROL SYSTEMS 112
5. Non repeated pair on imaginary axis
J x J1 x -J2
OR J X J2
x J1 x –J1 x – J2
Two non repeated pairs on imaginary axis.
C(t) t C(t) t Sustained oscillations with two frequencies 1 & 2
Marginally or critically stable Marginally or critically stable
6. Repeated pair on imaginary axis
J x x J1
x x -J1
C(t) ------------------------ t oscillation of increasing amplitude
Unstable
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IV– CONTROL SYSTEMS 113
Routh – Hurwitz Criterion:
This represents a method of determining the location of poles of a characteristics
equation with the respect to the left half & right half of the s-plane without actually solving the
equation.
The T.F.of any linear closed loop system can be represented as,
C(s) b0 sm + b1 sm-1 +….+ bm = R(s) a0 sn + a1 sn-1 + …. + an
Where ‘a’ & ‘b’ are constants.
To find the closed loop poles we equate F(s) =0. This equation is called as Characteristic
Equation of the system.
F(s) = a0 sn + a1 sn-1 + a2 sn-2 + ….. + an = 0.
Thus the roots of the characteristic equation are the closed loop poles of the system which decide
the stability of the system.
Necessary Condition to have all closed loop poles in L.H.S. of s-plane. In order that the above characteristic equation has no root in right of s-plane, it is necessary
but not sufficient that,
1. All the coefficients off the polynomial have the same sign.
2. Non of the coefficient vanishes i.e. all powers of ‘s’ must be present in descending order
from ‘n’ to zero.
These conditions are not sufficient.
Hurwitz’s Criterion :
The sufficient condition for having all roots of characteristics equation in left half of s-plane is
given by Hurwitz. It is referred as Hurwitz criterion. It states that:
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IV– CONTROL SYSTEMS 114
The necessary & sufficient condition to have all roots of characteristic equation in left half of s-
plane is that the sub-determinants DK, K = 1, 2,………n obtained from Hurwitz determinant ‘H’
must all be positive.
Method of forming Hurwitz determinant:
a1 a3 a5 …….. a2n-1 a0 a2 a4 ..…… a2n-2 0 a1 a3 .……. a2n-3 0 a0 a2 …….. a2n-4 H = 0 0 a1 ……... a2n-5 - - ……... - - - - ……... - 0 - - ……. an
The order is n*n where n = order of characteristic equation. In Hurwitz determinant all
coefficients with suffices greater than ‘n’ or negative suffices must all be replaced by zeros.
From Hurwitz determinant subdeterminants, DK, K= 1, 2, ….n must be formed as follows:
a1 a3 a5 D1 = a1 D2 = a1 a3 D3 = a 0 a2 a4 DK = H a0 a2 0 a1 a3 For the system to be stable, all above determinants must be positive. Determine the stability of the given characteristics equation by Hurwitz,s method.
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Ex 1: F(s)= s3 + s2 + s1 + 4 = 0 is characteristic equation. a0 = 1, a1 = 1, a2 = 1, a3 = 4, n = 3 a1 a3 a5 1 4 0 H = a 0 a2 a4 = 1 1 0 0 a1 a3 0 1 4 D1 = 1 = 1 1 4 D2 = 1 1 = -3 1 4 0 D3 = 1 1 0 = 4 –16 = -12. 0 1 4 As D2 & D3 are negative, given system is unstable. Disadvantages of Hurwitz’s method:
1. For higher order system, to solve the determinants of higher order is very complicated &
time consuming.
2. Number of roots located in right half of s-plane for unstable system cannot be judged by
this method.
3. Difficult to predict marginal stability of the system.
Due to these limitations, a new method is suggested by the scientist Routh called Routh’s
method. It is also called Routh-Hurwitz method.
Routh’s Stability Criterion:
It is also called Routh’s array method or Routh-Hurwitz’s method
Routh suggested a method of tabulating the coefficients of characteristic equation
in a particular way. Tabulation of coefficients gives an array called Routh’s array.
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Consider the general characteristic equation as,
F(s) = a0 sn + a1 sn-1 + a2 sn-2 + ….. + an = 0. Method of forming an array :
Sn a0 a2 a4 a6 ………. Sn-1 a1
a3 a5 a7 Sn-2 b1 b2 b3 Sn-3 c1 c2 c3 - - - - - - - - S0 an
Coefficients of first two rows are written directly from characteristics equation.
From these two rows next rows can be obtained as follows.
a1 a2 – a0 a3 a1 a4 – a0 a5 a1 a6 – a0 a7 b1 = , b2 = , b3 =
a1 a1 a1 From 2nd & 3rd row , 4th row can be obtained as b1 a3 – a1 b2 b1 a5 – a1 b3 C1 = , C2 = b1 b1
This process is to be continued till the coefficient for s0 is obtained which will be an. From this
array stability of system can be predicted.
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Routh’s criterion : The necessary & sufficient condition for system to be stable is “All the terms in the first
column of Routh’s array must have same sign. There should not be any sign change in first
column of Routh’s array”.
If there do sign changes exist then?
1. System is unstable.
2. The number of sign changes equals the number of roots lying in the right half of the
S-plane.
Examine the stability of given equation using Routh’s method: Ex.2: s3+6s2 + 11s + 6 =0
Sol: a0 = 1, a1 = 6, a2 =11, a3 = 6, n = 3
S3 1 11 S2 6 6 S1 11 * 6 – 6 =10 0 6 S0 6 As there is no sign change in the first column, system is stable. Ex. 3 s3 + 4s2 + s + 16 = 0 Sol: a0 =1, a1 = 4, a2 = 1, a3 = 16 S3 1 1 S2 +4 16 S1 4 - 16 = -3 0 4 S0 +16 As there are two sign changes, system is unstable.
Number of roots located in the right half of s-plane = number of sign changes = 2.
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Special Cases of Routh’s criterion: Special case 1 :
First element of any of the rows of Routh’s array is zero & same remaining rows contains at
least one non-zero element.
Effect : The terms in the new row become infinite & Routh’s test fails.
e.g. : s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 S5 1 3 2 S4 2 6 1 Special case 1 Routh’s array failed S3 0 1.5 0 S2 …. … Following two methods are used to remove above said difficulty. First method: Substitute a small positive number ‘’ in place of a zero occurred as a first element in the row. Complete the array with this number ‘’. Then examine lim Sign change by taking . Consider above Example. 0 S5 1 3 2 S4 2 6 1 S3 1.5 0 S2 6 - 3 1 0 S1 1.5(6 - 3) 0 - (6 - 3) S0 1
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To examine sign change, Lim Lim = 6 - 3 = 6 - 3 0 0 = 6 - = - sign is negative. Lim 1.5(6 – 3) - 2 = Lim 9 - 4.5 - 2 0 6 -3 0 6 - 3 = 0 – 4.5 – 0 0 -3 = + 1.5 sign is positive Routh’s array is, S5 1 3 2 S4 2 6 1 S3 + 1.5 0 S2 - 1 0 S1 +1.5 0 0 S0 1 0 0 As there are two sign changes, system is unstable.
Second method:
To solve the above difficulty one more method can be used. In this, replace‘s’ by ‘1/Z’ in
original equation. Taking L.C.M. rearrange characteristic equation in descending powers of ‘Z’.
Then complete the Routh’s array with this new equation in ‘Z’ & examine the stability with this
array.
Consider F(s) = s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 Put s = 1 / Z
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1 + 2 + 3 + 6 + 2 + 1 = 0 Z5 Z4 Z3 Z2 Z Z5 + 2Z4+ 6Z3+3Z2+2Z+ 1 = 0 Z5 1 6 2 Z4 2 3 1 Z3 4.5 1.5 0 Z2 2.33 1 0 Z1 - 0.429 0 Z0 1 As there are two sign changes, system is unstable. Special case 2 : All the elements of a row in a Routh’s array are zero.
Effect : The terms of the next row can not be determined & Routh’s test fails.
S5 a b c
S4 d e f
S3 0 0 0 Row of zeros, special case 2
This indicates no availability of coefficient in that row.
Procedure to eliminate this difficulty :
1. Form an equation by using the coefficients of row which is just above the row of zeros.
Such an equation is called an Auxillary equation denoted as A(s). For above case such
an equation is,
A(s) = ds4 + es2 + f
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IV– CONTROL SYSTEMS 121
Note that the coefficients of any row are corresponding to alternate powers of ‘s’ starting
from the power indicated against it.
So ‘d’ is coefficient corresponding to s4 so first term is ds4 of A(s).
Next coefficient ‘e’ is corresponding to alternate power of ‘s’ from 4 i.e. s2 Hence the term es 2 & so on.
2. Take the derivative of an auxillary equation with respect to ‘s’.
i.e. dA(s)
= 4d s3 + 2e s
ds
3. Replace row of zeros by the coefficients of dA(s) ds
S5 a b c
S4 d e f S3 4d 2e 0
4. Complete the array of zeros by the coefficients. Importance of auxiliary equation: Auxiliary equation is always the part of original characteristic equation. This means the roots of
the auxiliary equation are some of the roots of original characteristics equation. Not only is this
but roots of auxillary equation the most dominant roots of the original characteristic equation,
from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather
than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the
stability point of view. The remaining roots of the characteristic equation are always in the left
half & they do not play any significant role in the stability analysis.
e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5.
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Let A(s) = 0 be the auxiliary equation for the system due to occurrence of special case 2 of
the order m = 2.
Then out of 5 roots of F(s) = 0, the 2 roots which are most dominant (dominant means very
close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability
point of view are the 2roots of A(s) = 0. The remaining 5 –2 = 3 roots are not significant from
stability point of view as they will be far away from the imaginary axis in the left half of s-plane.
The roots of auxillary equation may be,
1. A pair of real roots of opposite sign i.e.as shown in the fig. 8.10 (a).
j j x x x x Fig. 8. 10 (b) Fig 8. 10(a)
2. A pair of roots located on the imaginary axis as shown in the fig. 8.10(b).
3. The non-repeated pairs of roots located on the imaginary axis as shown in the
fig.8.10 (c).
j j x xx x x
x xx
Fig. 8.10(c) Fig. 8.10(d).
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 123
4. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.10
(d).Hence total stability can be determined from the roots of A(s) = 0, which can be out
of four types shown above.
Change in criterion of stability in special case 2 :
After replacing a row of zeros by the coefficients of dA(s) , complete the Routh’s array.
ds
But now, the criterion that, no sign in 1st column of array for stability, no longer remains
sufficient but becomes a necessary. This is because though A(s) is a part of original
characteristic equation, dA(s) is not, which is in fact used to complete the array.
ds
So if sign change occurs in first column, system is unstable with number of sign changes
equal to number of roots of characteristics equation located in right half of
s-plane.
But there is no sign changes, system cannot be predicted as stable. And in such case
stability is to be determined by actually solving A(s) = 0 for its roots. And from the location
of roots of A(s) = 0 in the s-plane the system stability must be determined. Because roots
A(s) = 0 are always dominant roots of characteristic equation.
Application of Routh’s of criterion: Relative stability analysis: If it is required to find relative stability of system about a line s = - . i.e. how many roots
are located in right half of this line s = - , the Routh’s method can be used effectively.
To determine this from Routh’s array, shift the axis of s – plane & then apply Routh’s
array i.e. substitute s = s 1 - , ( = constant) in characteristic equation. Write polynomial in
terms of s1. Complete array from this new equation. The number of sign changes in first
column is equal to number of roots those are located to right of the vertical line
s = - .
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Imaginary j - 0 Determining range of values of K :
In practical system, an amplifier of variable gain K is introduced.
The closed loop transfer function is
C(s) KG(s) = R(s) 1+ KG(s) H(s)
Hence the characteristic equation is
F(s) = 1+ KG(s) H(s) = 0 So the roots of above equation are dependent on the proper selection of value of ‘K’.
So unknown ‘K’ appears in the characteristic equation. In such case Routh’s array is to be
constructed in terms of K & then the range of values of K can be obtained in such away that it
will not produce any sign change in first column of the Routh’s array. Hence it is possible to
obtain the range of values of K for absolute stability of the system using Routh’s criterion. Such
a system where stability depends on the condition of parameter K, is called conditionally stable
system.
Advantages of Routh’s criterion: Advantages of routh’s array method are:
1. Stability of the system can be judged without actually solving the characteristic equation.
2. No evaluation of determinants, which saves calculation time.
3. For unstable system it gives number of roots of characteristic equation having
Positive real part.
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4. Relative stability of the system can be easily judged.
5. By using the criterion, critical value of system gain can be determined hence
Frequency of sustained oscillations can be determined.
6. It helps in finding out range of values of K for system stability.
7. It helps in finding out intersection points of roots locus with imaginary axis.
Limitation of Routh’s criterion:
1. It is valid only for real coefficients of the characteristic equation.
2. It does not provide exact locations of the closed loop poles in left or right half of s-plane.
3. It does not suggest methods of stabilizing an unstable system.
4. Applicable only to linear system.
Ex.1. s6 + 4s5 +3s4 – 16s2- 64s – 48 = 0 Find the number of roots of this equation with positive
real part, zero real part & negative real part
Sol: S6 1 3 -16 -48 S5 4 0 -64 0
S4 3 0 -48 0 S3 0 0 0 dA A(s) = 3S4 – 48 = 0 = 12s3
ds
S6 1 3 -16 -48 S5 4 0 -64 0
S4 3 0 -48 0 S3 12 0 0 0 S2 ( )0 -48 0 0 S1 576 0 0 0 S0 -48
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Lim 576
0 + Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s – plane
i.e. with positive real part. Now
solve A(s) = 0 for the dominant roots
A(s) = 3s4 – 48 =0 Put S2 = Y 3Y2 = 48 Y2 =16, Y = 16 = 4 S2 = + 4 S2 = -4 S = 2 S = 2j So S = 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated
by a sign change is S = 2 as obtained by solving A(s) = 0. Total there are 6 roots as n = 6.
Roots with Positive real part = 1
Roots with zero real part = 2
Roots with negative real part = 6 –2 – 1 = 3
Ex.2 : For unity feed back system, k G(s) = , Find range of values of K, marginal value of K S(1 + 0.4s) ( 1 + 0.25 s) & frequency of sustained oscillations. Sol : Characteristic equation, 1 + G (s) H (s) = 0 & H(s) = 1 K 1 + = 0 s(1 + 0.4s) ( 1 + 0.25s) s [ 1 + 0.65s + 0.1s2} + K = 0
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0.1s3 + 0.65s2 +s + K = 0
S3 0.1 1 From s0, K > 0 S2 0.65 K from s1, S1 0.65 – 0.1K 0 0.65 – 0.1K > 0 0.65 0.65 > 0.1 K S0 K 6.5 > K Range of values of K, 0 < K < 6.5
Now marginal value of ‘K’ is that value of ‘K’ for which system becomes marginally stable. For
a marginal stable system there must be row of zeros occurring in Routh’s array. So value of ‘K’
which makes any row of Routh array as row of zeros is called marginal value of ‘K’. Now K = 0
makes row of s0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant
term in characteristic equation becomes zeros ie one coefficient for s0 vanishes which makes
system unstable instead of marginally stable.
Hence marginal value of ‘K’ is a value which makes any row other than s0 as row of zeros.
0.65 – 0.1 K mar = 0
K mar = 6.5
To find frequency, find out roots of auxiliary equation at marginal value of ‘K’
A(s) = 0.65 s2 + K = 0 ;
0.65 s2 + 6.5 = 0 Because K = 6.5
s2 = -10
s = j 3.162
Comparing with s = j
= frequency of oscillations = 3.162 rad/ sec.
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Ex : 3 For a system with characteristic equation
F(s) = s5 + s4 + 2s3 + 2s2 + 3s +15 = , examine the stability
Solution : S5 1 2 3
S4 1 2 15 S3 0 -12 0 S2 S1
S0
S5 1 2 3 S4 1 2 15 S3 -12 0 S2 (2 + 12) 15 0 S1 (2 + 12)( -12 ) – 15 0 2 + 12 S0 15 Lim 2 + 12 12 0 = 2 + = 2 + = +
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Lim (2 + 12)( -12 ) – 15 Lim -24 - 144 – 15 2
0 = 0 2 + 12 2 + 12 0 – 144 - 0 = = - 12 0 + 12
S5 1 2 3 S4 1 2 15 S3 -12 0 There are two sign changes, so system is unstable. S2 + 15 0 S1 - 12 0 S0 15 Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -sT G(s) = s ( s + 1 ) Sol : The characteristic equation is 1 + G(s) H(s) = 0 e -sT 1 + = 0 s ( s + 1 ) s2 + s + e –sT = 0 Now e – sT can be Expressed in the series form as s2 T2 e –sT = 1 – sT + + …… 2!
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Truncating the series & considering only first two terms we get esT = 1 – sT s2 + s + 1 – sT = 0 s2 + s ( 1- T ) + 1 = 0
So routh’s array is S2 1 1 S 1-T 0 S0 1 1 – T > 0 for stability T < 1 This is the required condition for stability of the system.
Ex: 5 Determine the location of roots with respect to s = -2 given that F(s) = s4 + 10 s3 + 36s2 + 70s + 75 Sol: shift the origin with respect to s = -2 s = s1 – 2 (s – 2) 4 + 10 (s – 2)3 + 36(s – 2)2 + 70 (s –2) + 75 = 0 S4 + 2s3 + 0s2 + 14s + 15 = 0 S4 1 0 15 S3 2 14 0 S2 -7 15 0 S1 18.28 0 0 S 0 15 Two sign change, there are two roots to the right of s = -2 & remaining ‘2’ are to the left of
the line s = -2. Hence the system is unstable.
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NYQUIST STABILITY CRITERIA CAN BE STATED AS FOLLOWS “ If The G(S) H(S) Contour In The G(S)H(S) Plane Corresponding To Nyquist Contour In The
S-Plane Encircles The Point -1+J0 In The Anticlockwise Direction As Many Times As The
Number Of Right Half S-Plane Poles Of G(S)H(S). Then the Closed Loop System Is Stable”
1. No of encirclement of -1+j0 point: This implies that the system is stable if there are no
poles of G(S)H(S) in the right half S-Plane. If there are poles on right half s-plane then
the system is unstable
2. Anticlockwise encirements of -1+j0 point: In this case the system is stable if the number
of anticlockwise encirclement is same as the number of poles of G(S)H(S) in the right
half of S-Plane. If the number of anticlockwise encirclement is not equal to the number of
poles on right half of S-plane then the system is unstable.
3. clockwise encirclement of the -1+j0 point: In this case the system is always unstable.
Also this incase if n poles of G(S)H(S) in right half of S-Plane, then the number of
clockwise encirclement is equal to no of poles of closed loop system on right half S-Plane
NOTE: If the Nyquist contour must not endure the poles (or) zeros lying on imaginary axis.
1+G(S)H(S)
PROBLEMS
(-1,0)
G(S) Contour
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 132
SOLUTION: Given that G(S) H(S) = K/S(S+2) (S+10) = K/(Sx2(S/2+1)x10(S/10+1))
= 0.05 K/S(1+0.5S)(1+0.1S)
The open loop transfer function has a pole at origin. Hence choose the Nyquist contour on S-
plane enclosing the entire right half plane except the origin
The Nyquist contour has four sections C1, C2, C3, C4 the mapping of each section is
performed separately and the overall Nyquist plot is obtained by combining the individual
sections
Mapping of section C1 In section c1,ω varies from 0 to ∞. The mapping of section C1 is given by the locus
G(jω)H(jω ) as ω is varied from 0 to ∞. This polar plot of G(jω)H(jω)
G(S)H(S)= 0.05K/S(1+0.5S)(1+0.1S)
Let S=jω
G(jω)H(jω S)= 0.05K/ jωS(1+0.5 jω )(1+0.1 jω )
= 0.05K/-0.6ω2+jω(1-0.05ω2)
when the locus of G(jω) H(jω ) crosses real axis the imaginary term will be zero and the
corresponding frequency is the phase crossover frequency ‘
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 133
ωPC.
At ω=ωpc ωpc (1-0.05 ωpc2)=0 1-0.05 ωpc
2=0
At ω=ωpc = 4.427 rad/sec
G(jω)H(jω) = 0.05K/-0.6ω2 = -0.00417 K
The open loop system is type -1 and third order system. Also it is a minimum phase system with
all poles. Hence the polar plot of G(jω)H(jω) starts at -90˚axis at infinity crosses real axis at -
0.00417 K and ends at origin in second quadrant. The section C1 and its mapping show below
diagram A and B
R∞
C1
C3
C4
σ
ω=+∞
ω=0
ω=-∞
ω=∞
ω=0
σ
ω=∞
-0.00417K
jv
Mapping of section c1 in G(S)H(S)
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Mapping of section C2 The mapping of section C2 from S-plane to G(S)H(S) –Plane is obtained by letting
Lt
S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the
G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ ST ]
G(S)H(S) = 0.05K/S(1+0.5S)(1+0.1S) ≈ 0.05K/S x 0.5S x 0.1S = K/S3
Lt
Let S=R∞ Rejθ
G(S)H(S)| S=R∞ Rejθ K/S3| = = 0e-j3θ
when θ=+∏/2 , G(S)H(S) = 0e-j3∏/2-----------------1a
when θ=-∏/2 , G(S)H(S) = 0e+j3∏/2--------------------------1b
Section c1 in s-plane
K Lt R∞ (Rejθ)3
M.I.E.T ENGG COLLEGE DEPT OF EEE
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From the equation 1a and 1b we can study that section c2 in S-plane fig1a
is mapped as circular are of zero radius around origin in G(S)H(S) plane with argument (phase
varying from -3∏/2 and +3∏/2 as shown in fig 1b
Mapping of section C3
In Section C3, ω varies from -∞ to 0. The mapping of section C3 is given by the locus of
G(jω)H(jω) as a ω is varied from -∞ to 0. This locus is the inverse polar plot of G(jω)H(jω)
The inverse polar plot is given by the mirror image of polar plot with respect to real axis.
The section C3 in S-Plane and its corresponding contour in G(S)H(S) plane shown in fig 2a and
fig 2b
R∞ ∞
C2 S=Plane jω
Section c2 in S-Plane
jv
u
G(S)H(S) plane
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 136
Mapping of section C4
The mapping of section c4 from s-plane to G(S)H(S) plane is obtained by letting
Lt
S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the
G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ 1 ]
0.05K
G(S)H(S) = ----------------------- = 0.05K/S
S(1+0.5S)(1+0.1S)
Lt
Let S= R0 Rejθ
G(S)H(S) = αe-jθ
when θ=+∏/2 , G(S)H(S) = αe-j3∏/2-----------------2a
when θ=∏/2 , G(S)H(S) = αe+j3∏/2--------------------------2b
From equation above we can say that sec c4 in S-Plane fig 3a is mapped as a circular are of
infinity radius with argument(phase) varying from +∏/2 to-∏/2 shown in figure
σ
jω
ω=0
ω=-∞
S-Plane
jv
ω=0
ω=-∞ -0.00417k
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 137
Complete Nyquist plot The entire Nyquist plot in G(S)H(S) plane can be obtained by combining the mappings of individual sections shown
R->∞
jv
R->0
C4
S-Plane
G(S)H(S) plane
Mapping of section c4
R0
-0.00417K
-1+j0 for K<240 -1+j0K>240
G(S) H(S) Contour
K G(S)H(S) = ---------------- S(S+2)(S+10)
Section C4 in s-plane
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 138
STABILITY ANALYSIS When -0.00417K =-1, the contour passes through (-1+j0) point and corresponding value of K
is the limiting value for K for stability
Limiting value of K = 1/0.00417 = 240
When K<240 When K is less than 240, the contour crosses real axis at appoint between 0 & -1+j0 . On
traveling through Nyquist plot along the indicated direction if it found that the point -1+j0 is not
encircled. Also the open loop transfer function has no poles on the right half of S-Plane.
Therefore the closed loop system is stable.
When K>240 When K is graeter than 240, the contour crosses real axis at a point B/W -1+j0 and -∞. On
traveling through Nyquist plot along the indicated direction it is found that the point -1+j0 is
encircled in clockwise direction two time. [ since there are two clockwise encirclement and no
right half open loop poles, the closed loop system has 2 poles on right half of S-Plane. Therefore
the closed loop system is unstable
RESULT
The value of K for stability is 0<K<240
RELATIVE STABIITY The relative stability indicates the closeness of the system to stable region. It is an indication
of strength or degree of stability
in time domain, the relative stability may be measured by relative settling times of each root
or pair of roots. The settling time is inversely proportional to the location of roots of
characteristic equation. If the root is located for away from the imaginary axis, then the transients
dies out faster and so the relative stability of system will improve. the transient response and so
the relative stability for various of roots in S-Plane is shown below fig
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 139
ROOT LOCUS INTRODUCTION The characteristics of the transient response of a closed loop control system is related to location
of the closed loop poles. If the system has a variable loop gain, then the location of the closed
loop poles depends on the value of the loop gain chosen. It is important, that the designer knows
how the closed loop poles move in the s-plane as the loop gain is varied. W. R. Evans introduced
a graphical method for finding the roots of the characteristic equation known as root locus
method. The root locus is used to study the location of the poles of the closed loop transfer
function of a given linear system as a function of its parameters, usually a loop gain, given its
open loop transfer function. The roots corresponding to a particular value of the system
parameter can then be located on the locus or the value of the parameter for a desired root
location can be determined from the locus. It is a powerful technique, as an approximate root
locus sketch can be made quickly and the designer can visualize the effects of varying system
parameters on root locations or vice versa. It is applicable for single loop as well as multiple loop
system.
t t
t t
jω
σ
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 140
ROOT LOCUS CONCEPT
To understand the concepts underlying the root locus technique, consider the second
order system shown in Fig. 1.
Fig. 23 Second order control system
The open loop transfer function of this system is
)1(a)s(s
KG(s)
Where, K and a are constants. The open loop transfer function has two poles one at origin s = 0
and the other at s = -a. The closed loop transfer function of the system shown in Fig.1 is
(2)Kass
K
G(s)H(s)1
G(s)
R(s)
C(s)2
The characteristic equation for the closed loop system is obtained by setting the
denominator of the right hand side of Eqn.(2) equal to zero. That is,
(3)0 KassG(s)H(s)1 2
The second order system under consideration is always stable for positive values of a and
K but its dynamic behavior is controlled by the roots of Eqn.(3) and hence, in turn by the
magnitudes of a and K, since the roots are given by
(4)K2
a
2
a
2a
4K)(a
2
as,s
22
21
R(s) E(s)
− a)s(s
K
C(s)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 141
From Eqn.(4), it is seen that as the system parameters a or K varies, the roots change.
Consider a to be constant and gain K to be variable. As K is varied from zero to infinity, the two
roots s1 and s2 describe loci in the s-plane. Root locations for various ranges of K are:
1) K= 0, the two roots are real and coincide with open loop poles of the system s1 =
0, s2 = -a.
2) 0 K < a2/4, the roots are real and distinct.
3) K= a2/4, roots are real and equal.
4) a2/4 < K < , the rots are complex conjugates.
The root locus plot is shown in Fig.2
Fig. 24 Root loci of s2+as+K as a function of K
Figure 2 has been drawn by the direct solution of the characteristic equation. This
procedure becomes tedious. Evans graphical procedure helps in sketching the root locus quickly.
The characteristic equation of any system is given by
(5)0Δ(s)
Where, (s) is the determinant of the signal flow graph of the system given by Eqn.(5). ∆=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations of
two no touching loops – sum of gain products of all possible combination of three no
touching loops) + ∙∙∙
Or
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 142
(6)PPP1Δ(s)m m3m m2m m1
Where, Pmr is gain product of mth possible combination of r no touching loops of the graph.
The characteristic equation can be written in the form
(7)0B(s)
KA(s)1
0P(s)1
For single loop system shown in Fig.3
(8)G(s)H(s)P(s)
Where, G(s)H(s) is open loop transfer function in block diagram terminology or transmittance in
signal flow graph terminology.
Fig. 25 Single loop feedback system
From Eqn.(7) it can be seen that the roots of the characteristic equation (closed loop
poles)occur only for those values of s where
(9)1P(s)
Since, s is a complex variable, Eqn.(9) can be converted into the two Evans conditions
given below.
)10(1)( sP
)11(2,1,0);12(180)( qqsP
Roots of 1+P(s) = 0 are those values of s at which the magnitude and angle condition
given by Eqn.(10) and Eqn.(11). A plot of points in the complex plane satisfying the angle
−
R(s) E(s) G(s)
C(s)
H(s)
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 143
criterion is the root locus. The value of gain corresponding to a root can be determined from the
magnitude criterion.
To make the root locus sketching certain rules have been developed which helps in
visualizing the effects of variation of system gain K ( K > 0 corresponds to the negative feed
back and K < 0 corresponds to positive feedback control system) and the effects of shifting
pole-zero locations and adding in anew set of poles and zeros.
GENERAL RULES FOR CONSTRUCTING ROOT LOCUS
1) The root locus is symmetrical about real axis. The roots of the characteristic equation are
either real or complex conjugate or combination of both. Therefore their locus must be
symmetrical about the real axis.
2) As K increases from zero to infinity, each branch of the root locus originates from an
open loop pole (n nos.) with K= 0 and terminates either on an open loop zero (m nos.)
with K = along the asymptotes or on infinity (zero at ). The number of branches
terminating on infinity is equal to (n – m).
3) Determine the root locus on the real axis. Root loci on the real axis are determined by
open loop poles and zeros lying on it. In constructing the root loci on the real axis choose
a test point on it. If the total number of real poles and real zeros to the right of this point
is odd, then the point lies on root locus. The complex conjugate poles and zeros of the
open loop transfer function have no effect on the location of the root loci on the real axis.
M.I.E.T ENGG COLLEGE DEPT OF EEE
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4) Determine the asymptotes of root loci. The root loci for very large values of s must be
asymptotic to straight lines whose angles are given by
)12(1-mn0,1,2,;mn
1)(2q180asymptotesofAngle
A
q
5) All the asymptotes intersect on the real axis. It is denoted by a , given by
)13(mn
)zz(z)pp(pmn
zerosofsumpolesofsumσ
m21n21
a
6) Find breakaway and break-in points. The breakaway and break-in points either lie on the
real axis or occur in complex conjugate pairs. On real axis, breakaway points exist
between two adjacent poles and break-in in points exist between two adjacent zeros. To
calculate this polynomial 0ds
dK must be solved. The resulting roots are the breakaway /
break-in points. The characteristic equation given by Eqn.(7), can be rearranged as
)z(s)z)(szK(sA(s)
and )p(s)p)(sp(s B(s) where,
(14)0KA(s)B(s)
m21
n21
The breakaway and break-in points are given by
)15(0Bds
dABA
ds
d
ds
dK
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 145
Note that the breakaway points and break-in points must be the roots of Eqn.(15), but
not all roots of Eqn.(15) are breakaway or break-in points. If the root is not on the root
locus portion of the real axis, then this root neither corresponds to breakaway or break-
in point. If the roots of Eqn.(15) are complex conjugate pair, to ascertain that they lie on
root loci, check the corresponding K value. If K is positive, then root is a breakaway or
break-in point.
7) Determine the angle of departure of the root locus from a complex pole
)16()zerosotherfromquestioninpolecomplexatovectorsofanglesof(sum
poles)otherfromquestioninpolecomplexatovectorsofanglesof(sum
180pcomplexafromdepartureofAngle
8) Determine the angle of arrival of the root locus at a complex zero
(17)poles)otherfromquestioninzerocomplexatovectorsofanglesof(sum
zeros)otherfromquestioninzerocomplexatovectorsofanglesof(sum
180zerocomplexatarrivalofAngle
9) Find the points where the root loci may cross the imaginary axis. The points where the
root loci intersect the j axis can be found by
a) use of Routh’s stability criterion or
b) letting s = j in the characteristic equation , equating both the real part and
imaginary part to zero, and solving for and K. The values of thus found give
the frequencies at which root loci cross the imaginary axis. The corresponding K
value is the gain at each crossing frequency.
10) The value of K corresponding to any point s on a root locus can be obtained using the
magnitude condition, or
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 146
)18(zerostopointsbetweenlengthofproduct
polestopointsbetweenlengthsofproductK
PHASE MARGIN AND GAIN MARGIN OF ROOT LOCUS Gain Margin
It is a factor by which the design value of the gain can be multiplied before the closed
loop system becomes unstable.
(19)KofvalueDesign
overcrossimaginaryatKofValueMarginGain
The Phase Margin
Find the point j1 on the imaginary axis for which 1jHjG for the design value
of K i.e. design
Kj/AjB .
The phase margin is (20))H(jωjωargG180φ11
Problem No 1 Sketch the root locus of a unity negative feedback system whose forward path transfer function
is s
KG(s) .
Solution:
1) Root locus is symmetrical about real axis.
2) There are no open loop zeros(m = 0). Open loop pole is at s = 0 (n = 1). One branch of
root locus starts from the open loop pole when K = 0 and goes to asymptotically when
K .
3) Root locus lies on the entire negative real axis as there is one pole towards right of any
point on the negative real axis.
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 147
4) The asymptote angle is A = .01,)12(180
mnq
mn
q
Angle of asymptote is A = 180.
5) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
0.01
0
6) The root locus does not branch. Hence, there is no need to calculate the break points. 7) The root locus departs at an angle of -180 from the open loop pole at s = 0.
8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross
over. The root locus plot is shown in Fig.1
Figure 26 Root locus plot of K/s Comments on stability: The system is stable for all the values of K > 0. Th system is over damped. Problem No 2
The open loop transfer function is 21)(s
2)K(sG(s)
. Sketch the root locus plot
Solution:
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IV– CONTROL SYSTEMS 148
1) Root locus is symmetrical about real axis. 2) There is one open loop zero at s=-2.0(m=1). There are two open loop poles at
s=-1, -1(n=2). Two branches of root loci start from the open loop pole when K= 0. One branch goes to open loop zero at s =-2.0 when K and other goes to (open loop zero) asymptotically when K .
3) Root locus lies on negative real axis for s ≤ -2.0 as the number of open loop poles plus
number of open loop zeros to the right of s=-0.2 are odd in number.
4) The asymptote angle is A = .01,)12(180
mnq
mn
q
Angle of asymptote is A = 180.
5) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
0.01
)2()11(
6) The root locus has break points.
The root loci brakes out at the open loop poles at s=-1, when K =0 and breaks in onto the real axis at s=-3, when K=4. One branch goes to open loop zero at s=-2 and other goes to along the asymptotically.
7) The branches of the root locus at s=-1, -1 break at K=0 and are tangential to a line s=-
1+j0 hence depart at 90. 8) The locus arrives at open loop zero at 180. 9) The root locus does not cross the imaginary axis, hence there is no need to find the
imaginary axis cross over. The root locus plot is shown in Fig.2.
4K3,s0;K1,s
02)(s
1)(s2)1)(s2(s
0ds
dKbygivenispointBreak
2)(s
1)(sK
21
2
2
2
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 149
Figure 27 Root locus plot of K(s+2)/(s+1)2
Comments on stability:
System is stable for all values of K > 0. The system is over damped for K > 4. It is critically damped at K = 0, 4.
Problem No 3
The open loop transfer function is 2)s(s
4)K(sG(s)
. Sketch the root locus.
Solution:
1) Root locus is symmetrical about real axis. 2) There are is one open loop zero at s=-4(m=1). There are two open loop poles at s=0, -
2(n=2). Two branches of root loci start from the open loop poles when K= 0. One branch goes to open loop zero when K and other goes to infinity asymptotically when K .
3) Entire negative real axis except the segment between s=-4 to s=-2 lies on the root locus.
4) The asymptote angle is A = .01,1,0,)12(180
mnq
mn
q
Angle of asymptote are A = 180.
5) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
0.21
)4()2(
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 150
6) The brake points are given by dK/ds =0. 7) Angle of departure from open loop pole at s =0 is 180. Angle of departure from pole at
s=-2.0 is 0. 8) The angle of arrival at open loop zero at s=-4 is 180 9) The root locus does not cross the imaginary axis. Hence there is no imaginary cross over.
The root locus plot is shown in fig.3.
Figure 3 Root locus plot of K(s+4)/s(s+2) Comments on stability: System is stable for all values of K. 0 > K > 0.343 : > 1 over damped K = 0.343 : = 1 critically damped 0.343 > K > 11.7 : < 1 under damped K = 11.7 : = 1 critically damped
11.7K6.828,s
0.343;K1.172,s
04)(s
2s)(s4)2)(s(2s
ds
dK
4)(s
2)s(sK
2
1
2
2
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 151
K > 11.7 : >1 over damped. Problem No 4
The open loop transfer function is 3.6)(ss
0.2)K(sG(s)
2 . Sketch the root locus.
Solution:
1) Root locus is symmetrical about real axis. 2) There is one open loop zero at s = -0.2(m=1). There are three open loop poles at
s = 0, 0, -3.6(n=3). Three branches of root loci start from the three open loop poles when
K= 0 and one branch goes to open loop zero at s = -0.2 when K and other two go to
asymptotically when K .
3) Root locus lies on negative real axis between -3.6 to -0.2 as the number of open loop
poles plus open zeros to the right of any point on the real axis in this range is odd.
4) The asymptote angle is A = 1,01,)12(180
mnq
mn
q
Angle of asymptote are A = 90, 270.
5) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
7.12
)2.0()6.3(
6) The root locus does branch out, which are given by dK/ds =0.
The root loci brakeout at the open loop poles at s = 0, when K =0 and breakin onto the
real axis at s=-0.432, when K=2.55 One branch goes to open loop zero at s=-0.2 and
other goes breaksout with the another locus starting from open loop ploe at s= -3.6. The
break point is at s=-1.67 with K=3.66. The loci go to infinity in the complex plane with
constant real part s= -1.67.
ly.respective 3.662.55,0,Kand1.670.432,0,s
01.44s4.8s2s
0.2)(s
)3.6s(s0.2)7.2s)(s(3s
ds
dK
0.2s
)3.6s(s-K
23
2
232
23
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IV– CONTROL SYSTEMS 152
7) The branches of the root locus at s=0,0 break at K=0 and are tangential to imaginary axis
or depart at 90. The locus departs from open loop pole at s=-3.6 at 0. 8) The locus arrives at open loop zero at s=-0.2 at 180. 9) The root locus does not cross the imaginary axis, hence there is no imaginary axis cross
over. The root locus plot is shown in Fig.4.
Figure 4 Root locus plot of K(s+0.2)/s2(s+3.6)
Comments on stability:
System is stable for all values of K. System is critically damped at K= 2.55, 3.66. It is under
damped for 2.55 > K > 0 and K >3.66. It is over damped for 3.66 > K >2.55.
Problem No 5
The open loop transfer function is 25)6ss(s
KG(s) . Sketch the root locus.
Solution:
1) Root locus is symmetrical about real axis. 2) There are no open loop zeros (m=0). There are three open loop poles at s=-0,
M.I.E.T ENGG COLLEGE DEPT OF EEE
IV– CONTROL SYSTEMS 153
-3j4(n=3). Three branches of root loci start from the open loop poles when K= 0 and all the three branches go asymptotically when K .
3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the real
axis.
4) The asymptote angle is A = .2,1,01,1,0,)12(180
mnq
mn
q
Angle of asymptote are A = 60, 180, 300.
5) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
0.23
)33(
6) The brake points are given by dK/ds =0.
j18.0434K
j2.0817and2s
02512s3sds
dK
25s)6s(s25)6ss(sK
1,2
1,2
2
232
For a point to be break point, the corresponding value of K is a real number greater than or equal to zero. Hence, S1,2 are not break points.
7) Angle of departure from the open loop pole at s=0 is 180. Angle of departure from
complex pole s= -3+j4 is
zeros) from inquestion polecomplex a to vectorsof angles theof (sum
poles)other fromquestion in polecomplex a to vectorsof angles theof (sum
180p
87.36)90
3
4tan180(180 1
p
Similarly, Angle of departure from complex pole s= -3-j4 is 36.87or323.13)270(233.13180φp
8) The root locus does cross the imaginary axis. The cross over point and the gain at the
cross over can be obtained by Rouths criterion The characteristic equation is 0K25s6ss 23 . The Routh’s array is
6s6
K150s
K6s
251s
0
1
2
3
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IV– CONTROL SYSTEMS 154
For the system to be stable K < 150. At K=150 the auxillary equation is 6s2+150=0. s = ±j5. or substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve for and K.
The plot of root locus is shown in Fig.5.
Figure 5 Root locus plot of K/s(s2+6s+25) Comments on stability: System is stable for all values of 150 > K > 0. At K=150, it has sustained oscillation of 5rad/sec. The system is unstable for K >150. Problem No 1 Sketch the root locus of a unity negative feedback system whose forward path transfer function
is j)3j)(s31)(s(s
2)K(sG(s)H(s)
. Comment on the stability of the system.
1500,Kj50,ω
025ωjωK)6ω(
0Kjω25jω6jω
0K25s6ss
22
23
23
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Solution:
9) Root locus is symmetrical about real axis. 10) There is one open loop zero at s = -2 (m = 1). There are three open loop poles at
s = -1, -3 ± j (n=3). All the three branches of root locus start from the open loop poles when K = 0. One locus starting from s = -1 goes to zero at s = -2 when K , and other two branches go to asymptotically (zeros at ) when K .
11) Root locus lies on the negative real axis in the range s=-1 to s= -2 as there is one pole to the right of any point s on the real axis in this range.
12) The asymptote angle is A = .0,11mnq,mn
1)(2q180
Angle of asymptote is A = 90, 270.
13) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
2.51
2)(3)31(
14) The root locus does not branch. Hence, there is no need to calculate break points. 15) The angle of departure at real pole at s=-1 is 180. The angle of departure at the complex
pole at s=-3+j is 71.57.
zeros) from inquestion polecomplex a to vectorsof angles theof (sum
poles)other fromquestion in polecomplex a to vectorsof angles theof (sum
180p
The angle of departure at the complex pole at s=-3-j is -71.57. 16) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross
over. The root locus plot is shown in Fig.1
57.71135)90(153.43180
900
2tanθ,135or -45
1-
1tan
153.43)atan2(-2,1θ
153.43or 57.262-
1tanθ
p
1
3
1
1
1
1
57.71522)270(206.57180p
M.I.E.T ENGG COLLEGE DEPT OF EEE
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Figure 1 Root locus plot of K(s+2)/(s+1)(s+3+j)(s+3-j)
Comments on stability: The system is stable for all the values of K > 0.
Problem No 2
The open loop transfer function is10)0.6s0.5)(ss(s
KG(s)H(s)
2 Sketch the root locus
plot. Comment on the stability of the system. . Solution:
10) Root locus is symmetrical about real axis. 11) There are no open loop zeros (m=0). There are four open loop poles (n=4) at s=0,
-0.5, -0.3 ± j3.1480. Four branches of root loci start from the four open loop poles when K= 0 and go to (open loop zero at infinity) asymptotically when K .
12) Root locus lies on negative real axis between s = 0 to s = -0.5 as there is one pole to the right of any point s on the real axis in this range.
13) The asymptote angle is A = .3,2,1,01,)12(180
mnq
mn
q
Angle of asymptote is A = 45, 135, 225, ±315.
14) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
M.I.E.T ENGG COLLEGE DEPT OF EEE
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275.04
)3.03.05.0(
The value of K at s=-0.275 is 0.6137. 15) The root locus has break points.
K = -s(s+0.5)(s2+0.6s+10) = -(s4+1.1s3+10.3s2+5s) Break points are given by dK/ds = 0
0520.6s3.3s4sds
dK 23
s= -0.2497, -0.2877 j 2.2189 There is only one break point at -0.2497. Value of K at s = -0.2497 is 0.6195.
16) The angle of departure at real pole at s=0 is 180 and at s=-0.5 is 0. The angle of
departure at the complex pole at s = -0.3 + j3.148 is -91.8
zeros) from inquestion polecomplex a to vectorsof angles theof (sum
poles)other fromquestion in polecomplex a to vectorsof angles theof (sum
180p
The angle of departure at the complex pole at s = -0.3 - j3.148 is 91.8 17) The root locus does cross the imaginary axis, The cross over frequency and gain is
obtained from Routh’s criterion.
The characteristic equation is s(s+0.5)(s2+0.6s+10)+K =0 or s4+1.1s3+10.3s2+5s+K=0 The Routh’s array is
Ks5.75
1.1K-28.75s
K5.75s
51.1s
K10.31s
0
1
2
3
4
8.91)9086.4(95.4180
900
6.296tanθ, 4.68
0.2
3.148tan
4.95or6.840.3-
3.148tanθ
p
1
3
1
2
1
1
91.8 )270273.6(264.6180p
M.I.E.T ENGG COLLEGE DEPT OF EEE
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The system is stable if 0 < K < 26.13 The auxiliary equation at K 26.13 is 5.75s2+26.13 = 0 which gives s = ± j2.13 at imaginary axis crossover. The root locus plot is shown in Fig.2.
Figure 28 Root locus plot of K/s(s+0.5)(s2+0.6s+10)
Comments on stability: System is stable for all values of 26.13 >K > 0. The system has sustained oscillation at = 2.13 rad/sec at K=26.13. The system is unstable for K > 26.13.
Problem No 3
The open loop transfer function is 20)4s)(s4s(s
KG(s)
2 . Sketch the root locus.
Solution:
10) Root locus is symmetrical about real axis.
M.I.E.T ENGG COLLEGE DEPT OF EEE
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11) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0, -4, -2 j4. Three branches of root loci start from the three open loop poles when K= 0 and to infinity asymptotically when K .
12) Root locus lies on negative real axis between s = 0 to s = -4.0 as there is one pole to the
right of any point s on the real axis in this range.
13) The asymptote angle is A = 3,2,1,01,)12(180
mnq
mn
q
Angle of asymptote are A = 45, 135, 225, 315.
14) Centroid of the asymptote is mn
zeros)of(sumpoles)of(sumσA
0.24
)0.40.20.2(
15) The root locus does branch out, which are given by dK/ds =0.
The root loci brakeout at the open loop poles at s = -2.0, when K = 64 and breakin and breakout at s=-2+j2.45, when K=100
16) The angle of departure at real pole at s=0 is 180 and at s=-4 is 0. The angle of departure at the complex pole at s = -2 + j4 is -90.
zeros) from inquestion polecomplex a to vectorsof angles theof (sum
poles)other fromquestion in polecomplex a to vectorsof angles theof (sum
180p
100Kj2.45,2.0s
64;K2.0,s
40)16s2)(4s(s
08040s32s16s8s4s
08072s24s4s
0ds
dKbygivenispointBreak
80s)36s8s(s
20)4s4)(ss(sK
2
1
2
223
23
234
2
90)90.463(116.6-180
900
8tanθ,4.36
2
4tanθ
116.6)atan2(4,-2θ
116.6or63.42-
4tanθ
p
1
3
1
2
1
1
1
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The angle of departure at the complex pole at s = -2 – j4 is 90
17) The root locus does cross the imaginary axis, The cross over point and gain at cross over
is obtained by either Routh’s array or substitute s= j in the characteristic equation and
solve for and gain K by equating the real and imaginary parts to zero.
Routh’s array
The characteristic equation is 0K80s36s8ss 234
For the system to be stable K > 0 and 2080-8K > 0. The imaginary crossover is given by 2080-8K=0 or K = 260. At K = 260, the auxiliary equation is 26s2+260 = 0. The imaginary cross over occurs at s= j10.
or
Ks26
8K2080s
K26s
808s
K361s
isarrayRouthsThe
0
1
2
3
4
260K0K36ωω
10js;10j0,ω080ω8ω
zerotopartsimaginaryandrealEquate
080ω8ωjK36ωω0Kjω80jω36jω8jω
jωsput
0K80s36s8ss
24
3
324
234
234
90270
)270296.6(243.4-180p
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The root locus plot is shown in Fig.3.
Figure 29 Root locus plot of K/s(s+4)(s2+4s+20)
Comments on stability: For 260 > K > 0 system is stable
K = 260 system has stained oscillations of 10 rad/sec.
K > 260 system is unstable.
UNIT-V
COMPENSATOR DESIGN Basic ideas of compensator design
In the following some analytical design methods will be discussed, which will lead directly to the
design solution in a strongly systematic way. In contrary to these direct design methods, the
methods hitherto discussed, e.g. the design method using frequency-domain characteristics or the
root-locus method, are indirect methods based more on systematic trial and error techniques
iterating through some design steps. The success depends strongly on the experience and skill of
the designer. The starting point was always the open loop, which was modified iteratively by
M.I.E.T ENGG COLLEGE DEPT OF EEE
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adding lead and lag elements until the closed loop shows the desired behavior. Whereas in the
direct design methods one will always start from the behavior of the closed loop. Mostly a
desired transfer Function
is given. In general this follows from the specification of some performance indices, which, for
example are required for the step response . For a series of appropriate transfer
functions a table of numerator and denominator polynomials of the associated transfer function
and its distribution of zeros and poles to yield a specific response are given. Then for
known plant behavior the required controller can be directly calculated.
The controllers designed in this way are not always optimal. They guarantee the compliance of
the desired specification, e.g. maximum overshoot and settling time. A drawback of theses
methods is that they cannot be applied directly to systems with dead time.
Design by specifying the closed-loop transfer function
The desired transfer function of the closed loop is given by
(10.1)
where and are polynomials in . In the following design methods, the distribution of
poles and zeros of will be chosen, such that the performance indices for the step response
are fulfilled. Often a detailed investigation of the distribution of poles and zeros of the
desired transfer function is not necessary, especially when the transfer function does not contain
zeros and when - because of the requirement - just yields and in the simplest
case
(10.2)
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For a closed loop with the transfer function according to Eq. (10.2) different possibilities exist,
the so called standard forms, which can be used by a table lookup for the step response ,
distribution of poles of and the coefficients of the denominator polynomial .
A first possibility is a distribution of poles with a real multiple pole at . Here and in the
following sections, the term is a relative frequency, not the natural frequency. Thus one
obtains for the step response of the desired behavior
(10.3)
This is a series connection of elements with the same time constant . This
representation is also called a binomial form. The standard polynomials of different order
are given in Table A.2. As this table further shows, the normalized step response will
become slower with increasing order. A design using this binomial form is only considered
when the step response is required to have no overshoot.
A further possibility of a standard form for of Eq. (10.2) is the Butterworth form. In this
form, all poles of are equally distributed on a semicircle with radius in the left-half
plane and centered at the origin. Table A.2 contains the standard polynomials and the
associated normalized step responses .
Numerous further possibilities for the development of standard forms of Eq. (10.2) can be
derived from the integral criteria given in Table 7.2. For example, the minimum performance
index is the basis of a standard form that is also shown in Table A.2. Further, often the
M.I.E.T ENGG COLLEGE DEPT OF EEE
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minimum settling time is used as the criterion. This table contains for the
corresponding standard form.
Furthermore, for pole assignment the Weber method can be used. This specifies the desired closed-loop transfer function
(10.4)
by a real pole with multiplicity and a pair of complex poles. Table A.1 contains for
different values of and the normalized step responses . By a proper choice of , and
a closed-loop transfer function can be found that fulfils in many instances the desired performance.
The method of Truxal and Guillemin
For the closed loop shown in Figure 10.1
Figure 10.1: Block diagram of the closed loop to be designed
the behavior is described by the transfer function
(10.5)
Where the numerator and denominator polynomials and must have no common roots.
Furthermore, is normalized to and must be valid.
It is assumed that is stable and minimum phase. For the controller to be designed, the
transfer function
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(10.6)
is chosen and normalized to . Because of the reliability of the controller the relation
must be valid. Now, the controller must be designed such that
the closed loop behaves like a given transfer function for Eq. (10.1), whereby should be
freely chosen under the condition of the reliability of the controller. From the closed-loop
transfer function
(10.7)
one obtains the controller transfer function
(10.8)
or with the numerator and denominator polynomials given above
(10.9)
The condition of reliability for the controller is
or
(10.10)
The pole excess ( ) of the desired closed-loop transfer function must be larger than or
equal to the pole excess ( ) of the plant. Within these constraints the order of is free.
According to Eq. (10.8) the controller contains the inverse plant transfer function . This
is a total compensation of the plant as shown in the block diagram of Figure 10.2. For the
realization
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Figure 10.2: Compensation of the plant
of the controller Eq. (10.9) is used, not the controller structure as shown in this figure with the
plant inverse . As the controller implicitly contains the plant inverse, i.e. the plant zeros
are in the set of the controller poles and the plant poles are in the set of the controller zeros, the
plant must be stable and minimum phase as mentioned at the beginning. Otherwise, the
manipulated variable and/or the controlled variable will show unstable behavior.
Example 10.3.1 The plant transfer function is given as
(10.11)
The pole excess of the plant is . According to (10.10) the pole excess of the desired
closed-loop transfer function must be
The coefficients of the transfer function that obeys the reliability condition (10.10) are
subjected to practical constraints, like the maximum range of the manipulated variable, plant
parameter errors and measurement noise in the controlled variable, which is disturbing the
controller output. The procedure for the design of will be demonstrated by the following
example.
Example 10.3.2 For a plant with the transfer function
(10.12)
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a controller should be designed such that the closed loop shows optimal behavior in the sense of
the performance index and has a rise time of .
First, it follows from the reliability condition Eq. (10.10) and from that the
pole excess of the desired transfer function is
Inspecting Table A.2 one obtains from the form for and the standard
polynomial
(10.13)
From the associated step response it follows from Table A.2 that the normalized rise
time
and with this value from the specified rise time the relative frequency is . Eq. (10.13) is now
As for the chosen standard form for the numerator polynomial is , so it follows
from Eq. (10.9) that the compensator transfer function is
or
This controller contains an integrator. The time responses are shown in Figure 10.3.
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Figure 10.3: Closed-loop behavior for the example 10.3.2: step response of the controlled
variable on step in the set point, step response of the associated controlled variable,
step response of the uncontrolled plant
If as a further example the plant given by Eq. (10.11) instead of Eq. (10.12) is taken, then for the
same the controller is
For these two very different plants the same closed-loop behavior for the controlled variable can
be achieved.
In the considerations of this section it has been hitherto assumed that is stable and
minimum phase. For plants that do not have these properties this design method cannot be
applied in this form. The method must be extended to the following:
A direct compensation of the plant poles and zeros by the controller must be avoided, otherwise
stability problems would arise. In these cases, the closed-loop transfer function cannot be
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arbitrary. For a stable non-minimum phase plant the transfer function must be given such
that the zeros of contain the right-half-plane zeros of . Whereas for an unstable
plant the zeros of the transfer function must contain the right-half-plane poles of
. Of course, this restricts the choice of as the following examples demonstrate.
Example 10.3.3 For an all-pass plant with the transfer function
a controller is to be designed such that the closed loop has the desired transfer function
Using Eq. (10.9) one gets for the controller transfer function
This controller gives a direct compensation (cancellation) of the plant zero. This is undesirable as
already discussed above, and must be selected as
With Eq. (10.9) one obtains the controller transfer function as
Because of this choice of , the closed loop shows also all-pass behavior. This effect is
more intense the smaller the time constant. Figure 10.4 shows the time responses of this
control system.
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Figure 10.4: Closed-loop behaviour of the example 10.3.3: step response of the controlled
variable on step in the set point, step response of the associated controlled variable,
step response of the uncontrolled plant ( ; )
Example 10.3.4 The transfer function of the unstable plant
is given and a controller is required for which fulfills the reliability condition
and for which the zeros of must contain the plant pole . This is expressed by the approach
Whereby is chosen such that
M.I.E.T ENGG COLLEGE DEPT OF EEE
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is valid. In the present case should be chosen such that . From this
follows . In order to have a stable one can take
and obtain
Observing the reliability condition, it follows that
Comparing the coefficients and taking one obtains
and
and finally
and
The parameters and are still free and may now be chosen such that taking an acceptable
behavior of the manipulated variable into account, a given damping ratio and natural frequency
for can be reached. Without going into details,
and
will be chosen for the present case and from this it follows that
and
The desired closed-loop transfer function will be
M.I.E.T ENGG COLLEGE DEPT OF EEE
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and
The conditions for the design are fulfilled and for the controller transfer function one obtains
from Eq. (10.8) or Eq. (10.9)
This design obviously produces a PI controller. The time responses of this control system are
shown in Figure 10.5 for . The relatively large maximum overshoot cannot be avoided
with an acceptable behavior of the manipulated variable.
Figure 10.5: Closed-loop behavior of the example 10.3.4: step response of the controlled
variable on step in the set point, step response of the associated controlled variable Generalized compensator design method
The basic idea
With the following method a control system according to Figure 10.1 using the controller given
by Eq. (10.6) will be designed for a plant described by Eq. (10.5) such that the closed loop
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behaves like the desired transfer function Eq. (10.1). Hereby the orders of the controller
numerator and denominator polynomials are equal, i.e. . The
closed-loop poles are the roots of the characteristic equation, which one obtains from
With respect to the polynomials defined in Eqs. (10.5) and (10.6) this gives
(10.14)
On the other hand it follows from Eq. (10.1) that
(10.15)
This polynomial has order , the coefficients depend linearly on the plant and controller parameters. Comparing both equations, the first coefficient is
(10.16)
and the last because of and
(10.17)
A general representation is given by
(10.18)
whereby
for and
for and
and . The coefficients are obtained from the poles. For the first, second last and last one gets
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(10.19)
(10.20)
(10.21)
While the coefficients according to Eqs. (10.19), (10.20) and (10.21) are directly given by the
closed-loop poles, the coefficients of Eq. (10.16), (10.17) and (10.18) contain the required
controller parameters. Comparing both sides of the latter equation one obtains the synthesis
equation, which is a system of linear equations for unknown controller
coefficients . The number of equations is . A unique solution
exists if .
A detailed analysis shows, however, that a controller obtained in this way does not usually
achieve the desired goals. Because of its small gain a finite steady-state error may occur. This
must be taken into consideration during the design. For plants with integral behavior an order
for the controller is sufficient; for proportional behavior or when disturbances at the
input of an integral plant are taken into consideration, the gain must be influenced so that an
integral behavior of the controller can be obtained. This happens if the order of the controller is
increased by one, i.e. , such that the system of equations is of lower rank. This gives an
additional degree of freedom and allows one to choose the controller gain , which is usually
introduced as a reciprocal gain factor:
(10.22)
Indeed, the order of the closed loop will be increased; it is now double that of the plant order.
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Zeros of the closed loop
In the method presented above, the zeros of the closed loop transfer function for command
changes
(10.23)
are obtained automatically. In fact, the zeros of the plant, i.e. the roots of , can be
considered during the choice of the pole distribution and may be compensated, but the
polynomial arises not in the design and must possibly be compensated after this step. This
can be done by introducing a pre-filter in the feed-forward path according to 10.6a with a
transfer function
Figure 10.6: Compensation of the plant zeros (a) with a controller in the feed-forward path and (b) in the feedback path
The zeros of the controller and plant are compensated in this way. For stability reasons, this is
only possible for left-half-plane zeros. If and are polynomials with only left-half-
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plane zeros and and the corresponding polynomials with only right-half-plane zeros
including the imaginary axis, the polynomials of and can be factorized as
(10.24)
(10.25)
with
(10.26)
(10.27)
and
(10.28)
(10.29)
For the case that and , and and do not have common divisors, i.e. the controller does not compensate plant poles and zeros, the denominator polynomial of the pre-filter can be determined as
(10.30)
The transfer function for a command input is then
(10.31)
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If both, the controller and the plant, show minimum phase behaviour and their transfer functions
do not have zeros on the imaginary axis, all zeros of the closed loop can be compensated, such
that one obtains instead of Eq. (10.31)
(10.32)
If the closed-loop transfer function also contains given zeros, the transfer function should
have a corresponding numerator polynomial. The coefficient in the numerator is used to make
the gain of the closed-loop transfer function equal to 1. From Eq. (10.31) it therefore
follows that
(10.33)
The expression in the denominator is the first coefficient of the characteristic polynomial
, and therefore with Eq. (10.33)
(10.34)
For a controller with integral action the coefficient is zero and according to Eq. (10.22)
. From Eqs. (10.33) and (10.24) to (10.28) and (10.29) it follows directly that
(10.35)
When the controller is inserted into the feedback path according to Figure 10.6b the inherent
closed-loop dynamics will not be changed compared with the configuration according to
Figure 10.6a, because the denominator polynomial of the transfer function, and therefore the
characteristic equation of the closed loop, are preserved. Indeed, the zeros of the controller
transfer function do no longer arise, but their poles as zeros in the closed-loop transfer function.
Analogous considerations for lead to
(10.36)
Whereby the polynomial contains the poles of the controller and the plant zeros in
the left-half plane. The transfer function
M.I.E.T ENGG COLLEGE DEPT OF EEE
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(10.37)
is the same as for the case of a stable controller and a minimum-phase plant according to Eq. (10.32).
The constant for a proportional controller is
(10.38)
For an integral controller in the feedback loop a feed-forward path is not realizable.
The synthesis equations
The system of equations described by Eq. (10.18) can be rewritten in matrix notation. Thus the
required controller parameters are combined into one parameter vector. The matrix of the plant
parameters applies for both the cases, controller order and .
For integral plants
( ) with controller order
and normalized
the system is:
(10.39)
and
(10.40)
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For proportional plants or in the case of disturbances at the input of integral plants, where the
order of the controller is increased by one to , with Eqs. (10.22) and (10.16) it follows that
(10.41)
(10.42)
Here the system is:
(10.43)
and
(10.44)
The matrices on the left side of Eqs. (10.39) and (10.43) are equal for . This matrix is always regular and therefore the solution is always unique.
Application of the method Example 10.4.1 An integral plant has the transfer function
As we do not consider a disturbance at the input of the plant, the controller coefficients can be calculated by Eqs. (10.39) and (10.40).
According to section 10.4.1 one obtains for this second-order plant the order of the
controller. The closed loop has therefore order . The step response of this loop
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should comply with the binomial form given in Table A.2, and should be met. This
corresponds to a value of . The associated characteristic polynomial is
and Eqs. (10.39) and (10.40) provide the synthesis equation
with
from which the controller coefficients follow as
The transfer function of the controller is
It can be seen that this design leads to an unstable controller, which is in addition non-minimum
phase. The closed-loop transfer function is
and it contains in the numerator polynomial, besides the zero of the plant, the right-half-plane
zero of the controller. According to the considerations in section 10.4.2 this zero cannot be
compensated by a pre-filter for stability reasons. Also this would not be necessary, as the plant
zero is dominant and has a stronger influence on the closed-loop step response (see Figure 10.7).
The denominator polynomial
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Figure 10.7: Step responses of the controlled variable for the case: (a) without pre-filter
, (b) with pre-filter , and the associated manipulated variable and , and the
step response of the given closed-loop transfer function .
for the transfer function of the pre-filter is determined as
and the numerator is
The closed-loop transfer function including the pre-filter is thus
This in fact still contains in the numerator polynomial the controller zero, but as can be seen
from Figure 10.7, the corresponding step response does not show a large deviation from
the step response of the given transfer function
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Conspicuous is the fact that the step response of the closed loop without pre-filter has a
large overshoot and does not show any similarity with the other step responses, though all three
other systems have the same inherent behavior. Here, the dominant behavior of the plant zero
has a noticeable effect.
Example 10.4.2 Given the third-order plant transfer function
The step response of this plant is shown in Figure 10.8. A controller should be designed
such that the step response of the closed loop has a desired standard form chosen from
Table A.1. In the current case of a plant the order of the controller must be chosen equal
to the order of the plant that is
Thus one obtains a sixth-order closed-loop transfer function . The desired transfer
function according to Eq. (10.4), which is the basis for Table A.1, has then exactly the total order
, if
is chosen. For this case the step response with
(see Table A.1) will be desired. The last unspecified parameter of the transfer function is
the relative frequency . All step responses in Table A.1 are normalised by this value such that
the time scale can still be chosen. Consequently, by a proper choice of the normalised step
response from Table A.1 can be scaled to the desired time scale. E.g., for a value of a
rise time would follow, for this would be . If a large value is
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taken for to obtain a small rise time, this would result in controller coefficients of very
different order such that this controller would not be realisable from a numerical point of view.
Therefore here
is a good choice. For the determined values of and one obtains for the desired closed-loop
transfer function from Eq. (10.4)
The Eqs. (10.41) and (10.42) deliver at first the absolute coefficients of the controller transfer
function
and
The remaining coefficients are obtained from Eq. (10.43) by solving
The solution is
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If the reciprocal gain is chosen according to Eq. (10.22) as
for
then the controller coefficients have the same order of magnitude and the controller transfer
function is
From Eq. (10.34) one obtains for the coefficient of the pre-filter
The transfer function of the pre-filter is
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Figure 10.8: (a) Step responses of the controlled variable for the plant in the uncontrolled
case and in the controlled case for the designed compensator and for the optimal PI
controller ; (b) Step responses of the associated manipulated variables and
The step response of the controlled variable of the closed loop is shown in Figure 10.8.
For comparison this figure also shows the corresponding step response of the closed loop
using a PI controller, which is optimal in the sense of the performance index according to
section 7.3.3. Both, the maximum overshoot and the rise time of this control system with a PI
controller are clearly worse than for the controller designed here. From the behaviour of the
manipulated variables and , respectively, it can be seen that in general a smaller rise
time must be bought by a larger amplitude of the controller output. Because of the always
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existing limitation on the value of the manipulated variable, too demanding specifications for the
transfer function cannot be realised.
DESIGN OF PHASE-LEAD COMPENSATOR Phase-lead compensator, which corresponds to PD controller, is appropriate, when speed is
required, because it will speed up the original response. Typical applications are in
servos.Transfer function of phase-lead compensators is given by
Gain at high frequencies
and at low frequencies
Upper cut-off frequency
and lower cut-off frequency
Maximum phase shift f max of phase-lead compensator
occurs at frequency
The value of gain at that frequency is
A typical Bode diagram of a phase-lead controller is given in Figure 4.1.
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WLead(s) is leading ahead the phase of the input signal. As can be seen from Figure 4.1 the gain
increases at higher frequencies. The gain at higher frequencies is a2. Since a2 > 1, this means
that WLead ( ) dB (a2 )dB is positive, which is clear from Figure 4.1. Let us study more
carefully how much phase margin can be added for various values of a2 .
Steps in design of phase-lead compensator STEP 1.
Choose gain K to satisfy steady-state requirements.
STEP 2.
Draw Bode-diagram of KG(s).
STEP 3.
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Determine the new crossover frequency wc, i.e.., the frequency at which the uncompensated
system has phase (-180o+PM’), where PM’ = desired phase margin by choosing a few values of
a2 starting with a2 =2 and going up to 10-20. Construct a table showing how much this choice
will add to phase margin and also to magnitude. In this way a2 and
Can be determined.This procedure also reveals if one phase lead circuit is enough.Remember
also that the bigger a2 is, the more susceptible it is for noise.
STEP 4.
The new crossover frequency
This completes the design of the phase-lead compensator.
STEP 5.
Check the design by simulating step response. Draw also the open-loop Bode diagram of the
compensated system for comparison purposes.
STEP 6.
If requirements are met, stop. Otherwise go back to STEP 1.
EXAMPLE
Open-loop transfer function of a unity feedback system of Fig.4.5 is given by
Fig.4. 5. The block diagram of the overall system. Here WLead(s) represents the phase-lag
controller or compensator and G(s) the open-loop system transfer function. The system has
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Unity feedback, H(s) = 1.
Specifications for the system are:
• Accuracy for a unit ramp input < 2%, i.e. steady-state error < 0.02.
• Maximum percent overshoot = PO < 20%.
Design a phase-lead compensator that satisfies the requirements.
SOLUTION: STEP 1. (is the same as in phase lag)
Compute the steady-state for unit ramp input ess
This implies that
K > 1/0.02 = 50, choose K = 50. (4- 12)
Gain is now sufficient. The controller is at this point a P (proportional) controller.
STEP 2. (As in phase lag)
From Fig.2.5 percent overshoot PO <20% corresponds to > 48° phase margin, PM. This
holds for second order systems and therefore we will make the dominant pole assumption.
Draw the Bode diagram of the transfer function KG(s) = 50 / s(1 + 0.2s) as before. It is
redrawn in Fig.4.6.
» kg=zpk([] ,[0 -5],250)
Zero/pole/gain:
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STEP 3
Recall that maximum phase shift
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Let us compare the Bode diagrams of the uncompensated and compensated systems by
Drawing them in the same figure, Figures 4.8.
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The step responses are plotted in the same figure, Fig.4.10:
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The zero is not very close to the pole p = -17.4439 but seems to cancel most of its
effect, but not quite enough. This is probably the reason why the PO requirement was not
satisfied right away.
Let us finally draw the Bode diagram of the phase-lead compensator.
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UNIT-V
STATE SPACE ANALYSIS
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