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CONTROL SYSTEM NOTES (For Bachelor of Engineering) of Engineering) Notes by:
PROF. SHESHADRI G. S
Soft Copy material designed by:
KARTHIK KUMAR H P Your feedbacks can be mailed to:
[email protected]
INDEX
(1) Introduction to Control system (2) Mathematical model of linear of linear systems (3) Transfer functions (4) Block diagram (5) Signal Flow Graphs (6) System Stability (7) Root Locus Plots (8) Bode Plots
Control Systems
1
CONTROL SYSTEM NOTES (For Bachelor of Engineering) of Engineering) Notes by:
PROF. SHESHADRI G. S
Soft Copy material designed by:
KARTHIK KUMAR H P Your feedbacks can be mailed to:
[email protected]
INDEX
(1) Introduction to Control system (2) Mathematical model of linear of linear systems (3) Transfer functions (4) Block diagram (5) Signal Flow Graphs (6) System Stability (7) Root Locus Plots (8) Bode Plots
Control Systems
1
Control Systems
Introduction to Control Systems By: CIT, Gubbi.
Control Syste System m means any quanti quantity ty of inte i nterest in in a machine achine or mechanism chanism is maintai aintaine ned or alte alterred iin n accordance accordance with desir sired manner. OR A syste systemwhich hich controls control s the output quanti quantity ty is call cal led a control control syste system.
Definitions: 1. Cont Co ntro roll lled ed Variab Var iab le: le : I t is is the the quantity quantity or condi condition tion that that is is measured asured & controll controlled.
2. Cont Co ntro roll ller er:: means measuring asuri ng the value value of the control controllled variabl variable e of the system & applyi applying ng the Controller m manipul anipulate ated d variabl vari able e to the syste system m to corr correct or to li limit the deviati viation on of the measured val value ue to the desired value.
3. Plant: A plant is is a piec piece e of equi equipm pment, which is i s a set of machine achine parts functi functioni oning ng together. The The purpose purpose of which hich is to perform forma particular particular operation. ation. Example ple: Furnace, Space Space craft craft etc.,
4. System: A system is is a combinati bination on of components that works orks toget togethe her & perform forms certain tain obje objective ctive.
5. Dist Di stur urba banc nce: e: A disturbance is a signal signal that tends tends to to affe affect the value of the outp output ut of a syste system m. If a disturbance disturbance is created inside the system, it is called internal. While an external disturbance disturbance is is gener nerated ated outside outside the the system.
6. Feedback Control: I t is is an operation operati on that, in i n the pres presence of distur disturbance bance tends tends to reduce the diff difference erence betwe between the output of a system& some reference reference input. input.
7. Servo Mechanism: A servo mechanism is is a feedback control controllled system in whi which ch the output is is some mechanical chanical position, velocity or acceleration. ration.
8. Open loop System: I n an an Open loop System, the control action action is independent of the desi desirred output. output. OR When When the output quantity quantity of the control control sys system temis not fed back to the input input quantity, quantity, the control control system sys temis call called an Open loop System.
9. Clos Cl osed ed loop lo op Syst Sy stem em:: I n the the Closed loop Control System the control action action is is depe dependent on the desir sired output, where the output quantity quantity is is considerably considerably controll controlled by sending nding a command signal to inpu inputt quanti quantity. ty. By: HPK Kumar
2
Introduction to Control System
1
2
Introduction to Control System
10. 10 . Feed Back: Normally, the feed back signal has opposite polarity to the input signal. This is called negative feed back. The The advantage advantage is the resultant signal si gnal obtaine obtained fr from the comparator being diff diffe erence of the two two signal signals s is is of small aller magnitude agnitude. It I t can be be handled easil asily b by y the control control syste system m. The The resul esulti ting ng si signal is called Actuating Signal This Thi s signal signal has zero value value when the desir sired output output is obtaine obtained. d. I n that condition, control system will not operate. Effects of Feed Back:
Let the syste system has open loop gain gain I nput nput signal signal . The Then the feed back sign ignal is, is,
E(S)
R(S) B(S)
feed back loop gain gain
Output signal signal
&
G(S)
C(S)
-
&
–
Hence,
H(S) = With this this eqn.
(a) (a )
(1) , we can write the effe ffects of fee feed back as foll ollows. ows.
Overall Gain:
Eqn. shows shows that the gain gain of the open loop sys system temis reduced educed by a factor in a fe feed back syste system. Here the the feed back signal signal is negati negative ve. I f the feed feed back gain has positi positive ve val value ue,, the overall all gain gain will be reduced. If I f the feed back gain gain has ne negative gativevalue value,, the overall all gain gain may increase.
(b) (b )
Stability:
I f a systemis able able to foll fol low the input comm command signal signal,, the system stemis said said to be Stable. A system is said said to be Unstable , if its output output is out of contr control. ol. I n eqn. qn. , if the output output of the syste system is infi infini nite te for any fini finite te input input.. This his shows that a stable stable system may become unstab unstablle for cer certain tain value value of a fe feed back gain. gain. The Therefor fore if the feed feed back is is not prope properly used, the syste ystem mcan be harmful. ul.
(c) (c )
Sensitivity: This This depends on the systemparam rameters. rs. For For a good control rol system, it is desirab irable that the system should be inse insensitive nsiti ve to its its param parameter ter changes. Sensitivity, SG =
This This fun function ion of the system can be reduced by inc increasing ing the valu of . This This can be done by selecti ecting ng proper feed feed back. By:
HPK Kumar
Control Systems
3
Control Systems
(d) (d )
Noise:
Examples are brush & commutation noise in electrical machines, Vibrations in in moving oving system etc.,. tc.,. The The effe ffect of feed feed back on the these noise noise signal signals s will wil l be greatly greatly infl influe uence nced by the the poi point nt at whi which ch these these signal signals s are are introduce introduced in in the syste system. I t is is possi possible ble to reduce reduce the the effe ffect of noise noi se by pr proper design sign of fe feed back system.
Classification of Control Control Systems The The Con Control rol Sy Systemcan be clas lassified mainly inly depending ing upon, (a) Method of of anal analysi ysis & design, design, as Linear & Non- Linear Systems. (b) The The type of the sign ignal, as Time Varying, Time Invariant, Continuous data, Discrete data systems etc., (c) The The type of systemcomponents, as Electro Mechanical, Hydraulic, Thermal, Pneumatic Control systems etc., (d) The The main purpo rpose, as Position control & Velocity control Systems.
1. Linear & Non -Linear Systems: I n a li l i near near syste system m, the principl pri nciple e of supe superposition rpositi on can be be app appll i ed. I n non- l i near near syste system m, this principle can’t be applied. Therefore a linear system is that which obeys superposition principle & homogeneity.
2. Time Varying & Time Invariant Systems: Whi Whi l e operating operati ng a control control system system, i f the param parameters are are unaffe unaff ected by by the tim ti me, then then the system system i s call call ed Time Invariant Control System. Most physical physical systems have param parameter eters changing changing wi wi th tim time. If I f this thi s var varii ation ati on is is measur easurable abl e during duri ng the system systemoperation operati on then then the system system is called Time Varying System. I f ther there i s no non-l non-l i near nearii ty in in the tim ti me var varyi ying ng syste system m, then then the system may be call cal l ed as Linear Time varying System.
3. Disc Di scre rete te Data Da ta Syst Sy stem em s: I f the signal signal is i s not continuou continuousl sly y varying varying wi wi th time time but it it is i s in the form for m of pulse pulses. s. Then Discrete Data Control System. the control syste system m i s call cal l ed Discrete I f the signal signal is i s in in the form for m of pulse pul se data, then then the syste system m i s call call ed Sampled Data Control System. Here the information supplied intermittently at specific instants of time. This has the advantage of Time sharing system. On the other hand, if the signal is in the form of digital code, the system is called Digital Coded System. Here use of Digital computers, µp, µc is made use use of such system systems are are analyze anal yzed d by the Z- transf transfor orm m theory. theory.
4. Cont Co ntin inuo uous us Data Da ta Syst Sy stem ems: s: I f the signal signal obtained at vari various ous parts of the syste system m are are var varying ying continuous continuousll y wi wi th tim time, then the system system i s call cal l ed Continuous Data Control Systems.
5. Adap Ad apti tive ve Cont Co ntro roll syst sy stem ems: s: I n some control control system systems, certai n param parameters are ar e ei ther not constant or vary vary in an unknow unknown n manner anner. I f the parame parameter ter variati vari ations ons are lar large ge or rapid, apid, it i t may be be de desir sirable able to to desi design gn for the capability of continuously measuring them & changing the compensation, so that the syste system m per perform for mance criteri cri teria a can alw always satisfi satisfie ed. Thi This s is call call ed Adaptive Control Systems. By: HPK Kumar
4
Introduction to Control System Identification & Parameter
3
4
Introduction to Control System Identification & Parameter adjustment
R(s)
E(s)
Compensator
System
+
C(s)
B(s) H(S) 6. Optimal Control System: Optimal Control System is obtained by minimizing and/or maximizing the performance
index. This index depends upon the physical system & skill.
7. Single Variable Control System: In simple control system there will be One input & One output such systems are called Single variable System (SISO – Single Input & Single Output).
8. Multi Variable Control System: In Multivariable control system there will be more than one input & correspondingly more output’s (MIMO - Multiple Inputs & Multiple Outputs).
Comparison between Open loop & Closed loop Gain Open Loop System
Closed Loop System
1. An open loop system has the ability to perform accurately, if its calibration is good. If the calibration is not perfect its performance will go down.
1. A closed loop system has got the ability to perform accurately because of the feed back.
2. It is easier to build.
2. It is difficult to build.
3. In general it is more stable as the feed back is absent.
3. Less Stable Comparatively.
4. If non- linearity’s are present; the system operation is not good.
4. Even under the presence of nonlinearity’s the system operates better than open loop system.
5. Feed back is absent. Example: (i) Traffic Control System. (ii) Control of furnace for coal heating. (iii) An Electric Washing Machine.
5. Feed back is present. Example: (i) Pressure Control System. (ii) Speed Control System. (iii) Robot Control System. (iv) Temperature Control System.
Note:
Any control system which operates on time basis is an Open Loop System.
By: HPK Kumar
Control Systems
5
Control Systems
5
Block Diagram of Closed Loop System: Controller
E(S) Ref. i/p
Control Elements
Plant
Controlled o/p
Feed Back elements
Actuator
Thermometer Block Diagram of Temperature Control System: Thermo meter
A/D Converter
Interface
Electric Furnace Heater
Relay
Amplifier
Interface
Programmed i/p
Temperature Control of Passenger Compartment Car: Sun
Sensor
Ambient Temperature
Radiation Heat Sensor
Desired
Controller
Air Conditioner
Passenger Car
O/p
Temperature i/p
Sensor
************ ******** ********** ************* *********** ******* ***** ******
By: HPK Kumar
Control Systems
Mathematical Models of Linear Systems
1
Control Systems
Mathematical Models of Linear Systems By: CIT, Gubbi.
A physical system is a collection of physical objects connected together to serve an objective. An idealized physical system is called a Physical model. Once a physical model is obtained, the next step is to obtain Mathematical model. When a mathematical model is solved for various i/p conditions, the result represents the dynamic behavior of the system. An al og ou s System:
The concept of analogous system is very useful in practice. Since one type of system may be easier to handle experimentally than another. A given electrical system consisting of resistance, inductance & capacitances may be analogous to the mechanical system consisting of suitable combination of Dash pot, Mass & Spring. The advantages of electrical systems are, 1. 2.
Many circuit theorems, impedance conc epts can be applicable. An Electrical engineer familiar with electrical systems can easily analyze the system under study & can predict the behavior of the system. 3. The electrical analog system is easy to handle experimentally. Translational System:
It has 3 types of forces due to elements. 1. Inertial Force: Due to inertial mass,
2 Fmt M.a 2 Where, .
F(t)
M
. .
2. Damping Force [Viscous Damping]: Due to viscous damping, it is proportional to velocity & is given by, D
.
Damping force is denoted by either D or B or F
3.
Spring Force: Spring force is proportional to displacement.
. . Fk
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Mathematical Models of Linear Systems Where, Rotational system:
HPK Kumar
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2
Mathematical Models of Linear Systems Where, Rotational system:
1. Inertial Torque: 2. Dampi ng Torque : . 3. Spring Torque :
Analogous quantities in translational & Rotational system: The electrical analog of the mechanical system can be obtained by, (i) (ii)
Sl. No.
Force Voltage analogy: (F.V) Force Current analogy: (F.I)
Mechanical Translational System
Mechanical Rotational System
F.V Analogy
F.I Analogy
1.
Force (F)
Torque (T)
Voltage (V)
Current (I)
2.
Mass (M)
Moment of Inertia (M)
Inductance (L)
Capacitance (C)
3.
Viscous friction (D or B or F)
Viscous friction (D or B or F)
Resistance (R)
Conductance (G)
4.
Spring stiffness (k)
5.
Linear displacement ( )
Torsional spring stiffness Reciprocal of Capacitance (1/C) ( ) Charge (q) Angular displacement ( )
Reciprocal of Inductance (1/L) Flux ( )
6.
Linear velocity ( )
Angular Velocity (w)
Voltage (v)
Current (i)
D’Alemberts Principle: The static equilibrium of a dynamic system subjected to an external driving force obeys the following principle, “For any body, the algebraic sum of externally applied forces resisting motion in any given direction is zero”.
Example Problems: (1) Obtain the electrical analog (FV & FI analog circuits) for the Machine system shown & also write the equations.
Free Body diagram
2 2
D2
D1
1 1
Ft
M1
Ft
M1
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Control Systems
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Transfer Functions By: CIT, Gubbi.
The input- output relationship in a linear time invariant system is defined by the transfer function.
The features of the transfer functions are, (1) It is applicable to Linear Time Invariant system. (2) It is the ratio between the Laplace Transform of the o/p variable to the Laplace Transform of the i/p variable. (3) It is assumed that initial conditions are zero. (4) It is independent of i/p excitation. (5) It is used to obtain systems o/p response. An equation describing the physical system has integrals & differentials, the step involved in obtaining the transfer function are; (1) Write the differential equation of the system.
by ‘S’ &
(2)
Replace the terms
by 1/S.
(3)
Eliminate all the variables except the desired variables. Impulse Response of the Linear System:
G(S)
R(S)
C(S)
In a control system, when there is a single i/p of unit impulse function, then there will be some response of the Linear System. The Laplace Transform of the i/p will be R(S) = 1
G(S) =
. .1
1
i.e., the Laplace Transform of the system o/p will be simply the ‘Transfer function’ of the system. Taking L-1
Here G(t) will be impulse response of the Linear System. This is called Weighing Function. Hence LT of the impulse response is the Transfer function of the system itself.
R
P ROBLEMS: (1) Obtain the Transfer ‐Function(TF) of the circuit shown in circuit 1.0
R
C
i
Laplace Transformed network
Solution:
i(S)
1
Contd……
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Transfer Functions
Circuit 1.0
2
Transfer Functions
.., 1 & 1 . . 1 1 1 Where, = RC (2) Obtain the TF of the mechanical system shown in circuit 2.
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Control Systems
3
(3) Transfer Function of an Armature Controlled DC Motor in circuit 3.0:
La
Ra
I f = Constant
Let,
ia
V i
‘Ra’ Resistance of armature in
ia
E b
F
V f
‘La’ Inductance of armature in H’s. ‘ia’ Armature current. & ‘i f ’ Field current.
Ω’s.
‘V i’ Applied armature voltage.
circuit 3.0
T m
J,
‘E b’ Back e.m.f in volts.
F
Angular
‘T m’ Torque developed by the motor in N-m. displacement of motor shaft in radians
J Equivalent moment of inertia of motor & load referred to the motor shaft. F Equivalent Viscous friction co-efficient of motor & load referred to the motor shaft.
i.e., .
The air gap flux ‘ ’ is proportional to the field current.
Where, ‘K f ’ is a constant.
The torque developed by the motor ‘T m’ is proportional to the product of the arm current & the air gap flux.
Since the field current is constant,
Where, K a & K f are the constants.
Where, K T is Motor – torque constant.
The motor back e.m.f is proportional to the speed & is given by,
Where, K b is back e.m.f constant.
The differential equation of the armature circuit is,
22
The torque equation is, Taking LT for above equation, we get
------------------------------- (1)
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ (A)
Taking LT for the torque equation & equating, we get
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Transfer Functions
4
Transfer Functions
2
------------------------------ (2)
-------------------------- (B)
Taking LT for back e.m.f equation, we get
------------------------------- (C)
Substituting the values of I a (S) & E b (S) from equation (C) & (2) in equation (1), we get
. The block diagram representation of armature controlled DC Motor can be obtained as follows,
From equation (A),
V i(S)
-
1
I a(S)
E b(S) From equation (B),
I a(S)
(S)
From equation (C),
(S)
E b(S)
The complete block diagram is as shown below,
V i(S)
E b(S)
1
I a(S)
2
(S)
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Control Systems
5
(4) Transfer function of Field Controlled DC Motor in circuit 4.0:
Let,
L f
R f
I a = Constant
R f Field winding resistance. L f Field winding inductance.
V f
i f
V f Field control voltage.
T m
I f Field current.
T m Torque developed by motor. J Equivalent moment of inertia of motor & load referred to the motor shaft.
circuit 4.0 J, F
F Equivalent Viscous friction co-efficient of motor & load referred to the motor shaft.
Angular displacement of motor
shaft. In the field controlled DC motor, the armature current is fed from a constant current source.
Where, K a & K f are the constants.
The KVL equation for the field circuit is,
On Laplace Transform,
. The torque equation is ,
---------------------------------- (1) --------------------------------- (A)
Where, K T is Motor – torque constant.
On Laplace Transform,
.
2 .
Substituting the value of
--------------------------- (2)
-------------------------
from equation (2) in equation (1), we get
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Transfer Functions
(B)
6
Transfer Functions
2 .
2 The block diagram representation of field controlled DC Motor can be obtained as follows,
From equation (A),
1
V f (S)
I f (S)
From equation (B),
I f (S)
(S)
The complete block diagram is as shown below,
1
V f (S)
(5) Obtain the TF
Solution:
I f (S)
(S)
R
for the network shown in circuit 5.0:
Laplace Transformed network
V i
R
1
R I(S)
V i (S)
C
C
V o
circuit 5.0 I 1(S)
1
R
Applying KVL to this circuit,
I 2(S) V o (S)
----------------
------------------ (1)
. 1 1 1 . 1 2 1 τ 2 1τ τ Designed By:
Let,
τ
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100 k
Control Systems
1M
Ω
7
Control Systems
Ω
1M
100 k (6) Find the TF
for the network shown in circuit 6.0: V i
Laplace Transformed network
10
10
Solution:
Ω
1 F
10 F
Circuit 6.0
V i (S)
Loop 1
10 V (S)
10
0
Loop 2
Writing KVL for loop (1), we get
110 5 10 5 2 10 5 . 2 1 105 1 1
------------------------------------------- (1)
Writing KVL for loop (2), we get
10
0
2 10111 -------------------------------------------- (2) 2. 106 2 106. --------------------------------- (3) Substituting for I 1 (S) from equation (2) in (1), we get
. 2 2. 10 11 105 1 1 2 10 11 1 1 1 105 1 From equation (3) the above equation becomes,
105
106. 10 2 21 10 10 10 21 10
V i(S)
10 1
I 1(S) +
1 10 11
I 2(S)
106
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Transfer Functions
V 0(S)
7
8
Transfer Functions
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2
Block Diagrams P ROBLEMS :
Reduce the Block Diagrams shown below:
(1)
+ -
-
-
Solution: By eliminating thefeed-back paths, we get + -
Combining the blocks in series, we get +
C(S)
-
Eliminating the feed back path, we get
Control Systems (2)
R(S)
C(S)
-
Solution:
-
Shifting thetake-off
‘
’
beyond the block
‘
’
, we get
R(S)
C(S)
-
-
Combining
and eliminating
(feed back loop), we get
R(S)
C(S)
-
-
Eliminating the feed back path
, we get
R(S)
C(S)
-
Combining all the three blocks, we get R(S)
C(S)
3
4
Block Diagrams
R(S)
-
-
(3)
Solution:
C(S)
-
Re-arranging the block diagram, we get R(S)
-
Eliminating
C(S)
-
-
loop & combining, we get C(S)
R(S)
-
-
Eliminating feed back loop
R(S)
C(S)
-
Eliminating feed back loop
R(S)
, we get
C(S)
Control Systems
Signal Flow Graphs By: CIT, Gubbi.
For complicated systems, Block diagram reduction method becomes tedious & time consuming. An alternate method is that signal flow graphs developed by S.J . Mason. In these graphs, each node represents a system variable & each branch connected between two nodes acts as Signal Multiplier. The direction of signal flow is indicated by an arrow.
Definitions: 1. Node: A nodeis a point representing a variable. 2. Transmittance: A transmittance is a gain between two nodes. 3. Branch: A branch is a line joining two nodes. The signal travels along a branch. 4. Input node [Source]: It is a node which has only out going signals. 5. Output node [Sink]: It is a node which is having only incoming signals. 6. Mixed node: It is a node which has both incoming & outgoing branches (signals). 7. Path: It is thetraversal of connected branches in the direction of branch arrows. Such that no node is traversed more than once. 8. Loop: It is a closed path. 9. Loop Gain: It is theproduct of the branch transmittances of a loop. 10. Non-Touching Loops: Loops are Non-Touching, if they do not possess any common node. 11. Forward Path: It is a path fromi/p node to the o/p node which doesn’t cross any node more than once. 12. Forward Path Gain: It is the product of branch transmittances of a forward path. MASON’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 formula for the determination of overall systemgain is given by, Where,
Path gain of
–
forward path.
Determinant of the graph.
The value of the T
for that part of the graph not touching the
Overall gain of the system.
forward path.
1
Control Systems
R(S)
5
C(S)
-
R(S)
C(S)
C(S)
R(S)
Signal flow graph:
R(S)
C(S)
No. of forward paths:
No. of individual loops:
(6) Reduce the Block Diagram shown.
-
R(S)
-
Solution: Shifting
C(S)
+
-
beyond
, weget
-
R(S)
-
-
C(S)
+
6
Signal Flow Graphs Eliminating feed back loop
, we get
R(S)
C(S)
-
Eliminating feed back loop
+
-
, we get
R(S)
C(S)
-
+
R(S)
C(S)
-
Eliminating the another feed back loop
+
, we get
R(S)
C(S)
+
R(S)
C(S)
Signal flow graph:
R(S)
C(S)
Contd......
8
Signal Flow Graphs Eliminating
loop, we get
-
(9) Using Mason’s gain rule, obtain the overall TF of a control system represented by the signal flow graph shown below.
Solution: No. of forward paths:
Individual loops:
Two non-touching loops = 0
(10)
Construct signal flow graph from the following equations & obtain the overall TF.
Contd......
Control Systems
Substituting ‘x’ value in the block diagram. The block diagrambecomes,
Signal flow graph:
No. of forward paths: No. of individual loops:
(16)
Obtain TF,
Two non-touching loops = 0
using block diagram algebra & also by using Masons Gain Formula. Hence Verify
the TF in both the methods.
Contd......
13
Control Systems
Hence,
17
Control Systems
System Stability By: CIT, Gubbi.
While considering the performance specification in the control system design, the essential & desirable requirement will be the system stability. This means that the system must be stable at all times during operation. Stability may be used to define the usefulness of the system. Stability studies include absolute & relative stability. Absolute stability is the quality of stable or unstable performance. Relative Stability is the quantitative study of stability. The stability study is based on the properties of the TF. In the analysis, the characteristic equation is of importance to describe the transient response of the system. Fromthe roots of the characteristic equation, some of the conclusions drawn will be as follows, (1) When all the roots of the characteristic equation lie in the left half of the S-plane, the system response due to initial condition will decrease to zero at time Thus the system will be termed as stable. (2) When one or more roots lie on the imaginary axis & there are no roots on the RHS of Splane, the response will be oscillatory without damping. Such a systemwill be termed as critically stable. (3) When one or more roots lie on the RHS of S-plane, the response will exponentially increase in magnitude; there by the systemwill be Unstable. Some of the Definitions of stability are, (1) A system is stable, if its o/p is bounded for any bounded i/p.
(2) A system is stable, if it‟s response to a bounded disturbing signal vanishes ultimately as time „t‟ approaches infinity. (3) A system is unstable, if it‟s response to a bounded disturbing signal results in an o/p of infinite amplitude or an Oscillatory signal. (4) If the o/p response to a bounded i/p signal results in constant amplitude or constant amplitude oscillations, then the system may be stable or unstable under some limited constraints. Such a systemis called Limitedly Stable system. (5) If a system response is stable for a limited range of variation of its parameters, it is called Conditionally Stable System. (6) system.
If a systemresponse is stable for all variation of its parameters, it is called Absolutely Stable
Routh-Hurwitz Criteria:
A designer has so often to design the systemthat satisfies certain specifications. In general, a system before being put in to use has to be tested for its stability. Routh-Hurwitz stability criteria may be used. This criterion is used to know about the absolute stability. i.e., no extra information can be obtained regarding improvement. As per Routh-Hurwitz criteria, the necessary conditions for a systemto be stable are, (1) None of the co-efficient‟of the Characteristic equation should be missing or zero. (2) All the co-efficient‟ should be real & should have the same sign.
1
2
System Stability A sufficient condition for a systemto be stable is that each & every term of the column of the Routh array must be positive or should have the same sign. Routh array can be obtained as follows. The Characteristic equation is of the form, ……………………… Where, 0 0
: 0
0
0
0
0
:
Similarly wecan evaluate rest of the elements,
The following are the limitations of Routh-Hurwitz stability criteria, (1) It is valid only if the Characteristic equation is algebraic. (2) If any co-efficient of the Characteristic equation is complex or contains power of „ ‟, this criterion cannot be applied. (3) It gives information about how many roots are lying in the RHS of S-plane; values of the roots are not available. Also it cannot distinguish between real & complex roots. Special cases in Routh-Hurwitz criteria:
(1) When the termin a row is zero, but all other terms are non-zeroes then substitute a small positive number for zero & proceed to evaluate the rest of the elements. When the column term is zero, it means that there is an imaginary root. (2) All zero row: In the case, write auxiliary equation from preceding row, differentiate this equation & substitute all zero row by the co-efficient‟ obtained by differentiating the auxiliary equation. This case occurs when the roots are in pairs. The systemis limitedly stable. Problems: C OMMENT ON THE STABILITY OF THE SYSTEM WHOSE CHARACTERISTIC EQUATION IS GIVEN BELOW :
(1) 1
21
20
6
36
0
15
20
0
28
0
0
20
0
0
The no. of sign changes in the column = zero. No roots are lying in the RHS of S-plane. The given Systemis Absolutely Stable.
Control Systems (10)
5
The open-loop TF of a unity feed back system is given by the above expression. Find the value of ‘K’ for which the system is just stable.
Solution: The characteristic equation is
1
23
2K
9
(15+K)
0
2K
0
0
0
0
0
2K
(i) (ii)
K>0 192 – K > 0 K < 192
(iii)
(192 – K)(15+K) – 162K > 0 (for themax. value of K)
From this evaluate for K, Using, Considering the positive value of „K ‟, So, 0 < K < 61.68
When the value of „K‟ is 61.68 the systemis just stable. (11)
Using Routh-Hurwitz criteria, find out the range of ‘K’ for which the system is stable. The characteristic equation is
Solution: 1
(2K+3)
5K
10
(i) (ii)
K >0
0
Considering the positive value of „K ‟, 10
The range of K is „ (12)
0
‟
A proposed control system has a system & a controller as shown. Access the stability of the system by a suitable method. What are the ranges of ‘K’ for the system to be stable?
Solution:
The characteristic equation is
(i) 16
(1+K)
8
K 0
K
(13)
0
K >0
(ii)
The range of K is,
Control Systems (17) 1
4
6
2
5
2
1.5
5
0
-1.666
2
0
6.8
0
0
2
0
0
(i) (ii) (iii)
No. of sign changes = 2. Two roots lie on RHS of S-plane. The systemis Unstable.
(18) Solution: 1
11
6
6
10
0
6
0
No sign changes. The systemis Absolutely Stable.
(19) Solution:
No. of sign changes = 1 The systemis Unstable.
1
-5
2
-6
-2
0
-6
0
(20) +ve
1
2
4
+ve
1
2
1
+ve
3
0
-ve
1
0
+ve
0
0
0
0
+ve
1
No. of sign changes = 2. The systemis Unstable.
(21) +ve
2
6
1
+ve
1
3
1
-1
0
1
0
0
0
0
0
+ve +ve -ve +ve
(22)
1
No. of sign changes = 2. The systemis Unstable.
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Control Systems (36)
The open-loop transfer function of a unity feed back control system is given by, , using Routh-Hurwitz criteria. Discuss the stability of the closed loopcontrol system. Determine the value of ‘K’ which will cause sustained oscillations in the closed loop
system. What are the corresponding oscillating frequencies?
Solution: The characteristic equation is 1
69
12
198
(i) (ii)
0
52.5
0 0
0
0
0
(iii)
The Auxiliary equation for the
row is
When
Hence, (37)
A feed back system has open-loop transfer function
Determine the
maximum value of ‘K’ for stability of the closed-loop system.
Solution: Generally control systems have very low Band width which implies that it has very low frequency range of operations. Hence for low frequency ranges, the term can be replaced by . i.e.,
The characteristic equation is
,
1 5
K 0
K
0
(i) (ii) The range of K is stable.
for the systemto be
11
Control Systems
1
Root Locus Plots By: CIT, Gubbi.
It gives complete dynamic response of the system. It provides a measure of sensitivity of roots to the variation in the parameter being considered. It is applied for single as well as multiple loop system. It can be defined as follows, It is the plot of the loci of the root of the complementary equation when one or more parameters of the open-loop Transfer function are varied, mostly the only one variable available is the gain ‘K’ The negative gain has no physical significance hence varying ‘K’ from ‘0’ to ‘ ∞’ , the plot is obtained called the “Root Locus Point”. Rules for the Construction of Root Locus
(1) The root locus is symmetrical about the real axis. (2) The no. of branches terminating on ‘ ∞’ equals the no. of open-loop pole-zeroes. (3) Each branch of the root locus originates from an open-loop pole at ‘K = 0’ & terminates at open-loop zero corresponding to ‘K = ∞’. (4) A point on the real axis lies on the locus, if the no. of open-loop poles & zeroes on the real axis to the right of this point is odd. (5) The root locus branches that tend to ‘ ∞’, do so along the straight line. 0
180 , Asymptotes making angle with the real axis is given by P = No. of poles & Z =No. of zeroes.
(6) The asymptotes cross the real axis at a point known as Centroid.
Where, n=1,3,5,…………………
i.e.,
∑ ∑
(7) The break away or the break in points [Saddle points] of the root locus or determined from the roots of the
equation
0.
(8) The intersection of the root locus branches with the imaginary axis can be determined by the use of Routh Hurwitz criteria or by putting ‘ ’ in the characteristic equation & equating the real part and imaginary to zero. To solve for ‘’ & ‘K’ i.e., the value of ‘’ is intersection point on the imaginary axis & ‘K’ is the value of gain at the intersection point. (9) The angle of departure from a complex open-loop pole( ) is given by,
180
Designed By:
HPK Kumar
(
[email protected])
Control Systems
Bode Plots
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Control Systems
Bode Plots By: CIT, Gubbi.
Sinusoidal transfer function is commonly represented by Bode Plot. It is a plot of magnitude against frequency. i.e., angle of transfer function against frequency. The following are the advantages of Bode Plot, (1) Plotting of Bode Plot is relatively easier as compared to other methods. (2) Low & High frequency characteristics can be represented on a single diagram. (3) Study of relative stability is easier as parameters of analysis of relativestability are gain & phase margin which are visibly seen on sketch. (4) If modification of an existing systemis to be studied, it can be easily doneon a Bode Plot. Initial Magnitude:
,
If
, , , , , , Phase Plot: GCF
Magnitude Plot: PCF
+ve PM
-ve GM line GCF
0 dB line PCF
GCF
PCF line
-ve PM GCF
0 dB line +ve GM PCF
1
Control Systems Examination Problem (Mar/Apr’ 99):
(7) The sketch given shows the Bode Magnitude plot for a system. Obtain the Transfer function.
dB D
40 A B
(Z) E
(P)
(P) (DZ)
C
Solution:
Since the initial slope is
there must be zero at the origin. &
Examination Problem (Sep/Oct’ 99):
(8) Estimate the Transfer function for the Bode Magnitude plot shown in figure.
dB
(Z) Solution:
5
