Control System Notes by HPK Kumar JWFILES

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CONTROL SYSTEM

NOTES(ForBachelorofEngineering)

Notesby:PROF. SHESHADRI G. S

SoftCopymaterialdesignedby:KARTHIK KUMAR H PYourfeedbackscanbemailedto:

[email protected]

INDEX(1)IntroductiontoControlsystem(2)Mathematicalmodeloflinearsystems(3)Transferfunctions(4)Blockdiagram(5)SignalFlowGraphs(6)SystemStability(7)RootLocusPlots(8)BodePlots

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1Control Systems

By: Sheshadri.G.S.HPK Kumar

Introduction to Control Systems

By: Sheshadri.G.S.CIT, Gubbi.

Control System means any quantity of interest in a machine or mechanism is maintained or altered inaccordance with desired manner. ORA system which controls the output quantity is called a control system.

Definitions:

1. Controlled Variable:It is the quantity or condition that is measured & controlled.

2.Controller:Controllermeans measuring the value of the controlled variable of the system & applying the

manipulated variable to the system to correct or to limit the deviation of the measured value to the

desired value.

3. Plant:A plant is a piece of equipment, which is a set of machine parts functioning together. The

purpose of which is to perform a particular operation. Example: Furnace, Space craft etc.,

4. System:A system is a combination of components that works together & performs certain objective.

5. Disturbance:A disturbance is a signal that tends to affect the value of the output of a system. If a disturbance

is created inside the system, it is called internal. While an external disturbance is generated outside the

system.

6. Feedback Control:It is an operation that, in the presence of disturbance tends to reduce the difference between the

output of a system & some reference input.

7. Servo Mechanism:A servo mechanism is a feedback controlled system in which the output is some mechanical

position, velocity or acceleration.

8. Open loop System:In anOpen loop System, the control action is independent of the desired output. OR

When the output quantity of the control system is not fed back to the input quantity, the control

system is called an Open loop System.

9.Closed loop System:In theClosed loop Control System the control action is dependent on the desired output, where

the output quantity is considerably controlled by sending a command signal to input quantity.

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2 Introduction to Control System


10. Feed Back:Normally, the feed back signal has opposite polarity to the input signal. This is called negative

feed back. The advantage is the resultant signal obtained from the comparator being difference of the

two signals is of smaller magnitude. It can be handled easily by the control system. The resulting signal

is called Actuating Signal This signal has zero value when the desired output is obtained. In that

condition, control system will not operate.

Effects of Feed Back:

Let the system has open loop gain feed back loop gain Output signal &

Input signal .

Then the feed back signal is,

With this eqn. , we can write the effects of feed back as follows.

(a) Overall Gain:Eqn. shows that the gain of the open loop system is reduced by a factor in

a feed back system. Here the feed back signal is negative. If the feed back gain has positive value, the

overall gain will be reduced. If the feed back gain has negative value, the overall gain may increase.

(b) Stability:If a system is able to follow the input command signal, the system is said to beStable.

A system is said to beUnstable, if its output is out of control. In eqn. , if the output of the

system is infinite for any finite input. This shows that a stable system may become unstable for certainvalue of a feed back gain. Therefore if the feed back is not properly used, the system can be harmful.

(c) Sensitivity:This depends on the system parameters. For a good control system, it is desirable that the system

should be insensitive to its parameter changes.

Sensitivity, SG= This function of the system can be reduced by increasing the valuof . This can be done by selecting proper feed back.

G(S)

H(S)

R(S)B(S)

E(S)

C(S)-

&

Hence,

= (1)

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3Control Systems


(d) Noise:Examples arebrush & commutation noise in electrical machines, Vibrations in moving system

etc.,. The effect of feed back on these noise signals will be greatly influenced by the point at which these

signals are introduced in the system. It is possible to reduce the effect of noise by proper design of feed

back system.

Classification of Control Systems

The Control System can be classified mainly depending upon,

(a)Method of analysis & design, asLinear & Non- Linear Systems.(b)The type of the signal, asTime Varying, Time Invariant, Continuous data, Discrete data systems etc.,(c)The type of system components, asElectro Mechanical, Hydraulic, Thermal, Pneumatic Control systems etc.,(d)The main purpose, asPosition control & Velocity control Systems.1. Linear & Non-Linear Systems:

In a linear system, the principle of superposition can be applied. In non- linear system,

this principle cant be applied. Therefore a linear system is that which obeys superposition

principle & homogeneity.

2. Time Varying & Time Invariant Systems:While operating a control system, if the parameters are unaffected by the time, then the

system is called Time Invariant Control System. Most physical systems have parameters

changing with time. If this variation is measurable during the system operation then the system

is called Time Varying System.

If there is no non-linearity in the time varying system, then the system may be called as

Linear Time varying System.

3.Discrete Data Systems:If the signal is not continuously varying with time but it is in the form of pulses. Then

the control system is called Discrete Data Control System.

If the signal is in the form of pulse data, then the system is called 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, p, c is

made use of such systems are analyzed by the Z- transform theory.

4. Continuous Data Systems:If the signal obtained at various parts of the system are varying continuously with time,

then the system is called Continuous Data Control Systems.

5. Adaptive Control systems:In some control systems, certain parameters are either not constant or vary in an

unknown manner. If the parameter variations are large or rapid, it may be desirable to design

for the capability of continuously measuring them & changing the compensation, so that the

system performance criteria can always satisfied. This is called Adaptive Control Systems.

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4 Introduction to Control System


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 outputs (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 toperform accurately, if its calibration is good.

If the calibration is not perfect its performancewill go down.

2. It is easier to build.3. In general it is more stable as the feed

back is absent.

4. If non- linearitys are present; thesystem operation is not good.

5. Feed back is absent.Example:

(i) Traffic Control System.(ii) Control of furnace for coal heating.(iii) An Electric Washing Machine.

1. A closed loop system has got the ability toperform accurately because of the feed back.

2. It is difficult to build.3. Less Stable Comparatively.4. Even under the presence of non-

linearitys the system operates better than openloop system.

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.

Compensator System

H(S)

Identification & Parameter

adjustment

R(s) E(s)

+

-

B(s)

C(s)

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1Control Systems

Designed By: HPK Kumar([email protected])

Mathematical Models of Linear Systems


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.

AnalogousSystem: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. Many circuit theorems, impedance concepts can be applicable.2. 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.TranslationalSystem:

It has 3 types of forces due to elements.

1. Inertial Force:Due to inertial mass,

2. Damping Force [Viscous Damping]: Due to viscous damping, it is proportional to velocity & is given by,

3. Spring Force: Spring force is proportional to displacement.

. Damping force is denoted by either D or B or F

D

Fk

.

M F(t) F

m

t

M.a

2

2

Where, . .

. .

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2 Mathematical Models of Linear Systems

Designed By: HPK Kumar

Rotationalsystem:1. Inertial Torque:2. Damping Torque:

3.

Spring Torque :

Analogousquantitiesintranslational&Rotationalsystem:The electrical analog of the mechanical system can be obtained by,

(i) Force Voltage analogy: (F.V)(ii) Force Current analogy: (F.I)Sl.

No.

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) Torsional spring stiffness

()Reciprocal

of Capacitance (1/C)

Reciprocal

of Inductance (1/L)

5. Linear displacement () Angular displacement () Charge (q) Flux ()6. Linear velocity () Angular Velocity (w) Current (i) Voltage (v)

DAlembertsPrinciple: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.

ExampleProblems:(1) Obtain the electrical analog (FV & FI analog circuits) for the Machine system shown & also write the equations.

.

Where,

D2

D1

22

11

Ft

Free Body diagram

M1

FtM1

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1Control Systems


G(S)R(S) C(S)

G(S) = 1

Transfer Functions


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.(2) Replace the terms by S & by 1/S.(3) Eliminate all the variables except the desired variables.

ImpulseResponseoftheLinearSystem:

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.

PROBLEMS:

(1) ObtaintheTransferFunction(TF)ofthecircuitshownincircuit1.0Solution:

i.e., the Laplace Transform of the system o/p will be simply the Transfer function of the system.

. . 1

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

R

Ci

Circuit 1.0

R

i(S)

Laplace Transformed network

1

Contd

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2 Transfer Functions


.., 1 & 1

(2) ObtaintheTFofthemechanicalsystemshownincircuit2.

. .

1

1

1 Where, = RC

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3Control Systems


(3) TransferFunctionofanArmatureControlledDCMotorincircuit3.0:

The air gap flux is proportional to the field current.i.e.,

. Where, Kf is a constant.The torque developed by the motor Tm is proportional to the product of the arm current & the air gap flux.

Where, Ka & Kf are the constants.

Since the field current is constant, Where, KT is Motor torque constant.The motor back e.m.f is proportional to the speed & is given by,

Where, Kb is back e.m.f constant.The differential equation of the armature circuit is,

The torque equation is, 2

2 Taking LT for above equation, we get

------------------------------- (1)

(A) Taking LT for the torque equation & equating, we get

Ra La

ia

iaVi Eb F

If = Constant

Vf

Tm

Let,

Ra Resistance of armature in s.

La Inductance of armature in Hs.

ia Armature current. & ifField current.

Vi Applied armature voltage.

Eb Back e.m.f in volts.

Tm Torque developed by the motor in N-m.

Angular displacement of motor shaft in radiansJ

Equivalentmoment of inertia of motor & load referred to the motor shaft.

FEquivalentViscous friction co-efficient of motor & load referred to the motor shaft.

circuit 3.0

J,

F

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2

------------------------------ (2)

-------------------------- (B)Taking LT for back e.m.f equation, we get

------------------------------- (C)Substituting the values ofIa(S) & Eb (S) from equation (C) & (2) in equation (1), we get

.

Theblock

diagram

representation

of

armature

controlled

DC

Motor

can

be

obtained

as

follows,

From equation (A),

From equation (B),

From equation (C),

The complete block diagram is as shown below,

1

Vi(S)

Eb(S)

-

Ia(S)

(S)Ia(S)

Eb(S)(S)

2 (S)

1

Vi(S)

Eb(S)-

Ia(S)

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5Control Systems


(4) TransferfunctionofFieldControlledDCMotorincircuit4.0:

In the field controlled DC motor, the armature current is fed from a constant current source.

Where, Ka & Kf are the constants.The KVL equation for the field circuit is,

On Laplace Transform,

. ---------------------------------- (1)

--------------------------------- (A)

The torque equation is , Where, KT is Motor torque constant.

On Laplace Transform,

.

2 .

--------------------------- (2)

------------------------- (B)Substituting the value of from equation (2) in equation (1), we get

Rf Lf

ifVf

Tm

Ia = ConstantLet,

Rf Field winding resistance.

Lf

Field winding inductance.

VfField control voltage.

If Field current.

TmTorque developed by motor.

J,

F

JEquivalentmoment of inertia of motor & load referred to the motor shaft.

FEquivalentViscous friction co-efficient of motor & load referred to the

motor shaft.

Angular displacement of motor shaft.

circuit 4.0

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2 .

TheblockdiagramrepresentationoffieldcontrolledDCMotorcanbeobtainedasfollows,

From equation (A),

From equation (B),

The complete block diagram is as shown below,

(5) ObtaintheTF forthenetworkshownincircuit5.0:

Solution:

Applying KVL to this circuit,

---------------------------------- (1)

. 1

1 1 . 1

12

2

(S)If(S)

1

Vf(S) If(S)

1

Vf(S) If(S) (S)

circuit 5.0

R C

R CVi Vo

Laplace Transformed network R

R 1

1

Vi (S) Vo (S)

I(S) I1(S) I2(S)

1 2 Let,

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7Control Systems


(6) FindtheTF forthenetworkshownincircuit6.0:

Solution:

Writing KVL for loop (1), we get

1 105 105

2 105

1 .

105 1 2

1 ------------------------------------------- (1)

Writing KVL for loop (2), we get

10

0

2 110 11 -------------------------------------------- (2) 2. 10

6 2

. 106 --------------------------------- (3)

Substituting for I1 (S) from equation (2) in (1), we get

2. 10 11 . 105 1 21

105 1 2 10 111 1

1

From equation (3) the above equation becomes,

105

. 106 10

2 21 10

Laplace Transformed network

Circuit 6.0

100 k 1M

10 F 1 FVi

Vi (S)

1010

10

10 V0 (S)

Loop 1 Loop 2

10

10

21 10

106

V0(S)110 11

I1(S)

+

I2(S)

1 1

10 1

Vi(S)

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2 Block Diagrams

PROBLEMS:

Reduce the Block Diagrams shown below:

(1)

Solution: By eliminating the feed-back paths, we get

Combining the blocks in series, we get

Eliminating the feed back path, we get

+

- - -

+

-

+C(S)

-

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3Control Systems

(2)

Solution: Shifting the take-off beyond the block , we get

Combining and eliminating (feed back loop), we get

Eliminating the feed back path , we get

Combining all the three blocks, we get

R(S)- -

-

C(S)

R(S) - -

-

C(S)

R(S) - -C(S)

R(S) C(S)

R(S)-

C(S)

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4 Block Diagrams

(3)

Solution: Re-arranging the block diagram, we get

Eliminating loop & combining, we get

Eliminating feed back loop

Eliminating feed back loop , we get

C(S)R(S)- - -

C(S)R(S)

- -

R(S) C(S)

-

C(S)R(S)

C(S)R(S)-- -

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1Control Systems

Signal Flow Graphs


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 node is 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 the traversal 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 the product 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 apath from i/p node to the o/p node which doesnt cross any node more than

once.

12.Forward Path Gain: It is the product of branch transmittances of a forward path.MASONS 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.

Masons gain formula for the determination of overall system gain is given by,

Where, Path gain of forward path.

Determinant of the graph.

The value of the for that part of the graph not touching the forward path.

T Overall gain of the system.

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5Control Systems

Signal flow graph:

No. of forward paths:

No. of individual loops:

(6) Reduce the Block Diagram shown.

Solution:

Shifting beyond , we get

C(S)R(S)

C(S)R(S)

-

-R(S)

- +

C(S)

-

-R(S)

- +

C(S)

R(S)-

C(S)

R(S) C(S)

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6 Signal Flow Graphs



Eliminating the another feed back loop , we get

Signal flow graph:

-

R(S)

- +

C(S)

-

R(S)

+

C(S)

-

R(S)

+

C(S)

R(S)

+

C(S)

C(S)R(S)

R(S) C(S)

Contd......

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8 Signal Flow Graphs

Eliminating loop, we get

(9) Using Masons gain rule, obtain the overall TF of a control system represented by the signal flow graphshown below.

Solution:


Individual loops:

Two non-touching loops = 0

(10) Construct signal flow graph from the following equations & obtain the overall TF.

-

Contd......

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13Control Systems

Substituting x value in the block diagram. The block diagrambecomes,

Signal flow graph:


No. of individual loops: Two non-touching loops = 0

(16) Obtain TF, using block diagram algebra & also by using Masons Gain Formula. Hence Verifythe TF in both the methods.

Contd......

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17Control Systems

Hence,

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1Control Systems

System Stability


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 includeabsolute & 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. From the 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 systemresponse 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 S-plane, the response will be oscillatory without damping. Such a system will be termed as critically stable.

(3) When one or more roots lie on the RHS of S-plane, the response will exponentially increasein magnitude; there by the system will 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 its response to a bounded disturbing signal vanishes ultimately as time

t approaches infinity.

(3) A system is unstable, if its response to a bounded disturbing signal results in an o/p ofinfinite amplitude or an Oscillatory signal.

(4) If the o/p response to a bounded i/p signal results in constant amplitude or constantamplitude oscillations, then the system may be stable or unstable under some limited constraints. Such a

system is called Limitedly Stable system.

(5) If a system response is stable for a limited range of variation of its parameters, it is calledConditionally Stable System.

(6) If a system response is stable for all variation of its parameters, it is called Absolutely Stablesystem.

Routh-Hurwitz Criteria:

A designer has so often to design the system that 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 system to be stable are,

(1)None of the co-efficient of theCharacteristic equation should be missing or zero.(2)All the co-efficient should be real & should have the same sign.

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2 System Stability

A sufficient condition for a system to 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,

0

0

0 0

0 0 0

Similarly we can 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 arenot available. Also it cannot distinguish between real & complex roots.

Special cases in Routh-Hurwitz criteria:

(1)When the term in a row is zero, but all other terms are non-zeroes then substitute a small positivenumber 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 system is limitedly stable.

Problems:

COMMENT 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

Where,

The no. of sign changes in the column = zero.

No roots are lying in the RHS of S-plane.

The given System is Absolutely Stable.

: :

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5Control Systems

(10) The open-loop TF of a unity feed back system is given by the aboveexpression. Find the value of K for which the system isjust stable.

Solution:

The characteristic equation is

1 23 2K

9 (15+K) 0

2K 0

0 0

2K 0 0

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. Thecharacteristic equation is

Solution:

1 (2K+3)

5K 10

0

10 0

The range of K is

(12) A proposed control system has a system & a controller as shown. Access the stability of the systemby a suitable method. What are the ranges of K for the system to be stable?

Solution: The characteristic equation is

16 (1+K)

8 K

0

K 0

(13)

(i) K > 0(ii) 192 K > 0

K < 192

(iii) (192 K)(15+K) 162K > 0(for the max. value of K)

From this evaluate for K,

Using,

Considering the positive value ofK,

So, 0 < K < 61.68

(i) K > 0(ii)

Considering the positive value ofK,

(i) K > 0(ii)

The range of K is,

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7Control Systems

(17)1 4 6

2 5 2

1.5 5 0

-1.666 2 0

6.8 0 02 0 0

(18)Solution:

1 11

6 6

10 0

6 0

(19)Solution:

No. of sign changes = 1

The system is Unstable.

(20)+ve 1 2 4

+ve 1 2 1

+ve 3 0

-ve 1 0

+ve 0 0

+ve 1 0 0

(21)+ve 2 6 1

+ve 1 3 1

+ve -1 0

+ve1 0

-ve 0 0

+ve 1 0 0

(22)

1 -5

2 -6

-2 0

-6 0

(i) No. of sign changes = 2.(ii) Two roots lie on RHS of

S-plane.

(iii) The system is Unstable.

No sign changes.

The system is Absolutely Stable.

No. of sign changes = 2.


No. of sign changes = 2.


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11Control 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 loop-

control 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 0

52.5 0

0 0

0 0

Hence,

(37) A feed back system has open-loop transfer function Determine themaximum 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)(iii) The Auxiliary equation for the row is

When

(i)(ii)

The range of K is for the system to be

stable.

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1Control Systems


Root Locus Plots


It gives complete dynamic response of the system. It provides a measure of sensitivity of roots to thevariation 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.

RulesfortheConstructionofRootLocus(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 thispoint is odd.

(5) The root locus branches that tend to , do so along the straight line.Asymptotes making angle with the real axis is given by

1800

, Where, n=1,3,5,

P = No. of poles & Z =No. of zeroes.

(6) The asymptotes cross the real axis at a point known as Centroid. i.e.,

(7) The break away or the break in points [Saddle points] of the root locus or determined from the roots of theequation

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

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1Control Systems

Bode Plots


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 relative stability are gain & phase

margin which are visibly seen on sketch.

(4)If modification of an existing system is to be studied, it can be easily done on a Bode Plot.Initial Magnitude:

If ,

,

,

,

,

,

,

Phase Plot: Magnitude Plot:

GCF

+ve PM

GCF

GCF

GCF

-ve PM

line

line

PCF

PCF

PCF

PCF

+ve GM

-ve GM

0 dB line

0 dB line

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5Control SystemsExamination Problem (Mar/Apr 99):

(7)The sketch given shows the Bode Magnitude plot for a system. Obtain the Transfer function.

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.

Solution:

dB

A

B

40

(Z)C

D

E

(P)

(P)(DZ)

dB

(Z)

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6 Bode PlotsExamination Problem (97):

(9)The bode plot (magnitude) of a unity feed back control system is as shown in the fig. Obtain thephase plot.

Solution:

dB

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Control System Notes by HPK Kumar JWFILES

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