Some important tips for control systems

CONTROL SYSTEM

NOTES (For Bachelor of Engineering)

Notes by:

PROF. SHESHADRI G. S

(8) Bode Plots

(7) Root Locus Plots

(6) System Stability

(5) Signal Flow Graphs

(4) Block diagram

(3) Transfer functions

(2) Mathematical model of linear systems

(1) Introduction to Control system

Soft Copy material designed by:

KARTHIK KUMAR H P

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Index

1 Control Systems

Introduction to Control Systems

Control System means any quantity of interest in a machine or mechanism is maintained or altered in accordance with desired manner. OR A 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: Controller means 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 an Open 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 the Closed 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.

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 be Stable.

A system is said to be Unstable, 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 certain value 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 value of . This can be done by selecting proper feed back.

G(S)

H(S)

R(S) B(S)

E(S)

C(S) -

& –

Hence,

= (1)

3 Control Systems

(d) Noise: Examples are brush & 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, as Linear & Non- Linear Systems. (b) The type of the signal, as Time Varying, Time Invariant, Continuous data, Discrete data systems etc., (c) The type of system components, as Electro Mechanical, Hydraulic, Thermal, Pneumatic Control systems etc., (d) The main purpose, as Position control & Velocity control Systems.

1. Linear & Non-Linear Systems: In a linear system, the principle of superposition can be applied. In non - linear system,

th is p rin cip le ca n ’t b e a p p lied . T h erefo re a lin ea r system is th a t w h ich o b eys su p erp o sitio n 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.

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, µp, µc 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.

3. Discrete Data Systems:

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 output’s (M IM O - 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.

2. It is easier to build.

3. In general it is more stable as the feed back is absent.

4. If non- lin ea rity’s a re present; the system 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 to perform accurately because of the feed back.

2. It is difficult to build.

3. Less Stable Comparatively.

4. Even under the presence of non- lin ea rity’s th e system o p era tes b etter th a n o p en loop 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)

5 Control Systems

Block Diagram of Closed Loop System:

Thermometer Block Diagram of Temperature Control System:

Temperature Control of Passenger Compartment Car:

Control Elements

Feed Back elements

Ref. i/p

E(S)

Controlled o/p

E(S) Plant

Controller

Actuator

Desired Temperature

i/p

Controller

Sensor

Radiation Heat Sensor

Air Conditioner Passenger Car

Sensor

O/p

Sun Ambient Temperature

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

* * * * * * * * * * * * * * * *

A/D Converter Interface

Relay Interface Amplifier Programmed i/p

Electric Furnace

Heater

Thermo meter

1Control Systems

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.

Analogous 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. 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.

Translational System: 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)

Fm t M . a2

2

Where, . .

. .

2 Mathematical Models of Linear Systems

Rotational system:

1. Inertial Torque:

2. Damping Torque:

3. Spring Torque :

Analogous quantities in translational & Rotational system:

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)

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.

.

Where,

D2

D1

22

11

F t

Free Body diagram

M1

F tM1

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.

Impulse Response of the Linear 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.

PROBLEMS:

(1) Obtain the Transfer‐Function(TF) of the circuit shown in circuit 1.0

Solution:

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

2 Transfer Functions

. ., 1 & 1

(2) Obtain the TF of the mechanical system shown in circuit 2.

.

. 1

1

1 Where, = RC

3Control Systems

(3) Transfer Function of an Armature Controlled DC Motor in circuit 3.0:

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

ia Vi Eb F

If = Constant

Vf

Tm

Let,

‘Ra’ Resistance of armature in Ω’s.

‘La’ Inductance of armature in H’s.

‘ia’ Armature current. & ‘if’ Field 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 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.

circuit 3.0

J,

F


2

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

-------------------------- (B)

Taking LT for back e.m.f equation, we get

------------------------------- (C)

Substituting the values of Ia (S) & Eb (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),

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)

5Control Systems

(4) Transfer function of Field Controlled DC Motor in circuit 4.0:

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

if Vf

Tm

Ia = Constant Let,

Rf Field winding resistance.

Lf Field winding inductance.

Vf Field control voltage.

If Field current.

Tm Torque developed by motor.

J, F

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.

Angular displacement of motor shaft.

circuit 4.0


2 .

The block diagram representation of field controlled DC Motor can be obtained as follows,

From equation (A),

From equation (B),

The complete block diagram is as shown below,

(5) Obtain the TF for the network shown in circuit 5.0:

Solution:

Applying KVL to this circuit,

----------------

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

. 1

1 1 . 1

1 2

2

(S) If (S)

1

Vf (S) If (S)

1

Vf (S) If (S)

(S)

circuit 5.0

R C

R C Vi Vo

Laplace Transformed network R

R 1

1

Vi (S) Vo (S)

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

τ 1 τ 2 τ

Let, τ

7Control Systems

(6) Find the TF for the network shown in circuit 6.0:

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 . 106 2 .

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

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

2 . 10 11 . 105 1

21

105 1

2 10 11 1 11

From equation (3) the above equation becomes,

105

. 106

10 2 21 10

Laplace Transformed network

Circuit 6.0

100 k Ω 1M Ω

10 F 1 F Vi

Vi (S)

1010

10

10 V0 (S)

Loop 1 Loop 2

1010 2 1 1 0

106

V0(S) 110 11

I1(S)

+

I2(S)

11

10 1

Vi(S)


1 Control Systems

Block Diagrams

It is a representation of the control system giving the inter-relation between the transfer function of various components. The block diagram is obtained after obtaining the differential equation & Transfer function of all components of a control system. The arrow head pointing towards the block indicates the i/p & pointing away from the block indicates the o/p.

After obtaining the block diagram for each & every component, all blocks are combined to obtain a complete representation. It is then reduced to a simple form with the help of block diagram algebra.

The following block diagram reduction algebra is used, (1) Blocks in Cascade [Series] :

(2) Combining blocks in Parallel:

(3) Eliminating a feed back loop:

(4) Moving a take-off point beyond a block:

(5) Moving a Take-off point ahead of a block:

If is the TF,

G2(S)

G(S) C(S)

(2)

(2)

(3) (6)

(2)

(2)

(3) (6)

G(S)

(3)

(6)

(2) (6)

(3)

(2)

(2) (6)

(6)

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) -

3 Control 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)

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) - - -

1 Control Systems

Signal Flow Graphs

By: Sheshadri.G.S. 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 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 a path from i/p node to the o/p node w hich doesn’t cross any node m ore than

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

M ASON’S GAIN FORM ULA: 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,

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.

2 Signal Flow Graphs

Problems:

(1) Obtain the closed loop TF, by using M ason’s gain form ula.

Solution: M ason’s gain form ula is, No. of forward paths:

No. of individual loops:

No. of three non-touching loops = 0.

(2) Obtain the closed loop TF, by using M ason’s gain form ula.

Solution: M ason’s gain form ula is,

Forward Paths:

Gain Products of all possible combinations of two non-touching loops:

Contd......

R(S)

C(S)

R(S) C(S)

3 Control Systems

No. of individual loops: Two Non-touching loops:

(3) Construct a signal flow graph from the following equations. Obtain overall TF using M ason’s gain form ula.

Where is i/p variable & is o/p variable.

Solution:

No. of forward paths:

Individual loops: Two non-touching loops:

Three non-touching loops = 0 M ason’s gain form ula is,

(4) Obtain by Block Diagram Reduction method & verify the result by signal flow graph.

Contd......

+

R(S)

+

+

C(S) -


Solution: Re-arranging the summing points, Signal flow graphs:



(5) Obtain the TF & Verify by signal flow graph. Solution: Shifting the take-off point ahead of the block . The BD reduces to,

M ason’s gain form ula is,

-

- R(S) C(S)

R(S)

-

-

C(S)

R(S)

C(S)

Contd......

R(S) C(S)

C(S)

+

- - R(S)

+

5 Control Systems

Signal flow graph:



(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)




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......

7 Control Systems



(7) Obtain the closed loop TF by using M ason’s gain form ula.

Solution: No. of forward paths:

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

(8) Obtain the TF of the closed loop control system represented by the Block Diagram shown below using block diagram reduction method. Solution: Shifting the take off point of beyond block & Simplifying for the blocks , we get

R(S) C(S)

- -

-

- -


Eliminating loop, we get

(9) Using M ason’s gain rule, obtain the overall TF of a control system represented by the signal flow graph shown below.

Solution:


Individual loops:

Two non-touching loops = 0

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

-

Contd......

9 Control Systems



Three non-touching loops:

Four non-touching loops = 0

(11) Obtain the TF using M ason’s gain form ula.

Two non-touching loops:

Contd......




Three non-touching loops = 0







Contd......

11 Control Systems



(14) Draw the signal flow graph for the Block Diagram shown in fig. Hence obtain , Using M ason’s gain form ula. Solution:


– –





Contd......



(15) Obtain TF, using block diagram algebra & also by using Masons Gain Formula. Hence Verify

the TF in both the methods. Solution:

Same block diagram can be re-arranged as shown below.

Shifting the take-off points beyond we get

-

13 Control Systems

Substituting ‘x’ value in the block diagram . T he block diagram becom es, Signal flow graph: No. of forward paths:

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

(16) Obtain TF, using block diagram algebra & also by using Masons Gain Formula. Hence Verify

the TF in both the methods.

Contd......


Solution: Same Block Diagram can be written as,

Substituting the value of ‘x’ Signal flow Graph:

15 Control Systems


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

(17) Find the output Solution:

(i) Let then we can find No. of forward paths:

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



(ii) Let Determine



‘ ’ rem ains sam e.

(iii) Let Determine



(iv) Let Determine (i.e., R esponse at ‘2’ w hen source ‘2’ is acting). (figure is in next page) No. of forward paths:


17 Control Systems

Hence,

1 Control 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 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. 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 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 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 increase in 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 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 system is 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) If a system response is stable for all variation of its parameters, it is called Absolutely Stable system.

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 the Characteristic equation should be missing or zero.

(2) All the co-efficient‟ should be real & should have the sam e sign.

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 are not 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 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 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.

: :

3 Control Systems

(2)

4 2

3 5

-4.66 0

5 0

(3)

1 1 5

2 4 0

-1 5 0

14 0 0

5 0 0

(4)

1 2 3 +ve

2 4 1 +ve

0 2.5 0 +ve

1 0 -ve

0 0 +ve

1 0 0 +ve

(5)

1 2 3

2 4 6

2 6 0

-16 0 0

6 0 0

(6)

1 8 20 16

2 12 16 0

2 12 16 0

6 16 0 0

2.66 0 0 0

16 0 0 0

The no. of sign changes in the column = 2 Two roots are lying in the RHS of S-plane.

The given System is unstable.

The no. of sign changes in the column = 2 Two roots are lying in the RHS of S-plane.


The no. of sign changes in the column = 2

Two roots are lying in the RHS of S-plane.


All Zero row. A.E. is The no. of sign changes in the column = 2

Two roots are lying in the RHS of S-plane. The given System is unstable & limitedly stable.

A.E. is The no. of sign changes in the column = 0

The System is limitedly stable. (Because, in the row all the elements are zero).

4 System Stability

(7) The open loop TF of a unity feed back system is . Find the restriction on ‘K’, So that

the closed loop is stable. Solution:

10 1

6.5 K

0

K 0

For, the closed loop system is stable.

(8) Solution:

1 4 K

3 1 0

K 0

0 0

K 0 0

For, the closed loop system is stable

(9) Determ ine the value of ‘K & b’, So that the system open loop T.F. oscillates at a

frequency of 2 radians. ( Assuming unity feed back i.e., H(S)=1 ) Solution:

The characteristic equation is

1 (3+K)

b (1+K)

0

(1+K) 0

From equations (9.1) & (9.2)

Either from (9.1) or (9.2)

For Stable system,

(i)

(ii)

For Stable system,

(i)

(i)

5 Control Systems

(10) The open-loop TF of a unity feed back system is given by the above

expression. Find the value of ‘K’ for w hich the system is just stable. Solution:


1 23 2K

9 (15+K) 0

2K 0

0 0

2K 0 0

When the value of „K ‟ is 61.68 the system is just stable. (11) Using Routh-Hurw itz criteria, find out the range of ‘K’ for w hich the system is stable. The

characteristic equation is Solution:

1 (2K+3)

5K 10

0

10 0

T he range of K is „ ‟ (12) A proposed control system has a system & a controller as shown. Access the stability of the system

by a suitable m ethod. W hat 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 of „K‟,

So, 0 < K < 61.68

(i) K > 0 (ii)

Considering the positive value of „K‟,

(i) K > 0 (ii)

The range of K is,

6 System Stability

(14)

(15)

(16) The open-loop TF of a control system is given by Find the range of

the gain constant ‘K’ for stability using Routh-Hurwitz criteria. Solution:


No. of sign changes = 2,

Two roots lie on RHS of S-plane.

The system is Unstable.

Auxiliary Equation:

No sign changes in the column.

T he system is “Limitedly Stable” .

(i) (ii)

For the system to be stable the range of the K is,

(i) (ii)

(iii)

(iv)

The Range of K is for the system to be stable.

7 Control Systems

(17)

1 4 6

2 5 2 1.5 5 0

-1.666 2 0

6.8 0 0

2 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 +ve 1 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. The system is Unstable.

No. of sign changes = 2. The system is Unstable.

8 System Stability

1 -2 -7 -4

1 -3 -4 0

1 -3 -4 0

-1.5 -4 0 0

-16.66 0 0 0

-4 0 0 0 (23)

1 5 8 4

2 8 8 0

1 4 4 0

2 4 0 0

4 0 0 0

(24)

1 2 0.5

0(4) 0(4) 0(0) 1 0.5 0

2 0 0

0.5 0 0 (25)

1 2 2 0.8 1.6 0 - 4 2 0 2 0 0 2 0 0

(26)

1 7500 34.5 7500K 0

7500K 0

Auxiliary equation:

No. of sign changes = 1. The system is Unstable &

Limitedly Stable.

Auxiliary equation:

Auxiliary equation:

No. of sign changes = 0 The system is Limitedly Stable.

Auxiliary Equation:

No. of sign changes = 0 The system is Limitedly Stable.

Auxiliary Equation:

No. of sign changes = 2 The system is Unstable.

(i) (ii)

For a stable closed loop system there should not be any sign change among the elements of the column in the Routh- Hurwitz table. This requires & T he range of „K ‟ is for the system to be stable.

9 Control Systems

(27) Given , .

Solution: The characteristic equation is

1 (21+K)

10 13K 0

13K 0

(28) ,

Solution: The Characteristic equation is

1 15 2K 7 25+K 0

2K 0

0 0

2K 0 0 (29) Solution:

1 5 15 20K 10 0

15 0

0 0

15 0 0 (30) Determ ine the values of ‘K’ & ‘a’. Such that the system oscillates at a frequency of 2 rad/sec. Solution: The characteristic equation is

1 (2+K)

a (1+K)

0

(1+K) 0

(31)

(i) (ii)

The range of K is , for the system to be stable.

(i) (ii)

(iii)

To solve for the K value, By simplifying,

Hence, the range of K is

(i) (ii)

(iii)

T he gain „K ‟ alw ays should be a real quantity. Hence this system is always Unstable.

, becomes

From (1) & (2), &

10 System Stability

1 2 3 1 -2 0

5 0

0 0

5 0 0 (32)

9 10 -9 0

0

0 0

0 0 (33)

1 24 24 23 9 24 24 15 21.33 21.33 21.33 0 15 15 15

7.5 15 0 0

0 0 0

15 0 0 0 (34)

1 9 4 36 5 9 20 36 7.2 0 28.8 0

36 0 0

0 0 0

36 0 0 0

(35) ,

Solution: The Characteristic equation is

1 2 3 K

0

K 0

No. of sign changes = 2


No. of sign changes = 3


Auxiliary equation: No. of sign changes = 2


Auxiliary equation: No. of sign changes = 2


(i) (ii)

The range of K is

9 0 36

11 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 loop-

control system . Determ ine the value of ‘K’ w hich w ill 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 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)

(iii) The Auxiliary equation for the row is

When

(i) (ii)

The range of K is for the system to be stable.

1Control Systems

Root Locus Plots

By: Sheshadri.G.S.

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.

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 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

1 Control Systems

Bode Plots

By: Sheshadri.G.S.

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 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

2 Bode Plots

Problems:

(1) To find the angle for the quadratic term

Solution: Put,

(2) Determine the transfer function. Whose approximate plot is as shown.

Solution:

The corner frequencies are & at , the slope changes from to .

Therefore there must be a factor . Since the initial slope is , There must be a pole at the

origin i.e., at

The slope at changes from to due to a factor . Open-loop TF is,

To find the value of ‘K’ :

(3) Find the open-loop TF of a system, whose approximate plot is as shown.

dB B

A 40 dB C

D

A

B

C

20

dB

A

B

C

3 Control Systems

Solution:

Corner frequencies are


but

Since the has a slope of , the corresponding factor is

Plot between & , It is having a slope of . Therefore at , there must be a zero & the factor is in the numerator.

At ,the slop is changing from to & the corresponding factor is in the Numerator.

At , the slope has changed from to due to a factor in the denominator.

Therefore the open-loop transfer function is,


Solution:

(1) Slope of the first line = indicating the term

(2) At slope changes from to indicating a term in the numerator.

(3) At slope changes to indicating a term in the Numerator.

(4) At slope changes to indicating a term in the Denominator.



dB

4 Bode Plots

Calculation of :

Calculation of :

The Open-loop transfer function is,


Solution:


Let , be the origin.

(6) Derive the Transfer function of the system from the data given on the Bode diagram given below.

[[

Solution: Between there is a decrease of there is a decrease of dB,


Therefore, the Open-loop transfer function is,

dB

A

B

dB

5 Control Systems

Exam ination Problem (M ar/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)

6 Bode Plots

Examination Problem (97):

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

Solution:

dB

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Some important tips for control systems