EC2255- Control System Notes( solved problems)

EC2255- Solved Problems in Control System

IV Semester ECE

Control Systems PHYSICAL SYSTEMS: INTRODUCTION:         



First step towards analysis of a control system is preparation of a mathematical model which is linear over a satisfactory range operating conditions giving us the property of linearity and superposition. A model may be defined as a representation of the essential aspects of the system which presents knowledge of the system in a usable manner. To be useful, the model must not so complicated that it cannot be understood and thereby be unsuitable for analysis. The components of a control system are diverse in nature and may include electrical, mechanical, thermal and fluidic devices. At the same time, it must not so complicated that it cannot be understood and thereby be unsuitable for analysis. At the same time, it must not be oversimplified and trivial. While dealing with control systems, we shall be concerned mostly with dynamic systems. The behavior of such systems is described in the form of differential equations. Although these will normally be nonlinear, it is customary to linearize them about an operating point to obtain linear differential equations. The components of a control system are diverse in nature and may include electrical, mechanical, thermal and fluidic devices. The differential equations relating the input and output quantities for these devices are obtained using the basic law of physics. These include balancing forces, energy and mass. In practice, some simplifying assumptions are often made to obtain linear differential equations. With constant coefficients, although in most cases exact analysis would lead to nonlinear partial differential equations. For most physical devices one may classify the variables as either THROUGH or ACROSS variables, in the sense that the former refer to a point while the latter are measured between two points. The input , output relations of various physical components of a system is governed by differential equation. The mathematical model of a control system constitutes a set of differential equation.

Solved by A.Devasena., Associate Professor., Dept/ECE

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SYSTEM Electrical Mechanical (translational) Mechanical (rotational) Thermal Fluidic

THROUGH VARIABLE Current I Force, F Torque, T Rate of flow of heat energy,q Volumetric rate of fluid flow,Q

IV Semester ECE

ACROSS VARIABLE Potential difference or voltage V. Relative velocity,v Relative angular velocity ,ω Difference in temp. T Difference in pressure,P

A mathematical model will be linear if the differential equations describing the system have constant coefficients. If the coefficients of the differential equations describing the system are constants, then the model is linear time invariant. If the coefficients of the differential equations governing the system are functions of time, then the model is linear time varying. The differential equations of a linear time invariant system can be reshaped into different form for the convenience of analysis. One such model for single input and single output system analysis is transfer function of the system. The transfer function of a system analysis is transfer function of the system. The transfer function of a system is defined as the ratio of laplace transfer function of the system. The transfer function of a defined as the ratio of laplace transfer function of a system is defined as the ratio of laplace transform of output to the laplace transform of input with zero initial conditions.

An equation describing a physical has integrals and differentials. The response can be obtained by solving such equations. The steps involved in obtaining the transfer function are:  

Write differential equation of the system. Replace terms involving d/dt by s and integral of dt by 1/s.

Advantages of open loop system. The advantages of open loop system are 1. Such systems are simple in construction. 2. Very much convenient when input is difficult to measure. 3. Such systems are easy for maintenance point of view. 4. Generally these are not troubled with problems of stability. 5. Such systems are simple to design and hence economical. Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

Disadvantages of open loop system. The disadvantages of open loop system are 1. Such systems are inaccurate and unreliable because accuracy of such systems are totally dependent on the accurate precalibration of the controller. 2. Such systems give inaccurate results if there are variation in the external environment i.e. systems cannot sense environmental changes. 3. Similarly they cannot sense internal disturbances in the system, after the controller stage.

Advantages of closed loop system. The advantages of closed loop system are: 1.accuracy of such system is always very high because controller modifies and manipulates the actuating signal such that error in the system will be zero. 2. Such systems senses environmental changes, as well as internal disturbances and accordingly modifies the error. 1. In such system, there is reduced effect of non-linearites and distortions. Comparison between open loop system and closed loop system OPEN LOOP SYSTEM Output measurement is not required for operation of the system. Highly affected by non-lineariteis Highly sensitive to the disturbances and environmental changes Feedback element and error detector are absent Generally stable in nature

CLOSED LOOP SYSTEM Output measurement is necessary. Reduced effect of nonlinearities. Less sensitive to disturbances and environmental changes. Feedback element and error detector are absent Stability is the major consideration while designing.

Linear system. A system is said to be linear if it obeys the principle of superposition and homogeneity. The principle of superposition states that the response of the system to a weighted sum of the responses of the system to each individual input signals. The system is said to be linear, if it satisfies the following two properties:  Adaptive property that is for any x and y belonging to the domain of the function f, we have F(x+y) = f(x) +f(y) Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE



Homogeneous property that is for any x belonging to the domain of the function f and for any scalar constant α We have  F(αx) = αf(x) These two properties together constitute a principle of superposition. Most of the systems are non-linear in nature because of different non-linearities such as saturation, friction, dead zone etc. present in the system.

Definition of Transfer Function Transfer function of a given system is defined as the ratio of the laplace transform of output variable to laplace transform of input variables at zero input conditions.

Properties of transfer functions. The properties of transfer function are as follows:  The transfer function of a system is the laplace transform of its impulse response. I.e. if the input to a system with transfer function P(s) is an impulse and all initial conditions are zero, the transform of the output is P(s).  The roots of the denominator are the system poles and the roots of the numerator are system zeros. The system stability can be described in terms of the location of the roots of the transfer function. Advantages of transfer function.  It helps in the stability analysis of the system.  It helps in determining the important information about the system Poles, zeros, characteristic equation etc.  Once transfer function is known, output response for any type of reference input can be calculated.  The system differential equation can be easily obtained by replacing variable ‘s’ by d/dt. Disadvantages of transfer function. The disadvantages of transfer function approach are:  Only applicable to linear time invariant systems.  It does not provide any information concerning the physical structure of the system. From transfer function, physical nature of the system, whether it is electrical, mechanical, thermal or hydraulic cannot be judged. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System 

IV Semester ECE

Effects arising due to initial conditions are totally neglected. Hence initial conditions loose importance.

Important features of feedback.  Reduced effects of non-linearities and distortion.  Increased accuracy.  Reduced sensitivity of the ratio of the output to input to variations in system characteristics.  Tendency toward oscillation or instability.

The basic elements used for modeling mechanical translational system: The model of mechanical translational system can be obtained by using three basic elements mass spring and dashpot. Translational system Consider a mechanical system in which motion is taking place along a straight line. Such systems are of translational type. These systems are characterized by displacement, linear velocity and linear acceleration. Definition of torque. This is the motion about a fixed axis. In such systems, the force gets replaced by a moment about the fixed axis. I.e. {force x distance from fixed axis} which is called torque. Definition of friction. Whenever there is a motion, there exists a friction. Friction may be between moving element and fixed support or between two moving surfaces. Friction is also nonlinear in nature. The types of friction. Friction can be divided into three types. They are  Viscous friction.  Static friction.  Coulomb friction. The two types of analogies for mechanical system are force-voltage analogy and force-current analogy. Analogous systems Systems whose differential equations are identical form are called analogous systems.


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IV Semester ECE

Two types of variables in physical system. The two variables in physical systems are through variables and across variable. Through variables refer to a point, the across variable is measured between two points. The other names for force voltage analogy and force current analogy are Force voltage analogy- Loop analysis. Force current analogy- nodal analysis.

BLOCK DIAGRAM ALGEBRA; INTRODUCTION: In block diagram, the system consists of so many components. These components are linked together to perform a particular function. Each component can be represented with the help of individual block. NEED FOR BLOCK DIAGRAM REDUCTION: Block diagrams of some of the systems turn out to be complex, such that the evaluation of their performance required simplification (or reduction) of block diagrams which is carried out by block diagram rearrangements. DEFINITION: A block diagram of a system is a pictorial representation of the functions performed by each component of the system and shows the flow of signals. ADVANTAGES OF BLOCK DIAGRAM:  Very simple to construct the block diagram for complicated systems.  Individual as well as overall performance of the system can be studied by using transfer functions shown in the block diagram.  Overall closed loop transfer function can be easily calculated using block diagram rules.  The function of the individual element can be visualized from the block diagram. DISADVANTAGES OF BLOCK DIAGRAM: The disadvantages of block diagram are:  Block diagram does not include any information about the physical construction of the system.  Source of energy is generally not shown in the block diagram, so block diagram for a given system is not unique. The basic components of block diagram are block, branches, summing point, arrows. Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

BLOCK: It indicates the function of particular system. R(s) is the reference or controlling variable. G(s) is the transfer function of the particular system.C(s) is output or controlled variable.

R(S)

C(S) G(S)

SUMMING POINT: It is used to add or subtract one or more signals. + indicates the add or subtract function. + indicates the signal is added to reference signal. -indicates the signal is subtracted to reference signal. It is called negative feedback. The signal which is added or subtracted to the reference signal is called feedback signal. Summing point R(S)

C(S) +

TAKE OF POINT Some of the signal at the top are bypass and it is given to other block for further performance. Take off point R(S)

C(S)

G(s)

The steps to reduce the block diagram.  Reduce the series blocks.  Reduce the parallel blocks. Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System   

IV Semester ECE

Reduce minor feedback loops. As for as possible shift summing point to the left and take-off point to the right. Repeat the above steps till canonical form is obtained.

BLOCK DIAGRAM FOR A CANONICAL SYSTEM: R(s) = reference or controlling variable. C(S) = Output or controlled variable. E(S) = Error actuating signal. B(S) = Feedback signal H(S) = Feedback element. G(S) = transfer function for the system. G(S) = C(S) E(S) Closed loop transfer function is C(S) . R(S) C(S) = G(S) E(S). Consider a negative feedback signal is applied Actuating signal E(S) = R(S) – B(S)

(1)

(2)

Feedback signal C(S)H(S) = B(S) Put equation (3) in eq.(2) E(S) = R(S)- C(S)H(S) Put eq (4) in eq.(1) C(S) = G(S){R(S) – C(S)H(S)} C(S)[ 1+ G(S) H(S)] = G(S)R(S)

(3) (4) (5) (6)

(transfer function for negative feedback signal C(S) = G(S)______ R(S) 1+G(S) R(S) (transfer function for positive feedback signal.) C(S) = G(S)______ R(S) 1-G(S) R(S)

Rules for reduction of block diagram Rule 1: If the blocks are in cascade then Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

R(S)

R(S)

C(S )

G2

G1

C(S)

=

G1G2

Rule 2: if the blocks are in parallel then, the blocks are added or subtracted depending on the summing point signal.

G2

R(S)

G1+G2

+

C(S)

=

G1 R(S)

+ +

C(s)

Rule 3: Moving the take-off point after the block

R(S)

G

R(S)

C (S)

G

C(S)

= 1/G

Rule 4: moving the take-off point ahead of the block

G G G Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

Rule 5: Moving summing point after the block, then

Rule 6: Moving the summing point ahead (before) the blockff

Rule 7: Eliminating feedback loop, then


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IV Semester ECE

1. Using block diagram reduction techniques find C(s) / R(s) as in fig. Fig1.2.1.1 R(s)

G1 -

C(s)

-

I H1

II H2 Fig1.2.1.1

Step 1: eliminating feedback loop I R(s)

C(s)

G1

-

1+G1H1

II H1

Step2: eliminating feedback loop II

G(s) = G1 / 1 +G1H1 H(s) = H2 G1 G1 / 1 +G1H1 C(s) / R(s) =

= 1 +G1H1 + G1H2

1 + (G1 / 1 +G1H1 )(H2)


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IV Semester ECE

G1 R(s)

C(s) 1 + G1H1+G1H2 Fig1.2.1.2 Answer.

2. Using block diagram reduction technique find closed loop transfer function C(s) / R(s) shown in fig 1.2.2.1

G3 + C(S)

R(S) G1 -

G4

G2

H1

H2 Fig 1.2.2.1


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IV Semester ECE

Step 1: Combine the blocks G1 &G2 which are in cascade and combine the blocks G2 &G3 which are in parallel as shown in fig 1.2.2.2.

R(S)

C(S) G1 G4 -

-

G2 +G3 I

H1 II H2 Fig 1.2.2.2

Step 2: Eliminate feedback loopI and combine the blocks (G1G4 / 1+ G1G4H1) & (G2 + G3) which are in parallel as shown in fig. Fig 1.2.2.3 C(S)

G1G4 G2+G3

R(S) 1+ G1G4H1

II H2


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IV Semester ECE

Step 3: eliminate feedback loop II

G1G4 G2 +G3 C(s)

1 +G1G4H1

= R(s) 1+

G1G4

G2 +G3 1 + G1G4H1

G1G4

G2 +G3

= 1 + G1G4H1

+


G1G4 (G2 +G3)

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IV Semester ECE

3.Find the transfer function C(s) / R(s) for the block diagram shown below as shown in (fig 1.2.3.1) (A.U.2004)

R(s)

C(s) G1(s) -

G2(s) I

II

(fig 1.2.3.1)

Step 1: Eliminating feedback loopI (as in fig.1.2.3.2) R(s)

C(s) G1(s)

G2(s)__ 1 + G2 (s)

II

(fig 1.2.3.2)


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IV Semester ECE

Step 2: Combining blocks G1(s) & {G2(s) / 1+ G2(s)} which are in cascade (as in fig 1.2.3.3) C(s)

R(s)

G1(s)G2(s)__ 1 + G2 (s) II

Fig 1.2.3.3

Step 3: Eliminating feedback loop II (as in fig 1.2.3.4) G1(S) G2(S) 1 + G2(S) C(s) R(S)

= 1 +

G1(S) G2(S) 1 + G2(S)

G1(S) G2(S) C(s) R(S)

R(S)

= (1 + G2(S)) + G1(S) G2(S)

C(S) G1(S) G2(S) (1 + G2(S)) + G1(S) G2(S) Answer.


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IV Semester ECE

4. Determine the transfer function C(s) / R(s) of the system shown in fig.1.2.4.1 by block diagram reduction method. (AU: Dec.2003)

H3 T1

R(s) S1 +

+

G1

-

+

C(s)

S3

G2

-

G3

S2

G4 T2

-

H1

H2 Fig.1.2.4.1.

Step:1. Shifting the summing point S2 before the block G1 and shifting the take off point T2 after the block G4

H

1 / G1

1 / G4

3

R(s) S1 +

+

-

+ G2

G1 -

C(s)

S3 G3

G4

-

S2 H1

H2 Fig.1.2.4.2.


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IV Semester ECE

Step:2. Exchange the summing points and take off points using associative law and combining the series blocks we get

H3 / G1G4 III R(s) +

-

S2

+ G1 G2

+ S1

-

C(s)

S3 G3 G4

I

-

H1

II H2

Fig.1.2.4.3.

Step:3 Eliminating inner feedback loops I, II H3 / G1G4 III R(s) + +

G1 G2______ 1 + G1G2H1

C(s)

G3 G4______ 1 + G3G4H2

S1 Fig.1.2.4.4.


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IV Semester ECE

Step:4 Combine the blocks in series H3 / G1G4 III R(s) + +

-

C(s)

G1 G2 G3 G4____________ (1 + G1G2H1)( 1 + G3G4H2) S1 Fig.1.2.4.5.

Step 5: Eliminate the feedback loop III

G1 G2 G3 G4_________ (1 + G1G2H1)( 1 + G3G4H2) C(s) = R(s) 1 + G1 G2 G3 G4____________ (1 + G1G2H1)( 1 + G3G4H2)

C(s) = R(s)

__H3___ G1G4

______G1G2G3G4_________________ (1 + G1G2H1)( 1 + G3G4H2) +G2G3H3 Answer.


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IV Semester ECE

5. Using block diagram reduction technique, find the transfer function from each input to the output C(s) for the system shown in fig.1.2.5.1. (AU: 2005) X(s)

R(s)

G4

+

+ G1

G2 -

G3

G5 +

-

C(s)

H5

Fig.1.2.5.1.

With X(s) = 0, the given block diagram reduces as

R(s)

+

+ G1

G2 -

G3

G5 +

-

C(s)

-

II

I III

H5

Fig.1.2.5.2.


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IV Semester ECE

Step 1: eliminate feedback loops I, II

R(s)

C(s)

+ __G2____ (1 + G2)

G1

G5______ (1 + G5H5)

G3

-

III

Fig.1.2.5.3.

Step 2: Combine the blocks in series

R(s)

+

C(s) __G2 G3 G5_____ (1 + G2) (1 + G5H5)

G1 -

III

Fig.1.2.5.4.


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IV Semester ECE

Step 3: Eliminate feedback loop III

R(s) G1

__G2 G3 G5_________________ { (1 + G2) (1 + G5H5)}+ {G2G3G5)

C(s)

Fig.1.2.5.5.

Step 4: Combine the blocks in series.

R(s) _ G1G2 G3 G5_________________ { (1 + G2) (1 + G5H5)}+ {G2G3G5)

C(s)

Fig.1.2.5.6.


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IV Semester ECE

With R(s) = 0, G1 vanishes, but minus sign at summing point must be considered by introducing block of -1 as shown G4

X(s)

C(s)

+

-1

G2

G3

G5 +

-

-

II

I H5

Fig.1.2.5.6.

Step 5: eliminate feedback loops I, II G4

X(s) -1

C(s)

+ G2____ ( 1+ G2 )

G3 +

G5_______ ( 1+ G5H5)

Fig.1.2.5.7.


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IV Semester ECE


X(s)

C(s)

+

G5_______ ( 1+ G5H5)

G4 +

III - G2G3____ ( 1+ G2 )

Fig.1.2.5.8.


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IV Semester ECE


X(s)

C(s)

+

G5_______ ( 1+ G5H5)

G4 +

III - G2G3____ ( 1+ G2 )

Fig.1.2.5.8.

Step 8: eliminate feedback loop III and combine the blocks in series

X(s) G4G5(1 +G2)_____________ (1+ G5H5 )(1 +G2)+ G2G3H5

C(s)

Answer.


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IV Semester ECE

6. Reduce the block diagram shown in fig.1.2.6.1. and obtain C(s) / R(s) (AU: may 2007)

G5 R(s)

S1

+

S2

+

S3

C(s) +

G1 +

-

G2

G3

G4

-

H1

H2 Fig. 1.2.6.1.


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IV Semester ECE

Step :1 Shifting the take off point between the blocks G2 and G3 to after the block G3 1/G3 R(s)

+

S1

+

G5

S2

+

S3

C(s) +

G1

G2

-

G3

G4

-

H1

H2 Fig. 1.2.6.2

Step :2 Combine the blocks G2 and G3, 1/G3 and G5 which are in series

G5/G3 R(s)

+

S1

+

S2

+

S3

C(s) +

G2 G3

G1 -

G4

-

I H1 H2 Fig. 1.2.6.3


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IV Semester ECE

Step :3 Eliminate feed back loop I, and then combine blocks G4 and G5 / G3 which are in parallel

R(s)

C(s)

S1

+

G1

G4 + (G5/G3)

G2 G3_____ 1 + G2G3H1

-

H2 Fig. 1.2.6.4

Step :4 Combine the blocks which are in series R(s)

+

S1

C(s) -

(G1G2 G3)( G3G4 + G5)__ G3(1 + G2G3H1) II H2 Fig. 1.2.6.5


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IV Semester ECE

Step :5 Eliminate the feed back loop II

C(s) R(s)

(G1G2 G3)( G3G4 + G5)__ G3(1 + G2G3H1) =

1+

C(s) = R(s)

G1G2G3(G3G4 + G5 ) G3(1 + G2G3H1)

H2

G1G2G3 (G3G4 + G5 )___________ G3(1 +G2G3H1) + [G1G2G3 (G3G4 + G5)] H2 Answer.

7. Determine the overall transfer function of the block diagram shown in fig. 1.2.7.1. (AU: Nov. 2005) G3

R(s)

+

S1

S2

+

G1 -

-

C(s)

+

G2 I

S3

H1

Fig.1.2.7.1.


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IV Semester ECE

Step :1 eliminate feed back loop I

G3

R(s)

+

S1

+

C(s)

+

G1____ 1 +G1H1

G2

S3

Fig.1.2.7.2.

Step :2 Shifting the summing point S3 before the block (G1 / 1+G1H1) and Combine the blocks G2 & (G1 / 1 +G1H1) which are in series. G3

R(s)

1 + G1H1 G1 +

C(s)

+ + S3

G2G1___ 1 +G1H1 S1

II

Fig.1.2.7.3.


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IV Semester ECE

Step :3 Eliminate feedback loop II and combine the blocks in parallel

R(s)

{G3(1 + G1H1)} 1+ G1

C(s) G1G2________ (1 +G1H1) +G1G2

Fig.1.2.7.4.


C(s) = { G1 +G3(1 + G1H1)} R(s) G1

___G1G2________ 1+G1H1 +G1G2

C(s) = { G1 +G3(1 + G1H1)}G2 R(s) 1+G1H1 +G1G2 Answer.


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8. Use block diagram reduction technique obtain the equivalent transfer function C(s) /R(s)

G2 R(s)

+ +

C(s)

+

G5

G4

G3

G1

-

-

I II

H2

H1

Fig.1.2.8.1

Step 1: Eliminate the feed back loops I ,II and combine the blocks G2 and G3 which are in parallel R(s)

C(s)

+

G5

G4____ 1 +G4H2

(G2+G3)

G1____ 1 +G1H1

III

Fig.1.2.8.2


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IV Semester ECE

Step 2: Combine the blocks in series R(s)

C(s)

+

G1(G2+G3) G4 G5___ ( 1 +G4H2) (1 +G1H1) III

Fig.1.2.8.3

Step 2: Eliminate the feed back loop III

C(s) = R(s)

G1(G2+G3) G4 G5___ ( 1 +G4H2) (1 +G1H1) 1+ __G1(G2+G3) G4 G5___ ( 1 +G4H2) (1 +G1H1)

C(s) = ______ G1(G2+G3) G4 G5_________________ R(s) ( 1 +G4H2) (1 +G1H1) + G1(G2+G3) G4 G5 Answer.


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IV Semester ECE

Signal flow graph: The graphical representation of the variables of a set of linear algebraic equations representing the system is called signal flow graph. The properties of signal flow graph  The signal in the system flows along the branches and along the arrows associated with the branches.  The value of variable represented by any node is an algebraic sum of all the signals entering at the node.  The signals gets multiplied by the branch gain or branch transmittance when it travels along it.  Applicable only to linear time invariant systems. Need for signal flow graph: Block diagrams are very successful for representing control systems, but for complicated systems, the block diagram reduction process is tedious and time consuming. So signal flow graphs are needed which does not require any reduction process because of availability of a flow graph formula, which relates the input and output system variables. The transmittance is the gain acquired by the signal when travels from one node to another node in the signal flow graph. Node represents a system variable, which is equal to the sum of all incoming signals at the node, outgoing signals from the node do not affect the value of the node variable. A signal travels along a branch from one node to another in the direction indicated by the branch arrow and in the process gets multiplied by the gain or transmittance of the branch. A node having incoming and outgoing branches is known as chain node. A feedback loop consisting of only one node is called self loop. The product of all the gains of the branches forming a loop is called loop gain. A path from the input to output node is defined as forward path.


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IV Semester ECE

1..Using Mason’s gain formula obtain C(s)/R(s) for the signal flow graph shown in fig1.3.1

By Mason’s Gain formula K ∑ PK ∆K 1

C(S) = R(S)

∆

Here K = 2 Forward paths P1 = G1G2 P2 = G1G3

Individual loops L1 = - G1H1 L2 = - G1G2H2 L3 = - G1G3H2

There are no two non touching loops. ∆ = 1 – (L1 +L2+L3) = 1 – (- G1H1 – G1G2H2 - G1G3H2) = 1 + G1H1+ G1G2H2 + G1G3H2


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IV Semester ECE

∆1 = 1 ∆2 = 1

P1 ∆1 + P2∆2

C(S)

=

R(S)

∆

(G1G2)(1) + (G1G3) = = 1 + G1H1+ G1G2H2 + G1G3H2 Ans.


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2. Using Mason’s gain formula, determine C(S) / R(S) for the figure 1.3.5 2.Using Mason’s gain formula obtain C(s)/R(s) for the signal flow graph shown in fig1.3.2

K ∑ PK ∆ K 1

By Mason’s Gain formula C(S) = R(S)

∆

Here K =1 Forward paths P1 = G1G2G3 Individual loops L1 = - G1H2 L2 = - G2G3H3 L3= - G1G2H1 There are no two non touching loops. ∆ = 1 – (L1 +L2+L3) = 1 – (- G1H2 - G2G3H3 - G1G2H1) ∆1 = 1 C(S) = P1∆1 R(S) ∆

=

____ (G1G2G3)(1)______________ 1 + G1H2 + G2G3H3 + G1G2H1 Ans.


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3. Using Mason’s gain formula, determine C(S) / R(S) for the figure 1.3.3

By Mason’s Gain formula

K ∑ PK ∆ K 1

C(S) = R(S)

∆

Here K =2

Forward paths P1 = G1G2G3 P2 = G4

Individual loops L1 = - G2H1 L2 = G1G2H1

There are no two non touching loops ∆ = 1 – (L1 +L2+L3) = 1 – (- G1H2 + G1G2H1) = 1 + G1H2 - G1G2H1 ∆1 = 1 ∆2 = 1


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C(S)

R(S)

(G1G2G3)(1) + (G4)(1)

P1 ∆1 + P2∆2 =

IV Semester ECE

= ∆

1 + G1H2 - G1G2H1 Ans.



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IV Semester ECE


K ∑ PK ∆ K 1

C(S) =

∆

R(S) Here K =2 Forward paths P1 = G1G2G3G4 P2 = G5G6G7G8

Individual loops L1 = G2H2 L2 = G3H3 L3 = G6H6 L4 = G7H7

Two pairs of two non touching loops are there. They are L1 L3 = G2H2 G6H6 L2 L4 = G3H3 G7H7 ∆ = 1 – (L1 +L2+L3 + L4 ) +( L1 L3 + L2 L4) = 1 – (G2H2 + G3H3 + G6H6 + G7H7) + (G2H2 G6H6+ G3H3 G7H7) ∆1 = 1 – (G6H6 + G7H7) = 1 - G6H6 -G7H7 ∆2 = 1 – (G2H2+ G3H3) = 1 – G2H2- G3H3 P1 ∆1 + P2∆2

C(S) =

∆

R(S)

(G1G2G3G4)( 1 –G6H6 -G7H7) + (G5G6G7G8)( 1 – G2H2- G3H3) = 1 – (G2H2 + G3H3 + G6H6 + G7H7) + (G2H2 G6H6+ G3H3 G7H7) Ans. Solved by A.Devasena., Associate Professor., Dept/ECE

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Fig.1.3.5 By Mason’s Gain formula K ∑ PK ∆ K 1

C(S) = R(S)

∆

Here K = 2 Forward paths P1 = G1G2G4G5 P2 = G1G2G3

Individual loops L1 = - G4 G5H2 L2 = - G2G4G5H1 L3 = - G1 G2G4 G5 L4 = - G1 G2G3 L5 = - G2G3H1 L6 = - G3H2 There are no two non touching loops ∆ = 1 - ( L1 + L2 + L3 + L4 + L5+ L6 ) = 1 – (- G4 G5H2 - G2G4G5H1 - G1 G2G4 G5 - G1 G2G3 - G2G3H1 - G3H2) = 1 + G4 G5H2 + G2G4G5H1 + G1 G2G4 G5 + G1 G2G3 + G2G3H1 + G3H2) ∆1 = 1 Solved by A.Devasena., Associate Professor., Dept/ECE ∆2 = 1

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IV Semester ECE

P1 ∆1 + P2∆2

C(S) =

∆

R(S)

(G1G2G4G5)(1) + (G1G2G3)(1) = 1 + (G4 G5H2 + G2G4G5H1 + G1 G2G4 G5 + G1 G2G3 + G2G3H1 + G3H2) G3H2) Ans.

6.Find the transfer function for the signal flow graph as shown in fig 1.3.6


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IV Semester ECE


K ∑ PK ∆ K 1

C(S) = R(S) Here K =3 Forward paths P1 = G1G2G3G4 G5G6 P2 = G1G2G7G6 P3 = G1G2G3G4 G8

∆

Individual loops L1 = - G4H4 L2 = - G5 G6H1 L3 = -G2G3G4 G5G6 H2 L4 = - G1G2G3G4 G5G6 H3 L5 = - G8H1 L6 = - G1G2G7 G6H3 L7 = -G1G2G3 G4G8 H3 L8 = -G2G3 G4G8 H2 L9 = - G2G7 G6H2

There are two pairs of two non touching loops. They are L1L6 = G1G2 G4G6G7H3H4 L1L9 = G4 G2G6G7H2H4 ∆ = 1 – (L1 +L2+L3 + L4 + L5 + L6 + L7 +L8 + L9) + (L1L6 + L1L9) = 1 –( - G4H4 - G5 G6H1 - G2G3G4 G5G6 H2 - G1G2G3G4 G5G6 H3 - G8H1 - G1G2G7 G6H3 - G1G2G3 G4G8 H3 - G2G3 G4G8 H2 - G2G7 G6H2) +( G1G2 G4G6G7H3H4 + G4 G2G6G7H2H4) ∆1 = 1 ∆2 = 1 – ( - G4H4) ∆3= 1


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IV Semester ECE

P1∆ 1 +P1∆2 + P3∆3

C(S) = R(S)

∆

= (G1G2G3G4 G5G6)(1) + (G1G2G7G6)( 1 + G4H4)+ (G1G2G3G4 G8)(1) 1 + G4H4 + G5 G6H1 +G2G3G4 G5G6 H2 + G1G2G3G4 G5G6 H3 +G8H1 + G1G2G7 G6H3 +G1G2G3 G4G8 H3 + G2G3 G4G8 H2 +G2G7 G6H2 + G1G2 G4G6G7H3H4 + G4 G2G6G7H2H4 Ans.

7.Determine the transfer function C(S)/R(S) for the signal flow graph shown in fig.1.3.7.


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IV Semester ECE


K ∑ PK ∆ K 1

C(S) = R(S) Here K =2

∆

Forward paths

Individual loops

P1 = G1G2G3

L1 = - G1 G2H1

P2 = G1G4

L2 = - G2 G3H2 L3 = -G1G2G3 L4 = - G4 H2 L5 = - G1G4

There are no two non touching loops.

∆ = 1 – (L1 +L2+L3 + L4 + L5 ) = 1 –( - G1 G2H1 - G2 G3H2- G1G2G3 - G4 H2- G1G4) = 1 + G1 G2H1 + G2 G3H2+ G1G2G3 + G4 H2+G1G4 ∆1 = 1 ∆2 = 1


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IV Semester ECE

P1∆ 1 +P1∆2

C(S) = R(S)

=

∆

(G1G2G3)(1) + (G1G4)( 1 ) 1 + G1 G2H1 + G2 G3H2+ G1G2G3 + G4 H2+G1G4

Ans.

8.Find C(S)/R(S) for the following system using mason’s gain formula for the signal flow graph shown in fig.1.3.8.


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IV Semester ECE


K ∑ PK ∆ K 1


∆

Forward paths

Individual loops

P1 = G1G2

L1 = G9

P2 = G4

L2 = G3

P3= G7G8

L3 = G5G6

P4 = G1G5 G8 P5 = G7G6 G2 There is one pair of two non touching loops. It is L1L2 = G9 G3 ∆ = 1 – (L1 +L2+L3 ) + (L1L2) = 1 – ( G9 + G3 + G5G6) + (G9 G3) = 1 - G9 - G3 - G5G6 + (G9 G3) ∆1 = 1 – L1 = 1 – G9 ∆2 = 1 – (L1 +L2+L3 ) + (L1L2) = 1 - G9 - G3 - G5G6) + (G9 G3) ∆3 = 1 – L2 = 1 – G3 ∆4 = 1 ∆5 = 1


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IV Semester ECE

P1∆ 1 +P1∆2 +P3∆ 3 +P4∆4 +P5∆ 5

C(S) = R(S)

∆

(G1G2)( 1 – G9) + (G4)(( 1 - G9 - G3 - G5G6) + (G9 G3) ) + (G7G8) (1 – G3) + (G1G5 G8)(1) + (G7G6 G2)(1) = 1 - G9 - G3 - G5G6 + (G9 G3) Ans.

9. Determine the transfer function C(s)/ R(s) for the signal flow graph shown in fig. 1.3.9.


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IV Semester ECE


K ∑ PK ∆ K 1

C(S) = R(S)

∆

Here K =2 Forward paths

Individual loops

P1 = G1G2 G3

L1 = - G1G2H1

P2 = G4

L2 = - G2G3H2 L3 = - G1G2 G3 L4 = - G4 L5 = G2G4 H1 H2


∆ = 1 – (L1 +L2+L3 L4 +L5) = 1 – (- G1G2H1 - G2G3H2- G1G2 G3 - G4 + G2G4 H1 H2) = 1 + G1G2H1 + G2G3H2 + G1G2 G3 + G4 - G2G4 H1 H2

∆1 = 1 ∆2 = 1


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IV Semester ECE

P1∆ 1 +P1∆2

C(S) = R(S)

∆

(G1G2 G3)(1) + (G4) (1)

1 + G1G2H1 + G2G3H2 + G1G2 G3 + G4 - G2G4 H1 H2 Ans.

10.Determine the transfer function C(S) /R(S) using Mason’s gain formula for the signal flow graph shown in fig.1.3.10


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IV Semester ECE


K ∑ PK ∆ K 1

C(S) = R(S)

∆

Here K =2 Forward paths

Individual loops

P1 = G1G2 G3 G4

L1 = - G1G2 G3 G4H1 L2 = - G3G4H2

P2 = G1 H3 G4 L3 = - G1G4 H3 H1 L4 = - G3G4


∆ = 1 – (L1 +L2+L3 L4) = 1 – (- G1G2 G3 G4H1 - G3G4H2 - G1G4 H3 H1 - G3G4 ) = 1 + G1G2 G3 G4H1 + G3G4H2 + G1G4 H3 H1 + G3G4 ∆1 = 1 ∆2 = 1


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IV Semester ECE

P1∆ 1 +P1∆2

C(S) = R(S)

=

∆

(G1G2 G3 G4)(1) + (G1 H3 G4)( 1 ) 1 + G1G2 G3 G4H1 + G3G4H2 + G1G4 H3 H1 + G3G4

Ans.


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IV Semester ECE

13.For the signal flow graph given in fig.1.3.13. Valuate the closed loop transfer function of the system. (A.U.April.2006)


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IV Semester ECE


K ∑ PK ∆ K 1


∆

Forward paths P1 = a12a23a34a45

Individual loops L1 = a23a32 L2 = a23a34a45 a52

P2 = a12a23a35

L3= a23a34a45 a42 L4 = a44 L5= a23a35


∆ = 1 – (L1 +L2+L3 L4 + L5 ) = 1 – (a23a32 + a23a34a45 a52 + a23a34a45 a42 + a44 + a23a35) = 1 – a23a32 - a23a34a45 a52 - a23a34a45 a42 - a44 - a23a35

∆1 = 1 ∆2 = 1 – L4 = 1- a44


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IV Semester ECE

P1∆ 1 +P1∆2

C(S) = R(S)

=

∆

(a12a23a34a45)(1) + (a12a23a35)( 1- a44 ) 1 – a23a32 - a23a34a45 a52 - a23a34a45 a42 - a44 - a23a35

Ans.

14. Using Mason’s gain formula find C(S) /R(S) for the fig. 1.3.14.


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IV Semester ECE


K ∑ PK ∆ K 1


∆

Forward paths

Individual loops

P1 = G1G2 G3 G4 G5

L1 = H1 L2 = G3H2

P2 = G1 G2 G3 G4 G6 L3 = G4 H3 L4 = H4 The combination of two non touching loops: L1 L2 = H1G3H2 L1 L3 = H1 G4 H3 L1 L4 = H1H4 L2 L4 = G3H2 H4

The combination of three non touching loops: L1 L3 L4 = H1 G4 H3 H4 L1 L2 L4 = H1G3H2 H4


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IV Semester ECE

∆ = 1 – (L1 +L2+L3 L4) + (L1 L2 + L1 L3 + L1 L4 + L3 L4) – (L1 L3 L4 + L1 L2 L4) = 1 – ( H1 + G3H2+ G4 H3 + H4 ) + (H1G3H2 + H1 G4 H3 + H1H4 + G3H2 H4) – (H1 G4 H3 H4 + H1G3H2 H4 )

∆1 = 1 ∆2 = 1

P1∆ 1 +P1∆2

C(S) = R(S)

=

∆

(G1G2 G3 G4 G5)(1) + (G1 G2 G3 G4 G6 )(1)

1 – ( H + G H2+ G4 H3 + H4 ) )(1) + (a12a23a351)( 1- 3a44 ) + (H1G3H2 + H1 G4 H3 + H1H4 + G3H2 H4) – (H1 G4 H3 H4 + H1G3H2 H4 )

Ans.


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IV Semester ECE

1. Obtain the analogous electrical network for the system shown in fig.5. (AU:Nov./Dec.2007)

The Mass M1 is under the displacement x1(t). The friction B1 is responsible to change the displacement from x1(t) to x2(t) The Mass M2 is under the displacement x2(t). The friction B2 and spring K1 are responsible to change the displacement from x2(t) to x3(t) The Mass M3 and spring K2 are under the influence of displacement x3(t). The equivalent Mechanical system is shown in fig.5.a.


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IV Semester ECE

The equilibrium equations are F(t) = M1(d2x1(t)/ dt2) + B1d(x1(t)- x2(t))/dt ------------------------------------------------(1) 0 = B1d(x2(t)- x1(t))/dt + M2(d2x2(t)/ dt2) + K1(x2(t)- x3(t)) + B2d(x2(t)- x3(t))/dt --------(2) 0 = K1(x3(t)- x2(t)) + B2d(x3(t)- x2(t))/dt + M3(d2x3(t)/ dt2) + K3x3(t) ---------------------(3)

Using force- voltage analogy , Mass is replaced by inductance, friction or dashpot is replaced by resistance, spring is replaced by reciprocal of capacitance, displacement is replaced by charge. Rate of change of displacement is replaced by current, force is replaced by voltage.

V(t) = L1di1(t)/dt + R1 (i1(t) – i2(t) ) ---------------------------------------------------------------(4) 0 = R1 (i2(t) – i1(t) ) + L2di2(t)/dt + 1/C1∫(i2(t) – i3(t))dt + R2 (i2(t) – i3(t) )--------------------(5) 0 = 1/C1∫(i3(t) – i2(t)) dt + R2 (i3(t) – i2 +(t) ) + L3di3(t)/dt + 1/C2∫(i3(t)dt.---------------------(6) The analogous electrical network is shown in fig.5.c.


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IV Semester ECE

2. Draw the equivalent mechanical system of the system shown in fig.6. write the set of equilibrium equations for it and obtain electrical analogous circuits using i) F-V analogy ii)F-I analogy. ( AU: May2009)

As shown in fig.6. M1,K1,and B1 are under the displacement x1 as K1 and B1 are with respect to rigid support. K2 is between x1 and x2 as it is responsible for the change in displacement. While M2 , K3 and B2 are under the displacement x2. Hence the equivalent mechanical system is as shown in fig.6.a.


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IV Semester ECE

The equilibrium equations are F(t) = M1(d2x1(t)/ dt2) + B1dx1(t)/dt + K1x1(t) + K2(x1(t)- x2(t)) -----------------------------(1) 0 = M2(d2x2(t)/ dt2) + B2dx2(t)/ dt + K2(x2(t)- x1(t)) + K3x2(t) -----------------------------(2)

Using force- voltage analogy , Mass is replaced by inductance, friction or dashpot is replaced by resistance, spring is replaced by reciprocal of capacitance, displacement is replaced by charge. Rate of change of displacement is replaced by current, force is replaced by voltage.

V(t) = L1di1(t)/dt + R1 i1(t) + 1/C1∫(i1(t) dt + 1/C2∫(i1(t) – i2(t)) dt -----------------------------(3) 0 = L2di2(t)/dt + R1 i2(t) +1/C2∫(i2(t) – i1(t))dt + 1/C3∫(i2(t) dt---------------------( 4) The analogous system for force voltage analogy is shown in fig.6.c.


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IV Semester ECE

Using force- current analogy , Mass is replaced by capacitance, friction or dashpot is replaced by reciprocal of resistance, spring is replaced by reciprocal of inductance, displacement is replaced by flux. Rate of change of displacement is replaced by voltage, force is replaced by current.

The analogous system for force current analogy is shown in fig.6.d.


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IV Semester ECE

I(t) = C1dV1(t)/dt + 1/R1 V1(t) + 1/L1∫(V1(t) dt + 1/L2∫(V1(t) – V2(t)) dt --------------------------(5) 0 = C2dV2(t)/dt + 1/R2 V2(t)+1/L2∫(V2(t) – V1(t))dt + 1/L3∫(V2(t) dt------------------------------(6)

3.Write the equations for mechanical system shown in figure 7.a. ( AU: April 2005, Dec 2005)

Due to force f(t) applied to M1, it will displace by the displacement x1(t). As K1 and B1 are between M1 and fixed support, both are under same displacement x1(t) Due to friction B3, the force transferred to M2 is different than f(t), hence M2 will displace by the displacement x2(t). As K2 and B2are between M2 and fixed support, both are under same displacement x2(t) The equivalent system is shown in fig.7.b.

The node equations for the system are At node x1 f(t) = M1(d2x1(t)/ dt2) + B1dx1(t)/dt + K1x1(t) + B3d(x1(t)- x2(t))/dt ----------------------(1) At node x2


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IV Semester ECE

0 = M2(d2x2(t)/ dt2) + B2dx2(t)/dt + K2x2(t) + B3d(x2(t)- x1(t))/dt ----------------------(2) Taking laplace transforms of equations(1) and (2) we get F(s) =M1s2X1(s) + B1sX1(s) + K2X1(s) + B3s(X1(s) – X2(s)) -----------------------(3) 0 =M2s2X1(s) + B2sX2(s) + K2X2(s) + B3s(X2(s) – X1(s)) -----------------------(4) 4.Obtain the mathematical model of the following mechanical system shown in figure 8.a ( AU:April 2004) The displacement is x(t) as shown if fig.8.a.

Both B1 and B2 are between mass M and fixed support. Hence under the influence of x(t). the spring k between mass M and fixed support. Hence under the influence of x(t). The equivalent mechanical model is shown in fig.8.b.

f(t) = M(d2x(t)/ dt2) + (B1 + B2)dx1(t)/dt + Kx (t) -------------------(1) This equation represents the mathematical model.


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IV Semester ECE

5. Obtain the differential equations of the following mechanical systems as shown in fig.9.a.

The equivalent nodal diagram for the given mechanical system is shown in fig.9.b.

At node x1 0 = M2(d2x1(t)/ dt2) + K1x1(t) + K1x1(t) + K2(x1(t)- x2(t)) -----------------------------(1) F(t) = M1(d2x2(t)/ dt2) + K2(x2(t)- x1(t)) ----------------------------(2) Taking Laplace transforms of equations(1) and (2) we get 0 =M2s2X1(s) + K1X1(s) + K2(X1(s) – X2(s)) -----------------------(3) F(s) =M2s2X1(s) + K2 (X2(s) – X1(s)) -----------------------(4)


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IV Semester ECE

6. For the spring, damper and mass system shown in fig.10.a. Obtain the differential equations governing the system. F(t) is the force applied (Au:2005).

The equivalent nodal diagram for the given mechanical system is shown in fig.10.b.

The differential equations governing the system are f(t) = M2(d2x2(t)/ dt2) + B2d(x1(t)- x2(t))/dt + K1(x2(t)- x1(t)) + B1d(x2(t)- x1(t))/dt ---(1)

0 = M1(d2x1(t)/ dt2) + B1d(x1(t)- x2(t))/dt +K1(x1(t)- x2(t))


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IV Semester ECE

7. Show that the systems shown in fig.11.a and fig.b are analogous system. (AU: April-2007)

There are three displacements xi(t) xo(t) and xy(t). The input is xi(t) and output is xo(t). So transfer function of the mechanical system is X0(s) / Xi(s). The equilibrium equations are B1d(xo(t)- xi(t))/dt + K1(xo(t)- xi(t)) + B1d(xo(t)- xy(t))/dt = 0 ---------(1) B2d(xy(t)- xo(t))/dt + K2xy(t) = 0 ------------------------------------ (2) Taking Laplace transform of both equations, neglecting initial conditions, B1sXo(s) - B1sXi(s) + K1Xo(s) - K1Xi(s) + B2sXo(s) – B2sXy(s) = 0 Therefore Xo(s)[sB1 + k1 +s B2] + Xi(s)[sB1 + k1] – B2sXy(s) = 0 --------------------- (3) B2sXy(s) - B2sXo(s) + K2Xy(s) = 0 Therefore Xy(s)[sB2 + k2 ] = B2s Xo(s) Therefore Xy(s) = {B2s / [sB2 + k2 ]} Xo(s)} ------------------------------------------(4) Substitute equation 4 in equation 3, Xo(s)[sB1 + k1 +s B2] - Xi(s)[sB1 + k1] - B2s{B2s / [sB2 + k2 ]} Xo(s)} Therefore Xo(s)[ sB1 + k1 +s B2 – s2B22 / (sB2 + k2) = Xi(s)[sB1 + k1] Xo(s) / Xi (s) = [sB1 + k1][ sB2 + k2] / {s2B1B2 + sB1K2 + sK1B2 + K1K2 + s2B22 + sK2B2 – s2B22} Xo(s) / Xi (s) = ___________ k2K1 ( 1 + B1/K1) ( 1 + B2/K2) ___________ k2K1 [( 1 + s2B1B2/K1K2 +s B1/K1 + s B2/K2) + s B2/K1 Xo(s) = __________ ( 1 + B1/K1) ( 1 + B2/K2) ____ ------------Required transfer function. Xi (s) ( 1 + B1/K1) ( 1 + B2/K2) + s B2/K1 The s domain network for the given electric network is


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Now I (s) = ____ Ei(s) ________ Z(s) + R1 + 1/sC1 And Eo(s) = I(s)[ R1 + 1 /sC1] Therefore Eo(s) = Eo(s) =

IV Semester ECE

with Z(s) = R2 parallel 1 /sC2

Ei(s) [ R1 + 1 /sC1] Z(s) + R1 + 1/sC1 Ei(s)( 1 + sR1C1) sC1Z(s) + sC1R1 +1

substituting Z(s) = R2 x 1/sC2 R2 + 1/sC2 Eo(s) = __________ ( 1 + sR1C1) ( 1 + sR2C2) ____ Ei(s) ( 1 + sR1C1) ( 1 + sR2C2) + sC1 R2 As spring is replaced by reciprocal of capacitance, friction is replaced by Resistance, the two transfer functions are identical , hence the two systems are analogous in nature.


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IV Semester ECE

22. Find the impulse response of the second order system whose open loop transfer function G(S)=

____9_____ S2+4S+9

(AU may 2009)

Comparing the denominator term with S2+2ZωnS+ωn2 ωn2 =9

2Zωn = 4.

ωn=3

Z= __4_ _ = 2ωn

___4__ = 0.667 2×3

The impulse response for Z<1 is given by C(t)= _ωn___ e-Z ωn t sin(ωn√1-Z2 )t √1-Z2 = __3_____ e-2t sin(3√1-(0.667)2)t √1-(0.667)2 =4.035 e-2tsin2.235t

Obtain the unit step response of a second order system [under damped condition] [AU: nov/dec 06]

Response of the second order system __C(S) = R(S)

___ωn2_______ S2+2ZωnS+ωn2

For unit step input R(S)=1 For under damped systems Z<1 S2+2ZωnS+ωn2 =0 has two roots S1,2 = - Zωn ± jωn√1-Z2 Now let And

Zωn=α ωn√1-Z2 = ωd


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S1,2= α+j ωd For unit step input R(S)=1/s C(S)= R(S) [ ___ωn2_______ ] = [ ___ωn2_______ ] 2 S +2ZωnS+ωn2 S(S2+2ZωnS+ωn2) C(S)=A + __BS+C____ S S2+2ZωnS+ωn2 ___ωn2_______ =A(S2+2ZωnS+ωn2)+(BS+C)S S2+2ZωnS+ωn2 S2+2ZωnS+ωn2 ωn2= AS2+2AZωnS+Aωn2+BS2+CS ωn2=S2(A+B)+S( 2AZωn+C)+A ωn2 equvating constant term ωn2= A ωn2 A=1 Equvating S term 0= 2AZωn +C C=2AZωn C=2Zωn 2 Equvating S term 0=A+B 0=1+B B=-1 Zωn=α C(S)= 1 + ___-S-2α___ S S2+2αS+ωn2

C(S)= 1 - { ___S+2α___} S S2+2αS+ωn2

Now consider S2+2αS+ωn2

Add and subtract α2 term to the above eq S2+2αS+ α2 + ωn2- α2 2

2

2

2

=(S+ α )+ ωn -Z ωn

(Zωn=α

α2= Z2 ωn2

=(S+ α2 )+ ωn2(1 -Z2)

WKT


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IV Semester ECE

ωd = ωn√1-Z2 ωd2 = ωn(1-Z2) = (S+ α) 2+ ωd2―> put this expression in the denominator ofC(S) we get C(S)= 1 - { ___S+2α___} 2 2 S (S+α) +ωd

Now L-1 { ___S+α___} = e-atcosωt (S+α)2 +ω2 L-1 { ___ω___ } = e-atsinωt (S+α)2 +ω2 Adjusting for C(s) we get C(S)= 1 - { ___S+α_ _ + ___α_ ____ } S (S+α)2 +ωd2 (S+α)2 +ωd2 Multiply and divide by ωd to the last term C(S)= 1 - { ___S+α_ _ + _α _ _ ωd ___ } S (S+α)2 +ωd2 ωd (S+α)2 +ωd2

Taking inverse laplace C(t)=1- e-αtcosωd t- α e-αtsinωd t ωd

Now put

Zωn=α And

ωn√1-Z2 = ωd C(t)=1- e-αtcosωd t- α e-αtsinωd t ωd C(t)=1- e- Zωn tcosωd t+ e-Zωn t sinωd t ωn√1-Z2 C(t)=1- e-Zωn t

(√1-Z2 ωd t+Zsin ωd t)


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IV Semester ECE

ωn√1-Z2 by trigonometry sin ( ωd t +θ)= sin ( ωd t)cos θ+cos + cos ( ωd t) sinθ

compare this equation with the above expression, sin θ= √1-Z2

we can write

cos θ =Z

√1-Z2 cosθ Z θ= tan-1√1-Z2 radians. Z Hence using this in the expression sinθ = tanθ=

C(t)=1- e-Zωn t (sin (ωd t+θ) ωn√1-Z2


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Find kp , kv , ka and steady state error for a system with open loop transfer function as G(S)H(S)= 15(S+4)(S+7)__ S(S+3)(S+6)(S+8) Where the input is r(t)=4+t+t2 The given open loop transfer function in time constant form G(S)H(S)= 15(S+4)(S+7)__ S(S+3)(S+6)(S+8) =15× 4× 7(1+S) (1+s) 4 7 S ×3× 6× 8(1+s_) (1+s_) (1+s_) 3 6 8 = 35(1+0.25s)(1+0.14s)______ 12s(1+0.3s)(1+0.16s)(1+0.125s)

=2.9(1+0.25S)(1+0.14S) s(1+0.3s)(1+0.16s)(1+0.125s) Now

kp= Lt S->0

G(S)H(S)=∞

KV= Lt S G(S)H(S)= 2.9(1+0.25S)(1+0.14S) S-> 0 s(1+0.3s)(1+0.16s)(1+0.125s) KV=2.9 Ka= = Lt S2 G(S)H(S)=0 S->0 Input is =4+t+t2=4+t+2.t2 2 Now input combination of three standard inputs A1=4 step of 4 A2=1 ramp of 1 A3=2, parabolic input of 2. a) For step 3 the error is Ess 1=A1___ = 4___ = 0 1+kp 1+∞ Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

b) For ramp of 1 the error is Ess 2=A2___ = 1___ KV 2.9 c) For parabolic of 2, the error is Ess 3=A3___ = 2___ = ∞ Ka 0 Then the steady state error is Ess=e ss1+ess2+ess3 =0+1___ +∞ 2.9 =∞


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Steady state analysis: Steady state error is an important measure of the accuracy of a control systems. Basically these errors occur from the nature of inputs, non-lineraities present in the system Etc. the steady state error ess is the difference between the input (or desired value )and the Output of a closed loop system input as t->∞ Mathematically Ess =Lt e(t) =Lt [ r(t)-c(t)] t->∞ t->∞ by using the final value theorem ess = Lt e(t) =Lt SE(S) t->∞ s->0 This is valid provided that SE(S) has no poles on the jω axis, No multiple poles on the jω axix and is in the right half of S-plane. Consider is a closed loop signal shown below E(s)=error signal B(S)=feed back signal Closed loop transfer function = C(S) = G(S)___ R(S) 1+G(S)H(S) C(S) =

G(S)___. R(S) 1+G(S)H(S)

E(S)=R(S)-B(S)----- >1 WKT B(S)=R(S)-B(S)------ >2 Put eq 2 in eq1 E(S)=R(S)-C(S)H(S)------ >3 WKT C(S)=E(S)G(S)------- >4 Put 4 in eq 3 E(S)=R(S)-E(S)G(S)H(S) E(S)+E(S)G(S)H(S)=R(S) Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

[1+G(S)H(S)]E(S) =R(S) E(S)=___R(S)___ 1+G(S)H(S)

E(S)=___R(S)__ 1+G(S)

for non unity feed back

For unity feed back

ESS= Lt SR(S)_____ s->0 1+G(S)H(S) Steady state error deoends on (i)

R(S) i.e reference i/p its type and magnitude.

(ii)

G(S)H(S) i.e open loop transfer function.

(iii)

Dominant non linearities if any. Effect of input on steady state error: (static error co efficient method) (a) Reference i/p is step of magnitude A R(S) = __A__ S ESS= Lt SR(S)_____ s->0 1+G(S)H(S)

= Lt S.A/S_____ s->0 1+G(S)H(S) = Lt A_____ s->0 1+G(S)H(S) ESS=_____A___ 1+ Lt G(S)H(S) s->0 For a system selected Lt

G(S)H(S) is constant and called


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s->0 positional error co_efficient of the system denoted as Kp. Kp= Lt G(S)H(S) =positional error co-efficient s->0 and corresponding error is ess=__A__ 1+kp Reference i/p is ramp of magnitude A R(S)=A/S2. ESS= Lt SR(S)_____ s->0 1+G(S)H(S) = Lt S.A/S2_____ s->0 1+G(S)H(S) = Lt A_____ s->0 S[1+G(S)H(S)] ESS =_____A___ Lt S G(S)H(S) s->0 for a seleted system Lt S G(S)H(S) is constant an called velocity error co efficient as Kv. s->0 ESS=A/Kv Kv= Lt S G(S)H(S) =velocity error effect. s->0 reference input is parabolic of magnitude A R(t)=__A__ t2 2 R(S)= _A__ Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

S3

ESS= Lt SR(S)_____ s->0 1+G(S)H(S) = Lt S.A/S3_____ s->0 1+G(S)H(S) = Lt A_____ s->0 S2[1+G(S)H(S)] ESS =_____A___ Lt S2 G(S)H(S) s->0 for a seleted system Lt S2 G(S)H(S) is constant an called velocity error co efficient as Ka. s->0

ESS=A/Ka Ka= Lt S2 G(S)H(S) =acceleration error effect. s->0

types of feed back control systems: The open loop transfer function G(S) of a unity feed back system can be written in two standard forms namely the time constant form and polr zero form. G(S)=K(TZ1S+1) (TZ2S+1) (TZ3S+1)…….. (TZmS+1) Sn(TP1S+1) (TP2S+1) (TP3S+1)…….. (TPmS+1) (TIME CONSTANT FORM) = _K’(S+Z1) (S+Z2) (S+Z3)…….. (S+Zm) Sn (S+P1) (S+P2) (S+P3)…….. (S+Pm)

(POLE ZERO FORM)

Where the relation between K and K’ is Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

K=K’(product of all zeros) Product of all poles. In both respresentatiom, the terms Sn in the denominator represent the n poles at the origin, which represent the type number of the system. If n=0, the system is known as type 0 system and if n=1 the system is known as type 1 system and so on.

The type number of a system is defined as the number of poles of G(S) at the origin or number of integrators present in the systems. For example

G(S)=(S+5)____ is a type 0 system (S+4)(S+8) G(S)=(S+9)___ is a type 1 system S(s+3)(s+7) G(s)=(s+10)____ is a type 2 system S2(S+2)(S+9)


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Steady state errors of type 0 unity feedback system. Let G(S) be an open loop transfer function of a system with no poles at the origin (n=0).the steady state errors for standard inputs are as follows. Positional error essp(t)= Lt s->0

s.1/s__ 1+G(S)

= ____1____ 1+ Lt G(S) s->0 ____1__ = __1___ 1+kp 1+k Velocity error essv(t) = Lt s->0

s.1/s2__ 1+G(S)

= ____1____ 1+ Lt s G(S) s->0

= 1/0=∞

acceleratiom error essa(t) = Lt s.1/s3__ s->0 1+G(S) = ____1____ = 1/0=∞ 1+ Lt s2 G(S) s->0 steady state error of type1 unity feed back system: Let G(S) be an open loop transfer function of a system with one poles at the origin (n=1).the steady state errors for standard inputs are as follows.

Positional error essp(t)= Lt s->0

s.1/s__ 1+G(S)

= ____1____ 1+ Lt G(S) s->0 ____1__ = 0 1+∞ Solved by A.Devasena., Associate Professor., Dept/ECE

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Velocity error essv(t) = Lt s->0

IV Semester ECE

s.1/s2__ 1+G(S)

= ____1____ Lt s+ s G(S) s->0

= ____1____ Lt s G(S) s->0

= 1/k

acceleratiom error essa(t) = Lt s.1/s3__ s->0 1+G(S) = ____1____ = 1/0=∞ 2 2 Lt s +s G(S) s->0 for tupe 1 system

Lt s->0

G(S)=k(Tz1S+1) (Tz2S+1)……. S(Tp1S+1) (Tp2S+1)……. G(S) =∞=kp

Lt s G(S) =k=kv s->0 Lt s2 G(S) =0=ka s->0

steady state errors of type 2 unity feedback system. G(S) be an open loop transfer function of a system with two poles at the origin (n=2).the steady state errors for standard inputs are as follows. Positional error essp(t)= Lt s->0

s.1/s2__ 1+G(S)

= ____1____ 1+ Lt G(S) s->0 ____1__ = 0 1+∞


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EC2255- Solved Problems in Control System Velocity error essv(t) = Lt s->0

IV Semester ECE

s.1/s2__ 1+G(S)

= ____1____ Lt s G(S) s->0

= 1/∞=0

acceleratiom error essa(t) = Lt s.1/s3__ s->0 1+G(S) = ____1____ = 1/k 2 2 Lt s +s G(S) s->0 for tupe 2 system

Lt s->0

G(S)=k(Tz1S+1) (Tz2S+1)……. S2(Tp1S+1) (Tp2S+1)……. G(S) =∞=kp

Lt s G(S) =∞=kv s->0 Lt s2 G(S) =k=ka s->0

generalized error coefficient method (or dynamic error co efficients) E(S)=___R(S)__ 1+G(S)H(S) Let us assume that is the product of two polynomial of s E(S)=F1(S)F2(S) Where f1(s)=__1___ . 1+G(S)H(S)

F2(S) =R(S)

F(S)=F1(S).F2(S) then using convolution integral t -1 L { F(S)}= F(t)= ∫F1(ι) F2(t-ι)dι 0 t t Similarly e(t)= ∫F1(ι) F2(t-ι)dι = ∫F1(ι) R(t-ι)dι Solved by A.Devasena., Associate Professor., Dept/ECE

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EC2255- Solved Problems in Control System 0

IV Semester ECE

0

R(t-ι) can be expanded by using taylor series form as R(t-ι)=R(t)-ιR’(t)+ι2 R”(t)+ι2 R’”(t)+…..

2!

3!

Substituting e(t)= t ∫F1(ι) [R(t)-ιR’(t)+ι2 R”(t)+ ι2 R’”(t)+…..]dι 0 2! 3!

t =

0

t ∫F1(ι) R(t) dι - ∫F1(ι) ιR’(t)dι+……. 0

Ess=lt e(t) t->∞ =

lt t t t->∞ [ ∫F1(ι) R(t) dι - ∫F1(ι) ιR’(t)dι+…] 0

0

∞ ∞ ∞ = R(t) ∫F1(ι) dι - R’(t) ∫F1(ι) ι dι+ R”(t) ∫F1(ι) ι2 dι]

0

0

0

2!

∞ Where ko= = ∫F1(ι) dι 0 ∞ K1= - ∫F1(ι) ι dι

0 ∞ K2= - ∫F1(ι) ι2 dι

0 Substituting these values we have Ess= k0R(t)+k1R’(t)+k2 R”(t)+……. 2!

Where ko, k1,k2…… are called dynamic error coeff To calculate these values co efficient use the following method According to the definition of laplace transform Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

∞ F(s)= ∫ F1(ι)e-sιdι 0

∞ Now k0=∫ F1(ι)dι 0 Multiplying by e-sι to both sides

∞ e-sι k0=∫ F1(ι) e-sι dι = F1(s) 0 Taking limit as s->0 of both sides k0 e-sι = lt F1(s)

Lt

s->0

s->0

where F(S)= ___1__ 1+G(S)H(S) TAKING derivative of k0 e-sι w.r.t ‘s’ we get -sι

-ι k0 e

substituting

= dF1(S) ds ∞ k0 = ∫ F1(ι) dι 0

∞ ∫ F1(ι) e-sι dι= dF1(S) 0 ds Taking limit as s->0 of both side

K1=

lt

s->0

dF1(S) ds

in general Kn=

lt

dn F1(S)


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EC2255- Solved Problems in Control System s->0

IV Semester ECE

ds

Introduction to PID controllers: A controller is a device which when introduced in feed back or forward path of system, controls the steady state and transient response as per the requirement. This controller converts the applied input to some other form of error which is proportional to the error due to which steady state and transient response gets improved. The output of the controller is proportional to the amount of error generated by that device. The performance of this controlling phenomena may be done by means of electrical, mechanical, pneumatic or hydraulic medium. Classification of controllers: 1.ON-OFF controller 2. proportional controller 3. integral controller 4. derivative controller 5. proportional plus integral controller 6. proportional plus derivative controller 7. proportional +integral+derivativecontroller ON-OFF controller: This type of controller is the simplest and cheapest typehere the actuating device (controller) Is the capable of assuming only one two positions, with either zero(or) maximum input to the process. Proportional controller: It is defined as the action of a controller in which the output signal m(t) is proportional to the measured actuating error signal e(t). M(t) α e(t) M(t) =kpe(t) Where kp is proportional sensitivity or the gain Taking laplace transform we get M(s)=kpE(S)


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IV Semester ECE

Kp= M(S) = change in controller output E(S) change in derivation So the proportional controller is essentially an amplifier with an adjustable gain. Integral controller: It is denoted by the term ‘I’ in the PID controller. In a controller with integral action, the value of the output m(t) is propprtional to the measured actuating error signal e(t). dm(t) = kie(t) ds t m(t) = ki ∫e(t) dt 0 Where ki is an adjustable constant the transfer function of the integral controller is M(S) = ki E(S) s The integral controller increases the type number of system by one. This integral control overcomes the drawback of proportional control by reducing the steady state error to zero without the use of excessive large control gain. The integral control action is also called as reset control action. Derivative controller: Derivative controller do not affect state error but effect transient response. The output of derivative controller depends on the rat e of change of error signal. The main drawback of derivative controller is that it amplifies the noise signal. Therefore it is not possible for us to use the derivative control action alone. We can use this derivative controller along with integral or proportional controller In time domain m(t)= kd dl(t) Dt In laplace domain m(S)=kds E(S) The output of the derivative controller is zero when the error signal constant. That is the derivative control has no input when it is acting on constant signal.

Proportional plus integral controller (PI controller) In PI controller the controller output m(t) is proportional to a linear combination of actuating signal e(t) plus integral of the error signal.


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IV Semester ECE

M(t) =kpe(t) +ki ∫e(t) dt. Taking laplace transform M(s) = kpE(S)+kI E(S) S

=E(S)[kp+ki_] S Assuming kp=1 we can write G(S)=(1+ki ) ωn2 = ___s_

S(s+2z ωn)

(ki+s) ωn2 s2(s+s2z ωn)

i.e the system becomes type 2 in nature and C(S) = R(S)

(ki+s) ωn2 s3+2z ωns2+s ωn2+ki ωn2

i.e it becomes third order that is it integral is included along with proportional controller means the system relatively becomes less stable as ki must be designed in such a way that system will remain in stable condition. Second order system will always stable. The transient response of of system gets affected badly if PI controller is not designed properly. By including ‘I’ controller along with ‘p’ controller, steady state the system gets improved and we can get the accurate output from the system. In general PI controller improves steady state part but it affects transient part of the system. Proportional plus derivative controller (PD controller) A controller in a forward path which changes the controller output corresponding to the proportional plus derivative of error signal is known as PD controller M(T) =kpe(t) +kpTd de(t) Dt Taking laplace transform M(S) = kpE(S) +kpTd s E(S) =E(S)[kp+TdS] Assuming kp=1 G(S) = (1+sTd) ωn2 S(s+2z ωn)


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IV Semester ECE

C(S) = (1+sTd) ωn2 R(S) s2+2[2zωn+ ωn2Td] + ωn2 C0mparing the denominator with standard form s2+2z ωns+ ωn2 ωn2 term is same as in the standard form but 2z ωn =2z ωn+ ωn2Td z= 2z ωn + ωn2Td z ωn z ωn because of this derivative controller the damping ratio increases by a factor ωnTd 2

Kp= lt G(S)H(S) =∞ S->0 KV= LT S G(S)H(S) = ωn s->0 2z as there is no change in co efficient, the error will also remain same. PD controller has following effects on the system. 1. 2. 3. 4. 5.

It reduces settling time It reduces peak overshoot Steady state error remains unchanged. Type of the system remains unchanged It increases damping ratio

PD controller improves transient part without affecting steady state. Proportional+integral +derivative controller(PID controller) A PID controller is the combination of proportional control action and derivative control action with an adjustable gain for each action. t M(t)= kpe(t)+kpTdde(t) + kp ∫ e(t) dt dt TI 0 taking laplace transform M(S)= kpE(S)+kpTdSE(S) + kp E(S) TiS M(S)= kpE(S)[1+Td + kp ] Ti S


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IV Semester ECE

Pd controller improves transient portion, pi controller improve steady state portion combination of two may be used to improve overall time response of the system.

G(s) =(K+STd+KI ) ωn2 S____ S(S+2Z ωn) C(S) = R(s)

(K+STd+KI ) ωn2/ S(S+2Z ωn) S_________ 1+ (K+STd+KI ) ωn2 S____ S(S+2Z ωn)

Transient response and steady state response. The time response of a system consists of two parts. Transient response Steady state response This can be denoted by equation as Total time response =transient response +steady state response C(t)=ctr(t)+css(t) Transient response of the system is the portion of total time response during which output Changes from one value to another value. In other words, it is the response before the output reaches the steady state value. Steady state response of the system for a given input after a very long time. In steady state, the output response settles to its final steady state value or steady oscillations. Derivation of steady state error: Definition: steady state error : it is the difference between the actual output and the desired output. Consider a simple closed loop system using negative feedback us shown. Where E(s)=error signal and B(s)=feed back signal. Now E(S)=R(S)-B(S) BUT B(S)=C(S)H(S)


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IV Semester ECE

E(S)=R(S)-C(S)H(S) C(S)=E(S)G(S) E(S)=R(S)-E(S)G(S)H(S) E(S)+E(S)G(S)H(S)=R(S). E(S)=R(S)___ 1+G(S)H(S) for non unity feed back . E(S)=R(S)___ 1+G(S) for unity feed back. This E(S) is the error in laplace domain and is expression in ‘s’. we want to calculate the error value. In the time domain corresponding error will be e(t). now steady state of the system is that state. Which remains as t-> ∞. Steady state error ess=Lt e(t) t->∞ standard test signals: the impulse signal has zero amplitude every where expect at the origin as shown in fig. mathematically the impulse signal can be represented by A δ(t)=0 for t#0 and ε ∫ A δ(t) dt= A -ε Where ε tends to zero. Here the value a represents the area of the signal or energy content of the signal and the laplace transform of the impulse is given by. L[ A δ(t)] =A

Unit impulse signal: If A=1, then the signal is called unit impulse signal that is for a unit impulse signal. δ(t)=0 for t#0 and ε ∫ A δ(t)dt = 1 -ε The laplace transform of the unit impulse signal is


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IV Semester ECE

L[ δ(t)] =1.

Step signal: Step signal of size A is a signal that changes from zero level to another level A is a zero time and stays there for ever as shown fig. the step signal is applied to the system to study the behavior of the system for a sudden change in input. Mathematically r(t) = A for t> 0 = 0 for t< 0 The laplace transform of the the step signal is R(S)=L[r(t)]=A/s.

Unit step signal: If the the magnitude a of the step signal is unity, then the step signal is shown and is denoted bt u(t). that is U(t) =1 for t > 0 =0 for < 0. The laplace transform of the unit step msignal u(t) is L[u(t)] =1/s.

Ramp signal: The ramp signal increases linearly with time from an initial value of zero at t=0. Mathematically representation of the ramp signal is given by R(t) = At for t> 0 = 0 for t=0 Where A represents the slope of the line The laplace transform of the ramp signal is known as unit ramp signal. Parabolic signal: The instantaneous value of a parabolic signal varies as square the time from an initial value of zero at t=0. Mathematically representation of the


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IV Semester ECE

Parabolic signal is r(t)=At2 t>0. = 0 t<0. The laplace transform of the parabolic signal is given by R(s)= L[At2 ] =2A/s3 To have a convinent form for laplace transform of parabolic signal is A/s3 . the parabolic signal in time domain is often defined as At2/2. Steady state erroer and error constants The stady state error ess is the difference between the input (or desired value ) and the output of a closed loop system for a known input as t->∞. Mathematically Ess = Lt e(t) t->∞ = Lt [r(t)-c(t)] t->0 bu using final value theorem

Ess = Lt e(t) = Lt s E(S) t->∞ s->0 this is valid provided that SE(S) has no poles on the jω axis no multiple poles on the jω axis and is the rignt half of s plane. The steady state error is a measure of system accuracy. With the help of step ramp and paroboic inputs it is passible for us to judge te steady state behavior of the system.

Analysis of first order system: Order: order of system is the highest power of s in the denominator of a closed loop transfer funttion. Consider a simple system shown in fig. find v0(t) i.e response if it is excited by unit step input. Vi(t)=1 t>0 =0 t<0

Vi(s)=1/s.


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IV Semester ECE

Now first calculate system T.F using laplace T.f Vi(t) = R i(t)+ 1 ∫ i(t) dt C

V0(t) = 1 ∫ i(t) dt C

Vi(s) = R I(s)+ 1 I(s) Cs V0(s) = 1 I(s) Cs I(S)=CSV0(S) Vi(S)=I(S)[R + 1 ] SC Vi(S)=CSV0(S)[R + 1 ] SC

V0(S) = ___1__ VI(S) 1+SRC V0(S) = ___1__ VI(S) 1+TS where T=RC

The step response of first order system . Let input applied vi(t) is the unit step voltage substituting vi(s)=1/s is the transfer function V0(S) = ___1__ S( 1+SRC)

= A + __b__ s 1+SRC

1= A(1+SRC) +BS 1=A+ASRC+BS 1=A+S(B+ARC) Equate constant term 1=A Equate s term 0=S(B+ARC) Put A=1 Solved by A.Devasena., Associate Professor., Dept/ECE

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IV Semester ECE

B+RC=0 B=-RC V0(S) = 1 + __-RC__ S 1+SRC

S

= 1 + __-RC__ RC (S+1/RC)

= 1 + __-1_____ S (S+1/RC) Taking laplace inverse V0(t) =1- e-t/RC =css+ct(t) Closed loop transfer function of a system is given by C(S) = G(S)___ R(S) 1±G(S)R(S) The equation which gives poles of system is called characteristics equation which is 1+G(S)H(S)=0 For the first order system this equation is also first order is general of the form 1+TS=0 AS closed loop poles are the roots of characteristics equation so for first order system there is only one closed loop pole i.e s=-1/T The time taken for the step response of a system to reach 63.2% of the final value is known as the time constant of the system. Unit ramp response of a first order system: The transfer function of a first order system is given by C(S) = ___K__ R(S) (TS+1) For a ramp input r(t)=t; t>0; R(S)=1/S2. There fore C(S) = ___K__ S2 (TS+1)


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C(S)= K [A +B + __C__] S S2 TS+1 __K__ = K[AS(TS+S)+BTS+B+CS2 ] S2(TS+1) S2(TS+1) __K__ = K[AS(TS+S)+BTS+B+CS2 ]

1=AS(TS+S)+BTS+B+CS2

Equate constant term A=1

equate s term A+BT=0 1+BT=0 BT=-1 B=-1/T

equate s2 term AT+C=0 C=-AT C=-T

C(S)= K [1 +-1/T + __-T__] S S2 TS+1

Taking inverse laplace transform C(s)=k [-T+t+e=t/T]; t>0 22) A unity feed back heat treatment system has open loop transfer function G(s)= ____10000______ . the output of set point is 5000c. what is the steady state temperature. (1+S)(1+0.5S)(1+0.02S) Here h(s)=1 R(S)=500/s Steady state error =Lt S.R(S) S->0 1+G(S) =Lt S.500/S S->0 1+ ____10000______ . (1+S)(1+0.5S)(1+0.02S) 500/1+10000 =0.040 STEADY STATE ERROR ESS=0.040C STEADY state temperature c(t)=r(t)-ess =5000c- 0.040c = 499.960c Solved by A.Devasena., Associate Professor., Dept/ECE

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(

1. Determine the stability of the following system using Routh’s criterion. (AU:May/ June 2008) (a) G(s)H(s) =______1______ (s +2)(s +4) (b) G(s)H(s) =______9______ s2(s +2)


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1.a.)G(s)H(s) =______1______ (s +2)(s +4) The characteristic equation is 1 + G(s)H(s) = 0 1 + ___1______ = 0 (s +2)(s +4) ( s + 2 ) ( s+4 ) + 1 =0 s2 + 6s + 9 = 0 a0 =1 a1 = 6 a2 =9 Routh’s array s2

1

9

s1

6

0

s0

9

No sign change in the first column of the Routh’s array, hence the system is stable.


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1.b.)G(s)H(s) =_____9______ s2(s +2) The characteristic equation is 1 + G(s)H(s) = 0 1 + ___9______ = 0 s2(s +2) s2(s +2) + 9 =0 s3 + 2s2 +9 = 0 a0 =1 a1 = 2 a2 =0 a3 =9 Routh’s array s3 1

0

b1 = 2 x 0 – 1x 9 = - 4.5 s

2

s

1

2

2

9

b2 = 0

S0

- 4.5 9

c1 = - 4.5 x 9

=9

- 4.5

There are two sign changes in the first column of the Routh’s array. Hence the system is unstable.

2. Utilize the Routh table to determine the number of roots of the following polynomials in the right half of the s-plane. Comment about the stability of the system. (AU:May/June 2006) ( i ) s5 + 2s4 + 3s3 + 6s2 + 10s + 15 ( ii ) s5 + 6s4 + 15s3 + 30s2 + 44s + 24


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2) i) s5 + 2s4 + 3s3 + 6s2 + 10s + 15 = 0 a0 =1 a1 = 2 a2 = 3 a3 = 6 a4 =10 a5 = 15

Routh’s array: s5

1

3

10

15

s4

2

6

s3

0

2.5

s3

ε

s2

6ε-5 ε

s1

2.5

Replace the 0 by small positive number ε

15

6 ε – 5 2.5 ____ε _________ - 15 ε 6ε-5 ε

s0

15

as ε tends to zero, the routh’s array becomes

s5

1

3

s4

2

6

s3

0

2.5

2

-∞

15

s1

2.5

0

s0

15

s

10 15


There are two sign changes in the first column of the routh’s array. There fore there are two roots on the right half of the s- plane.

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2) ii ) s5 + 6s4 + 15s3 + 30s2 + 44s + 24 = 0 a0 =1 a1 = 6 a2 = 15 a3 = 30 a4 =44 a5 = 24

Routh’s array: s5

1

15

44

24

s4

6

30

s3

10

40

s2

6

s1

24 Replace the 0 by small positive number ε

0

s1

ε

s0

24

as ε tends to zero, the routh’s array becomes s5

1

15

44

s4

6

30

24

s3

10

2

6

s

s1

0

s0

24

40 24


There is no sign change in the first column of the routh’s array. No roots of the polynomial are in the right half of the s- plane. Hence the system is stable.

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3. for a system with, F(s) = s4 + 22s3 + 10 s2 + s + K = 0. obtain the marginal value of K, and the frequency of oscillations of that value of K (AU May/june 2007)

F(s) = s4 + 22s3 + 10 s2 + s + K = 0 a0 = 1 a1 = 22 a2 = 10 a3 = 1 a4 = K Routh’s array

s4

1

10

K

s3

22

1

0

s2

9.95

K

s1

9.95 – 22K 9.95 s0

K

For a stable system, the first column of the routh’s array must be positive. Therefore K > 0 , And 9.95 – 22K > 0 0 < K < 0.452. The marginal value of K is 0.452. Consider 9.95s2 + K = 0. when K = 0.452, to find the frequency of oscillations s= + j 0.913 or jω = + j 0.913. ω = 0.213 rad/sec.


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4. Using Routh’s criterion for stability, discuss the stability of the system whose characteristic equation is s3 + 10s2 + 50s + 500 = 0 (AU: April/ may 2005)

s3 + 10s2 + 50s + 500 = 0 a0 = 1 a1 = 10 a2 = 50 a3 = 500 Routh’s array s3

1

50

s2

10

500

s1

0

0

0

by special case 2, replace zero row by the derivative of the auxiliary equation ( the auxiliary equation is the row, just above the zero row) Auxilary equation is A(s) = 10s2 + 500 dA(s) = 20s ds s3

1

50

s2

10

500

s1

20

0

s0

500

0

No sign change in the first column of the Routh’s array. But due to special case 2, system , may not be stable. For sufficient conditions solve A(s) = 0.


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10s2+ 500 = 0 s2 = - 50 s = + j √ 50. as dominant roots are on imaginary axis, and no root in right half of s-plane because no sign change in first column of routh’s array. the system is marginally stable. It oscillates with the frequency √ 50 rad./ sec.

5. Using Routh –Hurwitz criterion, determine the relation between K and T so that unity feedback control system whose open loop transfer function given below is stable.


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1.Sketch the root locus diagram of the following open loop transfer function G(s)H(s) =________K_____ s(s+2)(s+5)

(AU: Nov/DEC:2005)

Step 1: Number of poles P = 3 At s 0,-2,-5 Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 3

Step 3: The starting point of the root locus branches are from poles 0,-2,-5. Terminating point of the root locus branches are at ZEROs. But here we have no ZEROs. So The root locus branches terminates at infinity.

+jω

+σ -σ

P2

P3

-5

P1

-2

0

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 the sum of poles and zeros are even. So P2 is not the part of root locus. Consider P3. to the R.H.S. of P3 the sum of poles and zeros are odd. So P3 is the part of root locus. Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 4: Number of asymptotes Na = P-Z = 3- 0 = 3 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-0 = 3 n = 0,1,2. For n =0 θ0 = (2(0)+1)π = 60ο 3–0 For n =1 θ1 = (2(1)+1)π = 180ο 3–0 For n =2 θ2 = (2(2)+1)π = 300ο 3–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -2-5) – (0) 3–0 = -7 = -2.3 3 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+ K_________ = 0 s(s+2)(s+5) s(s2 + 7s + 10) +K= 0 K = - s(s2 + 7s + 10) K = - (s3 + 7s2 + 10s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - 3s2 - 14s – 10 = 0 ds i.e. 3s2+ 14s +10 = 0 now solve the above quadratic equation, we get s = - 0.88 , -3.78. The valid breakaway point is – 0.88. because breakaway point must be a part of root locus. Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 s3 + 7s2 + 10s + k = 0 a0 = 1 a1 = 7 a2 = 10 a3 = k Routh’s array

s3

1

10

s2

7

K

s1

7(10) – K 7

s0

K

0

To have intersection on imaginary axis, any one row should be zero. Let us make s1 row = 0 70 – K 7 70- K

= 0 = 0 K = 70

Now , consider auxiliary equation 7s2 +K = 0 Put

K = 70 in above equation.

7s2 +70 = 0 s2 +10 = 0 s = + j √10 = + j 3.2 The complete root locus is shown in fig.4.1 Solved by A.Devasena., Associate Professor., Dept/ECE

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fig. 4.1


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2.Sketch the root locus diagram of the following open loop transfer function G(s)H(s) =________K_____ s(s+1)(s+3)

(AU: Nov/DEC:2006)

Step 1: Number of poles P = 3 At s

0,-1,-3

Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 3

Step 3: The starting point of the root locus branches are from poles 0,-1,-3 Terminating point of the root locus branches are at ZEROs. But here we have no ZEROs. So The root locus branches terminates at infinity.

+jω

+σ -σ

P2

P3

-3

P1

-1

0

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2the sum of poles and zeros are even. So P2 is not the part of root locus. Consider P3. to the R.H.S. of P3 the sum of poles and zeros are odd. So P3 is the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 3- 0 = 3 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-0 = 3 n = 0,1,2. For n =0 θ0 = (2(0)+1)π = 60ο 3–0 For n =1 θ1 = (2(1)+1)π = 180ο 3–0 For n =2 θ2 = (2(2)+1)π = 300ο 3–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -1-3) – (0) 3–0 = -4 = -1.33 3 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+ K__________ = 0 s(s+1)(s+3) s(s+1)(s+3) +k = 0 K = - s(s2 + 4s + 3) K = - (s3 + 4s2 + 3s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - 3s2 - 8s – 3 = 0 ds i.e. 3s2+ 8s +3 = 0 now solve the above quadratic equation, we get s = - 0.451 , -2.22 The valid breakaway point is – - 0.451. because breakaway point must be a part of root locus. Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 s3 + 4s2 + 3s + k = 0 a0 = 1 a1 = 4 a2 = 3 a3 = k Routh’s array

s3

1

s2

4

s1

12 – K 4

s0

K

3

K 0

To have intersection on imaginary axis, any one row should be zero. Let us make s1 row = 0 12-K 4 12 - K

= 0 = 0

K = 12 Now , consider auxiliary equation 4s2 +K = 0 Put

K =12 in above equation.

4s2 +12 = 0 s2 +3 = 0 s = + j √3= + j 1.714 The complete root locus is shown in fig.4.2 Solved by A.Devasena., Associate Professor., Dept/ECE

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fig. 4.2


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3.Obtain the root locus for a unity feedback system with open loop transfer function G(s)H(s) =________K_____ (AU: may/june:2006 ) 2 s(s +6s+25)


0,-3+j4, -3 –j4

Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 3 Step 3: The starting point of the root locus are from poles 0, -3+j4, -3 –j4. the terminating point of the root locus is at ZERO. Here we have no ZEROs. So the root locus terminates at infinity.

+jω

-3+j4

+σ

P1

-σ 0

-jω

-3-j4 Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 3- 0 = 3 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-0 = 3 n = 0,1,2. For n =0 θ0 = (2(0)+1)π = 60ο 3–0 For n =1 θ1 = (2(1)+1)π = 180ο 3–0 For n =2 θ2 = (2(2)+1)π = 300ο 3–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -3-3) – (0) = -2 3–0 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+

K__________ = 0 s(s2+6s+25)

s(s2+6s+25)+K = 0 K = - s(s2+6s+25) K = - (s3 + 6s2 + 25s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - 3s2 - 12s – 25 = 0 ds i.e. 3s2+ 12s +25 = 0 now solve the above quadratic equation, we get s = - 2 + j0.5√39 this point is not on the root locus. Therefore there is no breakaway point. Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 1+

k__________ = 0 s(s2+6s+25)

s3 + 6s2 + 25s + K = 0 a0 = 1 a1 = 6 a2 = 25 a3 = K. Routh’s array s3

1

s2

6

K

s1

150 – K 6

0

s0

K

25

To have intersection on imaginary axis, any one row should be zero. Let us make s1 row = 0 150-K 6 150 - K

= 0 = 0



6s2 +150= 0 s2 +25 = 0 s= + j5 Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 8: We have complex poles. So angle of departure has to be calculated.

Angle of Departure calculation diagram.

P2

+j4

Φp1 -3

Φp2 -j4 P3 Consider point P2. Φp1 = -180ο – tan-(3/4) = - 143.13ο Φp3= -90ο ∑ Φp = Φp1 + Φp3 = - 143.13ο - 90ο = -233.13ο ∑ Φz = 0 ο Φ = ∑ Φp - ∑ Φz = -233.13ο - 0 ο = -233.13ο Angle of departure = 180ο- -233.13ο = - 53.13 ο Similarly for point P3. the angle of departure is - 53.13 ο The complete root locus is shown in fig.4.3. Solved by A.Devasena., Associate Professor., Dept/ECE

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Fig.4.3


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4.Determine the root locus of the system whose open loop gain is G(s)H(s) =________K_______ (AU: Nov/Dec:2006 ) 2 S(s+4)(s +4s+20)


0, -2+j4, -2 –j4 , -4

Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 4 Step 3: The starting point of the root locus are from poles 0, -2+j4, -2 –j4 , -4 . The terminating point of the root locus is at ZERO. Here we have no ZEROs. So the root locus terminates at infinity.

-2+j4

P2

+jω

+σ

P1

-4

-σ 0

-jω

-2-j4 Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2.To the R.H.S of the P2 the sum of poles and zeros are even. So p2 is not the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 4- 0 = 4 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 4-0 = 4 n = 0,1,2,3 For n =0 θ0 = (2(0)+1)π = 45ο 4– 0 For n =1 θ1 = (2(1)+1)π = 135ο 4–0 For n =2 θ2 = (2(2)+1)π = 225ο 4–0 For n =3 θ3 = (2(3)+1)π = 315ο 4–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -4-2 -2) – (0) 4–0 = -8 = - 4 4 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+

k__________ = 0 s(s+4) (s2+4s+20) s(s+4) (s2+4s+20) + K = 0

k = - (s 2 +4s)( (s2+4s+20) k = - (s4 +8s3 + 36s2 +80s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - ( 4s3 +24s2+72s +80) = 0 ds i.e. s3 +6s2+18s +20 = 0 Solved by A.Devasena., Associate Professor., Dept/ECE

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-2

1 0

6 -2

18 -8

1

4

10

IV Semester ECE

20 -20 0

When s= 0, dk/ds =80 (+v) When s =-4, dk/ds = -80 (-ve) When s=0, dk/ds = 0. the breakaway point is -2. The roots of s2 + 4s +10 = 0 are -2 + j2.45 The other break away points are -2 + j2.45


k__________ s(s+4)(s2+4s+20)

=0

s(s+4) (s2+4s+20) + K = 0 s4 +8s3 + 36s2 +80s + K = 0 a0 = 1 a1 = 8 a2 = 36 a3 = 80 a4 = K

.


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Routh’s array s4

1

36

K

s3

8

80

0

s2

26

K

s1

2080 –8 K 26

s0

K

To have intersection on jω axis, any one row should be zero. So make s1 row to be zero. 2080 –8K = 0 K = 260. Now , consider auxiliary equation 26s2 +K = 0 Put


26s2 +260= 0 s2 = - 10 s = + j √10 = + j 3.16 The complete root locus is shown in fig.4.3


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Step 8: Angle of departure:

Фp3

Фp2

Фp1

Фp4

Angle of departure Фd = 180ο + ∑ Фp Consider point P3 ∑ Фp = - Фp1 - Фp2 -Фp4 = - (180ο – tan-14/2 ) - tan-14/2 - 180ο = - 270ο Angle of departure Фd = 180ο + ∑ Фp = 180ο- 270ο = - 90ο The complete root locus is shown in fig.4.4.


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Fig.4.4.


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5. Sketch the root locus for the unity feedback system whose open loop transfer function is G(s)H(s) =________K_____ s(s+3)(s+8)


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IV Semester ECE


0,-3,-8

Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 3

Step 3: The starting point of the root locus branches are from poles 0,-3,-8. Terminating point of the root locus branches are at ZEROs. But here we have no ZEROs. So the root locus branches terminates at infinity.

+jω

P2

P3

P1

+σ

-σ -8

-3

0

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 the sum of poles and zeros are even. So P2 is not the part of root locus. Consider P3. to the R.H.S. of P3 the sum of poles and zeros are odd. So P3 is the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 3- 0 = 3 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-0 = 3 n = 0,1,2. For n =0 θ0 = (2(0)+1)π = 60ο 3–0 For n =1 θ1 = (2(1)+1)π = 180ο 3–0 For n =2 θ2 = (2(2)+1)π = 300ο 3–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -3-8) – (0) 3–0 = -11 = -3.66 3 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+ K__________ = 0 s(s+3)(s+8) s(s2 + 11s + 24) +k = 0 K = - s(s2 + 11s + 24) K = - (s3 + 11s2 + 24s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - 3s2 - 22s – 24 = 0 ds i.e. 3s2+ 22s +24 = 0 now solve the above quadratic equation, we get s = - 1.3, - 6. The valid breakaway point is – 1.3. Because breakaway point must be a part of root locus. But -6 is not the part of the root locus. Solved by A.Devasena., Associate Professor., Dept/ECE

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Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 s3 + 11s2 + 24s + K = 0 a0 = 1 a1 = 11 a2 = 24 a3 = K Routh’s array

s3

1

s2

11

K

s1

11(24) – K ) 11

0

s0

K

24

To have intersection on imaginary axis, any one row should be zero. Let us make s1 row = 0 264 – K 11 264 – K

= 0 = 0



11s2 +264 = 0 s2 +24 = 0 s = + j √24 = + j 4.9 The complete root locus is shown in fig.4.5 Solved by A.Devasena., Associate Professor., Dept/ECE

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Fig.4.5.


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6.Determine the root locus of the system whose open loop gain is G(s)H(s) =________K_______ ; S(s+4)(s2+8s+32) k > 0. Sketch the root locus of the system. Hence find the value of K so that system has a damping factor of 0.707


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IV Semester ECE


0, -4+j4, -4–j4 , -4

Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 4 The starting point of the root locus are from poles 0, -4+j4, -4–j4 , -4 The terminating point of the root locus is at ZERO. Here we have no ZEROs. So the root locus terminates at infinity.

-4+j4 +jω

P2

-4

+σ

P1

-σ 0

-jω

-4-j4 Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2.Tothe R.H.S of the P2 the sum of poles and zeros are even. So p2 is not the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 4- 0 = 4 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 4-0 = 4 n = 0,1,2,3 For n =0 θ0 = (2(0)+1)π = 45ο 4– 0 For n =1 θ1 = (2(1)+1)π = 135ο 4–0 For n =2 θ2 = (2(2)+1)π = 225ο 4–0 For n =3 θ3 = (2(3)+1)π = 315ο 4–0 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -4-4 -4) – (0) 4–0 = -12 = - 3 4 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+

K__________ = 0 s(s+4) (s2+8s+32) s(s+4) (s2+8s+32) + K = 0

K = - (s 2 +4s)( (s2+8s+32) K = - (s4 +12s3 + 64s2 +128s) to find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds dk = - ( 4s3 +36s2+128s +128) = 0 ds i.e. s3 +9s2+32s +32 = 0 Solved by A.Devasena., Associate Professor., Dept/ECE

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1 0

-2

1

IV Semester ECE

9 32 32 -1.57 -11.25 -31.92 7.43

20.35

0.08

If s =-1.57, then the value of k is nearly zero. So s=-1.57 is the valid breakaway point.


k__________ s(s+4)(s2+8s+32)

=0

s(s+4) (s2+8s+32) + K = 0 s4 +12s3 + 64s2 +128s + K = 0 a0 = 1 a1 = 12 a2 = 64 a3 = 128 a4 = K .


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Routh’s array s4

1

64

K

s3

12

128

0

s2

53.3

s1

6826.7 –12K 53.3

s0

K

K

To have intersection on jω axis, any one row should be zero. So make s1 row to be zero. 6826.7 -12K= 0 53.3 K = 568.9 Now , consider auxiliary equation 53.3s2 +K = 0 Put

K =568.9 in above equation.

53.3s2 +568.9= 0 s2 = - 568.9 53.3 s = + j 3.26


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Step 8 : Angle of Departure -4+4j P3

Фp2

P2

Фp1

-4 Фp4

P4 -4-4j Angle of departure Фd = 180ο + ∑ Фp Consider point P3 ∑ Фp = - Фp1 - Фp2 -Фp4 = - (135 ο + 90 ο + 90 ο ) = - 335ο Angle of departure Фd = 180ο + ∑ Фp = 180ο- 335ο = - 135ο Similarly at P4 Фd = + 135ο The complete root locus is shown in fig. 4.6.


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G(S)H(S) =________K_______ S(s+4)(s2+8s+32)

Fig.4.6.


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Calculation of the value of k at damping factor ζ = 0.707 δ = cos-1ζ = cos-10.707 = 45ο K = __│Mp1_ Mp2 Mp3 Mp4│____ 1 Where Mp1 = magnitude of pole P1 i.e. the distance of pole P1 from K

+j4

-4

-j4

K = 2.1 * 3 *3.6 *6.2 1 K = 140.62


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7.Sketch the root locus diagram of the following open loop transfer function G(s)H(s) =________K_(s+4)____ s(s2 +8s + 13)


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0,-3+2j,-3-2j

Number of Zeros Z= 1 s

-4

Step 2: Number of separate root loci (branches)= Number of poles (P) = 3 Step 3: The starting point of the root locus branch are from poles 0,-3+2j,-3-2j . the terminating point of the root locus at ZEROs. Here we have one ZERO. So one branch terminates at ZERO. The other poles terminate at infinity.

+jω

- 3+ 2j

+σ P2

P1

-σ -4 0

- 3 - 2j

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 the sum of poles and zeros are even. So P2 is not the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 3- 1 = 2 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-1 = 2 n = 0,1, For n =0 θ0 = (2(0)+1)π = 90ο 3–1 For n =1 θ1 = (2(1)+1)π = 270ο 3–1 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -3-3) – (-4) 3–1 = -2 = -1 2 Step 6: Breakaway point: In this problem, no poles are adjacent to each other. The pole at origin terminates at ZERO s = -5. the complex poles move to infinity along 90ο And 270ο asymptotes. Hence there is no valid breakaway point. Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 1+

K_(s+4)____ s(s2 +6s + 13)

=0

s3 +6s2 + 13s + K(s+4) = 0 s3 +6s2 + (13 + K)s + 4K = 0 a0 = 1 a1 = 6 a2 = 13 + k a3 =4K


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Routh’s array

s3

1

s2

6

s1

s0

78 + 2K 8

13 + K

4K

0

4K

To have intersection on jω axis, any one row should be zero. So make s1 row to be zero. 78 + 2 K = 0 6 K = -39. K value is negative. But to have intersection on jω axis, k value should be positive. So root locus does not intersect on jω axis.


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Step 8 : Angle of Departure for complex poles: Consider Point P2 Angle of departure Фd = 180ο - ∑ Фp + ∑ Фz = 180ο – ( - Фp1 - Фp3) + Фz1 = 180ο – (146 ο + 90 ο) + 63 ο = + 7 ο. Similarly for P3, the angle of departure is - 7 ο

-3 +j2 P2

Фz1

Фp1

-5 Фp3

P3 -3 - j2 The complete root locus is shown in fig.4.7.


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8. Sketch the root locus for the system and comment on stability G(s)H(s) = K(s + 4)(s + 5) , K > 0 (s+3)(s+1)

(AU: Nov/Dec. 2007)

Step :1 Number of poles P =2 At s

-3,-1

Number of ZEROs = 2 At s

-4 , -5

Step 2: Number of separate root loci (branches)= Number of poles (P) =2 Step 3 The starting point of the root locus are from poles -3,-1.terminating point of the root locus is at ZEROs i.e. -4,-5

.

P4

P3

-5

P2

-4

P1

-3

-1


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Consider P1. to the R.H.S. of P1 . the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 . the sum of poles and zeros are even. So P1 is not the part of root locus. Consider P3. to the R.H.S. of P3 . the sum of poles and zeros are odd. So P3 is the part of root locus. Consider P4. to the R.H.S. of P4 . the sum of poles and zeros are even. So P4 is not the part of root locus. Step 4: Number of asymptotes Na = P-Z = 2- 2 = 0 When number of asymptotes = 0, then the root locus plot is circular. So no need to find the intersection of asymptotes on real axis. Because asymptotes are required only when number of poles is not equal to number of zeros. i.e. P >z.

Step 5: There is no need to find centroid. Because P – Z = 2 – 2 = 0. Step 6: Breakaway point 1 + G(s)H(s) = 0 1 + K(s + 4)(s + 5) = 0 (s+3)(s+1) (s+3)(s+1) + K(s + 4)(s + 5) = 0 (s2 +4s+3)+ K(s2 + 9s +20) =0 K = - (s2 + 4s +3) (s2 + 9s +20) To find breakaway point, differentiate k with respect to s and equate to zero. dk = 0 ds


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dk = ( s2 + 9s + 20)(2s + 4) – (s2 + 4s + 3)(2s + 9) = 0 ds (s2 + 9s + 20)2 = 5s2 + 34s +53 = 0 Solving the above quadratic equation, we get Breakaway point at - 2.42 Breakin point - 4.38 Step: 7 Intersection on jω axis. Characteristic equation = 1 + G(s)H(s) = 0 1 + ___K_(s+4)(s + 5)____ ( s + 3)(s + 1)

=0

( s + 3)(s + 1) + K( s + 4)(s + 5) S2(1 + K) + s(9K + 4) + (20K + 3) = 0 a0 = 1 + K a1 = 9K + 4 a2 = 20K + 3 Routh’s array s2

1+K

s1

9K + 4

s0

20K + 3

20K + 3

To have intersection on jω axis, any one row should be zero. Make s1 row to be zero. (9K + 4)s = 0 9Ks + 4s = 0 9K = - 4 K = - (4 / 9) = - 0.44


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As K is negative, there is no intersection on jω axis. Mean of two breakaway points as centre, and mean of two break points as radius draw the root locus. Centre = -2.4 - 4.4 = -3.4 2 radius = -2.4 + 4.4 = 1 2 Entire root locus lies on the left half of the s plane. Hence the system is absolutely stable in nature. The complete root locus is shown in fig.4.8.


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Fig.4.8.


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9. The open loop transfer function G(S) = K(s + b) / s(s+a) and H(s) =1. for the system shown below, prove that a part of root locus is a circle. (AU: April:2004).

R(s)

+ G(s)

C(s)

-

H(s)

G(s)H(s) = K(s + b) s(s + a) For a complex point on the root locus s = σ + j ω G(s)H(s) = ____K(σ + j ω + b)_____ (σ + j ω) (σ + j ω + a) At that point


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G(s) H(s)

=

IV Semester ECE

tan -1 (ω / σ +b)___________ tan -1 (ω / σ) tan -1 (ω / σ +a)

Angle of positive K is 0ο


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G(s) H(s)

=

IV Semester ECE

tan -1 (ω / σ +b) - tan -1 (ω / σ) tan -1 (ω / σ +a)

= tan -1 (ω / σ +b) – tan-1 _ (ω / σ) +(ω / σ +a)_ 1 – (ω / σ) (ω / σ +a) i.e. tan-1A + tan-1B = tan-1

= tan -1 (ω / σ +b) - tan -1

_ A +B____ 1 – AB

ω(2 σ + a)______ σ (σ + a) – ω2

(ω / σ +b) - ω(2 σ + a)______ σ (σ + a) – ω2 = tan-1 1 + (ω / σ +b)

ω(2 σ + a)______ σ (σ + a) – ω2

This angle must be 180ο for a point on the root locus. (ω / σ +b) - ω(2 σ + a)______ σ (σ + a) – ω2 180ο =

tan-1 1 +(ω / σ +b) ω(2 σ + a)______ σ (σ + a) – ω2


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Taking tan on both sides and using tan (180ο) = 0, we get __ω___ - ω (2σ +a)__ σ +b σ (σ +a) - ω2

= 0

ω (σ2 + a σ - ω2) – ω ( 2 σ +a )( σ +b) = 0 ω{ (σ2 + a σ - ω2) - ( 2 σ +a ) (σ +b)} = 0 ω≠0 therefore { (σ2 + a σ - ω2) - ( 2 σ +a ) (σ +b)} = 0 i.e. σ2 + a σ - ω2 - 2σ2 - σ2 - 2σb - ab = 0 (adding and subtracting b2 we get) σ + 2bσ + b + ω - b - ab = 0 2

2

2

2

(σ +b)2 + (ω – 0)2 = (√ b(b – a) )2 This is the equation of a circle with centre as (-b , 0) which is the location of open loop zero and radius (√ b(b – a)). This proves that part of the root locus is a circle.

10. Prove that the breakaway points of the root locus are the solutions of dK/ds = 0. where K is the open loop gain of the system whose open loop transfer function is G(s) (AU: April-2004)

The characteristic equation 1 +G(s)H(s) = 0 is the combination of s terms and K terms. It can be arranged as F(s) = P(s) + KQ(s) = 0 -------------------------- (1) Where P(s) = polynomial containing s terms. KQ(s) = polynomial containing K and s terms. Taking K outside from KQ(s), it can be noted that both P(s) and Q(s) are polynomials in s.

Now at breakaway points, multiple roots occur. Mathematically this is possible if (dF(s) / ds ) = 0 From equation (1),


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dF(s) = dP(s) + KdQ(s) = 0 ds ds ds

therefore K = - {dP(s) / ds} ------------------------------------(2) dQ(s) / ds put equation(2) in equation (1), at breakaway point, we can write P(s)

- {dP(s) / ds} Q(s) = 0 dQ(s) / ds

dQ(s) P(s)+ KdP(s) Q(S) = 0 ----------------(3) ds ds solving equation(3), for value of s, breakaway point can be obtained. Now from equation(1) K = - P(s) Q(s)

Therefore dK(s) = ds

Q(s){dP(s) / ds} - P(s){dQ(s) / ds}------- (4) [Q(s)]2

if dK / ds is equated to zero, we get dK = P(s) dQ(s) - Q(s) dP(s) = 0 --------------(5) ds ds ds compare (3) and equation(5). Equation(5) is same as equation(3) which yields breakaway points. This proves that the roots of ( dK / ds ) = 0 are the actual breakaway points.


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EC2255- Solved Problems in Control System 11) An unity feedback system has G(s)

IV Semester ECE

G(s)) =________K__________ s(s + 1)(s + 3)(s + 4)

draw the complete root locus.

Step 1: Number of poles P = 4 At s 0,-1, -3, -4 Number of Zeros Z= 0 Step 2: Number of separate root loci (branches)= Number of poles (P) = 4

Step 3: The starting point of the root locus branches are from poles 0,-1,-3, -4. Terminating point of the root locus branches are at ZEROs. But here in this problem, we have no ZEROs. So the root locus branches terminates at infinity.

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 the sum of poles and zeros are even. So P2 is not the part of root locus. Consider P3. to the R.H.S. of P3 the sum of poles and zeros are odd. So P3 is the part of root locus. Consider P4. to the R.H.S. of P4 the sum of poles and zeros are even. So P4 is not the part of root locus.


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Step 4 : Number of asymptotes Na = P-Z = 4- 0 = 4 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 4-0 = 2 n = 0,1, 2,3. For n =0 θ0 = (2(0)+1)π 4–0

= 45ο

For n =1 θ1 = (2(1)+1)π 4–0

= 135ο

For n =2 θ2 = (2(2)+1)π 4–0

= 225ο

For n =3 θ3 = (2(3)+1)π 4–0

= 315ο

Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -1-3 – 4 ) - (0) = - 8__ = -2 3–1 4 Step 6: Breakaway point: 1 + G(s)H(s) = 0 1+

K__________ = 0 s(s+1)(s+3)(s + 4)

s(s+1)(s+3)(s + 4)+ K = 0


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K = - { s(s+1)(s+3)(s + 4)} = -{ s4 +7s3 +13s2 +7s +12} dK = 0 ds dK = -{4s3 +21s2 +26s +7} = 0 ds 4s3 +21s2 +26s +7= 0

-0.4

4

21

26

7

0

-1.6

7.76

-7.2

4

19.4

18.24

-0.2

If s = -0.4, then the value of s is nearly zero. So s = -0.4 is valid breakaway point. 4s2 +19.4s +18.24 = 0 On solving the above quadratic equation, we get s = -0.64,-1.78 so the valid breakaway points are -0.4, -0.64


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Step 7: Intersection on jω axis. Characteristic equation = 1 + G(s)H(s) = 0 1 + ___ K___________ = 0 s( s + 1)(s + 3) (s+4 s( s + 1)(s + 3) ( s + 4) + K = 0 s4 +7s3 +13s2 +7s +12 + K = 0 a0 = 1 a1 = 7 a2 = 13 a3= 7a4 = 12 + K Routh’s array

s4

1

13

12 + K

s3

7

7

0

s2

12

12+ K

s1

12*7 – 7(12+K) 12

s0 12 + K To have intersection on jω axis, any one row should be zero. Make s1 row to be zero. 12*7 – 7(12+K) = 0 12 84 – 7(12+K) = 0 K=0 Put K = 0 in s2 row. 12 s2 + 12 + K = 0 s2 = - 1; therefore s = + j1 the root locus intersects on jω axis at + j1. The complete root locus is shown in fig. 4.9.


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Fig.4.9.


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12. Sketch the root locus of the system having G(s)H(s) =______K(s + 2)_____ (s +1)(s + 3+j2)(s+3- j2) (AU:Nov/Dec:2007)


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-1,-3+2j,-3-2j

Number of Zeros Z= 1 s

-2

Step 2: Number of separate root loci (branches)= Number of poles (P) = 3 Step 3: The starting point of the root locus branch are from poles -1,-3+2j,-3-2j . the terminating point of the root locus at ZEROs. Here we have one ZERO. So one branch terminates at ZERO. The other poles terminate at infinity.

- 3+ 2j

+jω

P2

+σ

P1

-σ -2 0

- 3 - 2j

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2. to the R.H.S. of P2 the sum of poles and zeros are even. So P2 is not the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 3- 1 = 2 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z-1) P–Z Here n = Na = P-Z = 3-1 = 2 n = 0,1, For n =0 θ0 = (2(0)+1)π = 90ο 3–1 For n =1 θ1 = (2(1)+1)π = 270ο 3–1 Step 5: Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (-1-3-3) – (-2) 3–1 = -2.5 Step 6: Breakaway point: In this problem, no poles are adjacent to each other. So no breakaway point is there. Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 1 + ___K_(s+2)________ ( s+1)(s2 +6s + 13)

=0

( s+1)(s2 +6s + 13) + K(s+2) = 0 s3 +6s2 + 13s + s2 + 6s + 13 + Ks +2K = 0 s3 +7s2 + (19 + K)s + 13 + 2K = 0 a0 = 1 a1 = 7 a2 = 19 + k a3 = 13 +2K


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Routh’s array

s3

1

19 + K

s2

7

13+2K

s1

s0

7(19+K) - (13+2K)(1) 7

0

13 + 2K

To have intersection on jω axis, any one row should be zero. So make s1 row to be zero. 7(19+K) - (13+2K)(1) 7 K = -24.

= 0

K value is negative. But to have intersection on jω axis, K value should be positive. So root locus does not intersect on jω axis.


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Step 8 : Angle of Departure for complex poles: Consider Point P2 Angle of departure Фd = 180ο - ∑ Фp + ∑ Фz = 180ο – (Фp1 +Фp3) + Фz1 = 180ο – (135 ο +90 ο) + 116.5 ο = + 7 1.56ο. Similarly for P3, the angle of departure is - 7 1.56ο

-3 +j2 P2

Фz1

-2

Фp1

-1

-3 - j2 P3

The complete root locus is shown in fig.4.10.


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Fig.4.10.


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13. For a feedback system with loop transfer function G(s)H(s) =________K(s +1)______ draw the root locus. s(s+4)(s2+6s+10)


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0, -3+j1, -3 –j1 , -4

Number of Zeros Z= 1 At s

-1

Step 2: Number of separate root loci (branches)= Number of poles (P) = 4 Step 3: The starting point of the root locus are from poles 0, -3+j1, -3 –j1, -4. the simple pole at the origin terminates at ZERO (i.e at s= -1). Two complex poles go to infinity along 60ο ,300ο asymptotes. The simplepole at s = - 4 go to infinity along 180ο asymptote.

+jω

-3 + j P3

-4

P2

P1 +σ

-σ

-1 -3 - j

0

-jω

Consider P1. to the R.H.S. of P1 the sum of poles and zeros are odd. So P1 is the part of root locus. Consider P2.To the R.H.S of the P2 the sum of poles and zeros are even. So p2 is not the part of root locus. Consider P3.to the R.H.S. of P3, the sum of poles and zeros are odd. So P3 is the part of root locus.


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Step 4: Number of asymptotes Na = P-Z = 4- 1 = 3 Angle of asymptotes : θn = (2n+1)π ( n = 0,1,2,…….P-Z1) P–Z Here n = Na = P-Z = 4-1 = 3 n = 0,1,2, For n =0 θ0 = (2(0)+1)π = 60ο 4– 1 For n =1 θ1 = (2(1)+1)π = 180ο 4–1 For n =2 θ2 = (2(2)+1)π = 300ο 4–1 Centroid (σ ) = (∑ real parts of poles) - (∑ real parts of zeros) P–Z = (0 -4-3 -3) – (-1) 4–1 = -9 = - 3 3 Step 6: Breakaway point: In this problem, no poles are adjacent to each other. So no breakaway point is not there. Step 7: Intersection of root locus with imaginary axis (jω) Characteristic equation = 1 + G(s)H(s) = 0 1+

K(s +1)__________ s(s+4)(s2+6s+10)

=0

s(s+4) (s2+6s+20) + K = 0 s4 +10s3 + 34s2 + ( 40 + K )s + K = 0 a0 = 1 a1 = 10 a2 = 34 a3 = 40 + K a4 = K

.


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Routh’s array

s4

s3

1

10

34

40 + K

s2

( 300 - K) 10

s1

(300- K)(40-K) – 100K 300 - K

s0

K

K

K

To have intersection on jω axis, any one row should be zero. So make s1 row to be zero. (300 – K)(40 – K) – 100K = 0 300 - K 12000 + 300K – 40K – K2 -100K =0 K2 – 160K -12000 =0 On solving the above quadratic equation, we get K = +215.65 , K = -55.65


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1. The Transfer function of the system is used to determine: A. The output for a given input B. The type of the system C. The input for a given output D. The steady state gain Answer: A 2. The transfer function of a system is defined as: A. The laplace transform of the impulse response B. Laplace transform of the step response C. Laplace transform of the ramp response D. Laplace transform of the sinusoidal input Answer: A 3. The transfer function has the main application in the study of __ behavior of the system A. Steady B. Transient C. Both steady and transient D. None of the above Answer: A 4. Transfer function of the control system depends on A. Initial conditions of input and output B. System parameters alone C. Nature of the input D. Nature of the output Answer: B 5. The ON-OFF controller is a __ system A. Linear B. Non linear C. Discontinuous D. Digital Answer: B 6. The impulse function is a derivative of __ function: A. Parabolic B. Step C. Ramp D. Linear Answer: B


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7. Control Systems are normally designed with damping factor: A. Less than unity B. More than unity C. Zero D. Unity Answer: A 8. Error Constants of a system are measure of: A. Relative stability B. Transient state response C. Steady state response D. Steady state as well as transient state response Answer: C

1. A transfer function of a system is a Laplace transform of its: A. Square wave response B. Step response C. Ramp response D. Impulse response Answer: D 2. Laplace transform method has the advantage of: A. Corporation of initial conditions in the very first step B. Providing total solution more systematically C. Providing solutions in frequency domain only D. All the above E. Both (a) and (b) Answer: E 3. Transfer function of a control system depends on: A. Initial conditions of input and output B. System parameters alone C. Nature of input D. Nature of output Answer: B 5. Bimetallic thermostat is __ Controller A. Zero B. One term C. Two term D. On-off Answer: D Solved by A.Devasena., Associate Professor., Dept/ECE

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6. Two stator windings of the ac servo motors are oriented __ electrical degrees apart a) 90 b) 60 c) 120 d) 180 A 7. Which of the following is not true for an ac servo motor: a) Has low inertia rotor b) Is a single phase motor c) Has slip torque characteristics as straight line with negative slope d) Reference voltage frequency is high compared to signal frequency B 8. A gyroscope operates on the principle of: a) Law of conservation of energy b) Law of conservation of momentum c) Electro-mechanical conversion d) Newton's third law of motion B 9. AC Servo motor is basically a: a) Universal motor b) Capacitor motor c) 2 phase induction motor d) 3 phase induction motor C 10. The output of synchro error detector is a: a) Voltage signal of the receiver b) Voltage signal of constant amplitude c) Suppressed carrier modulated signal d) Angular displacement of the control transformer rotor C

Control Systems Objective Test Questions Answers


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1. __ can be used for converting the angular the angular position of a shaft into electrical signal: a) LVDT b) Synchros c) DC Servo motor d) AC Servo motor A 2. Synchros are widely used for transmission of __ data: a) Angular b) Digital c) Mathematical d) Computed A 3. The input to a controller is always a/an __ a) Error b) Servo c) Amplifier d) Sensor A 4. The thermocouple directly converts temperature into: a) dc voltage b) ac voltage c) direct current d) alternating current A 5. Potentiometers are used in control systems: a) To improve frequency response b) To improve time response c) To improve stability d) As error sensing transducer D 6. Tachometer is employed to measure a) Rotational speed b) Displacement c) Torque developed Solved by A.Devasena., Associate Professor., Dept/ECE

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d) Driven shaft speed A 7. In a control system the comparator measures the differences between: a) Output and input b) Actual and desired performance c) Controller and system d) None of the above B 8. A signal other than the reference input that tends to affect the value of controlled variable is termed as: a) Command b) Error signal c) Disturbance d) Actuating signal C 9. The weighting function of the control system is called the __ response a) Impulse b) Transient c) Steady state d) None of the above A 10. In a control system the comparator compares the output response and reference input and actuates the: a) Transducer b) Signal Conditioner c) Control element d) Primary sensing element C

Control Systems Objective Test Questions Answers 1. A Closed loop system is basically different from open loop control system due to: A. Feedback B. Servomechanism C. Actuating signal Solved by A.Devasena., Associate Professor., Dept/ECE

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D. Error Signal Answer: A 2. Introduction of the feedback reduces the effects of: A. Noise signal B. Disturbances C. Error signals D. Error and noise signals E. Noise signals and disturbances Answer: E 3. A control system in which control action depends on the output is called the __ control system A. Open loop B. Closed loop C. Stable D. Unstable Answer: A 4. In a control system the use of negative feedback: A. Eliminates the chances of instability B. Increases the reliability C. Reduces the effects of disturbance and noise signals in the forward path D. Increases the influence of variations of component parameters on the system performance Answer: C 5. A controller essentially is a: A. Comparator B. Sensor C. Amplfier D. Clipper Answer: B 6. In a control system, the controller output is given to: A. Sensor B. Comparator C. Amplifier D. Final Control element Answer: D 7. The deviation of the primary feedback signal from the reference input is called the: A. Actuating signal B. Error signal C. Command Solved by A.Devasena., Associate Professor., Dept/ECE

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D. Manipulated variable Answer: A

8. The transient response of a system is mainly due to: A. Friction B. Inertia forces C. Stored energy D. Internal forces Answer: C 9. Regenerative feedback: A. Implies feedback with positive sign B. Is sometimes used to increase the loop gain of the feedback system C. Has the transfer function with a negative sign in the denominator D. All the above Answer: D 10. In a hybrid feedback control systems carrier signals are: A. ac only B. dc only C. both ac and dc D. none of the above Answer: C

PID Controller Objective Questions Answers 1. Proportional band of a controller is defined as the range of: a) Measured variable to the set variable b) Air output as the measured variable varies from maximum to minimum c) Measured variables through which the air output varies from maximum to minimum d) None of the above C 2. Proportional band of the controller is expressed as: a) Gain b) Ratio c) Percentage d) Range of control variables C Solved by A.Devasena., Associate Professor., Dept/ECE

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3. The term reset control refers to: a) Proportional b) Integral c) Derivative d) None of the above B 4. The integral control: a) Increases the steady state error b) Decreases the steady state error c) Increases the noise and stability d) Decreases the damping coefficient A 5. In a proportional temperature controller, if the quantity under the heater increases the offset will: a) Increase b) Reduce c) Remain uneffected d) None of the above A 6. When derivative action is included in a proportional controller, the proportional band: a) Increases b) Reduces c) Remains unchanged d) None of the above C 7. The number of operational amplifiers require for designing of electronic PID controller is: a) 1 b) 2 c) 3 d) 6 A 8. Which of the following system provides excellent transient and steady state response: a) Proportional action Solved by A.Devasena., Associate Professor., Dept/ECE

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b) Proportional + Integral action c) Proportional + Differential action d) Proportional + Integral + Differential action D 9. In a PID controller, the offset has increased. The integral time constant has to be ___ so as to reduce offset: a) Reduced b) Increased c) Reduced to zero d) None of the above B 10. In a PID controller, the overshoots has increased. The derivative time constant has to be __ so as to reduce the overshoots: a) Increased b) Reduced c) Reduced to zero d) None of the above A Control System Objective Questions: Part-1 Control System Objective Questions From Competitive Exams [1] A system is described by the following state and output equations [dx1(t)/dt] =-3x1(t)+x2(t)+2u(t) [dx2(t)/dt]=-2x2(t)+u(t) y(t)=x1(t) where u(t) is the input and y(t) is the output The system transfer function is [GATE 2009] A. (s+2)/(s2+5s-6) B. (s+3)/(s2+5s+6) C. (2s+5)/(s2+5s+6) D. (2s-5)/(s2+5s-6)

[2] A two-port network is defined by the relation: [IES 2010] I=5V1+3V2 I2=2V1-7V2The value of Z12 is A. 3


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B. -3 C. 3/41 D. 2/31 Ans: C [3] The Z-transform of x(K) is given by [IES 2010] x(Z)= {(1-e-T)z-1} / {(1-z-1)(1-e-Tz-1)} The initial value of x(0) is A. Zero B. 1 C. 2 D. 3 Ans: [4] Consider the following statements with reference to the phase plane: 1. They are general and applicable to a system of any order. 2. Steady state accuracy and existence of limit cycle can be predicted. 3. Amplitude and frequency of limit cycle if exists can be evaluated. 4. Can be applied to discontinuous time system. Which of the above statements are correct? [IES2010] A. 1,2,3 and 4 B. 2 and 3 only C. 3 and 4 only D. 2,3 and 4 only

A

[5] For the circuit shown below, the natural frequencies at port 2 are given by s+2=0 and s+5=0,without knowing which refers to open-circuit and which to short-circuit. Then the impedences Z11 and Z22 are given respectively by [IES2010]

A. B. C. D.

K1{(s+5)/(s+2)}, K2{(s+2)/(s+5)} K1{(s+2)/(s+5)}, K2{(s+5)/(s+2)} K1{(s)/(s+2)}, K2{(s+2)/(s+5)} K1{(s+2)/(s+5)}, K2{(s+2)/(s+5)}

Ans: C [6] Consider the following statements in connection with two-position controller: 1.If the controller has a 4% neutral zone,its positive error band will be 2% and negative error band will be 8%. 2.The neutral zone is also known as dead band. 3.The controller action of a two-position controller is very similar to that of a pure on-off Solved by A.Devasena., Associate Professor., Dept/ECE

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controller. 4.Air-conditioning system works essentially on a two-position control basis. Which of the above statements are correct? [IES2010] A. 1,2 and 3 only. B. 2,3 and 4 only. C. 2 and 4 only. D. 1,2,3 and 4

[7] The polar plot of an open loop stable system is shown below. The closed loop system is [GATE 2009]

A.Always stable B.Marginally stable C.Unstable with one pole on the RH s-plane D.Unstable with two poles on the RH s-plane

[8] The open loop transfer function of a unity feedback system is given by G(s) =(e 0.1s )/s.The gain margin of this system is [GATE 2009] A. 11.95 dB


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EC2255- Control System Notes( solved problems)

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