33525999 Control Engg Notes

7/30/2019 33525999 Control Engg Notes

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Notes by B.K.Sridhar, NIE, Mysore

CHAPTER II

MATHEMATICAL MODELING

2.1 Introduction: In chapter I we have learnt the basic concepts of control systems such

as open loop and feed back control systems, different types of Control systems like

regulator systems, follow-up systems and servo mechanisms. We have also discussed a

few simple applications.

The requirements of an ideal control system are many and depend on the system under

consideration. Major requirements are 1) Stability 2) Accuracy and 3) Speed of

Response. An ideal control system would be stable, would provide absolute accuracy

(maintain zero error despite disturbances) and would respond instantaneously to a change

in the reference variable. Such a system cannot, of course, be produced. However, study

of automatic control system theory would provide the insight necessary to make the most

effective compromises so that the engineer can design the best possible system.

2.2 Modeling of Control Systems: The first step in the design and the analysis of control

system is to build physical and mathematical models. A control system being a collection

of several physical systems (sub systems) which may be of mechanical, electrical

electronic, thermal, hydraulic or pneumatic type. No physical system can be represented

in its full intricacies. Idealizing assumptions are always made for the purpose of analysis

and synthesis. An idealized representation of physical system is called a Physical Model.

1


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Control systems being dynamic systems in nature require a quantitative mathematical

description of the system for analysis. This process of obtaining the desired mathematical

description of the system is called Mathematical Modeling. The basic models of dynamic

physical systems are differential equations obtained by the application of appropriate

laws of nature. Having obtained the differential equations and where possible the

numerical values of parameters, one can proceed with the analysis.

Usually control systems are complex. As a first approximation a simplified model is built

to get a general feeling for the solution. However, improved model which can give better

accuracy can then be obtained for a complete analysis. Compromise has to be made

between simplicity of the model and accuracy. It is difficult to consider all the details for

mathematical analysis. Only most important features are considered to predict behaviour

of the system under specified conditions.

2.3 Modeling of Mechanical Systems: Mechanical systems can be idealized as spring-

mass-damper systems and the governing differential equations can be obtained on the

basis of Newtons second law of motion, which states that

F = ma: for rectilinear motion

T = I : for rotary motion

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Sign Convention: Forces and torques in the direction of motion are positive

Translational Systems:

(i) Spring mass system:

kx

(ii)

Friction less

(iii) Spring mass damper system:

3

m

F = ma

m x = - kx

mx + kx = 0

....

x

m

k

m

x

kx

mx + kx = 0..

m

xm

k

c

x

k x (Spring Force)

c x (damping Force).


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mx = - c x kx

(iv) Spring Mass Damper System

In the system shown a provision is made for an input displacement x at the top of the

spring corresponding to which the mass is responding with a displacement of y.

Let y > x

4

.. .

mx + c x + kx = 0.. .

m

X (Input)

Y (Response)

m

K(y)-x)

c y

.m

kx

c y

.

ky

From NSL

my = - k (y-x) - c y.. .

my + c y + ky = kx.. .

c

k


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CHAPTER VI

FREQUENCY RESPONSE

6.1 Introduction

We have discussed about the system response to step and ramp input in time domain

earlier in Chapter 3. When input signals are frequency dependent, frequency response

assume greater importance. In such systems the time domain analysis is difficult from the

design point of view.

Frequency response of a control system refers to the steady state response of a system

subject to sinusoidal input of fixed (constant) amplitude but frequency varying over a

specific range, usually from 0 to . For linear systems the frequency of input and output

signal remains the same, while the ratio of magnitude of output signal to the input signal

and phase between two signals may change. Frequency response analysis is a

complimentary method to time domain analysis (step and ramp input analysis). It deals

with only steady state and measurements are taken when transients have disappeared.

Hence frequency response tests are not generally carried out for systems with large time

constants.

The frequency response information can be obtained either by analytical methods or by

experimental methods, if the system exits. The concept and procedure is illustrated in

Figure 6.1 (a) in which a linear system is subjected to a sinusoidal input. I(t) = a Sin t

and the corresponding output is O(t) = b Sin ( t + ) as shown in Figure 6.1 (b).

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Figure 6.1 (a) Figure 6.1 (b)

The following quantities are very important in frequency response analysis.

M ( ) = b/a = ratio of amplitudes = Magnitude ratio or Magnification factor or gain.

( ) = = phase shift or phase angle

These factors when plotted in polar co-ordinates give polar plot, or when plotted in

rectangular co-ordinates give rectangular plot which depict the frequency response

characteristics of a system over entire frequency range in a single plot.

6.2 Frequency Response Data

The following procedure can be adopted in obtaining data analytically for frequency

response analysis.

1. Obtain the transfer function of the system

)()()(SISOSF = , Where F (S) is transfer function, O(S) and I(S) are the Laplace

transforms of the output and input respectively.

2. Replace S by (j ) (As S is a complex number)

)(

)()(

jI

jOjF =

)()()(

)(

jBA

jI

jO+==

(another complex number)

3. For various values of , ranging from 0 to determine M ( ) and .

jBAjBAjI

jOM +=+== )()(

)(

)()(

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22 BAM +=

jBAjI

jO+==

)(

)(

AB1tan =

4. Plot the results from step 3 in polar co-ordinates or rectangular co-ordinates.

These plots are not only convenient means for presenting frequency response data but

are also serve as a basis for analytical and design methods.

6.3 Comparison between Time Domain and Frequency Domain Analysis

An interesting and revealing comparison of frequency and time domain approaches is

based on the relative stability studies of feedback systems. The Rouths criterion is a time

domain approach which establishes with relative ease the stability of a system, but its

adoption to determine the relative stability is involved and requires repeated application

of the criterion. The Root Locus method is a very powerful time domain approach as it

reveals not only stability but also the actual time response of the system. On the other

hand, the Nyquist criterion (discussed later in this Chapter) is a powerful frequency

domain method of extracting the information regarding stability as well as relative

stability of a system without the need to evaluate roots of the characteristic equation.

6.4 Graphical Methods to Represent Frequency Response Data

Two graphical techniques are used to represent the frequency response data. They are: 1)

Polar plots 2) Rectangular plots.

6.5Polar Plot

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The frequency response data namely magnitude ratio M( ) and phase angle ( ) when

represented in polar co-ordinates polar plots are obtained. The plot is plotted in complex

plane shown in Figure 6.2. It is also called Nyquist plot. As is varied the magnitude

and phase angle change and if the magnitude ratio M is plotted for varying phase angles,

the locus obtained gives the polar plot. It is easier to construct a polar plot and ready

information of magnitude ratio and phase angle can be obtained.

Figure 6.2: Complex Plane Representation

A typical polar plot is shown in Figure 6.3 in which the magnitude ratio M and phase

angle at a given value of can be readily obtained.

8

M

Real

Img

0, + 360, -360

+90, -270

+180, -180

+270, -90

Positive angles

Negative angles


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Figure 6.3: A Typical Polar Plot

6.6 Rectangular Plot

The frequency response data namely magnitude ratio M( ) and phase angle ( ) can

also be presented in rectangular co-ordinates and then the plots are referred as Bode plots

which will be discussed in Chapter 7.

6.7 Illustrations on Polar Plots: Following examples illustrate the procedure followed in

obtaining the polar plots.

Illustration 1: A first order mechanical system is subjected to a input x(t). Obtain the

polar plot, if the time constant of the system is 0.1 sec.

9

x (t) (input)

y (t) (output)

C

K


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Transfer function F(S) = 1

1

)(

)(

+= SSX

SY

Given = 0.1 sec

SSSX

SY

1.01

1

11.0

1

)(

)(

+=

+=

To obtain the polar plot (i.e., frequency response data) replace S by j .

)1.0(1

1

11.0

1

)(

)(

jjjX

jY

+=

+=

Magnification Factor M =)1.0(1

)0(1

)1.0(1

1

)(

)(

j

j

jjX

jY

+

+=

+

=

2)1.0(1

1

+=M

Phase angle = )1.01()0(1)()(

)(

++=== jXjjY

jX

jY

)1.0(tan1

0tan 11

=

)1.0(tan 1 =

10

Governing Differential Equation:

KxKydt

dyC =+. by K

xydtdyKC =+.

xydt

dy=+. Take Laplace transform

SY(S) + Y(S) = X(S)

(S+1) Y(S) = X(S)


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Now obtain the values of M and for different values of ranging from 0 to asgiven in Table 6.1.

Table 6.1 Frequency Response Data

2)1.0(1

1+

=M )1.0(tan1

=

0 1.00 0

2 0.98 -11.13

4 0.928 -21.8

5 0.89 -26.6

6 0.86 -30.9

10 0.707 -45

20 0.45 -63.4

40 0.24 -76

50 0.196 -78.69

100 0.099 -84.29

0 -90

The data from Table 6.1 when plotted on the complex plane with as a parameter polar

plot is obtained as given below.

Illustration 2: A second order system has a natural frequency of 10 rad/sec and a

damping ratio of 0.5. Sketch the polar plot for the system.

11

= 6

= 50

= 20


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The transfer function of the system is given by

22

2

2)(

)(

nn

n

SSSX

SY

++

= Given n = 10 rad/sec and = 0.5

10010

100

)(

)(2 ++

=SSSX

SY

Replace S by j

10010

)0(100

10010)(

100

)(

)(22

++

+=

++==

j

j

jjjX

jYas 1=j

222

2

2)10()100(

100

)10()100(

)0(100

)(

)(

+=

+

+==

j

j

jX

jYM

Magnification Factor = 222 )10()100(

100

+=M

Phase angle = )10()100()0(100)10()100(

)0(100 22

jjj

j ++=+

+=

=

2

11

100

10tan

100

0tan

Now obtain the values of M and for various value of ranging from 0 to as givenin the following Table 6.2.

12

m

x (Input)

y (Response)

C

K


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Table 6.2 Frequency Response Data: Illustration 2

M( ) 0 1.00 0.0

2 1.02 -11.85 1.11 -33.7

8 1.14 -65.8

10 1.00 -90.0

12 0.78 -110.1

15 0.51 -129.8

20 0.28 -146.3

40 0.06 -165.1

70 0.02 -171.7

0.00 -180.0

The data from Table 6.2 when plotted on the complex plane with as a parameter polar

plot is obtained as given below.

Note: The polar plot intersects the imaginary axis at a frequency equal to the natural

frequency of the system = n = 10 rad/sec.

Illustration 3: Obtain the polar plot for the transfer function

)1(

10)(

+=S

SF Replace S by j

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1

10)(

+=

j

jF

1

10)()(

2 +==

jFM

( ) = F (j ) = 10 - (j +1)

1tan 1

=

Table 6.3 Frequency Response Data: Illustration 3

M( ) 0 1.00 0

0.2 9.8 -11

0.4 9.3 -21

0.6 8.6 -31

0.8 7.8 -39

2.0 4.5 -63

3.0 3.2 -72

4.0 2.5 -76

5.0 1.9 -79

10 0.99 -84

0.00 -90

Polar plot for Illustration 3

14

=

M( ) = 0


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6.8 Guidelines to Sketch Polar Plots

Polar plots for some typical transfer function can be sketched on the following guidelines.

I )(

)(

))((SFjBA

jfe

jdcjba=+=

+

++--- (Transfer function)

Magnitude Ratio =22

2222*

fe

dcba

jfe

jdcjbaM

+

++=

+

++=

Phase angle:

( )( )( )

)()()( jfejdcjbajfe

jdcjba++++=

+

++=

e

f

c

d

a

b 111tantantan

+=

II Values of tan functions

IQ: tan-1

(b/a) Positive:

IIQ: tan-1 (b/-a) Negative: 180 -

IIIQ: tan-1 (-b/-a) Positive: 180 +

IVQ: tan-1 (-b/a) Negative: 360 -

tan-1 (0) = 00 tan-1 (-0) =1800

tan-1 (1) = 450 tan-1 (-1) = 1350

tan-1 () = 900 tan-1 (-) = 2700

15

Real

ImgIQ

IIQ

IIIQ IVQ

(a+jb)

(-a-jb) (a-jb)

(-a+jb)


16/58

III Let K = Constant

= K + j(0)

Therefore KKK =+= 22 0 Applicable for both K>0 and K0

= tan-1 (-0) = 1800, if K


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= tan-1 ( /0) + tan-1 ( /0)+ . . . . . . . . . . . . n times

= 900 + 900 + . . . . . . . . . . . . n times

Sn = n * 900

Illustration 4: Sketch the polar plot for the system represented by the following open loop

transfer function.

)2)(10(

10)()(

++=

SSSSHSG , obtain M and for different values of

i) As 0,0 S jS=

==== 0

5.05.0

2*10*

10)()( 0

SSSHSG S

S= 5.0

0900=

090=

ii) As S,

010

))((

10)()(

3=== SSSS

SHSGS

310 S=

027090*30 ==

17


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Illustration 5: Sketch the polar plots for the system represented by the following open

loop transfer function.

M()

( )

0 -900

0 -2700

18

Real

Img

0

-90

-180

-270

= , M = 0

= 0, M =


19/58

)5()()(

2 +=SS

KSHSG

i) As 0,0 S jS=

, S is far lesser than 5 and can be neglected

22

5/

5 S

K

S

K==

===n

K

S

KM

2

5/)(

2

5/)(

S

K

= 2)5/( SK = = 0 2*90 = 0 180 = - 1800

ii) As S,

32 )5()()( S

K

SS

K

SHSG S =+=

0)(33===

K

S

KM

3)( SK =

= 0 - 3*90 = - 2700

19

)5()()(

200

+

=

SS

KSHSG

S


20/58

M()

( )

0 -1800

0 -2700

20

Real

Img

0

-90

-180

-270

= , M = 0 = 0, M =


21/58



)8)(5(

10)()( 2 ++= SSSSHSG

i) As 0,0 S jS=

2200

4/1

8.5.

1 0)()(

SSSHSG

S==

====0

4/14/14/1)(

22 S

M

24/1)( S=

= 0 - 2*90

= - 2*900 = 1800

ii) As S,

42

1

..

1 0)()(

SSSSSHSG

S==

44

1010)(

==S

M

as , M ( ) = 0

21


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410)( S=

= 0 4*90 = - 3600



)5)(4(

)2(10)()(

3

++

+=

SSS

SSHSG

i) As 0,0 S jS=

M()

( )

0 -1800

0 -3600

22

Real

Img

0, -360

-90

-180

-270

= , M = 0 = 0, M =


23/58

)5) (4(

)2(1 0)()(

300

++

+==

SSS

SSHSG

S

33

1

5.4.

2.10

SS==

===33

11)(

S

M

31)( S= = 0 3*90 = - 2700

ii) As S,

43

1

)..

.1 0)()(

SSSS

SSHSG

S===

01010

)(44===

S

M

410)( S= = 0 4*90 = -3600

23

Img-270


24/58

M()

( )

0 -2700

0 -3600

24

Real

0, -360

-90

-180

= , M = 0

= 0, M =


25/58



)4)(2(

10)()(

+=

SSSSHSG

i) As 0,0 S jS=

SSSHSG

S8/1 0

4) .2(1 0)()(

00

=

==

==

=

8/108/10)(S

M

S= 8/10)( = 1800 - 900 = 900

ii) As S,

3

1)()(

SSHSG

S==

01010

)(33===

S

M

310)( S= = 0 3*900 = - 2700

25

-270+90


26/58

Illustration 9: Sketch the polar plot for the system represented by the following transfer

function.

10010

100

)(

)(2

+=

SSSX

SY

M()

( )

0 900

0 -2700

26

Real

Img

0, -360

+270-90

+180-180

= , M = 0

= 0, M =


27/58

i) As 0,0 S jS=

11)( ==M , for both K positive and negative

01801)( ==

ii) As S,

2

100

S= 0

100)(

2== S

SM ,

2

100)(S

= = 0 2* =- 900*2

= -1800

6.9 Experimental determination of Frequency Response

Many a times the transfer function of a physical system may not be available in suchcircumstances it is necessary to obtain frequency response information experimentally.

Such data may then be used to establish the transfer function. This method requires the

actual system.

6.10 System Analysis using Polar Plots: Nyquist Criterion: Continued in Session 21

on 17.10.2006

CONTINUED IN SESSION 21: 17.10.2006

******************END****************

M()

( )0 1 1800

0 -1800

27

Real

Img

0, -360

+270

-90

+180

-180

-270

+90

= 0, M = 1 = , M = 0


28/58

CHAPTER VI

FREQUENCY RESPONSE

(Continued from Session 20)

SYSTEM ANALYSIS USING POLAR PLOTS: NYQUIST

CRITERION

6.10 System Analysis using Polar Plots: Nyquist Criterion

Polar plots can be used to predict feed back control system stability by the application of

Nyquist Criterion, and therefore are also referred as Nyquist Plots. It is a labor saving

technique in the analysis of dynamic behaviour of control systems in which the need for

finding roots of characteristic equation of the system is eliminated.

Consider a typical closed loop control system which may be represented by the simplified

block diagram as shown in Figure 6.4

Figure 6.4 Simplified System Block Diagram

The closed-loop transfer function or the relationship between the output and input of the

system is given by

28

H(S)

G(S)C(S)

R(S) +

-


29/58

)()(1

)(

)(

)(

SHSG

SG

SR

SC

+=

The open-loop transfer function is G(S) H(S) (the transfer function with the feedback

loop broken at the summing point).

1+ G(S) H(S) is called Characteristic Function which when equated to zero gives the

Characteristic Equation of the system.

1 + G(S) H(S) = 0 Characteristic Equation

The characteristic function F(S) = 1 + G(S) H(S) can be expressed as the ratio of two

factored polynomials.

Let).().........)()((

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

321

21

n

k

n

ZSPSPSPSS

ZSZSZSKSHSGSF

++++

+++=+=

The Characteristic equation in general can be represented as

F(S) = K (S+Z1) (S+Z2) (S+Z3) . (S+Zn) = 0

Then:

Z1, -Z2, -Z3 . Zn are the roots of the characteristic equation

at S= -Z1, S= -Z2, S= -Z3, 1+ G(S) H(S) becomes zero.

These values of S are termed as Zeros of F(S)

Similarly:

at S= -P1, S= -P2, S= -P3 . Etc. 1+ G (S) H (S) becomes infinity.

These values are called Poles of F (S).

6.10.1 Condition for Stability

For stable operation of control system all the roots of characteristic equation must

be negative real numbers or complex numbers with negative real parts. Therefore,

29


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for a system to be stable all the Zeros of characteristic equation (function) should be

either negative real numbers or complex numbers with negative real parts. These roots

can be plotted on a complex-plane or S-plane in which the imaginary axis divides the

complex plane in to two parts: right half plane and left half plane. Negative real numbers

or complex numbers with negative real parts lie on the left of S-plane as shown Figure

6.5.

Figure 6.5 Two halves of Complex Plane

Therefore the roots which are positive real numbers or complex numbers with positive

real parts lie on the right-half of S-plane.

In view of this, the condition for stability can be stated as For a system to be stable all

the zeros of characteristic equation should lie on the left half of S-plane.

Therefore, the procedure for investigating system stability is to search for Zeros on the

right half of S-plane, which would lead the system to instability, if present. However, it is

impracticable to investigate every point on S-plane as to which half of S-plane it belongs

to and so it is necessary to have a short-cut method. Such a procedure for searching the

30

2+j1

2-j1

-3-j2

-3+j2

Real

Img Right half of

S Plane

Left half of S

Plane


31/58

right half of S-plane for the presence of Zeros and interpretation of this procedure on the

Polar plot is given by the Nyquist Criterion.

6.10.2 Nyquist Criterion: Cauchys Principle of Argument:

In order to investigate stability on the Polar plot, it is first necessary to correlate the

region of instability on the S-plane with identification of instability on the polar plot, or

1+GH plane. The 1+GH plane is frequently the name given to the plane where 1+G(S)

H(S) is plotted in complex coordinates with S replaced by j . Likewise, the plot of G(S)

H(S) with S replaced by j is often termed as GH plane. This terminology is adopted in

the remainder of this discussion.

The Nyquist Criterion is based on the Cauchys principle of argument of complex

variable theory. Consider [F(S) = 1+G(S) H(S)] be a single valued rational function

which is analytic everywhere in a specified region except at a finite number of points in

S-plane. (A function F(S) is said to be analytic if the function and all its derivatives

exist). The points where the function and its derivatives does not exist are called singular

points. The poles of a point are singular points.

Let CS be a closed path chosen in S-plane as shown Figure 6.6 (a) such that the function

F(S) is analytic at all points on it. For each point on CS represented on S-plane there is a

corresponding mapping point in F(S) plane. Thus when mapping is made on F(S) plane,

the curve CG mapped by the function F(S) plane is also a closed path as shown in Figure

6.6 (b). The direction of traverse of CG in F(S) plane may be clockwise or counter

clockwise, depending upon the particular function F(S).

31


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Then the Cauchy principle of argument states that: The mapping made on F(S) plane

will encircle its origin as many number of times as the difference between the

number of Zeros and Poles of F(S) enclosed by the S-plane locus CS in the S-plane.


32

S-plane+j

-j

S5

S1

S2

S3S4

CS-

+j

-j

-

GS1

GS2

GS3

GS4

GS5

(0+j )CG

F(S) = 1+G(S) H(S) plane


33/58

Figure 6.6 Mapping on S-plane and F(S) plane

Thus N = Z P

N0+j0 = Z P

Where N0+j0: Number of encirclements made by F(S) plane plot (CG) about its

origin. Z and P: Number of Zeros and Poles of F(S) respectively enclosed by the locus

CS in the S-plane.

Illustration: Consider a function F(S)

)25)(25)(5)(3(

)22)(22)(1()(

jSjSSSS

jSjSSKSF +++++ ++++=

Zeros: -1, (-2-j2), (-2+j2) indicated by O (dots) in the S-plane

Poles: 0, -3, -5, (-5 j2), (-5 +j2) indicated by X (Cross) in S-plane: As shown in Figure

6.6 (c)

33


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Figure 6.6 (c) Figure 6.6 (d)

Now consider path CS1 (CCW) on S-plane for which:

Z: 2, P =1

34

S-plane+j

-j

CS2

-

CS1

O: ZEROS

X: POLES

+j

-j

CG2

-

CG1

CG2

CG2

(0+j0)

F(S) = 1+G(S) H(S) plane

CS1

CS2


35/58

Consider another path CS2 (CCW) in the same S plane for which:

Z = 1, P = 4

CG1 and CG2 are the corresponding paths on F(S) plane [Figure 6.6 (d)].

Considering CG1 [plot corresponding to CS1 on F (S) plane]

N0+j0 = Z P = 2-1 = +1

CG1 will encircle the origin once in the same direction of CS1 (CCW)

Similarly for the path CG2

N0+j0 = Z P = 1 4 = - 3

CG2 will encircle the origin 3 times in the opposite direction of CS2 (CCW)Note: The mapping on F(S) plane will encircle its origin as many number of times as the

difference between the number of Zeros and Poles of F(S) enclosed by the S-plane locus.

From the above it can be observed that

In the expression

N= Z - P,

N can be positive when: Z>P

N = 0 when: Z = P

N can be negative when: Z


36/58

N is negative the map CG encircles the origin N times in the opposite direction as

that of CS

6.10.3 Nyquist Path and Nyquist Plot

The above Cauchys principle of argument can be used to investigate the stability of

control systems. We have seen that if the Zeros of characteristic function lie on the right

half of S-plane it will lead to system instability. Now, to encircle the entire right half of

S-plane, select a closed path as shown in Figure 6.6 (e) such that all the Zeros lying on

the right-half of S-plane will lie inside this path. This path in S-plane is known as

Nyquist path. Nyquist path is generally taken in CCW direction. This path consists of

the imaginary axis of the S-plane (S = 0+j , - < < ) and a closing semicircle of

infinite radius. If the system being tested has poles of F(S) on the imaginary axis, it is

customary to modify the contour as shown Figure 6.6 (f) excluding these poles from the

path.

36


37/58

Figure 6.6 (e): Nyquist Path Figure 6.6 (f)

37

S-plane

+j

-j

-

-j

+j

0+j00-j0

r =

+j

-j

-

S=+j0

S= -j0

r =

S= -j

r 0


38/58

Corresponding to the Nyquist path a plot can be mapped on F(S) = 1+G(S) H(S) plane as

shown in Figure 6.6 (g) and the number of encirclements made by this F(S) plot about its

origin can be counted.

Figure 6.6 (g)

Now from the principle of argument

N0+j0 = Z-P

N0+j0 = number of encirclements made by F(S) plane plot

Z, P: Zeros and Poles lying on right half of S-plane

For the system to be stable: Z = 0

N0+j0 = - P Condition for StabilityApart from this, the Nyquist path can also be mapped on G(S) H(S) plane (Open-loop

transfer function plane) as shown in Figure 6.6 (h).

Now consider

F(S) = 1+ G(S) H(S) for which the origin is (0+j0) as shown in Figure 6.6 (g).

Therefore G(S) H(S) = F(S) 1

= (0+j0) 1 = (-1+j0) Coordinates for origin on G(S) H(S) plane as shown

in Figure 6.6(h)

38

ImgG(S) H(S) plane

Real

Img

Real

Img

0+j0

1+ G(S) H(S) plane


39/58

Figure 6.6 (h)

Thus a path on 1+ G(S) H(S) plane can be easily converted to a path on G(S) H(S) plane

or open loop transfer function plane. This path will be identical to that of 1+G(S) H(S)

path except that the origin is now shifted to the left by one as shown in Figure 6.6 (h).

This concept can be made use of by making the plot in G(S) H(S) plane instead of 1+

G(S) H(S) plane. The plot made on G(S) H(S) plane is termed as the Niquist Plot and its

net encirclements about (-1+j0) (known as critical point) will be the same as the number

of net encirclements made by F(S) plot in the F(S) = 1+G(S) H(S) plane about the origin.

Now, the principle of argument now can be re-written as

N-1+j0 = Z-P

Where N-1+j0 = Number ofnet encirclements made by the G(S) H(S) plot (Nyquist Plot)

in the G(S) H(S) plane about -1+j0

For a system to be stable Z = 0

N-1+j0 = -P

Thus the Nyquist Criterion for a stable system can be stated as The number of net

encirclements made by the Nyquist plot in the G(S) H(S) plane about the critical

point (-1+j0) is equal to the number of poles of F(S) lying in right half of S-plane.

39

0+j0

(-1+j0)

Origin of the plotfor G(S) H(S)


40/58

[Encirclements if any will be in the opposite direction. Poles of F(S) are the same as

the poles of G(S) H(S)].

Thus the stability of closed-loop control system is determined from its open-loop transfer

function.

CONTINUED IN SESSION 22: 18.10.2006

******************END****************

CHAPTER VI

FREQUENCY RESPONSE(Continued from Session 21)

6.10.4 System Analysis using Nyquist Criterion: Illustrations

Illustration 1: Sketch the Nyquist plot for the system represented by the open loop

transfer function and comment on its stability.

0,0)(

)()( >>+

aKaSS

KSHSG Poles: S = 0 (on imaginary axis) and S = -a

Step 1: Define Nyquist path. Let the Nyquist path be defined as given below.

40

+j

-j

S=+j0

S= -j0

r =

S= -j

r 0-

S= -j

Nyquist Path

S= +j


41/58

Section I: S = +j to S = +j0; Section II: S = +j0 to S = -j0

Section III: S = -j0 to S = -j; Section IV: S = -j to S = +j

2. Corresponding to different sections namely I, II, III, and IV Obtain polar plots on G(S)

H(S) plane, which are nothing but Nyquist Plots.

Nyquist Plot for Section I: In S-plane section I runs from S= + j to S = +j0To obtain polar plot in G(S) H(S) plane:

0,0)(

)()( >>+

aKaSS

KSHSG

(i) 2)()( S

KSHSG jS = ,

0)()(2==

S

KSHSG

2

2)()( SK

S

KSHSG ==

= 0 2*900 = - 1800

(ii) ====S

K

S

aK

aS

KSHSG jS

1

0.

)()( ,

G(S) H(S) = K/a S = 0 - 900 = - 900

41

Img G(S) H(S) Plane


42/58

Nyquist Plot for Section II: In S-plane section II runs from S= + j0 to S = -j0

In this region 0S

In S-plane section II is a semicircle from S = + j0 to S = -j0 of radius r 00, covering an

angle of 1800 in clockwise direction

S

K

aS

KSHSG

1

)()()( == where K1 = K/a

But S = re+j equation of a circle in exponential form

wherere

KSHSG j

j,Re)()(

1

+==

r

KR

1

=

G(S) H(S) = R.e-j

=r

KR

1

(as r is very small)

S M()

( )

S 0 -1800

0S -900

42

S=j0

Real

-900

S=j

-1800

Section I


43/58

This shows that G(S) H(S) plot of section II of Nyquist path is a circle of radius = starting from S = +j0 and ending at a point S = -j0 covering an angle of 180 0 in oppositedirection of section II of Nyquist path (CCW direction i.e., negative sign)

In general if G (S) H (S) =nS

K1

jn

jnneR

er

KSHSG

+== .)()(

1

Where R = nr

K1

The G (S) H (S) plot will be a portion circle (part) of radius R , starting at a point S= +j and ending at a point S = -j covering an angle of (n*180 0) in the opposite direction

(CCW) (since sign is negative)

Nyquist Plot for Section III: In S-plane section III runs from S= -j0 to S = j

)()()( 0

aSS

KSHSG jS

+=

SKaS

K/

.

1== where Kl = (K/a)

== SKSHSG /)()(1

G(S) H(S) = Kl S, Kl is negative

= -Kl S

G(S) H(S) = 1800 900 = 900

=+

= jSjSaSS

KSHSG

)()()( = K/S2

43

S=j0

S= -j0

Real

Img

R=

G(S) H(S) Plane

Section II


44/58

0)()(2==

S

KSHSG

G(S) H(S) = K - S2 = 0 2 (-900) = 1800 ; (S is negative)

This Section III is the mirror image of section of the Section I.

Nyquist Plot for Section IV: In S-plane section IV runs from S= -j to S = j

In this region S

In the S-plane it is a semicircle of radius R from S = - j to S = + j covering anangle of 1800 in the counter clockwise direction.

In the G(S) H(S) plane

2)()(

S

KSHSG

S =

But S = R e+j

Equation in circle in exponential form

2

22.

.)()( j

jer

eR

KSHSG == where 02 =

R

Kr

Thus G(S) H(S) plot for section IV is also a circle of radius 0r starting at S = - j andending at S = + j covering an angle of 2* 1800 (2 ) in the opposite direction (CW).

44

S= -j

S= -j0

+900

+1800

Section III

G(S) H(S) Plane

Real

Img

ImgG(S) H(S) Plane


45/58

Now assemble the Nyquist plots of all the sections as given below to get the overall

Nyquist plot

From Nyquist Criterion:

No. of encirclements made by Nyquist plot about ( 1 +j0) = N-1+j0 = Z P

P = No. of poles lying in the right half of S plane

45

S= +j

S= -j

RealSection IV

r 0

Real

ImgG(S) H(S) Plane

(-1+j0)

j0

j

-j0

-j


46/58

For the function)(

)()(aSS

KSHSG

+= , the Poles are: S = 0, S = - a = 0, which lie on

the left half of S-plane

Therefore P = 0: Number of poles on the right half of S-plane

N-1+j=0= 0 as counted from the Nyquist plot

N-1+j0 = Z P

0 = Z 0

Therefore Z = 0

Number of zeros lying on the right half of S-plane is 0 and hence the system is stable

Illustration 2: Obtain the Nyquist diagram for the system represented by the block

diagram given below and comment on its stability

)1()(

2 +=SS

KSG

SSH =)(

,)1(

*)1(

)()(2 +

=+

=SS

KS

SS

KSHSG

Poles are S = 0, -1/ P =

0

Solution is same as that of Illustration 1

CONTINUED IN THE NEXT PART OF THE NOTES

******************END****************

46

R (S) +C (S)

)1(2 +SS

K

S

-


47/58

CHAPTER II

MATHEMATICAL MODELING (Continued)(Continued from session III: 30.08.2006)

2.3.2 Models for Translational Systems:

Illustration 1: Figure 2.7 shows a spring mass system that represents the simplest

possible mechanical system. It is a single DOF system since one coordinate (x) is

sufficient to specify the position of mass at any time.

Figure 2.7: Spring Mass System Figure 2.7 a: Free Body Diagram

Using Newtons second law of motion, we will consider the derivation of the equation of

motion in this section. The procedure we will use can be summarized as follows:

1) Select a suitable coordinate (x) to describe the position of the mass or rigid body

in the system. Use a linear coordinate to describe the linear motion of a point mass or

the centroid of a rigid body, and an angular coordinate ( ) to describe the angular

motion of a rigid body.

2) Determine the static equilibrium configuration of the system and measure the

displacement of the mass or rigid body from its static equilibrium position.

47

mmx

KKx (Spring force)

Static equilibrium position


48/58

3) Draw the free-body diagram of the mass or rigid body when a positive

displacement and velocity are given to it. Indicate all the active and reactive forces

acting on the mass or rigid body.

4) Apply Newtons second law of motion to the mass or rigid body shown by the

free-body diagram. Newtons second law of motion can be stated as follows:

Note:

1. Spring Force: Stiffness*displacement = K*x

2. Sign Convention: Forces and torques in the direction of motion are positive.

Therefore in the above example spring force (Kx) is to be taken negative as it is in the

direction opposite to that of displacement (motion).

Illustration 2: Here the mass m slides on a frictionless surface. Therefore there is no

damping

Figure 2.8:

From NSL

Kx

m

x

Friction less surfacem

K

48

F = ma ---- Newtons Second Law of Motion

m x = - Kx

mx + Kx = 0

..

..


49/58

F = ma

m x = - Kx

mx + Kx = 0

Points to Note:

The displacement of the mass is to be considered zero with the system at equilibrium i.e.,

the spring is neither stretched nor compressed

Sign Convention: Adopted sign convention must not be changed during the course of the

problem

All the forces / torque in the direction of the displacement are considered positive,

otherwise, negative

Behaviour of the system is independent of the sign convention

Illustration 3: The weight of the mass was not factor in the system just analyzed.

However, weight is a factor in the spring mass system shown in figure 2.9

Figure 2.9:

From NSL

F = ma

m x = k ( x) - mg

k ( x)

x

k

m mg

x

49

..

..

Position with sprinrelaxed

Reference position

(x = 0)

..


50/58

= k kx mg

= (mg / k) = Static deflection

mx = - kx mx + kx = 0

The weight of the mass has no effect on the differential equation when the reference is at

the equilibrium position. Hence, the differential equation is same as that of the previous

system in which weight was not a factor.

Illustration 4: Figure 2.10 (a) shows a spring mass damper system and a corresponding

free body diagram is shown in figure 2.10 (b) which assumes positive (downward)

displacement and velocity. If the reference x = 0 is chosen at the equilibrium position, the

mass can be considered weight less. Both the spring force and the damping force act

vertically upwards.

Figure 2.10 (b)

Figure 2.10 (a)

From NSL

F = ma

mx = - c x kx

50

xm

k

c

x

k x (Spring Force)

c x (damping Force).

.

mx + c x + kx = 0.. .

.. ..

..


51/58

Illustration 5: A variation of the previous system is shown in figure 2.11 (a) where a

provision is made for input displacement x at the top of spring. Figure 2.11 (b) shows the

free body diagram of the mass responding to a positive input at x. It is assumed that x is

displaced upward and that the mass is responding with a positive displacement y (less

than x) and with a positive velocity. With this assumption the spring force acts upward

with a magnitude k (x-y) and the damping force is acts downwards as shown in the free

body diagram.

Figure 2.11 (b)

Figure 2.11 (a)

--- GDE

51

m

X (Input)

y (Response)

m

k (x-y)

c y.

m

kx

c y.

ky

From NSLmy = + k (x-y) - c y.. .

my + c y + ky = kx.. .

c

k

x > y


52/58

2.3.3 Models for Rotational or Torsional Systems: The principles presented previously

can be extended to obtain the mathematical model for torsional or rotational systems.

Illustration 6: Consider a system with rotating inertia J with torsional spring of stiffness

kt and a rotary damper with damping coefficient b as shown in figure 2.12 (a). If J is

displaced by the corresponding free body diagram is as shown in figure 2.12 (b).


From NSL for rotating system

= TJ

bkJ t =

0=++ tkbJ

Illustration 7: A variation of the previous system is shown in figure 2.13 (a) where a

provision is made for input displacement i at the end of torsional spring. Figure 2.13

(b) shows the free body diagram of the inertia responding to a positive input i. It is

assumed that i is displaced clockwise (looking from left) and that the inertia is

responding with a clockwise displacement 0 (less than i) and a positive velocity.

kt

bJ

Jkt. b.

52

.. .

.. .


53/58

With this assumption the spring torque is acting clockwise with a magnitude kt ( i - 0)

and the damping torque is acting counter clockwise as shown in the free body diagram.

Figure 2.13 (a)

Figure 2.13 (b)

Let i: Input

And 0: ResponseLet (i > 0)


= TJ

Illustration 8: A torsional system with torque T as input and angular displacement as

output.

kt

bJ

0

i

J bKti

Kt(

i-

0) J

b

kt

0

.

.

53

J0 = kt (i - 0) - b

J 0 + b0 + kt0 = kt i --- Model

.. .

.. . .


54/58

Figure 2.14 (a)

Figure 2.14 (b)


= TJ

J = T - b

J + b = T --- Model

2.3.4 System with more than one mass: System involving more than one mass is

discussed below.

Illustration 9: For a two DOF spring mass damper system obtain the mathematical

model where F is the input x1 and x2 are responses.

b

T

J

b

T

.J

54

.. .

.. .


55/58

Figure 2.15 (a)

Figure 2.15 (b)

From NSL F= ma

For mass m1

m1x1 = F - b1 (x1-x2) - k1 (x1-x2) --- (a)

For mass m2

m2 x2 = b1 (x2-x1) + k1 (x2-x1) - b2 x2 - k2 x2 --- (b)

m2

m1

k2

b2

(Damper)

x2 (Response)

x1

(Response)

k1

F

b1

m2

m1

F

k2

x2

b2

x2

k1

x2

k1

x1

b1

x1

b1

x2

x2

k1

x2

k1

x1

b1

x1 b1 x2

.

. .

.

m2

m1

F

k2

x2

b2

x2

k1

(x1-x

2)

.

.b

1(x

1-x

2)

.

x1

.

55

.. . .

.. . . .

Draw the free body diagram for mass m1and m2 separately as shown in figure 2.15

(b)

Apply NSL for both the masses separatelyand get equations as given in (a) and (b)


56/58

Illustration 10: For the system shown in figure 2.16 (a) obtain the mathematical model if

x1 and x2 are initial displacements.

Let an initial displacement x1 be given to mass m1 and x2 to mass m2.

Figure 2.16 (a)

K1

K2

K3

m2

m1

X1

X2

56


57/58

Figure 2.16 (b)

Based on Newtons second law of motion: F = maFor mass m1

m1x1 = - K1x1 + K2 (x2-x1)

m1x1 + K1x1 K2 x2 + K2x1 = 0

m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

For mass m2

m2x2 = - K3x2 K2 (x2 x1)

m2x2 + K3 x2 + K2 x2 K2 x1

m2 x2 + x2 (K2 + K3) = K2x1 ----- (2)

Mathematical models are:

m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

K1

X1

K2

X2

K2

X2

K2

X1

K2

X1

K3

X2 K

3X

2

K1

X1

X1 X

1

m1

m1

m2 m

2X

2 X2

K2

(X2

X1)

K2

(X2

X1)

57

..

..

..

..

..

..

..

..


58/58

m2x2 + x2 (K2 + K3) = K2x1 ----- (2)

CONTINUED IN SESSION V: 05.09.2006

******************END****************

Documents

33525999 Control Engg Notes