Control System Engg. EE-304-F

TABLE OF CONTENTS

• Syllabus

• Section-C and D

SYLLABUS

• TIME DOMAIN ANALYSIS :Typical test signals, time response of first order systems to various standard inputs, time response of 2nd order system to step input, relationship between location of roots of characteristics equation, w and wn, time domain specifications of a general and an under-damped 2nd order system, steady state error and error constants, dominant closed loop poles, concept of stability, pole zero configuration and stability, necessary and sufficient conditions for stability Hurwitz stability criterion Routh stability criterion and relative stability. Root locus concept, development of root loci for various systems, stability considerations.

Section-D

• FREQUENCY DOMAIN ANALYSIS , COMPENSATION & CONTROL COMPONENT :Relationship between frequency response and time-response for 2nd order system, polar, Nyquist, Bode plots, stability, Gain-margin and Phase Margin, relative stability, frequency response specifications. Necessity of compensation, compensation networks, application of lag and lead compensation, basic modes of feedback control, proportional, integral and derivative controllers, illustrative examples. Synchros, AC and DC techo-generators, servomotors, stepper motors, & their applications, magnetic amplifier.

Section-C &D

Stability and its different types

i. Root Locus

ii. Polar Plot

iii. Nyquist Plot

iv. Bode Plot

v. Compensators

Stability of a closed loop system

Definition of stability of linear system:

Energised the closed-loop system input a short impulse which move the system from it’s steady-state position and wait. When the transient response has died and the system remove the original position or remain inside a predicted defined small area of the original position the system is stable and linear.

Definition of stability of non-linear system:

Energised the closed-loop system input a short impulse which move the system from it’s steady-state position and wait. If the system remove and remain inside a predicted defined small area of the original position the system is stable.

If the transient response hasn’t died the system is unstable.

)(sGc ( )PFG s)s(G)s(G1

)s(G)s(G)s(W

)s(r

)s(y

PFC

PFCyr

Stability examination from closed loop transfer function

)(sGC )(sGPF

nm

m

m

mn

n

ndt

trdb

dt

tdrbtrbtya

dt

tdya

dt

tyda

)()()()(

)()(1001

01

1

1

01

1

1

)()(

)(

)(

)(1

)(

)(

)()(

)(

asasasa

bsbsbsb

sNsD

sN

sD

sN

sD

sN

sWsr

syn

n

n

n

m

m

m

myr

r(s) y(s)

Unit feedback model

)(

)()()()(0

sD

sNsGsGsG PFC

)()()()(

)()()()(

01

1

1

01

1

1

srbssrbsrsbsrsb

syassyasysasysa

m

m

m

m

n

nm

n

n

Continuation of stability examination from closed loop

)(sGc)(sGp

n

i

i

m

j

jm

ps

zsb

1

1

)(

)(

n

i

t

icTransientieKtx

1

)(

The transient solution is given bz the roots of

the differential equation’s characteristic equation.

001

1

1

aaa n

n

n

The roots are equal the denominator of closed loop transfer function and so it gives transient solution too.

001

1

1

asasas n

n

n

Definition of stability in frequency domain

)(sGc)(sGp

The system is stable if the all real part of the roots of denominator of the transfer function is negative.

0)()(1 sGsG pc

n

1i

i

m

1j

jm

)ps(

)zs(b

)s(r

)s(y

The pi are named the poles and zj are named the zeros

Root Locus

Construction of root loci Step-1: The first step in constructing a root-locus plot is to locate the open-loop poles and zeros in s-plane.

)2)(1()()(

sss

KsHsG

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

)2()1( is angle The sss

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

Construction of root loci Step-2: Determine the root loci on the real axis.

p1

To determine the root loci on real axis we select some test points. e.g: p1 (on positive real axis).

The angle condition is not satisfied.

Hence, there is no root locus on the positive real axis.

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis


p2

Next, select a test point on the negative real axis between 0 and –1.

Then

Thus

The angle condition is satisfied. Therefore, the portion of the negative real axis between 0 and –1 forms a portion of the root locus.

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis


p4

Similarly, test point on the negative real axis between -2 and – ∞ satisfies the angle condition.

Therefore, the negative real axis between -2 and – ∞ is part of the root locus.

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imag

inar

y A

xis


Construction of root loci Step-3: Determine the asymptotes of the root loci. That is, the root loci when s is far away from origin.

Asymptote is the straight line approximation of a curve

Actual Curve Asymptotic Approximation

Construction of root loci

Step-3: Determine the asymptotes of the root loci.

where

n-----> number of poles

m-----> number of zeros

For this Transfer Function

mn

kasymptotesofAngle

)12(180

)2)(1()()(

sss

KsHsG

03

)12(180

k



Since the angle repeats itself as k is varied, the distinct angles for the asymptotes are determined as 60°, –60°, and 180°.

3 when 420

2 when 300

1 when 180

0n whe60

k

k

k

k03

)12(180

k



Before we can draw these asymptotes in the complex plane, we need to find the point where they intersect the real axis.

Point of intersection of asymptotes on real axis (or centroid of asymptotes) is

mn

zerospoles



For

03

0)210(

)2)(1()()(

sss

KsHsG

13

3



-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

60

60

180

180 , 60, 60

1

Construction of root loci Step-4: Determine the breakaway/break-in point.

The breakaway/break-in point is the point from which the root locus branches leaves/arrives real axis.

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis


Step-4: Determine the breakaway point or break-in point.

The breakaway or break-in points can be determined from the roots of

It should be noted that not all the solutions of dK/ds=0 correspond to actual breakaway points.

If a point at which dK/ds=0 is on a root locus, it is an actual breakaway or break-in point.

0ds

dK



Set dK/ds=0 in order to determine breakaway point.

)2)(1( sssds

d

ds

dK

sssds

d

ds

dK23 23

263 2 ssds

dK

0263 2 ss

0263 2 ss

5774.1

4226.0

s



Since the breakaway point needs to be on a root locus between 0 and –1, it is clear that s=–0.4226 corresponds to the actual breakaway point.

Point s=–1.5774 is not on the root locus. Hence, this point is not an actual breakaway or break-in point.

5774.1

4226.0

s

)2)(1()()(

sss

KsHsG


Step-4: Determine the breakaway point.

-5 -4 -3 -2 -1 0 1 2-1

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

60

60

1804226.0s

Construction of root loci Step-5: Determine the points where root loci cross the imaginary axis.

Let s=jω in the characteristic equation, equate both the real

part and the imaginary part to zero, and then solve for ω and K.

For present system the characteristic equation is

023 23 Ksss

02)(3)( 23 Kjjj

0)2()3( 32 jK

Construction of root loci Step-5: Determine the points where root loci cross the imaginary axis.

Equating both real and imaginary parts of this equation to zero

Which yields

0)2()3( 32 jK

0)3( 2 K

0)2( 3

-7 -6 -5 -4 -3 -2 -1 0 1 2-5

-4

-3

-2

-1

0

1

2

3

4

5

Root Locus

Real Axis

Imagin

ary

Axis

Polar Plot & Nyquist Plot

Nyquist stability criteria

Topics to be covered include:

• Nyquist stability criteria.

• Minimum phase systems.

• Simplified Nyquist stability criterion.

33

( )kf s

Nyquist path Nyquist plot

0111 kNPZ

000 NPZ

Nyquist fundamental

0111 kNPZ

Example: Discuss about the RHP roots of following system.

)1(

)1(2)(

ss

ssf

1800

22)(

ssf

22545

22)(

ssf

27090

22)(

ssf

9090

22)(

ssf

4545

22)(

ssf

00

22)(

ssf

Ver

y im

po

rtan

t p

art

?

0)1(

)1(21

ss

sk

Example: Discuss about the RHP roots of the system.

0)1(

)1(21

ss

sk

)1(

)1(2)(

ss

ssf

)1(

)1(2)(

jj

jjf

)( jf

?

?

Which point is very important?

dennumjf )( )tan90(tan180 11 1tan290

10)( jf02

)1(

)11(2)1(

jj

jjf

1

2

1

2

0Im(f)let ly equivalentOr

Example: Discuss about the RHP roots of the following system for different values of k.

0)1(

)1(21

ss

sk

1

21 1

1110 NPZk 101 Z

Unstable( one RHP root) 11 Z

1110 NPZk 11 NZ

2

0

1Z

05.0 k

5.0k

Stable

Unstable (two RHP root)

More Study:

000 NPZ 101

Bode Plot

Bode plot

• 1st order or higher terms that can be written as a product of 1st order terms

– 4 cases:

• 2nd order terms

– 2 cases:

ss

asas

1,,

)(

1),(

22

22

2

1,2

nn

nnss

ss

)()( assG

)1()()( a

jaajjG

ω = 0

aM

ajG

log20log20

)(

ω >> a

log20log20

90)()(

M

ja

jajG

phase = 0

phase = 90

ω = a

3log202log20log20

)()(

aaM

ajajG phase = 45

First order terms Case I: one zero at -a

Asymptotes (approximation)

Break frequency

= freq. at which mag. has changed by 3 db

12 10

3 dB at break frequency

)(

1)(

assG

)1(

1

)(

1)(

a

ja

ajjG

ω = 0

)/1log(20log20

/1)(

aM

ajG

ω >> a

log20log20

9011

)(

1)(

M

j

aja

jG

phase = 0

phase = -90

ω = a 3)/1log(20)2/1log(20log20 aaM

phase = -45

First order terms Case II: one pole at -a

G(s) = s

G(s) = 1/s

G(s) = s+a

G(s) = 1/(s+a)

Bode Plots Find magnitude and phase of each term

and sum them up!!!

)()()()()(

)(log20)(log20log20

)(log20)(log20log20)(log20

)()(

)()()(

))((

))(()(

2121

21

21

21

21

21

21

pspsszszsKsG

pspss

zszsKsG

pspss

zszsKsG

pspss

zszsKsG

m

m

m

m

mag(num)-mag(den)

phase(num)-phase(den)

Example

sketch bode plot of

)2)(1(

)3()(

sss

ssG

break frequency at 1,2,3

Frequency small 1 2 3

s -20 -20 -20 -20

1/(s+1) 0 -20 -20 -20

1/(s+2) 0 0 -20 -20

(s+3) 0 0 0 20

Total Slope -20 -40 -60 -40

Slope at each break frequency for magnitude plot

Magnitude Plot

Frequency small 0.1 0.2 0.3 10 20 30

s 0 0 0 0 0 0 0

1/(s+1) 0 -45 -45 -45 0 0 0

1/(s+2) 0 0 -45 -45 -45 0 0

(s+3) 0 0 0 45 45 45 0

Total Slope 0 -45 -90 -45 0 45 0

Slope at each point for phase plot

Phase Plot

Stability examination from transfer function of opened-loop

)(sGc)(sGp

)s(G)s(G)s(G 0pc

Opened-loop transfer function

01)(0)(1 jjGsG olol From definition of stability: 180j)(j

0 e1e)(A0j1)j(G

Gain cross over Frequeny:The frequency at which the open-loop system gain is unity is termedthe gain-crossover frequency. Phase cross over Frequeny:The frequency at which the open-loop system phase-shift is -180° is termed the phase-crossover frequency. Note: The system is stable if plots of the opened-loop transfer function at the gain-crossover frequency the phase-shift less than -180° and at the phase-crossover frequency the amplitude gain less than unity.

Gain margin and phase margin

Opened-loop transfer function: )()(01)( j

ol eAjjG

The gain margin is the reciprocal of open-loop gain at the phase-crossover frequency. Expressed in decibel the gain margins is: -20log(open-loop’s gain at the phase-crossover frequency) dB The phase margin is the difference of the phase-shift of the system and -180° at the gain-crossover frequency. Note: The system is stable if plots of the opened-loop transfer function at the gain-crossover frequency has a positive the phase margin and at the phase-crossover frequency has a positive the gain margin.

Stability examination from transfer function of opened-loop

)(j

0 e)(A0j1)j(G

Plots on j (s) plane is named Nyquist diagram. This is a point on the s plane. If the area of the s plane which is bordered by the positive real axis and the Nyquist plots doesn’t cover this point than the system is stable.

Independent plots the amplitude gain and phase-shift on lg axis is named Bode diagram. It is the x axis line on the amplitude plots and the -180 phase-shift line on the phase-shift plots. If the gain and phase margin criterion is true the system is stable.

Advantages of working with Bode plots

n

1i

i

m

1j

j

i0

)sT1(

)s1(

s

K)s(G

Bode form of opened-loop transfer function

Bode plots of systems in series simply add. Bode’s important phase-gain relationship is given. A much wider range of system behaviour can be displayed on a single plot. Dynamic compensator design can be based on Bode plots.

State Space Analysis

)()()(

)()()(

tDutCxty

tButAxtxdt

d

State equation

Output equation

Dynamic equation

1

2

1

)(

)(

)(

)(

nn tx

tx

tx

tx

1

2

1

)(

)(

)(

)(

rr tu

tu

tu

tu

1

2

1

)(

)(

)(

)(

pp ty

ty

ty

ty

State space

State variable

r- input p- output

nnA

rnB

npC

rpD

1

2

1

)0(

)0(

)0(

)0(

nnx

x

x

x

Motivation of state space approach

3

1

)(

)(

s

s

sU

sY

+ -

)(tu )(ty

)1(

1

s

s

1

2

s +

)(tn noise

)3)(1(

)1(2

)(

)(0

ss

s

sN

sEu

BIBO stable

unstable

Transfer function

Example 1 )(te

1

1

s

s

1

1

s

)(tu )(ty

1

1

su

yBIBO stable, pole-zero cancellation

-2 s

1

s

1

)(tu )(ty

+ +

+

+ +

-

1x 2x1x 2x)(tv

us

uv

ss

s

u

v

1

2

1

21

1

1

2

212

11 2

xy

uxxx

uxx

202

101

)0(

)0(

xx

xx

)(2

1)

2

1()()(

)(2)(

1010202

101

tuexeexxtxty

tuexetx

ttt

tt

2010

2010

2.2

0.1

xxif

xxif

then system stable

State-space description Internal behavior description

)]()[()0(])[()( 1111 sBUAsILxAsILtx

Example

M2 M1

B3 B1 B2

K 1y2y

)(tf

0)()(

)()()(

121222322

212121111

yyKyyByByM

tfyyKyyByByM

23

12

211

yx

yx

yyx

let

)(

0

10

)(

)(110

1

3

2

1

2

32

2

2

2

1

2

1

21

1

3

2

1

tfM

x

x

x

M

BB

M

B

M

K

M

B

M

BB

M

K

x

x

x

aR aL

ai

tea

fR

fL

teb

m

mJLT

Armature circuit Field circuit

te f

dt

diLeiRte a

abaaa )(

mbb ke

a

a

m

a

ba

a

aa eLL

ki

L

R

dt

di 1

L

m

m

m

m

aim

Tdt

dB

dt

dJ

ikT

2

2

m

m

L

m

m

m

m

a

m

im

dt

d

TJdt

d

J

Bi

J

k

dt

d

1

Example

Using MATLAB

Start of stability examination K=1

First using root’s loci. Enabled damping ratio: 0.72, 0.45, 0,23

Secondly using Bode plots. Enabled phase shift: -120, -135, -150

Define the time domain performance specifications.

K)101)(51)(1(

1

sss

Summary questions

• Definition of stability of linear and non-linear systems.

• Which methods have used the closed-loop transfer function to examine the stability a control system?

• Which methods have used the opened-loop transfer function to examine the stability a control system?

• Explain the following terms: phase margin, gain margin.

• Explain what is meant by the term “dead time” in control and how it may affect the stability of a control loop.

NPTEL/other online links

• https://nptel.ac.in/courses/108101037/

• https://www.tutorialspoint.com/control_systems/control_systems_introduction.htm

• https://en.wikibooks.org/wiki/Control_Systems/Introduction

• https://lecturenotes.in/subject/52/control-system-engineering-cse

• https://www.researchgate.net/publication/265168969_Control_Systems_Engineering

Control System Engg. EE-304-F

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Transcript of Control System Engg. EE-304-F