MODERN CONTROL SYSTEMS - aast.edu

MODERN CONTROL SYSTEMS

Emam Fathy

Department of Electrical and Control Engineering

email: [email protected]

http://www.aast.edu/cv.php?disp_unit=346&ser=68525

Lecture 8

Root Locus

1

mailto:[email protected]

http://www.aast.edu/cv.php?disp_unit=346&ser=68525

Introduction

• What is root locus?– Root locus is the locus (graphical presentation) of the

closed-loop poles as a specific parameter (usually gain, K)is varied from 0 to infinity.

• Why do we need to use root locus?– We use root locus to analyze the transient response

qualitatively. (E.g. the effect of varying gain upon percentovershoot, settling time and peak time).

– We can also use root locus to check the stability of thesystem.

Introduction

we use root locus to analyze the feedback control system.

K in the feedback system is called a gain. Gain is used to vary thesystem in order to get a different output response.

Example

How the dynamics of the system (camera) changes as K is varied ?

Drawing the root locus

Characteristic equation:

𝑺𝟐 + 𝟏𝟎𝑺 + 𝑲 = 𝟎


Next step is to plot the poles values on the s-plane by varying the gain, K, value.


Join the poles with solid lines and you will get the shape of the locus (path)


• The process of drawing a root locus is time consuming. If the system is complex.

• An alternative approach is to sketch the root locus instead of drawing the root locus.

Sketching the root locus

In order to sketch the root locus we must follow these six rules.

Example

1) Sketch the root locus of the following system:

2) Determine the value of K such that the damping ratio ζ of a pair of dominant complex conjugate closed-loop is 0.5.

Construction of root loci

• Step-1: The first step in constructing a root-locus plot is tolocate the open-loop poles and zeros of G(s)H(s) 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

12

n=3 -----> number of polesp1 = 0; p2 = -1; p3 = -2

• m=0----> number of zeros

Construction of root loci• Step-2: Determine the root loci on the 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

13

The loci on the real axis areto the left of an ODDnumber of REAL polesand/or REAL zeros ofG(s)H(s)

Red lines on s-plane areparts of the real-axis wherethe root locus exists.

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

14

Number of asymptotes = n - m

Intersection point of asymptotes with real axis:

Angles of asymptotes with real axis:

, k=0,1,2,…,(n-m-1) mn

ko

12180

03

0)210(

1

3

3

mn

zpm

i

n

i

2 when 300

1 when 180

0n whe60

k

k

k

15

03

12180

ko

180 , 60, 60

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

-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

16

implicit step

17

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

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

18

Break Away Point

Construction of root loci

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

• The breakaway or break-in points are the closed-loop poles thatsatisfy:

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

• The characteristic equation of the system is

0ds

dK

19

)2)(1()()(

sss

KsHsG

0)2)(1(

1)()(1

sss

KsHsG

• The breakaway point can now be determined as

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

1)2)(1(

sss

K

)2)(1( sssK

)2)(1( sssds

d

ds

dK

20

sssds

d

ds

dK23 23

263 2 ssds

dK

4226057741

0263

21

2

.s,.s

ss

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

-0.5

0

0.5

1

Pole-Zero Map

Real Axis

Imagin

ary

Axis

21

Break Away Point

4226.0s

Construction of root loci• Step-5: Intersection points with the imaginary axis.

• An alternative approach is to let s=jω in the characteristicequation, 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

26

Construction of root loci• Step-5: Determine the points where root loci cross the

imaginary axis.

• Equating both real and imaginary parts of this equationto zero

• Which yields

0)2()3( 32 jK

0)3( 2 K

0)2( 3

27

-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

29

Example#1

• Consider following unity feedback system.

• Determine the value of K such that the damping ratio ofa pair of dominant complex-conjugate closed-loop polesis 0.5.

)2)(1()()(

sss

KsHsG

30

Example#1

• The damping ratio of 0.5 corresponds to

cos

1cos

60)5.0(cos 1

31

Example#1

• The value of K that yields such poles is found from themagnitude condition

1)2)(1(

5780.03337.0

jssss

K

33

Example#1

• The third closed loop pole at K=1.0383 can be obtainedas

0)2)(1(

1)()(1

sss

KsHsG

0)2)(1(

0383.11

sss

00383.1)2)(1( sss

35

Home Work


• Determine the value of K such that the naturalundamped frequency of dominant complex-conjugateclosed-loop poles is 1 rad/sec.

)2)(1()()(

sss

KsHsG

37

Example#2

• Sketch the root locus of following system anddetermine the location of dominant closed looppoles to yield maximum overshoot in the stepresponse less than 30%.

38

Example#2

• Step-1: Pole-Zero Map

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-Zero Map

Real Axis

Imagin

ary

Axis

39

Example#2

• Step-2: Root Loci on Real axis

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-Zero Map

Real Axis

Imagin

ary

Axis

40

Example#2

• Step-3: Asymptotes

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-Zero Map

Real Axis

Imagin

ary

Axis

90

2

41

Example#2

• Step-4: breakaway point

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-Zero Map

Real Axis

Imagin

ary

Axis

-1.55

42

Example#2

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5-8

-6

-4

-2

0

2

4

6

8

Root Locus

Real Axis

Imagin

ary

Axis

43

Example#2

• Mp<30% corresponds to

10021

eM p

100%3021

e

35.0

44

Example#2

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5-8

-6

-4

-2

0

2

4

6

8

0.35

0.35

6

6

Root Locus

Real Axis

Imagin

ary

Axis

45

Example#2

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5-8

-6

-4

-2

0

2

4

6

8

0.35

0.35

6

6

System: sys

Gain: 28.9

Pole: -1.96 + 5.19i

Damping: 0.354

Overshoot (%): 30.5

Frequency (rad/sec): 5.55

Root Locus

Real Axis

Imagin

ary

Axis

46

Root Locus of 1st Order System

• 1st order systems (without zero) are represented by followingtransfer function.

• Root locus of such systems is a horizontal line starting from -αand moves towards -∞ as K reaches infinity.

s

KsHsG )()(

jω

σ-α

-∞

47

Home Work

• Draw the Root Locus of the following systems.

2)()(

s

KsHsG

1)()(

s

KsHsG

s

KsHsG )()(

1)

2)

3)

48

jω

σ-α-β

Root Locus of 1st Order System

• 1st order systems with zero are represented by followingtransfer function.

• Root locus of such systems is a horizontal line starting from -αand moves towards -β as K reaches infinity.

s

sKsHsG

)()()(

49

Home Work


2)()(

s

KssHsG

1

)5()()(

s

sKsHsG

s

sKsHsG

)3()()(

1)

2)

3)

50

Root Locus of 2nd Order System

• Second order systems (without zeros) have two poles and thetransfer function is given

• Root loci of such systems are vertical lines.

))(()()(

21

ss

KsHsG

jω

σ-α1-α2

51

Home Work


)2()()(

ss

KsHsG

2)()(

s

KsHsG

)3)(1()()(

ss

KsHsG

1)

2)

3)

103)()(

2

ss

KsHsG4)

52

Root Locus of 2nd Order System

• Second order systems (with one zero) have two poles and thetransfer function is given

• Root loci of such systems are either horizontal lines or circulardepending upon pole-zero configuration.

))((

)()()(

21

ss

sKsHsG

jω

σ-α1-α2 -β

jω

σ-α1-α2-β

jω

σ-α1-α2 -β

53

Home Work


)2(

)1()()(

ss

sKsHsG

2

)2()()(

s

sKsHsG

)3)(1(

)5()()(

ss

sKsHsG

1)

2)

3)

54

Example

• Sketch the root-locus plot of following systemwith complex-conjugate open loop poles.

55

Example

• Step-1: Pole-Zero Map

• Step-2: Determine the root loci on real axis

• Step-3: Asymptotes

56

Example• Step-4: Determine the angle of departure from the

complex-conjugate open-loop poles.

– The presence of a pair of complex-conjugate open-looppoles requires the determination of the angle ofdeparture from these poles.

– Knowledge of this angle is important, since the rootlocus near a complex pole yields information as towhether the locus originating from the complex polemigrates toward the real axis or extends toward theasymptote.

57

Example• Step-4: Determine the angle

of departure from thecomplex-conjugate open-looppoles.

58

Example

• Step-5: Break-in point

59

Root Locus of Higher Order System

• Third order System without zero

))()(()()(

321

sss

KsHsG

61

Root Locus of Higher Order System

• Sketch the Root Loci of following unity feedback system

)4)(2)(1(

)3()()(

ssss

sKsHsG

62

• Let us begin by calculating the asymptotes. The real-axis intercept isevaluated as;

• The angles of the lines that intersect at - 4/3, given by

63

• The Figure shows the complete root locus as well as the asymptotesthat were just calculated.

64

Example: Sketch the root locus for the system with the characteristic equationof;

a) Number of finite poles = n = 4.

b) Number of finite zeros = m = 1.

c) Number of asymptotes = n - m = 3.

d) Number of branches or loci equals to the number of finite poles (n) = 4.

e) The portion of the real-axis between, 0 and -2, and between, -4 and -∞, lieon the root locus for K > 0.

• Using Eq. (v), the real-axis asymptotes intercept is evaluated as;

• The angles of the asymptotes that intersect at - 3, given by Eq. (vi), are;

σ𝑎 =−2 + 2 −4 − (−1)

𝑛 − 𝑚=

−10 + 1

4 − 1= −3

θ𝑎 =(2𝑘 + 1)π

𝑛 − 𝑚=

(2𝑘 + 1)π

4 − 1

For K = 0, θa = 60o

For K = 1, θa = 180o

For K = 2, θa = 300o65

• The root-locus plot of the system is shown in the figure below.

• It is noted that there are three asymptotes. Since n – m = 3.

• The root loci must begin at the poles; two loci (or branches) must leave the double pole

at s = -4.

• Using Eq. (vii), the breakaway point, σ, can be determine as;

• The solution of the above equation is 𝜎 = −2.59.

66

Example: Sketch the root loci for the system.

• A root locus exists on the real axis between points s = –1 and s = –3.6. • The intersection of the asymptotes and the real axis is determined as,

• The angles of the asymptotes that intersect at – 1.3, given by Eq. (vi), are;

• Since the characteristic equation is

• We have

σ𝑎 =0 + 0 + 3.6 − 1

𝑛 − 𝑚=

2.6

3 − 1= −1.3

θ𝑎 =(2𝑘 + 1)π

𝑛 − 𝑚=

(2𝑘 + 1)π

3 − 1

For K = 0, θa = 90o

For K = 1, θa = -90o or 270o

(a)67

• The breakaway and break-in points are found from Eq. (a) as,

From which we get,

• Point s = 0 corresponds to the actual breakaway point. But points are neither breakaway nor break-in points, because the corresponding gain values Kbecome complex quantities.

68

• To check the points where root-locus branches may cross the imaginary axis, substitute 𝑠= 𝑗𝜔 into the characteristic equation, yielding.

• Notice that this equation can be satisfied only if𝜔 = 0, 𝐾 = 0.

• Because of the presence of a double pole at theorigin, the root locus is tangent to the 𝑗𝜔axis at𝑘 = 0.

• The root-locus branches do not cross the 𝑗𝜔axis.

• The root loci of this system is shown in theFigure.

69

Home Work


• Determine

– Root loci on real axis

– Angle of asymptotes

– Centroid of asymptotes70

MODERN CONTROL SYSTEMS - aast.edu

Documents

Transcript of MODERN CONTROL SYSTEMS - aast.edu