1 سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر حسين بلندي - دکتر سید مجید اسما عیل زاده

1

سیستمهای کنترل خطی

1389پاییز

بسم ا... الرحمن الرحيم

دکتر حسين بلندي- دکتر سید مجید اسما عیل زاده

2

Recap.

• Steady State Error• Stability• Routh Horowitz Method

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Root Locus

The root locus is a plot of the roots of the characteristic equation of the closed-loop system as a function of the gain of the open-loop transfer function.

This graphical approach yields a clear indication of the effect of gain adjustment.

INTRODUCTION:

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• Performance (stability and response) of a closed-loop system (feedback system) depends on its pole locations.

• In feedback control system design, it is usually necessary to adjust one or more parameters to achieve desired pole locations for the closed-loop system.

• Root locus is a plot of closed-loop system pole locations when the gain K is varied.

• A graphical technique exists for plotting the root locus which does not involve calculating the roots of a closed-loop polynomial.

• Root locus is a powerful graphical tool for analysis and design of feedback control systems.

Root Locus

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i.e; A specific technique which shows how changes in one of a system’s parameter

(usually the controller gain, K)

will modify the location of the closed-loop poles

in the s-domain.

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Graphical Evaluation )Hostetter):

A rational function

1 2

1 2

( ) m

n

s z s z s zF s K

s p s p s p

When evaluated at a specific value of the variable , is:0s s

0 1 0 2 00

0 1 0 2 0

( ) m

n

s z s z s zF s K

s p s p s p

7

On a pole-zero plot, suppose a directed line segment is drawn from the position of a pole, say to the value at which the function is to be evaluated.

0s1p

The segment has length and makes the angle With the real axis.

0 1s p

As shown below, at a root s1

10 ps

8

Thus:

0 1 0 2

0 1 0 2

0 1 0 2

0

0 1 0 2

...( )

...

j s z j s z

j s p j s p

s z e s z eF s K

s p e s p e

00

0

( )

product of the length of the directed

line segments from the zero to sF s K

product of the length of the directed

line segments from the poles to s

00

sum of the angles of the directed line segments from zeros to s( )

sum of the pole angles 180 if k is negativeF s

9

Consider the system:

Calculation of Magnitude and Angles (Ogata)

1

1 2 3 4

2 3

( ) ( )

where and are complex-conjugate poles.

s zG s H s K

s p s p s p s p

p p

1 1 2 3 4

1

1 2 3 4

( ) ( )

( ) ( )

G s H s

KBG s H s

A A A A

10

Definition

01 sHsKG

• The closed-loop poles of the negative feedback control:

are the roots of the characteristic equation:

01 sHsKG

The root locus is the locus of the closed-loop poles when a specific parameter (usually gain, K)

is varied from 0 to infinity.

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Root Locus Method Foundations

• The values of s in the s-plane that make the loop gain KG(s)H(s) equal to -1 are the closed-loop poles

(i.e. )

• KG(s)H(s) = -1 can be split into two equations by equating the magnitudes and angles of both sides of the equation.

101 sHsKGsHsKG

12

Root Locus for Feedback Systems

magnitude conditions angle conditions

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Angle and Magnitude Conditions

Independent of K

,,,l 210

,,,l 210

12180

1

12180

1

1

0

lsHsG

KsHsG

lsHsKG

sHsKG

sHsKG

o

14

APPLICATION OF THE MAGNITUDE AND ANGLE CONDITIONS:

Once the open-loop transfer function G(s)H(s) has been determined and put into the proper form, the poles and zeros of this function are plotted in the S plane.

As an example, consider:

with the damping ratio < 1, the complex-conjugate poles are:

15

The plot of the poles and zeros is shown in bellow:

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17

magnitude conditions:

angle conditions:

All angles are considered positive, measured in the counter clock wise sense.

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PLOTTING ROOTS OF A CHARACTERISTIC EQUATION:

Example1 : position-control system

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Example1 (Continue) : position-control system

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21


22


These curves are defined as the root-locus plot of

23


Corresponding to the selected roots there is a required value of K that can be determined from the plot . When the roots have been selected, the time response can be obtained.

The value of K is normally considered to be positive. However, it is possible for K to be negative.

For an increase in the gain K of the system, analysis of the root locus reveals the following characteristics:

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25

simple second-order system:

A zero is added to the simple second-order system:

QUALITATIVE ANALYSIS OF THE ROOT LOCUS:

Compare

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If a pole, instead of a zero, is added to the simple second-order system,the resulting transfer function is:

root locus of the closed-loop control system

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root-locus configurations for negative feedback control systems

?

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Example2:

Determine the locus of all the closed-loop poles of C(s)/R(s), for

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Root Locus

Real Axis

Imag

inar

y A

xis

-6 -5 -4 -3 -2 -1 0 1-10

-8

-6

-4

-2

0

2

4

6

8

10

System: sysGain: 0.35Pole: -1.51 + 1.72e-008iDamping: 1Overshoot (%): 0Frequency (rad/sec): 1.51

System: sysGain: 0Pole: -5Damping: 1Overshoot (%): 0Frequency (rad/sec): 5

System: sysGain: InfPole: -4Damping: 1Overshoot (%): 0Frequency (rad/sec): 4

System: sysGain: 44Pole: -1.93 + 6.42iDamping: 0.288Overshoot (%): 38.8Frequency (rad/sec): 6.71

>> sys = zpk( -4, [-1 -2 -5],1) Zero/pole/gain: (s+4)----------------------(s+1) (s+2) (s+5)

>> rlocus(sys)

Rlocus in MATLAB:

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GEOMETRICAL PROPERTIES (CONSTRUCTION RULES):

To facilitate the application of the root-locus method, the following rules are established for K > 0.

These rules are based upon the interpretation of the angle condition and an analysis of the characteristic equation.

These rules can be extended for the case where K < 0.

These rules provide checkpoints to ensure that the solution is correct.

which provides a qualitative idea of achievable closed-loop system performance.

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Rule 1: Number of Branches of the Locus

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• RL starts from n poles (k=0)• RL ends at m zeroes (k=infinity)• # of branches= max( n, m)

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36

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Rule 2: Real-Axis Locus

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• The locus includes all points along the real axis to the left of an odd number of poles plus zeros of GH

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42

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Rule 3: Asymptotes of Locus as s Approaches Infinity

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Rule 3(continue):

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Rule 4: Real-Axis Intercept of the Asymptotes

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Rule 5: Breakaway Point on the Real Axis

breakaway pointbreakin point

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The breakaway and break-in points can easily be calculated for an open-loop pole-zero combination for which

the derivatives of W(s)=-K

As an example, if

then

Multiplying the factors together gives

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by taking the derivative of this function and setting it equal to zero, the points can be determined

or

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Rule 6: Complex Pole (or Zero): Angle of Departure

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In a similar manner the approach angle to a complex zero can be determined

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Rule 7: Imaginary-Axis Crossing Point

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Rule 7(continue):

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Learning by doing Example

Sketch the root locus of the following system:

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#1Assuming n poles and m zeros for G(s)H(s):

• The n branches of the root locus start at the n poles.

• m of these n branches end on the m zeros• The n-m other branches terminate at infinity

along asymptotes.

First step: Draw the n poles and m zeros of G(s)H(s) using x and o respectively

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Draw the n poles and m zeros of G(s)H(s) using x and o respectively.

• 3 poles: p1 = 0; p2 = -1; p3 = -2

• No zeros

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1

ssssHsG

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Draw the n poles and m zeros of G(s)H(s) using x and o respectively.

• 3 poles: p1 = 0; p2 = -1; p3 = -2

• No zeros

21

1

ssssHsG

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#2

• The loci on the real axis are to the left of an ODD number of REAL poles and REAL zeros of G(s)H(s)

Second step: Determine the loci on the real axis. Choose a arbitrary test point. If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus

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Determine the loci on the real axis:

• Choose a arbitrary test point.

• If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus

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Determine the loci on the real axis:

• Choose a arbitrary test point.

• If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus

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• The root loci for very large values of s must be asymptotic to straight lines originate on the real axis at point:

radiating out from this point at angles:

Third step: Determine the n - m asymptotes of the root loci. Locate s = α on the real axis. Compute and draw angles. Draw the asymptotes using dash lines.

mn

lo

l

12180

mn

zps m

in

i

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Determine the n - m asymptotes:• Locate s = α on the real axis:

• Compute and draw angles:

• Draw the asymptotes using dash lines.

13

210

03321

ppp

s

mn

ll

12180

0

0

1

00

0

18003

112180

6003

102180

,,,l 210

63

Determine the n - m asymptotes:• Locate s = α on the real axis:

• Compute and draw angles:

• Draw the asymptotes using dash lines.

13

210

03321

ppp

s

mn

ll

12180

0

0

1

00

0

18003

112180

6003

102180

,,,l 210

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Breakpoint Definition

• The breakpoints are the points in the s-domain where multiples roots of the characteristic equation of the feedback control occur.

• These points correspond to intersection points on the root locus.

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#4Given the characteristic equation is KG(s)H(s) = -1

• The breakpoints are the closed-loop poles that satisfy:

Fourth step: Find the breakpoints. Express K such as:

Set dK/ds = 0 and solve for the poles.

0ds

dK

.sHsG

K1

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Find the breakpoints.

• Express K such as:

• Set dK/ds = 0 and solve for the poles.

4226057741

0263

21

2

.s,.s

ss

sssK

sss)s(H)s(G

K

23

211

23

67

Find the breakpoints.

• Express K such as:

• Set dK/ds = 0 and solve for the poles.

4226057741

0263

21

2

.s,.s

ss

sssK

sss)s(H)s(G

K

23

211

23

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• The n branches of the root locus start at the n poles.

• m of these n branches end on the m zeros• The n-m other branches terminate at infinity

along asymptotes.

Last step: Draw the n-m branches that terminate at infinity along asymptotes

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Applying Last Step

Draw the n-m branches that terminate at infinity along asymptotes

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Points on both root locus & imaginary axis?

• Points on imaginary axis satisfy:

• Points on root locus satisfy:

• Substitute s=jω into the characteristic equation and solve for ω.

js jω?

- jω

01 sHsKG

20 or

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EXAMPLE : ROOT LOCUS

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EXAMPLE (continue): ROOT LOCUS

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74


75


76


77


78


79

Example of an application of the root-locus method

The systematic application of the for the construction of a root locus is shown in the following non-trivial example for the open-loop transfer function :

40122

1)()(

20

ssss

sksHsG

The degree of the numerator polynomial is m=1. This means that the transfer function has one zero (s=-1). The degree of the denominator polynomial is n=4 and we have the four poles (s=0, s=-2 s=-6+2j, s=-6-2j). First the poles (x) and the zeros (o) of the open loop are drawn on the S plane as

80

40122

1)()(

20

ssss

sksHsG

Values of are in red 0k

81

We have (n-m=3) branches that go to infinity and the asymptotes of these three branches are lines which intercept the real axis according to rule. the crossing is at :

33.4

3

16620

a

the slopes of the asymptotes are:

3,,3

3

12180

pk

The asymptotes are shown in Figure as blue lines:

82

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Using Rule 2 it can be checked which points on the real axis are points on the root locus. The points -1<s<0 and s<-2 belong to the root locus, because to the right of them the number of poles and zeros is odd.

According to rule 4 breakaway and break-in points can only occur pairwise on the real axis to the left of -2.

68.31

Bs

47.51

Bs

866.076.04,3

jsB

The real roots and are the

positions of the breakaway and the break-in point

68.31

Bs 47.51

Bs

0ds

dK

84

The angle of departure of the root locus from the complex pole at can be determined from

jsp 263

8.661808.2463 p

)12(1802.1586.1614.153903 kp

85

40122

1)()(

20

ssss

sksHsG

0k

Breakaway Point Example

Consider the following loop transfer function.

Real axis loci exist for thefull negative axis.

Asymptotes: angles

jw

sXX

–4X

–2

–2j

2j

+1

60°

asymptotes

23)()(

ss

ksHsG

5,,3

3

12ppp

pk

5p

23

0)033(

a

Breakaway Point Example

Determine the breakaway points from

then

jw

sXX

–4X

–2

–2j

2j

+1

3,1

0)3)(1(342

--=

=++=++

s

ssss

0)96(

)9123(

96)3(

223

2

232

=++

++-=

úû

ùêë

é

++=ú

û

ùêë

é

+

sss

ssK

sss

K

ds

d

ss

K

ds

d

Root Locus Plot Example

Loop Transfer function:

Roots:s = 0, s = –4, s = –2 4j

Real axis segments:between 0 and –4 .

Asymptotes:

angles = 2

4)0224( -=----=as

47

,4

5,

43,

404)12( =

-+k ppppp

asymptotes

jw

sX

–4X

–2

–2j

2j

+1

45°

4j

–4jX

X

2044)()(

2

ssss

ksHsG


Breakaway points:

Three points that breakaway at 90° .

solving,

jw

sX

–4X

–2

–2j

2j

+1

45°

4j

–4jX

X

2

0

80368

8072244

80368

2234

23

234

ssss

sssk

ssss

k

ds

d

jsb 45.22,2


The imaginary axis crossings:

Characteristic equation

Routh table

s4 1 36 K

s3 8 80 0

s2 26 K 0

s 80-8K/26 0 0

s0 K 0 0

Condition for critical stability80-8K/26 > 0 or K<260

The auxiliary equation26 s 2 + 260 = 0solving

080368 234 =++++ Kssss

jjs 16.310 ±=±=

The final plot is shown on the right.

What is the value of the gain K corresponding to the breakaway point at


sX

–4X

–2

–2j

2j

4j

–4jX

X

3.16j

jw

?45.22 jsb

From the general magnitude condition the gain corresponding to the point s1 on the loci is

For the point s1 = –2 + 2.45j

K = |–2 + 2.45j| ·|–2 + 2.45j + 4| ·|–2 + 2.45j + 2 + 4j| ·|–2 + 2.45j + 2 – 4j| / 1.0

= 3.163 · 3.163 · 6.45 · 1.55 = 100.0

Root Locus Plot ExampleGain Calculations

ÕÕ==

++=m

ii

n

ii zspsK

11

11

Is there a gain corresponding to a damping ratio of 0.707 or more for all system modes?z = 0.707 = cos( q )q = 45°


sX

–4X

–2

–2j

2j

4j

–4jX

X

3.16j

jw

45°

Examine the responses for the various gains and relate them to the location of the closed-loop roots.

K = 64, roots are –2, –2, –2±3.46j K = 100, roots are –2±2.45j, –2±2.45j K = 260, roots are ±3.16j, –4±3.16j

sX

–4X

–2

–2j

2j

4j

–4jX

X

3.16j

jw

K=64 K=100

K=260

Root Locus Plot ExampleTime Responses


Time (sec.)

Am

plit

ud

eStep Response, K = 64

0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

s = –2, –2

s = –2±3.46j

whole response


Time (sec.)

Am

plit

ud

eStep Response, K = 100

0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

whole response

s = –2±2.45j (repeated)


Time (sec.)

Am

plit

ud

eStep Response K = 260

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

whole response

s = ±3.16j

s = –4±3.16j

Root Locus with Two Parameters

The root locus method focuses on the roots of the characteristic equation.

There can be several different loop transfer functions that have the same closed-loop characteristic equation.

To construct the root locus for a characteristic equation which has two parameters, we construct fictitious loop transfer functions and apply the normal methods.

Root Locus with Two Parameters

Consider the following characteristic equation.s3 + s2 + b s + a = 0 .

Write this in the general form of 1 + GH(s) = 0 with b as a multiplying gain.

This will allow the plotting of the root locus with respect to the gain b for any given value of a.

The roots of the characteristic equation of GH´(s) define the starting points for the root loci. Consider the loci of these roots.

01 23=

+++

absss

GH´(s)

Root Locus with Two Parameters The characteristic

equation of GH´(s) is s3 + s2 + a = 0 , which may be written as

Now construct the root locus of GH´´(s) in terms of the gain a .

0)1(

1 2=

++

ss

a

GH´´(s)

jw

sX

–2 –1X X

a1a1a2

a2

a1 = 0.3s = –1.2, 0.1±0.5j

a2 = 1.8s = –1.66, 0.33±1.0j

Root Locus with Two Parameters Now construct the

root locus for

where the open-loop poles correspond to the previous root locus for varying a .

Asymptotes: ± 90°

a = 0.3 sa = (-1.2 +0.1 +0.1)/2

= -0.5 a = 1.8

sa = (-1.66 +0.33 +0.33)/2 = -0.5

23=++ a

bsss

GH´(s)

Root Locus with Two Parameters Imaginary axis

crossings:s3 + s2 + b s + a = 0

s3 1 b 0s2 1 a 0s -b a 0 0s0 a 0 0 a = , b s2 + = 0a

ajs -=

jw

s

–2 –1

a1a1a2

a2

O

103

104

105

106

SUMMARY OF ROOT-LOCUS CONSTRUCTION RULESFOR NEGATIVE FEEDBACK

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1 سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر حسين بلندي - دکتر سید مجید اسما عیل زاده