me322 module 6i

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ME322Module 6, Slide 1

Northern Illinois UniversitySummer 2005

Peter A. LinCopyright 2001-2005

Root-locus Design Method

Outline

Build dynamic models

Study the dynamic responses of a system

Basic properties of feedback control

How to design control systems

Root-locus method

Frequency-response method

State-space method

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Input, ROutput, Y

+_

KGG(s)KA

For a basic feedback control system, the

closed loop transfer function is:

)(1

)(

)(

)(

sGKK

sGKK

sR

sY

GA

GA

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The characteristic equation is:

Definition: The graph of all possible roots ofthe characteristic equation relative to someparticular variable is called the root locus.

Purpose: to show graphically the generaltrend of the roots of the closed-loop systemwhen some parameters are varied.

0)(1 sGKK GA

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Assumptions:

The plant transfer function can be expressed

in general terms as the ratio of twopolynomials (m

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Assumptions:

The plant gainKGis positive. n > m.

The root-locus parameter is defined as

Then, the root-locus form of the

characteristic equation is:

KsG

orsKG

1)(

0)(1

GAKKK

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Root-locus for a DC-motor (Example)

DC Motor transfer function:

Characteristic equation of the closed loopsystem is [1+KG(s)]

)1)(1()(

21

ss

AsG

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Roots of the closed loop system

characteristic equation is

When K=0, open loop roots:

21

21

2

21212,1

21

2)1(4)()(

0)1)(1(

KAs

KAss

2

2

1

1

1;

1

ss

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Example

j

0-1/t1-1/t2= pole

= zero

K=0 K=0

21

21

2)(

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Observations:

- One root locus branch per root

- Branches begin at open loop roots for K=0

since the system is open loop for K=0.

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Observations: when K increases,

- poles moving towards each other and intoimaginary conjugate poles at breakaway point(where roots move away from the real-axis)

- Decreasing time constant (faster response)

- Decreasing rise time- Decreasing damping ratio, more overshoot, less

stable

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Phase condition

From the root-locus form of the characteristic

equation (assuming K is real and positive):

we can define that the root locus of G(s) is theset of points in the s-plane where thephase of G(s) is 180.

KsG

1)(

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Guidelines for sketching a Root Locus:

Step 1: Draw the pole and zero plot in the s-

plane.

]16)4[(

1)(

2

ss

sG

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Pole and Zero Plot

j

0= pole

= zero

4

4

4

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Step 2: Find the real axis portion of the locus.

The angles of complex poles cancel each other (theangles are all measured in a counterclockwisedirection from a horizontal line)

The angle of G(s0) for the test point s0 on the real

axis is given by the angles from poles and zeroson the real axis only.

The angle is 180 when the test point is to the left ofan odd number of poles plus zero.

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Pole and Zero Plot

j

0= pole

= zero

4

4

4

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Step 3: For large K, the branches of the root

locus go to either zeros or infinity. Drawthe asymptotes for large values of K.

At point

draw radial lines at the (n-m) distinct anglesmn

zps

ii

0

mnlmn

ll

,....,2,1,

)1(360180

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Step 3: Draw the asymptotes for large values

of K.

At point

draw radial lines at the (n-m) distinct angles

67.203

)044(0

mn

zps

ii

;300;180;60

,....,2,1,)1(360180

321

mnlmn

ll

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Pole and Zero Plot

j

0= pole

= zero

4

4

4

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Step 4: Compute the departure angles from the poles

(q is the order of the multiple poles):

where i is the sum of the angles to the remainingpoles and i is the sum of the angles to all thezeros. lare positive or negative integers.

lq iidep 360180

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Step 4: Compute the arrival angles at zeros (q is the

order of the multiple zeros):

where i is the sum of the angles to all the poles

and i is the sum of the angles to all theremaining zeros. lare positive or negativeintegers.

lq iiarr 360180

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Root-locus Design MethodGuidelines for sketching a Root Locus:

Step 4: take the pole #2 at4 +4j, the angles to theremaining poles are 90 and 135, respectively.

Set l= - 1, we obtain:

Based on the complex conjugate symmetry, thedeparture angle at pole4-4j is 45. Thedeparture angle for the pole in the origin is 180.

(No zero here, hence no arrival angle calculation)

45180225

)1(*360180)13590(0

360180_2

liidep

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Pole and Zero Plot

Test point S0 is close to

the pole.

j

0= pole

s0

4

4

4

1

2

3

= zero

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Pole and Zero Plot

j

0= pole

= zero

4

4

4

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Step 5: estimate the cross-over points on the

imaginary axis.The characteristic equation is:

0328

0]16)4[(

1

23

2

Ksss

ss

K

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Root-locus Design MethodGuidelines for sketching a Root Locus:

Step 5: estimate the cross-over points on theimaginary axis.

Build Routh array:

Cross-over at K=2560:

08

256:

8:

321:

0

1

2

3

Ks

Ks

Ks

s

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Step 5: estimate the cross-over points on theimaginary axis.

Asymptote cross-over points based on

trigonometric relationships:

(2.67*tan60)j=4.62j. Conjugate: -4.62j

66.5

0256)(32)(8)(

0

0

2

0

3

0

jjj

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Pole and Zero Plot

j

0= pole

= zero

4

4

4

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Step 6: estimate the locations of multiple

closed-loop roots, and find the arrival anddeparture angles at these locations.

We can find K at breakaway point by taking a

derivative:

0)(1

)(

;0)1

(

2

0

ds

dba

ds

dab

bb

a

ds

d

Gds

dss

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ME322Module 6, Slide 29 Northern Illinois UniversitySummer 2005 Peter A. LinCopyright 2001-2005



Step 6:

Rule of thumb: Two locus segments comingtoward each other will break away with a+/- 90 change of direction. 3 locussegments will approach at 120 wrt each

other and depart with a 60 changerelative to the arrival angles.

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Step 7: complete the sketch. The root locus

branches start at poles and end at zeros orinfinity.

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Pole and Zero Plot

j

0

= pole

= zero

4

4

4

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Applications of the Guidelines

Consider the root locus of the characteristic

equation 1+KG(s)=o, where

Construct the root locus for various values ofP.

)(

1)(

2 pss

ssG

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Applications of the Guidelines

Case 1: Large values of P.

2

1)(

s

ssG

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Step 1: mark poles & zeros

j

0

= pole

= zero

1

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Step 2: draw locus on real axis to the left of an

odd number of poles plus zeros:j

0

= pole

= zero

1

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Step 3: Find the number of asymptotes and draw

them

number of asymptotes = n-m=2 - 1=1

does not matter for there is only one asymptote.

(its the root locus on the real axis)

180

12

)11(360180

,....,2,1,)1(360180

1

mnlmn

ll

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Step 4: Calculate departure and arrival angles

j

0

= pole

= zero

1

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Draw a small circle around the

double poles.

180360180)180180(

360180

;90;90;360180)00(2

360180

_2_1

l

lq

l

lq

arr

iiarr

depdepdep

iidep

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j

0

= pole

= zero

1

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Root-locus Design MethodStep 5: Cross-over on the imaginary axis

The characteristic equation is

Build Routh array:

For K>0, all roots are in LHP and do not crossimaginary axis.

02 KKss

0:

0:

1:

0

1

2

Ks

Ks

Ks

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Root-locus Design MethodStep 6: estimate the locations of multiple roots, and find the

arrival and departure angles at these locations.

2;0

;0)2(;01*2*)1(

1;1)(

2;)(

0)(1

2

2

2

ss

sssssds

dbssb

sds

dassa

ds

dba

ds

dab

b

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Step 7: Sketch the locus

j

0

= pole

= zero

1

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Basic Properties of Feedback

Homework Assignment

Draw the root locus for the following:

1).

2).

)5)(1(

4)(

ss

sG

2

)3(

)( s

s

sG

me322 module 6i

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