MCLECTURE11.PPT

7/27/2019 MCLECTURE11.PPT

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10/30/2013 (C) 2001, Ernest L. Hall, University of Cincinnati 1

ManufacturingControls

FALL 2001

Lecture 11


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Syllabus

DATE TOPIC NOTES 1. Sep. 20 Mechatronics Design Process Ch. 1

2. Sep. 25 System Modeling and Simulation Ch. 2

3. Sep. 27 Laplace Transforms and Transfer Functions Ch. 2

4. Oct. 2 Electrical Examples Ch.2, Notes 5. Oct. 4 Mechanical Examples Ch.2, Notes

6. Oct. 9 More Examples, Thermal and Fluid Examples, QUIZ 1(Take Home)

7. Oct. 11 Sensors and Transducers Ch. 3

8. Oct. 16 Digital control, Advanced MATLAB

9. Oct. 18 Analog and Digital Sensing Ch. 3,

Notes 10. Oct. 23 Actuating Devices, time and frequency response Ch. 4

11. Oct. 25 DC Motor Model Ch. 4,Notes

12. Oct. 30 Boolean Logic Ch. 5

13. Nov. 1 Programmable Logic Controllers Ch. 5, Notes

14. Nov. 6 Stability and Compensators, P, PI and PD Ch. 6

15. Nov. 8 PID Controllers Ch. 7

16. Nov. 13 QUIZ 2(In Class - Open Book)

17. Nov. 15 Practical and Optimal Compensator Design Ch. 8 18. Nov. 20 Frequency Response Methods Ch. 9, Notes

19. Nov. 22 THANKSGIVING HOLIDAY Ch. 9, Notes

20. Nov. 27 Optimal Design of a Motion Control System Ch. 9, Notes

21. Nov. 29 QUIZ 3(In Class - Closed Book)

22. Dec. FINAL EXAM (In Class - Closed Book) Comprehensive


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Todays objective

To continue the introduction tosystems theory by continuingthe concept of frequencyresponse for digital control forcompensation of a feedback

control for the motorized arm. By the end of this class you

will be able to describe theadvantage of using both timeand frequency response fordescribing both the actuatingdevice and the control

compensation of a motorizedsystem.

Understand first and secondorder systems.


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Example

Suppose a DC motor is used to drive a robot

arm horizontally.

Mg

x

y

L

Figure 12. A single joint robot arm driven by an armature-controlled DC motor

horizontally

z


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Frequency Response and Time

response

Permit descriptions with greater clarity

Time response also important

Need both


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Sinusoids are eigenvectors of

linear systems

That is, if a sinusoid is put into a linear system

a sinusoid will be the output. It may be changed

only in magnitude and phase.

Linear Systemsin Msin(


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o get frequency response Substitute s=jw into

transfer function to get

frequency response

H(s=jw)sinwt Msin(wt


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Time response

The output of a system is the sum of two

responses:

Forced response or steady state response or

particular solution

Natural response or homogeneous solution


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Second order systems

general form

G(s) = a/(s2+as+b)R(s)=1/s C(s) 2

4

0

rootsforequationsticcharacteriSolve

2

2

baas

bass


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Stable forms of second order

system

Look at signs of

coefficients with a>0

and b> 0

Which of these areabsolutely stable, i.e.

have poles in the left

half plane?

0

0

0

0

0

0

0

0

2

2

2

2

2

2

2

2

bass

bass

bass

bass

bass

bass

bass

bass


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Routh-Hurwitz Criterion

A systematic method for determining if the

characteristic equation has poles in the rhp,

lhp or on the jwaxis is the Routh Hurwitz

criteria.

It requires making a table

Examining the table for sign changes in the

first colums


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Example for

a4s4+a3s

3+a2s2+a1s+a0

s4 a4 a2 a0

s3 a3 a1 0

s2 b1 b2 0

s1 c1 0 0

s0 d1 0 0


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Second column

1

1

21

1

1

21

13

1

3

03

24

1

0

c

c

bb

d

b

bb

aa

c

a

aa

aa

b


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Third column

1

1

1

2

1

1

3

2

3

3

04

2

0

0

0

0

0

c

c

b

d

b

b

a

c

a

a

aa

b


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Exampleunity feedback system

G(s) = 1000/((s+2)(s+3)(s+5))R(s) C(s)


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First step- form closed loop response

G(s) = 1000/

(s3+10s2+31s+1030))R(s) C(s)


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Example for s3+10s2+31s+1030

s3 1 31 0

s2 10 1030 0

s1 -72 0 0

s0

103 0 0


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Any row can be multiplied by a positive

constant without changing the results

s3 1 31 0

s2 1 103 0

s1 -72 0 0

s0

103 0 0


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Computation

10372

072

1031

721

1031

311

1

1

c

b


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Basic rule

The number of roots of the polynomial that

are in the right half plane is equal to the

number of sign changes in the first column.

In this example, there are two sign changes

indicating two poles in the right half plane and

an unstable system.


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Check with Matlab

num = [0,0,0,1];

den = [1,10,31,1030];

sys=tf(num,den);

pzmap(sys)

Real Axis

ImagAxis

Pole-zero map

-14 -12 -10 -8 -6 -4 -2 0 2-10

-8

-6

-4

-2

0

2

4

6

8

10


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Back to second order system


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Stable forms of second order

system

Look at signs of

coefficients with a>0

and b> 0

Which of these areabsolutely stable, i.e.

have poles in the left

half plane?

0

0

0

0

0

0

0

0

2

2

2

2

2

2

2

2

bass

bass

bass

bass

bass

bass

bass

bass


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Construct Routh Hurwitz table

s2 1 b 0

s1 a 0 0

s0 b 0 0

02 bass


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Conclusion

Either a or b or both < 0

will cause at least one

sign change in the first

column and lead to anunstable system

Only stable second

order systems are 02 bass

02 bass


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Response determined by pole locationsroots of characteristic equation

Real unequal polesoverdamped system

Complex roots on jwaxisUndampedsystem

Complex roots not on jwaxisUnderdamped

system Real and equal rootsCritically damped

system


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Consider the step responsepoles

real and unequal - overdamped from Nice pp. 168

G(s) = 9/(s2+9s+9)R(s)=1/s C(s)

Real Axis

ImagAxis

Pole-zero map

-8 -7 -6 -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

1

num = [0,0,1];

den = [1,9,9];sys=tf(num,den);

pzmap(sys)

Poles at: -7.854

and -1.146

Time (sec.)

Amplitude

Step Response

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.02

0.04

0.06

0.08

0.1

0.12From: U(1)

To:Y(1)


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Nowpoles complex conjugates -

underdampedfrom Nice pp. 168

G(s) = 9/(s2+2s+9)R(s)=1/s C(s)

Real Axis

ImagAxis

Pole-zero map

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-3

-2

-1

0

1

2

3

num = [0,0,1];

den = [1,2,9];

sys=tf(num,den);

pzmap(sys)

Time (sec.)

Amplitude

Step Response

0 1 2 3 4 5 60

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16From: U(1)

To:Y(1)


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Nowpoles complex conjugates on jaxis- undampedfrom Nice pp. 168

G(s) = 9/(s2+9)R(s)=1/s C(s)

Real Axis

ImagAxis

Pole-zero map

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-3

-2

-1

0

1

2

3

num = [0,0,1];

den = [1,0,9];

sys=tf(num,den);

pzmap(sys)

Time (sec.)

Amplitude

Step Response

0 5 10 15 200

0.05

0.1

0.15

0.2

0.25From: U(1)

To:Y(1)


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Nowpoles real and equal- critically

dampedfrom Nice pp. 168

G(s) = 9/(s2+6s+9)R(s)=1/s C(s)

Real Axis

ImagAxis

Pole-zero map

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

num = [0,0,1];

den = [1,6,9];

sys=tf(num,den);

pzmap(sys)

Time (sec.)

Amplitude

Step Response

0 0.5 1 1.5 2 2.5 3 3.5 4 4.50

0.02

0.04

0.06

0.08

0.1

0.12From: U(1)

To:Y(1)


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Summary for second order

systems

Overdamped response

sluggish

Poles: two real poles at

s1, -s2

Natural response: two

exponentials with time

constants equal to the

reciprocal of the polelocations

ttKKtc 21 21)(

ss


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Underdamped response

Poles: Two complex poles

at: -sd+ andjwd

Natural response: Damped

sinusoid with an exponential

envelope whose timeconstant is equal to the

reciprocal of the poles real

part

The radian frequency of thesinusoid is equal to the

imaginary part of the poles

)cos()( w s

tAtc dtd


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Undamped response

Poles: Two imaginary

poles at + and = wl

Natural response:

Undamped sinusoidwith radian frequency

equal to the imaginary

part of the poles

)cos()( w tAtc l


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Critically damped response

Poles: Two real, equalpoles atsl

Natural response: One termis an exponential whosetime constant is the

reciprocal of the polelocation. Another term is theproduct of time, t, and anexponential whose timeconstant is equal to thereciprocal of the pole

location. Note this is the fastest

response without overshoot.

tt tKKtc 11 21)( ss


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General second order system


are so common that a

special notation has

been adopted todescribe them.

ratiodampingtheis

frequencynaturaltheis

where

bb

aa

So

ssbass

n

nn

n

n

nn

w

ww

ww

ww

,

2,2

2

2

222


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Homework due next Thursday,

Nov. 1, 2001

For the following systems,

Determine the pole-zero plot

Step response

And determine if the systems are: Overdamped

Underdamped

Undamped

Critically damped Determine the damping ratio and natural frequency

when appropriate for each


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Homework systems

208

20

)(

)(

168

16

)(

)(

128

12

)(

)(

2

2

2

sssR

sC

sssR

sC

sssR

sC


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MCLECTURE11.PPT