MODERN CONTROL SYS-LECTURE V.pdf

22
MODERN CONTROL SYSTEMS ENGINEERING COURSE #: CS421 INSTRUCTOR: DR. RICHARD H. MGAYA Date: October 25 th , 2013

Transcript of MODERN CONTROL SYS-LECTURE V.pdf

Page 1: MODERN CONTROL SYS-LECTURE V.pdf

MODERN CONTROL SYSTEMS ENGINEERING

COURSE #: CS421

INSTRUCTOR: DR. RICHARD H. MGAYA

Date: October 25th, 2013

Page 2: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement

Example: Continuous control system (a) is to be replaced by

digital control system (b)

a) Find the value of K that give damping ratio, ζ, of 0.5 for the continuous

system

b) Find the closed-loop bandwidth ωb and make the sampling frequency ωs a

factor 10 time higher. What is the sampling period, T?

Dr. Richard H. Mgaya

Page 3: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

c) Find the open-loop impulse transfer function G(z) when the sample and

hold device is cascaded with the plant

d) With D(z) = K from system (a), plot the responses of the systems

e) Map the closed-loop poles from the s-plane to the z-plane and design the

compensator D(z) such that both continuous and discrete system responses are

identical, i.e., ζ = 0.5

Solution:

a) Root locus

Dr. Richard H. Mgaya

sradjs /866.0 ,5.0or 866.05.0

336.0K

Page 4: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

b) Plotting closed-loop frequency response

• Bandwidth, ωb = 1.29 rad/s

• 10 time higher, i.e., 12.9 rad/s

• Therefore,

Dr. Richard H. Mgaya

sec5.0T

10-2

10-1

100

101

102

-180

-135

-90

-45

0

Phase (

deg)

Bode Diagram

Gm = Inf dB (at Inf rad/sec) , Pm = 89.5 deg (at 1.01 rad/sec)

Frequency (rad/sec)

-80

-60

-40

-20

0

20

System: f

Frequency (rad/sec): 1.28

Magnitude (dB): -3.07

Magnitu

de (

dB

)

Page 5: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

c) Cascaded system

Using transformation pair

Simplify

Dr. Richard H. Mgaya

1

31)(

2

1

ssZzzG

)(1

5.115.03)(

5.0

5.05.0

ezz

ezezG

)6065.0)(1(

)8467.0(3196.0)(

zz

zzG

Page 6: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

d) D(z) = K = 0.336

Dr. Richard H. Mgaya

Page 7: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

e) Mapping closed-loop poles from s to z-plane

• T = 0.5

Magnitude:

Phase:

• Polar to Cartesian

• Characteristic equation:

Dr. Richard H. Mgaya

Tez

779.05.05.0 ez

8.24433.05.0866.0 radTz

327.0707.0 jz

izz ... 0607.0414.12

Page 8: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

• Characteristic equation for cascade compensator D(z) and the plant G(z) is

given as:

• If equation i and ii are to be identical then the compensator is given in the

form:

• Select a such that non unity pole in G(z) is cancelled

• Characteristic equation:

Dr. Richard H. Mgaya

iizGzD ... 0)()(1

bz

azKzD

)()(

)6065.0)(1(

)8467.0(3196.0)6065.0()()(

zz

z

bz

zKzGzD

)1)((

)8467.0(3196.01

zbz

zK

Page 9: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

• Simplify:

• Equating i and iii:

Dr. Richard H. Mgaya

iiibKzbKz ... 0)2706.0()1319.0(2

807.0 15902.0

607.0 2706.0

414.113196.0

K

bK

bK

519.0

607.0)327.02706.0(

b

b

327.0

193.05902.0

K

K

Page 10: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

• Compensator D(z):

• Similarly:

• Difference equation for the control algorithm:

Dr. Richard H. Mgaya

TkuTkekTekTu )1(519.0)1(1983.0)(327.0)(

519.0

)6065.0(327.0

)(

)()(

z

z

zE

zUzD

1

1

519.01

)6065.01(327.0)(

z

zzD

Page 11: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Design Using Pole Placement Cont…

Solution:

Dr. Richard H. Mgaya

Page 12: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane

• Generally, for plant dynamics with slow pole, i.e., near z = 1, a

compensator with a zero is used to cancel the slow pole

Placing a pole with faster response

Hardware may limit the control effort - saturation

• Let the z-domain transfer function of the plant have a pole at

p1, which we wish to cancel, the compensator will be of the

form:

The numerator cancel the pole of the plant and pc is the selected pole

place far away from z = 1

Integrator pole at z = 1 reduces the steady-state error

Dr. Richard H. Mgaya

c

c

pz

pzKzD

)()( 1

Page 13: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

Example: Consider the plant for thermal control system

• Compensate the plant in s-domain and then convert the design

to z-plane by employing the method of bilinear transformation

Solution:

Place a zero p1 at s = - 0.1957 and pc at – 4

Closed-loop characteristic equation

Dr. Richard H. Mgaya

)554.2)(1957.0(

1)(

sssG

4

)1957.0()(

s

sKsD

0)544.2)(1957.0)(4(

)1957.0(1)()(1

sss

sKsDsG

Page 14: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

Closed-loop characteristic polynomial

Undamped natural frequency:

Damping ratio

Let ζ = 0.707

Characteristic equation:

Dr. Richard H. Mgaya

0216.10544.62 Kss

K

216.102

544.6

544.62

216.10

n

n K

259.11K

0475.21544.62 ss

Page 15: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

From the characteristic polynomial undamped and damped natural

frequency are given as:

Proportional Compensator:

Kp : compensator D(s) = Kp:

Since ζ = 0.707, then Kp = 3.28, ωn = 1.94 rad/s and ωd = 1.37 rad/s

Dr. Richard H. Mgaya

rad/s274.3

rad/s63.4

d

n

)( 4

)1957.0(259.11)( a

s

ssD

05.07495.22 pKss

Page 16: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

Sampling interval T = 0.25:

Discrete compensator D(z) from eqn (a):

Simplify:

Zero-order-hold plant transfer function:

Dr. Richard H. Mgaya

1

18

1

12

z

z

z

z

Ts

4)1/()1(8

)1957.0)1/()1(8259.11)(

zz

zzzD

333.0

952.069.7)(

z

zzD

)528.0)(952.0(

)816.0(025.0)(

zz

zzG

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Digital Compensator Design

Compensator Design via z-plane Cont…

Closed-loop discrete time characteristic equation:

Note: The roots are too low for acceptable performance

Alternatively,

Root locus:

Dr. Richard H. Mgaya

333.0

)952.0()( 1

z

zKzD

469.03345.0

0332.0699.0

2,1

2

jz

zz

333.0

)952.0(4)(

z

zzD

Page 18: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

First-Order Compensator for discrete time system:

• Compensated versus proportional control

Dr. Richard H. Mgaya

333.0

)952.0(4)(

z

zzD

Page 19: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Compensator Design via z-plane Cont…

• Alternatively

Cancel the poles of the plant G(z) and insert integrator pole and the

pole for faster response

Compensator gain Kc for ζ = 0.7

Root locus: Kc = 7

Second-Order Compensator for discrete time system:

Dr. Richard H. Mgaya

)2.0)(1(

)952.0)(528.0()(

zz

zzKzD c

)2.0)(1(

)952.0)(528.0(7)(

zz

zzzD

Page 20: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Implementation

• Controller Gc(z) placed in a forward path can be implemented in a

computer through numerical algorithm

• Consider a second-order compensator:

• Cross multiply and solving for terms with highest power of z on X(z):

Dr. Richard H. Mgaya

)(zE )(zX )(zC)(zGp )(zG

Computer

Emulating Compensator

Plant with

Sample-and-Hold

01

2

2

01

2

2

3

3

)(

)()(

bzbzb

azazaza

zE

zXzGc

)()()(

)()(

0101

2

2

3

3

2

2

01

2

2

3

301

2

2

zXbzbzEazazazazXzb

zEazazazazXbzbzb

Page 21: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Implementation Cont…

Division by coefficients of X(z):

Inverse Laplace:

• Present sample of the compensator is the function of future, present and past samples of e*(t) and past output values of x*(t)

Dr. Richard H. Mgaya

)()()( 2

2

01

2

12

2

01

2

1

2

2

2

3 zXzb

bz

b

bzEz

b

az

b

a

b

az

b

azX

)2()(

)2()()()()(

*

2

0*

2

1

*

2

0*

2

1*

2

2*

2

3*

Ttxb

bTtx

b

b

Tteb

aTte

b

ate

b

aTte

b

atx

Page 22: MODERN CONTROL SYS-LECTURE V.pdf

Digital Compensator Design

Digital Compensator Implementation Cont…

• Flow chart of a digital compensator for a3 equals zero

The output can not be dependent on the future value of the input

Dr. Richard H. Mgaya