MODERN CONTROL SYS-LECTURE V.pdf
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Transcript of MODERN CONTROL SYS-LECTURE V.pdf
MODERN CONTROL SYSTEMS ENGINEERING
COURSE #: CS421
INSTRUCTOR: DR. RICHARD H. MGAYA
Date: October 25th, 2013
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
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
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
)
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
Digital Compensator Design
Digital Compensator Design Using Pole Placement Cont…
Solution:
d) D(z) = K = 0.336
Dr. Richard H. Mgaya
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
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
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
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
Digital Compensator Design
Digital Compensator Design Using Pole Placement Cont…
Solution:
Dr. Richard H. Mgaya
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
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
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
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
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
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
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
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
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
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
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