Topic #58 Lead Compensators Reference textbook : Control Systems, - - PowerPoint PPT Presentation
ME 779 Control Systems Topic #58 Lead Compensators Reference textbook : Control Systems, Dhanesh N. Manik, Cengage Publishing, 2012 1 Lead Compensators Magnitude and phase Bode plot of a lead compensator ( ) s z 1 G s ( )
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Reference textbook:
Control Systems, Dhanesh N. Manik, Cengage Publishing, 2012
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Magnitude and phase Bode plot of a lead compensator
1 1
( ) ( ) ( )
c
s z G s s p
1 1 1 1 1 1
tan tan , p z z p
1 1
(j ) (j ) (j )
c
z G p
High pass filter Comparable to a PD controller
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Electrical lead compensator
1 1 1 1
1 R Z jR C
1 2 1 1 1 2 1 2 1 1 1
1 1
eq
R Z R jR C jR R C R R jR C
in eq
E i Z
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Electrical lead compensator
2 1 1 2 1 2 1 1 2
1
in
jR C E iR jR R C R R
1 1 1 2 1 2 1 2
R C R R R
2 1 1 2
1 1
E j E j
1 1 1 2
tan ( ) tan ( )
1 2
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Mechanical lead compensator
1 1 1 1 1 1 1 1 2
( ) ( ) ( ) ( ) c c k k x t x t c c k k k y t y t
1 1 1 1 1 1 1 2 1
( ) ( ) k c s k c s X s k c s k k c s Y s
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X(s)=Y(s): rigid body motion
1 1 1 2 1
( ) ( ) k c s Y s X s k k c s
Mechanical lead compensator 1 1 1 1 1 2 1 2
c k k k k
2 1 1 2 1 2
1 ( ) , ( ) 1 s Y s X s s
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2 1 1 2 1 2
1 ( ) , ( ) 1 j Y X j
Mechanical lead compensator
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Frequency response
2 1
1
E E
Low frequencies
1
E E
High frequencies
Between lower and higher frequencies, the following variations take
1
1 , the magnitude of the lead
compensator transfer function in decibels is given by
2 1
20log , which is
2
1
, and then becomes zero decibels.
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Assume τ1=10 and τ2=1, for a lead compensator and plot magnitude and phase of the transfer function Determine the frequency at which the phase angle becomes a maximum and the value
Example
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Frequency response
Magnitude of the transfer function of a lead compensator
1 2
2 1
1 20log 20log 20 10 dB
Example
20log 1 0dB
2 1
1 10log 10log 10 10 dB
maximum Minimum
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Phase angle of the transfer function of a lead compensator
Example
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Compare a lead compensator with a PD
and τ2=1 for the lead compensator and Kp=1 and Td=1 for the PD controller.
( ) (1 )
PD p d
G s K T s
Example
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Example
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Maximum phase angle
1 2 2 2 1 2
1 1 d d
1 1 1 2
tan ( ) tan ( )
2 1 2
1
m
1 1 1 2 1 2 1 2 1 1 1 2 2 1
1 1 tan tan tan tan
m
1 1 1 1 2 1 1 2 2 1
tan tan 1 tan tan
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Maximum phase angle
1 2 m
1 2 1 2
sin sin cos cos sin
m
1 2 2 2 1 2
1 1 sin 1 1
m
1 2
1 sin 1 sin
m m
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Lead compensator design (phase margin)
function, whose phase margin has to be increased.
application dependent and give an additional margin of 5o.
that has to be introduced by the lead compensator:
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m e u
1 2
1 sin 1 sin
m m
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5.
The magnitude of the lead compensator varies from
2 1
20log
to 0 between
1
1 and
2
1 rad/s. Therefore, using the ratio
which the decibel gain of the uncompensated system is
1 2
10log and use this as ωm. This will be the frequency
at which the compensated system crosses the unit circle. Since this is also the frequency at which maximum phase angle of the lead compensator is obtained, it will result in maximum increase in phase angle.
Lead compensator design (phase margin)
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6. Using equation
2 1 2
1
m
, determine the actual values
Lead compensator design (phase margin)
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The open-loop transfer function of a unity feedback system is given by
100 ( ) ( 2)( 4) G s s s s
(a) Draw a Bode plot of the above open-loop transfer function and determine the phase margin (b) Use a lead compensator and change the phase margin of the compensated system to 25o (c) Draw root-locus plots of the uncompensated and compensated systems (d) Determine the bandwidth for compensated and uncompensated systems
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2 2
100 ( ) ( 2)( 4) 100 ( ) ( 2)( 4) 100 6 8 G s s s s G j j j j j
6 4 2
100 ( ) 20 60 G j
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Bode plot for the compensated and uncompensated
The uncompensated system is unstable
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Phase margin of the uncompensated system, φu can be obtained as -18o Since the gain margin is negative, the given uncompensated system is unstable 1 2
1 sin 1 sin 48 6.79 1 sin 1 sin 48
m m
Since the expected value of gain margin for the compensated system is 25o, the maximum phase angle φm of the lead compensator is 25+5-(-18)=480.
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1 2
10log 10log6.79 8.3dB
6 4 2
100 20log 20 64
6 4 2
100 20log 8.3 20 64
m m m
ωm =2.52 rad/s
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2 2 2 1 2 1
1 1 0.15 2.52 6.79 0.15 6.79 1.03
m
s s
2 1 2
1
m
25
2 1 1 2
1 100 ( ) ( ) 1 ( 2)( 4) 100 (1 0.15 ) 6.79 ( 2)( 4)(1 1.03 )
c
s G s G s s s s s s s s s s
( ) ( 2)( 4) K G s s s s
1 2
1 ( ) ( ) 1 ( 2)( 4)
c
K s G s G s s s s s
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Root-locus plot of the uncompensated unity feedback system
( ) ( 2)( 4) K G s s s s
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Root-locus of the system compensated by a lead compensator
1 2
1 ( ) ( ) 1 ( 2)( 4)
c
K s G s G s s s s s
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( 2)( 4) ( ) ( 2)( 4) 1 ( 2)( 4) K K s s s T s K s s s K s s s
1 2 1 1 2 1 2
1 1 1 ( 2)( 4) ( ) 1 1 1 ( 2)( 4) 1 1 ( 2)( 4)
c
K s s s s s K s T s K s s s s s K s s s s s
uncompensated Compensated
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Closed-loop frequency response of uncompensated and lead compensated systems
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The open-loop transfer function of a unity feedback system is given by
( ) ( 2)( 4) K G s s s s
Design a lead compensator so the dominant closed-loop response has a damping ratio of 0.47 with an undamped natural frequency of 1.7 rad/s. Assume that these values could not have been obtained just by changing the open-loop gain K. That means the closed-loop poles having the above damping ratio and undamped natural frequency are not on the root locus of the above system
(1)
31 2 1,2
0.47 1.7 1 0.47 0.8 1.5 s j
Expected close-loop poles
( 2.5) ( ) ( 2)( )( 4)
c
K s G s s s s p s
Compensated system
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Open-loop poles and zeros of the compensated system and the desired closed-loop poles
Available space for compensator
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41 (118 51 25 ) 180
c
27
c
1.5 0.8 3.75 tan 27
c
p
( 2.5) ( ) ( 2)( 3.75)( 4) K s G s s s s s
0.8 1.5 0.8 1.5 2 0.8 1.5 3.75 0.8 1.5 4 17 0.8 1.5 2.5 j j j j K j
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Root locus of the compensated system to obtain the desired closed-loop poles
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