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 ( )  c ( s p ) 1   (j z )   1 G (j )   c (j p ) 1 High pass filter Comparable to a PD controller             1   1   tan tan , p z 1 1     z p 1 1 2

Lead Compensators R Electrical lead compensator  1 Z jR C   1 1 1 1 R   1 Z R   eq 2 jR C 1 1 1    jR R C R R  1 2 1 2 1   jR C 1 1 1 E  in i Z eq 3

Lead Compensators Electrical lead compensator     E R jR C 1   in 2 1 1 E iR    o 2   jR R C R R R C 1 2 1 1 2 1 1 1  R          E 1 j 2 1   o   2 1  2 R R        E 1 j 1 2 i 1 2             1 1 tan ( ) tan ( ) 1 2 1 2 4

Lead Compensators Mechanical lead compensator           c c x t ( ) k k x t ( )   1 1 1 1         0            c c y t ( ) k k k y t ( ) 1 1 1 1 2         k c s k c s X s ( ) 1 1 1 1     0         k c s k k c s Y s ( ) 1 1 1 2 1 5

Lead Compensators Mechanical lead compensator X(s)=Y(s ): rigid body motion  Y s ( ) k c s  1 1 c     1 X s ( ) k k c s 1 1 2 1 k 1  k   1 1       Y s ( ) 1 s  2     k k  2  1  , 1 2    1 2    X s ( ) 1 s 1 2 6

Lead Compensators Mechanical lead compensator         Y ( ) 1 j      2  1  ,      1 2    X ( ) 1 j 1 2 7

Lead Compensators Frequency response  E E    o 2 1 o 1 High frequencies Low frequencies  E E in 1 in Between lower and higher frequencies, the following variations take 1  , the magnitude of the lead place. Up to a frequency given by 1 20log  2 compensator transfer function in decibels is given by  , which is 1 1  negative. Then it keeps on increasing till a frequency , and then 2 becomes zero decibels . 8

Lead Compensators Example Assume τ 1 =10 and τ 2 =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 of maximum phase angle. 9

Lead Compensators Frequency response Magnitude of the transfer Example function of a lead     compensator 10 , s 1 s 1 2    20log 1 0 dB maximum      1     2    10log 10log 10 dB      10 1      1     2    20log 20log 20 dB      10 1 Minimum 10

Lead Compensators Phase angle of the transfer Example function of a lead compensator 11

Lead Compensators Example Compare a lead compensator with a PD controller. Use the following values: τ 1 =10 and τ 2 =1 for the lead compensator and K p =1 and T d =1 for the PD controller.   G ( ) s K (1 T s ) PD p d 12

Lead Compensators Example 13

Lead Compensators Maximum phase angle          1 1 tan ( ) tan ( ) 1 2     1 1            1 1 tan tan        m 1   2   d       1 2 1 2 1 2         2    2 d 1 1       1 2         1 1 1 2 tan tan           2 1            1 1 1 tan tan   1  1     2 2   m    1 2 1      1   1 2 tan tan     2   1 14

Lead Compensators Maximum phase angle      m 1 2        sin sin cos cos sin 1 2 1 2 m      1 1 1 sin    2 1      1 m 2 sin    m 2   1  1 sin 1 1  2 m 2 15

Lead Compensators Lead compensator design (phase margin) 1. Determine the phase margin φ u of the uncompensated feedback transfer function, whose phase margin has to be increased. 2. Decide on the expected value of the compensated system φ e that is application dependent and give an additional margin of 5 o . 3. Based on the first two steps, select a maximum value of phase angle φ m that has to be introduced by the lead compensator:       0 5 m e u 4. Using the φ m from step 3, determine the ratio of time constants    1 sin   1 m   1 sin 2 m 16

Lead Compensators Lead compensator design (phase margin) 20log  2 5. The magnitude of the lead compensator varies from  1 1 1  and to 0 between  rad/s. Therefore, using the ratio 1 2 of time constants from step 4, determine the frequency at which the decibel gain of the uncompensated system is 10log  1  and use this as ω m. This will be the frequency 2 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. 17

Lead Compensators Lead compensator design (phase margin) 6. 1   2   Using equation m , determine the actual values 1 2 of time constants and then design other the elements of the lead compensator 18

Lead Compensators Example The open-loop transfer function of a unity feedback system is given by 100  G s ( )   s s ( 2)( s 4) (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 25 o (c) Draw root-locus plots of the uncompensated and compensated systems (d) Determine the bandwidth for compensated and uncompensated systems 19

Lead Compensators 100  G s ( )   s s ( 2)( s 4) 100   G j ( )      j ( j 2)( j 4) 100          2 2 6 j 8 100 G j   ( )      6 4 2 20 60 20

Lead Compensators Bode plot for the compensated and uncompensated Example open-loop transfer function of equation The uncompensated system is unstable 21

Lead Compensators Example Phase margin of the uncompensated system, φ u can be obtained as -18 o Since the gain margin is negative, the given uncompensated system is unstable Since the expected value of gain margin for the compensated system is 25 o , the maximum phase angle φ m of the lead compensator is 25+5-(-18)=48 0.     1 sin 1 sin 48    1 m 6.79     1 sin 1 sin 48 2 m 22

Lead Compensators Example     1 10log 10log6.79 8.3 dB 2     100   20log          6 4 2 20 64      100 ω m =2.52 rad/s   20log 8.3       6 4 2  20 64   m m m 23

Lead Compensators Example 1   2   m 1 2 1 1     0.15 s   2 2 2.52 6.79   2 1 m 2     0.15 6.79 1.03 s 1 24

Lead Compensators K Example  G s ( )   s s ( 2)( s 4)       1 s 100   2  1  G s G s ( ) ( )      c    1 s s s ( 2)( s 4) 1 2    100 (1 0.15 ) s         6.79 s s ( 2)( s 4)(1 1.03 ) s   1    K s     1 G s G s ( ) ( )   c 1      s s ( 2)( s 4) s    2 25

Lead Compensators K Example  G s ( )   s s ( 2)( s 4) Root-locus plot of the uncompensated unity feedback system 26

Lead Compensators Example Root-locus of the system compensated by a lead compensator   1    K s     1 G s G s ( ) ( )   c 1      s s ( 2)( s 4) s    2 27