Topic #10 Higher-order Systems Reference textbook : Control - PowerPoint PPT Presentation

ME 779 Control Systems Topic #10 Higher-order Systems Reference textbook : Control Systems, Dhanesh N. Manik, Cengage Publishing, 2012 1

Control Systems: Higher-order Systems Learning Objectives • General closed-loop response function • Pole-zero map -overdamped closed-loop poles -critically damped closed-loop poles -undamped closed-loop poles -negatively underdamped closed-loop poles -negatively overdamped closed-loop poles • Transient response using residues • Example 2

Control Systems: Higher-order Systems General closed-loop response function C ( s ) G ( s )   R ( s ) 1 G ( s ) H ( s ) Open-loop    K s ( z )( s z )...( s z ) transfer function  1 2 m G s H s ( ) ( )    ( s p )( s p )...( s p ) 1 2 n    G s s ( )( p )( s p )...( s p ) C s ( )  1 2 n        R s ( ) ( s p )( s p )...( s p ) K s ( z )( s z )...( s z ) 1 2 n 1 2 m 3

Control Systems: Higher-order Systems Pole-zero map Overdamped closed-loop poles Pole-zero map Normalized response (b) (a) 4

Control Systems: Higher-order Systems Pole-zero map Critically damped closed-loop poles Pole-zero map Normalized response (a) (b) 5

Control Systems: Higher-order Systems Pole-zero map Underdamped closed-loop poles Normalized response Pole-zero map (b) (a) 6

Control Systems: Higher-order Systems Pole-zero map Undamped closed-loop poles Pole-zero map Normalized response (a) (b) 7

Control Systems: Higher-order Systems Pole-zero map Negatively underdamped closed-loop poles Pole-zero map Normalized response (a) (b) 8

Control Systems: Higher-order Systems Pole-zero map Negatively overdamped closed-loop poles Pole-zero map 9

Control Systems: Higher-order Systems Transient response using residues    R s G s s ( ) ( )( p )( s p )...( s p )  1 2 n C s ( )        ( s p )( s p )...( s p ) K s ( z )( s z )...( s z ) 1 2 n 1 2 m n b a    i Step response C s ( )  c s s p  i 1 i 10

Control Systems: Higher-order Systems Transient response using residues    G s s ( )( p )( s p )...( s p )    c 1 2 n a ( s p )          i i s ( s p )( s p )...( s p ) K s ( z )( s z )...( s z ) c  1 2 n 1 2 m s pi    G s s ( )( p )( s p )...( s p )   1 2 n b s          s ( s p )( s p )...( s p ) K s ( z )( s z )...( s z )  1 2 n 1 2 m s 0 c p t n    i c t ( ) b a e i  i 1 11

Control Systems: Higher-order Systems Example K K   2 H s ( ) 1 K 1 =1 K 2 =5 G s ( )   ( s 1) ( s 5) K K K   1 2 G s H s ( ) ( )     ( 5)( 1) ( 5)( 1) s s s s     C s ( ) K s ( 1) c p 3 j  1 1    R s ( ) ( s 5)( s 1) K K 1 2    c p 3 j   C s ( ) ( s 1) s 1 2        2 R s ( ) ( s 5)( s 1) 5 s 6 s 10 12

Control Systems: Higher-order Systems Example Unit step input  s 1   ( 1) s C s ( )      b s 0.1     2 6 10 s s s   2 6 10 s s s  s 0    ( s 1) ( 2 j )         a ( s 3 j ) 0.05 0.35 j     1   2 ( 3 j )(2 ) j s s 6 s 10    s 3 j    ( s 1) ( 2 j )         a ( s 3 j ) 0.05 0.35 j      2   2 ( 3 j )( 2 ) j s s 6 s 10    s 3 j 13

Control Systems: Higher-order Systems Example     0.1 ( 0.05 0.35 ) ( 0.05 0.35 ) j j    C s ( )     s s 3 j s 3 j      3 t c t ( ) 0.1 e ( 0.1cos t 0.7sin ) t 14

Control Systems: Higher-order Systems Example K 1  1 ( s 5)   G s ( )      1 2 K K K s 5 s 9    1  2  1 1       5 1 ( 5) s s s 1 1 1       s ( ) 0.9 K lim G s ( )  e p 1 1  9 1 K s 0  1 p 9      3 t c t ( ) 0.1 e ( 0.1cos t 0.7sin ) t 15