Assignment 5 Solutions 2009

2 c k 1 3600 2 1 25 2 w w 2 1 3600 1 25 2 w w

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Unformatted text preview: OL (iw = 1.2 c e -10 iw ) k iw (1 - 60 ) iw (1 - 5 ) ´ (1 + 3600 w ) (1 + 25 w ) 2 2 Now determining the amplitude ratio: 2 2 AR = Re g OL ) + Im g OL ) ( ( AR = 1.2 c k 1 + 3600 2 1 + 25 2 w w 2 1 + 3600 1 + 25 2 w w ( )( ) Now calculat ing the phase angle: ( OL æ Im g ) ö F = tan -1 ç ç Re g ) ÷ ÷ è ( OL ø which can be wr itten as, ) ) ) F = tan - 1 (- t p1 w + tan -1 (- t p 2 w + (- t d w Where t p1 = 60 , t p 2 = 5 and t d = 10 . F = tan -1 (- 60 + tan -1 (- 5 + (- 10 w) w) w) When AR = 1 and F = -p , which gives cross over frequency (wco). Equations for AR and F beco me: 1 = 1 2 c . k 1 + 3600 2 1 + 25 2 w w 2 1 + 3600 1 + 25 2 w w ( )( ) - p = tan -1 (- 60 + tan -1 (- 5 + (- 10 w) w) w) Solving these simultaneously for kc and w, we get: kc = 7.3 (this is the ult imate kc) and wco = 0.123 Now we can determine the ult imate period: 2p 2 p Pu = = = 51 w 0 123 . co The next step is to determine the Z­N controller settings using the ult imate gain and the ult imate period: kc t I ­ 42.5 25.5 3.65 3.32 4.29 P PI PID t d ­ ­ 6.375 We o ffer another solut ion using direct substitution. We have for the closed­loopsystem, g CL = 1 + g c g p = 0 g CL = 1 + k c 1 2 -10 s . e = 0 (60 s + 1 (5 s + 1 ) ) (60 s + 1)(5 s + 1) + k c 1. 2 e -10 s = 0 Using a first­order Pade approximat ion for the deadtime, e -10 s » 1 - 5 s 1 + 5 s Subst ituting this in the main equation and rearranging, we get: (60 s + 1 (5 s + 1 (5 s + 1 + k c (1 - 5 s ) = 0 ) ) ) (300 s 2 + 65 s + 1) (5 s + 1) + 1. 2 k c (1 - 5 s ) = 0 (1500 s 3 + 625 s 2 + 70 s + 1) + 1. 2 k c (1 - 5 s ) = 0 Using the direct substitution method to solve the above, we apply s = iw (1500 (iw ) 3 ) + 625 iw 2 + 70 iw + 1 + 1 2 c (1 - 5 iw ) = 0 . k () () () - 1500iw - 625 2 + 70iw + 1 + 1. c - 6 c iw = 0 3 w 2k k ( ) - 625 2 + 1 + 1. c + - 1500 + 70 - 6 c w i = 0 w 2 k w3 w k We can now extract the...
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