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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 Fall '08
 Hjortso,M
 WCO, Closedloop response

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