Assignment 5 Solutions 2009

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern