hwsln-421-s06-09

hwsln-421-s06-09 - ECE 421, Spring 2006 Solutions to HW...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 421, Spring 2006 Solutions to HW Assignment #9 Problem #B-7-9 G p ( s ) = 5 s (0 . 5 s + 1) = 10 s ( s + 2) (1) The phase shift of G p ( s 1 ) is − 210 ◦ = +150 ◦ , so the compensator’s phase shift at s 1 must be 6 G c ( s 1 ) = 30 ◦ . Since this angle is positive, a lead compensator is needed, as indicated in the problem statement. The pole/zero placement procedure presented in the text will be used, which yields the following calculation. The speci f ed point for the dominant closed-loop pole is s 1 = − 2 + j 2 √ 3 = − 2 + j 3 . 4641 . 6 ( s 1 ) = tan − 1 μ 3 . 4641 − − 2 − ¶ = 120 ◦ (2) 6 ( s 1 − z c ) = 120 + 30 2 = 75 ◦ , 6 ( s 1 − p c ) = 120 − 30 2 = 45 ◦ d zc = Im ( s 1 ) tan ( 6 ( s 1 − z c )) = 0 . 9282 , d pc = Im ( s 1 ) tan ( 6 ( s 1 − p c )) = 3 . 4641 The compensator zero is placed at s = − 2 . 9282 , and the compensator pole is placed at s = − 5 . 4641 . The compensator gain is computed so that the series combination of the plant and lead compensator satis...
View Full Document

This note was uploaded on 10/25/2010 for the course ECE 421 taught by Professor Cook,g during the Spring '08 term at George Mason.

Page1 / 3

hwsln-421-s06-09 - ECE 421, Spring 2006 Solutions to HW...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online