This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: o , by Î¶ = 0 . 707. Lead compensator is needed. We compute Î± value for the lead compensator as Î± = 1sin(70 o ) 1 + sin(70 o ) = 0 . 0311 . The zero and pole of lead compensator are z lead = Ï‰ c âˆš Î± = 8 . 8176 , p lad = Ï‰ c / âˆš Î± = 283 . 524 . 2 Hence we obtain the lead compensator as D lead ( z ) = 1 + s/ 8 . 8176 1 + s/ 283 . 524 . The magnitude increase at Ï‰ c = 50 is âˆ† G =10log 10 ( Î± ) =15dB . To compute gain K , we have (recall Ï‰ c = 50) K  D lead ( jÏ‰ c ) G ( jÏ‰ c )  â‰ˆ . 02 K/ âˆš Î± = 1 . Hence K = 50 âˆš Î± = 8 . 8176. Now the position constant is: K p = 10 K = 88 . 176 â‰¥ 50 . No lad compensator is necessary. 3...
View
Full Document
 Fall '10
 GU
 Harshad number, lead compensator, plad

Click to edit the document details