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Unformatted text preview: o , by = 0 . 707. Lead compensator is needed. We compute value for the lead compensator as = 1sin(70 o ) 1 + sin(70 o ) = 0 . 0311 . The zero and pole of lead compensator are z lead = c = 8 . 8176 , p lad = c / = 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 c = 50 is G =10log 10 ( ) =15dB . To compute gain K , we have (recall c = 50) K  D lead ( j c ) G ( j c )  . 02 K/ = 1 . Hence K = 50 = 8 . 8176. Now the position constant is: K p = 10 K = 88 . 176 50 . No lad compensator is necessary. 3...
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 Fall '10
 GU

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