hw5_sol - Lead-Lag Homework Solution 3 a) Gc ( s ) K 1.5 1...

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Lead-Lag Homework Solution a) 3 G c s () K 1.5 1 0 2 2 2 ) ( n n d s s s ϖ δϖ + + = 9 3 ) ( 2 + + = s s s d 2 2 ) 2 3 3 ( ) 2 / 3 ( + + = s Root Locus does not go through closed loop poles b) Ys () Rs = K10 ()sa ss b s 1 ( ) 10 K s a a 27 13 10 ) 10 ( ) 1 ( ) ( 2 3 aK s K b s b s s + + + + + = -4 -3 -2 -1 0 1 2 -5 -4 -3 -2 -1 0 1 2 3 4 5 Real Axis Imag Axis c b
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= ) ( s d sc () s 2 3s 9 = ) ( s d s 3 s 2 3c s9 3c 9c a 27 13 b5 c3 K 1.3 10 aK 9 c b1 0 K 93 c c) Ys = 13 s 27 ss 3 s 2 9 = 1 s + 4 9 s3 + As B s 2 9 1 ()s 3 s 2 9 4 9 s 3 2 9s As B s 2 13 s 27 s 3 1 4 9 A 0 A 13 9 s 2 6 4 3 3A B 0 B3 s 18 4 3B 13 = 1 s + 4 9 - 13 9 s 27 13 s 3 2 2 3 3 2 . 2 . = 1 s + 4 9 - 13 9 s 3 2 s 3 2 2 3 3 2 . 2 . - 13 9 2 33 544 . 3 . 3 2 s 3 2 2 3 3 2 . 2 .
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yt () = 1 + 4 9 exp 3 t . - 13 9 exp 1.5 t cos 3 3 2 t . - .302 exp 1.5 t ( ) sin 3 3 2 t d) Steady state error Es = 1 1G c s ()G p s Rs . = 1 s 0 lim s SSE = sR s 1 c s p . s . = 1 c 0 p s = 1 1 10 s1 1.3 27 13 . 5 . + 1 1 = = 0 -10 -8 -6 -4 -2 0 2 0 2 4 6 8 10 Real Axis Imag Axis 2) Angle Condition o 180 } ) 8 )( 1 ( 10 2 { 2 25 . 1 - = + + + + + - = j s s s s b s s angle
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o 14 2 2 tan 1 + = - - b b1 0 Magnitude Condition 1 ) 8 )( 1 ( ) 10 ( 10 ) 2 ( 2 25 . 1 = + + + + + - = j s s s s s s K K = 14 3) Neglect Log when designing Lead angle angle condition to find "b" o 180 } ) 8 )( 2 ( 10 3 { 3 2 2 - = + + + + + - = j s s s s b s s angle 877 . 15 ) 2 ( 249 . 46 . 3 14 2 3 2 tan 1 = - = - =
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This note was uploaded on 03/18/2008 for the course EML 4312 taught by Professor Dixon during the Fall '07 term at University of Florida.

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hw5_sol - Lead-Lag Homework Solution 3 a) Gc ( s ) K 1.5 1...

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