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# hw5_sol - Lead-Lag Homework Solution 3 a Gc s K 1.5 1 0 d s...

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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) Y s ( ) R s ( ) = K 10 ( ) s a ( ) s s 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 s c ( ) s 2 3 s 9 = ) ( s d s 3 s 2 3 c ( ) s 9 3 c ( ) 9 c a 27 13 b 5 c 3 K 1.3 10 aK 9 c b 10 K 9 3 c b 1 3 c c) Y s ( ) = 13 s 27 s s 3 ( ) s 2 3 s 9 Y s ( ) = 1 s + 4 9 s 3 ( ) + As B s 2 3 s 9 1 ( ) s 3 ( ) s 2 3 s 9 4 9 s 3 3 s 2 9 s As B ( ) s 2 3 s 13 s 27 s 3 1 4 9 A 0 A 13 9 s 2 6 4 3 3 A B 0 B 3 s 18 4 3 B 13 B 3 Y s ( ) = 1 s + 4 9 s 3 ( ) - 13 9 s 27 13 s 3 2 2 3 3 2 . 2 . Y s ( ) = 1 s + 4 9 s 3 ( ) - 13 9 s 3 2 s 3 2 2 3 3 2 . 2 . - 13 9 2 3 3 544 ( ) . 3 . 3 2 s 3 2 2 3 3 2 . 2 .
y t ( ) = 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 E s ( ) = 1 1 G c s ( ) G p s ( ) R s ( ) . R s ( ) = 1 s 0 lim s SSE = sR s ( ) 1 1 G c s ( ) G p . s ( ) . = 1 1 G c 0 ( ) G p s ( ) = 1 1 10 s 1 ( ) 1.3 27 13 . 5 . + 1 1 = = 0 -10 -8 -6 -4 -2 0 2 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Axis Imag Axis 2) Angle Condition 180 } ) 8 )( 1 ( 10 2 { 2 25 . 1 - = + + + + + - = j s s s s b s s angle

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14 2 2 tan 1 + = - - b b 10 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" 180 } ) 8 )( 2 ( 10 3 { 3 2 2 - =
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hw5_sol - Lead-Lag Homework Solution 3 a Gc s K 1.5 1 0 d s...

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