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Solutions_F11_ECH157_Quiz2

Solutions_F11_ECH157_Quiz2 - 11.15 Quiz II Solution Key a...

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a) 1 1 ) 1 /( 1 1 1 1 ) ( ) ( + + τ - + = + τ - = τ - + τ - = s K K K K K K K K s K K s K K s K K s Y s Y c c c c c c c sp For stability 0 1 + τ - K K c Since τ is positive, the denominator must be negative, i.e., 0 1 < + K K c 1 - < K K c K K c / 1 - < Note that K K K K K c c CL + = 1 b) K = 10 and τ = 20 and we want 10 1 = + τ - K K c or c K ) 10 )( 10 ( 10 20 + = - c K 100 30 = - 3 . 0 - = c K Offset: 5 . 1 2 3 ) 10 )( 3 . 0 ( 1 ) 10 )( 3 . 0 ( = - - = - + - = CL K Offset = +1 - 1.5 = - 50% (Note this result implies overshoot)
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c) K K s s K K s s K K s s K K s Y s Y c m c m c m c sp + + τ τ - = + τ τ - + + τ τ - = ) 1 )( 1 ( ) 1 )( 1 ( 1 ) 1 )( 1 ( ) ( ) ( K K s s K K c m m c + + τ - τ + ττ - = 1 ) ( 2 1 1 1 ) 1 /( 2 + + τ - τ + + ττ - + = s K K s K K K K K K c m c m c c (standard form) For stability, (1) 0 1 + ττ - K
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