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Lecture11 - 1 Root Locus Characteristic Equation...

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1 Root Locus - Characteristic Equation: 1 + G ( s ) D ( s ) = 0 Closed-loop poles Stablity and Performance How are the characteristic roots (closed-loop poles) influenced by a system parameter? Example 1.1 . (a) D ( s ) = K . 1 + KG ( s ) = 0 How do the roots move as K changes? (b) D ( s ) = KD o ( s ) . 1 + G ( s ) D ( s ) = 0 1 + KG ( s ) D o ( s ) = 0 1 + KL ( s ) = 0 , where L ( s ) = G ( s ) D o ( s ) (c) D ( s ) = 1 s + K . 1 + G ( s ) s + K = 0 s + K + G ( s ) = 0 1 + KL ( s ) = 0 , where L ( s ) = 1 s + G ( s ) PROBLEM: Given L ( s ) , plot the roots of 1 + KL ( s ) = 0 on the complex plane as K : 0 → ∞ . The resulting lines are called the ROOT LOCUS. Example 1.2 D ( s ) = K, G = 1 s - 1 Characteristic Equation: 1 + G ( s ) D ( s ) = 0 1 + K 1 s - 1 = 0 ( s - 1) + K = 0 closed-loop poles : s = - K + 1 . We assume L ( s ) in the form of L ( s ) = b ( s ) a ( s ) = ( s - z 1 )( s - z 2 ) · · · ( s - z m ) ( s - p 1 )( s - p 2 ) · · · ( s - p n ) , n m. Fact : A complex number s is on the root locus 1 + KL ( s ) = 0 for some K > 0 L ( s ) is real negative. 1
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Example 1.3 L ( s ) = s - z 1 ( s - p 1 )( s - p 2 ) s - p i = A i e j φ i s - z i = B i e j ψ i L ( s ) = B 1 e ψ 1 ( A 1 e φ 1 ) ( A 2 e φ 2 ) = B 1 A 1 A 2 e j ( ψ 1 - φ 1 - φ 2 ) L ( s ) is real negative iff ψ 1 - φ 1 - φ 2 = (2 - 1) × 180 o for some integer . In general, L ( s ) = ( s - z 1 )( s - z 2 ) · · · ( s - z m ) ( s - p 1 )( s - p 2 ) · · · ( s - p n ) = ( B 1 e ψ 1 )( B 2 e ψ 2 ) · · · ( B m e ψ m ) ( A 1 e φ 1 ) ( A 2 e φ 2 ) · · · ( A n e φ n ) = re j θ where r = B 1 B 2 · · · B m A 1 A 2 · · · A n , θ = m i =1 ψ i - n i =1 φ i .
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