Lecture11

Therefore ls 1 1 2 180o s 1 is part of the

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Unformatted text preview: L(s) = Σ∠(s − zi ) − Σ∠(s − pi ) = (2 − 1) × 180 o 4 Consider s on the real axis: (a) If s > −1, then φ1 + φ2 = 360o , ψ1 = 0o . Therefore, ∠L(s) = ψ1 − (φ1 + φ2 ) = 360o = (2 − 1) × 180o . s > −1 is not part of the root locus. (b) If s < −1, then φ1 + φ2 = 360o , ψ1 = 180o . Therefore, ∠L(s) = ψ1 − (φ1 + φ2 ) = −180o . s < −1 is part of the root locus. 3 2 1 0 −1 −2 −3 −4 −8 −6 s+3 (s + 1)(s + 2) Consider s on the real axis: 1 : ψ1 = 0, φ1 = 0, φ2 = 0, ∠L = 0. 1 ∈ root locus. / 2 : ψ1 = 0, φ1 = 180o , φ2 = 0, ∠L = −180o . 2 ∈ root locus. 3 : ψ1 = 0, φ1 = 180o , φ2 = 180o , ∠L = −360o . 3 ∈ root locus. / 4 : ψ1 = 180o , φ1 = 180o , φ2 = 180o , ∠L = −180o . 4 ∈ root locus. Example 1.6 L( s ) = Break in point Break away point. Rule 3: n − m branches of the root locus approach the following asymptotes: s = α + M ej θ α= p−z , n−m 2 −1 θ= × 180o , n−m M >0 n p= m pi , z= i=1 zi i=1 = 1, 2, · · · , n − m. 3 −4 −2 0 Root Locus 15 Example 1.7 n = 3, 10 s+1 L( s ) = (s + 2)(s2 + 9) p1,2,3 = −2, ±3j, m = 1, −2 + 3j − 3j − (−1) 1 α= =− 3−1 2 5 0 z1 = −1. −5 −10 −15 −2.5 −2 −1.5 −1 −0.5 0 0.5 Root Locus 10 Example 1.8 n = 3, L( s ) = s+1 (s + 21)(s2 + 9) p1,2,3 = −21, ±3j, m = 1, −21 + 3j − 3j − (−1) α= = −10 3−1 5 z1 = −1. 0 −5 −10 −25 Derivation of Rule 3 1 + KL(s) = 0 K→∞ ⇒ L( s ) → 0 s→∞ Go to ze...
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This note was uploaded on 01/27/2014 for the course MAE 171A taught by Professor Idan during the Winter '09 term at UCLA.

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