Lecture11

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online