Lecture11

9 n 3 6 1 s 2s2 9 l s 4 p123 2 3j m

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Unformatted text preview: ros Rule 3 The following approximation will be used. for |x| (1 + x)β ≈ 1 + β x −K = = = 1. 1 (s − p1 )(s − p2 ) · · · (s − pn ) = L(s ) (s − z1 )(s − z2 ) · · · (s − zm ) sn − psn−1 + · · · sn − psn−1 ≈m sm − zsm−1 + · · · s − zsm−1 n s 1 − p/s p z · ≈ sn−m (1 − )(1 + ) m 1 − z/s s s s p−z pz p−z + 2 ≈ sn−m 1 − s s s p−z 1 α 1 − ( n − m) · · ≈ sn−m 1 − n−m s s = sn− m 1 − = sn− m = (s − α)n−m 4 n−m −20 −15 −10 −5 0 So ⇒ (s − α)n−m ≈ Kej (2 −1)×180o 1 2 −1 o 1 2 −1 for integer o ⇒ s − α ≈ K n−m ej n−m ×180 ⇒ s ≈ α + K n−m ej n−m ×180 Root Locus Example 1.9 n = 3, α= 6 1 (s + 2)(s2 + 9) L( s ) = 4 p1,2,3 = −2, ±3j m = 0, 2 −2 + 3j − 3j 2 =− 3 3 0 −2 −4 −6 Rule 4: Departure angle at pk is q φk,dep = −Σ ∠(pk − pi ) + Σ ∠(pk − zi ) + (2 − 1) × 180o Arrival angle at zk is q ψk,arr = Σ ∠(zk − pi ) − Σ ∠(zk − zi ) + (2 − 1) × 180o where q is the multiplicity. Derivation of Rule 4 Consider the case pk is not repeated. Phase condition: ∠(s − zi ) − (s − pi ) = (2 − 1) × 180o For s in the root locus, if s ≈ pk , then φk,dep ≈ ∠(s ...
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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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