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lecture10 - MAE 171A: Dynamic Systems Control Lecture 10:...

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MAE 171A: Dynamic Systems Control Lecture 10: Root Locus Goele Pipeleers
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Situation Feedback Control D ( s ) G ( s ) u y w r v In the Previous Lecture P, D, I, PD, PID controllers analysis of their advantages and disadvantages 1
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In the Subsequent Lectures systematic ways to design such controllers root locus (L 11-12) Bode and Nyquist plot (L 13-16) In This Lecture: Root Locus problem formulation basic characterization 4 root locus drawing rules 2
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Root Locus: Problem Formulation Feedback Control D ( s ) G ( s ) u y w r v closed-loop stability and performance closed-loop poles closed-loop poles = roots of the characteristic equation: 1 + D ( s ) G ( s ) = 0 3
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how are the characteristic roots influenced by a system parameter? example 1: D ( s ) = k how do the roots of 1 + kG ( s ) change as a function of k example 2: D ( s ) = 1 s + k characteristic roots: 1 + G ( s ) s + k = 0 s + k + G ( s ) = 0 1 + kL ( s ) = 0 where L ( s ) = 1 s + G ( s ) general formulation: given L ( s ) , plot the roots of 1 + kL ( s ) on the complex plane as k : 0 → ∞ the resulting plot is called the root locus 4
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Root Locus: Basic Characterization a complex number s is on the root locus 1 + kL ( s ) = 0 for some k > 0 L ( s ) is real and negative L ( s ) = (2 l - 1) π , for some integer l 5
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example: L ( s ) = s - z 1 ( s - p 1 )( s - p 2 ) polar coordinates: ± s - p i = a i e i s - z i = b i e i L ( s ) = b 1 e 1 a 1 e 1 a 2 e 2 = b 1 a 1 a 2 e j ( ψ 1 - φ 1 - φ 2 ) L ( s ) is real and negative ψ 1 - φ 1 - φ 2 = (2 l - 1) π ℑ{ s } ℜ{ s } ψ 1 φ 1 φ 2 s z 1 p 1 p 2 6
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in general: L ( s ) = ( s - z 1 )( s - z 2 ) ··· ( s
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This note was uploaded on 08/08/2011 for the course MAE 171 taught by Professor Pipeleers during the Summer '11 term at UCLA.

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lecture10 - MAE 171A: Dynamic Systems Control Lecture 10:...

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