EE3530-5 - Chapter 5 The Root-Locus Design Method r 6 e C(s...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 5 The Root-Locus Design Method - - C ( s ) - P ( s ) - 6 r e u y - Closed-loop poles: 1 + P ( s ) C ( s ) = 0 . Denote L ( s ) := P ( s ) C ( s ) L ( s ) = K ( s - z 1 )( s - z 2 ) · · · ( s - z m ) ( s - p 1 )( s - p 2 ) · · · ( s - p n ) where z 1 , . . . , z m are the open-loop zeros, p 1 , . . . , p n are the open- loop poles, and K is a variable gain. Objective: study how the closed-loop poles change when K varies from 0 to . 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Find all points satisfying L ( s ) = - 1 . magnitude condition : | L ( s ) | = 1 phase condition : 6 L ( s ) = (2 k + 1)180 o , k = 0 , ± 1 , . . . Magnitude condition can always be satisfied by a suitable K 0. Phase condition does not depend on the value of K (but does depend on the sign of K ): 6 L ( s ) = m X i =1 6 ( s - z i ) - n X j =1 6 ( s - p j ) = (2 k + 1)180 o . Thus the key is to find all those points that satisfy the phase condition. 2
Image of page 2
Example : Consider L ( s ) = K ( s - z 1 )( s - z 2 ) ( s - p 1 )( s - p 2 )( s - p 3 ) , p 2 = p 1 . a pole is represented by a “ × a zero is represented by a “ ”. The phase of L ( s ) at a point s in the complex plane is 6 L ( s ) = 6 ( s - z 1 ) + 6 ( s - z 2 ) - 6 ( s - p 1 ) - 6 ( s - p 2 ) - 6 ( s - p 3 ) = φ 1 + φ 2 - α 1 - α 2 - α 3 . - 6 * × O y × × 9 Y K M 0 z 2 z 1 p 3 p 1 p 2 s α 1 α 2 φ 1 α 3 φ 2 σ 3
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Root Locus Rules: 0 K ≤ ∞ 1 . The root locus is symmetric with respect to the real axis. 2 . The root loci start from n poles p i (when K = 0) and approach the n zeros ( m finite zeros z i and n - m infinite zeros when K → ∞ ). 3 . The root locus includes all points on the real axis to the left of an odd number of open loop real poles and zeros. 4 . As K → ∞ , n - m branches of the root locus approach asymp- totically to n - m straight lines (called asymptotes) with angles θ = (2 k + 1)180 o n - m , k = 0 , ± 1 , ± 2 , . . . and the starting point of all asymptotes is on the real axis at σ = n X i =1 p i - m X j =1 z j n - m = X poles - X zeros n - m . 5. The breakaway points (where the root loci meet and split away, usually on real axis) and the breakin points (where the root loci meet and enter the real axis) are among the roots of the equation: dL ( s ) ds = 0. (On real axis only those roots satisfy Rule 3 are breakaway or breakin points.) 4
Image of page 4
6. The departure angle φ k (from a pole, p k ) is given by φ k = m X i =1 6 ( p k - z i ) - n X j =1 ,j 6 = k 6 ( p k - p j ) ± 180 o . (In the case p k is l repeated poles, the departure angle becomes φ k /‘ .) The arrival angle ψ k (at a zero, z k ) is given by ψ k = - m X i =1 ,i 6 = k 6 ( z k - z i ) + n X j =1 6 ( z k - p j ) ± 180 o . (In the case z k is l repeated zeros, the arrival angle becomes ψ k /‘ .) 5
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example : Asymptotes and breakaway points are illustrated. Con- sider an open loop transfer function L ( s ) = K s ( s + 4)( s + 5) . - 6 × × × - 7 w ² I N 0 60 - 60 180 - 5 - 4 - 3 breakaway point the angles of the three asymptotes: θ = (2 k + 1)180 o 3 = 60 o , - 60 o , 180 o for k = 0 , - 1, and 1 and the intersection of the asymptotes with the real axis is given by σ = 0 - 4 - 5 3 = - 3 . Note that one could have set k = 0 , 1 , 2 to get θ = 60 o , 180 o , 300 o which are the same angles.
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern