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EE 4314

EE 4314 . The Nyquist Stability Criterion and Stability Margins Spring 2011. Stability. No RHP poles for a stable system. How to determine the number of RHP based on K(s)G(s) without calculating the roots of 1+K(s)G(s)? Argument principle and Nyquist stability criterion.

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EE 4314

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  1. EE 4314 The Nyquist Stability Criterionand Stability Margins Spring 2011

  2. Stability • No RHP poles for a stable system. • How to determine the number of RHP based on K(s)G(s) without calculating the roots of 1+K(s)G(s)? • Argument principleand Nyquist stability criterion

  3. Argument Principle • Let H(s) be a tf: • Evaluate H(s) along contour Cand get the resultant contour H(C) • # of CW encirclementsof point (0,0) by H(C): N = Z−P, Z = # of zeros in CP = # of poles in C

  4. Argument Principle • Evaluate H(s) • N = Z−P (CW)Z = # of zerosP = # of poles • http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/Freq/Nyquist2.html H(C) H(C)

  5. Application of Argument Principle • Pick C such that it encircles the entire RHP in CW direction • Plot H(C) and count the # of encirclements (N) of the origin (0,0) • Find the # of zeros (Z) of H(s) • # of RHP poles of H(s) is P = Z−N • http://www.math.ucla.edu/~tao/java/Argument.html

  6. Application of Argument Principle • We are interested inthe poles of the closedloop system for stabilityi.e. roots of 1+KG=0 • Let G(s) = b(s)/a(s),then we are interested in the zeros of • We can find the poles of H(s) = poles of G(s)

  7. Nyquist Stability Criterion • P = # of poles of G(s)in RHP • N = # of CW encirclementsof −1 • # of zeros of H(s) in RHP: Z = N+P H(s) =

  8. Precedure for Nyquist Plot

  9. Nyquist Plot: Example 6.8, p.324 • P = # of poles of KG(s) in RHP=> P = 0 • N = # of CW encirclements of −1 => N = 0 • # of zeros of H(s) in RHP: Z=N+P=0Stable always

  10. Nyquist Plot: Example 6.9 , p.327 • P = # of poles of G(s) in RHP=> P = 0 • N = # of CW encirclements of −1/K • For largeK, −1/K isin the loop => N=2 • Z=N+P=2=>Unstable

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