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# 17 - PoissonSensitivityConstraint17 filled - ME 575...

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Unformatted text preview: ME 575 Handouts Poisson Integral Constraint on Sensitivity • Lemma 9.5 (Nonminimum Phase Systems or Unstable Zeros) Consider the stable CL system with one-DOF controller configuration and open loop TF given by L(s) = G(s)C(s) = e − sτL(s), τ≥0 where L(s) is a rational TF. Assume that L(s) has open-loop unstable zeros at z1,…,zM , where zk = γk + jδk . Then, when L(s) has no unstable open-loop poles , the sensitivity function satisfies ∫ ∞ −∞ ln S ( jω) γk 2 γk + ( ω − δk ) 2 dω = 0, ∀k = 1 …,M , When L(s) has unstable open-loop poles at p1,… p N , we have ∫ ∞ −∞ where ln S ( jω) Bp ( s ) γk dω = −π ln Bp ( zk ) > 0 γ + ( ω − δk ) −p i* s−p ∏ s + p *i , Blaschke product i =1 i 2 k 2 zk pi N ME575 Session 17 – Poisson Integral Constraint on Sensitivity School of Mechanical Engineering Purdue University p i* Slide 1 Poisson Integral Constraints on Sensitivity • Remarks γk = 2 – Independent of controller design, a weighted compensation of low and weighted δk = 3 high sensitivity regions has to be achieved – The weighting function γk W ( z k , ω) , zk = γ k + jδk 2 2 γk + ( ω − δk ) decays with frequency, meaning the above compensation has essentially to be achieved over a finite frequency band – Weighted length of frequency axis: ωc ⎛ ω − δk ⎞ −1 ⎛ ωc + δk ⎞ Ω ( zk , ωc ) ∫ W ( zk , ω) dω = tan−1 ⎜ c ⎟ + tan ⎜ ⎟ −ωc ⎝ γk ⎠ ⎝ γk ⎠ ω2 −ω1 ∫ W ( ) dω + ∫ W ( ) dω = Ω ( zk , ω2 ) − Ω ( zk , ω1 ) and ω1 −ω2 ME575 Session 17 – Poisson Integral Constraint on Sensitivity School of Mechanical Engineering Purdue University ∫ ∞ −∞ W ( zk , ω ) dω = π Slide 2 1 ME 575 Handouts Poisson Integral Constraints on Sensitivity • Remarks – The appearance of unstable open-loop poles adds more openpositive value to the total integral, since ln Bp ( zk ) < 0, ∀ RHP zk and thus makes the allocation of sensitivity in frequency domain more difficult – When one RHP zero approaches an unstable open-loop openpole, |ln |B p ( z k )|| grows without bound, which would make the allocation almost impossible. ME575 Session 17 – Poisson Integral Constraint on Sensitivity School of Mechanical Engineering Purdue University Slide 3 Example of Design Implications • Typical Performance Specification S ( jω ) < ε S , for ω < ωl T ( jω ) < ε T , for ω > ωh ME575 Session 17 – Poisson Integral Constraint on Sensitivity School of Mechanical Engineering Purdue University Slide 4 2 ME 575 Handouts Example of Design Implications • Poisson Integral Constraint for NMP Zeros ∫ ∞ −∞ ln S ( jω) W ( zk , ω) dω = −π ln Bp ( zk ) , zk = γk + jδk ⇒ Lower bound for sensitivity peak Smax ≥ S ( jω ) : ln Smax ≥ 1 ⎡ π ln Bp ( zk ) + Ω ( zk , ωh ) − Ω ( zk , ωl ) ⎣ ME575 Session 17 – Poisson Integral Constraint on Sensitivity (ln εS ) Ω ( zk , ωl ) − ( π − Ω ( zk , ωh ) ) ln (1 + ε T )⎤ ⎦ School of Mechanical Engineering Purdue University Slide 5 Example of Design Implications Example • Observations – When CL bandwidth is large when compared to the speed of NMP zero (e.g., ωl = 2γ k ), even without considering the effect of any effect possible open-loop unstable poles and the performance constraints openon T(s), it is easy to verify that there will be a huge sensitivity peak sensitivity Example: Assume ωl = 2γ k , δk =0, and εS = 0.3 1 ⇒ ln S ≥ Ω z , ∞ − Ω z , ω (ln ε ) Ω ( z , ω ) ( m ax = k ) ( k l ) S k 1 (ln 0.3 ) Ω ( z k , 2 γ k ) , π − Ω (zk , 2γk ) = 2.86 ⇒ S m ax ≥ 17.5 & l Ω ( z k , 2 γ k ) = 2.21 Tm ax ≥ S m ax − 1 = 16.5 – Sharp transitions in the sensitivity frequency response, i.e., ωl close to ωh , will contribute to large sensitivity peaks. ME575 Session 17 – Poisson Integral Constraint on Sensitivity School of Mechanical Engineering Purdue University Slide 6 3 ...
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