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8 - sensitivity8 filled - ME 575 Handouts Analysis of...

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Unformatted text preview: ME 575 Handouts Analysis of Feedback Systems • Typical Classical Feedback Controller Structure • Nominal Sensitivity Functions • Stability of Nominal Feedback System – Internal Stability – Routh’s Stability Test – Root Locus Method – Nyquist Stability Criterion ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 1 Two-Degree-of-Freedom Closed Loop Di (s) Reference Value R(s) H(s) R(s) E(s) C(s) U (s ) − Ym(s) U D (s ) x0 G(s) Do (s) Y (s) Plant N (s ) Noise Controller ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Sensor Slide 2 1 ME 575 Handouts Closed-Loop Transfer Functions ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 3 Sensitivity Functions NG (s)NC (s) G(s)C(s) Complementary = : sensitivity function 1 + G(s)C(s) DG (s)DC (s) + NG (s)NC (s) DG (s)DC (s) 1 S(s) = = : Sensitivity function 1+ G(s)C(s) DG (s)DC (s) + NG (s)NC (s) NG (s)DC (s) G(s) Input disturbance Si (s) = = : sensitivity function 1+ G(s)C(s) DG (s)DC (s) + NG (s)NC (s) DG (s)NC (s) C(s) Control Su (s) = = : sensitivity function 1 + G(s)C(s) DG (s)DC (s) + NG (s)NC (s) T (s) = ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 4 2 ME 575 Handouts Sensitivity Function • The sensitivity function S describes how error signal E is related to the reference input R. • The sensitivity function S also describes how much a change in the plant G (perhaps due to parameter errors) affects the closed-loop transfer function T (a.k.a. complementary sensitivity). • A controller C with high gain will make CG large, CG therefore making S small, exactly as desired. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 5 Sensitivity Function • Relationship Between Sensitivities S ( s) + T ( s) = 1 Si (s) = S (s)G(s) Su (s) = S (s)C (s) • Practical Implications ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 6 3 ME 575 Handouts Internal Stability ⎡ ⎢ ⎡ E ( s) ⎤ ⎢ ⎥ 1⎢ ⎢ ⎢U D ( s ) ⎥ = ⎢ ⎥ 1 + CG ⎢ ⎢ Ym ( s ) ⎥ ⎢ ⎣ ⎦ ⎢ ⎣ ⎤ ⎡ R (s) ⎤ ⎥ ⎥⎢ ⎢ Di ( s ) ⎥ ⎥ ⎥ ⎥⎢ ⎥ ⎢ Do ( s ) ⎥ ⎥ ⎥⎢ ⎥ ⎢ N (s) ⎥ ⎦⎣ ⎦ This system is internally stable if the internal signals decay to zero when any of the inputs are impulses. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 7 Internal Stability Theorem: The feedback system is internally stable if and only if • the sensitivity function S has no poles in the RHP (s ≥ 0) • there is no cancellation of unstable poles between the controller C and the plant G. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 8 4 ME 575 Handouts Routh-Hurwitz Stability Criterion Allows checking for stability without having to solve for the poles. QUICK TEST: Hurwitz Necessary Condition n Given the char. eq. p( s) = a s k = 0 ∑ k =0 k a stable system must have: (i) (ii) ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 9 Hurwitz Necessary Condition Ex: p(s) = s2 − 3s + 2 = 0 stable? Ex: p(s) = s3 + s2 + 2s + 8 = 0 stable? We need a sufficient condition. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 10 5 ME 575 Handouts Routh Array n For p( s ) = ∑ ak s k = 0 k =0 n s s n−1 s n−2 s n −3 an an−1 b1 c1 an−2 an−3 b2 c2 an−4 an−5 b3 c3 L L 0 0 b1 = b2 = an an −1 an − 2 an − 3 −an −1 an an −1 an − 4 an −5 −an −1 , , M M s0 c1 = an −1 b1 an −3 b2 −b1 M ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 11 Routh Sufficient Condition The number of roots of p(s) having positive real parts (in the RHP) equals the number of sign changes in the 1st column of the Routh array. Ex: p(s) = s3 + s2 + 2s + 8 ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 12 6 ME 575 Handouts Root Locus • Definition Root Locus is the plot of the roots of the following equation on the complex s plane when the parameter K varies from 0 → ∞: varies 1+ K ⋅ N (s) = 0 or D ( s ) + K ⋅ N ( s ) = 0 D( s) where N(s) and D(s) are known polynomials in factorized form: N ( s ) = ( s − z1 )( s − z2 )L ( s − z N z ) D( s ) = ( s − p1 )( s − p2 )L ( s − pN P ) z1 , z2 , …, zNz , are called the finite open-loop zeros. are openp1 , p2 , …, pNp , are called the finite open-loop poles. are open- ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 13 Root Locus Methods of obtaining root locus: • Given a value of K, numerically solve the characteristic equation for a K, numerically set of roots. Repeat this for a set of K values and plot the corresponding roots on the complex plane. • Use MATLAB. In MATLAB use the commands rlocus and rlocus rlocfind. You can use on-line help to find the usage for these rlocfind oncommands. 1+ K ⋅ 800 =0 s ( s + 50) >> G = tf(800,[1 50 0]); >> rlocus(G); >> [K, poles]=rlocfind(G); • Apply the following root locus sketching rules to obtain an approximate approximate root locus plot. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 14 7 ME 575 Handouts Root Locus Magnitude and Phase Conditions Magnitude The necessary and sufficient conditions for a point on the s-plane to be on the root locus (or a value of s that satisfies the characteristic eq.) are that s satisfies the following phase and magnitude conditions: Phase Condition (or Angle Criterion): ( ∠( s − z1 ) + L + ∠ s − zNz ) ( ) − ∠( s − p1 ) −L− ∠ s − pN p = 180o Sum of phase angles to all open-loop zeros − Sum of phase angles to all open-loop poles = 180o Magnitude Condition: K= K= D(s) N ( s) = s − p1 s − p2 L s − pN p s − z1 s − z2 L s − z N z Multiplication of lengths to all open-loop poles Multiplication of lengths to all open-loop zeros ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 15 Root Locus Sketching Rules Rule 1: There is a branch of the root locus for each root of the characteristic equation. The number of branches is equal to the number of open-loop poles or open-loop zeros, whichever is openopengreater. Rule 2: Root locus starts at open-loop poles (when K = 0) and ends at 0) openopen-loop zeros (when K → ∞). If the number of poles is greater openthan the number of zeros, roots start at the excess poles and terminate at zeros at infinity. If the reverse is true, branches will branches start at poles at infinity and terminate at the excess zeros. Rule 3: Root locus is symmetric about the real axis, i.e., closed-loop closedpoles appear in complex conjugate pairs. Rule 4: Along the real axis, the root locus includes all points to the left of left an odd number of real poles and zeros. odd ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 16 8 ME 575 Handouts Root Locus Sketching Rules Rule 5: If number of poles NP exceeds the number of zeros NZ , then as K → ∞, (NP − NZ) branches will become asymptotic to straight branches lines. These straight lines intersect the real axis with angles θk at σ0 . σ0 = θk = ∑ p −∑z i NP − NZ i = Sum of open-loop poles − Sum of open-loop zeros # of open-loop poles − # of open-loop zeros (2k + 1)π (2k + 1)180° [rad] = [deg] , k = 0, 1, 2, L NP − NZ NP − NZ If NZ exceeds NP , then as K → 0, (NZ − NP) branches behave as branches above. Rule 6: Breakaway and/or break-in (arrival) points can be obtained by breaksolving s in the following equations: d ( K (s) ) = 0 ds ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 17 Root Locus Sketching Rules Rule 7: The departure (arrival) angle for a pole pi (zero zi) can be calculated by slightly modifying the following equation: ∠( s − z1 ) + ∠( s − z2 ) + L + ∠( s − z N Z ) − ∠( s − p1 ) − ∠( s − p2 ) − L − ∠( s − pN p ) = 180° The departure angle θn from the pole pn can be calculated by replacing the term ∠( s − pn ) with θn and replacing all the s’s with pn in the other terms. Rule 8: If the root locus passes through the imaginary axis (the stability stability boundary), the crossing point jω and the corresponding gain K can be found as follows: – Replace s in the left side of the closed-loop characteristic equation closedwith jω to obtain the real and imaginary parts of the resulting complex number. – Set the real and imaginary parts to zero, and solve for ω and K. This will tell you at what values of K and at what points on the jω axis the roots will cross. ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 18 9 ME 575 Handouts Root Locus Example A feedback control system is proposed. The corresponding block diagram is: R(s) + − Controller C (s ) Plant G (s) U(s) 1 s(s + 2) Y(s) Let C(s) = K. Sketch the root locus of the closed-loop poles closedas the controller gain K varies from 0 to ∞. Find closed-loop characteristic equation: closed- ME575 Session 8 – Sensitivity Functions and Stability School of Mechanical Engineering Purdue University Slide 19 Root Locus Example C (s ) = K C(s ) = K (s + 3) Im s j2 j1 −6 −4 −2 Im s j2 j1 Re s −6 −4 −2 Re s −j1 −j2 ME575 Session 8 – Sensitivity Functions and Stability −j1 −j2 School of Mechanical Engineering Purdue University Slide 20 10 ...
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This note was uploaded on 12/09/2011 for the course ME 575 taught by Professor Meckl during the Fall '10 term at Purdue.

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