6.241 Dynamic Systems and Control Lecture 16: Bode’s Sensitivity Integral Readings: DDV, Chapter 18 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology April 4, 2011 E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 1 / 7Cauchy’s integral theorem Let Ω ⊂ C be an open, simply connected set. Let f : Ω C be a holomorphic function. In other words, the limit → f �(s0) = lim f (s) − f (s0) s s0→s − s0 exists (and is continuous) for all s0 ∈ Ω. Note that a complex function is holomorphic if and only if it is analytic. Let γ : [0, 1] Ω be a differentiable function, such that γ(0) = γ(1). → Then, � 1 � f (γ(t))γ�(t) dt = f (z) dz = 0, 0 Γ where Γ is the closed contour traced by γ(t) as t ranges from 0 to 1. E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 2 / 7� ����� � ����� �Cauchy’s integral formula Let Ω, f , γ, and Γ be as in the previous slide. Furthermore, Let p be a point that is encircled counter-clockwise exactly once by γ(t) as t ranges from 0 to 1. Then, � f (z) dz = 2πj f (p). Γ z − p Proof sketch: Using Cauchy’s integral theorem, show that the value of the integral does not dep end on the enclosing path. In particular, we can choose a circle C� of radius � around p. 12πjt1 e Note, for γ(t) = p + �e 2πjt Then, , dz = 2πj� dt = 2πj. C� z − p 0 �e2πjt ≤ f (z) z − p dz − 2πj f (p) |f (z) − f (p)||z − p|dz| ≤ |C� C� 2π maxz∈C� |f � (z) − f (p)| � dt → 0. 0 E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 3 / 7Bode’s Sensitivity Integral for open-loop stable systems Let L(s) be a proper, scalar rational transfer function, of relative degree nr , i.e., nr is the difference between the degrees of the denominator and of the numerator. Define G (s) = (1 + L(s))−1, and assume that G has neither poles nor zeros in the closed right half plane (i.e., both L and G are stable). Then, �� ∞ 0 if nr > 1, log G (jω) dω =| | −κ π if nr = 1,02 where κ = lims→∞ sL(s). E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 4 / 7� � � � � � ��� ���Proof sketch G (s)| is analytic on the RHP, then Ci log |G (s)| ds + C∞ log |G (s)| ds = 0, i.e., Since log |log G (s)| ds| = D ∞ 1 j log |G (jω)| dω = 2 log |1 + L(s)| ds. 0 C∞ For large s, log |1 + L(s)| ≈ log |1 + as−nr | ≈ |as−nr |, so on CE , � π/2 · Ejejt dt = 1 a e−jnr t CE log |1 + L(s)| ds ≈ − 2 E nr 0 � π/2 − E naj r −10 ejt dt = − E naj r −1 π 2 In other words, for nr > 1, the integral on CE converges to 0, and for nr = 1, it converges to κ π 2 j. E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 5 / 7� � �Bode’s Sensitivity Integral Let L(s) be a proper, scalar rational transfer function, of relative degree nr . Define G (s) = (1 + L(s))−1, and assume that G has no poles in the closed right half plane (i.e., G is stable), and has q ≥ 0 zeros in the closed RHP plane (i.e., L can be unstable), at location z1, z2, . . . , zq , with Re(zi ) ≥ 0. Then, ∞ π iq =1 zi if nr > 1,log |G (jω)| dω = −κ π �q 02 + π i=1 zi if nr = 1, where κ = lims→∞ sL(s). E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 6 / 7�� � � � � ���� ����� ���� ����� ���� ����� ������� ������� � Proof sketch The function log Gˆ(s) = G (s) |G (s)| is not analytic on the RHP. Define . q s+zi i=1 s−zi ˆGˆ(s)Since log G (s)| is analytic on the RHP, then log ds = 0, and hence | | |D qlog G (s) ds + log G (s) ds + log s + zi s − zi ds = 0.| | | |Ci C D∞ i=1 Proceeding as in the basic case, and noting that log s + zi s − zi ds = 0, Ci and log s + zi s − zi zi zids = log 1 + ds − log 1 − ds = −jπzi , s sC∞ C∞ C∞ We get the desired result. Notice also that i Re(zi ) = i zi . E. Frazzoli (MIT) Lecture 16: Bode Integral April 4, 2011 7 / 7MIT OpenCourseWare http://ocw.mit.edu 6.241J / 16.338J Dynamic Systems and Control Spring 2011 For information about citing these materials or our Terms of Use, visit: