MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 191 Reading: • Nise: Chapter 4. 1 System Poles and Zeros Consider a system with transfer function N(s)H(s) = . D(s) If we factor the numerator and denominator polynomials and write H(s) = K (s − z1)(s − z2) . . . (s − zm) (s − p1)(s − p2) . . . (s − pn) where p1, p2, . . . , pn — are the roots of the characteristic polynomial D(s), and are known as the system poles, z1, z2, . . . , xm — are the roots of the numerator polynomial N(s), and are known as the system zeros. Note that because the coefficients of N(s) and D(s) are real (they come from the modeling parameters), the system poles and zeros must be either (a) purely real, or (b) appear as complex conjugates and in general we write pi, or zi = σi + jωi. Example 1 Find the poles and zeros of the system 5s2 + 10s G(s) = s3 + 5s2 + 11s + 5 5s(s + 2) = (s + 3)(s2 + 2s + 5) 5s(s + 2) = (s + 3)(s + (1 + j2))(s + (1 − j2)) 1copyright cD.Rowell 2008 19–1s - p l a n e {s}Á {s}jwsooxxx- 3 - 2 - 1- j 2j 2s = s + jwso that we have (a) a pair of real zeros at s = 0, −2 and (b) three poles at s = −3, −1 + j2, and s = −1 − j2. The system poles and zeros completely characterize the transfer function (and there-fore the system itself) except for an overall gain constant K: �m (s − zi)G(s) = K i=1�ni=1(s − pi) 1.1 The Pole-Zero Plot The values of a system’s poles and zeros are often shown graphically on the complex s-plane in a pole-zero plot. For example, the poles and zeros of the previous example are drawn: where the pole p ositions are denoted by an x, and the zeros are drawn as an o. The figure shows zeros at s = 0, −2, and poles at s = −3, −1 ± j2. The pole-zero plot is used extensively throughout control theory and system dynamics to provide a qualitative indication of the dynamic behavior of systems. Aside: In MATLAB a system may be specified by its poles and zeros using the function zpk(zeros, poles, gain), for example sys = zpk([0 2], [-3, -1+i*2, -1-i*2], 5) step(sys) will plot the step response of the system in the previous example. 19–2s - p l a n ejwsxx- 3- 4s = s + jwg e n e r a t e s a c o m p o n e n t e - 4 tg e n e r a t e s a c o m p o n e n t e - 3 tThe characteristic equation of a system is D(s) = (s − p1)(s − p2) . . . (s − pn) = 0 so that the poles are the system eigenvalues, and the form of the homogeneous response is dictated by the poles: npit yh(t) = � Ciei=1 (when the poles are distinct), and the constants Ci are determined by the initial condi-tions. Example 2 Find the poles and hence the homogeneous response components of the system 12 G(s) = s2 + 7s + 12 The characteristic equation is D(s) = (s + 4)(s + 4) = 0 and the poles are s = −3, −4 The homogeneous response components are therefore y1(t) = C1e−3t and y2(t) = C2e−4t, where C1 and C2 are defined by the initial conditions. Note: The p oles do not specify the amplitude of the modal components in the response. They simply indicate the nature of the response components. 19–3s - p l a n ejwsooxxx- 4- j 4- 1- j 2j 2s = s + jwj 4x1.2 Complex Poles and Zeros We have noted that in general s = σ + jω, and that poles and zeros may appear as complex conjugate pairs: pi,i+1 = σi ± jωi zk,k+1 = σk ± jΩk For example the pole-zero plot corresponds to the transfer function G(s) = K (s − j4)(s + j4) s(s + 4)(s + (1 + j2))s + (1 − j2)) s2 + 16 = K s(s + 4)(s2 + 2s + 5) s2 + 16 = K s4 + 16s3 + 13s2 + 2 s The homogeneous response we will have a pair of complex exponential terms associated with each pair of conjugate pair of poles, such as . . . + Cie(σi+jωi)t + Ci+1e(σi−jωi)t . . . but Ci and Ci+1 are also complex (say a ± jb), so we can write Cie(σi+jωi)t + Ci+1e(σi−jωi)t = (a + jb)e σit ejωit + (a − jb)e σit e−jωit = ae σit �ejωit + ejωit� − jbeσit �ejωit − ejωit� Euler’s formulas state jωt + e−jωt)cos(ωt) = 21 (e� � ejωt = cos(ωt) + j sin(ωt) sin(ωt) = 1 (ejωt − e−jωt) or e−jωt = cos(ωt) − j sin(ωt)2j 19–4y ( t )tA eA e s i n (wt )ststs - p l a n ejwsooxxx- 2 - 1- j 2j 2s = s + jwso that the contribution of the complex conjugate pole pair to the homogeneous response may be written yi,i+1(t) = 2ae σit cos(ωit) + 2ae σit sin(ωit)) = 2√a2 + b2e σit � √aa cos(ωit) + √ab sin(ωit)� 2 + b2 2 + b2 = Aie σit sin(ωit + φi) where Ai = 2√a2 + b2 and φi = tan−1 �a� . b which is shown below (for σi < 0). Example 3 Find and plot the poles and zeros of 7s + 14 G(s) = s3 + 3s2 + 7s + 5 and then determine the modal response components of this system. s + 2 s + 2 G(s) = 7 = 7 (s + 1)(s2 + 2 s + 5 (s + 1)(s + (1 + j2))(s + (1 − j2)) The pole-zero plot is 19–50u n s t a b l e r e g i o ns t a b l e r e g i o nsjwand the modal components are (1) Ce−t (corresponding to the pole at s = −1), and (2) Ae−t sin(2t + φ) (corresponding to the complex conjugate pole pair at s = −1 ± j2), and where the constants C, A, and φ are determined from the initial conditions. Note: A pair of purely imaginary poles (on the imaginary axis of the s-plane) implies σ = 0 and there will be no decay. The system will act as a pure oscillator. The effect of pole locations in the s-plane on the modal response components is summa-rized in the figure below: We note (a) Poles in the left-half of the s-plane (the lhp), that is σ < 0, generate components that decay with time. (b) Conversely, poles in the right-half s-plane (the rhp), that is …
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