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.a s + a s + . . . + a s + annn - 1n - 110b s + b s + . . . + b s + bm mm - 1m - 110U ( s )Y ( s )a s + a s + . . . + a s + annn - 1n - 110b s + b s + . . . + b s + bm mm - 1m - 110U ( s )Y ( s )1X ( s )a s + a s + . . . + a s + annn - 1n - 110b s + b s + . . . + b s + bm mm - 1m - 110U ( s )Y ( s )1F ( s )� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 171 Reading: • Nise: Chapter 4 1 System Response Our systems are • Linear and Time-Invariant (LTI), that is they are linear, and their properties do not change with time, and • are Single-Input Single-Output (SISO). and are usually represented by a transfer function which is equivalent to an ordinary differential equation (ODE) with constant coefficients. dyn dyn−1 dy dum dum−1 du an dtn + an−1 dtn−1 + . . . + a1 dt + a0y = bm dtm + bm−1 dtm−1 + . . . + b1 dt + b0u where the coefficients are determined by the system components. Block diagram algebra allows us to redraw the blo ck diagram as or alternatively Consider first the action of the block containing the numerator polynomial 1copyright cD.Rowell 2008 17–1a s + a s + . . . + a s + annn - 1n - 110U ( s )1Y ( s )� �b s + b s + . . . + b s + bm mm - 1m - 110P ( s )Q ( s )then dpm dpm−1 dp q(t) = bm dtm + bm−1 dtm−1 + . . . + b1 dt + b0p, which is not a differential equation. It simply shows that the output of this block is a weighted sum of the input and its derivatives. Initially therefore, we concentrate on solving the sub-system represented by the block or as a differential equation dyn dyn−1 dy an dtn + an−1 dtn−1 + . . . + a1 dt + a0y = u. From 18.03 we know that the solution to an ODE of this form may be considered to have two components: y(t) = yh(t) + yp(t) where yh(t) is the solution to the homogeneous equation, that is the solution when u(t) ≡ 0,• and • yp(t) is a particular solution that satisfies the ODE for the given u(t). a) The Homogeneous Solution yh(t): Assume that u(t) ≡ 0 and that the system has initial conditions y(0) = C0, y˙(0) = C1, y¨(0) = C2, . . . y(n)(0) = Cn, then conjecture a solution of the form yh(t) = K e λt so that dmy(m)(t) = yh = Kλm e λt . dtm Substituting into the homogeneous differential equation gives λnK �an+ an−1λn−1 + . . . + a1λ + a0� e λt = 0 Because K = 0, and eλt = 0 for finite t, we therefore require λn an+ an−1λn−1 + . . . + a1λ + a0 = 0, which is the characteristic equation of the system. 17–2U ( s )Y ( s )H ( s )a sU ( s )Y ( s )N ( s )D ( s )n u m e r a t o r p o l y n o m i a ld e n o m i n a t o r p o l y n o m i a lu ( t ) = 01y ( t )s + 5 s + 62The assumed form of the solution yh(t) = K e λt is a solution for any value of λ satisfying the chatacteristic equation. Since the order of the characteristic polynomial is n, there will be n such roots, and the most general form of yh(t) will be nλit yh(t) = � Ki e i=1 where the constants Ki are determined from the initial conditions, and the λi are known as the eigenvalues (characteristic values) of the system. We note that if we write the the transfer function N(s)H(s) = D(s) then the characteristic equation (in terms of s is D(s) = 0. Example 1 Find the homogeneous response of the system when y(0) = 2, and ˙y(0) = 1. The characteristic equation is s 2 + 5s + 6 = 0 (s + 3)(s + 2) = 0 so that s1, s2 = −3, −2 17–30 0 . 5 1 1 . 5 2- 6- 4- 2024687 e- 2 t- 5 e- 3 tt o t a l r e s p o n s etr e s p o n s eand the response is yh(t) = K1 e−3t + K2 e−2t . K1 and K2 are found from the initial conditions. When t = 0: y(0) = 2 = K1 + K2 y˙(0) = 1 = −3K1 − 2K2 from which K1 = −5, and K2 = 7. The complete solution is yh(t) = −5 e−3t + 7 e−2t . Notice that this system is stable, that is all exponential components in the ho-mogeneous response decay to zero as time t increases. a) The Particular Solution yp(t): The particular solution is often found using the method of undetermined coefficients, in which a solution of a given form, with unknown coefficients, is assumed for the given input function u(t), and the coefficients are found to satisfy the differential equation. The following table summarizes the assumed solution forms for some common input functions: 17–4Term in u(t) Assumed form for yp(t) Test value k ktn (n = 1, 2, 3 . . .) keλt kejωt k cos(ωt) k sin(ωt) K1 Kntn + Kn−1tn−1 + · · · + K1t + K0 K1eλt K1ejωt K1 cos(ωt) + K2 sin(ωt) K1 cos(ωt) + K2 sin(ωt) 0 0 λ jω jω jω Example 2 Find the particular solution, and the complete solution for the system 1 H(s) = s2 + 5s + 6 when the input u(t) = 4 (for t > 0) and y(0) = 0 and ˙y(0) = 0. The differential equation is d2y dy+ 5 + 6y = u(t). dt2 dt From the above table, for a constant input we assume yp(t) = K and substituting into the differential equation d2yp dyp+ 5 + 6yp = K dt2 dt 0 + 5 × 0 + 6K = 4 or yp(t) = K = 2/3. From the previous example, yh(t) = K1 e−3t + K2 e−2t, so the complete solution is 2 y(t) = yh(t) + yp(t) = K1 e−3t + K2 e−2t + . 3 With the stated initial conditions (y(0) = 0 and ˙y(0) = 0), y(0) = 0 = K1 + K2 + 2/3, y˙(0) = 0 = −3K1 − 2K2 …
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