By Greg Knowles (Eds.)

ISBN-10: 0124169600

ISBN-13: 9780124169609

**Read Online or Download An Introduction to Applied Optimal Control PDF**

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**Additional info for An Introduction to Applied Optimal Control**

**Sample text**

Example 2 (Bushaw [lJ; Lee and Marcus [5J) Consider the mini mal time control to the origin for X + 2bx + k 2 x = u, x(O) (the damped linear oscillator) where b > lu(t)j S 1. First, (2) is equivalent to = x o, ° x(O) and k > ° = Yo (2) are constants and (3) To begin with, we shall suppose b2 - k2 ~ 0; (2) is then critically or over damped. j, = - t/JA, u*(t) = sgn(t/J(t)b) = sgn(t/J2(t». Writing this out, we see that .. [0 -lJ (t/J 1> t/J 2) = (t/J 1, t/J 2) k 2 that is, If; 2 - 2b + k2t/J 2 = 0, which has solutions if b2 - k 2 = e bt( 1X + f3t) t/J 2(t) = { «e" sinh(ll t + f3) if b2 - k 2 > 2bifJ 2 ° 0, 24 N X 25 26 II.

E. E. = const. E. E. is kinetic energy), so tmv 2 where v = dsjdt and s so at any time mgy = E - = arc length. Initially v = 0, y = 0, hence E = 0, tmv 2 = mgy, ds dt = J2gy, dt =~ = 'J2gy In this problem f( y,y ') = J1 + (dyjdx)2 dx. JY - (y) JY y I that is, JYJ1 + (y')2 = A, where A is a positive constant, or Y' - j9 . - y y -- I = const, 44 III. O). e) sin(te)' that is or x = B + tA(e - sine) and y = tA(l - cos e) (these curves are called cycloids). We solve for A and B by finding the solution passing through y = P a when x = a, y = P b when x = b.

Jb 2 + 1 in the lower right the solution of(3) passing through (0,0) with u = hand quadrant and by the solution of (3) passing through (0,0) with u = - 1 in the upper left hand quadrant, then the switching locus X2 = W(x 1 ) is as pictured in Fig. 11. The optimal control synthesizer is then for for X2 X2 > W(xd and on (I' _) < W(x 1) and on (r +). The verification of these details is exactly the same as in Section 2 Example 1. u ~-1 -------------t-------------~Xl u ~ ----_ ... +1 Fig. 11 27 4.

### An Introduction to Applied Optimal Control by Greg Knowles (Eds.)

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