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MIT 18 03 - Problem Set 8 - Part II Solutions

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18.03 Problem Set 8: Part II Solutions Part I points: 29. 4, 30. 0, 32. 8, 33. 4. Part II solutions: 29. (a) [8] (1) {1, i, −i}: For large t, these functions have exponential growth rate et . (This means that for any a < 1 < c, eat < |f(t)| < ect for large t.) They also show a small oscillation of approximately constant amplitude and circular frequency 1. a b Examples: f(t) = aet + b sin(t) (a, b = 0) with F (s) = + .≥s − 1 s2 + 1 (2) {−1 + 4i, −1 − 4i}: For large t, these functions show exponential decay like e−t , and oscillate with circular frequency 4. Examples: f(t) = ae−t sin(4t) (a = 0) with ≥4a F (s) = . (s + 1)2 + 16 (3) {−1}: For large t, these functions decay like e−t and do not oscillate. Examples: a f(t) = ae−t (a = 0) with ≥ F (s) = s + 1 . (4) No poles: For large t, these functions decay to zero faster than any exponential. Examples: f(t) = aω(t − b) (a = 0≥−, b bs� 0) with F (s) = ae−bt, or f(t) = a(u(t) − u(t − b)) (a = 0≥ , b > 0) with F (s) = a 1 − se. (b) (i) [4] Method I: For t > 0, w(t) is a solution to the homogeneous equation. The roots must be −21 ± 23 i, so p(s) = m � s − (−21 + 23 i) �� s − (−21 − 23 i) � = m(s2 +s+ 25 ). To find m, remember that w˙ (0+) = 1 (for a second order system). w˙ (t) = u(t)e−t/2(3 cos(3t/2) −m 2 (−12 sin(3t/2)), so w˙ (0+) = 32 , m = 23 , and p(s) = 32 (s2 + s + 52 ). 3/2 1 Method II: W (s) = L[w(t)] = (s + (1/2))2 + (9/4) = (2/3)(s2 + s + (5/2)), so p(s) = 23 (s2 + s + 52 ). (ii) [4] {−12 ± 3 i}.2 3/2 (iii) [4] This is a throwback problem. The complex gain is W (i�) = , so ((5/2) − �2) + i� 3/2 the gain is W (i�) = � . This is maximized when the denominator, | | ((5/2) − �2)2 + �2 or its square (52 − �2)2 + �2, is minimized. The derivative with respect to � is 2(52 − �2)(−2�) + 2�, which has roots at � = 0 and � = ±→2. So �r = →2 is the worst frequency. (iv) [6] The yellow box lies in the plane above the imaginary axis. The base is the amplitude response curve. The green box lies in the plane above the real axis. Its top lies on the graph of W (s) , and its base is the real axis. The red arrows lie above the | |poles of W (s). The yellow diamonds are located at (±i�, |W (i�)|), and represent the chosen value of the input circular frequency and the corresponding gain. 32. (a) [8] p(s) = s2 +4s+3 = (s+2)2 −1 has roots r1 = −1 and r2 = −3. Basic solutions are given by x1 = e−t and x2 = e−3t . (The order is not determined, and in fact any other pair of linearly independent solutions count as “basic.”) x˙1 = −e−t , x˙2 = −3e−3t .� � � � � � � � � � � � � � � � � � � � p(s) = s2 + s + 5 = (s + 1 )2 + 9 has roots r = −1 3 i. Basic solutions are given by x1 = 2 2 4 22 ± −t/2(−1 e−t/2 cos(32 t) and x2 = e−t/2 sin(32 t). (Same caveats as above.) x˙1 = e2 cos(32 t) −3 sin(3 t)). x˙2 = e−t/2(−1 sin(3 t) + 3 cos(3 t)).2 2 2 2 2 2 0 1 0 1 (b) [4] , 5 . −3 −4 −2 −1 1 The ray containing corresponds to −� 1 �(c) [4] 1 x1; the ray containing corresponds −3 to x2. 1 x (d) [4] The solution passing through at t = 0 is where x is the solution to 0 x˙x¨ + 4 ˙x + 3x = 0 with x(0) = 1, x˙(0) = 0. The general solution x(t) = c1e−t + c2e−3t has x(0) = c1 + c2 and x˙(0) = −c1 − 3c2, so c1 + c2 = 1 and −c1 − 3c2 = 0. Thus c2 = −21 and c1 = 23 and x(t) = 12 (3e−t − e−3t), so u(t) = xx˙((tt)) = 21 −33ee−−tt −+ 3e−e3−t 3t . Description of the graph of x(t): For t << 0 it is very negative and increasing. x(t) = 0 for t = −ln 3 . It reaches a maximum x(0) = 1, and then falls back through an inflection 2 point to become asymptotic to x = 0 as t � �. 3e−(t−a) −3(t−a) (e) [4] u(t) = 12 −3e−(t−a) −+ 3ee−3(t−a) . (f) [8] The trajectory of interest is the spiral pass-ing through (0, 1). x(t) 0 The solution u(t) = passing through at t = 0 is given by the solution x(t)x˙(t) 1 of x¨ + ˙x + 52 x = 0 with x(0) = 0, ˙x(0) = 1. The general solution x(t) = e−t/2(c1 cos(32 t) + c2 sin(32 t)) has x(0) = c1 and x˙ (0) = 12 c1 + 32 c2, so c1 = 0 and � c2 = 23 . Thus x(t) � =−2 2 1 −t/23 sin(32 t) 3 e−t/2 sin(23 t), ˙x(t) = e−t/2(cos(23 t) − 3 sin(23 t)), and u(t) = ecos(3 t) − 1 sin(3 t). 2 3 2 This passes through the y axis when x(t) = 0, i.e. when sin( 32 t) = 0, which is when t is an integral multiple of 2�/3.� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � Both x(t) and y(t) are damped sinuosoids, and x(0) = 0 and x˙(0) = 1, y(0) = 1, y˙(0) = −1. x(t) x˙(t)(g) [4] The velocity vector of u(t) = is u˙ (t) = . When x˙(t) = 0, x˙(t) x¨(t) 0 0 1 then, u˙ (t) = , which is vertical. Alternatively, u˙ = Au and A = , so if x¨(t) c d x 0 1 x 0 u = then u˙ = = x , which is vertical. 0 c d 0 c 33. (a) [8] A = 0 1 has characteristic polynomial pA(�) = det −� 1= −3 −4 −3 −4 − � −�(−4−�)+3 = �2 +4�+3 (the same as the characteristic polynomial of D2 +4D+3I!), which has roots �1 = −1, �2 = −3. A vector v is an eigenvector for eigenvalue � when 1 1 (A − �I)v = 0. With � = −1 this gives v1 = 0, so a nonzero eigen-� � −3 −3 1 value for � = −1 is v1 = or any …


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MIT 18 03 - Problem Set 8 - Part II Solutions

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