� � 18.03 Problem Set 4: Part II Solutions Part I points: 13. 3, 14. 8, 15. 5, 16. 4. (3+4i)t 5+4i13. (a) [4] ˙z + 2z = e has solution zp = e(3+4i)t/((3 + 4i) + 2) = 25+16 e3t(cos(4t) + i sin(4t)) so xp = Re(zp) = e3t( 5 cos(4t) + 4 sin(4t)). The homogeneous equation has 41 41 general solution xh = ce−2t, so x = e3t( 3 cos(4t) + 4 sin(4t)) + ce−2t .25 25 (b) [4] z¨ + ˙z + 2z = eit has solution zp = eit/p(i); p(i) = −1 + i + 2 = 1 + i so 1 it 1−i 1 1 zp = 1+i e = 2 (cos t + i sin t) and xp = 2 cos t + 2 sin t. We can also write p(i) = 1 + i = √2eiπ/4, so zp = 1e−iπ/4eit = √2 ei(t−π/4) and xp = Re(zp) = √2 cos(t − π ). The √22 2 4 characteristic polynomial is p(s) = s2 + s + 2 = (s + 1 )2 + 7 with roots −1 √7 , so e2 −t/2 � 4 �√7 � 2 ±�√27 �� the homogeneous equation has general solution xh = a cos 2 t + b sin 2 t or xh = Ae−t/2 cos √27 t − φ . The general solution is x = xp + xh. (c) [4] I measure the first six zeros to be 2.36, 4.76, 7.12, 9.48, 11.89, 14.19. The successive differences are 2.40, 2.36, 2.36, 2.41, 2.30, with an average of 2.366. My me asured pseu-√7doperiod is twice this, 4.73. The damped circular frequency of this system is ωd = 2 so the pseudoperiod is 2π = 4π� 4.7496. Not bad agreement. ωd √7 (d) [4] I measure the amplitude as 0.71. It looks like xp(0) � 0.50. Computed amplitude √2 √2 π 1is � 0.707. Computed value is xp(0) = cos(− ) = . Good agreement! x˙p(t) = − √22 sin(t − π ) so x˙p(0) = − √2 sin(−π ) = 1 . 2 4 2 2 4 2 4 2 (e) [5] We need to select xh so that xh(0) = −xp(0) = −1 and ˙xh(0) = −x˙p(0) = −1 . It’s 2 2 more convenient to use the rectangular expression for xh for this. xh(0) = a so a = −1 .�� � � � � �� 2 x˙h = e−t/21 a + √7 b cos √7 t + ( ) sin √7 t so x˙h(0) = −1 a + √7 b. Thus b = 2 2 2 2 2 2 � −� � � ···� � � � �� √27 x˙h(0) + 12 a = √27 −12 − 14 = −2√37 . xh = e−t/2 −12 cos √27 t − 2√37 sin √27 t . 14. (a) [8] Since the input signal has amplitude 1, the gain is the amplitude of the system response. The equation is x¨ + 12 x˙ + 4x = 4 cos(2t). The complex replacement is z¨ + 12 z˙ + 4z = 4e2it . Since p(2i) = (2i)2 + 12 (2i) + 4 = i, zp = 4e2it/i, so xp = Re(zp) = 4 sin(2t): the gain is 4. Also the phase lag of the sine behind the cosine is φ = π 2 . The time lag is t0 = φ = π � 0.7854. ω 4 4 4 4 (b) [8] H(ω) = = 1 . The gain is H(ω) = � . p(iω)4 − ω2 + 2 iω | | (4 − ω2)2 + 1 ω2 4 Im(p(iω)) ω/2 The phase lag φ is −Arg(H(ω)) = Arg(p(iω)) so tan φ = Re(p(iω)) =4 − ω2 . 15. (a) [6] p(s) = s + 1 and p(−1) = 0, so we are in resonance. p�(s) = 1 so the ERF/Resonant gives xp = te−t . (b) [6] We can’t apply undetermined coefficients directly since p(0) = 0. Let u = x˙, so u¨− u = t2 + 1. Try u = at2 + bt + c, so u¨ = 2a and t2 + 1 = u¨ − u = −at2 − bt + (2a − c) implies a = −1, b = 0, 2a − c = 1 or c = −3: so up = −t2 − 3. Then xp is the integral of up: xp = −31 t3 − 3t. To solve the homogeneous equation, factor p(s) = s(s − 1)(s + 1) so xh = c1 + c2et + c3e−t . x = xp + xh.MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
View Full Document
Unlocking...