� � ���� � � 18.03 Hour Exam III Solutions: April 23, 2010 1. (a) The miminal period is 2. (b) f(t) is even. 1 cos(πt) cos(2πt) cos(3πt)(c) xp(t) = ωn 2 + 2(ωn 2 − π2) + 4(ωn 2 − 4π2) + 8(ωn 2 − 9π2)+ · · · (d) There is no p e riodic solution when ωn = 0, π, 2π, 3π, . . .. 2. (a) � � � � � � or � � � � � � (b) � � 1 1 or � � −1 −1 (c) f�(t) = (u(t + 1) − u(t − 1)) − δ(t + 1) − δ(t − 1); f�(t) = u(t + 1) − u(t − 1), rf�(t) = −δ(t + 1) − δ(t − 1). sw(t − τ)u(τ ) dτ = 0 0 t t (e−(t−τ ) − e−3(t−τ )) dτ3. (a) v(t) = w(t) ∗ u(t) = t3τ 1 − e−3t = 23 − e−t + e−3t = e−t τt ee |0 − e−3t = (1 − e−t) − . 3 3 3 0 1 1 (b) W (s) = L[w(t)] = s + 1 − s + 3 . (c) W (s) = (s + 3) − (s + 1) = 2 , so p(s) = 12 (s2 + 4s + 3). (s + 1)(s + 3) s2 + 4s + 34. (a) s − 1= 1 − 1 � δ(t) − u(t), so e−s(s − 1) � δ(t − 1) − u(t − 1). s s s s + 10 a b(s + 1) + c 10(b) F (s) = = + . By coverup, a = = 1. By complex s3 + 2s2 + 10s s (s + 1)2 + 9 10 coverup (multiply through by (s + 1)2 + 9 and set s to be a ro ot, say −1 + 3i), b(3i) + c = 1 s + 1 9+3i −1+3i = −3i, so b = −1, c = 0, and F (s) = s − (s + 1)2 + 9, which is the Laplace transform of 1 − e−t cos(3t). 5. (a) {0, −1 + 3i, −1 − 3i}. �� 2 s + 10 2 (b) X(s) = W (s)F (s). F (s) = s2 + 4, so X(s) = s3 + 2s2 + 10s s2 + 4 .MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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