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18.03 Problem Set 4: Part II Solutions
Part I points: 13. 3, 14. 8, 15. 5, 16. 4.
(3+4i)t 5+4i
13. (a) [4] ˙z + 2z = ehas 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
e
2
−t/2 � 4 �√7 � 2 ±
�√
2
7 ��
the homogeneous equation has general solution xh = a cos 2 t + b sin 2 t
or xh = Ae−t/2 cos √
2
7 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 measured pseu-
√7
doperiod 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 1
is � 0.707. Computed value is xp(0) = cos(−) = . Good agreement! x˙p(t) =
− √2
2 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
� −
� � � ··· � � � � ��
√
2
7 x˙h(0) + 1
2 a = √
2
7 −1
2 − 1
4 = −2√
3
7 . xh = e−t/2 −1
2 cos √
2
7 t − 2√
3
7 sin √
2
7 t .
14. (a) [8] Since the input signal has amplitude 1, the gain is the amplitude of the
system response. The equation is x¨ + 1
2 x˙ + 4x = 4 cos(2t). The complex replacement is
z¨ + 1
2 z˙ + 4z = 4e2it
. Since p(2i) = (2i)2 + 1
2 (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 = −3
1 t3 − 3t. To solve the homogeneous equation, factor p(s) = s(s − 1)(s + 1) so
xh = c1 + c2et + c3e−t
. x = xp + xh.
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