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The solutions to problem set 4: part ii in complex analysis. It includes the calculations for finding the complex roots, the homogeneous solutions, and the comparison of measured values with computed values.
Typology: Exercises
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Part I points: 13. 3, 14. 8, 15. 5, 16. 4.
(3+4i)t 5+4i
(3+4i)t /((3 + 4 i) + 2) = 25+
e
3 t (cos(4t) +
i sin(4t)) so xp = Re(zp) = e
3 t (
5 cos(4t) +
4 sin(4t)). The homogeneous equation has 41 41
general solution xh = ce
− 2 t , so x = e
3 t (
3 cos(4t) +
4 sin(4t)) + ce
− 2 t . 25 25
(b) [4] z¨ + z˙ + 2z = e
it has solution zp = e
it /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 =
2 e
iπ/ 4 , so zp =
1 e
−iπ/ 4 e
√ 2 e
i(t−π/4) and xp = Re(zp) =
√ 2 cos(t −
π √ ). The 2 2 2 4
characteristic polynomial is p(s) = s
2
1 )
2
7 with roots −
1
√ 7 , so
e
2
−t/ 2
7
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
4 π � 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 x˙h(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/ 2 1 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.
system response. The equation is x¨ +
1 2
x˙ + 4x = 4 cos(2t). The complex replacement is
z¨ +
1 2
z˙ + 4z = 4e
2 it
. Since p(2i) = (2i)
2
1 2
(2i) + 4 = i, zp = 4e
2 it /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 t 0 =
π � 0 .7854. ω 4
(b) [8] H(ω) = = 1
. The gain is H(ω) = �. p(iω) 4 − ω 2
2
iω
(4 − ω
2 )
2
1 ω
2 4
Im(p(iω)) ω/ 2
The phase lag φ is −Arg(H(ω)) = Arg(p(iω)) so tan φ =
Re(p(iω))
4 − ω 2
� (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 = t
2
2
2
2 − bt + (2a − c)
implies a = −1, b = 0, 2 a − c = 1 or c = −3: so up = −t
2 − 3. Then xp is the integral of
up: xp = − 3
1 t
3 − 3 t. To solve the homogeneous equation, factor p(s) = s(s − 1)(s + 1) so
xh = c 1 + c 2 e
t
−t
. x = xp + xh.