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Solutions to Problem Set 4: Part II in Complex Analysis, Exercises of Differential Equations

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

2011/2012

Uploaded on 08/07/2012

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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 = ce2t, so x = e3t( 3 cos(4t) + 4 sin(4t)) + ce2t
.
25 25
(b) [4] z¨ + ˙z + 2z = eit has solution zp = eit/p(i); p(i) = 1 + i + 2 = 1 + i so
1 it 1i 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 = 1eiπ/4eit = 2 ei(tπ/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 = Aet/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 = et/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 = et/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()4 ω2 + 2 | |
(4 ω2)2 + 1 ω2
4
Im(p()) ω/2
The phase lag φ is Arg(H(ω)) = Arg(p()) so tan φ = Re(p()) =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 = tet
.
(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 + c3et
. x = xp + xh.
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Download Solutions to Problem Set 4: Part II in Complex Analysis and more Exercises Differential Equations in PDF only on Docsity!

18.03 Problem Set 4: Part II Solutions

Part I points: 13. 3, 14. 8, 15. 5, 16. 4.

(3+4i)t 5+4i

  1. (a) [4] z˙ + 2z = e has solution zp = e

(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

it

√ 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

  • s + 2 = (s +

1 )

2

7 with roots −

1

√ 7 , so

e

2

−t/ 2

7

�√^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

2 π

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.

  1. (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 = 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 =

φ

π � 0 .7854. ω 4

(b) [8] H(ω) = = 1

. The gain is H(ω) = �. p(iω) 4 − ω 2

2

(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

  1. (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 = t

2

    1. Try u = at

2

  • bt + c, so u¨ = 2a and t

2

  • 1 = u¨ − u = −at

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

  • c 3 e

−t

. x = xp + xh.

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