Tài liệu Bài giải mạch P9 docx

Chapter 9, Solution 1.

(a) angular frequency ω = 103

rad/s

(b) frequency f =

π

ω

2

= 159.2 Hz

(c) period T = f =

1

6.283 ms

(d) Since sin(A) = cos(A – 90°),

vs = 12 sin(103

t + 24°) = 12 cos(103

t + 24° – 90°)

vs in cosine form is vs = 12 cos(103

t – 66°) V

(e) vs(2.5 ms) = 12sin((10 )(2.5 10 ) 24 ) 3 -3 × + °

= 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°)

= 2.65 V

Chapter 9, Solution 2.

(a) amplitude = 8 A

(b) ω = 500π = 1570.8 rad/s

(c) f =

π

ω

2

= 250 Hz

(d) Is = 8∠-25° A

Is(2 ms) = 8cos((500 )(2 10 ) 25 ) -3 π × − °

= 8 cos(π − 25°) = 8 cos(155°)

= -7.25 A

Chapter 9, Solution 3.

(a) 4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°)

(b) -2 sin(6t) = 2 cos(6t + 90°)

(c) -10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)

Chapter 9, Solution 4.

(a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

(b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

Chapter 9, Solution 5.

v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°)

v2 = 60 cos(ωt − 10°)

This indicates that the phase angle between the two signals is 20° and that v1 lags

v2.

Chapter 9, Solution 6.

(a) v(t) = 10 cos(4t – 60°)

i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)

Thus, i(t) leads v(t) by 20°.

(b) v1(t) = 4 cos(377t + 10°)

v2(t) = -20 cos(377t) = 20 cos(377t + 180°)

Thus, v2(t) leads v1(t) by 170°.

(c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)

X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04°

x(t) = 13.928 cos(2t – 21.04°)

y(t) = 15 cos(2t – 11.8°)

phase difference = -11.8° + 21.04° = 9.24°

Thus, y(t) leads x(t) by 9.24°.

Chapter 9, Solution 7.

If f(φ) = cosφ + j sinφ,

-sin jcos j(cos jsin ) jf( ) d

df = φ + φ = φ + φ = φ φ

= jdφ f

df

Integrating both sides

ln f = jφ + ln A

f = Aejφ

= cosφ + j sinφ

f(0) = A = 1

i.e. f(φ) = ejφ

= cosφ + j sinφ

Chapter 9, Solution 8.

(a) 3 j4

15 45

−

∠ °

+ j2 = ∠ °

∠ °

5 - 53.13

15 45

+ j2

= 3∠98.13° + j2

= -0.4245 + j2.97 + j2

= -0.4243 + j4.97

(b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°

(2 j)(3 - j4)

8 - 20

+

∠ ° + - 5 j12

10

+

= ∠ °

∠ °

11.18 - 26.57

8 - 20

+

25 144

(-5 j12)(10)

+

−

= 0.7156∠6.57° − 0.2958

− j0.71

= 0.7109 + j0.08188 −

0.2958 − j0.71

= 0.4151 − j0.6281

(c) 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°

= 109.25 – j31.07

Chapter 9, Solution 9.

(a) 2 +

5 j8

3 j4

−

+

= 2 +

25 64

(3 j4)(5 j8)

+

+ +

= 2 +

89

15 + j24 + j20 − 32

= 1.809 + j0.4944

(b) 4∠-10° +

∠ °

−

3 6

1 j2

= 4∠-10° +

∠ °

∠ °

3 6

2.236 - 63.43

= 4∠-10° + 0.7453∠-69.43°

= 3.939 – j0.6946 + 0.2619 – j0.6978

= 4.201 – j1.392

(c) ∠ °− ∠ °

∠ °+ ∠ °

9 80 4 50

8 10 6 - 20

=

1.5628 j8.863 2.571 j3.064

7.879 j1.3892 5.638 j2.052

+ − −

+ + −

=

1.0083 j5.799

13.517 j0.6629

− +

− = ∠ °

∠ °

5.886 99.86

13.533 - 2.81

= 2.299∠-102.67°

= -0.5043 – j2.243

Chapter 9, Solution 10.

(a) z1 = 6 − j8, z 2 = 8.66 − j5, and z 3 = −4 − j6.9282

z1 + z2 + z3 = 10.66 − j19.93

(b) 9.999 7.499

3

1 2 j

z

z z = +

Chapter 9, Solution 11.

(a) = (-3 + j4)(12 + j5) 1 2 z z

= -36 – j15 + j48 – 20

= -56 + j33

(b) ∗

2

1

z

z

=

12 j5

- 3 j4

−

+

=

144 25

(-3 j4)(12 j5)

+

+ +

= -0.3314 + j0.1953

(c) = (-3 + j4) + (12 + j5) = 9 + j9 1 2 z + z

1 2 z − z = (-3 + j4) – (12 + j5) = -15 – j

1 2

1 2

z z

z z

−

+

= -(15 j)

9(1 j)

+

+

= 2 2 15 1

- 9(1 j)(15 - j)

−

+

=

226

- 9(16 + j14)

= -0.6372 – j0.5575