chap09

chap09 - Chapter 9, Solution 1. (a) angular frequency ω =...

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Unformatted text preview: Chapter 9, Solution 1. (a) angular frequency ω = 10 3 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 ° ), v s = 12 sin(10 3 t + 24 ° ) = 12 cos(10 3 t + 24 ° – 90 ° ) v s in cosine form is v s = 12 cos(10 3 t – 66 ° ) V (e) v s (2.5 ms) = 12 ) 24 ) 10 5 . 2 )( 10 sin(( 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 = 500 π = 1570.8 rad/s (c) f = π ω 2 = 250 Hz (d) I s = 8 ∠-25 ° A I s (2 ms) = ) 25 ) 10 2 )( 500 cos(( 8 3- ° − × π = 8 cos( π − 25 ° ) = 8 cos(155 ° ) = -7.25 A Chapter 9, Solution 3. (a) 4 sin( ω t – 30 ° ) = 4 cos( ω t – 30 ° – 90 ° ) = 4 cos( ω t – 120 ° ) (b)-2 sin(6t) = 2 cos(6t + 90 ° ) (c)-10 sin( ω t + 20 ° ) = 10 cos( ω t + 20 ° + 90 ° ) = 10 cos( ω t + 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. v 1 = 20 sin( ω t + 60 ° ) = 20 cos( ω t + 60 ° − 90 ° ) = 20 cos( ω t − 30 ° ) v 2 = 60 cos( ω t − 10 ° ) This indicates that the phase angle between the two signals is 20 ° and that v 1 lags v 2 . 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) v 1 (t) = 4 cos(377t + 10 ° ) v 2 (t) = -20 cos(377t) = 20 cos(377t + 180 ° ) Thus, v 2 (t) leads v 1 (t) by 170 ° . (c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90 ° ) X = 13 ∠ ° + 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 ) = cos φ + j sin φ , ) ( f j ) sin j (cos j cos j-sin d df φ = φ + φ = φ + φ = φ φ = d j f df Integrating both sides ln f = j φ + ln A f = Ae j φ = cos φ + j sin φ f(0) = A = 1 i.e. f( φ ) = e j φ = cos φ + j sin φ Chapter 9, Solution 8. (a) 4 j 3 45 15 − ° ∠ + j2 = ° ∠ ° ∠ 53.13- 5 45 15 + 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 ° j4)- j)(3 (2 20- 8 + ° ∠ + j12 5- 10 + = ° ∠ ° ∠ 26.57- 11.18 20- 8 + 144 25 ) 10 )( 12 j 5- ( + − = 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 + 8 j 5 4 j 3 − + = 2 + 64 25 ) 8 j 5 )( 4 j 3 ( + + + = 2 + 89 32 20 j 24 j 15 − + + = 1.809 + j0.4944 (b) 4 ∠-10 ° + ° ∠ − 6 3 2 j 1 = 4 ∠-10 ° + ° ∠ ° ∠ 6 3 63.43- 236 . 2 = 4 ∠-10 ° + 0.7453 ∠-69.43 ° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392 (c) ° ∠...
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chap09 - Chapter 9, Solution 1. (a) angular frequency ω =...

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