chap09

chap09 - Chapter 9 Solution 1(a(b angular frequency...

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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 ° )
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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 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 ) = 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
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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
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= 4 -10 ° + 0.7453 -69.43 ° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392 (c) ° ° ° + ° 50 4 80 9 20 - 6 10 8 = 064 . 3 j 571 . 2 863 . 8 j 5628 . 1 052 . 2 j 638 . 5 3892 . 1 j 879 . 7 + + + = 799 . 5 j 0083 . 1 6629 . 0 j 517 . 13 + = ° ° 86 . 99 886 . 5 81 . 2 - 533 . 13 = 2.299 -102.67 ° = -0.5043 – j2.243 Chapter 9, Solution 10. (a) z 9282 . 6 4 z and , 5 66 . 8 z , 8 6 3 2 1 j j j = = = 93 . 19 66 . 10 3 2 1 j z z z = + + (b) 499 . 7 999 . 9 3 2 1 j z z z + = Chapter 9, Solution 11. (a) = (-3 + j4)(12 + j5) 2 1 z z = -36 – j15 + j48 – 20 = -56 + j33 (b) 2 1 z z = 5 j 12 4 j 3 - + = 25 144 ) 5 j 12 )( 4 j 3 (- + + + = -0.3314 + j0.1953 (c) = (-3 + j4) + (12 + j5) = 9 + j9 2 1 z z + 2 1 z z = (-3 + j4) – (12 + j5) = -15 – j 2 1 2 1 z z z z + = ) j 15 ( - ) j 1 ( 9 + + = 2 2 1 15 j) - 15 )( j 1 ( 9 - + = 226 ) 14 j 16 ( 9 - + = -0.6372 – j0.5575
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Chapter 9, Solution 12. (a) = (-3 + j4)(12 + j5) 2 1 z z = -36 – j15 + j48 – 20 = -56 + j33 (b) 2 1 z z = 5 j 12 4 j 3 - + = 25 144 ) 5 j 12 )( 4 j 3 (- + + + = -0.3314 + j0.1953 (c) = (-3 + j4) + (12 + j5) = 9 + j9 2 1 z z + 2 1 z z = (-3 + j4) – (12 + j5) = -15 – j 2 1 2 1 z z z z + = ) j 15 ( - ) j 1 ( 9 + + = 2 2 1 15 j) - 15 )( j 1 ( 9 - + = 226 ) 14 j 16 ( 9 - + = -0.6372 – j0.5575 Chapter 9, Solution 13.
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