Chapter 9, Solution 1. (a) (b) angular frequency frequency f = ω = 103 rad/s ω = Hz 2π (c) period T = 1 = ms f (d) Since sin(A) = cos(A – 90°), vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V vs( ms) = 12 sin((10 3 )( × 10 -3 ) + 24°) = 12 sin( + 24°) = 12 sin(° + 24°) = V (e) Chapter 9, Solution 2. (a) (b) (c) (d) amplitude = 8 A ω = 500π = rad/s f = ω = 250 Hz 2π Is =. | Chapter 9 Solution 1. a angular frequency 103 rad s b frequency f - Hz 1 c period T f ms d Since sin A cos A - 90 vs 12 sin 103t 24 12 cos 103t 24 - 90 vs in cosine form is vs 12 cos 103t - 66 V e vs ms 12sin 103 x 10 3 24 12 sin 24 12 sin 24 V Chapter 9 Solution 2. a amplitude 8 A b 500n rad s c f 250 Hz 2n d Is 8Z-25 A Is 2 ms 8cos 500n 2 x 10-3 - 25 8 cos n - 25 8 cos 155 A Chapter 9 Solution 3. a 4 sin t - 30 4 cos t - 30 - 90 4 cos t - 120 b -2 sin 6t 2 cos 6t 90 c -10 sin t 20 10 cos t 20 90 10 cos 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. v1 20 sin Z 60 20 cos at 60 - 90 20 cos at - 30 v2 60 cos t - 10 This indicates that the phase angle between the two signals is 20 and that v1 lags v . 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 vi t by 170 . c x t 13 cos 2t 5 sin 2t 13 cos 2t 5 cos 2t - 90 X 13Z0 5Z-90 13 - j5 x t cos 2t - y t 15 cos 2t - phase difference Thus y t leads x t by . Chapter 9 Solution 7. If f cos j sin df -smo j cos j cos jsin jf d0 df f jd Integrating both sides In f j In A f Ae cos j sin f 0 A 1 . f ej cos j sin Chapter 9 Solution 8. 15Z45 . _ 15Z45 . a 3 - j4 j2 j2 j2 j2 b 2 j 3 - j4 6 - j8 j3 4 10 - j5 8Z -20 10 _ 8Z -20 -5 - j12 10 2 j 3-j4 -5 j12 25 144 - - - - - c 10 8Z50 10 - Chapter 9 Solution 9. a 3 j4 2 ---- 5 - j8 2 3 j4 5 j8 25 64 2 15 j24 j20 - 32 89 b o 1 - j2 _ o - 3Z6 - 3Z6
