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Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)
1.
Engineering Circuit Analysis,
7th Edition 1. Chapter Eightenn Solutions 10 March 2006 (a) 5, 10, 15, 20, 25 (all rad/s) (b) 5, 10, 15, 20, 25 (all rad/s) (c) 90, 180, 270, 360, 450 (all rad/s) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
2.
Engineering Circuit Analysis,
7th Edition 2. (a) ωo = 2π rad/s, f = 1 Hz, 10 March 2006 therefore T = 1 s. (b) ωo = 5.95 rad/s = 2π f rad/s, (c) ) ωo = 1 rad/s = 2πf rad/s, Chapter Eightenn Solutions f = 0.947 Hz, f = 1/2π Hz, therefore T = 1.056 s. therefore T = 2π s. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
3.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 3. v(t ) = 3 − 3cos(100π t − 40°) + 4sin(200π t − 10°) + 2.5cos 300π t V (a) Vav = 3 − 0 + 0 + 0 = 3.000 V (b) 1 Veff = 32 + (32 + 42 + 2.52 ) = 4.962 V 2 (c) T= (d) v(18ms ) = 3 − 3cos(−33.52°) + 4sin(2.960°) + 2.5cos(19.440°) = −2.459 V 2π ωo = 2π = 0.02 s 100π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 4. (a) t t v 0 2 0.55 -0.844 0.05 2.96 0.6 0.094 0.1 3.33 0.65 0.536 0.15 2.89 0.7 0.440 0.2 1.676 0.75 0 0.25 0 0.8 -0.440 0.3 -1.676 0.85 -0.536 0.35 -2.89 0.9 -0.094 0.4 -3.33 0.95 0.844 0.45 -2.96 1 2 0.5 (b) v -2 v′ = −4π sin 2π t + 7.2π cos 4π t = 0 ∴ 4sin 2π t = 7.2(cos 2 2π t − sin 2 2π t ) −4 ± 16 + 414.72 = 0.5817, − 0.8595 = sin 2π t 28.8 = 3.330 (0.5593 for smaller max) ∴ 4sin 2π t = 7.2(1 − 2sin 2 2π t ) ∴ x = ∴ t = 0.09881, 0.83539 ∴ vmax (c) vmin = 3.330 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
5.
Engineering Circuit Analysis,
7th Edition 5. Chapter Eightenn Solutions 10 March 2006 (a) a0 = 0 (b) a0 = 0 (c) a0 = 5 (d) a0 = 5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
6.
Engineering Circuit Analysis,
7th Edition 6. Chapter Eightenn Solutions 10 March 2006 (a) a0 = 0 (b) a0 = 0 (c) a0 = 100 (d) a0 = 100 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
7.
Engineering Circuit Analysis,
7th Edition 7. Chapter Eightenn Solutions 10 March 2006 (a) a0 = 3, a1 = 0, a2 = 0, b1 = 0, b2 = 0 (b) a0 = 3, a1 = 3, a2 = 0, b1 = 0, b2 = 0 (c) a0 = 0, a1 = 0, a2 = 0, b1 = 3, b2 = 3 (d) 3 cos(3t − 10o ) = 3cos 3t cos10o + 3sin 3t sin10o a0 = 0, a1 = 3cos10o = 2.954, a2 = 0, b1 = 3sin10o = 0.521, b2 = 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
8.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions ao = 1 T b1 = 2 T 2 2 5 10 f (t ) sin ω0tdt = ∫0 ∫1 5sin π tdt = − 2π cos π t 1 = − π T 2−0 b2 = 8. 10 March 2006 2 T 2 2 5 ∫0 f (t ) sin 2ω0tdt = 2 − 0 ∫1 5sin 2π tdt = − 2π cos 2π t 1 = 0 T ∫ T 0 f (t )dt = 2.5 . a1 = a2 = 0 since function has odd symmetry 2 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 2 ao = 1 T a1 = 9. 2 T 2 2 ⎛ 2π f (t ) cos ω0tdt = ∫ 2 cos ⎜ ∫0 0 T 3 ⎝ 3 ∫ T 0 f (t )dt = 1 2 2 ∫0 2dt = 3 t 0 = 3 10 March 2006 4 . 3 4⎛ 3 ⎞ t ⎟ dt = ⎜ 3 ⎝ 2π ⎠ ⎞ ⎛ 2π ⎟ sin ⎜ ⎠ ⎝ 3 2 ⎞ t ⎟ = − 0.551 ⎠0 2 2 T 2 2 ⎛ 4π a2 = ∫ f (t ) cos 2ω0tdt = ∫ 2 cos ⎜ T 0 3 0 ⎝ 3 ⎞ t ⎟ dt = ⎠ 4⎛ 3 ⎜ 3 ⎝ 4π ⎞ ⎛ 4π ⎟ sin ⎜ ⎠ ⎝ 3 ⎞ t ⎟ = 0.276 ⎠0 2 T 2 2 ⎛ 6π a3 = ∫ f (t ) cos 3ω0tdt = ∫ 2 cos ⎜ T 0 3 0 ⎝ 3 ⎞ t ⎟ dt = ⎠ 4⎛ 3 ⎜ 3 ⎝ 6π ⎞ ⎛ 6π ⎟ sin ⎜ ⎠ ⎝ 3 ⎞ t⎟ = 0 ⎠0 2 T 2 2 ⎛ 2π b1 = ∫ f (t ) sin ω0tdt = ∫ 2sin ⎜ 0 0 T 3 ⎝ 3 2 2 ⎞ ⎛4⎞ 3 ⎛ 2π ⎞ t ⎟ dt = − ⎜ ⎟ cos ⎜ t ⎟ = 0.955 ⎠ ⎝ 3 ⎠ 2π ⎝ 3 ⎠0 2 T 2 2 ⎛ 4π b 2 = ∫ f (t ) sin 2ω0tdt = ∫ 2sin ⎜ T 0 3 0 ⎝ 3 ⎞ ⎛4⎞ 3 ⎛ 4π t ⎟ dt = − ⎜ ⎟ cos ⎜ ⎠ ⎝ 3 ⎠ 4π ⎝ 3 2 ⎞ t ⎟ = 0.477 ⎠0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10.
Engineering Circuit Analysis,
7th Edition 10. Chapter Eightenn Solutions 10 March 2006 h(t) = –3 + 8 sin πt + f(t) Use linearity and superposition. T = 2 s. ao = − 3 + 1 T 1 ∫0 f (t )dt = −3 + 2 = − 2.5 . T a2 = 0 b1 = 8 + b2 = 2 T 2 2 2 ∫0 f (t ) sin ω0tdt = 8 + 2 ∫0 (1) sin π tdt = 8 − π = 7.36 T 2 T 2 2 ∫0 f (t ) sin 2ω0tdt = 2 ∫1 (1) sin 2π tdt = 0 T PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 11. (a) T = 10 s, Fav = ao = 0.1(2 × 4 + 2 × 2) = 1.200 (b) Feff = 10 March 2006 2 2 1 (4 − t ) 2 dt = 0.2 ∫ (16 − 8t + t 2 ) dt 5∫ 0 0 2 2 ⎡ 2 1 3 ⎤ 8⎞ ⎛ 2 = 0.2 ⎢16t −4t + t ⎥ = 0.2 ⎜ 32 − 16 + ⎟ = 1.9322 0 3 0⎥ 3⎠ ⎝ ⎢ 0 ⎣ ⎦ 2 2π t a3 = × 2 ∫ (4 − t ) cos 3 × dt = 0.4 ∫ 4 cos 0.6π t dt − 0.4 ∫ t cos 0.6π t dt 10 10 0 0 0 2 (c) 2 2 2 2 1 t ⎛ 1 ⎞ = 1.6 sin 0.6π t −0.4 ⎜ cos 0.6π t + sin 0.6π t ⎟ 2 0.6π 0.6π ⎝ 0.36π ⎠0 0 = 8 10 4 sin1.2π − 2 (cos1.2π − 1) − sin 1.2π = −0.04581 3π 9π 3π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12.
Engineering Circuit Analysis,
7th Edition 12. (a) Chapter Eightenn Solutions 10 March 2006 T=8−2=6s 1 Hz 6 (b) fo = (c) ω o = 2π f o = (d) 1 ao = (10 × 1 + 5 × 1) = 2.5 6 (e) b2 = π 3 rad/s 3 4 2⎡ 2π t 2π t ⎤ 10sin dt + ∫ 5sin dt ⎥ ⎢∫ 6 ⎣2 3 3 3 ⎦ 3 1 ⎡ 30 2π t 15 2π t cos cos = ⎢− − 3 ⎢ 2π 3 2 2π 3 ⎣ 1 ⎡ 15 ⎛ 4π ∴ b2 = ⎢ − ⎜ cos 2π − cos 3⎣ π ⎝ 3 ⎤ ⎥ 3⎥ ⎦ 4 8π 7.5 ⎞ 7.5 ⎛ ⎞ ⎤ 1 ⎡ 15 ⎤ − cos 2π ⎟ ⎥ = ⎢ − (1.5) − (−1.5) ⎥ = −1.1937 ⎟− ⎜ cos 3 π ⎠ π ⎝ ⎠⎦ 3 ⎣ π ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 13. a3 = = 4 3 3 4 ⎤ 2⎡ 6π t 6π t ⎤ 1 ⎡10 5 10 cos dt + ∫ 5cos dt ⎥ = ⎢ sin π t − sin π t ⎥ ⎢∫ 6 ⎣2 6 6 3⎥ 2 π 3 ⎦ 3⎢π ⎣ ⎦ 10 ⎛ 1 1 ⎞ ⎜ sin 3π − sin 2π + sin 4π − sin 3π ⎟ = 0 3π ⎝ 2 2 ⎠ 4 3 3 4 ⎤ ⎤ 1 ⎡ 10 1⎡ 5 b3 = ⎢ ∫ 10sin π tdt + ∫ 5sin π t dt ⎥ = ⎢ − cos π t − cos π t ⎥ 3 ⎣2 2 π 3⎥ 3 ⎦ 3⎢ π ⎣ ⎦ =− 10 ⎛ 1 1 10 ⎞ ⎜ cos 3π − cos 2π + cos 4π − cos 3π ⎟ = − (−1) = 1.0610 3π ⎝ 2 2 3π ⎠ 2 a3 + b32 = 1.0610 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14.
Engineering Circuit Analysis,
7th Edition 14. Chapter Eightenn Solutions 10 March 2006 2π = 12.5 ms, ave value = 1.9 160π (a) 3.8cos 2 80πt = 1.9 + 1.9 cos160πt , T = (b) 3.8cos3 80πt = (3.8cos80πt )(0.5 + 0.5cos 160πt ) = 1.9 cos80πt + 0.95cos 240πt + 0.95cos80πt = 2.85cos80πt + 0.95cos 240πt 2π = 25 ms, ave value = 0 T= 80π (c) 3.8cos 70πt − 3.8sin 80πt; ωot = πt , ωo = π, T = 2π = 2 s; ave value = 0 π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 15. T = 2 s 1 21 4 × 2πt 1 b4 = ∫ sin dt = − cos 4πt 20 2 4π 0 t (a) t 1 (1 − cos 4πt1 ) 4π π max when 4πt1 = , t1 = 0.125 s 2 ∴ b4 = (b) b4 = 1 4π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 16. g (t ) = 5 + 8cos10t − 5cos15t + 3cos 20t − 8sin10t − 4sin15t + 2sin 20t 2π = 1.2566 s 5 (a) ωo = 5 ∴ T = (b) fo = (c) G av = −5 (d) 1 G eff = (−5) 2 + (82 + 52 + 32 + 82 + 42 + 22 ) = 116 = 10.770 2 5 10 β = 4 fo = = 3.183 Hz 2π π (e) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 17. T = 0.2, f (t ) = Vm cos 5πt , −0.1 < t < 0.1 0.1 an = 0.1 2 Vm cos 5πt cos10 nπt dt = 5Vm ∫ [ cos(5π + 10nπ)t + cos(10nπ − 5π)t ] dt 0.2 −∫ −0.1 0.1 0.1 1 1 ⎡ ⎤ = 5Vm ⎢ sin(10nπ + 5π) t + sin(10nπ − 5π)t ⎥ 10nπ − 5π ⎣10nπ + 5π ⎦ −0.1 V ⎡ 2 2 ⎤ = m⎢ sin(10nπ + 5π) 0.1 + sin(10nπ − 5π) 0.1⎥ π ⎣ 2n + 1 2n − 1 ⎦ Vm ⎡ 2 2 ⎤ ⎢ 2n + 1 sin(nπ + 0.5π) + 2n − 1 sin(nπ − 0.5π) ⎥ π ⎣ ⎦ V ⎡ 2 2 1 ⎞ ⎤ 2V ⎛ 1 cos nπ + (− cos nπ) ⎥ = m cos nπ ⎜ = m⎢ − ⎟ 2n − 1 π ⎣ 2n + 1 π ⎦ ⎝ 2n + 1 2n − 1 ⎠ 2V 2n − 1 − 2n − 1 4V cos nπ = m cos nπ =− m 2 2 4n − 1 π π 4n − 1 0.1 1 1 ⎡ π ⎛ π ⎞ ⎤ 2V ao = ∫ Vm cos 5πt dt = 5Vm 5π ⎢sin 2 − sin ⎜ − 2 ⎟⎥ = πm 0.2 −0.1 ⎠⎦ ⎝ ⎣ = ∴ v(t ) = 2Vm 4Vm 4V 4V 4V + cos10πt − m cos 20πt + m cos 30πt − m cos 40πt + ... π 3π 15π 35π 63π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
18.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 18. 1 − wave 2 (a) even, (b) bn = 0 for all n; aeven = 0; ao = 0 (c) b1 = b2 = b3 = 0, a2 = 0 nπt nπt nπ nπ ⎞ 8 10 6 20 ⎛ ∫ 5cos 6 dt = 3 nπ sin 6 1 = nπ ⎜ sin 3 − sin 6 ⎟ 12 1 ⎝ ⎠ 2 an = ∴ a1 = 2 20 ⎛ π π⎞ 20 ⎛ π⎞ 20 = −2.122 ⎜ sin − sin ⎟ = 2.330, a3 = ⎜ sin π − sin ⎟ = − π⎝ 3 6⎠ 3π ⎝ 2⎠ 3π PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
19.
Engineering Circuit Analysis,
7th Edition 19. (a) Chapter Eightenn Solutions 10 March 2006 ao = an = 0 ∴ y (t ) = 0.2sin1000πt + 0.6sin 2000πt + 0.4sin 3000πt (b) Yeff = 0.5(0.22 + 0.62 + 0.42 ) = 0.5(0.56) = 0.5292 (c) y (2ms) = 0.2sin 0.2π + 0.6sin 0.4π + 0.4sin 0.6π = 1.0686 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
20.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 20. (a) (b) (c) (d) [a]b5 = 0, a5 = 4 2π5t 32 6 5πt 3.2 ⎛ 15π 10π ⎞ ∫ 8cos 6 dt = 6 10π sin 3 2 = π ⎜ sin 3 − sin 3 ⎟ = 0.8821 62 ⎝ ⎠ [b]a5 = 0, b5 = 4 2π5t 32 ⎛ −6 ⎞ ⎛ 15π 10π ⎞ 3.2 ∫ 8sin 6 dt = 6 ⎜ 10π ⎟ ⎜ cos 3 − cos 3 ⎟ = − π (−0.5) = 0.5093 62 ⎝ ⎠⎝ ⎠ 3 3 (e) 3 10π ⎞ 8 2π5t 64 12 ⎛ 15π dt = − sin [c]b5 = 0, a5 = ∫ 8cos ⎜ sin ⎟ = 3.801 6 6 ⎠ 12 2 12 12 10π ⎝ 3 8 10πt 64 ⎛ 12 ⎞⎛ 15π 10π ⎞ ∫ 8sin 12 dt = 12 ⎜ − 10π ⎟⎜ cos 6 − cos 6 ⎟ = 1.0186 12 2 ⎠ ⎝ ⎠⎝ 3 [d ]a5 = 0, b5 = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
21.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 21. T = 4 ms (a) 1000 ao = 4 =− 0.004 ∫ 0 0.004 250 × 8 8sin125πt dt = cos125πt −125π 0 16 ⎛ π ⎞ 16 ⎜ cos − 1⎟ = = 5.093 2 ⎠ π π⎝ 0.004 a1 = 4000 (b) ∫ sin125πt cos 0 2πt dt 0.004 0.004 ∴ a1 = 4000 ∫ 0.004 sin125πt cos 500πt dt = 2000 0 0.004 = 0.004 ∫ (sin 625πt − sin 375πt ) dt 0 ⎛ cos 625πt cos 375πt ⎞ = 2000 ⎜ − + ⎟ 625π 375π ⎠0 ⎝ b1 = 4000 ∫ 3.2 5.333 (1 − cos 2.5π) − (1 − cos1.5π) = −0.6791 π π 0.004 sin125πt sin 500πt dt = 2000 0 ∫ (cos 375πt − cos 625πt ) dt 0 1 1 ⎞ ⎡ 1 ⎤ ⎛ −1 (sin1.5π) − (sin 2.5π) ⎥ = 2000 ⎜ = 2000 ⎢ − ⎟ = −2.716 625π ⎣ 375π ⎦ ⎝ 375π 625π ⎠ (c) −4 < t < 0 : 8sin125πt (d) b1 = 0, a1 = 4000 8 0.004 ∴ a1 = 2000 ∫ 0 = 0.004 ∫ 8sin125πt cos 250πt dt 0 ⎡ cos 375πt cos125πt ⎤ + [sin 375πt − sin125πt ] dt = 2000 ⎢− 375π 125π ⎥ 0 ⎣ ⎦ 0.004 π 5.333 16 (1 − cos1.5π ) + ⎛ cos − 1⎞ = −3.395+ ⎜ ⎟ π π⎝ 2 ⎠ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
22.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 22. 1 − wave ∴ ao = 0, an = 0, beven = 0 2 T = 10ms = 0.01 s odd and 8 ⎡ = ⎢ 0.01 ⎣ ⎤ ⎛ −1 ⎞ 10sin 200nπt dt ⎥ = 8000 ⎜ bodd ⎟ cos 200nπt ∫ ⎝ 200nπ ⎠ 0 ⎦ 40 40 ∴ bodd = − (cos 0.2nπ − 1) = (1 − cos 0.2nπ) nπ nπ ∴ b1 = 2.432, b3 = 5.556, b5 = 5.093, b7 = 2.381, b9 = 0.2702 0.001 0.001 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
23.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 23. odd and 1 8 − wave, T = 8 ms ∴ bn = 2 T 2π ωo = = 250π ∴ bn = 1000 T Now, 1 ∫ x sin ax dx = a ( sin a 2 T /4 ∫ f (t ) sin nωot dt 0 0.001 ∫ 1000 t sin 250πnt dt 0 x − ax cos ax ) , a = 250 nπ 106 0.001 sin 250nπt − 250nπt cos 250nπt )0 2 2 2 ( 250 n π nπ nπ 16 ⎛ π π π⎞ 16 ⎛ nπ ⎞ + 0 ⎟ ∴ b1 = 2 ⎜ sin − cos ⎟ = 0.2460 ∴ bn = 2 2 ⎜ sin − 0 − cos 4 4 π ⎝ 4 4 4⎠ 4 nπ ⎝ ⎠ f (t ) = 103 t ∴ bn = b3 = beven 16 ⎛ 3π 3π 3π ⎞ 16 ⎛ 5π 5π 5π ⎞ sin − cos ⎟ = 0.4275− ; b5 = sin − cos ⎟ = 0.13421 2 ⎜ 2 ⎜ 9π ⎝ 4 4 4 ⎠ 25π ⎝ 4 4 4 ⎠ =0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
24.
Engineering Circuit Analysis,
7th Edition 24. (a) even, T = 4: (c) odd, (d) even, 10 March 2006 odd, T = 4 (b) Chapter Eightenn Solutions 1 − wave: T = 8 2 1 − wave, T = 8 : 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
25.
Engineering Circuit Analysis,
7th Edition 25. (a) vs = 5 + Chapter Eightenn Solutions 10 March 2006 20 ∞ 1 2πnt 20 20 ∑ n sin 0.4π ∴ vsn = nπ sin 5nt , Vsn = nπ (− j1) π 1,odd Zn = 4 + j 5n 2 = 4 + j10n, I fn = Vsn − j 20 j5 = =− Zn nπ(4 + j10n) 1 + j 2.5n 12.5 + j 5 j 5 1 − j 2.5n =− 2 nπ 1 + 6.25n nπ(1 + 6.25n 2 ) 12.5 1 5 1 ∴ i fn = − cos 5nt + sin 5nt 2 π 1 + 6.25n nπ 1 + 6.25n 2 ∞ 1 5 ⎡ 12.5 ⎤ ∴ i f = 1.25 + ∑ − cos 5nt + sin 5nt ⎥ 2 ⎢ π nπ ⎦ 1,odd 1 + 6.25n ⎣ ∴ I fn = − (b) in = Ae −2t , i = i f + in , i (0) = 0, i f (0) = 1.25 + ∞ 1 ∑ 1 + 6.25n 1, odd ∴ i f (0) = 1.25 − 2 π ∞ ∑ 1,odd 2 ⎛ 12.5 ⎞ ⎜− ⎟ ⎝ π ⎠ 1 2 π = 1.25 − tanh 0.2π = 0.55388 n + 0.16 π 4 × 0.4 2 ∴ A = −0.55388, i = −0.55388e−2t + 1.25 + ∞ 1 ∑ 1 + 6.25n 1,odd 2 5 ⎡ 12.5 ⎤ ⎢ − π cos 5nt + nπ sin 5nt ⎥ ⎣ ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
26.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 26. (a) 0 < t < 0.2π : i = 2.5(1 − e −2t ) ∴ i (0.2π) = 2.5(1 − e−0.4 π ) = 1.78848 A (b) 0.2π < t < 0.4π : i = 1.78848 e−2( t −0.2 π ) ∴ i (0.4π) = 0.50902 A (c) 0.4π < t < 0.6π : i = 2.5 − (2.5 − 0.50902)e−2( t −0.4 π ) , i (0.6π) = 1.9335− PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
27.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 27. (a) vs = 5 + 20 ∞ 1 ∑ sin 5nt π 1,odd n 20 sin 5nt nπ 20 Vsn = − j nπ 1 1 1 − j 20 / nπ − j 20 / nπ 1 − j 20n Zn = 2 + = 2+ ∴ Vcn = × = × 2 + 1/ j10n j10n 1 + j 20n 1 − j 20n j 5n 2 j10n 20 1 −20n − j1 20 ∴ Vcn = × , vcn = ( −20n cos 5nt + sin 5nt ) 2 1 + 400n nπ nπ 1 + 400n 2 20 ∞ 1 ⎛1 ⎞ ∴ vcf = 5 + ∑ 1 + 400n2 ⎜ n sin 5nt − 20 cos 5nt ⎟ π 1,odd ⎝ ⎠ vsn = (b) vn = Ae −t / 4 (c) vc (0) = A + 5 + ∞ ∑ 1,odd 20 ∞ −20 1 ∑ 1 + 400n2 = A + 5 − π π 1,odd ∞ ∑ 1,odd 1 n + (1/ 20) 2 2 π π π 1 = = 5π tanh = 1.23117 tanh 2 4(1/ 20) 20 × 2 40 n + (1/ 20) 2 1 ∴ A = 0 − 5 + × 1.23117 = −4.60811 π 20 ∞ 1 ⎛1 ⎞ ∴ vc (t ) = −4.60811e− t / 4 + 5 + ∑ 1 + 400n2 ⎜ n sin 5nt − 20 cos 5nt ⎟ π 1,odd ⎝ ⎠ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
28.
Engineering Circuit Analysis,
7th Edition 28. Chapter Eightenn Solutions 10 March 2006 At the frequency ω = 10nπ ( ) 10 ⎡10 + j10nπ 5 ×10−3 ⎤ ⎦ Ω and I = 8 − j Zn = ⎣ ( ) Sn πn 20 + j10nπ 5 ×10−3 ( Therefore Vn = ) ⎡ 10 + j 0.05nπ ⎤ 80 (− j) ⎢ ⎥. nπ ⎣ 20 + j 0.05nπ ⎦ In the time domain, this becomes v1 (t ) = 2 ⎛ 40 ⎞ 1 + (0.005nππ ) cos 10nπ − 90o + tan −1 0.005nπ − tan −1 0.0025nπ ∑ ) ⎜ nπ ⎟ 2 ⎠ 1 + (0.0025nππ ) n =1 ( odd ⎝ ∞ ( PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. )
29.
Engineering Circuit Analysis,
7th Edition 29. Chapter Eightenn Solutions 10 March 2006 At the frequency ω = nπ I Ln = n −1 10 32 I Sn and I Sn = − j ( −1) 2 2 20 + jnπ 5 × 10−3 (π n ) ( ) Thus, in the time domain, we can write iL (t ) = n −1 ⎤ ⎡ 320 1 cos nπ t − 90o − tan −1 0.00025nπ ⎢ ( −1) 2 ⎥ ∑ ⎢ nπ 2 2 n =1 (odd) ( ⎥ 20 1 + ( 0.00025nπ ) ) ⎣ ⎦ ∞ ( ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
30.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 30. 0.001 0.005 ⎤ −3 103 ⎡ 100e − j 3×2 πt / 6×10 − ∫ 100e− j100 πt ⎥ ⎢ ∫ 6 ⎣ 0 0.003 ⎦ 0.001 0.005 ⎤ 105 ⎡ −1 1 − j1000 πt − j1000 πt = + e e ⎢ ⎥ 6 ⎢ j1000π j1000π 0 0.003 ⎥ ⎣ ⎦ 100 − jπ 100 = (1 + 1 − 1 + 1) = − j10.610 e + 1 + e − j 5 π − e − j 3π = j 6π j 6π c3 = ( ) ∴ c−3 = j10.610; c3 = 10.610 0.001 0.005 ⎤ 2 ×103 ⎡ a3 = ⎢ ∫ 100 cos100πt dt − ∫ 100 cos1000πt dt ⎥ 6 ⎣ 0 0.003 ⎦ 5 2 ×10 1 = ( sin π − 0 − sin 5π + sin 3π ) = 0 6 1000π 1 1 2 c3 = (a3 − jb3 ) = − j b3 ∴ b3 = 21.22 and a3 + b32 = 21.22 2 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
31.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 31. (a) (b) ⎤ 100e− j 400 πnt dt ⎥ ∫ ∫ 0 0.001 ⎦ 0.001 0.002 ⎡ ⎤ ∴ cn = 20, 000 ⎢ ∫ 1000t e − j 400 πnt dt + ∫ e − j 400 πnt dt ⎥ 0.001 ⎣ 0 ⎦ 0.002 ⎡ e − j 400 πnt ⎤ 1 0.001 − j 400 πnt e ( j 400πnt + 1)0 + ∴ cn = 20, 000 ⎢ ⎥ 2 2 − j 400πn ⎢160n π 0.001 ⎥ ⎣ ⎦ 1 ∴ co = ao = (50 × 10−3 + 100 × 10−3 ) = 0.15 × 200 = 30 0.005 ⎡ 1 ⎤ 1 1 c1 = 20, 000 ⎢ e − j 0.4 π (1 + j 0.4π) − e− j 0.8 π − e − j 0.4 π ⎥ − 2 2 j 400π 160π ⎣160π ⎦ 125 = 2 (1∠ − 72°) (1.60597∠51.488°) − 12.66515 + 15.91548 ∠90°(1∠ − 144° − 1∠ − 72°) π = 12.665(1∠ − 72°) (1 + j1.2566) − 12.665 + j15.915(1∠ − 144° − 1∠ − 72°) = 20.339∠ − 20.513° − 12.665 + 18.709∠ − 108° = 24.93∠ − 88.61° c2 = 3.16625∠ − 144° (1 + j 2.5133) − 3.16625 + j 7.9575(1∠ − 288° − 1∠ − 144°) T = 5 ms cm = 1 ⎡ ⎢ 0.005 ⎣ 0.001 105 te − j 400 πnt dt + 0.002 ( ) = 8.5645 ∠ − 75.697° − 3.16625 + 15.1361∠144° = 13.309∠177.43° PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
32.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 32. Fig. 17-8a: Vo = 8 V, τ = 0.2 μ s, f o = 6000 pps 1 1 , f o = 6000, τ = 0.2 μ s ∴ f = = 5 MHz τ 6000 (a) T= (b) f o = 6000 Hz (c) 6000 × 3 = 18, 000 (closest) ∴ c3 = 8 × 0.2 × 10−6 sin(1/ 2 × 3 ×12, 000π × 0.2 × 10−6 1/ 6000 0.0036 π ∴ c3 = 9.5998 mV (d) 2 ×106 8 × 0.2 ×10−6 sin(1/ 2 × 333 ×12, 000π × 0.2 × 10−6 = 333.3 ∴ c333 = = 7.270 mV 6 × 103 1/ 6000 1/ 2 × 333 ×12, 000π × 0.2 × 10−6 (e) β = 1/ τ = 5 MHz (f) 2 < ω < 2.2 Mrad/s ∴ (g) 2000 2200 < f < kHz or 318.3 < f < 350.1 kHz 2π 2π f o = 6 kHz ∴ f = 6 × 53 = 318; 324,330,336,342,348 kHz ∴ n = 5 c227 8 × 0.2 ×10−6 sin(1/ 2 × 227 × 12, 000π × 0.2 × 10−6 = = 8.470 mV 1/ 6000 (′′) f = 227 × 6 = 1362 kHz PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
33.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 33. T = 5 ms; co = 1, c1 = 0.2 − j 0.2, c2 = 0.5 + j 0.25, c3 = −1 − j 2, cn = 0, n ≥ 4 (a) an = − jbn = 2cn ∴ ao = co = 1, a1 − jb1 = 0.4 − jb1 = 0.4 − j 0.4, a2 − jb2 = 1 + j 0.5, a3 − jb3 = −2 ∴ v(t ) = 1 + 0.4 cos 400π t + cos800π t − 2 cos1200π t + 0.4sin 400π t − 0.5sin 800π t + 4sin1200π t (b) v (1 ms ) = 1 + 0.4 cos 72° + cos144° − 2 cos 216° + 0.4 sin 72° − 0.5sin144° + 4 sin 216° ∴ ∴ v (1 ms ) = −0.332V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
34.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 34. 0.6×10−6 (a) 106 t T = 5 μ s ∴ cn = × 2 ∫ 1cos 2π n dt 5 5 × 10−6 0.4×10−6 ∴ cn = 4 × 105 ∴ cn = 5 × 10−6 ( sin 43.2°n − sin 28.8°n ) 2π n 1 ( sin 43.2°n − sin 28.8°n ) nπ 1 (sin172.8° − sin115.2°) = −0.06203 4π (b) c4 = (c) co = ao = (d) a little testing shows co is max ∴ cmax = 0.08 (e) 0.01× 0.08 = 0.8 × 10−3 ∴ 0.2 × 10−6 + 0.2 ×10−6 = 0.08 5 × 10−6 1 ( sin 43.2°n − sin 28.8°n ) ≤ 0.8 ×10−3 nπ 125 ( sin 43.2°n − sin 28.8°n ) ≤ 1 nπ ok for n > 740 ∴ (f) β = 740 f o = 740 ×106 = 148 MHz 5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
35.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 35. T = 1/16, ω o = 32π 1/ 96 (a) c3 = 16 ∫ 40e 0 ∴ c3 = j (b) − j 96π t 16 × 40 − j 96π dt − e − j 96π 1/ 96 0 20 − jπ 40 (e − 1) = − j = − j 4.244 V 3π 3π Near harmonics are 2f o = 32 Hz, 3f o = 48 Hz Only 32 and 48 Hz pass filter an − jbn = 2cn a3 − jb3 = 2c3 = − j8.488 ∴ a3 = 0, b3 = 8.488 V I3 = 8.488 1 = 1.4536 ∠ − 31.10° A; P3 = × 1.45362 × 5 = 5.283 W 5 + j 0.01× 96π 2 c2 = 1 1/16 1/ 96 ∫ 0 40e − j 64π t dt = 640 (e − j 64π / 96 − 1) = 2.7566 − j 4.7746 V − j 64π a2 − b2 = 2c2 = 5.5132 − j9.5492 = 11.026 ∠ − 60° 11.026∠ − 60° = 2.046∠ − 65.39° A 5 + j 0.01× 64π 1 ∴ P2 = × 2.0462 × 5 = 10.465 W ∴ Ptot = 15.748 W 2 ∴ I2 = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
36.
Engineering Circuit Analysis,
7th Edition 36. Chapter Eightenn Solutions 10 March 2006 f (t ) = 5[u (t + 3) + u (t + 2) − u (t − 2) − u (t − 3)] f(t) (a) t -3 -2 ∞ (b) F( jω) = ∫ -1 0 1 2 3 f (t ) e − jωt dt −∞ −2 ∴ F( jω) = ∫ 5e −3 − j ωt 2 dt + ∫ 10e −2 − jωt 3 dt + ∫ 5e − jωt dt 2 5 10 − j 2 ω 5 (e j 2 ω − e j 3ω ) + (e (e − j 3ω − e − j 2 ω ) − e j 2ω ) + − jω − jω − jω 5 5 10 ( −e j 3ω + e − j 3ω ) + (e j 2 ω − e − j 2 ω ) + (−e j 2 ω + e − j 2 ω ) = − jω − jω − jω 5 5 10 (− j 2) sin 3ω + ( j 2) sin 2ω + = (− j 2) sin 2ω − jω − jω − jω 10 10 20 10 (sin 3ω + sin 2ω) ∴ F( jω) = sin 3ω − sin 2ω + sin 2ω = ω ω ω ω ∴ F( jω) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
37.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 37. (a) f (t ) = e − at u (t ), a > 0 ∴ F( jω ) = ∞ ∫ −∞ ∴ F( jω ) = −1 − ( a + jω )t e a + jω ∞ = 0 ∞ f (t )e − jωt dt = ∫ e− at e− jω t dt 0 1 a + jω ∞ (b) f (t ) = e at6 e − at u (t − to ), a > 0 ∴ F( jω ) = e ato ∫ e − ( a + jω )t dt to ∴ F( jω ) = e ato −1 − ( a + jω ) t e a + jω ∞ = e ato to 1 −1 ⎡ − ( a + jω )to ⎦ = ⎤ e − jω to ⎣ −e a + jω a + jω ∞ (c) f (t ) = te − at u (t ), a > 0 ∴ F( jω ) = ∫ te− ( a + jω ) t dt 0 ∴ F( jω ) = − ( a + jω ) t 1 1 e ∞ −(a + jω )t − 1]0 = 0 − [−1] = 2 [ 2 (a + jω ) 2 (a + jω ) (a + jω ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
38.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 38. −4 < t < 0 : f (t ) = 2.5(t + 4); 0 < t < 4 : f (t ) = 2.5(4 − t ) 0 ∴ F( jω) = ∫ −4 4 2.5(t + 4) e− jωt dt + ∫ 2.5(4 − 5)e− jωt dt 0 0 ln 1st , let t = τ ∴ I1 = ∫ 2.5(4 − τ)e jωτ (− d τ) 4 4 4 0 0 ∴ I1 = ∫ 2.5(4 − τ)e jωτ d τ ∴ F( jω) = 2.5∫ (4 − t )(e jωt + e− jωt ) dt 4 4 4 1 ∴ F( jω) = 5 ∫ (4 − t ) cos ωt dt = 20 × sin ωt − 5∫ cos ωt dt ω 0 0 0 20 5 4 sin 4ω − 2 (cos ωt + ωt sin ωt )0 ω ω 20 5 5 5 = sin 4ω − 2 (cos 4ω − 1) − 2 4ω sin 4ω = 2 (1 − cos 4ω) ω ω ω ω ∴ F( jω) = 2×5 ⎛ sin 2ω ⎞ or, F( jω) = 2 sin 2 2ω = 10 ⎜ ⎟ ω ⎝ ω ⎠ 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
39.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 39. π f (t ) = 5sin t, − π < t < π ∴ F( jω) = ∫ 5sin t e − jωt dt −π π ∴ F( jω) = 5 − jt − j ωt jt ∫ (e − e ) e dt j 2 −π π = 5 − jt (1+ω ) jt (1−ω ) ] dt ∫ [e − e j 2 −π F( jω) = ⎤ 5 ⎡ 1 1 j π (1−ω ) − e− jπ (1−ω) ) − (e− jπ (1+ω) − e jπ (1+ω) ) ⎥ ⎢ j (1 − ω) (e j2 ⎣ − j (1 + ω) ⎦ 2.5 −2.5 ( −e − jπω + e jπω ) − ( −e − jπω + e jπω ) 1− ω 1+ ω 2.5 1 1 ⎞ −2.5 ⎛ ( j 2sin πω) − ( j 2sin πω) = j 5sin πω ⎜ − − = ⎟ 1− ω 1+ ω ⎝ 1− ω 1+ ω ⎠ = j10sin πω j10sin πω ⎛ 1+ ω +1− ω ⎞ = j 5sin πω(−1) ⎜ = ⎟=− 2 1 − ω2 ω2 − 1 ⎝ 1− ω ⎠ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
40.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 40. f (t ) = 8cos t [u (t + 0.5π) − u (t − 0.5π)] π/ 2 ∴ F( jω) = ∫ 8cos te − jωt dt = 4 −π / 2 π/ 2 =4 ∫ −π / 2 π/ 2 ∫ (e jt + e − jt ) e− jωt dt −π / 2 ⎡e jt (1−ω) + e − jt (1+ω) ⎤ dt ⎣ ⎦ ⎧ 1 π/2 1 ⎪ − j ωt = 4⎨ e jt e − π / 2 − e− jt e− jωt j (1 + ω) ⎪ j (1 − ω) ⎩ ⎫ ⎪ ⎬ −π / 2 ⎪ ⎭ π/ 2 ⎧ 1 ⎫ 1 − j πω / 2 ⎡ je − jπω / 2 − (− j ) e jπω / 2 ⎤ − − je jπω / 2 ⎤ ⎬ = 4⎨ ⎦ ⎣ ⎦ j (1 + ω) ⎡ − je ⎣ ⎩ j (1 − ω) ⎭ 1 ⎞ πω πω ⎫ πω ⎛ 1 1 ⎧ 1 = 4⎨ × 2 cos + × 2 cos + ⎬ = 8cos ⎜ ⎟ 2 1+ ω 2 ⎭ 2 ⎝ 1− ω 1+ ω ⎠ ⎩1 − ω πω 2 cos πω / 2 = 8cos = 16 2 2 1− ω 1 − ω2 (a) ω = 0 ∴ F( j 0) = 16 (b) ω = 0.8, F( j 0.8) = 16 cos 72° = 13.734 0.36 (c) ω = 3.1, F( j 3.1) = 16 cos(3.1× 90°) = −0.2907 1 − 3.12 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
41.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 41. (a) F( jω) = 4 [u (ω + 2) − ω (ω − 2) ] ∴ f (t ) = 2 4 2 1 j ωt jωt ∴ f (t ) = ∫2 e d ω = π jt e 2π − ∴ f (t ) = (b) ( 2 e j 2t − e− j 2t jπt = −2 ) 2 4 5 j 2sin 2t = sin 2t ∴ f (0.8) = sin1.6 rad = 1.5909 πt π 2πt F( jω) = 4e ∞ −2 ω ∴ f (t ) = 4 −2 ω + j ω t dω ∫e 2π −∞ ∞ 0 ∴ f (t ) = 2 ∞ 1 j ωt ∫ e F( jω)d ω 2π −∞ 2 2 (2 + jt ) ω d ω + ∫ e( −2+ jω) t d ω ∫e π −∞ π 0 ⎤ 2⎛ 1 2⎡ 1 1 1 ⎞ 2 4 ⎢ 2 + jt (1 − 0) + −2 + jt (0 − 1) ⎥ = π ⎜ 2 + jt + 2 − jt ⎟ = π 4 + t 2 π⎣ ⎦ ⎝ ⎠ 8 8 ∴ f (t ) = ∴ f (0.8) = = 0.5488 2 π(4 + t ) π× 4.64 = (c) F( jω) = 4 cos πω [u (ω + 0.5) − u (ω − 0.5) ] ∴ f (t ) = 4 2π 1 = π = 0.5 ∫ cos πω× e jωt d ω = −0.5 0.5 ∫ −0.5 2 π 0.5 ∫ 2 (e 1 j πω ) + e − jπω e jωt d ω −0.5 ⎡ e( jπ+ jt ) ω + e( − j 0.5 π− j 0.5t ) ω ⎤ d ω ⎣ ⎦ ⎤ 1⎡ 1 1 j 0.5 π+ j 0.5 t e− j 0.5π+ j 0.5t − e j 0.5π− j 0.5t ⎥ − e − j 0.5 π− j 0.5t + ⎢ j (π + t ) e j (−π + t ) π⎣ ⎦ ( ) ( ) ⎤ 1⎡ 1 1 j 0.5t + je − j 0.5t + − je j 0.5t − je− j 0.5t ⎥ ⎢ j (π + t ) je π⎣ j (−π + t ) ⎦ 1⎡ 1 1 1 ⎞ ⎤ 2 cos 0.5t ⎛ 1 = ⎢ − 2 cos 0.5t − 2 cos 0.5t ⎥ = ⎜ ⎟ π ⎣π+t −π + t π ⎦ ⎝ π + t −π + t ⎠ = ( ) ( ) 4 ⎛ −2 ⎞ = 2 cos 0.5t ⎜ 2 = 2 2 cos 0.5t ∴ f (0.8) = 0.3992 2 ⎟ ⎝ t −π ⎠ π −t PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
42.
Engineering Circuit Analysis,
7th Edition 42. Fv ( jω) = 10 March 2006 v(t ) = 20e1.5t u (−t − 2) V (a) Chapter Eightenn Solutions ∞ ∫ 20e1.5t u (−t − 2)e − jωt dt = −∞ = (b) 20 e(1.5− jω)t 1.5 − jω −2 ∫ 20e 1.5 t − jωt dt −∞ −2 = −∞ 20 20 −3 e −3+ j 2 ω ∴ Fv ( j 0) = e = 0.6638 1.5 − jω 1.5 Fv ( jω) = A v (ω) + Bv (ω) = 20 e −3e j 2 ω 1.5 − jω 20 e −3 e j 4 = 0.39830 ∠282.31° = 0.08494 − j 0.38913 1.5 − j 2 ∴ A v (2) = 0.08494 ∴ Fv ( j 2) = (c) Bv (2) = −0.3891 (d) Fv ( j 2) = 0.3983 (e) φv(j2) = 282.3o or -77.69o PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
43.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 43. I( jω) = 3cos10ω [u (ω + 0.05π) − u (ω − 0.05π) ] (a) W = 4× 10 March 2006 = (b) 9 π 18 π ωx 1 2π ∞ ∫ −∞ I( jω) d ω = 2 2 π 0.05 π ∫ 9 cos 2 10ω d ω −0.05 π π / 20 π / 20 9 9 1 ⎛1 1 ⎞ ∫/ 20 ⎜ 2 + 2 cos 20 ω ⎟ d ω = π × 0.1π + π 20 sin 20ω −π / 20 = 0.9 J ⎝ ⎠ −π 9⎡ ∫ (1 + cos 20ω) d ω = 0.45 = π ⎢ 2ω ⎣ −ωx x + 1 ⎤ × 2sin 20 ωx ⎥ 20 ⎦ ∴ 0.05π = 2ωx + 0.1sin 20ωx , ωx = 0.04159 rad/s PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
44.
Engineering Circuit Analysis,
7th Edition 44. (a) ∞ ∞ ∞ W1Ω = ∫ f (t ) dt = ∫ 100t e 2 −8 t 2 = 0 e−8t (64t 2 + 16t + 2) dt = 100 × (−512) 0 100 × 2 = 0.3906 J 512 ∞ F( jω) = F {10te u (t )} = 10 ∫ t e −4 t ∞ − (4 + jω ) t 0 = (c) 10 March 2006 f (t ) = 10te −4t u (t ) 0 (b) Chapter Eightenn Solutions 10e − (4+ jω)t dt = [−(4 + jω)t − 1 (4 + jω) 2 0 10 10 ∴ F( jω) = 2 2 (4 + jω) ω + 16 F ( jω ) 2 = (ω 100 2 + 16) 2 F ( jω ) ω = 0 = 390.6 mJ/Hz , F ( jω ) ω = 4 = 97.66 mJ/Hz 2 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
45.
Engineering Circuit Analysis,
7th Edition 45. v(t ) = 8e (a) W1Ω = −2 t ∞ v 2 (t ) dt = 2 × 64 −∞ (b) − j ωt ∫ e v(t ) dt = 8 −∞ 0 ∴ Fv ( jω) = 8 ∫ e −∞ 8 = e(2− jω)t 2 − jω ω (c) ∫e −4 t dt = 32 J ∫e e− jωt dt 0 ∞ Fv ( jω) = 10 March 2006 V ∞ ∫ Chapter Eightenn Solutions (2 − jω) t ∞ −2 t −∞ ∞ dt + 8∫ e− (2+ jω) t dt 0 0 8 − e − (2+ jω)t 2 + jω −∞ 1 1 322 322 0.9 × 32 = dω = ∫ 2π −ω1 (ω2 + 4) 2 2π ∞ = 0 8 32 8 = = Fv ( jω) + 2 − jω 2 + jω 4 + ω2 ⎡ 1 ω ω ⎤ + tan −1 1 ⎥ ⎢ 2 2⎦ ⎣ 8(ω1 + 4) 16 16 ⎡ ω1 1 ω1 ⎤ 2 ⎡ 2ω1 ω ⎤ ×2⎢ + + tan −1 1 ⎥ ⎥= ⎢ 2 2 2⎦ π ⎣ 8(ω1 + 4) 16 2 ⎦ π ⎣ ω1 + 4 2ω ω ∴ 0.45π = 2 1 + tan −1 1 ∴ω1 = 2.7174 rad/s (by SOLVE) 2 ω1 + 4 ∴ 0.9 = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
46.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 46. (a) ∞ Prove: F { f (t − to )} = e − jωto F { f (t )} = ∫ f (t − to )e − jωt dt Let t − to = τ −∞ ∴F { f (t − to )} = ∞ ∫ f (τ)e − jωτ e − jωto dt = e − jωto F { f (t )} −∞ (b) Prove: F { f (t )} = jωF { f (t )} = ∞ ∫e − jωt −∞ dv = df , v = f ∴F { f (t )} = f (t )e− jωt ∞ −∞ df dt Let u = e − jωt , du = − jωe− jωt , dt ∞ + ∫ jωf (t )e− jωt dt −∞ We assume f (±∞) = 0 ∴F { f (t )} = jωF { f (t )} (c) Prove: F { f (kt )} = ∴F { f (kt )} = ∞ ∫ ∞ 1 ⎛ jω ⎞ − j ωt F⎜ dt Let τ = kt , k > 0 ⎟ = ∫ f (kt )e k ⎝ k ⎠ −∞ f (τ)e − jωτ / k −∞ 1 1 ⎛ jω ⎞ dτ = F⎜ ⎟ k k ⎝ k ⎠ If k < 0, limits are interchanged and we get: − ∴F { f (kt )} = 1 ⎛ jω ⎞ F⎜ ⎟ k ⎝ k ⎠ 1 ⎛ jω ⎞ F⎜ ⎟ k ⎝ k ⎠ (d) Prove: F { f (−t )} = F(− jω) Let k = 1 in (c) above (e) d Prove: F {tf (t )} = j F( jω) Now, F( jω) = ∫ f (t )e − jωt dt dω −∞ ∞ ∞ ∴ dF( jω) = ∫ f (t )(− jt )e − jωt dt = − j F {tf (t )} ∴F {tf ( f )} = jωF f (t )} dω −∞ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
47.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 47. (a) f (t ) = 4[sgn(t )δ(t − 1)] ∴F {4[sgn(t )δ(t − 1)] = F {4sgn(1) δ(t − 1)} = F {4δ(t − 1)} = 4e − jω (b) f (t ) = 4[sgn(t − 1) δ(t )] ∴F {4sgn(−1)δ(t )} = F {−4δ(t )} = −4 (c) ⎧4 ⎫ f (t ) = 4sin(10t − 30°) ∴F {4sin(10t − 30°) = F ⎨ ⎡e j (10t −30°) − e − j (10t −30°) ⎤ ⎬ = ⎣ ⎦ ⎩ j2 ⎭ − j 30° j10 t j 30° − j10 t − jπ / 6 jπ / 6 F {− j 2e e + j 2e e } = − j 2e 2πδ(ω − 10) + j 2e 2πδ(ω + 10) = − j 4π [e − jπ / 6 δ(ω − 10) − e jπ / 6δ(ω + 10)] PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
48.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 48. (a) f (t ) = A cos(ωo t + φ) ∴ F( jω) = F {A cosφ cos ωo t − A sin φ sin ωo t} = ⎧π ⎫ A cos φ{π[δ(ω + ωo ) + δ(ω − ωo )]} − A sin φ ⎨ [δ(ω − ωo ) − δ(ω + ωo )]⎬ = ⎩j ⎭ πA{cos φ [δ(ω + ωo ) + δ(ω − ωo )] + j sin φ [δ(ω − ωo ) − δ(ω + ωo )]} ∴ F( jω) = πA[e jφ δ(ω − ωo ) + e − jφ δ(ω + ωo )] (b) f (t ) = 3sgn(t − 2) − 2δ(t ) − u (t − 1) ∴ F( jω) = e − j 2 ω × 3 × ∴ F( jω) = − j (c) ⎡ 2 1 ⎤ − 2 − e − jω ⎢πδ(ω) + jω jω ⎥ ⎣ ⎦ 6 − j 2ω 1⎤ ⎡ − 2 − e − jω ⎢ πδ(ω) − j ⎥ e ω ω⎦ ⎣ ⎧1 ⎫ f (t ) = sinh kt u (t ) ∴ F( jω) = F ⎨ [e kt − e− kt ] u (t ) ⎬ ⎩2 ⎭ k + jω + k − jω −k 1 1 1 1 ∴ F( jω) = − = = 2 2 2 ω + k2 2 − k + jω 2 k + jω 2(− k − ω ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
49.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 49. ∞ (a) F( jω) = 3u (ω + 3) − 3u (ω − 1) ∴ f (t ) = 1 3 3 1 j ωt j ωt ∴ f (t ) = ∫3 e dt = 2π jt e 2π − ∴ f (5) = − j (b) 1 = −3 1 jωt ∫ [3u(ω + 3) − 3u(ω − 1)] e d ω 2π −∞ 3 (e+ jt − e− j 3t ) j 2πt 3 (1∠5rad − 1∠ − 15rad ) = 0.10390 ∠ − 106.48° 10π F( jω) = 3u (−3 − ω) + 3u (ω − 1) → ∴ F( jω) = 3 − Fa ( jω) 3 (e jt − e − j 3t ) ∴ f (5) = 0 − 0.10390 ∠ − 106.48° j 2πt so f(5) = 0.1039∠73.52o f (t ) = 3δ(t ) − (c) F( jω) = 2δ(ω) + 3u (−3 − ω) + 3u (ω − 1) Now, F {2δ(ω)} = ∴ f (t ) = 2 1 = 2π π ⎤ 1 ⎡ 3 1 (e jt − e − j 3t ) ⎥ ∴ f (5) = − 0.10390 ∠ − 106.48° = 0.3618 ∠15.985+° + ⎢− π ⎣ j 2πt π ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
50.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 50. (a) F( jω) = 3 3 + + 3 + 3δ(ω − 1) 1 + jω jω ∴ f (t ) = 3e − t u (t ) + 1.5sgn(t ) + 3δ(t ) + (b) (c) 1.5 jt e π 1 sin ω 8 / 2 5sin 4ω = 8 × 2.5 ω ω8 / 2 ∴ f (t ) = 2.5[u (t + 4) − u (t − 4)] F( jω) = F( jω) = 6(3 + jω) 6(3 + jω) = ∴ f (t ) = 3−3t cos 2t u (t ) 2 2 2 (3 + jω) + 4 (3 + jω) + 2 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
51.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 51. T = 4, periodic; find exp′l form 1 1 ∴ cn = ∫ 10te − jnπt / 2 dt 4 −1 1 ⎡ ⎛ ⎞⎤ t 1 ∴ cn = 2.5 ⎢ e − jnπt / 2 ⎜ − 2 2 ⎟⎥ ⎝ − jnπ / 2 −n π / 4 ⎠ ⎦ −1 ⎣ ⎡ ⎛ 1 ⎛ 1 1 ⎞ 1 ⎞⎤ ∴ cn = 2.5 ⎢ e − jnπ / 2 ⎜ + 2 2 ⎟ − e jnπ / 2 ⎜ + 2 2 ⎟⎥ ⎝ − jnπ2 n π / 4 ⎠ ⎝ jnπ / 2 n π / 4 ⎠ ⎦ ⎣ ⎡ 1 ⎤ 4 (−e − jnπ / 2 − e jnπ / 2 ) + 2 2 (e − jnπ / 2 − e jnπ / 2 ) ⎥ = 2.5 ⎢ nπ ⎣ jnπ / 2 ⎦ = j5 nπ 10 ⎛ nπ ⎞ × 2 cos + 2 2 ⎜ − j 2sin ⎟ nπ 2 nπ ⎝ 2 ⎠ ∞ nπ 20 nπ ⎤ ⎡ j10 cos ∴ f (t ) = ∑ ⎢ − j 2 2 sin ⎥ e jnπt / 2 2 nπ 2⎦ −∞ ⎣ nπ ∞ nπ nπ ⎤ nπ ⎞ 20 ⎡ j10 ⎛ ∴ F( jω) = ∑ ⎢ − j 2 2 sin ⎥ 2πδ ⎜ ω − ⎟ cos nπ 2 2⎦ 2 ⎠ ⎝ −∞ ⎣ nπ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
52.
53.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 53. ∞ F( jω) = 20∑ −∞ 1 δ(ω − 20n) n !+ 1 1 1 1 1 ⎡ 1 = 20 ⎢ δ(ω) + δ(ω + 20) + δ(ω − 20) + δ(ω + 40) + δ(ω − 40) 1+1 1+1 2 +1 3 ⎣1 + 1 1 1 ⎤ + δ(ω + 60) + δ(ω − 60) + ...⎥ 7 7 ⎦ 20 20 = 10δ(ω) + [πδ(ω + 20) + πδ(ω − 20)] + [πδ(ω + 40) + πδ(ω − 40)] + 2π 3π 20 20 [πδ(ω + 60) + πδ(ω − 60) + [πδ(ω + 80) + πδ(ω − 80)] + ... 7π 25π 10 20 20 20 20 cos 20t + cos 40t + cos 60t + cos80t + ... ∴ f (t ) = + 2π 2π 3π 7π 25π 20 ⎡ 1 1 1 1 = ⎢0.25 + 2 cos 20t + 3 cos 40t + 7 cos 60t + 25 cos80t + ... π ⎣ ∴ f (0.05) = 20 ⎡ 1 1 1 1 ⎤ rad ⎢0.25 + 2 cos1 + 3 cos 2 + 7 cos 3 + 25 cos 4 + ...⎥ = 1.3858 π ⎣ ⎦ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
54.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 t Input = x(t ) = 5[u (t ) − u (t − 1)] 54. ∫ x( z ) h(t − z ) dz −∞ h(t ) = 2u (t ) (a) y (t ) = (b) h(t ) = 2u (t − 1) x(t – z) (c) h(t ) = 2u (t − 2) x(t – z) x(t – z) 5 t-1 z t h( z) t-1 z t h( z) 2 t-1 2 2 z y(t) z 1 z 2 y(t) y(t) 10 z t h( z) 10 10 t t t 1 1 2 2 t < 0: y(t) = 0 t < 1: y (t ) = 0 t < 2 : y (t ) = 0 0 < t < 1: 1< t < 2: 3 2 < t < 3: t t t y (t ) = ∫ 10dz = 10t y (t ) = ∫ 10dz = 10(t - 1) y (t ) = ∫ 10dz = 10(t - 2) t > 1: t > 2: t > 3: 0 1 t y (t ) = ∫ 10dz = 10 t -1 2 t y (t ) = ∫ 10dz = 10 t -1 t y (t ) = ∫ 10dz = 10 t -1 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
55.
Engineering Circuit Analysis,
7th Edition 55. Chapter Eightenn Solutions 10 March 2006 x(t ) = 5[u (t ) − u (t − 2)]; h(t ) = 2[u (t − 1) − u (t − 2)] t y (t ) = ∫ x( z ) h(t − z ) dz −∞ t < 1: y (t ) = 0 t −1 1 < t < 2 : y (t ) = ∫ 10 dz = 10(t − 1) 0 2 < t < 3 : y (t ) = 10 2 3 < t < 4 : y (t ) = ∫ 10 dz = 10(2 − t + 2) = 10(4 − t ) t −2 t > 4 : y (t ) = 0 ∴ y (−0.4) = 0; y (0.4) = 0; y (1.4) = 4 y (2.4) = 10; y (3.4) = 6; y (4.4) = 0 ∞ or…. y (t ) = ∫ x(t − z ) h( z ) dz 0 t < 1: y (t ) = 0 t 1 < t < 2 : y (t ) = ∫ 10 dz = 10(t − 1) 1 2 < t < 3 : y (t ) = 10 2 3 < t < 4 : y (t ) = ∫ 10 dz = 10(2 − t + 2) = 10(4 − t ) t −2 t > 4 : y (t ) = 0 same answers as above PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
56.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 56. h(t ) = 3[e − t − e −2t ], x(t ) = u (t ) t y (t ) = ∫ x( z )h(t − z) dz −∞ t = ∫ 3[e − ( t − z ) − e −2( t − z ) ] dz 0 t −t = 3e [e ] − 3e z t 0 −2 t ⎡ 1 2Z ⎤ ⎢2 e ⎥ ⎣ ⎦0 = 3e −t (et − 1) − 1.5e−2t (e 2t − 1) ∴ y (t ) = 3(1 − e − t ) − 1.5(1 − e −2t ) = 1.5 − 3e− t + 1.5e−2t , t > 0 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
57.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 57. ∞ y (t ) = ∫ x(t − 2)h( z )dz 0 (a) 2 h(t ) = (5 − t ), 2 < t < 5 3 5 5 2 20 y (t ) = ∫ 10 × (5 − z ) dz = (5 − z ) dz 3 3 ∫ 2 2 Note: h( z ) is in window for 4 < t < 6 5 (b) 20 ⎛ 1 ⎞ y (t ) = ⎜ − ⎟ (5 − z ) 2 3 ⎝ 2⎠ 2 =− 10 (0 − 9) = 30 at t = 5 3 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
58.
Engineering Circuit Analysis,
7th Edition 58. x(t ) = 5e − ( t − 2) Chapter Eightenn Solutions 10 March 2006 ∞ u (t − 2), h(t ) = (4t − 16) [u (t − 4) − u (t − 7)], y (t ) = ∫ x(t − z ) h( z ) dz 0 (a) t < 6 : y (t ) = 0 ∴ y (5) = 0 (b) t = 8 : y (8) = ∫ 5e − (8− z − 2) (4 z − 16) dz 6 4 ∴ y (8) = 20e −6 6 ∫ze z dz − 80e 4 −6 6 ∫e z dz 4 6 ⎡ ez ⎤ = 20e ⎢ ( z − 1) ⎥ − 80e −6 (e6 − e4 ) ⎣1 ⎦4 −6 = 20e −6 (5e6 − 3e 4 ) − 80 + 80e −2 = 20 + 80e−2 − 60e−2 = 20 (1 + e −2 ) = 22.71 7 (c) t = 10 : y (10) = ∫ 5e− (10− z − 2) (4 z − 16) dz 4 7 ∴ y (10) = ∫ 20e−8e z ( z − 4)dz 4 7 7 ∴ y (10) = 20e −8 ∫ ze z dz − 80e−8 ∫ e z dz = 20e−8 [e z ( z − 1)]7 − 80e−8 (e7 − e 4 ) 4 4 −8 4 = 20e (6e − 3e ) − 80(e−1 − e −4 ) = 40e−1 + 20e −4 = 15.081 7 4 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
59.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 59. h(t ) = sin t , 0 < t < π; 0 elsewhere, Let x(t ) = e −t u (t ) ∞ y (t ) = ∫ x(t − z ) h( z ) dz 0 t < 0 : y (t ) = 0 t t 0 < t < π : y (t ) = ∫ sin z × e − t + z dz = e − t ∫ e z sin z dz 0 0 t ⎡1 ⎤ ∴ y (t ) = e −t ⎢ e z (sin z − cos z ) ⎥ ⎣2 ⎦0 1 = e − t [et (sin t − cos t ) + 1] 2 1 = (sin t − cos t + e − t ) 2 (a) y (1) = 0.3345+ (b) y (2.5) = 0.7409 (c) y > π : y (t ) = e − t ∫ e z sin z dz π 0 π 1 ⎡1 ⎤ y > π : y (t ) = e ⎢ e z (sin z − cos z ) ⎥ = e − t (e π + 1) = 12.070e − t ⎣2 ⎦0 2 ∴ y (4) = 0.2211 −t PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
60.
Engineering Circuit Analysis,
7th Edition 60. Chapter Eightenn Solutions 10 March 2006 x(t ) = 0.8(t − 1)[u (t − 1) − u (t − 3)], h(t ) = 0.2 (t − 2)[u (t − 2) − u (t − 3)] ∞ y (t ) = ∫ x(t − z ) h( z ) dz, 0 t < 3 : y (t ) = 0 t −1 ∫ (a) 3 < t < 4 : y (t ) = 0.8(t − z − 1) 0.2( z − 2) dz 2 t −1 ∴ y (t ) = 0.16 ∫ (tz − 2t − z 2 + 2 z − z + 2) dz 2 t −1 t −1 1 ⎡ 1 ⎤ = 0.16 ∫ [− z 2 + (t + 1) z + 2 − 2t ] dz = 0.16 ⎢ − z 3 + (t + 1) z 2 + (2 − 2t ) z ⎥ 2 ⎣ 3 ⎦2 2 8 1 1 ⎡ 1 ⎤ = 0.16 ⎢ − (t − 1)3 + + (t + 1) (t − 1) 2 − (t + 1) 4 + (2 − 2t ) (t − 1 − 2) ⎥ 3 2 2 ⎣ 3 ⎦ 1 8 1 ⎡ 1 ⎤ ∴ y (t ) = 0.16 ⎢ − t 3 + t 2 − t + + + (t 2 − 1) (t − 1) − 2t − 2 + 2t − 6 − 2t 2 + 6t ⎥ 3 3 2 ⎣ 3 ⎦ ⎡1 1 1 ⎤ 3 9 9⎞ ⎛ 1 ⎞ ⎛ ⎞ ⎛1 = 0.16 ⎢ t 3 + t 2 ⎜1 − − 2 ⎟ + t ⎜ −1 − + 6 ⎟ + 3 + − 8⎥ = 0.16 ⎜ t 3 − t 2 + t − ⎟ 2 2 ⎦ 2 2 2⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎝6 ⎣6 ∴ y (3.8) = 13.653 × 10−3 3 1 ⎡ 1 ⎤ 4 < t < 5 : y (t ) = ∫ 0.16 (t − z − 1) ( z − 2) dz = 0.16 ⎢ − z 3 + (t + 1) z 2 + (2 − 2t ) z ⎥ 2 ⎣ 3 ⎦2 2 3 (b) 1 ⎡ 1 ⎤ ∴ y (t ) = 0.16 ⎢ − (27 − 8) + (t + 1) 5 + (2 − 2t )1⎥ 2 ⎣ 3 ⎦ 11 ⎞ ⎡ 19 ⎤ ⎛ = 0.16 ⎢ − + 2.5t + 2.5 + 2 − 2t ⎥ = 0.16 ⎜ 0.5t − ⎟ 6⎠ ⎣ 3 ⎦ ⎝ ∴ y (4.8) = 90.67 × 10−3 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
61.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 61. x(t ) = 10e −2t u (t ), h(t ) = 10e −2t u (t ) ∞ y (t ) = ∫ x(t − z ) h( z ) dz 0 t ∴ y (t ) = ∫ 10e −2( t − z ) 10e −2 z dz 0 t = 100e −2t ∫ dz = 100 e−2t × t 0 −2 t ∴ y (t ) = 100t e u (t ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
62.
Engineering Circuit Analysis,
7th Edition 62. W1Ω = 25 ∫ e−8t dt = 10 March 2006 h(t ) = 5e −4t u (t ) (a) Chapter Eightenn Solutions 0.8 0.1 25 −0.8 −6.4 (e − e ) = 1.3990 J 8 ⎛ 25 ⎞ ∴ % = 1.3990 / ⎜ ⎟ ×100% = 44.77% ⎝ 8 ⎠ (b) 5 1 25 25 1 ω ∴ W1Ω = ∫ 2 dω = H( jω) = tan −1 π 0 ω + 16 π 4 40 jω + 4 2 ∴ W1Ω = 2 25 1 0.9224 tan −1 = 0.9224 J ∴ % = × 100% = 29.52% 4π 2 25 / 8 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
63.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 63. F( jω) = 2 2 2 = − ∴ f (t ) = (2e− t − 2e−2t ) u (t ) (1 + jω)(2 + jω) 1 + jω 2 + jω ∞ (a) W1Ω = ∫ (4e−2t − 8e−3t + 4e −4t ) dt = 0 (b) 4 8 4 1 − + = J 2 3 4 3 f (t ) = −2e −t + 4e−2t = 0, − 2 + 4e− t = 0, et = 2, t = 0.69315 ∴ f max = 2(e −0.69315 − e−2×0.69315 ) = 0.5 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
64.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 64. (a) (b) (c) (d) 1 1/ 6 1/ 2 1/ 3 = − + jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω 1 1 1 ∴ f (t ) = sgn(t ) − e−2t u (t ) + e−3t u (t ) 12 2 3 F( jω) = 1 + jω 1/ 6 1/ 2 2/3 = + − jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω 1 1 2 ∴ f (t ) = sgn(t ) + e−2t u (t ) − e −3t u (t ) 12 2 3 F( jω) = (1 + jω) 2 1/ 6 1/ 2 4/3 = − + F( jω) = jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω 1 1 4 ∴ f (t ) = sgn(t ) − e−2t u (t ) + e −3t u (t ) 12 2 3 (1 + jω)3 1/ 6 1/ 2 8/3 = 1+ + − jω(2 + jω)(3 + jω) jω 2 + jω 3 + j ω 1 1 8 ∴ f (t ) = δ(t ) + sgn(t ) + e−2t u (t ) − e −3t u (t ) 12 2 3 F( jω) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
65.
Engineering Circuit Analysis,
7th Edition 65. H( jω) = 2 × (b) 1 1 1 Vo 1/ jω = = H( jω) = 2 1 + jω 2 Vi 1 + 1/ jω (c) 10 March 2006 h(t ) = 2e −t u (t ) (a) Chapter Eightenn Solutions Gain = 2 1 2 = 1 + jω 1 + jω PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
66.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 66. 1 1 jω + ( jω) 2 + 2 2 jω = Vo ( jω) = 1 1 ( jω) 2 + 2( jω) + 2 1 + jω + 2 jω −2( jω) ( jω) 2 + 2( jω) + 2 − 2( jω) ∴ Vo ( jω) = = 1+ 2 2 ( jω) + 2( jω) + 2 ( jω) + 2( jω) + 2 −2 ± 4 − 8 2x = −1 ± j1 ; x= 2 x + 2x + 2 A B A B ∴ Vo ( x) = 1 + + = Let x = 0 ∴ + =0 1 + j1 1 − j1 x + 1 + j1 x + 1 − j1 A B B + j2 B Let x = −1 ∴ + = 2 ∴ A − B = j 2, A = B + j 2 ∴ + =0 j1 − j1 1 + j1 1 − j1 ∴ B − jB + j 2 + 2 + B + jB = 0 ∴ B = −1 − j1 ∴ A = −1 + j1 −1 + j1 −1 − j1 1 − j1 1 + j1 ∴ Vo ( x) = 1 + + , Vo ( jω) = 1 − − ( jω) + 1 + j1 ( jω) + 1 − j1 x + 1 + j1 x + 1 − j1 Let jω = x ∴ Vo ( x) = 1 − ∴ vo (t ) = δ(t ) − (1 − j1) e( −1− j1) t u (t ) − (1 + j1)e( −1+ j1)t u (t ) = δ(t ) − 2 e − j 45°− jt −t u (t ) − 2 e j 45°+ jt −t u (t ) = δ(t ) − 2 2 e −t cos(t + 45°) u (t ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
67.
Engineering Circuit Analysis,
7th Edition Chapter Eightenn Solutions 10 March 2006 67. 5 / jω 10 / jω = 5 / jω + 35 + 30( jω) 1/ jω + 7 + 6( jω) 10 10 / 6 ∴ Vc ( jω) = = 2 6( jω) + 7( jω) + 1 ( jω) 2 + 7 ( jω) + 1 6 6 ⎛ 49 24 ⎞ 1 10 / 6 2 2 ∴ jω = ⎜ −7 / 6 ± − ⎟ / 2 = − , − 1 ∴ Vc ( jω) = = − ⎜ ⎟ 36 36 ⎠ 6 ( jω + 1/ 6)( jω + 1) jω + 1/ 6 jω + 1 ⎝ ∴ vc (t ) = 2(e −t / 6 − e− t ) u (t ) Vc ( jω) = 10 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
68.
Engineering Circuit Analysis,
7th Edition 68. 10 March 2006 f (t ) = 5e −2t u (t ), g (t ) = 4e−3t u (t ) (a) Chapter Eightenn Solutions f ∗ g = ∫ f (t − z ) g ( z ) dz ∞ 0 t t = ∫ 5e −2t e 2 z 4e −3 z dz = 20e−2t ∫ e− z dz 0 0 −2 t = −20 e (e − 1) V t ∴ f ∗ g = (e −2t − e−3t ) u (t ) (b) 5 4 20 , G( jω) = ∴ F( jω)G( jω) = jω + 2 jω + 3 ( jω + 2)( jω + 3) 20 20 ∴ F( jω)G( jω) = − ∴ f ∗ g = 20(e −2t − 2−3t ) u (t ) jω + 2 jω + 3 F( jω) = PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
69.
Engineering Circuit Analysis,
7th Edition H ( jω ) = 24 jω 10 March 2006 j 2ω 4 + j 2ω Vi ( jω ) = 69. Chapter Eightenn Solutions from Table 18.2 ⎡ j 2ω ⎤ ⎛ 24 ⎞ 24 Therefore Vo ( jω ) = ⎢ ⎥ ⎜ jω ⎟ = 2 + jω ⎣ 4 + j 2ω ⎦ ⎝ ⎠ In the time domain, then, we find vo (t ) = 24e −2t u (t ) V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
70.
Engineering Circuit Analysis,
7th Edition 70. h(t ) = 2e − t cos 4t H ( jω ) = 10 March 2006 so fromTable 18.2, 2 (1 + jω ) (1 + jω ) Chapter Eightenn Solutions 2 + 16 . Define output function f(t). (a) I ( jω ) = 4πδ (ω ) ⎡ 8π (1 + jω ) ⎤ 8π Therefore F(ω) = ⎢ ⎥ δ (ω ) = 17 δ (ω ) . 2 ⎣ (1 + jω ) + 16 ⎦ The time domain output is then given by f(t) = 4/17. (b) I ( jω ) = 2e − jω ⎡ 4(1 + jω ) ⎤ − jω Therefore F(ω) = ⎢ ⎥e . 2 ⎣ (1 + jω ) + 16 ⎦ The time domain output is then given by f(t) = 4e −(t −1) cos ⎡ 4 ( t − 1) ⎤ u (t − 1) ⎣ ⎦ (c) We find the response due to a unit step u (t ) and treat i (t ) as two unit steps, each shifted appropriately. R ( jω ) = r (t ) = 2(1 + jω ) ⎡ 1 ⎤ ⎢πδ (ω ) + jω ⎥ 2 (1 + jω ) + 16 ⎣ ⎦ 1 1 e−t + sgn(t ) − 2 [ cos 4t − 4sin 4t ] u (t ) 17 17 17 Therefore the system response is 2 ⎡ −( t + 0.25 ) {cos 4(t + 0.25) − 4sin 4(t + 0.25)}⎤ u (t + 0.25) ⎣1 − e ⎦ 17 2 − t − 0.25 ) − ⎡1 − e ( {cos 4(t − 0.25) − 4sin 4(t − 0.25)}⎤ u (t − 0.25) ⎣ ⎦ 17 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
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