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- 1. Engineering Circuit Analysis, 7th Edition 1. Chapter Sixteen Solutions 10 March 2006 We have a parallel RLC with R = 1 kΩ, C = 47 μF and L = 11 mH. (a) Qo = R(C/L)½ = 65.37 (b) fo = ωo/ 2π = (LC)-½ / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source: 10-3∠0o A jωL -j/ ωC The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω2 + jω/RC). Since V = 10-3 Z, we note that |V| > 0 as ω → 0 and also as 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.
- 2. Engineering Circuit Analysis, 7th Edition 2. Chapter Sixteen Solutions (a) R = 1000 Ω and C = 1 μF. Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 = (b) L = 12 fH and C = 2.4 nF R = Qo (L/ C)½ = (c) R = 121.7 kΩ and L = 100 pH C = (Qo / R)2 L = 10 March 2006 25 μH 447.2 mΩ 270 aF 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 3. Chapter Sixteen Solutions 10 March 2006 We take the approximate expression for Q of a varactor to be Q ≈ ωCjRp/ (1 + ω2 Cj2 Rp Rs) (a) Cj = 3.77 pF, Rp = 1.5 MΩ, Rs = 2.8 Ω (b) dQ/dω = [(1 + ω2 Cj2 Rp Rs)(Cj Rp) - ωCjRp(2ωCj2 Rp Rs)]/ (1 + ω2 Cj2 RpRs) Setting this equal to zero, we may subsequently write CjRp (1 + ω2 Cj2 Rp Rs) - ωCjRp(2ωCj2 Rp Rs) = 0 Or 1 – ω2 Cj2 Rp Rs = 0. Thus, ωo = (Cj2 RpRs)–½ = 129.4 Mrad/s = 21.00 MHz Qo = Q(ω = ωo) = 366.0 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.
- 4. Engineering Circuit Analysis, 7th Edition 4. Chapter Sixteen Solutions 10 March 2006 Determine Q for (dropping onto a smooth concrete floor): (a) A ping pong ball: Dropped twice from 121.1 cm (arbitrarily chosen). Both times, it bounced to a height of 61.65 cm. Q = 2πh1/ (h1 – h2) = 12.82 (b) A quarter (25 ¢). Dropped three times from 121.1 cm. Trial 1: bounced to 13.18 cm Trial 2: bounced to 32.70 cm Trial 3: bounced to 16.03 cm. Quite a bit of variation, depending on how it struck. Average bounce height = 20.64 cm, so Qavg = 2πh1/ (h1 – h2) = 7.574 (c) Textbook. Dropped once from 121.1 cm. Didn’t bounce much at all- only 2.223 cm. Since the book bounced differently depending on angle of incidence, only one trial was performed. Q = 2πh1/ (h1 – h2) = 6.4 All three items were dropped from the same height for comparison purposes. An interesting experiment would be to repeat the above, but from several different heights, preferrably ranging several orders of magnitude (e.g. 1 cm, 10 cm, 100 cm, 1000 cm). 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 Chapter Sixteen Solutions 10 March 2006 5. α = 80Np/s, ω d = 1200 rad/s, Z(−2α + jω d ) = 400 Ω ω o = 12002 + 802 = 1202.66 rad/s ∴ Qo = Now, Y( s ) = C ( s + α − jω d )( s + α + jω d ) (−α )(−α + j 2ω d ) ∴ Y(−2α + jω d ) = C s −2α + jω d ∴ Y(−160 + j1200) = C ∴C = ωo = 7.517 2α −80(−80 + j 2400) 1 −1 + j 30 ∴ Y(−160 + j1200) = = 80C −160 + j1200 400 −2 + j15 1 229 1 1 = 15.775− μ F; L = 2 = 43.88 mH; R = = 396.7 Ω ωo C 32, 000 901 2α C 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 Chapter Sixteen Solutions 10 March 2006 6. Yin = 1 1 2 − j 0.1ω jω + 0.2 + = + 0.2 + 2 2 + j 0.1ω 1 + 1000 / jω 4 + 0.01ω 1000 + j10 2 − j 0.1ω ω 2 + j1000ω 1000ω −0.1ω + 0.2 + ∴ + 2 =0 2 6 2 2 4 + 0.1ω 10 + ω 4 + 0.01ω ω + 106 ∴ 0.1ω 3 + 105ω = 4000ω + 10ω 3 ∴ 9.9ω 2 = 96, 000 ∴ω = 98.47 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.
- 7. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 7. Parallel: R = 106 , L = 1, C = 10−6 , Is = 10∠0° μ A (a) ωo = (b) 10 March 2006 1⎞ I ⎡ 1000 ⎞ ⎤ ⎛ ⎛ ω − Y = 10−6 + j ⎜ 10−6 − ⎟ , V = = 10−5 /10−3 ⎢10−3 + j ⎜ ω⎠ ω ⎟⎥ Y ⎝ ⎝ 1000 ⎠⎦ ⎣ 1 = 1000 rad/s; Qo = ω o RC = 103+ 6−6 = 1000 LC ∴V = 10−2 10−2 , V = 2 1000 ⎞ ⎛ ω 1000 ⎞ ⎛ ω 0.001 + j ⎜ − 10−6 + ⎜ − ω ⎟ ⎝ 1000 ⎠ ω ⎟ ⎝ 1000 ⎠ ω V 995 996 997 998 999 1000 1001 1002 1003 1004 1005 999.5 1000.5 0.993 1.238 1.642 2.423 4.47 10.0 4.47 2.428 1.646 1.243 0.997 7.070 7.072 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 Sixteen Solutions 10 March 2006 8. (a) 5(100 / jω ) j 0.1ω +2+ 5 + (100 / jω ) 10 + j 0.01ω 500 100 100(20 − jω ) j10ω j10ω j10ω (1000 − j ) = +2+ = +2+ = +2+ 2 ω + 400 ω 2 + 106 100 + j 5ω 1000 + jω 20 + jω 1000 + jω Zin = 104 ω −100ω + 2 = 0 ∴ω 2 + 106 = 100ω 2 + 40, 000, 99ω 2 = 960, 000 2 6 ω + 400 ω + 10 ∴ω o = 960, 000 / 99 = 98.47 rad/s ∴ (b) Zin (ω o ) = 10ω 2 2000 + 2 + 2 o 6 = 2.294 Ω 2 ω o + 400 ω o + 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.
- 9. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 9. (a) 2 2 α = 50 s −1 , ω d = 1000 s −1 ∴ω o = α 2 + ω d = 1, 002,500 ∴ω o = 1001.249 1 106 1 106 + L= 2 = = 0.9975 H; R = = = 10 k Ω 2α C 100 ω o C 1, 002,500 (b) 1 1 ⎛ ⎞ Y = 10−4 + j ⎜ 10−6 ω − ⎟ , ω = 1000 ∴ Z = = 9997∠1.4321° Ω 0.9975ω ⎠ Y ⎝ 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 Chapter Sixteen Solutions 10 March 2006 10. f min = 535 kHz, f max = 1605 kHz, Qo = 45 at one end and Qo ≤ 45 for 535 ≤ f ≤ 1605 kHz f o = 1/ 2π LC ∴ 535 × 103 = 1 1 ,1605 × 103 = 2π L max C 2π L min C 2 1 ⎛ ⎞ ∴ L max / L min = 3; L max C = ⎜ = 8.8498 × 10−14 3 ⎟ ⎝ 2π × 535 ×10 ⎠ ω o RC ≤ 45,535 × 103 ≤ ωo ≤ 1605 × 103. Use ω o max 2π ∴ 2π × 1605 ×103 × 20 × 103 C = 45 ∴ C = 223.1pF ∴ L max = 8.8498 ×10−14 L = 397.6 μ H, L min = max = 44.08 μ H −12 223.1× 10 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.
- 11. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 11. (a) Apply ± 1V. ∴ IR = −10−4 A 1 + 10−4 + (1 − [105 (−10−4 )])10−8 s 4.4 × 10−3 s 1000 48.4 × 10−8 s 2 + 4.4 × 10−4 s + 1000 −4 −8 ∴ Yin = + 10 + 11× 10 s = 4.4 s 4.4 s −8 2 −4 1000 − 48.4 × 10 ω + j 4.4 ×10 ω ∴ Yin ( jω ) = j 4.4ω ∴ Yin = Iin = (b) 2 At ω = ω o , 1000 = 48.4 × 10−8 ω o , ω o = 45.45− krad/s −1 ⎛ j 4.4 ×10−4 ω o ⎞ Zin ( jω o ) = ⎜ ⎟ = 10 k Ω j 4.4ω o ⎠ ⎝ 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. ω0 = Chapter Sixteen Solutions 10 March 2006 ω 1 = 24 = 4.9 rad/s or f0 = 0 = 780 mHz 2π LC 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 13. ω0 = 1 = LC 1 ( 1 25 × 10−6 1.01 ) Chapter Sixteen Solutions = 200 rad/s or f0 = 10 March 2006 ω0 = 31.99 Hz 2π 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 Chapter Sixteen Solutions 1 103 = =5Ω 2α C 200 (a) α = 1 2 RC ω0 = 14. 10 March 2006 ω 1 1 = = 1000 rad/s or f0 = 0 = 159.2 Hz 2π LC 10−6 ∴ R= Zin(ω0) = R = 5 Ω (b) We see from the simulation result that the ratio of the test source voltage to its current is 5 Ω at the resonant frequency; the small error is due to the series resistance PSpice required. 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 15. (a) α = Chapter Sixteen Solutions 10 March 2006 1 2 = 50 s -1 and ωd = ω0 − α 2 = 5000 rad/s 2RC Zin(ω0) = R so find R. ( ) 2 L ωd + α 2 1 1 = = 250 Ω C= 2 = 2 = 40 μ F . R = 2α C 2(50) ω0 L ωd + α 2 L 1 ( ) 1 = 5000 rad/s or f 0 = 795.8 Hz . LC We see from the simulation result that the ratio of the test source voltage to its current is 250 Ω at the resonant frequency; the small error is due to the series resistance PSpice required. (b) The resonant frequency is 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 Sixteen Solutions 16. ω o = 1000 rad/s, Qo = 80, C = 0.2 μ F (a) 1 106 80 = 5 H, Qo = ω o RC ∴ R = 3 = 400 k Ω L= 2 = 6 ω o C 0.2 ×10 10 × 0.2 ×10−6 (b) 10 March 2006 B = ω o / Q o = 1000 / 80 = 12.5 1 ∴ B = 6.25 rad/s 2 ω − ωo ⎛ ω − ωo ⎞ ∴ Z = R / 1+ j = 400 × 10 / 1 + ⎜ ⎟ B/2 ⎝ 6.25 ⎠ 2 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.
- 17. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 17. ω1 = 103rad/s, ω 2 = 118, Z( j105) = 10 Ω ω o2 = ω1ω 2 = 103 ×118 ωo 110.245+ ∴ω o = 110.245 , B = 118 − 103 = 15 rad/s, Qo = = = 7.350 B 15 7.350 1 1 ∴ 7.350 = ω o RC ∴ RC = = 66.67 × 10−3 , LC = 2 = + 110.2451 ω o 12,154 + 1 1 ⎞ 12,154 ⎞ ⎛ ⎛ + j ⎜ 105C − C ⎟ = 18.456 C ⎟ = 15C + j ⎜105C − R 105L ⎠ 105 ⎝ ⎝ ⎠ 0.1 1 1 ∴C = = 5.418 mF, R = C = 12.304 Ω, L = = 15.185− mH 18.456 15 12,154C Y( j105) = 0.1 = 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.
- 18. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 18. ω o = 30 krad/s, Qo = 10, R = 600 Ω, (a) B= ωo (b) N= ω − ωo (c) Zin(j28 000) = 600 / (1 – j1.333) = 360 ∠ 53.13o Ω (d) Qo 1 10 ⎡ 1 ⎤ Zin ( j 28, 000) = ⎢ + j 28, 000C − j ⎥ , C = ω R = 30, 000 × 600 28, 000L ⎦ ⎣ 600 o Qo = 3 krad/s B/ 2 = 28 − 30 = −1.3333 1.5 −1 R 600 1 30, 000 ×10 ⎡ 1 ⎛ 28 10 30 10 ⎞ ⎤ L= , = = ∴ Zin = ⎢ + j⎜ × − ⎟⎥ 600 ω o Qo 30, 000 ×10 L ⎝ 30 600 28 600 ⎠ ⎦ ⎣ 600 600 = 351.906∠54.0903°Ω Zin = ⎛ 28 30 ⎞ 1 + j10 ⎜ − ⎟ ⎝ 30 28 ⎠ (e) −1 approx-true 360 − 351.906 = 100% = 2.300% true 351.906 53.1301° − 54.0903° = −1.7752% angle: 100% 54.0903° magnitude: 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.
- 19. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 19. f o = 400 Hz, Qo = 8, R = 500 Ω, IS = 2 ×10−3 A ∴ B = 50 Hz (a) V = 2 × 10−3 × 500 / 1 + N 2 = 0.5 ∴1 + N 2 = 4, N = ± 3 = 10 March 2006 f − 400 50 / 2 ∴ f = 400 ± 25 3 = 443.3 and 356.7 Hz (b) IR = v R = 1 1+ N 2 × 1 = 0.5 × 10−3 ∴ 1 + N 2 = 4, N 2 = 15, N = ± 15 500 ∴ f = 400 ± 25 15 = 496.8 and 303.2 Hz 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 20. Qo = (b) Approx: 2 = 5 / 1 + N 10 March 2006 ω o = 106 , Qo = 10, R = 5 × 103 , p.r. (a) Chapter Sixteen Solutions R 5 ×103 = 0.5 mH ∴L = ωo L 10 × 106 2 ∴ N = 2.291 = ω − 106 106 / 20 ∴ω = 1.1146 Mrad/S 2 ⎛ ω ωo ⎞⎤ 1⎡ 1⎞ ⎛ Exact: Y = ⎢1 + jQo ⎜ − ⎟ ⎥ ∴ 0.5 = 0.2 1 + 100 ⎜ ω − ⎟ (ω in Mrad/S) R⎣ ω⎠ ⎝ ⎝ ωo ω ⎠⎦ 1 1 ∴ 6.25 = 1 + 100(ω 2 − 2 + 1/ ω 2 ), ω 2 − 2 + 2 = 0.0525, ω 2 + 2 = 2.0525 ω 4 − 2.0525ω 2 + 1 = 0, ω 2 = (c) ( ω ) 1 2.0525 + 2.05252 − 4 = 1.2569, ω = 1.1211 Mrad/s 2 Approx: ∠Y = 30° ∴ tan −1 N = 30°, N = 0.5774 = Exact: Y = ∴ω − 1 ω ω ω −1 1/ 20 , ω = 1.0289 Mrad/s 1 ⎡ 1 ⎞⎤ 1⎞ ⎛ ⎛ ⎢1 + j10 ⎜ ω − ω ⎟ ⎥ (in Mrad/s) ∴ tan 30° = 0.5774 = 10 ⎜ ω − ω ⎟ 5000 ⎣ ⎝ ⎠⎦ ⎝ ⎠ = 0.05774, ω 2 − 0.05774ω − 1 = 0, ω = 0.05774 + 0.057742 + 4 = 1.0293 Mrad/s 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.
- 21. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 21. (a) (b) C = 3 + 7 = 10 nF ∴ω o = 1 −4 10 10 −8 = 106 rad/s Q o = ω o CR = 10610−85 5 × 103 = 50 B = ω o / Q o = 20 krad/s Parallel current source is 1∠0° = jω 3 × 10−9 At ω o , I s = j106 −9 × 3 Z3 ∴ V1,0 = j 3 × 10−3 × 5 × 103 = 15∠90° V (c) ω − ω o = 15 × 103 ∴ N = 15 ×103 15∠90° = 1.5 ∴ V1 = = 8.321∠33.69° V 3 10 × 10 1 + j1.5 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.
- 22. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 22. (a) (5 + 0.01s )(5 + 106 / s ) (5 + 0.01s )(5s + 106 ) = 10 + 0.01s + 106 / s 0.01s 2 + 10 s + 106 0.05s 2 + 25s + 104 s + 5 × 106 Zin ( s ) = 0.01s 2 + 10 s + 106 5 ×106 − 0.05ω 2 + j10, 025ω ∴ Zin ( jω ) = 106 − 0.01ω 2 + j10ω 10, 025ω o 10ω o 2 2 At ω = ω o , = 6 , 10.025 × 109 − 100.25 ω o = 5 × 107 − 0.5 ω o 6 2 2 5 × 10 − 0.05ω o 10 − 0.01ω o Zin ( s ) = 2 ∴ 99.75ω o = 9.975 × 109 , ω o = 10, 000 rad/s (b) Zin ( jω o ) = (5 + j100) (5 − j100) = 25 + 10, 000 = 1002.5 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.
- 23. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 23. , f o = 1000 Hz, Qo = 40, Zin ( jω o ) = 2k Ω ∴ B = 25 Hz (a) Zin(jω) = 10 March 2006 2000 f − 1000 , N= , f = 1010, ∴ N = 0.8 1 + jN 12.5 Zin = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω (b) 0.9 f o < f < 1.1 f o ∴ 900 < f < 1100 Hz 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. Chapter Sixteen Solutions 10 March 2006 Taking 2–½ = 0.7, we read from Fig. 16.48a: 1.7 kHz – 0.6 kHz = 1.1 kHz Fig. 16.48b: 2×107 Hz – 900 Hz = 20 MHz 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. Chapter Sixteen Solutions 10 March 2006 Bandwidth = 2π f 0 = 2π 106 = ω2 − ω1 , where ω1 = 2π ( 5.5 )103 . (a) ω2 = ω1 + B , therefore f2 = 5.5 + 103 kHz = 1.0055 MHz (b) f 0 = (c) Q0 = f1 f 2 = ( 5.5 )(1005.5) = 74.37 kHz f 0 74.37 ×103 = = 0.074 106 B 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 26. Chapter Sixteen Solutions 10 March 2006 Bandwidth = 109 Hz = f 2 − f1 , where f1 = 75.3 ×106 Hz. (a) f 2 = f1 + B , therefore f2 = 1.0753 GHz (b) f 0 = (c) Q0 = f1 f 2 = ( 75.3 ×10 )(1.0753 ×10 ) = 6 9 284.6 MHz f 0 284.6 × 106 = = 0.2846 109 B 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 27. Chapter Sixteen Solutions 10 March 2006 (a) To complete the sketch, we need to first find ω0, which we obtain in part (b). (b) ω0 = ω1ω2 = 2000 rad/s or f 0 = 318.3 Hz (c) B = ω2 − ω1 = 3000 rad/s or (d) Q = ω0 ω2 − ω1 = 477.5 Hz 2000 = 0.667 3000 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 Sixteen Solutions 10 March 2006 (a) We begin by labelling the series string with the capacitor as string 1, and the other as string 2. We next find the parallel equivalent of each, and determine the frequency where Xp1 + Xp2 = 0. 2 2 R12 + X 12 R2 + X 2 , and similarly X p2 = . Then X p1 = X1 X2 For X p1 + X p2 = 0 we have 1 At ω0, X 1 = − ω0C At ω0, X 2 = ω0 L ∴ ∴ 2 2 R12 + X 12 R2 + X 2 + =0 X1 X2 R12 + X 12 = X1 52 + [1] 1024 ω02 ( 330 ) −1012 330ω0 2 . 2 2 2 R2 + X 2 52 + 10−4 ω0 = . 10−2 ω0 X2 Enforcing Eq. [1], then, leads to ω0 = 1022 − ( 25 ) (330)1012 (330)108 − 25(33) 2 = 550.5 krad/s or f0 = 87.61 kHz. (b) We see the simulation result agrees reasonably, with a resonant frequency of 87.6 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.
- 29. Engineering Circuit Analysis, 7th Edition 29. Chapter Sixteen Solutions 10 March 2006 (a) We design for a bandwidth of 5.5 kHz, a low-frequency cut-off of 500 Hz, and a resonant impedance of 1 kΩ (no value was specified). Thus, we need to specify values for R, L, and C. f 2 = f1 + B = 6 kHz f0 = f1 f 2 = ( 0.5 ) (6) = 3 kHz f0 3 × 103 Q0 = = B 5.5 × 103 Q0 = ω0 RC so C = L= ( Q0 1 = = 28.9 nF 3 ω0 R 5.5 × 10 ( 2π )103 ( ) ) 5.5 ×103 103 1 = = 292 mH ω02C 2π 3 × 106 ( ) and, of course, R = 1 kΩ (b) From the simulation, we observe a bandwidth of 5.5 kHz, a lower frequency cutoff of approximately 500 Hz, and a peak impedance of 1000 Ω, as desired. 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 30. (a) f 0 = (b) Q0 = 1 2π 1 1 = LC 2π ω0 L = R Chapter Sixteen Solutions 1 ( )( 400 × 10−6 3.3 × 10−6 ) 10 March 2006 = 4.38 kHz 1 L 1 400 = = 1.10 LC R 10 3.3 (c) Z at resonance = R = 10 Ω (d) Z at 0.438 kHz = ⎡ 1 10 + j ⎢ 2π ( 438 ) 400 × 10−6 − 2π ( 438 ) 3.3 × 10−6 ⎢ ⎣ ( ) ( (e) Z at 43.8 kHz = ⎡ 1 10 + j ⎢ 2π ( 438 ) 400 × 10−4 − 2π ( 438 ) 3.3 × 10−4 ⎢ ⎣ ( ) ( ) ⎤ ⎥ = 10 − j109.01 Ω ⎥ ⎦ ) ⎤ ⎥ = 10 − j108.98 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.
- 31. Engineering Circuit Analysis, 7th Edition 31. Chapter Sixteen Solutions 10 March 2006 Bandwidth = 3 MHz, f1 = 17 kHz. (a) f 2 = f1 + B = (b) f 0 = (c) Q0 = f1 f 2 = 3.017 MHz 226.5 kHz f0 = 0.0755 B 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 32. Chapter Sixteen Solutions 10 March 2006 (a) Z0 = 1 Ω by definition (b) ω0 = 1 103 = = 707 rad/s LC 2 = 112.5 Hz (c) PSpice simulation verifies an impedance of 1 Ω at f = 112.6 Hz. 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 33. Chapter Sixteen Solutions 10 March 2006 (a) Z0 = 1 kΩ by definition (b) ω0 = 1 106 = = 707 krad/s LC 2 = 112.5 kHz (c) PSpice simulation verifies an impedance of 1 kΩ at f = 112.8 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.
- 34. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 34. (a) 20A 6Ω, 3 6 = 2, 40 V in series with 2 + 1 = 3 Ω 1 ω L 60 = 10 rad/s, Qo = o = = 20 Ω R 3 LC 10 1 B= = 0.5, B = 0.25, Vout ( jω o ) = 40Qo = 800 V 20 2 ωo = ⎛ ω − 10 ⎞ ∴ Vout ( jω ) = 800 / 1 + ⎜ ⎟ ⎝ 0.25 ⎠ (b) 2 ω = 9 rad/s 800 = 194.03V 17 40 600 Exact: Vout = × 3 + j (6ω − 600 / ω ) jω 24, 000 ∴ Vout ( j 9) = = 204.86∠ − 13.325− V 9[3 + j (54 − 66.67)] (Approx: Vout ( j 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.
- 35. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 35. Series: R = 50 Ω, L = 4 mH, C = 10−7 (a) ω o = 1/ 4 × 10−3−7 = 50 krad/s (b) f o = 50 ×103 / 2π = 7.958 kHz (c) Qo = (d) B = ω o / Qo = 50 × 103 / 4 = 12.5 krad/s (e) ω1 = ω o ⎡ 1 + (1/ 2Qo ) 2 − 1/ 2Qo ⎤ = 50 ⎡ 1 + 1/ 64 − 1/ 8⎤ = 44.14 krad/s ⎣ ⎦ (f) ω 2 = 50 ⎡ 65 / 64 + 1/ 8⎤ = 56.64 krad/s ⎣ ⎦ (g) Zin ( j 45, 000) = 50 + j (180 − 107 −3 / 45) = 50 − j 42.22 = 65.44∠ − 40.18°Ω (h) Zc / Z R ωo L R = ⎣ 45,000 50 ×103 × 4 × 10−3 =4 50 ⎦ = 107 / j 45, 000 × 50 = 4.444 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. 10 March 2006 Apply 1 A, in at top. ∴ VR = 10 V (a) Chapter Sixteen Solutions Vin = Zin = 10−3 s + 10 + 108 1.2 ×108 (0.5 ×10 + 1) = 10−3 s + 10 + s 5s 8 8 −3 −3 Zin ( jω ) = 10 + j (10 ω − 1.2 × 10 / ω ) ∴10 ω o = 1.2 × 10 / ω o 2 ∴ω o = 1.2 × 1011 , ω o = 346.4 krad/s (b) Qo = ωo L R = 346.4 ×103−3 = 34.64 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.
- 37. Engineering Circuit Analysis, 7th Edition 37. Chapter Sixteen Solutions 10 March 2006 Find the Thévenin equivalent seen by the inductor-capacitor combination: ⎛ V ⎞ SC : 1.5 = V1 + 10 ⎜ 1 − 0.105 V1 ⎟ ∴ V1 = 50 V ⎝ 125 ⎠ 50 ∴↓ ISC = = 0.4 A 125 1.5 = 3.75 Ω OC :V1 = 0 ∴ VOC = 1.5 V ∴ R th = 0.4 1000 × 4 ∴ω o = 1/ 4 × 0.25 × 10−6 = 1000, Qo = = 1066.7 3.75 1000 1 = 0.9375, B = 0.4688 rad/s B = ω o / Qo = 1066.7 2 VC max = Qo Vth = 1066.7 ×1.5 = 1600 V Therefore, keep your hands off! To generate a plot of |VC| vs. frequency, note that VC(jω) = 1.5 − j ωC 3.75 + jωL − j ωC 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 Sixteen Solutions 38. Series, f o = 500 Hz, Qo = 10, X L ,0 = 500 Ω (a) 500 = ω o L = 2π (500)L ∴ L = 0.15915+ H, C = 10 March 2006 Qo = 10 = (b) X L ,0 R = 1 2π = = 0.6366 μ F 2 ω o L (2π × 500) 2 500 ∴ R = 50 Ω R ⎛ ⎛ 1 106 × 0.5π ⎞ 250, 000 ⎞ 1 = I ⎜ 50 + j 2π f × −j ⎟ = I ⎜ 50 + j f − j ⎟ f 2π 2π f ⎠ ⎝ ⎠ ⎝ 6 10 × 0.5π ∴ I = 1/ 50 + j ( f − 250, 000 / f ), Vc = I j 2π f − j 250, 000 / f ∴ Vc (2π × 450) = 4.757 V VC = 50 + j ( f − 250, 000 / f ) Vc (2π × 500) = 10, 000 V Vc (2π × 550) = 4.218 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.
- 39. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 39. X : s = 0, ∞, 0 : s = −20, 000 ± j80, 000 s −1 , Zin (−104 ) = −20 + j 0 Ω ∴ SERIES α = 20, 000, ω d = 80, 000 ∴ω o = (64 + 4)108 = 82, 462 rad/s, 1 2 = ω o = 68 × 108 LC R R 1 L 68 ×108 1 = α = 20, 000 ∴ = 40, 000, × = = 170, 000; Z(σ ) = R + σ L + 2L L1 LC R 40, 000 σC 1 170, 000 1 =R− R− R ∴ R = 1.2308 Ω 4 10, 000 10, 000C 1 1.2308 ∴L = = 30.77 μ H, C = = 4.779 μ F 170, 000 ×1.2308 40, 000 ∴−20 = R − 10, 000L − 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 40. ωo 1/ 10 −3− 7 Chapter Sixteen Solutions 10 March 2006 105−3 = 10 rad/s, Q L = = 100, R PL = 10, 000 Ω 1 5 1 = 500, R PC = 5002 × 0.2 = 50, 000 Ω 10 × 0.2 50 10 = 8.333 k Ω ∴ Q o = ωo CR = 105−7 × 8333 = 83.33 Qc = 5− 7 100, 000 = 1200 rad/s, Zin ( jωo ) = 8333 Ω 83.33 (99 − 100)103 8.333 ω = 99, 000 ∴ N = = −1.6667, Zin ( j 99, 000) = 600 1 − j1.667 = 4.287 ∠ 59.04o kΩ B= 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 41. Chapter Sixteen Solutions 10 March 2006 Req = Qo/ ωo C = 50 / 105-7 = 5000 Ω. Thus, we may write 1/5000 = 1/8333 + 1/Rx so that Rx = 12.5 kΩ. 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 Chapter Sixteen Solutions 10 March 2006 42. 3mH 1.5 mH = 1mH, 2 μF + 8 μF = 10 μF, ∴ωo = 1 10−3−5 = 10 krad/s 3 × 10−3 ×104 = 100, R p = 1002 × 0.3 = 3 k Ω 0.3 1.5 × 10−3 × 104 Q= = 60, R p = 60 × 0.25 = 900 Ω 0.25 692.3 900 3000 = 692.3 Ω ∴ Q L = 4−3 = 69.23 10 692.3 ∴ R LS = = 0.14444 Ω 69.232 106 Q= 4 = 125, R pc = 1252 × 0.1 = 1562.5 Ω 10 μF 10 × 0.1× 8 1562.5 ∴ Qc = 104 × 10−5 × 15625 = 156.25 ∴ R SC = = 0.064 Ω (156.25) 2 Q= ∴ R S ,tot = 0.14444 + 0.064 = 0.2084 Ω = Zin min , ωo = 10 krad/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.
- 43. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 43. (a) ωo 1/ 2 × 0.2 ×10−3 = 50 rad/s QleftL = 50 × 2.5 / 2 = 62.5, 2 × 62.52 = 7812.5 Ω 50 ×10 = 50, 10 × 502 = 25 k Ω 10 1000 Qc = = 100, 1002 ×1 = 10 k Ω, R p = 7.8125 25 10 = 3731Ω 50 × 0.2 ×1 50 1 Qo = 50 × 3731× 0.2 ×10 −3 = 37.31; B = = 1.3400, B = 0.6700 37.31 2 −3 ∴ V o = 10 × 3731 = 3.731V Q rightL = 3.731 V 2.638 V |V| (volts) ↔ 1.34 rad/s 50 (b) ω (rad/s) V = 10−3 [(2 + j125) (10 + j500) (1 − j100)] = 10−3 = 3.7321∠ − 0.3950+ ° V 1 1 1 + + 2 + j125 10 + j 500 1 − j100 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 Chapter Sixteen Solutions 10 March 2006 44. (a) 1000 = 2000 rad/s, Qc = 2000 × 2 ×10−6 × 25 ×103 = 100 0.25 R 20 ×104 ∴ R C , S = 25, 000 /1002 = 2.5 Ω; Q L = = = 40 ωo L 2000 × 0.25 ωo 20, 000 = 12.5 Ω ∴ R tot = 12.5 + 2.5 = 15 Ω 1600 2000 × 0.25 1 ∴ Qo = = 33.33 ∴ Vx = 1× 33.33 × = 16.667 V 15 2 ∴ R L,S = (b) 20, 000 × j 500 = 12, 4922 + j 499.688 Ω 20, 000 + j 500 25, 000(− j 250) = 2.4998 − j 249.975 25, 000 − j 250 = 25, 000 − j 250 20, 000 j 500 = ∴ Zin = 12.4922 + 2.4998 + j 499.688 − j 250 − j 249.975 = 14.9920 − j 0.2870 Ω ∴ I = 1/ 14.9920 − j 0.2870 = 66.6902 mA ∴ Vx = 250 × 66.6902 ×10−3 = 16.6726 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.
- 45. Engineering Circuit Analysis, 7th Edition 45. Q = ωCR, RS = XS = − Chapter Sixteen Solutions 10 March 2006 RP Q2 X P , and X S = 1 + Q2 1 + Q2 1 1 1 + Q2 , XP = − ∴ CS = C P Q2 ωCS ωCP (a) ω = 103 rad/s, Q = 5 Therefore, RS = 5/26 = 192 Ω, CS = 26/25 μF = 1.06 μF (b) ω = 104 rad/s, Q = 50 Therefore, RS = 5/2501 = 2 Ω, CS = 2501/2500 μF = 1.0004 μF (c) ω = 105 rad/s, Q = 500 Therefore, RS = 5000/250001 = 20 mΩ, CS = 250001/250000 μF = 1.0 μF 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 46. ( ) RP = RS 1 + Q 2 , and X P = X S C P = CS Chapter Sixteen Solutions 10 March 2006 1 + Q2 Q2 Q2 1 + Q2 (a) ω = 103 rad/s, Q = 0.2 Therefore, RP = 5(1 + 0.04) = 5.2 kΩ, CP = 38.5 nF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 5(1 + 0.0004) = 5.002 kΩ, CP = 400 pF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 5(1 + 4×10–6) = 5 kΩ, CP = 4 pF 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 47. Q= Chapter Sixteen Solutions 10 March 2006 RP Q2 X P R Q2 , RS = , and X S = . LS = LP 1 + Q2 ωL 1 + Q2 1 + Q2 (a) ω = 103 rad/s, Q = 142.4×103 Therefore, RS = 470/(1 + Q2) = 23.2 nΩ, LS = 3.3 μH (b) ω = 104 rad/s, Q = 14.24×103 Therefore, RS = 470/(1 + Q2) = 23.2 μΩ, LS = 3.3 μH (c) ω = 105 rad/s, Q = 1.424×103 Therefore, RS = 470/(1 + Q2) = 232 μΩ, LS = 3.3 μH 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 48. Chapter Sixteen Solutions 10 March 2006 ⎛ 1 + Q2 ⎞ RP = RS 1 + Q 2 , and X P = X S ⎜ ⎟ 2 ⎝ Q ⎠ ⎛ 1 + Q2 ⎞ LP = LS ⎜ ⎟ 2 ⎝ Q ⎠ ( ) (a) ω = 103 rad/s, Q = 7.02×10–6 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 67 mF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 670 μF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 6.70 μF 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 49. Chapter Sixteen Solutions 10 March 2006 R 470 ≈ 7 −6 = 47 . Since Q > 5, the series ω L 10 10 equivalent is a 10/47 Ω resistor in series with 1 μH. (a) For the left parallel circuit, Q = For the right parallel circuit, Q = ωCR ≈ 10710−8 ( 200 ) = 20 . Again, Q > 5, so the series equivalent is a 10/20 Ω = 500 mΩ resistor in series with 10 nF. We may therefore approximate the network as a 700 mΩ resistor in series with a 10 nF capacitor, in series with a 1 μH inductor, in series with the 10 μH inductor of interest. At the resonant frequency the network connected in series with the inductor has an impedance of 700 mΩ. The inductor present an impedance of 100 Ω. Thus, |Vx| = 1 V. (b) ZL = ( 470 ) ( j10710−6 ) 470 + j10 1 jωC2 = 0.213 + j9.995 Ω . Z L = = 0.499 − j9.975 Ω 1 R2 + jωC2 R2 Z3 = j100 Ω. Thus, Vx = Z3 j100 = 0.99745 + j 0.0071 V (1∠0 ) = 0.714 + j 0.02 Z1 + Z L + Z3 So that |Vx| = 0.99977 V . Our approximation was pretty accurate, at least at this frequency. 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 Sixteen Solutions 10 March 2006 50. (a) (b) 50 20 × 103 = 0.5 K f = = 0.02 100 106 1 0.5 ∴ 9.82 μH → 0.5 × 9.82 × = 24.55 μH, 31.8 μH → × 31.8 = 795 μH 0.02 0.02 2.57 = 257 nF 2.57 nF → 0.5 × 0.02 Km = same ordinate; divide numbers on abscissa by 50 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 Sixteen Solutions 10 March 2006 51. (a) Apply 1 V ∴ I1 = 10A ∴ 0.5 I1 = 5A ↓; 5A 0.2 Ω can be replaced by 1 V in series with 0.2 Ω ∴ Iin → = 10 + s + 10 1 − (−1) 2s 4s + 20 20( s + 5) = 10 + = = ∴ Zin ( s) = 20( s + 5) s + 10 0.2 + 2 / s 0.2s + 2 0.2s + 2 2( s / 5 + 10) 0.1( s + 50) = 20( s / 5 + 5) s + 25 (b) K m = 2, K f = 5 ∴ Zin ( s ) → (c) 0.1Ω → 0.2 Ω, 0.2 Ω → 0.4 Ω, 0.5F → 0.05 F, 0.5 I1 → 0.5 I1 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. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 52. (a) ωo 1/ (2 + 8)10−310−6 = 104 rad/s −3 Q L ,8 = 10 / 8 ×10 10 = 125 ∴ R L , S 4 4 2 + 8 = 10 mH ∴ Q L = 104 = = 0.64 Ω 1252 104 ×10 × 10−3 = 156.25 0.64 1 = 100, R C , P = 1002 × 1 = 10 k Ω 10 ×10−6 ∴ R P = 20 15.625 10 = 4.673 k Ω ∴ Qo = 104 × 10−6 × 4.673 × 103 = 46.73 ∴ R L , P = 0.64 × 156.252 = 15.625 k Ω; QC = 4 (b) K f = 106 /104 = 100, K m = 1 ∴ R ′s stay the same; 2 mH → 20 μH, 8mH → 80 μH,1μF → 10 nF (c) ωo = 106 rad/s, Qo stays the same, ∴ B = 106 = 21.40 krad/s 46.73 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.
- 53. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 53. (a) K m = 250, K f = 400 ∴ 0.1F → 5Ω → 1250 Ω, 2H → 0.1 = 1μF 250 × 400 2 × 250 = 1.25 H, 4 Ix → 103 Ix 400 1.25 H 1 μF (b) ω = 103. Apply 1 V ∴ I x = 10−6 s, ↓ I1250 = 1250 Ω 103 1 1250 1 − 10−3 s ∴1000 I x = 10 s ∴→ I L = 1.25s 1 0.8 0.8 (1 − 10−3 s ) = 10−6 s + ; s = j103 ∴ Iin = 10−6 s + + 1250 s s −3 0.8 ×10 1 1000 ∴ Iin = j10−3 + = j 0.2 × 10−3 ∴ Zth = = = − j 5 k Ω Voc = 0 Iin j j 0.2 −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.
- 54. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 54. (a) Is = 2∠0° A, ω = 50 ∴ Vout = 60∠25° V (b) Is = 2∠40° A, ω = 50 ∴ Vout = 60∠65° V (c) Is = 2∠40° A, ω = 200, ∴ OTSK (d) K m = 30, IS = 2∠40° A, ω = 50 ∴ Vout = 1800∠65° V (e) K m = 30, K f = 4, Is = 2∠40° A, ω = 200 ∴ Vout = 1800∠65° 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.
- 55. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 55. (a) H /( s) = 0.2 ∴ H dB = 20 log 0.2 = −13.979 dB (b) H( s ) = 50 ∴ H dB = 20 log 50 = 33.98dB (c) H( j10) = (d) H dB = 37.6 dB ∴ H( s) = 1037.6 / 20 = 75.86 (e) H dB = −8dB ∴ H( s ) = 10−8/ 20 = 0.3981 (f) H dB = 0.01dB ∴ H( s ) = 100.01/ 20 = 1.0012 12 26 6 13 292 + j 380 + ∴ H dB = 20 log + = 20 log = 6.451dB 2 + j10 20 + j10 1 + j 5 10 + j 5 −60 + j 220 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 56. (a) Chapter Sixteen Solutions 10 March 2006 (d) MATLAB verification- shown adjacent to Bode plots below. 20( s + 1) 0.2(1 + s ) H( s ) = = , 0.2 → −14 dB s + 100 1 + s /100 1 10 100 2000( s + 1) s 0.2 s(1 + s) = , 0.2 → −14 dB 2 ( s + 100) (1 + s /100) 2 (b) H( s ) = (c) 200 s 2 + 45s + 200 ( s + 5)( s + 40) 200(1 + s / 5)(1 + s / 40) H( s ) = s + 45 + = = = , 200 → 46 dB s s s 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.
- 57. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 57. H( s ) = VC (20 + 2 s)(182 + 200 / s) 200 / s = × IR 202 + 2 s + 200 / s 182 + 200 / s 400( s + 10) 200(10 + s ) = 2 2( s + 101s + 100) (1 + s)(100 + s ) 20(1 + s /10) H( s ) = , 20 → 26 dB (1 + s )(1 + s /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.
- 58. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 58. 5 ×108 s ( s + 100) 2.5s (1 + s /100) = , 2.5 → 8dB 3 ( s + 20)( s + 1000) (1 + s / 20)(1 + s /1000)3 (a) H( s ) = (b) Corners: ω = 20, 34 dB; ω = 100, 34 dB; ω = 1000, 54 dB Intercepts: 0 dB, 2.5ω = 1, ω = 0.4 2.5ω (ω /100) 2.5ω2 (20)109 ω = 1, 8dB; 0 dB, = = 1 ∴ω = 22,360 rad/s (ω / 20)(ω /1000)3 100ωω3 (c) Corners: ω = 20, 31.13dB ω = 100, 36.69 dB H dB = 20 log 2.5ω 1 + (ω /100) 2 [1 + (ω / 20) 2 ][1 + (ω /1000) 2 ]3 ω = 1000, 44.99 dB 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 Sixteen Solutions 10 March 2006 59. (a) (b) H( s ) = 5 × 108 s ( s + 100) 2.5s (1 + s /100) = , 3 ( s + 20)( s + 1000) (1 + s / 20)(1 + s /1000)3 ω = 2 : ∠ = 90° ⎛ ⎝ ω = 10 : ∠ = 90° − ⎜ 45° + 45° log 10 ⎞ ⎟ = 58.5° 20 ⎠ ⎛ ⎝ 100 ⎞ ⎛ 100 ⎞ ⎟ + ⎜ 45° + 45° log ⎟ = 58.5° 20 ⎠ ⎝ 100 ⎠ 200 ⎞ ⎛ 200 ⎞ ⎛ ω = 200 : ∠ = 90° − 90° + ⎜ 45° + 45° log ⎟ − 3 ⎜ 45° + 45° log ⎟ = 17.9° 100 ⎠ ⎝ 100 ⎠ ⎝ ω = 100 : ∠ = 90° − ⎜ 45° + 45° log 1000 ⎞ ⎛ ⎟ = −45° 1000 ⎠ ⎝ ω = 10, 000 : ∠ = 90° − 90° + 90° − 3 × 90° = −180° ω = 1000 : ∠ = 90° − 90° + 90° − 3 ⎜ 45° + 45° log (c) ω = 2 : ∠ = 90° + tan −1 0.02 − tan −1 0.1 − 3 tan −1 0.002 = 85.09° ω = 10 : ∠ = 90° + tan −1 0.1 − tan −1 0.5 − 3 tan −1 0.01 = 67.43° ω = 100 : ∠ = 90° + tan −1 1 − tan −1 5 − 3 tan −1 0.1 = 39.18° ω = 200 : ∠ = 90° + tan −1 2 − tan −1 10 − 3 tan −1 0.2 = 35.22° ω = 1000 : ∠ = 90° + tan −1 10 − tan −1 50 − 3 tan −1 1 = −49.56° ω = 10, 000 : ∠ = 90° + tan −1 100 − tan −1 500 − 3 tan −1 10 = −163.33° 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 Chapter Sixteen Solutions 10 March 2006 60. (a) 20 400 s 2 + 20 s + 400 + 2 = s s s2 1 + 2 × 0.5( s / 20) + ( s / 20) 2 = 400 s2 ∴ω o = 20, ζ = 0.5 H( s ) = 1 + Hdb 20 log 400 = 52dB Correction at ω o is 20 log 2 ζ = 0 dB (b) ω = 5 : H dB = 52 − 2 × 20 log 5 = 24.0 dB (plot) H dB = 20 log 1 − 16 + j 4 = 23.8dB (exact) ω = 100 : H dB = 0 dB (plot) H dB = 20 log 1 − 0.04 + j 0.2 = −0.170 dB (exact) (c) 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 Sixteen Solutions 10 March 2006 61. (a) (b) H( s ) = VR 25 25s = = = 2 V5 10 s + 25 + 1000 / s 10 s + 25s + 1000 ⎛ 1 ⎞⎛ s ⎞ ⎛ s ⎞ 1 + 2 ⎜ ⎟⎜ ⎟ + ⎜ ⎟ ⎝ 8 ⎠⎝ 10 ⎠ ⎝ 10 ⎠ 2 ⎛ 1⎞ ∴ω o = 10, ζ = 1/ 8 ∴ correction = −20 log ⎜ 2 × ⎟ = 12 dB ⎝ 8⎠ 0.025 → −32 dB HdB (c) 0.025s ω = 20, H( j 20) = ang(H) j 0.5 ∴ H dB = −15.68 dB ∠H( j 20) = −80.54° 1 − 4 + j 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.
- 62. Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006 62. 1st two stages, H1 ( s ) = H 2 ( s ) = −10; H 3 ( s ) = −1/(50 ×103 ×10−6 ) −20 = 3 −6 s + 1/(200 × 10 × 10 ) s + 5 ⎛ −20 ⎞ −400 ∴ H( s ) = (−10)(−10) ⎜ ⎟= ⎝ s + 5 ⎠ 1+ s / 5 −400 → 52 dB 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 63. (a) Chapter Sixteen Solutions 10 March 2006 1st stage: C1 A = 1 μ F, R1 A = ∞, R fA = 105 ∴ H A (S) = − R fA C1 A s = −0.1 s 2nd stage: R 1B = 105 , R fB = 105 , C fB = 1 μ F ∴ H B ( s ) = −1/ R1B C fB s + 1/ R fB C fB 1/(105 × 10−6 ) 10 =− 5 −6 s + 1/(10 × 10 ) s + 10 3rd stage: same as 2nd ∴ H B (s) = 0.1s ⎛ −10 ⎞ ⎛ −10 ⎞ ∴ H( s ) = (−0.1s ) ⎜ ⎟⎜ ⎟=− (1 + s /10) 2 ⎝ s + 10 ⎠ ⎝ s + 10 ⎠ 20log10(0.1) = -20 dB (b) (c) 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 64. Chapter Sixteen Solutions 10 March 2006 An amplifier that rejects high-frequency signals is required. There is some ambiguity in the requirements, as social conversations may include frequencies up to 50 kHz, and echolocation sounds, which we are asked to filter out, may begin below this value. Without further information, we decide to set the filter cutoff frequency at 50 kHz to ensure we do not lose information. However, we note that this decision is not necessarily the only correct one. Our input source is a microphone modeled as a sinusoidal voltage source having a peak amplitude of 15 mV in series with a 1-Ω resistor. Our output device is an earphone modeled as a 1-kΩ resistor. A voltage of 15 mV from the microphone should correspond to about 1 V at the earphone according to the specifications, requiring a gain of 1000/15 = 66.7. Rf = 66.7 - 1 = 65.7 If we select a non-inverting op amp topology, we then need R1 Arbitrarily choosing R1 = 1 kΩ, we then need Rf = 65.7 kΩ. This completes the amplification part. Next, we need to filter out frequencies greater than 50 kHz. Placing a capacitor across the microphone terminals will “short out” high frequencies. 1 We design for ωc = 2πfc = 2π(50×103) = . Since Rmic = 1 Ω, we require Rmic C filter Cfilter = 3.183 μF. 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. Chapter Sixteen Solutions 10 March 2006 We choose a simple series RLC circuit. It was shown in the text that the “gain” of the ωRC . circuit with the output taken across the resistor is AV = 1 2 2 2 2 2 2 1 - ω LC + ω R C This results in a bandpass filter with corner frequencies at [( ωc = L -RC + R 2 C 2 + 4 LC 2 LC and ωc = H ) ] RC + R 2 C 2 + 4 LC 2 LC If we take our output across the inductor-capacitor combination instead, we obtain the opposite curve- i.e. a bandstop filter with the same cutoff frequencies. Thus, we want -RC + R 2 C 2 + 4 LC 2π(20) = and 2π(20×103) = 2 LC RC + R 2 C 2 + 4 LC 2 LC Noting that ω cH – ω cL = R/L = 125.5 krad/s, we arbitrarily select R = 1 kΩ, so that L = 7.966 mH. Returning to either cutoff frequency expression, we then find C = 7.950 μF PSpice verification. The circuit performs as required, with a lower corner frequency of about 20 Hz and an upper corner frequency of about 20 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.
- 66. Engineering Circuit Analysis, 7th Edition 66. Chapter Sixteen Solutions 10 March 2006 We choose a simple RC filter topology: Vout 1 = Vin 1 + jωRC Vout 1 . We desire a cutoff = 2 Vin 1 + (ωRC ) frequency of 1 kHz, and note that this circuit does indeed act as a low-pass filter (higher frequency signals lead to the capacitor appearing more and more as a short circuit). Thus, 1 1 = = where ωc = 2πfc = 2000π rad/s. 2 2 1 + (ω c RC ) Where and hence A small amount of algebra yields 1 + [2π(1000)RC]2 = 2 or 2000πRC = 1. Arbitrarily setting R = 1 kΩ, we then find that C = 159.2 nF. The operation of the filter is verified in the PSpice simulation below: 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 67. Chapter Sixteen Solutions 10 March 2006 We are not provided with the actual spectral shape of the noise signal, although the reduction to 1% of its peak value (a drop of 40 dB) by 1 kHz is useful to know. If we place a simple high-pass RC filter at the input of an op amp stage, designing for a pole at 2.5 kHz should ensure an essentially flat response above 25 kHz, and a 3 dB reduction at 2.5 kHz. If greater tolerance is required, the 40 dB reduction at 1 kHz allows the pole to be moved to a frequency even closer to 1 kHz. The PSpice simulation below shows a 1 filter with R = 1 kΩ (arbitrarily chosen) and C = = 63.66 nF . 2π (2.5 × 10 3 )(1000) At a frequency of 25 kHz, the filter shows minimal gain reduction, but at 1 kHz any signal is reduced by more than 8 dB. We therefore design a simple non-inverting op amp circuit such as the one below, which with Rf = 100 kΩ and R1 = 1 kΩ, has a gain of 100 V/V. In simulating the circuit, a gain of approximately 40 dB at 25 kHz was noted, although the gain dropped at higher frequencies, reaching 37 dB around 80 kHz. Thus, to completely assess the suitability of design, more information regarding the frequency spectrum of the “failure” signals would be required. 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. Chapter Sixteen Solutions 10 March 2006 We select a simple series RLC circuit with the output taken across the resistor to serve as a bandpass filter with 500 Hz and 5000 Hz cutoff frequencies. From Example 16.12, we know that R 1 ωc L = + R 2C 2 + 4LC = 2π (500) 2L 2LC and R 1 ωcH = + R 2C 2 + 4LC = 2π (5000) 2L 2LC With ω cH - ω cL = 2p(5000 – 500) = R/L, we (arbitrarily) select R = 1 kΩ, so that L = 35.37 mH. Substituting these two values into the equation for the high-frequency cutoff, we find that C = 286.3 nF. We complete the design by selecting R1 = 1 kΩ and Rf = 1 kΩ for a gain of 2 (no value of gain was specified). As seen in the PSpice simulation results shown below, the circuit performs as specified at maximum gain (6 dB or 2 V/V), with cutoff frequencies of approximately 500 and 5000 KHz and a peak gain of 6 dB. 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 69. Chapter Sixteen Solutions 10 March 2006 For this circuit, we simply need to connect a low-pass filter to the input of a noninverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the cutoff frequency is 1 ωc = = 2π (3000) RC Selecting (arbitrarily) R = 1 kΩ, we find C = 53.05 nF. The PSpice simulation below shows that our design does indeed have a bandwidth of 3 kHz and a peak gain of 10 V/V (20 dB). 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. Chapter Sixteen Solutions 10 March 2006 We require four filter stages, and choose to implement the circuit using op amps to isolate each filter subcircuit. Selecting a bandwidth of 1 rad/s (no specification was given) and a simple RLC filter as suggested in the problem statement, a resistance value of 1 Ω leads to an inductor value of 1 H (bandwidth for this type of filter = ωH – ωL = R/L). The capacitance is found by designing each filter’s respective resonant frequency ( 1 LC ) at the desired “notch” frequency. Thus, we require CF1 = 10.13 μF, CF2 = 2.533 μF, CF3 = 1.126 μF and CF4 = 633.3 nF. The Student Version of PSpice® will not permit more than 64 nodes, so that the total solution must be simulated in two parts. The half with the filters for notching out 50 and 100 Hz components is shown below; an additional two op amp stages are required to complete the design. 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.
- 71. Engineering Circuit Analysis, 7th Edition 71. Chapter Sixteen Solutions 10 March 2006 Using the series RLC circuit suggested, we decide to design for a bandwidth of 1 rad/ s (as no specification was provided). With ωH – ωL = R/ L, we arbitrarily select R = 1 Ω so that L = 1 H. The capacitance required is obtained by setting the resonant frequency of the circuit ( 1 LC ) equal to 60 Hz (120π rad/s). This yields C = 7.04 μF. vin vout 1Ω 1H 7.04 μF 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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