YOMEDIA
ADSENSE
Bài giải mạch P14
84
lượt xem 11
download
lượt xem 11
download
Download
Vui lòng tải xuống để xem tài liệu đầy đủ
Tham khảo tài liệu 'bài giải mạch p14', kỹ thuật - công nghệ, điện - điện tử phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả
AMBIENT/
Chủ đề:
Bình luận(0) Đăng nhập để gửi bình luận!
Nội dung Text: Bài giải mạch P14
- Chapter 14, Solution 1. Vo R jωRC H (ω) = = = Vi R + 1 jωC 1 + jωRC jω ω 0 1 H (ω) = , where ω 0 = 1 + jω ω 0 RC ω ω0 π ω H = H (ω) = φ = ∠H (ω) = − tan -1 1 + (ω ω0 ) 2 2 ω0 This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that ω0 = 1 RC . Thus, the sketches of H and φ are shown below. H 1 0.7071 0 ω0 = 1/RC ω φ 90° 45° 0 ω0 = 1/RC ω
- Chapter 14, Solution 2. R 1 1 R H (ω) = = = , where ω0 = R + jωL 1 + jωL R 1 + jω ω 0 L 1 ω H = H (ω) = φ = ∠H (ω) = - tan -1 1 + (ω ω0 ) 2 ω0 The frequency response is identical to the response in Example 14.1 except that ω0 = R L . Hence the response is shown below. H 1 0.7071 0 ω0 = R/L ω φ ω0 = R/L 0° ω -45° -90° Chapter 14, Solution 3. (a) The Thevenin impedance across the second capacitor where Vo is taken is R Z Th = R + R || 1 sC = R + 1 + sRC 1 sC Vi VTh = Vi = R + 1 sC 1 + sRC ZTh + + 1 VTh Vo − sC −
- 1 sC Vi Vo = ⋅ VTh = Z Th + 1 sC (1 + sRC)(1 + sCZ Th ) Vo 1 1 H (s) = = = Vi (1 + sCZ Th )(1 + sRC) (1 + sRC)(1 + sRC + sRC (1 + sRC)) 1 H (s) = s R C + 3sRC + 1 2 2 2 (b) RC = (40 × 10 3 )(2 × 10 -6 ) = 80 × 10 -3 = 0.08 There are no zeros and the poles are at - 0.383 s1 = = - 4.787 RC - 2.617 s2 = = - 32.712 RC Chapter 14, Solution 4. 1 R (a) R || = jωC 1 + jωRC R Vo 1 + jωRC R H (ω) = = = Vi R R + jωL (1 + jωRC) jωL + 1 + jωRC R H (ω) = - ω RLC + R + jωL 2 R + jωL jωC (R + jωL) (b) H (ω) = = R + jωL + 1 jωC 1 + jωC (R + jωL) - ω 2 LC + jωRC H (ω) = 1 − ω 2 LC + jωRC
- Chapter 14, Solution 5. Vo 1 jωC (a) H (ω) = = Vi R + jωL + 1 jωC 1 H (ω) = 1 + jωRC − ω 2 LC 1 R (b) R || = jωC 1 + jωRC Vo jωL jωL (1 + jωRC) H (ω) = = = Vi jωL + R (1 + jωRC) R + jωL (1 + jωRC) jωL − ω 2 RLC H (ω) = R + jωL − ω 2 RLC Chapter 14, Solution 6. (a) Using current division, Io R H (ω) = = I i R + jωL + 1 jωC jωRC jω (20)(0.25) H (ω) = = 1 + jωRC − ω LC 1 + jω(20)(0.25) − ω2 (10)(0.25) 2 jω5 H (ω) = 1 + jω5 − 2.5 ω 2 (b) We apply nodal analysis to the circuit below. 0.5 Vx Vx Io 1/jωC + − Is R jωL
- Vx Vx − 0.5Vx Is = + R jωL + 1 jωC 0.5 Vx But Io = → Vx = 2 I o ( jωL + 1 jωC) jωL + 1 jωC Is 1 0 .5 = + Vx R jωL + 1 jωC Is 1 1 = + 2 I o ( jωL + 1 jωC) R 2 ( jωL + 1 jωC) I s 2 ( jωL + 1 jωC) = +1 Io R Io 1 jωRC H (ω) = = = I s 1 + 2 ( jωL + 1 jωC) R jωRC + 2 (1 − ω 2 LC) jω H (ω) = jω + 2 (1 − ω2 0.25) jω H (ω) = 2 + jω − 0.5 ω 2 Chapter 14, Solution 7. (a) 0.05 = 20 log10 H 2.5 × 10 -3 = log10 H H = 10 2.5×10 = 1.005773 -3 (b) - 6.2 = 20 log10 H - 0.31 = log10 H H = 10 -0.31 = 0.4898 (c) 104.7 = 20 log10 H 5.235 = log10 H H = 10 5.235 = 1.718 × 10 5
- Chapter 14, Solution 8. (a) H = 0.05 H dB = 20 log10 0.05 = - 26.02 , φ = 0° (b) H = 125 H dB = 20 log10 125 = 41.94 , φ = 0° j10 (c) H(1) = = 4.472∠63.43° 2+ j H dB = 20 log10 4.472 = 13.01 , φ = 63.43° 3 6 (d) H(1) = + = 3.9 − j1.7 = 4.254∠ - 23.55° 1+ j 2 + j H dB = 20 log10 4.254 = 12.577 , φ = - 23.55° Chapter 14, Solution 9. 1 H (ω) = (1 + jω)(1 + jω 10) H dB = -20 log10 1 + jω − 20 log10 1 + jω / 10 φ = - tan -1 (ω) − tan -1 (ω / 10) The magnitude and phase plots are shown below. HdB 0.1 1 10 100 ω 1 20 log 10 -20 1 + jω / 10 1 20 log10 -40 1 + jω φ 0.1 1 10 100 ω -45° 1 arg 1 + jω / 10 -90° 1 arg 1 + jω -135° -180°
- Chapter 14, Solution 10. 50 10 H( jω) = = jω(5 + jω) jω 1 jω1 + 5 HdB 40 20 log1 20 10 0.1 1 100 ω 1 -20 20 log 1 jω 20 log jω 1+ -40 5 φ 1 0.1 10 100 ω -45° 1 arg 1 + jω / 5 -90° 1 arg -135° jω -180° Chapter 14, Solution 11. 5 (1 + jω 10) H (ω) = jω (1 + jω 2) H dB = 20 log10 5 + 20 log10 1 + jω 10 − 20 log10 jω − 20 log10 1 + jω 2 φ = -90° + tan -1 ω 10 − tan -1 ω 2
- The magnitude and phase plots are shown below. HdB 40 34 20 14 0.1 1 10 100 ω -20 -40 φ 90° 45° 0.1 1 10 100 ω -45° -90°
- Chapter 14, Solution 12. 0.1(1 + jω ) T ( w) = , 20 log 0.1 = −20 jω (1 + jω / 10) The plots are shown below. |T| (db) 20 0 ω 0.1 1 10 100 -20 -40 arg T 90o 0 ω 0.1 1 10 100 -90o
- Chapter 14, Solution 13. 1 + jω (1 10)(1 + jω) G (ω) = = ( jω) (10 + jω) ( jω) 2 (1 + jω 10) 2 G dB = -20 + 20 log10 1 + jω − 40 log10 jω − 20 log10 1 + jω 10 φ = -180° + tan -1ω − tan -1 ω 10 The magnitude and phase plots are shown below. GdB 40 20 0.1 1 10 100 ω -20 -40 φ 90° 0.1 1 10 100 ω -90° -180°
- Chapter 14, Solution 14. 50 1 + jω H (ω) = 25 jω10 jω 2 jω1 + + 25 5 H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 jω − 20 log10 1 + jω2 5 + ( jω 5) 2 ω10 25 φ = -90° + tan -1 ω − tan -1 1 − ω2 5 The magnitude and phase plots are shown below. HdB 40 26 20 6 0.1 1 10 100 ω -20 -40 φ 90° 0.1 1 10 100 ω -90° -180°
- Chapter 14, Solution 15. 40 (1 + jω) 2 (1 + jω) H (ω) = = (2 + jω)(10 + jω) (1 + jω 2)(1 + jω 10) H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 1 + jω 2 − 20 log10 1 + jω 10 φ = tan -1 ω − tan -1 ω 2 − tan -1 ω 10 The magnitude and phase plots are shown below. HdB 40 20 6 0.1 1 10 100 ω -20 -40 φ 90° 45° 0.1 1 10 100 ω -45° -90° Chapter 14, Solution 16. jω G (ω) = 2 jω 100(1 + jω)1 + 10
- GdB 20 log jω 20 0.1 1 10 100 jω ω − 40 log -20 10 -40 20 log(1/100) -60 φ arg(jω) 90° ω 0.1 1 10 100 1 -90° 1 arg arg 1 + jω 2 jω -180° 1 + 10 Chapter 14, Solution 17. (1 4) jω G (ω) = (1 + jω)(1 + jω 2) 2 G dB = -20log10 4 + 20 log10 jω − 20 log10 1 + jω − 40 log10 1 + jω 2 φ = -90° - tan -1ω − 2 tan -1 ω 2 The magnitude and phase plots are shown below. GdB 20 0.1 1 10 100 -12 ω -20 -40
- φ 90° 0.1 1 10 100 ω -90° -180° Chapter 14, Solution 18. 4 (1 + jω 2) 2 G (ω) = 50 jω (1 + jω 5)(1 + jω 10) G dB = 20 log10 4 50 + 40 log10 1 + jω 2 − 20 log10 jω − 20 log10 1 + jω 5 − 20 log10 1 + jω 10 where 20 log10 4 50 = -21.94 φ = -90° + 2 tan -1 ω 2 − tan -1 ω 5 − tan -1 ω 10 The magnitude and phase plots are shown below. GdB 20 0.1 1 10 100 ω -20 -40 -60 φ 180° 90° 0.1 1 10 100 ω -90°
- Chapter 14, Solution 19. jω H (ω) = 100 (1 + jω 10 − ω2 100) H dB = 20 log10 jω − 20 log10 100 − 20 log10 1 + jω 10 − ω2 100 ω 10 φ = 90° − tan -1 1 − ω2 100 The magnitude and phase plots are shown below. HdB 40 20 0.1 1 10 100 ω -20 -40 -60 φ 90° 0.1 1 10 100 ω -90° -180°
- Chapter 14, Solution 20. 10 (1 + jω − ω2 ) N(ω) = (1 + jω)(1 + jω 10) N dB = 20 − 20 log10 1 + jω − 20 log10 1 + jω 10 + 20 log10 1 + jω − ω2 ω φ = tan -1 − tan -1 ω − tan -1 ω 10 1 − ω2 The magnitude and phase plots are shown below. NdB 40 20 0.1 1 10 100 ω -20 -40 φ 180° 90° 0.1 1 10 100 ω -90° Chapter 14, Solution 21. jω (1 + jω) T(ω) = 100 (1 + jω 10)(1 + jω 10 − ω2 100) TdB = 20 log10 jω + 20 log10 1 + jω − 20 log10 100
- − 20 log10 1 + jω 10 − 20 log10 1 + jω 10 − ω2 100 ω 10 φ = 90° + tan -1 ω − tan -1 ω 10 − tan -1 1 − ω2 100 The magnitude and phase plots are shown below. TdB 20 0.1 1 10 100 ω -20 -40 -60 φ 180° 90° 0.1 1 10 100 ω -90° -180° Chapter 14, Solution 22. 20 = 20 log10 k → k = 10 A zero of slope + 20 dB / dec at ω = 2 → 1 + jω 2 1 A pole of slope - 20 dB / dec at ω = 20 → 1 + jω 20
- 1 A pole of slope - 20 dB / dec at ω = 100 → 1 + jω 100 Hence, 10 (1 + jω 2) H (ω) = (1 + jω 20)(1 + jω 100) 10 4 ( 2 + jω) H (ω) = ( 20 + jω)(100 + jω) Chapter 14, Solution 23. A zero of slope + 20 dB / dec at the origin → jω 1 A pole of slope - 20 dB / dec at ω = 1 → 1 + jω 1 1 A pole of slope - 40 dB / dec at ω = 10 → (1 + jω 10) 2 Hence, jω H (ω) = (1 + jω)(1 + jω 10) 2 100 jω H (ω) = (1 + jω)(10 + jω) 2 Chapter 14, Solution 24. The phase plot is decomposed as shown below. φ 90° arg (1 + jω / 10) 45° 0.1 1 10 100 1000 ω -45° arg ( jω) 1 arg 1 + jω / 100 -90°
- k ′ (1 + jω 10) k ′ (10)(10 + jω) G (ω) = = jω (1 + jω 100) jω (100 + jω) where k ′ is a constant since arg k ′ = 0 . k (10 + jω) Hence, G (ω) = , where k = 10k ′ is constant jω (100 + jω) Chapter 14, Solution 25. 1 1 ω0 = = = 5 krad / s LC (40 × 10 -3 )(1 × 10 -6 ) Z(ω0 ) = R = 2 kΩ ω0 4 Z(ω0 4) = R + j L − 4 ω0 C 5 × 10 3 4 Z(ω0 4) = 2000 + j ⋅ 40 × 10 -3 − 4 (5 × 10 3 )(1 × 10 -6 ) Z(ω0 4) = 2000 + j (50 − 4000 5) Z(ω0 4) = 2 − j0.75 kΩ ω0 2 Z(ω0 2) = R + j L − 2 ω0 C (5 × 10 3 ) 2 Z(ω0 2) = 2000 + j (40 × 10 -3 ) − 2 (5 × 10 3 )(1 × 10 -6 ) Z(ω0 4) = 2000 + j (100 − 2000 5) Z(ω0 2) = 2 − j0.3 kΩ 1 Z(2ω0 ) = R + j 2ω0 L − 2ω0 C
- 1 Z(2ω0 ) = 2000 + j (2)(5 × 10 3 )(40 × 10 -3 ) − (2)(5 × 10 3 )(1 × 10 -6 ) Z(2ω0 ) = 2 + j0.3 kΩ 1 Z(4ω0 ) = R + j 4ω0 L − 4ω0 C 1 Z(4ω0 ) = 2000 + j (4)(5 × 10 3 )(40 × 10 -3 ) − (4)(5 × 10 )(1 × 10 ) 3 -6 Z(4ω0 ) = 2 + j0.75 kΩ Chapter 14, Solution 26. 1 1 (a) fo = = = 22.51 kHz 2π LC 2π 5 x10 −9 x10 x10 −3 R 100 (b) B= = = 10 krad/s L 10 x10 −3 ωo L 1 L 10 6 10 x10 −3 (c ) Q= = = 3 = 14.142 R LC R 50 0.1x10 Chapter 14, Solution 27. At resonance, 1 Z = R = 10 Ω , ω0 = LC R ω 0 ω0 L B= and Q= = L B R Hence, RQ (10)(80) L= = = 16 H ω0 50
ADSENSE
CÓ THỂ BẠN MUỐN DOWNLOAD
Thêm tài liệu vào bộ sưu tập có sẵn:
Báo xấu
LAVA
AANETWORK
TRỢ GIÚP
HỖ TRỢ KHÁCH HÀNG
Chịu trách nhiệm nội dung:
Nguyễn Công Hà - Giám đốc Công ty TNHH TÀI LIỆU TRỰC TUYẾN VI NA
LIÊN HỆ
Địa chỉ: P402, 54A Nơ Trang Long, Phường 14, Q.Bình Thạnh, TP.HCM
Hotline: 093 303 0098
Email: support@tailieu.vn