# Bài giải phần giải mạch P14

Chia sẻ: Tran Long | Ngày: | Loại File: PDF | Số trang:73

0
51
lượt xem
6

## Bài giải phần giải mạch P14

Mô tả tài liệu

Chapter 14, Solution 1. Vo R jωRC = = Vi R + 1 jωC 1 + jωRC jω ω 0 , 1 + jω ω 0 H (ω) = H (ω) = where ω 0 = 1 RC H = H (ω) = ω ω0 1 + (ω ω0 ) 2 φ = ∠H (ω) = ω π − tan -1   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...

Chủ đề:

Bình luận(0)

Lưu

## Nội dung Text: Bài giải phần giải mạch P14

1. 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
2. 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 −
3. 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
4. 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
5. 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
6. 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°
7. Chapter 14, Solution 10. 50 10 H( jω) = = jω(5 + jω)  jω  1 jω1 +   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
8. 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°
9. 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
10. 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°
11. Chapter 14, Solution 14. 50 1 + jω H (ω) = 25  jω10  jω  2  jω1 +  +     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°
12. 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 
13. 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
14. 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°
15. 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°
16. 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
17. − 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
18. 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°
19. 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 
20.  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