Bài giải mạch P19

Chia sẻ: Tran Thach | Ngày: | Loại File: PDF | Số trang:125

0
60
lượt xem
12
download

Bài giải mạch P19

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

Tham khảo tài liệu 'bài giải mạch p19', kỹ thuật - công nghệ, điện - điện tử phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả

Chủ đề:
Lưu

Nội dung Text: Bài giải mạch P19

  1. Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a). 1Ω 4Ω I2 = 0 + Io + I1 V1 6Ω 2Ω V2 − − (a) V1 z 11 = = 1 + 6 || (4 + 2) = 4 Ω I1 1 Io = I , V2 = 2 I o = I 1 2 1 V2 z 21 = = 1Ω I1 To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 1Ω 4Ω Io ' + + V1 6Ω 2Ω V2 I2 − − (b) V2 z 22 = = 2 || (4 + 6) = 1.667 Ω I2 2 1 Io' = I2 = I2 , V1 = 6 I o ' = I 2 2 + 10 6 V1 z 12 = = 1Ω I2
  2. Hence, 4 1  [z ] =  Ω  1 1.667  Chapter 19, Solution 2. Consider the circuit in Fig. (a) to get z 11 and z 21 . 1Ω Io ' 1Ω 1Ω 1Ω I2 = 0 + Io + I1 V1 1Ω 1Ω 1Ω V2 − − 1Ω 1Ω 1Ω 1Ω (a) V1 z 11 = = 2 + 1 || [ 2 + 1 || (2 + 1) ] I1  3 (1)(11 4) 11 z 11 = 2 + 1 ||  2 +  = 2 + = 2 + = 2.733  4 1 + 11 4 15 1 1 Io = Io' = Io' 1+ 3 4 1 4 Io' = I1 = I1 1 + 11 4 15 1 4 1 Io = ⋅ I1 = I1 4 15 15 1 V2 = I o = I 15 1 V2 1 z 21 = = = z 12 = 0.06667 I 1 15
  3. To get z 22 , consider the circuit in Fig. (b). I1 = 0 1Ω 1Ω 1Ω 1Ω + + V1 1Ω 1Ω 1Ω V2 I2 − − 1Ω 1Ω 1Ω 1Ω (b) V2 z 22 = = 2 + 1 || (2 + 1 || 3) = z 11 = 2.733 I2 Thus,  2.733 0.06667  [z ] =  Ω  0.06667 2.733  Chapter 19, Solution 3. (a) To find z 11 and z 21 , consider the circuit in Fig. (a). -j Ω I2 = 0 Io + + I1 V1 jΩ 1Ω V2 − − (a) V1 j (1 − j) z 11 = = j || (1 − j) = = 1+ j I1 j +1− j By current division, j Io = I = j I1 j +1− j 1
  4. V2 = I o = jI 1 V2 z 21 = =j I1 To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 -j Ω + + V1 jΩ 1Ω V2 I2 − − (b) V2 z 22 = = 1 || ( j − j) = 0 I2 V1 = j I 2 V1 z 12 = =j I2 Thus,  1+ j j  [z ] =  Ω  j 0 (b) To find z 11 and z 21 , consider the circuit in Fig. (c). jΩ -j Ω I2 = 0 + + 1Ω I1 V1 V2 -j Ω 1Ω − − (c)
  5. V1 -j z 11 = = j + 1 + 1 || (-j) = 1 + j + = 1.5 + j0.5 I1 1− j V2 = (1.5 − j0.5) I 1 V2 z 21 = = 1.5 − j0.5 I1 To get z 22 and z 12 , consider the circuit in Fig. (d). I1 = 0 jΩ -j Ω + + 1Ω V1 V2 I2 -j Ω 1Ω − − (d) V2 z 22 = = -j + 1 + 1 || (-j) = 1.5 - j1.5 I2 V1 = (1.5 − j0.5) I 2 V1 z 12 = = 1.5 − j0.5 I2 Thus,  1.5 + j0.5 1.5 − j0.5  [z ] =  Ω  1.5 − j0.5 1.5 − j1.5  Chapter 19, Solution 4. Transform the Π network to a T network. Z1 Z3 Z2
  6. (12)( j10) j120 Z1 = = 12 + j10 − j5 12 + j5 - j60 Z2 = 12 + j5 50 Z3 = 12 + j5 The z parameters are (-j60)(12 - j5) z 12 = z 21 = Z 2 = = -1.775 - j4.26 144 + 25 ( j120)(12 − j5) z 11 = Z1 + z 12 = + z 12 = 1.775 + j4.26 169 (50)(12 − j5) z 22 = Z 3 + z 21 = + z 21 = 1.7758 − j5.739 169 Thus,  1.775 + j4.26 - 1.775 − j4.26  [z ] =  Ω  - 1.775 − j4.26 1.775 − j5.739  Chapter 19, Solution 5. Consider the circuit in Fig. (a). 1 s I2 = 0 + Io + I1 V1 1 1/s 1/s V2 − − (a) 1  1  1  1 + s +  1  1 s  1   s + 1  s z 11 = 1 || || 1 + s +  = || 1 + s +  = s  s 1  s  1  1 1+   +1+ s + s  s + 1 s
  7. s2 + s +1 z 11 = s 3 + 2s 2 + 3s + 1 1 1 s 1 || s s +1 s +1 Io = I1 = I1 = I1 1 1 1 1 s 1 || + 1 + s + +1+ s + + s2 + s +1 s s s +1 s s +1 s Io = I1 s + 2s + 3s + 1 3 2 1 I1 V2 = I o = 3 s s + 2s 2 + 3s + 1 V2 1 z 21 = = 3 I 1 s + 2s + 3s + 1 2 Consider the circuit in Fig. (b). I1 = 0 1 s + + V1 1 1/s 1/s V2 I2 − − (b) V2 1  1 1  1  z 22 = = || 1 + s + 1 ||  = || 1 + s +  I2 s  s s  s + 1  1  1  1  1 + s +  1+ s +  s  s + 1 s +1 z 22 = = 1 1 s +1+ s + 1+ s + s2 + s s +1 s +1 s 2 + 2s + 2 z 22 = s 3 + 2s 2 + 3s + 1 z 12 = z 21 Hence,
  8.  s2 + s + 1 1   s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1  [z ] =    1 s 2 + 2s + 2   s 3 + 2s 2 + 3s + 1 s 3 + 2s 2 + 3s + 1  Chapter 19, Solution 6. To find z 11 and z 21 , connect a voltage source V1 to the input and leave the output open as in Fig. (a). I1 10 Ω Vo 20 Ω + + V1 0.5 V2 30 Ω V2 − − (a) V1 − Vo Vo 30 3 = 0.5 V2 + , where V2 = Vo = Vo 10 50 20 + 30 5 3  V V1 = Vo + 5  Vo  + o = 4.2 Vo 5  5 V1 − Vo 3.2 I1 = = V = 0.32 Vo 10 10 o V1 4.2 Vo z 11 = = = 13.125 Ω I 1 0.32 Vo V2 0.6 Vo z 21 = = = 1.875 Ω I 1 0.32 Vo To obtain z 22 and z 12 , use the circuit in Fig. (b). 10 Ω 20 Ω I2 + + V1 0.5 V2 30 Ω V2 − − (b)
  9. V2 I 2 = 0.5 V2 + = 0.5333 V2 30 V2 1 z 22 = = = 1.875 Ω I 2 0.5333 V1 = V2 − (20)(0.5 V2 ) = -9 V2 V1 - 9 V2 z 12 = = = -16.875 Ω I 2 0.5333 V2 Thus,  13.125 - 16.875  [z ] =  Ω  1.875 1.875  Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I2=0 I1 20 Ω 100 Ω + + + vx 50 Ω 60 Ω V1 - V2 - - 12vx - + V1 − Vx Vx Vx + 12Vx 40 = +  → Vx = V1 20 50 160 121 V − Vx 81 V1 V1 I1 = 1 = ( ) → z11 = = 29.88 20 121 20 I1
  10. 13Vx 57 57 40 57 40 20x121 V2 = 60( ) − 12Vx = − Vx = − ( )V1 = − ( ) I1 160 8 8 121 8 121 81 V = −70.37 I1  → z 21 = 2 = −70.37 I1 To get z12 and z22, we consider the circuit below. I2 I1=0 20 Ω 100 Ω + + + vx 50 Ω 60 Ω V1 - V2 - - 12vx - + 50 1 V2 V2 + 12Vx Vx = V2 = V2 , I2 = + = 0.09V2 100 + 50 3 150 60 V2 z 22 = = 1 / 0.09 = 11.11 I2 1 11.11 V V1 = Vx = V2 = I 2 = 3.704I 2  → z12 = 1 = 3.704 3 3 I2 Thus,  29.88 3.704 [z] =  Ω − 70.37 11.11
  11. Chapter 19, Solution 8. To get z11 and z21, consider the circuit below. j4 Ω I1 -j2 Ω 5Ω I2 =0 • • + j6 Ω j8 Ω + V2 V1 10 Ω - - V V1 = (10 − j2 + j6)I1  → z11 = 1 = 10 + j4 I1 V2 V2 = −10I1 − j4I1  → z 21 = = −(10 + j4) I1 To get z22 and z12, consider the circuit below. j4 Ω I1=0 -j2 Ω 5Ω I2 • • + j6 Ω j8 Ω + V2 V1 10 Ω - -
  12. V2 V2 = (5 + 10 + j8)I 2  → z 22 = = 15 + j8 I2 V V1 = −(10 + j4)I 2  → z12 = 1 = −(10 + j4) I2 Thus,  (10 + j4) − (10 + j4) [z] =  Ω − (10 + j4) (15 + j8)  Chapter 19, Solution 9. It is evident from Fig. 19.5 that a T network is appropriate for realizing the z parameters. R1 R2 6Ω 2Ω R3 4Ω R 1 = z 11 − z 12 = 10 − 4 = 6 Ω R 2 = z 22 − z 12 = 6 − 4 = 2 Ω R 3 = z 12 = z 21 = 4 Ω Chapter 19, Solution 10. (a) This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b). I1 z11 z22 I2 + + + + V1 z12 I2 − − z21 I1 V2 − − (a)
  13. (b) This is a reciprocal network and the two-port look like the one shown in Figs. (c) and (d). I1 z11 – z12 z22 – z12 I2 + + V1 z12 V2 − − (c) I1 25 Ω 10 Ω I2 + + + + V1 20 I2 − − 5 I1 V2 − − (b) 2 1 z 11 − z 12 = 1 + = 1+ s 0.5 s z 22 − z 12 = 2s 1 z 12 = s I1 1Ω 0.5 F 2H I2 + + V1 1F V2 − − (d)
  14. Chapter 19, Solution 11. This is a reciprocal network, as shown below. 1+j5 3+j 1Ω j5 Ω 3Ω j1 Ω 5-j2 5Ω -j2 Ω Chapter 19, Solution 12. This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a) and (b). I1 z11 – z12 z22 – z12 I2 + + V1 z12 V2 2Ω − − (a) I1 8Ω 2Ω I2 Io + + V1 4Ω V2 2Ω − − (b) From Fig. (b), V1 = (8 + 4 || 4) I 1 = 10 I 1
  15. By current division, 1 Io = I , V2 = 2 I o = I 1 2 1 V2 I1 = = 0 .1 V1 10 I 1 Chapter 19, Solution 13. This is a reciprocal two-port so that the circuit can be represented by the circuit below. 40 Ω 50 Ω 20 Ω + I1 + + I2 120∠0° V 10 I2 − − 30 I1 100 Ω − rms We apply mesh analysis. For mesh 1, - 120 + 90 I 1 + 10 I 2 = 0  → 12 = 9 I 1 + I 2 (1) For mesh 2, 30 I 1 + 120 I 2 = 0  → I 1 = -4 I 2 (2) Substituting (2) into (1), - 12 12 = -36 I 2 + I 2 = -35 I 2 → I 2 = 35 2 1 2 1  12  P = I 2 R =   (100) = 5.877 W 2 2  35 
  16. Chapter 19, Solution 14. To find Z Th , consider the circuit in Fig. (a). I1 I2 + + ZS V1 Vo = 1 − − (a) V1 = z 11 I 1 + z 12 I 2 (1) V2 = z 21 I 1 + z 22 I 2 (2) But V2 = 1 , V1 = - Z s I 1 - z 12 Hence, 0 = (z 11 + Z s ) I 1 + z 12 I 2  → I 1 = I z 11 + Z s 2  - z 21 z 12  1= + z 22  I 2  z 11 + Z s  V2 1 z z Z Th = = = z 22 − 21 12 I2 I2 z 11 + Z s To find VTh , consider the circuit in Fig. (b). ZS I1 I2 = 0 + + + VS V1 V2 = VTh − − − (b) I2 = 0 , V1 = Vs − I 1 Z s
  17. Substituting these into (1) and (2), Vs Vs − I 1 Z s = z 11 I 1  → I 1 = z 11 + Z s z 21 Vs V2 = z 21 I 1 = z 11 + Z s z 21 Vs VTh = V2 = z 11 + Z s Chapter 19, Solution 15. (a) From Prob. 18.12, z12z 21 80x 60 ZTh = z 22 − = 120 − = 24 z11 + Zs 40 + 10 ZL = ZTh = 24Ω z 21 80 (b) VTh = Vs = (120) = 192 z11 + Zs 40 + 10 V 2Th 192 2 Pmax = = = 192 W 8R Th 8x 24 Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a). 5Ω 10 – j6 Ω 4 – j6 Ω a + 15∠0° V j6 Ω j4 Ω − (a) b
  18. At terminals a-b, Z Th = (4 − j6) + j6 || (5 + 10 − j6) j6 (15 − j6) Z Th = 4 − j6 + = 4 − j6 + 2.4 + j6 15 Z Th = 6.4 Ω j6 VTh = (15∠0°) = j6 = 6∠90° V j6 + 5 + 10 − j6 The Thevenin equivalent circuit is shown in Fig. (b). 6.4 Ω + + 6∠90° V Vo j4 Ω − − (b) From this, j4 Vo = ( j6) = 3.18∠148° 6.4 + j4 v o ( t ) = 3.18 cos( 2t + 148°) V Chapter 19, Solution 17. To obtain z 11 and z 21 , consider the circuit in Fig. (a). 4Ω Io I2 = 0 Io ' + 2Ω + I1 V1 V2 8Ω − − 6Ω (a)
  19. In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, I o , passes through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current, I o ' , passes through them. V1 (12)(8) z 11 = = (4 + 8) || (2 + 6) = 12 || 8 = = 4 .8 Ω I1 20 8 2 3 Io = I1 = I1 Io' = I 8 + 12 5 5 1 But - V2 − 4 I o + 2 I o ' = 0 -8 6 -2 V2 = -4 I o + 2 I o ' = I1 + I1 = I 5 5 5 1 V2 - 2 z 21 = = = -0.4 Ω I1 5 To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 4Ω + 2Ω + V1 V2 I2 8Ω − − 6Ω (b) V2 (6)(14) z 22 = = (4 + 2) || (8 + 6) = 6 || 14 = = 4 .2 Ω I2 20 z12 = z 21 = -0.4 Ω Thus,  4.8 - 0.4  [z ] =  Ω  - 0.4 4.2  We may take advantage of Table 18.1 to get [y] from [z]. ∆ z = (4.8)(4.2) − (0.4) 2 = 20
  20. z 22 4.2 - z 12 0.4 y 11 = = = 0.21 y 12 = = = 0.02 ∆z 20 ∆z 20 - z 21 0.4 z 11 4.8 y 21 = = = 0.02 y 22 = = = 0.24 ∆z 20 ∆z 20 Thus,  0.21 0.02  [y ] =  S  0.02 0.24  Chapter 19, Solution 18. To get y 11 and y 21 , consider the circuit in Fig.(a). I1 6Ω 3Ω I2 + V1 + 6Ω 3Ω V2 = 0 − − (a) V1 = (6 + 6 || 3) I 1 = 8 I 1 I1 1 y 11 = = V1 8 -6 - 2 V1 - V1 I2 = I1 = = 6+3 3 8 12 I 2 -1 y 21 = = V1 12 To get y 22 and y 12 , consider the circuit in Fig.(b). I1 6Ω Io 3Ω I2 + + V1 = 0 6Ω 3Ω V2 − − (b)
Đồng bộ tài khoản