Bài giải phần giải mạch P19

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Bài giải phần giải mạch P19

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Chapter 19, Solution 1. To get z 11 and z 21 , consider the circuit in Fig. (a). 1Ω + I1 V1 − 4Ω Io 6Ω 2Ω I2 = 0 + V2 − (a) z 11 = V1 = 1 + 6 || (4 + 2) = 4 Ω I1 Io = z 21 = 1 I , 2 1 V2 = 1Ω I1 V2 = 2 I o = I 1 To get z 22 and z 12 , consider the circuit in Fig. (b). I1 = 0 + V1 − 1Ω Io ' 4Ω + 2Ω V2 − 6Ω I2 (b) z 22 = V2 = 2 || (4 + 6) = 1.667 Ω I2 Io' = z 12 = 2 1...

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Nội dung Text: Bài giải phần 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)
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