Bài giải phần giải mạch P19

Chapter 19, Solution 1. To get z and , consider the circuit in Fig. (a). 11 21 z 1 Ω 4 Ω I 2 = 0 + V 2 − I o + V 1 − 2 Ω 6 Ω I 1 (a) Ω=++== 4)24(||61 1 1 11 I V z 1o 2 1 II = , 1o2 2 IIV == Ω== 1 1 2 21 I V z To get z and , consider the circuit in Fig. (b). 22 12 z 1 Ω 4 Ω I 1 = 0 I o ' + V 2 − + V 1 − 2 Ω 6 Ω I 2 (b) Ω=+== 667.1)64(||2 2 2 22 I V z 22 ' o 6 1 102 2 III = + = , 2o1 '6 IIV == Ω== 1 2 1 12 I V z Hence, =][z Ω       667.11 14 Chapter 19, Solution 2. Consider the circuit in Fig. (a) to get and . 11 z 21 z + V 2 − I 2 = 0 1 Ω 1 Ω + V 1 − 1 Ω I o 1 Ω 1 Ω I o ' 1 Ω 1 Ω I 1 1 Ω 1 Ω 1 Ω 1 Ω (a) ])12(||12[||12 1 1 11 +++== I V z 733.2 15 11 2 4111 )411)(1( 2 4 3 2||12 11 =+= + +=       ++=z ' o ' oo 4 1 31 1 III = + = 11 ' o 15 4 4111 1 III = + = 11o 15 1 15 4 4 1 III =⋅= 1o2 15 1 IIV == 06667.0 15 1 12 1 2 21 ==== z I V z To get z , consider the circuit in Fig. (b). 22 1 Ω 1 Ω 1 Ω 1 Ω I 1 = 0 + V 2 − + V 1 − 1 Ω1 Ω1 Ω I 2 1 Ω 1 Ω 1 Ω 1 Ω (b) 733.2)3||12(||12 11 2 2 22 ==++== z I V z Thus, =][z Ω       733.206667.0 06667.0733.2 Chapter 19, Solution 3. (a) To find and , consider the circuit in Fig. (a). 11 z 21 z I o + V 2 − + V 1 − 1 Ωj Ω I 2 = 0 -j Ω I 1 (a) j1 j1j )j1(j )j1(||j 1 1 11 += −+ − =−== I V z By current division, 11o j j1j j III = −+ = 1o2 jIIV == j 1 2 21 == I V z To get z and , consider the circuit in Fig. (b). 22 12 z I 1 = 0 + V 2 − + V 1 − 1 Ω -j Ω j Ω I 2 (b) 0)jj(||1 2 2 22 =−== I V z 21 j IV = j 2 1 12 == I V z Thus, =][z Ω       + 0j jj1 (b) To find and , consider the circuit in Fig. (c). 11 z 21 z + V 1 − -j Ω 1 Ω + V 2 − 1 Ω I 2 = 0 j Ω -j Ω I 1 (c) 5.0j5.1 j1 j- j1-j)(||11j 1 1 11 += − ++=++== I V z 12 )5.0j5.1( IV −= 5.0j5.1 1 2 21 −== I V z To get z and , consider the circuit in Fig. (d). 22 12 z + V 2 − -j Ω 1 Ω + V 1 − 1 Ω I 1 = 0 j Ω -j Ω I 2 (d) j1.5-1.5(-j)||11-j 2 2 22 =++== I V z 21 )5.0j5.1( IV −= 5.0j5.1 2 1 12 −== I V z Thus, =][z Ω       −− −+ 5.1j5.15.0j5.1 5.0j5.15.0j5.1 Chapter 19, Solution 4. Transform the Π network to a T network. Z 1 Z 3 Z 2 5j12 120j 5j10j12 )10j)(12( 1 + = −+ =Z j512 j60- 2 + =Z 5j12 50 3 + =Z The z parameters are j4.26--1.775 25144 j5)--j60)(12( 22112 = + === Zzz 26.4j775.1 169 )5j12)(120j( 1212111 +=+ − =+= zzZz 739.5j7758.1 169 )5j12)(50( 2121322 −=+ − =+= zzZz Thus, = ][ z Ω       −− −+ 739.5j775.126.4j775.1- 26.4j775.1-26.4j775.1 Chapter 19, Solution 5. Consider the circuit in Fig. (a). s 1 I 2 = 0 I o 1/s1/s + V 2 − + V 1 − 1 I 1 (a) s 1 s1 1s 1 s 1 s1 1s 1 s 1 s1|| s 1 1 s 1 s 1 s1|| s 1 ||1 11 +++       +       ++       + =       ++ + =       ++=z 1s3s2s 1ss 23 2 11 +++ ++ =z 1 2 11o 1ss 1s s 1s s s 1 s1 1s 1 1s 1 s 1 s1 s 1 ||1 s 1 ||1 IIII +++ + + = +++ + + = +++ = 1 23 o 1s3s2s s II +++ = 1s3s2s s 1 23 1 o2 +++ == I IV 1s3s2s 1 23 1 2 21 +++ == I V z Consider the circuit in Fig. (b). s 1 I 1 = 0 1/s1/s 1 + V 1 − + V 2 − I 2 (b)       + ++=       ++== 1s 1 s1|| s 1 s 1 ||1s1|| s 1 2 2 22 I V z 1s s ss1 1s 1 s1 1s 1 s1 s 1 1s 1 s1 s 1 2 22 + +++ + ++ = + +++       + ++       =z 1s3s2s 2s2s 23 2 22 +++ ++ =z 2112 zz = Hence, = ][ z           +++ ++ +++ ++++++ ++ 1s3s2s 2s2s 1s3s2s 1 1s3s2s 1 1s3s2s 1ss 23 2 23 2323 2 Chapter 19, Solution 6. To find and , connect a voltage source to the input and leave the output open as in Fig. (a). 11 z 21 z 1 V V o + V 2 − 20 Ω I 1 30 Ω 0.5 V 2 + − 10 Ω V 1 (a) 50 5.0 10 o 2 o1 V V VV += − , where oo2 5 3 3020 30 VV = + V = o o oo1 2.4 55 3 5 V V VVV =+       += oo o1 1 32.0 10 2.3 10 VV VV I == − = Ω=== 125.13 32.0 2.4 o o 1 1 11 V V I V z Ω=== 875.1 32.0 6.0 o o 1 2 21 V V I V z To obtain and , use the circuit in Fig. (b). 22 z 12 z I 2 0.5 V 2 + − 10 Ω + V 1 − 30 Ω 20 Ω V 2 (b) 2 2 22 5333.0 30 5.0 V V VI =+= Ω=== 875.1 5333.0 1 2 2 22 I V z 2221 -9)5.0)(20( VVVV =−= Ω=== -16.875 5333.0 9- 2 2 2 1 12 V V I V z Thus, = ][ z Ω       1.8751.875 16.875-125.13 Chapter 19, Solution 7. To get z 11 and z 21 , we consider the circuit below. I 2 =0 I 1 20 100 Ω Ω + + + v x 50 Ω 60 Ω V 1 - V 2 - - 12v x - + 1x xxxx1 V 121 40 V 160 V12V 50 V 20 VV =→ + += − 88.29 I V z) 20 V ( 121 81 20 VV I 1 1 11 1x1 1 ==→= − = 37.70 I V zI37.70 I 81 121x20 ) 121 40 ( 8 57 V) 121 40 ( 8 57 V 8 57 V12) 160 V13 (60V 1 2 211 11xx x 2 −==→−= −=−=−=−= To get z 12 and z 22 , we consider the circuit below. I 2 I 1 =0 20 100 Ω Ω + + + v x 50 Ω 60 Ω V 1 - V 2 - - 12v x - + 2 x22 222x V09.0 60 V12V 150 V I,V 3 1 V 50100 50 V = + +== + = 11.1109.0/1 I V z 2 2 22 === 704.3 I V zI704.3I 3 11.11 V 3 1 VV 2 1 12222x1 ==→==== Thus, Ω       − = 11.1137.70 704.388.29 ]z[