chap19

# chap19 - Chapter 19 Solution 1 To get z 11 and z 21...

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Chapter 19, Solution 1. To get z and , consider the circuit in Fig. (a). 11 21 z 1 4 I 2 = 0 + V 2 I o + V 1 2 6 I 1 (a) = + + = = 4 ) 2 4 ( || 6 1 1 1 11 I V z 1 o 2 1 I I = , 1 o 2 2 I I V = = = = 1 1 2 21 I V z To get z and , consider the circuit in Fig. (b). 22 12 z 1 4 I 1 = 0 I o ' + V 2 + V 1 2 6 I 2 (b) = + = = 667 . 1 ) 6 4 ( || 2 2 2 22 I V z 2 2 ' o 6 1 10 2 2 I I I = + = , 2 o 1 ' 6 I I V = = = = 1 2 1 12 I V z

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Hence, = ] [ z 667 . 1 1 1 4 Chapter 19, Solution 2. Consider the circuit in Fig. (a) to get and . 11 z 21 z + V 2 I 2 = 0 1 1 + V 1 1 I o 1 1 I o ' 1 1 I 1 1 1 1 1 (a) ] ) 1 2 ( || 1 2 [ || 1 2 1 1 11 + + + = = I V z 733 . 2 15 11 2 4 11 1 ) 4 11 )( 1 ( 2 4 3 2 || 1 2 11 = + = + + = + + = z ' o ' o o 4 1 3 1 1 I I I = + = 1 1 ' o 15 4 4 11 1 1 I I I = + = 1 1 o 15 1 15 4 4 1 I I I = = 1 o 2 15 1 I I V = = 06667 . 0 15 1 12 1 2 21 = = = = z I V z
To get z , consider the circuit in Fig. (b). 22 1 1 1 1 I 1 = 0 + V 2 + V 1 1 1 1 I 2 1 1 1 1 (b) 733 . 2 ) 3 || 1 2 ( || 1 2 11 2 2 22 = = + + = = z I V z Thus, = ] [ z 733 . 2 06667 . 0 06667 . 0 733 . 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 j I 2 = 0 -j I 1 (a) j 1 j 1 j ) j 1 ( j ) j 1 ( || j 1 1 11 + = + = = = I V z By current division, 1 1 o j j 1 j j I I I = + =

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1 o 2 j I I V = = 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 -j j I 2 (b) 0 ) j j ( || 1 2 2 22 = = = I V z 2 1 j I V = j 2 1 12 = = I V z Thus, = ] [ z + 0 j j j 1 (b) To find and , consider the circuit in Fig. (c). 11 z 21 z + V 1 -j 1 + V 2 1 I 2 = 0 j -j I 1 (c)
5 . 0 j 5 . 1 j 1 j - j 1 -j) ( || 1 1 j 1 1 11 + = + + = + + = = I V z 1 2 ) 5 . 0 j 5 . 1 ( I V = 5 . 0 j 5 . 1 1 2 21 = = I V z To get z and , consider the circuit in Fig. (d). 22 12 z + V 2 -j 1 + V 1 1 I 1 = 0 j -j I 2 (d) j1.5 - 1.5 (-j) || 1 1 -j 2 2 22 = + + = = I V z 2 1 ) 5 . 0 j 5 . 1 ( I V = 5 . 0 j 5 . 1 2 1 12 = = I V z Thus, = ] [ z + 5 . 1 j 5 . 1 5 . 0 j 5 . 1 5 . 0 j 5 . 1 5 . 0 j 5 . 1 Chapter 19, Solution 4. Transform the Π network to a T network. Z 1 Z 3 Z 2

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5 j 12 120 j 5 j 10 j 12 ) 10 j )( 12 ( 1 + = + = Z j5 12 j60 - 2 + = Z 5 j 12 50 3 + = Z The z parameters are j4.26 - -1.775 25 144 j5) - -j60)(12 ( 2 21 12 = + = = = Z z z 26 . 4 j 775 . 1 169 ) 5 j 12 )( 120 j ( 12 12 1 11 + = + = + = z z Z z 739 . 5 j 7758 . 1 169 ) 5 j 12 )( 50 ( 21 21 3 22 = + = + = z z Z z Thus, = ] [ z + 739 . 5 j 775 . 1 26 . 4 j 775 . 1 - 26 . 4 j 775 . 1 - 26 . 4 j 775 . 1 Chapter 19, Solution 5. Consider the circuit in Fig. (a). s 1 I 2 = 0 I o 1/s 1/s + V 2 + V 1 1 I 1 (a) s 1 s 1 1 s 1 s 1 s 1 1 s 1 s 1 s 1 || s 1 1 s 1 s 1 s 1 || s 1 || 1 11 + + + + + + + = + + + = + + = z
1 s 3 s 2 s 1 s s 2 3 2 11 + + + + + = z 1 2 1 1 o 1 s s 1 s s 1 s s s 1 s 1 1 s 1 1 s 1 s 1 s 1 s 1 || 1 s 1 || 1 I I I I + + + + + = + + + + + = + + + = 1 2 3 o 1 s 3 s 2 s s I I + + + = 1 s 3 s 2 s s 1 2 3 1 o 2 + + + = = I I V 1 s 3 s 2 s 1 2 3 1 2 21 + + + = = I V z Consider the circuit in Fig. (b).

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