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• Mohammed Idrees Idrees

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Wednesday, June 29, 2011

CHAPTER 19 P.P.19.1

Comparing the network with that in Fig. 19.5, we observe that z 12 = z 21 = 3 Ω z 22 – z 12 = 0 or z 22 = z 12 = 3 Ω z 11 – z 12 = 4 or z 11 = 4 + 3 = 7 Ω 7 3  3 3

 z    P.P.19.2

V1  6 I 1  j4 I 2 V2  - j4 I 1  8 I 2

But

V2  0

and

2 30  V1  2 I 1

(1) (2)

 

V1  2 30  2 I 1

Substituting these into (1) and (2), 2 30  2 I 1  6 I 1  j4 I 2 2 30  8 I 1  j4 I 2

(3)

0  -j4 I 1  8 I 2 I 1 = –j2I 2

(4)

Substituting (4) into (3), 230   j16 I 2  j4 I 2   j20 I 2 230 I2   100120˚ mA 20  90 I1   j2 I 2  20030˚ mA

P.P.19.3

Consider the circuit in Fig. (a) for y 11 and y 21 .

I1

2

6

+ 1A

V1

I2 +

4

V2 = 0

V 1 = [2+(4x6/(4+6))]I 1 = 4.4 I 1 y 11 = (I 1 /V 1 ) = 1/4.4 = 0.2273 S By current division,

-4 I1  -0.4 I1 10 I  2  –0.4I 1 /(4.4I 1 ) = –0.09091 S V1

I2  y 21

For y 12 and y 22 , consider the circuit in Fig. (b).

I1

2

6

+

I2 +

4

V1 = 0 

V2 

(b) V 2 = [6 +(4x2/(4+2))]I 2 = 7.333I 2 I y 22  2 = 1/7.333 = 0.13636 S V2 By current division, I 1 = [–4/(4+2)]I 2 = –(2/3)I 2 I y 12  1 = –(2/3)I 2 /7.333I 2 = –0.09091 S V2 Therefore,  227 .3 - 90.91  [y ]    mS  - 90.91 136.36 

1A

P.P.19.4

Consider the circuit in Fig (a).

6

1 + I1

V1

2 Vo

2

Io

I2 +

3

2 Io



V2 = 0 

(a) At node 1,

V1 V1  Vo  3 6 6 I 1  3 V1  Vo

I1 

At node 2,

(1)

V1  Vo 2  I 2  2 I o  V1 6 3 0  Vo - Vo  2 2

But

I2 

Hence,

V1  Vo Vo 2   V1 6 2 3

  V1 

-4 V 3 o

From (1) and (2),

6 I1  -4 Vo  Vo  -5 Vo

Thus,

  I1 

y 11 

I 1 (- 5 6)Vo 5    0.625 S V1 (- 4 3)Vo 8

y 21 

I 2 (- 1 2)Vo 3    0.375 S V1 (- 4 3)Vo 8

-5 V 6 o

(2)

Consider the circuit in Fig. (b). The 3- resistor is short-circuited so that I o  0 . Consequently, the circuit is equivalent to that shown in Fig.(c).

6

I1

2

Vo

Io

+

+

3

V1 = 0

2 Io



I2

V2 

(b) I1

6

2 + V2

I2



(c) V2  8I 2 , I1  - I 2 I2 1 y 22    0.125 S V2 8

y 12 

I1 -I2   -0.125 S V2 8 I 2

Therefore,  0.625 - 0.125   625 - 125  [y ]   S or    mS  0.375 0.125   375 125 

P.P.19.5 To find h 11 and h 21 , we use the circuit in Fig. (a). Note that the 5- resistor is short-circuited. 2

I2

+ I1

V1

2



(a)

5

V 1 = [2x3/(2+3)]I 1 = 1.2I 1 V h 11  1  1.2 Ω I1

2 I  0.4 I 1 = (4/(4+6))I 1 = 0.4I 1 23 1 I  2  -0.4 I1

-I2  h 21

To get h12 and h 22 , we use the circuit in Fig. (b). I1 = 0

3

I2

+ V1

5

2



(b)

V 1 = [2/(2+3)]V 1 = 0.4V 1

h 12 

V1  0.4 V2

V 2 = [5x5/(5+5)]I 2 = 2.5I 2

h 22 

I2 1   0.4 S V2 2.5

Therefore, 0 .4   1 .2  [h]     - 0.4 400 mS 

+ 

V2

P.P.19.6

Our goal is to get Z in  V1 I 1 . I2

I1

+ V2 

+ V1 

ZL

V1  h 11 I 1  h 12 V2 I 2  h 21 I 1  h 22 V2

(1) (2)

But V2  - Z L I 2 Substituting this equation into (1) and (2),

I 2  h 21 I 1  Z L h 22 I 2

V2  - Z L I 2 

V1  h11 I1 

  I 2 

h 21 I 1 1  Z L h 22

- Z L h 21 I 1 1  Z L h 22

Z L h12 h 21 I1 1  Z L h 22

V1 Z h h  h11  L 12 21 I1 1  Z L h 22 (50  10 3 )(10 -4 )(100) Z in  2000   1.6667 kΩ 1  (50  10 3 )(10 -5 )

Z in 

P.P.19.7

We get g 11 and g 21 using the circuit in Fig. (a). I1

s

s

I2 = 0 Io

V1

+ 

1

1

+ V2 

(a) V1  [ s  1 || (s  1) ] I 1

V1 s  1 s 2  3s  1 s  I1 s2 s2 I1 s2 g 11   2 V1 s  3s  1 By current division, V1 1 I1  2 s2 s  3s  1 V1 V2  I o  2 s  3s  1 V2 1 g 21   2 V1 s  3s  1 Io 

We obtain g 12 and g 22 using the circuit in Fig. (b). s

I1

s

Io' +

1

1

V2 

(b)

  s   I V2  [1 || (s  1 || s) ] I 2   1 || s    s  1 2 s 2  2s V2 s (s  2) s  2s s 1  2  1 ||  2 s  2s s  3s  1 I2 s 1 1 s 1 V2 s (s  2) g 22   2 I 2 s  3s  1 2

By current division,

-1 ' I s 1 o 1 s 1 Io'  I2  2 I 2 s  2s s  3s  1 2 1 s 1 -1 I1  2 I s  3s  1 2 I1 

and

I2

g12 

I1 -1  2 I 2 s  3s  1

Therefore, s2   s 2  3s  1 [g]   1   s 2  3 s  1

P.P.19.8

-1  2 s  3 s  1  s ( s  2)  2 s  3 s  1 

To get A and C, we use the circuit in Fig. (a). I1

2

6

I2 = 0 +

V1

+ 

4

V2 

(a) A = V 1 /V 2 = (2+4)I 1 /(4I 1 ) = 1.5 C = I 1 /V 2 = I 1 /(4I 1 ) = 0.25 S To get B and D, we use the circuit in Fig. (b). I1

2

6

I2 +

V1

+ 

4

V2 = 0 

(b)

I2  Also,

-4 I1 10

  D 

- I1 10   2.5 I2 4

V 1 = [2+(4x6/(4+6))]I 1 = 4.4I 1 B = –V 1 /I 2 = –4.4I 1 /(–0.4I 1 ) = 11 Ω

Therefore,  1 .5 [T]    250 mS

P.P.19.9

11   2.5 

From Eq. (19.22), V1  5 V2  10 I 2 I 1  0 . 4 V2  I 2

(1) (2)

At the output port, V2  -10 I 2 . At the input port, V1  14  2 I 1 . Substituting these into (1) and (2), 14  2 I 1  -50 I 2  10 I 2 14  2 I 1  60 I 2 I 1  -4 I 2  I 2 I 1  -5 I 2

(4)

Substituting (4) into (3), 14  -10 I 2  60 I 2 - 14 I2   - 0.2 A 70 I 1  (-5)(-0.2)  1 A

P.P.19.10  6 4 [z]     4 6

 z  36  16  20

z 11  6  z 22

z 12  z 21  4

From Table 19.1,

z 22 6   0.3 S  z 20 - z 21   -0.2 S z

- z 12 - 4   -0.2 S z 20 z 11   0.3 S z

y 11 

y 12 

y 21

y 22

A

z 11 6   1.5 z 21 4

C

1 1   0.25 S z 21 4

(3)

 z 20   5 z 21 4 z 22 6 D   1.5 z 21 4

B

Therefore,  0.3 - 0.2  [y]   S  - 0.2 0.3 

 1 .5 5 [T]     0.25 S 1.5 