sadiku 5th ed chapter 19

8/11/2019 Sadiku 5th ed Chapter 19

1/190

PROPRIETARY MATERIAL . 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the priorwritten permission of the publisher, or used beyond the limited distribution to teachers and educators

permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,you are using it without permission.

Chapter 19, Problem 1.

Obtain the z parameters for the network in Fig. 19.65.

Figure 19.65

For Prob. 19.1 and 19.28.


2/190



Chapter 19, Solution 1.To get 11z and 21z , consider the circuit in Fig. (a).

=++== 4)24(||611

111 I

Vz

1o 2

1

II = , 1o2 2 IIV == == 1

1

221 I

Vz

To get 22z and 12z , consider the circuit in Fig. (b).

=+== 667.1)64(||22

222 I

Vz

22'

o 61

1022

III =+= , 2o1 '6 IIV ==

== 121

12 I

V

z

Hence, =][z

667.11

14

4

+

V1

+

V2

I1 = 0 1

(b)

I o'

I2 = 04

+

V1

1

I1

(a)

Io +

V2


3/190




* Find the impedance parameter equivalent of the network in Fig. 19.66.

Figure 19.66

For Prob. 19.2.

* An asterisk indicates a challenging problem.

Chapter 19, Solution 2.

Consider the circuit in Fig. (a) to get 11z and 21z .

I2 = 01

+

V1

1

I1

(a)

Io +

V2

1 I o'

1 1 1 1

1


4/190



])12(||12[||121

111 +++== I

Vz

733.21511

24111)411)(1(

243

2||1211 =+=++=

++=z

'o

'oo 4

131

1III =+=

11'

o 154

41111

III =+=

11o 151

154

41

III ==

1o2 151

IIV ==

06667.0151

121

221 ==== zI

Vz

To get 22z , consider the circuit in Fig. (b).

733.2)3||12(||12 112

222 ==++== zI

Vz

Thus,

=][z

733.206667.006667.0733.2

1 1

(b)

1

1 1 1 1

1

+

V1

+

V2

I1 = 0


5/190




Find the z parameters of the circuit in Fig. 19.67.

Figure 19.67

For Prob. 19.3.


12 216 z j z= = 11 12 11 124 4 4 6 z z z z j = = + = + 22 12 22 1210 10 4 z z j z z j j = = =

4 6 6[ ]

6 4 j j

z j j

+ = =

+

4 j6 j6 j6 j4


6/190




Calculate the z parameters for the circuit in Fig. 19.68.

Figure 19.68

For Prob. 19.4.

Chapter 19, Solution 4.Transform the network to a T network.

5 j12120 j

5 j10 j12)10 j)(12(

1 +=+=Z

j512 j60-

2 +=Z

5 j12

503 +=Z

The z parameters are

j4.26--1.77525144

j5)--j60)(12(22112 =+=== Zzz

26.4 j775.1169

)5 j12)(120 j(1212111 +=+

=+= zzZz

739.5 j7758.1169

)5 j12)(50(2121322

=+=+= zzZz Thus,

=][z

739.5 j775.126.4 j775.1-26.4 j775.1-26.4 j775.1

Z 3 Z 1


7/190


8/190



Consider the circuit in Fig. (b).

+++=

++==

1s1

s1||s1

s1

||1s1||s1

2

222 I

Vz

1ssss1

1s1

s1

1s1s1

s1

1s1

s1s1

222

+++++++

=++++

+++

=z

1s3s2s2s2s

23

2

22 +++++=z

2112 zz =

Hence,

=][z

1s3s2s2s2s

1s3s2s1

1s3s2s1

1s3s2s1ss

23

2

23

2323

2

+

V2

+

V1

I1 = 0 1

(b)

s

1/s


9/190




Compute the z parameters of the circuit in Fig. 19.70.

Figure 19.70

For Prob. 19.6 and 19.73.


To find z 11 and z 21 , consider the circuit below.

I1 5 10 4I1 I2=0Vo

++

V1 20 V2

+ _


10/190



1 111

1 1

(20 5)25

V I z

I I

+= = =

1 120

2025o

V V I = =

2 2 2 1 1 1 14 0 4 20 4 24o oV I V V V I I I I + = = + = + =

221

1

24V

z I

= =

To find z 12 and z 22, consider the circuit below.

I1=0 5 10 4I1 I2

++

V1 20 V2

2 2 2(10 20) 30V I I = + =

222

1

30V

z I

= =

1 220V I = 1

122

20V

z I

= =

Thus,25 20

[ ]24 30

z =

+ _


11/190


12/190



1xxxxx1 V

12140

V160

V12V50V

20VV = ++=

88.29I

Vz)

20

V(

121

81

20

VVI

1

1

11

1x1

1 == ==

37.70IV

zI37.70

I81

121x20)

12140

(8

57V)

12140

(8

57V

857

V12)160

V13(60V

1

2211

11xxx

2

== =

====

To get z 12 and z 22, we consider the circuit below.I2

I1=0 20 100

++ +

vx 50 60 V1

- V2 - -

12v x -

+

2x22

222x V09.060V12V

150V

I,V31

V50100

50V =++==

+=

11.1109.0/1IV

z2

222 ===

704.3I

VzI704.3I

3

11.11V

3

1VV

2

112222x1 == ====

Thus,

=

11.1137.70704.388.29

]z[


13/190




Find the z parameters of the two-port in Fig. 19.72.

Figure 19.72

For Prob. 19.8.


To get z 11 and z 21, consider the circuit below.

j4 I1 -j2 5 I2 =0

+ j6 j8

+V2

V1

10 --


14/190



4 j10IV

zI)6 j2 j10(V1

11111 +== +=

)4 j10(IV

zI4 jI10V

1

221112 +== =

To get z 22 and z 12, consider the circuit below.

j4 I1=0 -j2 5 I2

+

j6 j8 +

V2 V1

10 --

8 j15IV

zI)8 j105(V2

22222 +== ++=

)4 j10(IV

zI)4 j10(V2

11221 +== +=

Thus,

++++

= )8 j15()4 j10()4 j10()4 j10(

]z[


15/190




The y parameters of a network are:

[ ]

=

4.02.02.05.0y

Determine the z parameters for the network.


2211

0.42.50.16

y z

y = = = , 11 22 21 12 05 0.4 0.2 0.2 0.16y y y y y x x = = =

1212 21

0.21.25

0.16y

z z y

= = = =

1122

0.53.125

0.16y

z y

= = =

Thus,2.5 1.25

[ ]1.25 3.125

z =

125.325.1

25.15.2


16/190




Construct a two-port that realizes each of the following z parameters.

(a) [ ]

=105

2025z

(b) [ ]

+

+=

ss

s

ss1

21

131

z


(a) This is a non-reciprocal circuit so that the two-port looks like the oneshown in Figs. (a) and (b) .

z11 z22

z12 I 2

(a)

+

V2 +

I 1 I 2

+

V1 +

25 10

20 I 2

(b)

+

V2 +

I 1 I 2

+

V1 +


17/190



(b) This is a reciprocal network and the two-port look like the one shown inFigs. (c) and (d) .

s5.01

1s2

11211 +=+=zz

s21222 =zz

s1

12 =z

z11 z 12 z22 z 12

(c)

+V2

I1 I2

+V1

(d)

+

V2

I1 I 2

+

V1

2 H0.5 F1


18/190




Determine a two-port network that is represented by the following z parameters:

[ ]

+= j j

j j

825

2536z


This is a reciprocal network, as shown below.1+j5 3+j 1 j5 3

5-j2 5

-j2

j1


19/190




For the circuit shown in Fig. 19.73, let

[z]

=124

610

Find 121 ,, V I I , and 2V .

Figure 19.73

For Prob. 19.12.


20/190




1 1 210 6V I I = (1)2 2 24 12V I I = + (2)

2 210V I = (3)

If we convert the current source to a voltage source, that portion of the circuit becomeswhat is shown below.

4 2 I1

+

V112 V

1 1 1 112 6 0 12 6 I V V I + + = = (4)

Substituting (3) and (4) into (1) and (2), we get

1 1 2 1 212 6 10 6 12 16 6 I I I I I = = (5)2 1 2 1 2 1 210 4 12 0 4 22 5.5 I I I I I I I = + = + = (6)

From (5) and (6),

2 2 2 212 88 6 82 0.1463 A I I I I = = = 1 25.5 0.8049 A I I = =

2 210 1.463 VV I = = 1 112 6 7.1706 VV I = =

+ _


21/190




Determine the average power delivered to 45 j Z L += in the network of Fig. 19.74. Note: The voltage is rms.

Figure 19.74

For Prob. 19.13.


Consider the circuit as shown below.

10 I1 I2

++

500 V V 1V2 ZL

_

1 1 240 60V I I = + (1)2 1 280 100V I I = + (2)2 2 2 (5 4)LV I Z I j = = + (3)

1 1 1 150 10 50 10V I V I = + = (4)Substituting (4) in (1)

1 1 2 1 250 10 40 60 5 5 6I I I I I = + = + (5)

Substituting (3) into (2),2 1 2 1 2(5 4) 80 100 0 80 (105 4)I j I I I j I + = + = + + (6)

Solving (5) and (6) givesI2 = 7.423 + j3.299 A

[z]+ _


22/190



We can check the answer using MATLAB.

First we need to rewrite equations 1-4 as follows,

U

5000

0

X*A

IIV

V

010014 j5010

1008010

604001

2

1

2

1

=

==

+

>> A=[1,0,-40,-60;0,1,-80,-100;0,1,0,(5+4i);1,0,10,0]A =

1.0e+002 *0.0100 0 -0.4000 -0.6000

0 0.0100 -0.8000 -1.00000 0.0100 0 0.0500 + 0.0400i

0.0100 0 0.1000 0>> U=[0;0;0;50]U =

000

50>> X=inv(A)*UX =-49.0722 +39.5876i50.3093 +13.1959i

9.9072 - 3.9588i-7.4227 + 3.2990i

P = |I 2|25 = 329.9 W .


23/190




For the two-port network shown in Fig. 19.75, show that at the output terminals,

sZz

zzzZ

+=

11

211222Th

and

ss

VZz

zV

+=

11

21Th

Figure 19.75

For Prob. 19.14 and 19.41.


To find ThZ , consider the circuit in Fig. (a).

2121111 IzIzV += (1)2221212 IzIzV += (2)

+

(a)

+

V1

I1 I2

Z S


24/190



But12 =V , 1s1 - IZV =

Hence, 2s1112

12121s11

-)(0 IZz

z

IIzIZz += ++=

222s11

1221-1 IzZz

zz

++=

===22

2Th

1II

VZ

s11

122122 Zz

zzz

To find ThV , consider the circuit in Fig. (b).

02 =I , s1s1 ZIVV =

Substituting these into (1) and (2),

s11

s1111s1s Zz

VIIzZIV += =

s11

s211212 Zz

VzIzV +==

== 2Th VVs11

s21

Zz

Vz

(b)

+

V1

I1

VS

+

V2 = V Th

Z S I2 = 0

+


25/190




For the two-port circuit in Fig. 19.76,

[z]

=12080

6040

(a) Find LZ for maximum power transfer to the load.(b) Calculate the maximum power delivered to the load.

Figure 19.76

For Prob. 19.15.


(a) From Prob. 18.12,

241040

60x80120

Zz

zzzZ

s11

211222Th =

+=

+=

== 24ZZ ThL

(b) 192)120(1040

80V

Zzz

V ss11

21Th =+

=+

=

==

= 24x4R

R 2V

P 2Th

2

Th

Thmax 384W


26/190




For the circuit in Fig. 19.77, at 2= rad/s, z =1011 , z 12 = z = 621 j , z = 422 . Obtainthe Thevenin equivalent circuit at terminals a -b and calculate ov .

Figure 19.77

For Prob. 19.16.


27/190




As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).

At terminals a-b,)6 j105(||6 j)6 j4(Th ++=Z

6 j4.26 j415

)6 j15(6 j6 j4Th ++=+=Z

=ThZ 4.6

==++= 6 j)015(6 j1056 j6 j

ThV V906

The Thevenin equivalent circuit is shown in Fig. (b).

From this,

=+= 14818.3)6 j(4 j4.64 j

oV

=)t(v o V)148t2cos(18.3

4 j6 5

150 V

(a)

a

b

+

10 j6

6.4

690 V

(b)

+

+

Vo


28/190




* Determine the z and y parameters for the circuit in Fig. 19.78.

Figure 19.78

For Prob. 19.17.

* An asterisk indicates a challenging problem.


To obtain 11z and 21z , consider the circuit in Fig. (a).

In this case, the 4- and 8- resistors are in series, since the same current, oI , passesthrough them. Similarly, the 2- and 6- resistors are in series, since the same current,

'oI , passes through them.

===++== 8.420

)8)(12(8||12)62(||)84(

1

111 I

Vz

11o 52

1288

III =+= 1'

o 53

II =

I2 = 0

8

4

+

V1 I1

(a)

+

V2

I o'

Io

6


29/190



But 024- 'oo2 =+ IIV

111'

oo2 52-

56

58-

2-4 IIIIIV =+=+=

=== -0.45

2-

1

221

I

Vz

To get 22z and 12z , consider the circuit in Fig. (b).

===++== 2.420

)14)(6(14||6)68(||)24(222

2I

Vz

== -0.42112 zz

Thus,

=][z

4.20.4-

0.4-8.4

We may take advantage of Table 18.1 to get [ y] from [ z].20)4.0()2.4)(8.4( 2z ==

21.020

2.4

z

2211 ===

zy 02.0

204.0-

z

1212 ===

zy

02.020

4.0-

z

2121 ===

zy 24.0

208.4

z

1122 ===

zy

Thus,

=][y S24.002.0

02.021.0

I1 = 0

8

4

+

V1

(b)

+

V2

6


30/190




Calculate the y parameters for the two-port in Fig. 19.79.

Figure 19.79

For Prob. 19.18 and 19.37.


To get 11y and 21y , consider the circuit in Fig.(a).

111 8)3||66( IIV =+=

81

1

111 == V

Iy

12-

832-

366- 11

12

VVII ==+=

121-

1

221 == V

Iy

3 6

(a)

+

V2 = 0V1 +

I 1 I 2


31/190


32/190




Find the y parameters of the two-port in Fig. 19.80 in terms of s.

Figure 19.80

For Prob. 19.19.


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

1111 1s22

)s1(2s2

2||s1

IIIV +=+=

=

5.0s2

1s2

1

111 +=

+==V

Iy

2-

1s2-

2)s1()s1-( 11

12

VIII =+=+=

-0.51

221 == VIy

1

(a)

+

V2 = 0V1 +

I1 I2

1

1/s


33/190



To get 22y and 12y , refer to the circuit in Fig.(b).

222 2ss2

)2||s( IIV +==

s1

5.0s22s

2

222 +=

+==V

Iy

2-

s22s

2ss-

2ss- 2

221

VVII =

++=+=

-0.52

112 == V

Iy

Thus,

=][y Ss10.50.5-

0.5-5.0s

(b)

+

V1 = 0 +

I 1 I21

1

1/s


34/190




Find the y parameters for the circuit in Fig. 19.81.

Figure 19.81

For Prob. 19.20.


To get y 11 and y 21, consider the circuit below.

3ix

I1 2 I2

+ ix +

I1 V1 4 6 V2 =0- -

Since 6-ohm resistor is short-circuited, i x = 0

75.0

V

IyI

6

8)2//4(IV

1

111111 == ==

5.0VI

yV21

)V86

(32

I24

4I

1

2211112 == ==+

=


35/190



To get y 22 and y 12, consider the circuit below.

3ix

I1 2

+ ix +

V1=0 4 6 V2 I2 -

-

1667.061

VI

y6

V2

Vi3iI,

6V

i2

222

22xx2

2x === =+==

0VI

y02

Vi3I

2

112

2x1 == ==

Thus,

S1667.05.0075.0

]y[

=


36/190




Obtain the admittance parameter equivalent circuit of the two-port in Fig. 19.82.

Figure 19.82

For Prob. 19.21.


To get 11y and 21y , refer to Fig. (a).

At node 1,

4.04.02.05 1

11111

11 == =+= V

IyVV

VI

-0.2-0.21

22112 == = V

IyVI

0.2 V 1

(a)

+

V2 = 0V1 +

I1 I 2

10

V1


37/190



To get 22y and 12y , refer to the circuit in Fig. (b).

Since 01 =V , the dependent current source can be replaced with an open circuit.

1.0101

102

22222 === = V

IyIV

02

112 == V

Iy

Thus,

=][y S1.02.0-

04.0

Consequently, the y parameter equivalent circuit is shown in Fig. (c) .

+

V1 = 0+

I 1 I20.2 V 1

(b)

10

V1

+

V2

+

V1

I 1 I 2

(c)


38/190




Obtain the y parameters of the two-port network in Fig. 19.83.

Figure 19.83

For Prob. 19.22.


39/190




To obtain y 11 and y 21, consider the circuit below.

5 I2

+ +

I1 5 0.5V 2 2 V2=0V1

The 2- resistor is short-circuited.

= = = =1 11 111

25 0.4

2 5I I

V y V

12

2 1 211 1

11 2 0.22 2.5

I I I I y

V I = = = =

To obtain y 12 and y 22, consider the circuit below.

I1 5

+ +

5 0.5V 2 2 V2 I2V1=0

At the top node, KCL gives2 2 2

2 2 2 222

0.5 1.2 1.22 5

V V I I V V y

V = + + = = =

2 11 2 12

2

0.2 0.25

V I I V y

V = = = =

Hence, =

0.4 0.2

[ ] S0.2 1.2

y


40/190




(a) Find the y parameters of the two-port in Fig. 19.84.(b) Determine V ( )s2 for ( )t uvs 2= V.

Figure 19.84

For Prob. 19.23.


(a)

)1s(ys1s1

//1/1y 1212 += +=

=

2s1s1y1y1yy 12111211 +=++== =+

s1ss

1ss1

ys1

ysyys2

12221222++=++== =+

+++

++=

s1ss

)1s(

)1s(2s]y[ 2


41/190



(b) Consider the network below.

I1 I2 1

+ + +

Vs V1 V2 2- - -

11s VIV += or V s V 1 = I1 (1)

22 I2V = (2)

2121111 VyVyI += (3)

2221212 VyVyI += (4)

From (1) and (3)

212111s2121111s VyV)y1(VVyVyVV ++= += (5)

From (2) and (4),

22221

12221212 V)y5.0(y1VVyVyV5.0 += += (6)

Substituting (6) into (5),

++= =

+++=

)y5.0)(y1(y1

y

s/2V

s2

VyVy

)y5.0)(y1(V

221121

12

2

212221

2211s

( ) )2.1s8.1s()1s(8.0

s1ss

21

2s11s

1)1s(

s/2V

222 +++=

++++++

++=

[y]


42/190




Find the resistive circuit that represents these y parameters:

[y]

=

83

41

41

21


Since this is a reciprocal network, a network is appropriate, as shown below .

S41

41

21

12111 ==+= yyY , =1Z 4

S41

- 122 == yY , =2Z 4

S8

1

4

1

8

321223 ==+= yyY , =3Z 8

Y1

Y2

(a)

1/4 S

1/4 S

(b)

4

4

(c)


43/190




Draw the two-port network that has the following y parameters:

[y]

=5.15.0

5.01S


This is a reciprocal network and is shown below.

0.5 S

0.5S 1S


44/190




Calculate [ y] for the two-port in Fig. 19.85.

Figure 19.85

For Prob. 19.26.


To get 11y and 21y , consider the circuit in Fig. (a).

At node 1,

x1xx

xx1 -2

412

2VV

VVV

VV = +=+ (1)

But 5.15.122

2 11

11111x1

1 == =+

=

=V

IyV

VVVVI

Also,1x2x

x

2-3.575.12

4VVIV

VI == =+

-3.51

221 == V

Iy

2

(a)

+

V2 = 0V1 +

I21

2 V x

4

2

+

Vx


45/190



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

At node 2,

42 x2x2

VVVI

+= (2)

At node 1,

x2xxxx2

x -23

1242 VVV

VVVVV = =+=

+ (3)

Substituting (3) into (2) gives

2xxx2 -1.55.121

2 VVVVI ===

-1.52

222 == V

Iy

5.022

-

2

112

2x1 == == V

Iy

VVI

Thus,

=][y S5.1-5.3-

5.05.1

2

(b)

I21

2 V x

4

2

+

Vx +

I1


46/190




Find the y parameters for the circuit in Fig. 19.86.

Figure 19.86

For Prob. 19.27.


Consider the circuit in Fig. (a).

25.04

41

1

1

11111 === = I

IVI

yIV

55201

221112 == == V

IyVII

0.1 V 2+ V1

(a)

I 1 I 2

+

V2 = 0

4


47/190


48/190




In the circuit of Fig. 19.65, the input port is connected to a 1-A dc current source.Calculate the power dissipated by the 2 - resistor by using the y parameters. Confirmyour result by direct circuit analysis.


We obtain 11y and 21y by considering the circuit in Fig.(a).

4.34||61in =+=Z

2941.01

in1

111 === ZV

Iy

11

12 346-

4.3106-

106-

VV

II =

==

-0.176534

6-

1

221 === V

Iy

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

4

+

V1

1

I1

(a)

+

V2 = 0

I2

4

+

V1 = 0

+

V2

1

(b)

I oI 1


49/190



2

2

22 IV

2434

)734(2)734)(2(

76

4||2)1||64(||21 ==

+=

+=+=

y

7059.03424

22 ==y

o1 76-

II = 222o 347

4814

)734(22

VIII ==+=

-0.176534

6-34

6-

2

11221 === = V

IyVI

Thus,

=][y S0.70590.1765-0.1765-2941.0

The equivalent circuit is shown in Fig. (c). After transforming the current source to avoltage source, we have the circuit in Fig. (d).

5454.0167.149714.0

)5.8)(9714.0(667.55.8889.1||2)5.8)(889.1||2(V =+=++=

===2

)5454.0(R

VP

22

W1487.0

6/34 S

1 A

(c)

5.667 8.5

(d)

+ 8.5 V

+

V


50/190




In the bridge circuit of Fig. 19.87, 101 = I A and 42 = I A(a) Find 1V and 2V using y parameters.(b) -Confirm the results in part (a) by direct circuit analysis.

Figure 19.87

For Prob. 19.29.


(a) Transforming the subnetwork to Y gives the circuit in Fig. (a).

It is easy to get the z parameters22112 == zz , 32111 =+=z , 322 =z

54921122211z === zzzz

22z

2211 5

3y

zy === , 5

2--

z

122112 ===

zyy

1 1

10 A

Vo

+

V1

+

V2

(a)


51/190



Thus, the equivalent circuit is as shown in Fig. (b).

21211 235052

53

10 VVVVI = == (1)

21212 3-220-53

52-

-4 VVVVI += +==

2121 5.1105.110 VVVV += = (2)Substituting (2) into (1),

= += 222 25.43050 VVV V8

=+= 21 5.110 VV V22

(b) For direct circuit analysis, consider the circuit in Fig. (a).

For the main non-reference node,

122

410 oo = = VV

=+=

= o1o1 10

110 VV

VVV22

==

= 41

4- o2o2 VV

VVV8

2/5 S

10 A

+

V1

+

V2

(b)

1/5 S

I1 I2


52/190




Find the h parameters for the networks in Fig. 19.88.

Figure 19.88

For Prob. 19.30.


(a) Convert to z parameters; then, convert to h parameters using Table 18.1.=== 60211211 zzz , =10022z

24003600600021122211z === zzzz

241002400

22

z11 ==

=

zh , 6.0

10060

22

1212 === z

zh

-0.6-

22

2121 == z

zh , 01.0

1

2222 == zh

Thus,

=][h

S0.010.6-

6.024

(b) Similarly,= 3011z === 20222112 zzz

200400600z ==

1020

20011 ==h 120

2012 ==h

-121 =h 05.020

122 ==h

Thus,

=][h

S0.051-

110


53/190


54/190



Adding (1) and (2),

1441 6.3518 IVVI = = 1143 8.283 IIVV ==

1131 8.3 IIVV =+=

== 8.31

111 I

Vh

-3.6-3.61

-

1

2211

42 == == I

IhI

VI

To get 22h and 12h , refer to the circuit in Fig. (b). The dependent current source can bereplaced by an open circuit since 04 1 =I .

4.052

1222

2

112221 == =++= V

VhVVV

S2.051

5122 22

2222

2 === =++= VI

hVV

I

Thus,

=][h

S0.23.6-

4.038

1 1

(b)

+

V1

2 I2

+

I 1


55/190




Find the h and g parameters of the two-port network in Fig. 19.90 as functions of s.

Figure 19.90

For Prob. 19.32.


56/190



Chapter 19, Solution 32.(a) We obtain 11h and 21h by referring to the circuit in Fig. (a).

1211 1ss

s1s1

||ss1 IIV

+++=

++=

1ss

1s 21

111 +++== I

Vh

By current division,

1s1-

1s-

s1ss1-

21

221

112 +== +=+= I

Ih

III

To get 22h and 12h , refer to Fig. (b).

1s1

1ss1ss1

22

1122

221 +== +=+= V

Vh

VVV

1ss

s1s1

s1

s 22

22222 +=+==

+=V

IhIV

Thus,

=][h

1ss

1s1-

1s1

1ss

1s

22

22

s1

I 1

(a)

I 2

+

V2 = 0

+

V1

s

s1

(b)

I 2

+

V1

sI1 = 0

+


57/190



(b) To get 11g and 21g , refer to Fig. (c).

1sss

s1s11

s1

s1 21

11111 ++=++==

++=V

IgIV

1ss1

1sss1s1s1

21

2212

112 ++== ++=++= V

Vg

VVV

To get 22g and 12g , refer to Fig. (d).

222 s1s1s)1s(

s)1s(||s1

s IIV

+++

+=

++=

1ss1s

s 22

222 ++

++==I

Vg

1ss1-

1ss-

s1s1s1-

22

1122

221 ++== ++=++= I

Ig

III

Thus,

=][g

1ss1s

s1ss

11ss

1-1ss

s

22

22

s1

V1

(c)

+

V2

s I2 = 0

+

I1

s1

(d)

s I2

+

V1 = 0

+

V2

I 1


58/190




Obtain the h parameters for the two-port of Fig. 19.91.

Figure 19.91

For Prob. 19.33.


59/190




To get h 11 and h 21, consider the circuit below.4 j6 I2

+-j3 +

I1 V1 5 V2=0- -

2821.1 j0769.3IV

h6 j9

I)6 j4(5I)6 j4//(5V

1

111

111 +==+

+=+=

Also, 2564.0 j3846.0IIhI

6 j95I

1

22112 +== +

=

To get h 22 and h 12, consider the circuit below.

4 j6 I2 I1

+ +-j3 +

V1 5 V2 - -

2564.0 j3846.06 j9

5VV

hV6 j9

5V

2

11221 =+

== +

=

2821.0 j0769.0

)6 j9(3 j3 j9

)6 j9//(3 j1

VI

hI)6 j9//(3 jV 2

222

22+=

++=

+==

+=

Thus,

+++=

2821.0 j0769.02564.0 j3846.02564.0 j3846.02821.1 j077.3

]h[


60/190




Obtain the h and g parameters of the two-port in Fig. 19.92.

Figure 19.92

For Prob. 19.34.


Refer to Fig. (a) to get 11h and 21h .

At node 1,

x1xx

1 43003000

100VI

VVI =

+= (1)

11x 754300

IIV ==

But == =+= 8585101

1111x11 I

VhIVIV

10

I 1

1

+

Vx

+

V1

50 2

+

V2 = 0

300

I2

(a)


61/190



At node 2,

111xxxx

2 75.1430075

575

300530050100

IIIVVVV

I ===+

=

75.141

221 == I

Ih


At node 2,

x22x22

2 8094005010

400VVI

VVVI +=

++=

But 4400100 2

2x

VVV ==

Hence, 2222 29209400 VVVI =+=

S0725.040029

2

222 === V

Ih

25.041

4 21

122

x1 === == VV

hV

VV

=][h

S0725.075.1425.085

10 1

+

Vx

+

V1

50 2

300

(b)

I1 = 0

+

I 2


62/190



To get 11g and 21g , refer to Fig. (c).

At node 1,

x1xxx

1 5.1435035010

100 VIVVV

I = +

+= (2)

But x11x1

1 1010VVI

VVI =

=

or 11x 10 IVV = (3)


11111 5.144951455.14350 VIIVI = =

S02929.0495

5.14

1

111 ===

V

Ig

At node 2,

xxx2 -8.42861035011

)50( VVVV =

=

1111 4955.14

)286.84(-8.4286286.84-8.4286 VVIV

+=+=

-5.96-5.961

22112 == = V

VgVV

10 1+

Vx

+

V2

50 2

300

(c)

+

I2 = 0I 1

V1


63/190



To get 22g and 12g , refer to Fig. (d).

091.9100||10 =

091.93005010 2x2

2 +++

=VVV

I

x22 818.611818.7091.309 VVI += (4)

But 22x 02941.0091.309091.9

VVV == (5)


22 9091.309 VI =

== 34.342

222 I

Vg

091.30934.34

091.30922

o

IVI ==

)091.309)(1.1(34.34-

110100- 2

o1

III ==

-0.1012

112 ==

I

Ig

Thus,

=][g

34.345.96-0.101-S02929.0

10

+

Vx

50

300

(d)

+

V2

Io I o

+

V1 = 0

I 1


64/190


65/190



Chapter 19, Solution 35.To get 11h and 21h consider the circuit in Fig. (a).

144

n4

Z 2R ===

== =+= 22)11(1

111111

I

VhIIV

-0.521-

2- N N-

1

221

1

2

2

1 === ==II

hII


Since 01 =I , 02 =I .Hence, 022 =h .

At the terminals of the transformer, we have 1V and 2V which are related as

5.021

2n N N

2

112

1

2

1

2

=== === VV

hV

V

Thus,

=][h

05.0-

5.02

4 1

I 1

(a)

I 2

+

V2 = 0

+

V1

1 : 2

4 1

(b)

I 2

+

V1

1 : 2I 1 = 0

+


66/190




For the two-port in Fig. 19.94,

[h ]

S01.02

316

Find:

(a) 12 /V V (b) 12 / I I (c) 11 /V I (d) 12 / I V

Figure 19.94

For Prob. 19.36.


67/190




We replace the two-port by its equivalent circuit as shown below.

= 2025||100

112 40)2)(20( IIV == (1)

032010- 21 =++ VI 111 140)40)(3(2010 III =+=

141

1 =I , 1440

2 =V

14136

316 211 =+= VIV

708-

)2(125100

12 =

= II

(a) ==13640

1

2

VV

2941.0

(b) =1

2

I

I1.6-

(c) ==136

1

1

1

V

IS10353.7 -3

(d) ==1

40

1

2

IV 40

2 I 1

++

16

10 V 3 V 2

4 I 1 I 2

+

V1

+

V2


68/190




The input port of the circuit in Fig. 19.79 is connected to a 10-V dc voltage source whilethe output port is terminated by a 5 - resistor. Find the voltage across the 5- resistor

by using h parameters of the circuit. Confirm your result by using direct circuit analysis.


(a) We first obtain the h parameters. To get 11h and 21h refer to Fig. (a).

26||3 =

== =+= 88)26(1

111111

I

VhIIV

32-

32-

636-

1

221112 == =+= I

IhIII

3

+

V1

6

I1

(a)

I 2

+

V2 = 0


69/190



To get 22h and 12h , refer to the circuit in Fig. (b).

49

9||3 =

94

49

2

22222 == = V

IhIV

32

32

366

2

112221 == =+= V

VhVVV

=][h

S94

32-

32

8

The equivalent circuit of the given circuit is shown in Fig. (c).

1032

8 21 =+ VI (1)

1112 2930

2945

32

49

||532

IIIV =

=

=

21 3029

VI = (2)

3

+

V1

I1 = 0 6

(b)

+

I2

(c)

++

8

10 V 2/3 V 2

I 1 I 2

+

V2


70/190



Substituting (2) into (1),

1032

3029

)8( 22 =+

VV

== 252300

2V V19.1

(b) By direct analysis, refer to Fig.(d).

Transform the 10-V voltage source to a6

10-A current source. Since

= 36||6 , we combine the two 6- resistors in parallel and transform

the current source back to V536

10 = voltage source shown in Fig. (e).

815

8)5)(3(

5||3 ==

==+= 6375)5(8156 8152V 1.1905 V

3 6

+ 10 V

(d)

+

V2

3 3

+ 5 V

(e)

+

V2


71/190




The h parameters of the two-port of Fig. 19.95 are:

[h ]

=mS230

04.0600

Given the 2=s Z k and = 400 L Z , find in Z and out Z .

Figure 19.95For Prob. 19.38.


From eq. (19.75),

12 2111 3

22

0.04 30 400600 333.33

1 1 1 2 10 400re fe L L

in ieoe L L

h h R h h R x x Z h h

h R h R x x= = = =

+ + +

From eq. (19.79),

113

0 11 22 21 12

2, 000 600650

( ) ( ) 2600 2 10 30 0.04s ie s

out s ie e re fe s

R h R h Z

R h h h h R h h h h x x x+ + += = = =

+ +


72/190




Obtain the g parameters for the wye circuit of Fig. 19.96.

Figure 19.96

For Prob. 19.39.


73/190




We obtain g 11 and g 21 using the circuit below.I1 R 1 R 3 I2=0

+

R 2V1 V2

1 11 11

1 2 1 1 2

1V I I g

R R V R R= = =

+ +

By voltage division,2 2 2

2 1 211 2 1 1 2

R V RV V g R R V R R

= = =+ +

We obtain g 12 and g 22 using the circuit below.

I1 R 1 R 3

+ +

R 2 I2V1=0 V2

By current division,2 1 2

1 2 121 2 2 1 2

R I R I I g

R R I R R= = =

+ +

Also,

1 22 2 3 1 2 2 3

1 2

( // ) R R

V I R R R I R R R

= + = + + 2 1 222 3

2 1 2

V R Rg R

I R R= = +

+

21

21322

21

221

21

212

2111

R R R R

R g,R R

R g

R R R g,

R R 1g

++=

+=

+=

+=

+ _


74/190




Find the g parameters for the circuit in Fig. 19.97.

Figure 19.97

For Prob. 19.40.


75/190


76/190




For the two-port in Fig. 19.75, show that

g L +

=Zg

g

I

I

11

21

1

2

( )( ) s Ls L

s ZggZgZgZg

VV

12212211

212

1 ++=

where g is the determinant of [ g] matrix.


77/190


78/190




The h parameters of a two-port device are given by

h = 60011 , h 312 10= , h 12021 = , h 622 102 = S

Draw a circuit model of the device including the value of each element.


With the help of Fig. 19.20, we obtain the circuit model below.

I1 600 I2

+ +

V1 10 -3 V2 + 120I 1 500 k V2


79/190




Find the transmission parameters for the single-element two-port networks in Fig. 19.98.

Figure 19.98

For Prob. 19.43.


80/190




(a) To find A and C , consider the network in Fig. (a).

12

121 == = V

VAVV

002

11 == = VICI

To get B and D , consider the circuit in Fig. (b).

11 IZV = , 12 - II =

ZI

IZ

I

VB ===

1

1

2

1

---

1-

2

1 ==I

ID

Hence,

=][T

10

Z1

Z

+

I1 I2

+

V2

V1

(a)

+

V2 = 0

Z

+

I1 I2

V1

(b)


81/190



(b) To find A and C , consider the circuit in Fig. (c).

12

121 == = V

VAVV

YZV

ICVIZV === == 12

1211

To get B and D , refer to the circuit in Fig.(d).

021 == VV 12 - II =

0-

2

1 ==IV

B , 1-

2

1 ==II

D

Thus,

=][T

1Y

01

+

I1 I2

+V2

V1

(c)

+

V2 = 0

I2

I 1

(d)

+

V1


82/190




Determine the transmission parameters of the circuit in Fig. 19.99.

Figure 19.99

For Prob. 19.44.


To determine A and C , consider the circuit in Fig.(a).

11 ])20 j15 j(||)10 j-(20[ IV +=

111 310

j20 j15-

-j10)(-j5)(20 IIV

=

+=

1'

o II =

11o 32

5 j10 j-10 j-

III

=

=

'20-j20)( oo2 IIV += = 11 I20I340

j + = 1I340

j20

15

- 20

+

(a)

I o'

V1

I2 = 0-j10

+

V2

I1

Io

Io


83/190



=

==1

1

2

1

I3

40 j20

)310 j20( IVV

A 0.7692 + j0.3461

===3

40 j20

1

2

1VI

C 0.03461 + j0.023

To find B and D , consider the circuit in Fig. (b).

We may transform the subnetwork to a T as shown in Fig. (c).

10 j20 j10 j15 j

-j10))(15 j(1 =

=Z

340-j

j15-)-j10)(-j20(2 ==Z

20 j j15-

-j20))(15 j(3 ==Z

15

- 20

+

(b)

V1

-j10 I1 I 2

+

V2 = 0

j10

+

(c)

I1 I 2

+

V2 = 0V1

20


84/190



112 j32 j3

20 j340 j20340 j20

- III +=+

=

6923.0 j5385.02 j3

j3-

2

1 +=+==

I

ID

11 20 j340 j20)340 j20)(20 j(

10 j IV

+

+=

)18 j24( j])7 j9(210 j[ 111 =++= IIV

j55)-15(136

j3 j2)-(3-

)18 j24( j--

1

1

2

1 +=

+

==

I

I

I

VB

+= j25.385-6.923B

=][T

j0.69230.5385S j0.0230.03461

j25.3856.923-3461.0 j7692.0


85/190


86/190




To determine A and C, consider the circuit below.

I1 -j2 I2 =0

+

V1 4 V2

1 1 2 1(4 2) , 4V j I V I = =

= = = 1

2

4 21 0.54

V j A j V

1 1

2 1

0.254

I I C

V I = = =

To determine B and D, consider the circuit below.

I1 -j2 I2

+

V1 4 V2 =0

The 4- resistor is short-circuited. Hence,1

2 12

, 1I I I D I

= = =

= = = = = 1 21 1 22 2

22 2 2V j I V j I j I B j

I I

Hence,1 0.5 2

0.25 1A B j j

C D S

=

=

1S25.02 j5.0 j1

+ _

+ _


87/190




Find the transmission parameters for the circuit in Fig. 19.101.

Figure 19.101

For Prob. 19.46.

Chapter 19, Solution 46.To get A and C , refer to the circuit in Fig.(a).

At node 1,

2o12oo

1 23212VVI

VVVI =

+= (1)

At node 2,

2ooo

x2o -2

24

41

VVVV

IVV = ===

(2)

From (1) and (2),

S-2.525-

-522

121 === = V

ICVI

But 21o1

1 1VV

VVI +=

=

21212 -3.52.5- VVVVV = +=

-3.52

1 ==V

VA

1 1

+

1 2

(a)

I x

I 1

+

V2 V1

I2 = 0

+

Vo


88/190




At node 1,

o1oo

1 3212VI

VVI = += (3)

At node 2,04

1 xo

2 =++ IV

I

o2oo2 -302 VIVVI = =+= (4)

Adding (3) and (4),

2121 -0.502 IIII = =+ (5)

5.0-

2

1 ==I

ID

But o11o1

1 1VIV

VVI +=

= (6)

Substituting (5) and (4) into (6),

2221 65-

31-

21-

IIIV =+=

=== 8333.065-

2

1

I

VB

Thus,

=][T

0.5-S2.5-

0.83333.5-

1 1

+

1 2

(b)

I x

I 1

V1

+

Vo

+

V2 = 0

I 2


89/190




Obtain the ABCD parameters for the network in Fig. 19.102.

Figure 19.102

For Prob. 19.47.


To get A and C, consider the circuit below.

6

I1 1 4 I2=0

+ + + +Vx 2 5Vx V2

V1

x1xxxx1 V1.1V

10V5V

2V

1VV = +=

3235.04.3/1.1VV

AV4.3V5)V4.0(4V21xxx2 === =+=


90/190



02941.04.3/1.0VI

CV1.0VV1.11

VVI

2

1xxx

x11 === ==

=

To get B and D, consider the circuit below.

6

I1 1 4 I2

0V+ + + +

Vx 2 5Vx V2=0V1

- - - -

x1xxx1 V

610

V2

V6

V1

VV = += (1)

xxx

2 V1217

6V

4V5

I == (2)

x11 VIV += (3)From (1) and (3)

4706.0)1712

(64

II

DV64

VVI2

1xx11 === ==

176.1)1712

(6

10IV

B2

1 ===

=4706.002941.0176.13235.0

]T[


91/190




For a two-port, let A = 4, B = 30 , C = 0.1 S, and D = 1.5. Calculate the inputimpedance, 11 / IVZ =in when:

(a) the output terminals are short-circuited,(b) the output port is open-circuited,(c) the output port is terminated by a 10- load.


92/190


93/190




Using impedances in the s domain, obtain the transmission parameters for the circuit inFig. 19.103.

Figure 19.103

For Prob. 19.49.

Chapter 19, Solution 49.To get A and C , refer to the circuit in Fig.(a).

1s1

s11s1

s1

||1 +=+=

12 s1||1s1s1||1

VV +=

s1s2

1s1

1s1

s1

2

1 +=

+

++

==V

VA

++

+=

++

+= )1s(s1s2

||1s

11s

1s1

||1s

1111 IIV

)1s3)(1s(1s2

)1s(s1s2

1s1

)1s(s1s2

1s1

1

1

+++=

++++

++

+=

IV

1/s

+

(a)

I 1

+

V2

V1

I 2 = 0


94/190



Buts

1s221

+= VV

Hence,)1s3)(1s(

1s2s

1s2

1

2

+++

=+

I

V

s)1s3)(1s(

1

2 ++==IV

C


1s2s21

||1s1

||s1

||1 1111 +=

=

= IIIV

1

1

2 12ss-

s

1

1s

11s

1-

II

I +=

++

+=

s1

2s

1s2-

2

1 +=+==I

ID

s1-

s-s-1s2

1s21

2

1221 == =

+

+=

I

VB

IIV

Thus,

=][T

+++

+

s1

2s

)1s3)(1s(s1

s1s2

1/s

+

(b)

I 1

V1

I2

+

V2 = 0


95/190




Derive the s-domain expression for the t parameters of the circuit in F

sadiku 5th ed chapter 19

Documents