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CH 16-Quadripolos

The document provides solutions to example problems involving two-port network analysis. It calculates current and voltage values for various two-port networks given their y-parameters. It also calculates characteristic impedances and impedance parameters for two-port networks composed of series and parallel resistances. The solutions demonstrate applying the definitions of y-parameters, characteristic impedance, and impedance parameters to analyze basic two-port networks.

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Laura Magalhaes
Copyright
© © All Rights Reserved
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61 views89 pages

CH 16-Quadripolos

The document provides solutions to example problems involving two-port network analysis. It calculates current and voltage values for various two-port networks given their y-parameters. It also calculates characteristic impedances and impedance parameters for two-port networks composed of series and parallel resistances. The solutions demonstrate applying the definitions of y-parameters, characteristic impedance, and impedance parameters to analyze basic two-port networks.

Uploaded by

Laura Magalhaes
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Irwin, Basic Engineering Circuit Analysis, 9/E 1

SOLUTION:

The correct answer is a.

I1  2 A , I 2  0
I 1  y11V1  y12V2
I 2  y 21V1  y 22V2
1 1
V1  V2  2
14 21
1 1
 V1  V2  0
21 7
21V1  14V2  588
 V1  3V2  0
V1  36V
V2  12V

Chapter 16: Two-Port Networks Problem 16.FE-1


Irwin, Basic Engineering Circuit Analysis, 9/E 1

SOLUTION:

The correct answer is b.

I 1  y11V1  y12V2
I 2  y 21V1  y 22V2
y11V1  y12V2  0

y12
V1   V2
y11
 y 
I 2  y 21   12 V2   y 22V2
 y11 

 y y 
I 2  V2  21 12  y 22 
 y11 

V2 y11
Z TH  
I 2  y 21 y12  y11 y 22
1
Z TH  14  9
  1   1   1  1 
       
 21  21   14  7 

Chapter 16: Two-Port Networks Problem 16.FE-2


Irwin, Basic Engineering Circuit Analysis, 9/E 1

SOLUTION:

The correct answer is d.

3
V1
I 3
y11  1  56  S
V1 V2  0
V1 56
2
0 V1
V V 14   2 V
I2  2  1
4 4 56
2
 V1
I2 2 1
y 21   56   S   S
V1 V2  0
V1 56 28
16
16 8  
3
16
16 4
V  3 V2   V2  V2
16 28 7
4
3
4
0  V2
V1  V 7 1
I1     V2
16 16 28
1
 V2
I1 1
y12   28   S
V1 V 0 V2 28
1

Chapter 16: Two-Port Networks Problem 16.FE-3


2 Irwin, Basic Engineering Circuit Analysis, 9/E

4
V2  V2
V V 7 3
I2  2   V2
4 4 28
3
V2
I 3
y 22  2  28  S
V2 V2 28

Problem 16.FE-3 Chapter 16: Two-Port Networks


Irwin, Basic Engineering Circuit Analysis, 9/E 1

SOLUTION:

The correct answer is c.

The 3 and 9 are in series.

9(12) 36
z11   
9  12 7
12 4
I ( I1 )  I1
12  9 7
 4  12
V2  3I  3 I 1   I 1
7  7
12
I1
V 12
z 21  2  7  
I1 I 0 I1 7
2

3 3 1
I (I 2 )  I2  I2
3  18 21 7
 1  12
V1  12 I  12 I 2   I 2
7  7
12
I2
V1 12
z12   7  
I 2 I 0 I2 7
1

3(18) 18
z 22   
3  18 7

Chapter 16: Two-Port Networks Problem 16.FE-4


Irwin, Basic Engineering Circuit Analysis, 9/E 1

SOLUTION:

The correct answer is a.

h11  2 4   8 
28

3
4
I2   I 1 
42
2
I 2   I1
3
2
 I1
I 3 2
h21  2 
I1 V2  0
I1 3
V2 V
I2   2
24 6

V2
I 1
h22  2  6  S
V2 I1  0
V2 6

V  2
V1  4 I 2  4 2   V2
 6  3
2
V2
V1 2
h12   3 
V2 I 0 V2 3
1

Chapter 16: Two-Port Networks Problem 16.FE-5

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