Two Port Networks-Electrical Circuital Analysis-Lecture Slides, Slides of Electronic Circuits Analysis

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An Introduction To

Two – Port Networks

Electrical and Computer Engineering

wlg

Generalities: The standard configuration of a two port:

The Network

Input

Port

Output

Port

_ _

V 1 V 2

I 1 I 2

The network?

The voltage and current convention?

• notes

Z parameters:

11 I

V

z  0 2

I 

12 I

V

z  0 1

I 

21 I

V

z  0 2

I 

22 I

V

z  0 1

I 

z 11 is the impedance seen looking into port 1

when port 2 is open.

z 12 is a transfer impedance. It is the ratio of the

voltage at port 1 to the current at port 2 when

port 1 is open.

z 21 is a transfer impedance. It is the ratio of the

voltage at port 2 to the current at port 1 when

port 2 is open.

z 22 is the impedance seen looking into port 2

when port 1 is open.

• notes

Y parameters:

11 V

I

y  0 2

V 

12 V

I

y  0 1

V 

21 V

I

y  0 2

V 

22 V

I

y  0 1

V 

y 11 is the admittance seen looking into port 1

when port 2 is shorted.

y 12 is a transfer admittance. It is the ratio of the

current at port 1 to the voltage at port 2 when

port 1 is shorted.

y 21 is a transfer impedance. It is the ratio of the

current at port 2 to the voltage at port 1 when

port 2 is shorted.

y 22 is the admittance seen looking into port 2

when port 1 is shorted.

• notes

Z parameters: Example 1 (cont 1)

For z 11 :

Z 11 = 8 + 20||30 = 20 

For z 22 :

For z 12 :

Z 22 = 20||30 = 12 

12 I

V

z  0 1

I 

8

20 20 

 



10

_

_

V 1 V 2

I 1 I 2

2

2 1 8 20 30

20 20 xI

xI x V  



Therefore:

8

8

2

2 12 ^  I

xI z^ ^ = z 21

Z parameters: Example 1 (cont 2)

The Z parameter equations can be expressed in

matrix form as follows.













 



 

  











8 12

20 8

I

I

V

V























  











I

I

z z

z z

V

V

Z parameters: Example 2 (continue p2)

11 I

V

z  0 2

I 

1 

1 

4 

2 

2Vx

Vx

_ _

V 1 V 2

I 1 I 2

6

6 2

6

2

1

1

Vx Vx Vx Vx Vx V x I

  

  

1

V x

I ^ ;^ but Vx  V 1  I 1

Substituting gives;

 

2

(^3 ) 1

V I I



^ or

11 1

1

z

I

V

Z 21 = -0.667 

Z 12 = 0.222 

Z 22 = 1.111 

Other Answers

Transmission parameters (A,B,C,D):

The defining equations are:

 



 





 



 

   



 



2

2

1

1

I

V

C D

A B

I

V

2

1

V

V A 

I 2 = 0 2

1

I

V B 



V 2 = 0

2

1

V

I C 

I 2 = 0 2

1

I

I

D

V2 = 0

1

R 1

2

R

2

1 2

R

R  R

Transmission parameters (A,B,C,D):

Example (cont.)

V 1 = (R 1 + R 2 )I 1 + R 2 I 2

V 2 = R 2 I 1 + R 2 I 2

From these equations we can directly evaluate the A,B,C,D parameters.

2

1

V

V A 

I 2 = 0

2

1

I

V B 



V 2 = 0

2

1

V

I C 

I 2 = 0 2

1

I

I

D

V2 = 0

Later we will see how to interconnect two of these networks together for a final answer

• notes

Hybrid Parameters: The equations for the hybrid parameters are:

2

1

21 22

11 12

2

1

V

I

h h

h h

I

V

1

1 11 I

V h  V 2 = 0 2

1 12 V

V h  I 1 = 0

1

2 21 I

I h 

V 2 = 0 2

2 22 V

I h  (^) I 1 = 0

• notes

4

1 K

K 2

K 3

K 1

Hybrid Parameters:

D

V I CI

V AI BV

2 2 1

1 1 2

 

 

We want to evaluate the H parameters from the above set of equations.

1

1 11 I

V h 

V 2 = 0 2

1 12 V

V h  I 1 = 0

1

2 21 I

I h 

V 2 = 0 2

2 22 V

I h 

I 1 = 0

2

1

R

• 1

R 1

Hybrid Parameters: Another example with hybrid parameters.

Given the circuit below.

_

_

R 1

V 1 R 2 V 2

I 1 -I 2

The equations for the circuit are:

V 1 = (R 1 + R 2 )I 1 + R 2 I 2

V 2 = R 2 I 1 + R 2 I 2

The H parameters are as follows.

1

1 11

I

V

h 

2

1 12 V

V

h 

1

2 21

I

I

h 

2

2 22

V

I

h 

V 2 =

V 2 =

I 1 =

I 1 =

Modifying the two port network:

We modify the network as shown be adding elements outside the two ports

8

20 20 

 



10

_

_

V 1 V 2

I 1 I 2

_

10 v

6 

4 

We now have:

V 1 = 10 - 6I 1

V 2 = - 4I 2

Modifying the two port network:

We take a look at the original equations and the equations describing

the new port conditions.

 



 



 



 

   



 



2

1

2

1

8 12

20 8

I

I

V

V V

1 = 10 - 6I 1

V 2 = - 4I 2

So we have,

10 – 6I 1 = 20I 1 + 8I 2

-4I 2 = 8I 1 + 12I 2

• notes