This document discusses two-port networks and their admittance and impedance parameters. It provides examples of using the admittance parameters Y to find the current through a resistor in a network and the impedance parameters Z to solve for the current given input and output constraints. Learning extensions show finding the parameters and using them to solve additional example circuits.

1. TWO-PORT NETWORKS
PART II
(Topics)
Example on Admittance
Impedance Parameters

2. An application of the admittance parameters
2221212
2121111
VyVyI
VyVyI


212
211
6
5
2
1
2
1
2
3
VVI
VVI


221 4,2 IVAI 
The model plus the conditions at the
ports are sufficient to determine the
other variables.
22
4
1
VI 
Determine the current through the
4 Ohm resistor

3. 2221212
2121111
VyVyI
VyVyI


21
21
4
1
6
5
2
1
0
2
1
2
3
2
VV
VV








][
11
2
][
11
8
6
13
2
2
21
AI
VV
VV




4. LEARNING EXTENSION Find the admittance (Y) parameters
1I


1V
2I
2221212
2121111
VyVyI
VyVyI


1I


1V
2I


2V
111
42
3
)
42
1
21
1
( VVI 
12
4221
42
II

 ][
21
1
][
14
1
21
11
Sy
Sy



5. 1I 2I


2V
21
22
5.1021
5.10
21
1
21
2
II
VI









][
21
1
][
7
1
12
22
Sy
Sy



6. LEARNING EXTENSION
5
A10
OI
Use the admittance (Y) parameters to find the current Io
Conditions at I/O ports
AI 101 
2
22
5
1
II
VI
o 

1I


1V
2I


2V 2221212
2121111
VyVyI
VyVyI


][
21
1
][
14
1
21
11
Sy
Sy


][
21
1
][
7
1
12
22
Sy
Sy


Replace in model
)5(
7
1
21
1
)5(
21
1
14
1
10
1
1
oo
o
IVI
IV


Solve for variable of interest ][
98
420
AIo 

7. IMPEDANCE PARAMETERS
The network contains NO independent sources
2221212
2121111
IzIzV
IzIzV


The ‘z parameters’ can be derived in a manner similar to the Y parameters
01
2
21
01
1
11
22 

II
I
V
z
I
V
z
02
2
22
02
1
12
11 

II
I
V
z
I
V
z
z11 is called open circuit input impedance, z22=open circuit output impedance
Z21 and z12 is called open circuit transfer impedance

8. LEARNING EXAMPLE Find the Z parameters
Write the loop equations
)(42
)(42
1222
2111
IIjIjV
IIjIV




24
442
2221
1211
jzjz
jzjz
212
211
24
4)42(
IjIjV
IjIjV


rearranging
01
2
21
01
1
11
22 

II
I
V
z
I
V
z
02
2
22
02
1
12
11 

II
I
V
z
I
V
z
2221212
2121111
IzIzV
IzIzV



9. LEARNING EXAMPLE Use the Z parameters to find the current through the 4 Ohm
resistor
2221212
2121111
IzIzV
IzIzV


Output port constraint
22 4IV 
Input port constraint
11 )1(012 IV 
212
211
24
4)42(
IjIjV
IjIjV


21 )24(40 IjIj 
21 4)43(12 IjIj 
)43( j
4j
2))43)(24(16(48 Ijjj   73.13761.12I

10. LEARNING EXTENSION Find the Z parameters.
Find the current on a 4 Ohm load with a 24V input source
1I


1V
2I


2V
)(63
)(612
2122
2111
IIIV
IIIV




9,6
6,18
2221
1211
zz
zz
4


V24
212
211
96
618
IIV
IIV


1I


1V
2I


2V
22 4IV :constraintportoutput
][241 VV :constraintportinput
21
21
1360
61824
II
II


)3(
2)639(24 I
][
33
24
2 AI 