TWO-PORT NETWORKS
In many situations one is not interested in the internal organization of a
network. A description relating input and output variables may be sufficient
A two-port model is a description of a network that relates voltages and currents
at two pairs of terminals
LEARNING GOALS
Study the basic types of two-port models
Admittance parameters
Impedance parameters
Hybrid parameters
Transmission parameters
Understand how to convert one model into another
ADMITTANCE PARAMETERS
The network contains NO independent sources
The admittance parameters describe the currents in terms of the voltages
y21 determines the current
I1  y11V1  y12V2
The first subindex identifies
the output port. The second
the input port.
flowing into port 2 when the I 2  y21V1  y22V2
port is short - circuited and a
voltage is applied to port 1
The computation of the parameters follows directly from the definition
y11 
I1
V1 V
y12 
I2
V1 V
y22 
2 0
y21 
2 0
I1
V2 V 0
1
I2
V2 V 0
1
LEARNING EXAMPLE
Find the admittance parameters for the network
I1  y11V1  y12V2
I 2  y21V1  y22V2
Circuit used to determine y11, y21
 I2
Circuit used to determine y12 , y22
1
3
I1  (1  )V1  y11  [ S ]
2
2
1
1
1
 I2 
I1  I 2   V1  y21   [ S ]
1 2
2
2
5
 1 1
I 2    V2  y22  [ S ]
6
 2 3
3
3 5
1
 I1 
I2 
V2  y12  [ S ]
23
5 6
2
Next we show one use of this model
An application of the admittance parameters
Determine the current through the
4 Ohm resistor
I1  y11V1  y12V2
I 2  y21V1  y22V2
3
1
I1  V1  V2
2
2
1
5
I 2   V1  V2
2
6
1
I


V2
I1  2 A, V2  4 I 2 2
4
The model plus the conditions at the
ports are sufficient to determine the
other variables.
3
1
2  V1  V2
2
2
1
5 1
0   V1    V2
2
6 4
13
V2
6
8
V2  [V ]
11
2
I 2   [ A]
11
V1 
LEARNING EXTENSION
I1
Find the admittance (Y) parameters
I2

V1

V2


I1
I1  y11V1  y12V2
I 2  y21V1  y22V2
I2
1 1
3
 )V1  V1
21 42
42
42
I2  
I1
21  42
I1  (

V1

I1
1
[S]
14
1
y21   [ S ]
21
y11 
I2

V2

2 1
I 2    V2
 21 21 
10.5
I1  
I2
21  10.5
1
y22  [ S ]
7
1
y12   [ S ]
21
Use the admittance (Y) parameters to find the current Io
LEARNING EXTENSION
IO
I1
10 A

V1
I2

I1  y11V1  y12V2

5 V2

I 2  y21V1  y22V2
1
y22  [ S ]
7
1
y12   [ S ]
21
Conditions at I/O ports
I1  10 A
1
I 2   V2
5
Io   I2
Replace in model
1
1
V1  (5 I o )
14
21
1
1
 I o   V1  (5 I o )
21
7
10 
Solve for variable of interest
1
[S]
14
1
y21   [ S ]
21
y11 
Io 
420
[ A]
98
IMPEDANCE PARAMETERS
The network contains NO independent sources
V1  z11I1  z12 I 2
V2  z21 I1  z22 I 2
The ‘z parameters’ can be derived in a manner similar to the Y parameters
z11 
V1
I1 I
z12 
V1
I2
z21 
2 0
V2
I1
z22 
I1 0
I 2 0
V2
I2
I1 0
LEARNING EXAMPLE
Find the Z parameters
V1  z11I1  z12 I 2
V2  z21 I1  z22 I 2
z11 
z12 
Write the loop equations
V1  2 I1  j 4( I1  I 2 )
V1
I1 I
V1
I2
z21 
2 0
z22 
I1 0
V2  j 2 I 2  j 4( I 2  I1 )
rearranging
V1  (2  j 4) I1  j 4 I 2
V2   j 4 I1  j 2 I 2
 z11  2  j 4 
z12   j 4
z21   j 4
z22   j 2
V2
I1
I 2 0
V2
I2
I1 0
LEARNING EXAMPLE
Use the Z parameters to find the current through the 4 Ohm
resistor
V1  z11I1  z12 I 2
V2  z21 I1  z22 I 2
Output port constraint
V2  4I 2
Input port constraint
V1  120  (1) I1
V1  (2  j 4) I1  j 4 I 2
V2   j 4 I1  j 2 I 2
0   j 4 I1  (4  j 2) I 2
 (3  j 4)
12  (3  j 4) I1  j 4 I 2
 j4
48 j  (16  (4  j 2)(3  j 4)) I 2
 I 2  1.61137.73
LEARNING EXTENSION
I1
I2

V1

V2


Find the Z parameters.
Find the current on a 4 Ohm load with a 24V input source
V1  12 I1  6( I1  I 2 )
V2  3 I 2  6( I1  I 2 )
z11  18, z12  6
z21  6,
z22  9
I1

V1

I2

24V


V2

V1  18 I1  6 I 2
V2  6 I1  9 I 2
output port constraint: V2  4I 2
input port constraint: V1  24[V ]
24  18 I1  6 I 2
0  6 I1  13 I 2
 (3)
24  (39  6) I 2
I2  
24
[ A]
33
4
HYBRID PARAMETERS
The network contains NO independent sources
V1  h11 I1  h12V2
I 2  h21 I1  h22V2
h11 
V1
I1 V
h21 
V1
V2
h22 
2 0
h12 
I1  0
I2
I1 V
h11  short - circuit input impedance
I2
V2
h21  short - circuit forward current gain
2 0
I1  0
h12  open - circuit reverse voltage gain
h22  open - circuit output admittance
These parameters are very common in modeling transistors
Find the hybrid parameters for this circuit
LEARNING EXAMPLE
Non-inverting amplifier
V1  h11 I1  h12V2
Equivalent linear circuit
I 2  h21 I1  h22V2
I R2
I DS
V1  ( Ri  R1 || R2 ) I1  h11  Ri 
I 2   I R 2  I DS  
R1R2
R1  R2
R1
ARi I1
I1 
R1  R2
Ro
 AR
R1 

h21   i 
R
R

R
 o
1
2
V1 
R1
R1
V2  h12 
R1  R2
R1  R2
Vi  0  I 2 
h22 
V2
Ro || ( R1  R2 )
Ro  R1  R2
Ro ( R1  R2 )
LEARNING EXTENSION
I1
Find the hybrid parameters for the network
I2

V1

V2


V1  h11 I1  h12V2
I 2  h21 I1  h22V2
I2
I1

V1


V1
I2
V2  0
V1  (12  (6 || 3)) I1  h11  14
6
2
I2  
I1  h21  
3 6
3
I1  0


V2

V1 
6
2
V2  h12 
3 6
3
I2 
V2
1
 h22  [ S ]
9
9
Determine the input impedance of the two-port
LEARNING EXTENSION
I1
V1  h11 I1  h12V2
I2

V1
I 2  h21 I1  h22V2

V2

V1
I1
4 output port constraint: V2  4I 2

V1  h11 I1  h12 (4 I 2 )
2
3
1
h22  [ S ]
9
h11  14, h12 
2
h21  
3
Rin 
Rin  14 
I 2  h21 I1  h22 (4 I 2 )  I 2 

4h12 h21 

 I1
V1   h11 
1  4h22 

4(2 / 3)(2 / 3)
16
 14   15.23
1  4(1 / 9)
13
Verification
Rin  12  6 || 7  12 
42

13
h21
I1
1  4h22
TRANSMISSION PARAMETERS ABCD parameters
The network contains NO independent sources
V1  AV2  BI 2
I1  CV2  DI 2
A
V1
V2
B
C
I 2 0
V1
I2 V
2 0
I1
V2
D
A  open circuit voltage ratio
I 2 0
B  negative short - circuit transfer impedance
I1
I2 V
2 0
C  open - circuit transfer admittance
D  negative short - circuit current ratio
LEARNING EXAMPLE
Determine the transmission parameters
V1  AV2  BI 2
I1  CV2  DI 2
A
V1
V2
B
C
I 2 0
V1
I2 V
2 0
when I 2  0
1
j
V2 
V  A  1  j
1 1
1
j
V2 
1
I
I1  1  j
j
V2
when V2  0
1
1
j
I2  
I1  
I1
1
1  j
1
j
I1
V2
D
I 2 0
I1
I2 V
2 0
 D  1  j

 2  j 
1 
 (1  j )I 2
V1  1  (1 ||
) I1  

j 

 1  j 
B  2  j
LEARNING EXTENSION
I1

V1
Determine the transmission parameters
V1  AV2  BI 2
I2
I3


V2

when I 2  0
10.5
V1  A  3
10.5  21
42
V
I
1
I3 
I1  2  C  1  [ S ]
42  21
10.5
V2 7
V2 
I1  CV2  DI 2
A
V1
V2
B
C
I 2 0
V1
I2 V
I1
V2
D
2 0
I 2 0
I1
I2 V
2 0
When V2  0
I2  
42
3
I1  D 
42  21
2
3
V1  (42 || 21) I1  14 I1  14  ( I 2 )
2
B  21
PARAMETER CONVERSIONS
If all parameters exist, they can be related by conventional algebraic manipulations.
As an example consider the relationship between Z and Y parameters
V1  z11 I1  z12 I 2
V2  z21 I1  z22 I 2
V1   z11
V    z
 2   21
 y11
y
 21
1
z12   I1   I1   z11
 



z22   I 2   I 2   z21
y12   z11


y22   z21
z12 
z22 
1
1

Z
z12  V1   y11

z22  V2   y21
y12  V1 
y22  V2 
 z22  z12 
 z

 21 z11 
with  Z  z11z22  z21z12
In the following conversion table, the symbol  stands for the determinant of the
corresponding matrix
Z 
z11
z21
z12
y
, Y  11
z22
y21
y12
h
h
A B
,  H  11 12 , T 
y22
h21 h22
C D
INTERCONNECTION OF TWO-PORTS
Interconnections permit the description of complex systems in terms of simpler
components or subsystems
The basic interconnections to be considered are: parallel, series and cascade
PARALLEL: Voltages are the same.
Current of interconnection
is the sum of currents
The rules used to derive models
for interconnection assume that
each subsystem behaves in the
same manner before and after
the interconnection
SERIES: Currents are the same.
Voltage of interconnection is the sum
of voltages
CASCADE:
Output of first subsystem
acts as input for the
second
Parallel Interconnection: Description Using Y Parameters
Interconne ction
descriptio n
 I1   y11
I    y
 2   21
I  YV
y12  V1 
y22  V2 
In a similar manner
 I1a 
V1a 
 y11a y12a 
I a   ,Va   ,Ya  
  I a  YaVa
I b  YbVb
I
V
y
y
 2a 
 2a 
 21a
22 b 
Interconnection constraints :
 I  I a  I b  I  YaVa  YbVb  (Ya  Yb )V
I1  I1a  I1b , I 2  I 2a  I 2b

V  Va  Vb
V1  V1a  V1b , V2  V2a  V2b
Y  Ya  Yb
Series interconnection using Z parameters
SERIES: Currents are the same.
Voltage of interconnection is the sum
of voltages
Description of each subsystem
Va  Z a I a , Vb  Z b I b
Interconnection constraints
Ia  Ib  I
 V  Za I  Zb I  ( Za  Zb ) I
V  Va  Vb
Z  Za  Zb
Cascade connection using transmission parameters
CASCADE:
Output of first subsystem
acts as input for the
second
Interconnection constraints :
I 2a   I1b
V2a  V1b
V1  V1a
V2  V2b
I1  I1a
I 2  I 2b
V1a   Aa
 I   C
 1a   a
Ba   V2a 
Da   I 2a 
V1b   Ab
 I   C
 1b   b
Bb   V2b 
Db   I 2b 
V1   Aa
 I   C
 1  a
V1  AV2  BI 2
I1  CV2  DI 2
V1   A
 I   C
 1 
B   V2 
D   I 2 
Matrix multiplication does not commute.
Order of the interconnection is important
Ba   Ab
Da  Cb
Bb   V2 
Db   I 2 
Find the Y parameters for the network
LEARNING EXAMPLE
 j2
I1

V1


 y   1  j 1 S, y   j 1
11a
12 a
1
1 
 3
1
2
j
2
2
V2

j


j


 5
2
5
2

 [S]
1
1
Y 


y21a   j S ; y22a  j S
1
1
2
1


2
2
   j 

j
  5
2
5
2 
V1  V2   j 2 I1
I 2   I1
I1

V1
I2
I2
1
2
1

V2

V1  2 I1  I 2
V2  I1  3I 2
1
2 1
1  3  1
Yb  
  5  1 2 
1
3




Find the Y parameters for the network using a direct approach
I1
Vx
I2

V1

V2


Vx Vx  V1 Vx  V2
2V  V


 0  Vx  1 2
1
1
2
5
V1  V x V1  V2

1
 j2
V  V x V2  V1
I2  2

2
 j2
I1 
Replace Vx and rearrange
LEARNING EXAMPLE
Find the Z parameters of the network
Network A
Use direct method,
or given the Y parameters transform to Z
… or decompose the network in a series
connection of simpler networks
2  2 j
3  2 j
Za  
 2
 3  2 j
2 
3 2 j
2  4 j 
3  2 j 
1 1
Zb  

1 1
5  4 j
3  2 j
Z  Za  Zb  
5  2 j
 3  2 j
Network B
5 2 j
3 2 j
5  6 j 
3  2 j 
LEARNING EXAMPLE
Find the transmission parameters
By splitting the 2-Ohm resistor,
the network can be viewed as the
cascade connection of two identical
networks
 A B  1  j
C D    j

 
2  j  1  j
1  j   j
 A B  1  j
C D    j

 
2  j 
1  j 
2  j 
1  j 
(1  j )(2  j )  (2  j )(1  j )
 A B   (1  j ) 2  (2  j ) j


C D  
2
j (2  j )  (1  j )

  j (1  j )  (1  j )( j )

 A B  1  4 j  2 2
C D   
2

  2 j  2
4  6 j  2 2 

1  4 j  2 2 
LEARNING by APPLICATION
Given the demand at the receiving end, determine
the conditions on the sending end
V1  AV2  BI 2
I1  CV2  DI 2
Transmission parameters
are best suited for this
application
In the next slide we show how to determine the transmission parameters for the
line. Here we assume them known and use them for analysis
| V2 | VL  300kV (line voltage)  V2 | VL | 0
P
P  3 | VL || I L |  pf  | I || I |
L
2
3VL pf
P
1
 I2 
  cos pf
3VL pf
Conditions at the
sending end
V1  AV2  BI 2
Preceived


I1  CV2  DI 2
Psent
V1 
 pfsending  Psending
I1 
Ptotal  3 |Vline || I line | cos f
Qtotal  3 |Vline || I line | sin  f
Vline
f

Power factor angle
I line
Determining the transmission parameters for the line
V1  AV2  BI 2
I1  CV2  DI 2
I2  0  A 
V2  0  B  
V1 R  Z L  ZC

 0.95900.27
V2
ZC
V1
I2
V1  ( R  Z L ) I 2  B  R  Z L  100.0084.84
IT 
IT
ZC
V
I1  2
R  Z L  2 ZC
ZC
I2  0  C 
V2  0  D  
R  Z L  2 ZC
I1
C
 975.1090.13S
V2
Z C2
I1 R  Z L  ZC

 0.95900.27
I2
ZC
LEARNING EXAMPLE
Determine the effect of the load on the voltage gain
A  20,000, Ri  1M, Ro  500, R1  1k, R2  49k
Hybrid parameters are computed in next slide
Ideal gain  1 
G
R2
 50
R1
 h21
h11h22  h12 h21 
Using the hybrid parameters
V1  h11 I1  h12V2
I 2  h21 I1  h22V2
Eliminating I1 and solving for V2
V2 
 h21V1  h11I 2
h11h22  h12 h21
Constraint at output port : V2   RL I 2
Solving for V2
 h21
V2 
V
h11 1
h11h22  h12 h21 
RL
h11
RL

49.88
1.247
1
RL
Effect of load resistance
Computing the hybrid parameters for non-inverting amplifier (repeat earlier example)
Non-inverting amplifier
V1  h11 I1  h12V2
Equivalent linear circuit
I 2  h21 I1  h22V2
I R2
I DS
V1  ( Ri  R1 || R2 ) I1  h11  Ri 
I 2   I R 2  I DS  
R1R2
R1  R2
R1
ARi I1
I1 
R1  R2
Ro
 AR
R1 

h21   i 
R
R

R
 o
1
2
V1 
R1
R1
V2  h12 
R1  R2
R1  R2
Vi  0  I 2 
h22 
V2
Ro || ( R1  R2 )
Ro  R1  R2
Ro ( R1  R2 )
LEARNING BY DESIGN
Gain required = 10,000 on a load of 1kOhm
Amplifier parameters
R
Ideal gain  1  2  10,000
A  20,000
R1
Ri  1M
R  9.999 M
Ro  500
2
R1  1k
h11  1.001M, h12  1.0  104
h21  4.0  107 , h22  1.0mS
From the conversion table
  H h12 h21  h11h22
h
A

B   11
h21
h21
h21
C 
h22
h21
D
V1  AV2  BI 2
I1  CV2  DI 2
eliminating I1
1
h21
V2   RL I 2
V2
1
6667


V1 A  B 1  166.7
RL
RL
For the final solution we will need to cascade
amplifiers. Hence the transmission parameters
will prove very useful
Analysis of solution:
-Even with infinite load the maximum gain
is only 6,667
Likely causes:
-R2 is higher than input resistance Ri
-Desired gain is comparable to the maximum
gain, A, of the Op-Amp
Proposed solution:
-Cascade two stages, each with ideal gain of
100. This also lowers R2 to 99kOhm
Analysis of proposed solution
Since the two stages will be cascaded, the transmission
parameters of the proposed solution will be
V1   Aa
 I   C
 1  a
Ba   Ab
Da  Cb
1
A  Aa Ab  Ba Cb  V2 
V1 A  B
B  Aa Bb  Ba Db
RL
Bb  V2   Aa




Db   I 2  Ca
Ba   Ab
Da  Cb
 Bb   V2 
 Db   I 2 
Effect of load resistance. G=10,000
Identical stages
Two-Ports