Download Two – port network

TWO PORT
NETWORKS
1
SUB - TOPICS
Z – PARAMETER
Y – PARAMETER
T (ABCD) – PARAMETER
TERMINATED TWO PORT NETWORKS
2
OBJECTIVES
• TO UNDERSTAND ABOUT TWO – PORT
NETWORKS AND ITS FUNTIONS.
• TO UNDERSTAND THE DIFFERENT
BETWEEN Z – PARAMETER, Y –
PARAMETER, T – PARAMETER AND
TERMINATED TWO PORT NETWORKS.
• TO INVERTIGATE AND ANALYSIS THE
BEHAVIOUR OF TWO – PORT NETWORKS.
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TWO – PORT NETWORKS
• A pair of terminals through which a
current may enter or leave a network is
known as a port.
• Two terminal devices or elements (such as
resistors, capacitors, and inductors)
results in one – port network.
• Most of the circuits we have dealt with so
far are two – terminal or one – port
circuits.
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• A two – port network is an electrical
network with two separate ports for
input and output.
• It has two terminal pairs acting as
access points. The current entering
one terminal of a pair leaves the
other terminal in the pair.
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I
+
V
-
Linear network
I
One – port network
I1
+
V1
-
I2
+
V2
-
Linear network
I1
I2
Two – port network
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• Two (2) reason why to study two port –
network:
Such networks are useful in
communication, control system, power
systems and electronics.
Knowing the parameters of a two – port
network enables us to treat it as a
“black box” when embedded within a
larger network.
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• From the network, we can observe that
there are 4 variables that is I1, I2, V1and
V2, which two are independent.
• The various term that relate these
voltages and currents are called
parameters.
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Z – PARAMETER
• Z – parameter also called as impedance
parameter and the units is ohm (Ω)
• Impedance parameters is commonly used
in the synthesis of filters and also useful in
the design and analysis of impedance
matching networks and power distribution
networks.
• The two – port network may be voltage –
driven or current – driven.
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• Two – port network driven by voltage
source.
I1
V1
+

I2
+

Linear network
V2
• Two – port network driven by current
sources.
I1
+
V1
-
Linear network
+
V2
I2
-
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• The “black box” is replace with Z-parameter
is as shown below.
I2
I1
+
V1
-
Z11
Z12
Z21
Z22
+
V2
-
• The terminal voltage can be related to the
terminal current as:
V1  z11I1  z12 I 2
(1)
V2  z 21I1  z 22 I 2
(2)
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• In matrix form as:
V1   z11
V    z
 2   21
z12   I1 



z22   I 2 
• The Z-parameter that we want to
determine are z11, z12, z21, z22.
• The value of the parameters can be
evaluated by setting:
1. I1= 0 (input port open – circuited)
2. I2= 0 (output port open – circuited)
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• Thus,
V1
z11 
I1
V2
z 21 
I1
I 2 0
V1
z12 
I2
I1  0
I 2 0
V2
z 22 
I2
I1  0
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• Where;
z11 = open – circuit input impedance.
z12 = open – circuit transfer impedance from
port 1 to port 2.
z21 = open – circuit transfer impedance from
port 2 to port 1.
z22 = open – circuit output impedance.
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Example 1
Find the Z – parameter of the circuit below.
I2
I1
+
V1
+
240Ω
120Ω
_
V2
_
40Ω
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Solution
i)
I2 = 0(open circuit port 2). Redraw the
circuit.
Ia
I1
+
V1
240Ω
Ib
120Ω
_
+
V2
_
40Ω
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V1  120 I b .......(1) V2  240 I a .......(3)
280
120
Ib 
I1......(2) I a 
I1.......(4)
400
400
sub (1)  (2)
sub (4)  (3)
V1
V2
 Z11   84  Z 21 
 72
I1
I1
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ii) I1 = 0 (open circuit port 1). Redraw the
circuit.
Iy
I2
+
V1
+
120Ω
240Ω
_
Ix
V2
_
40Ω
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V2  240 I x .......(1)
V1  120 I y .......(3)
160
Ix 
I 2 .......( 2)
400
sub (1)  (2)
V2
 Z 22 
 96
I2
240
Iy 
I 2 .......(4)
400
sub (4)  (3)
V1
 Z12   72
I2
In matrix form:
84 72
Z   

72 96
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Example 2
Find the Z – parameter of the circuit below
I1
+
V1
_
2Ω
10Ω
j4Ω
+
_
I2
+
10I2
-j20Ω
V2
_
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Solution
i) I2 = 0 (open circuit port 2). Redraw the circuit.
V1  I1 (2  j4)
I1
2Ω
j4Ω
I2 = 0
+
+
V1
_
V2
_
V1
 Z11   (2  j4) 
I1
V2  0 (short circuit)
 Z 21  0
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ii) I1 = 0 (open circuit port 1). Redraw the circuit.
I1 = 0
10Ω
+
V1
_
+
_
I2
V1  10I 2
+
10I2
-j20Ω
V2
_
In matrix form;
0 
(2  j4)
Z  

10
(16
j8)


V1
 Z12 
 10
I2
V2
V2 - 10I 2
I2 

 j20
10
1
 j
2I 2  V2 
 
 20 10 
V2
 Z 22 
 (16 - j8) 
I2
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Y - PARAMETER
• Y – parameter also called admittance
parameter and the units is siemens (S).
• The “black box” that we want to replace
with the Y-parameter is shown below.
I2
I1
+
V1
-
Y11
Y12
Y21
Y22
+
V2
-
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• The terminal current can be expressed in
term of terminal voltage as:
I1  y11V1  y12V2
(1)
I 2  y21V1  y22V2
(2)
• In matrix form:
 I1   y11
I    y
 2   21
y12  V1 



y22  V2 
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• The y-parameter that we want to determine
are Y11, Y12, Y21, Y22. The values of the
parameters can be evaluate by setting:
i) V1 = 0 (input port short – circuited).
ii) V2 = 0 (output port short – circuited).
• Thus;
I1
Y11 
V1
Y21
I2

V1
V2  0
I1

V2
V1  0
V2  0
I2

V2
V1  0
Y12
Y22
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Example 1
Find the Y – parameter of the circuit shown
below.
I1
5Ω
I2
+
V1
_
+
20Ω
15Ω
V2
_
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Solution
i)
V2 = 0
5Ω
I1
+
V1
_
Ia
20Ω
I2
V1  20 I a .......(1)
5
Ia 
I1.......(2)
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sub (1)  (2)
I1
1
 Y11 
 S
V1
4
V1  5 I 2
I2
1
 Y21    S
V1
5
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ii) V1 = 0
I1
V2  15 I x .......(3)
5Ω
I2
+
15Ω
Ix
V2
_
In matrix form;
 1
 4
Y    1

 5
1
 
5 S
4 

15 
5
Ix 
I 2 .......( 4)
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sub (3)  (4)
I2
4
 Y22 
 S
V2 15
V2  5I1
I1
1
Y12 
 S
V2
5
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Example 2 (circuit with
dependent source)
Find the Y – parameters of the circuit
shown.
I1
+
V1
_
2Ω
10Ω
j4Ω
+
_
I2
+
10I2
-j20Ω
V2
_
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Solution
i) V2 = 0 (short – circuit port 2). Redraw the
circuit.
I1
2Ω
+
10Ω
j4Ω
+
V1
_
_
I2
10I2
I0
V1  (2  j4)I1
I1
1
 Y11 

 (0.1 - j0.2) S
V1 2  j4
I2
 Y21 
 0S
V1
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ii) V1 = 0 (short – circuit port 1). Redraw the circuit.
I1
2Ω
10Ω
j4Ω
+
_
- 10I 2
........(1)
2  j4
V2
V2 - 10I 2
I2 

- j20
10
I1 
1
1 
.......(2)
2I 2  V2  
 10 - j20 
I2
+
10I2
-j20Ω
V2
_
I2
 Y22 
 (0.05  j0.025) S
V2
sub (2)  (1)
I
Y12  1  (-0.1  j0.075) S
V2
In matrix form;
0.1  j0.2  0.1  j0.075
 Y   
S

0
0.05  j0.025  31

T (ABCD) PARAMETER
• T – parameter or ABCD – parameter is a
another set of parameters relates the
variables at the input port to those at the
output port.
• T – parameter also called transmission
parameters because this parameter are
useful in the analysis of transmission lines
because they express sending – end
variables (V1 and I1) in terms of the
receiving – end variables (V2 and -I2).
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• The “black box” that we want to replace with T –
parameter is as shown below.
I2
I1
+
V1
-
A11
B12
C21
D22
+
V2
-
• The equation is:
V1  AV2  BI 2 .......(1)
I1  CV2  DI 2 .......(2)
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• In matrix form is:
V1   A B   V2 
 I   C D  I 
 2 
 1 
• The T – parameter that we want
determine are A, B, C and D where A and
D are dimensionless, B is in ohm (Ω) and
C is in siemens (S).
• The values can be evaluated by setting
i) I2 = 0 (input port open – circuit)
ii) V2 = 0 (output port short circuit)
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• Thus;
V1
A
V2
I1
C
V2
I 2 0
V1
B
I2
V2  0
I 2 0
I1
D
I2
V2  0
• In term of the transmission parameter, a
network is reciprocal if;
AD - BC  1
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Example
Find the ABCD – parameter of the circuit
shown below.
I1
2Ω
4Ω
+
V1
_
I2
+
10Ω
V2
_
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Solution
i) I2 = 0,
I1
V2  10 I1
2Ω
+
V1
_
+
10Ω
V2
_
I1
C 
 0.1S
V2
V1  2 I1  V2
6
 V2 
V1  2   V2  V2
5
 10 
V1
A
 1.2
V2
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ii) V2 = 0,
I1
2Ω
4Ω
+
V1
10Ω
I1 + I2
_
I2
10
I2  
I1
14
I1
D  
 1.4
I2
V1  2 I1  10I1  I 2 
V1  12 I1  10 I 2
1.2 6.8
T   

0.1 1.4 
 14 
V1  12 
I 2   10 I 2
 10 
V1
B  
 6.8
I2
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TERMINATED TWO – PORT
NETWORKS
• In typical application of two port network,
the circuit is driven at port 1 and loaded at
port 2.
• Figure below shows the typical terminated
2 port model.
Zg
Vg
+

I1
+
V1
-
I2
Two – port
network
+
V2
-
ZL
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• Zg represents the internal impedance of
the source and Vg is the internal voltage of
the source and ZL is the load impedance.
• There are a few characteristics of the
terminated two-port network and some of
them are;
V1
i)
input impedance, Zi 
ii)
output impedance, Z o 
iii)
current gain, A i 
I2
I1
iv) voltage gain, A v 
V2
V1
v)
I1
V2
I2
overall voltage gain, A g 
V2
Vg
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• The derivation of any one of the desired
expression involves the algebraic
manipulation of the two – port equation. The
equation are:
1) the two-port parameter equation either Z
or Y or ABCD.
For example, Z-parameter,
V1  Z11I1  Z12I 2 .......(1)
V2  Z21I1  Z22I 2 .......(2)
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2) KVL at input,
V1  Vg  I1Zg .......(3)
3) KVL at the output,
V2  I2 ZL .......(4)
• From these equations, all the characteristic
can be obtained.
42
Example 1
For the two-port shown below, obtain the
suitable value of Rs such that maximum
power is available at the input terminal. The
Z-parameter of the two-port network is given
as
 Z11 Z12  6 2

Z


Z
4
4


22 
 21
V2
With Rs = 5Ω,what would be the value of
Vs
Rs
Vs
+

I1
+
V1
-
I2
Z
+
V2
-
4Ω
43
Solution
1) Z-parameter equation becomes;
V1  6 I1  2 I 2 .......(1)
V2  4 I1  4 I 2 .......(2)
2) KVL at the output;
V2  4I 2 .......(3)
Subs. (3) into (2)
I1
I 2   .......( 4)
2
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Subs. (4) into (1)
V1  5I1.......(5)
V1
 Z1   5
I1
For the circuit to have maximum power,
Rs  Z1  5
45
V2
Vs
To find
at max. power transfer, voltage
drop at Z1 is half of Vs
Vs
V1  .......(6)
2
From equations (3), (4), (5) & (6)
Overall voltage gain,
V2 1
Ag 

Vs 5
46
Example 2
The ABCD parameter of two – port network shown
below are.
20
 4
0.1S

2 

The output port is connected to a variable load for
a maximum power transfer. Find RL and the
maximum power transferred.
47
Solution
ABCD parameter equation becomes
(1)
V1 = 4V2 – 20I2
(2)
I1 = 0.1V2 – 2I2
At the input port, V1 = -10I
(3)
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(3) Into (1)
-10I1 = 4V2 – 20I2
I1 = -0.4V2 2I2
(4)
(2) = (4)
0.1V2 – 2I2 = -0.4V2 + 2I2
0.5V2 = 4I2
(5)
From (5);
ZTH = V2/I2 = 8Ω
(6)
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But from Figure (b), we know that
V1 = 50 – 10I1 and I2 =0
Sub. these into (1) and (2)
50 – 10I1 = 4V2
(7)
I1 = 0.1V2
(8)
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Sub (8) into (7)
V2 = 10
Thus, VTH = V2 = 10V
RL for maximum power transfer,
RL = ZTH = 8Ω
The maximum power
P = I2RL = (VTH/2RL)2 x RL = V2TH/4RL = 3.125W
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