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Chapter 8 Two Port Network & Network Functions
A pair of terminals through which a current may enter or leave a network is known as a port. A
port is an access to the network and consists of a pair of terminals; the current entering one
terminal leaves through the other terminal so that the net current entering the port equals zero.
For example, most circuits have two ports. We may apply an input signal in one port and obtain
an output signal from the other port. The parameters of a two-port network completely describes
its behaviour in terms of the voltage and current at each port. Thus, knowing the parameters of a
two port network permits us to describe its operation when it is connected into a larger network.
Two-port networks are also important in modeling electronic devices and system components.
For example, in electronics, two-port networks are employed to model transistors and Op-amps.
Other examples of electrical components modeled by two-ports are transformers and
transmission lines.
Four popular types of two-port parameters are examined here: impedance, admittance, hybrid,
and transmission. We show the usefulness of each set of parameters, demonstrate how they are
related to each other.
Learning Outcomes:
At the end of this module, students will be able to:
1.Differentiate one port and two port network devices.
2.Calculate two port network parameters such as z, y, ABCD and
h parameters for given electrical network.
3.Relate different two port network parameters.
4.Simplify the complex network such as cascade, parallel networks using
fundamental two port network parameters.
5.Find the various driving point & transfer functions of two port network.
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A Typical one port or two terminal network is shown in figure 1.1. For example resistor,
capacitor and inductor are one port network.
Fig.1.1
Fig. 1.2 represents a two-port network.A four terminal network is called a two-port network
when the current entering one terminal of a pair exits the other terminal in the pair. For example,
I1 enters terminal ‘a’ and exit terminal ‘b’ of the input terminal pair ‘a-b’. Example for four-
terminal or two-port circuits are op amps, transistors, and transformers.
Fig.1.2
To characterize a two-port network requires that we relate the terminal quantities
2121 ,, IandIVV .The various terms that relate these voltages and currents are called parameters.
Our goal is to derive four sets of these parameters.
1.3 Open circuit Impedance Parameter (z Parameter):
Let us assume the two port network shown in figure is a linear network then using superposition
theorem, we can write the input and output voltages as the sum of two components, one due to I1
and other due to I2:
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2221212
2121111
IzIzV
IzIzV
Putting the above equations in matrix form, we get
2
1
2
1
2221
1211
2
1][
I
Iz
I
I
zz
zz
V
V
the z terms are called the z parameters, and have units of ohms. The values of the parameters can
be evaluated by setting 01 I or 02 I .
The z parameters are defined as follows:
Thus
01
111
2
II
Vz
01
221
2
II
Vz
02
112
1
II
Vz
02
222
1
II
Vz
In the preceding equations, letting 01 I or 02 I is equivalent to open-circuiting the input or
output port. Hence, the z parameters are called open-circuit impedance parameters.
Here 11z is defined as the open-circuit input impedance, 22z is called the open-circuit output
impedance, and 12z and 21z are called the open-circuit transfer impedances.
If 12z = 21z , the network is said to be reciprocal network. Also, if 11z = 22z then the network is
called a symmetrical network.
We obtain 11z and 21z by connecting a voltage 1V (or a current source 1I ) to port 1 with port 2
open-circuited as in fig.
Similarly 12z and 22z by connecting a voltage 2V (or a current source 2I ) to port 2 with port 1
open-circuited as in fig.
A two-port is reciprocal if interchanging an ideal voltage source at one port with an ideal
ammeter at the other port gives the same ammeter reading.
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Example 8.1
Determine the z parameters for the circuit in the following figure and then compute the
current in a 4Ω load if a 0024 V source is connected at the input port.
To find 11z and 21z , the output terminals are open circuited. Also connect a voltage source 1V to
the input terminals. This gives a circuit diagram as shown in Fig
Applying KVL to the left-mesh, we get
111 612 VII
11 18IV
Hence
01
111
2II
Vz 18Ω
Applying KVL to the right-mesh, we get
0603 12 IV
12 6IV
Hence
01
221
2
II
Vz = 6 Ω
To find 12z and 22z , the input terminals are open circuited. Also connect a voltage source V2 to
the output terminals. This gives a circuit diagram as shown in Fig.
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Applying KVL to the left-mesh, we get
21
21
6
6012
IV
IV
02
112
1
II
Vz =6 Ω
Applying KVL to the right-mesh, we get
063 222 IIV
22 9IV
02
222
1
II
Vz =9 Ω
The equations for the two-port network are, therefore
211 618 IIV (1)
212 96 IIV (2)
The terminal voltages for the network shown in Fig.8.2 are 0
1 024V (3)
22 4IV (4)
Fig.8.2
Combining equations (1) and (2) with equations (3) and (4) yields
21
0 618024 II
21 1360 II
On Solving, we get 0
2 073.0 I A
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1.4 Admittance Parameter( y Parameter):
The terminal currents can be expressed in terms of the terminal voltages: The y terms are known
as the admittance parameters (or, simply, y parameters) and have units of siemens.
2221212
2121111
VyVyI
VyVyI
Putting the above equations in matrix form, we get
2
1
2
1
2221
1211
2
1][
V
Vy
V
V
yy
yy
I
I
The y terms are called the y parameters, and have units of seimens. The values of the parameters
can be evaluated by setting 01 V or 02 V .
The y parameters are defined as follows:
Thus
01
111
2
VV
Iy
01
221
2
VV
Iy
02
112
1
VV
Iy
02
222
1
VV
Iy
In the preceding equations, letting 01 V or 02 V is equivalent to short-circuiting the input or
output port. Hence, the y parameters are called short-circuit admittance parameters.
If 12y = 21y , the network is said to be reciprocal network. Also, if 11y = 22y then it is called a
symmetrical network.
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A reciprocal network ( 12y = 21y ) can be modeled by the equivalent circuit in Fig.8.3
Fig.8.3
Example 8.4
Determine the admittance parameters of the T network shown in Fig.8.4
Fig.8.4
To find 11y and 21y , we have to short the output terminals and connect a current source 1I to the
input terminals. The circuit so obtained is shown in Fig.
5
22
224
111
VVI
Hence SV
Iy
V5
1
01
111
2
Using the principle of current division,
222
2 112
III
Appliying KVL to first loop => 211 105 IIV
Hence SV
Iy
V10
1
01
221
2
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To find 12y and 22y , we have to short-circuit the input terminals and connect a current source
2I to the output terminals. The circuit so obtained is shown in Fig.
10
3
24
242
222
VVI
Hence SV
Iy
V10
3
02
222
1
3
42
2
21
21
II
II
Applying KVL to loop 2
1122 1024 IIIV
Hence SV
Iy
V10
1
02
112
1
It may be noted that, 12y = 21y therefore the network is reciprocal .
Thus, in matrix form we have
YVI
2
1
2
1
10
3
10
110
1
5
1
V
V
I
I
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Example 8.5
For the network shown in Fig.8,5 determine the y parameters.
Fig.8.5 Fig.8.6
To find 11y and 21y short the output terminals and connect a current source 1I to the input
terminals. The resulting circuit diagram is as shown in Fig. 8.6
KCL at node 1V :
2
11
111
aVVVI
11 23 IVV a
KCL at node aV :
021
0
2
1 11
V
VVV aa
0aV
Substituting this in above equation => 113 IV
SV
Iy
V
3
01
111
2
SV
Iy
V
0
01
221
2
To find 22y and 12y short-circuit the input terminals and connect a current source I2 to the
output terminals. The resulting circuit diagram is shown in Fig. 8.6
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Fig.8.6
KCL at node V2: 222
1
2
1I
VVV a
223 IVV a …(5)
KCL at node Va:
001
2
1
0 2
VVV aa
03 2 VVa
23
1VVa ………(6)
Substituting equation (6) in equation (5) yields
22
23
3 IV
V
223
8IV
Hence
SV
Iy
V3
8
02
222
2
We have 23
1VVa
Also 031 II
31 II
aa V
V2
2
1 ………(7)
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Making use of equation (7) in (6) yields
21
3
1
2V
I
Hence SV
Iy
V3
2
02
112
1
1.6 Transmission Parameters:
The transmission parameters are defined by the equations:
221
221
DICVI
BIAVV
Putting the above equations in matrix form, we get
2
2
2
2
1
1][
I
VT
I
V
DC
BA
I
V
02
1
2
IV
VA
02
1
2
IV
IC
02
1
2
VI
VB
02
1
2
VI
ID
A, B, C and D parameters represent the open-circuit voltage ratio, the negative short-circuit
transfer impedance, the open-circuit transfer admittance, and the negative short-circuit current
ratio, respectively.
Example 8.7
Determine the transmission parameters in the s domain for the network shown in Fig.8.7
Fig.8.7 Fig.8.8
The s domain equivalent circuit with the assumption that all the initial conditions are zero is
shown in Fig. 8.8
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To find the parameters A and C, open-circuit the output port and connect a voltage source 1V at
the input port as shown in Fig. below (Left side)
1
1
111
112
111
s
VI
sV
s
sV
s
VI
Therefore 1
02
1
2
sV
VA
I
And sV
IC
I
02
1
2
To find the parameters B and D, short-circuit the output port and connect a voltage source 1V to
the input port as shown in Fig. above (Right)
Applying the current division formula
1
112
sII
Hence )1(
02
1
2
sI
ID
V
The total impedance as seen by the source 1V is
)2(
)1()1(
1
2
11
11
1
121
1
1
s
sVsII
I
V
s
s
s
sZ
Hence )2(
02
1
2
sI
VB
V
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Example 8.9
Find the transmission parameters for the network shown in Fig.8.9
Fig.8.9
To find the parameters A and C, open the output port and connect a voltage source V1 to the
input port as shown in Fig.8.10
Fig.8.10
Applying KVL to the input loop, we get
2
3
1
3
1 10105.1 VIV
Also KCL at node a gives
01040
403
21
VI
2
6
3
21 1025.6
10160V
VI
Substitute the value of 1I in the preceding loop equation we get
2
3
2
63
1 10)1025.6(105.1 VVV
2
3
1 10375.8 VV
Hence 3
02
1 10375.8
2
IV
VA
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6
02
1 1025.6
2
IV
IC
To find the parameters B and D, refer the circuit shown in Fig.8.11
Fig.8.11
Applying KCL at node b, we find
21 040 II
12 40II
40
1
02
1
2
VI
ID
Applying KVL to the input loop, we get
40105.1
105.1
23
1
1
3
1
IV
IV
5.3740
105.1 3
02
1
2VI
VB
1.7 Hybrid Parameters ( h parameters):
2221212
2121111
VhIhI
VhIhV
Putting the above equations in matrix form, we get
2
1
2
1
2221
1211
2
1][
V
Ih
V
I
hh
hh
I
V
01
111
2
VI
Vh
01
221
2
VI
Ih
02
112
1
IV
Vh
02
222
1
IV
Ih
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The parameters 11h , 12h , 21h and 22h represent the short circuit input impedance, the open circuit
reverse voltage gain, the short-circuit forward current gain, and the open-circuit output
admittance respectively. Because of this mix of parameters, they are called hybrid parameters.
Example 8.12
For the network shown in Fig.8.12 determine the h parameters.
Fig.8.12
To find 11h and 21h short-circuit the output terminals so that 02 V .Also connect a current
Source 1I to the input port as in Fig.
Applying KCL at node x
00
11
IR
V
R
VI
C
x
B
x
CB
xRR
VI11
11
CB
CBx
RR
RRIV
1)1(
Hence,
01
111
2
VI
Vh
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A
CB
CB
A
CB
CB
V
Ax
RRR
RR
RIRR
RRI
I
RIV
)(
)1(
)(
)1(
1
1
01
1
2
KCL at node y
00
0
21
321
C
x
R
VII
III
Substitute for xV in the above equation and simplifying results in
)(
)(
01
221
2BC
BC
VRR
RR
I
Ih
To find 12h and 22h open-circuit the input port so that 1I . Also, connect a voltage source 2V
between the output terminals as shown in Fig.
KCL at node y:
01211
I
R
VV
R
V
CB
Since 01 I we get
CCB
CCB
R
V
RRV
R
V
R
V
R
V
21
211
11
0
CB
B
IRR
R
V
Vh
02
112
1
Applying KVL to the output mesh, we get
0)( 2212 IRIIRV BC
Since 01 I we get
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222 VIRIR BC
Hence, BCI
RRV
Ih
1
02
222
1
Example 8.13
Determine the h parameters of the circuit shown in Fig.8.13
Fig.8.13
Performing Δ to Y transformation, the network shown in Fig.8.13 takes the form as shown in
Fig.8.14
Since all the resistors are of same value RRY
3
1
Fig.8.14
To find h11 and h21, short-circuit the output port and connect a current source I1 to the input
port as in Fig. 8.15
Fig.8.15
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)]44(4[11 IV
11 6IV
Hence
6
01
111
2VI
Vh
Using the principle of current division
444
12
II
2
12
II
2
1
01
221
2
VI
Ih
To find h12 and h22, open-circuit the input port and connect a voltage sourceV2 to the output port
as shown in Fig. 8.16
Fig.8.16
Using the principle of voltage division, we get
444
21
VV
2
1
02
112
1
IV
Vh
222 8)44( IIV
Hence, SV
Ih
I8
1
02
222
1
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1.8 Relationship between the two port parameters:
If all the two-port parameters for a network exist, it is possible to relate one set of parameters to
another, since these parameters interrelate the variables 1V , 1I , 2V and 2I .
1.9 Relation between the z parameters and y parameters:
Or
2
11
2
1][
V
Vz
I
I
And y Parameters are related by
2
1
2
1
2221
1211
2
1][
V
Vy
V
V
yy
yy
I
I
Equating above two equations we see that
1 zy
To find 1z
z
zadjoz
)int(1 where 21122211
2221
1211zzzz
zz
zzz
And
1121
1222)int(
zz
zzzadjo
Hence,
z
z
z
zz
z
z
z
yy
yy
1121
1222
2221
1211
1.10 Relation between z and h parameters:
2221212
2121111
IzIzV
IzIzV
From second equation
2
22
1
22
212
1V
zI
z
zI
Substitute this in 1st equation
2
22
121
22
211222111 V
z
zI
z
zzzzV
Writing above equation in matrix form
2
1
2
1
2221
1211
2
1][
I
Iz
I
I
zz
zz
V
V
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2
1
2222
21
22
12
22
2
1
1 V
I
zz
z
z
z
z
z
I
V
Compare this with h parameter equation
2
1
2221
1211
2
1
V
I
hh
hh
I
V
2222
21
22
12
22
2221
1211
1
zz
z
z
z
z
z
hh
hh
Similarly all parameters can be related as shown in Table 1.11
Table 1.11
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Example 8.17
Determine the y parameters for a two-port network if the z parameters are:
95
510z
6555910 z
65
10
65
565
5
65
9
1121
1222
2221
1211
z
z
z
zz
z
z
z
yy
yy
Example 8.18
Following are the hybrid parameters for a network:
63
25h Determine the y parameters for the network.
Solution
242365 h
5
24
5
35
6
5
11
1111
21
11
22
11
2221
1211
h
h
h
h
h
h
h
yy
yy
Example 8.19
Determine the y and z parameters for a two-port network.
Applying KCL to the node 1, 121 5.05.1 IVV
Applying KCL to the node 2, 1221 35.0 IIVV
Writing above equations in the matrix form
2
1
2
1
13
01
15.0
5.05.1
I
I
V
V
2
1
1
2
1
13
01
15.0
5.05.1
I
I
V
V
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=
2
1
2.12.3
4.04.0
I
I
Therefore
2.12.3
4.04.0][z
Hence
5.04
5.05.1][ 1zy
1.12 Interconnection of Two port network:
A large, complex network may be divided into subnetworks for the purposes of analysis and
design. The subnetworks are modeled as two-port networks, interconnected to form the original
network. The two-port networks may therefore be regarded as building blocks that can be
interconnected to form a complex network. The interconnection can be in series, in parallel, or in
cascade. Although the interconnected network can be described by any of the six parameter sets,
a certain set of parameters may have a definite advantage. For example, when the networks are in
series, their individual z parameters add up to give the z parameters of the larger network. When
they are in parallel, their individual y parameters add up to give the y parameters of the larger
network. When they are cascaded, their individual transmission parameters can be multiplied
together to get the transmission parameters of the larger network.
The basic interconnections to be considered are: parallel, series and cascade.
PARALLEL: Voltages are the same.Current of interconnection is the sum of currents.
SERIES: Currents are the same. Voltage of interconnection is the sum of voltages.
CASCADE: Output of first subsystem acts as input for the second.
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1.13 Networks in parallel:
Two two-port networks are in parallel when their port voltages are equal and the port currents of
the larger network are the sums of the individual port currents. In addition, each circuit must
have a common reference and when the networks are connected together, they must all have their
common references tied together.
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Fig.8.20
YVI
V
V
yy
yy
I
I
2
1
2221
1211
2
1
aaa
aa
aa
a
a
a
a
a
a
a
VYI
yy
yyY
V
VV
I
II
2221
1211
2
1
2
1,,
bbb VYI
mannersimilar a In
baba
baba
VVVVVV
IIIIII
222111
222111
,
,
ba
ba
VVV
III
VYYVYVYI babbaa )(
ba YYY
Hence if two port sub networks are connected in parallel then the resultant network y parameters
can be calculated by adding individual y parameters of sub networks.
1.14 Neworks in series
The networks are regarded as being in series because their input currents are the same and their
voltages add. In addition, each network has a common reference, and when the circuits are
placed in series, the common reference points of each circuit are connected together.
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Fig.8.21
Description of subnetworks
bbbaaa IZVIZV ,
Where
bb
bb
b
aa
aa
azz
zzZ
zz
zzZ
2221
1211
2221
1211,
Since by definition III ba and ba VVV 111
IZZV ba )(1
ba ZZZ
1.15 Networks in cascade:
Output of first subsystem acts as input for the second
Fig.8.22
Interconnection constraints
baba IIVV 212
aa VVVV 1111 and
ba IIII 2211
a
a
aa
aa
a
a
I
V
DC
BA
I
V
I
V
2
2
1
1
1
1
b
b
bb
bb
b
b
a
a
I
V
DC
BA
I
V
I
V
2
2
1
1
2
2
Hence
2
2
1
1
I
V
DC
BA
DC
BA
I
V
bb
bb
aa
aa
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It is this property that makes the transmission parameters so useful. Keep in mind that the
multiplication of the matrices must be in the order in which the networks Na and Nb are
cascaded.
Example 8.23:Comptute the y parameter of the network shown in Fig.Q.8.23
Fig.Q.8.23
Let us refer to the upper network as Na and the lower one as Nb. The two networks are
connected in parallel. Comparing Na and Nb with the circuit shown below.
we obtain
324 122212112112 aaaaaa yyyyyjy
Or
43424 22112112 jyjyyjy aaaa
i.e. Sjj
jjYa
434
442
And
624 122212112112 jyyjyyyy bbbbbb
or
64244 22112112 jyjyyy bbbb
i.e. Sj
jYb
644
424
Therefore, overall y parameters are
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Sjj
jjYYY ba
2744
4426
1.16 Network Functions for One Port and Two Port Network
1.17 Driving Point Functions:
The impedance or admittance found at a given port is called a driving point impedance(or
admittance).
1.Driving Point Impedance:
Z11(s) = V1(s)/I1(s)
2. Driving Point Admittance:
Y11(s) = I1(s)/V1(s) = 1/Z11(s)
1.18 Transfer Function:
The transfer function relates the transform of a quantity at one port to the transform of another
quantity at another port.Thus transfer functions which relate voltages and currents have
following possible forms:
The ratio of one voltage to another voltage, or the voltage transfer ratio.
The ratio of one current to another current ,or the current transfer ratio.
The ratio of one current to another voltage or one voltage to another voltage.
Transfer function for the two port network:
Denominator Numerator
)(2 sV )(2 sI
)(1 sV )(12 sG )(12 sY
)(1 sI )(12 sZ )(12 s
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 28
Example 8.24
Find driving point impedance Z(s) for the network shown in Fig.Q.8.24
Solution
The driving point impedance for RLC series circuit is
Fig.Q.8.24
sC
sRCLCs
sCLsRsZ
11)(
2
Example 8.25
Find driving point impedance Z(s) for the network shown in Fig.Q.8.25
Solution
The driving point impedance for series RL network shunted by a capacitor is
Fig.8.25
LCs
LRs
LRs
CsLR
sCsZ
1)(
1
)(1
1)(
2
and driving point admittance function )(
1)(
sZsY
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 29
Example 8.26
For the network shown find the voltage ratio )(
)()(
1
212
sV
sVsG and
)(
)()(
1
11sV
sIsY by
considering no current in the output terminals.
Fig.8.26
Solution
The network acts as a voltage divider .
)()(1
)()(1
)(
2
1
sVsIsC
and
sVsIsC
sRI
Therefore the ratio of these equations is
RCs
RC
sIsCR
sIsC
sV
sVsG
/1
/1
)()/1(
)()/1(
)(
)()(
1
212
RCs
s
RsCRsV
sIsY
/1
1
/1
1
)(
)()(
1
11
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 30
Example 8.27
For the network shown in Fig.8.27 find G12(s)
Fig.8.27
Solution
LCs
LC
LCssCsL
sC
VV
V
sV
sVsG
CL
C
/1
/1
1
1
/1
/1
)(
)()(
22
1
212
Where LV and CV are the voltages across Inductor and capacitor respectively.
Example 8.28
For the network shown in Fig. 8.28 find G12(s)
Solution Fig.8.28
1R and sC/1 can be combined into an equivalent impedance having the value
1/1
1)(
1
1
1
CsR
R
RsCsZ eq
)()(
)()(
2
2
1
212
sZR
R
sV
sVsG
eq
CRRRRs
CRs
RRCRsR
RCRsRsG
2121
1
2121
22112
/)(
/1)(
Exercise: Q.101
Obtain the ABCD parameter representation of circuit shown in Fig.Q.101
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 31
Fig.Q.101
Ans:
32425.0
220025.29
ST
Q.102 for the bridge circuit shown in Fig.Q.102. Find the transmission parameters.
Fig.Q.102
Q.103 The network of Fig.Q.103 contains both a dependent current source and a
dependent voltage source. Determine y and z parameters.
Fig.Q.103
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 32
Q.104 Find the h parameters for the network shown in Fig.Q.104. Keep the result in s
domain.
Fig.Q.104
Ans : )2(2)2(2
)4(
)2(2
4
)2(2
4522211211
s
sh
s
sh
s
sh
s
sh
105. For the network shown in Fig.Q.105, determine the h parameters
Fig.Q.105
Ans:
CBCB
B
CB
BCA
CB
CB
RRh
RR
Rh
RR
RRhR
RR
RRh
1)()1(22122111
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 33
Q.106 For what value of ‘a’ is the circuit shown in Fig.Q.106 is reciprocal? Also find h
parameters.
Fig.Q.106
Ans: a=2 and
5.02
22h
Q107. What is the value of n for the network shown in Fig.Q.107 to be reciprocal? Also find
y12 and y21 for that n
Fig.Q.107
Ans: n= -0.01 and y12 = y21=-0.002
ECE -2103: Network Analysis
Department of Electronics and Communication Engineering, M.I.T. Manipal. Page 34
Summary
1. Various electronic devices such as transistors, transformers etc can be modelled
using various two port network parameters such as impedance, admittance,
transmission and hybrid parameters.
2. Relation between various two port network parameters studied and verified for
different two port parameters.
3. Complex two port network in cascade, parallel or any other form can be modeled
using simple two port network parameters.
4. Various driving point & transfer functions are defined and determined for the two
port networks.
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