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TWO PORT NETWORK
Port: It is a pair of terminals through which a current may enter or leave a network
Examples
Note: Reference direction and notations for port variable is shown in the figure (h).
Two-port Network Parameters:
S.No Name of Parameter Function Matrix Equation
1. Open Circuit Impedance Parameters or
Z-Parameters
( ) ( )2121 ,, IIfVV =
=
2
1
2221
1211
2
1
I
I
ZZ
ZZ
V
V
2. Short Circuit Admittance Parameters or
Y-Parameters
( ) ( )2121 ,, VVfII =
=
2
1
2221
1211
2
1
V
V
YY
YY
I
I
3. Transmission Parameters or
T-Parameters or A,B,C,D Parameters
( ) ( )2211 ,, IVfIV −=
−
=
2
2
1
1
I
V
DC
BA
I
V
4. Inverse Transmission Parameters or
T’-Parameters or A
’,B
’,C
’,D
’ Parameters
( ) ( )1122 ,, IVfIV −=
−
=
1
1
2
2
I
V
DC
BA
I
V
Fig .1
Network
(a) Two Port Network
Network
(c) One Port Network
Network
(b) Two Port Network
Network
(e) Three Port Network
Network
(d) Four Port Network
(g) T-Network(h) Π Network (f) T-Network
Network +
-
+
-
V1 V2
I1 I2
I1 I2
I/P Port O/P Port
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5. Hybrid Parameters or
h parameters
( ) ( )2121 ,, VIfIV =
=
2
1
2221
1211
2
1
V
I
hh
hh
I
V
6. Inverse Hybrid Parameters or
g parameters
( ) ( )2121 ,, IVfVI =
=
2
1
2221
1211
2
1
I
V
gg
gg
V
I
Method for calculation of different parameters
1. Open Circuit Impedance Parameters or Z-Parameters:
We know for z-parameters
( ) ( )2121 ,, IIfVV =
=
2
1
2221
1211
2
1
I
I
ZZ
ZZ
V
V
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
From above the equations the equivalent circuit will be
Case I: Output port is open circuited, i.e. I2 = 0
Put I2 = 0 in equation (1) & (2)
impedancepointdrivinginputcircuitopen
01
111
2
===I
I
VZ
impedancetransferforwardcircuitopen
01
221
2
===I
I
VZ
(b) Case I: I2 = 0
Network V1
I1 I2=0
V2
V1
I1
Z11
Z12I2 V2
I2
Z22
Z21I1
(a) Equivalent Circuit for Z-Parameters
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Case II: Input port is open circuited, i.e. I1 = 0
Put I1 = 0 in equation (1) & (2)
mpedancetransferreversecircuitopen
02
112
1
iI
VZ
I
===
impedancepointdrivingoutputcircuitopen
02
222
1
===I
I
VZ
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )211 , IIfV =
( )212 , IIfV =
Step 3: Compare the above two equations with the standard equations of z-parameters.
Example 1: Determine Z-parameters for the network shown in figure 3a.
Solution: We know for z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
Case I: When I2 = 0, i.e. output port is open circuited, fig.3b
Ω==⇒=+= 1010551
1111111
I
VZIIIV
Ω==⇒= 551
22112
I
VZIV
Case II: When I1 = 0, i.e. input port is open circuited, fig.3c
(c) Case II: I1 = 0
Network V1
I1=0 I2
V2
Fig. 2
5Ω 5Ω V1
I1
V2
I2
(a)
5Ω5Ω V1
I1
V2
I2=0
(b)
5Ω 5Ω V1
I1=0
V2
I2
(c)
Fig.3
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Ω==⇒= 552
11221
I
VZIV
Ω==⇒= 552
22222
I
VZIV
Hence
=
55
510
2221
1211
ZZ
ZZ Answer
Alternate Method:
Apply KVL in LHS loop of fig.3a
)3(510)(55 2112111 −−−−−+=⇒++= IIVIIIV
Apply KVL in RHS loop of fig.3a
)4(55)(5 212212 −−−−−+=⇒+= IIVIIV
Comparing equation (3) with equation (1) and equation (4) with equation (2)
=
55
510
2221
1211
ZZ
ZZ Answer
Note: Alternate method is easer, so we will apply only this method for rest of the examples in
these notes.
Example 2: Determine Z-parameters for the network shown in figure 4.
Solution: We know for z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
Apply KVL in LHS loop
)3(3
2
3
42)(22 21112111 −−−−−+=⇒−++= IIVVIIIV
2Ω 2Ω V1
I1
V2
I2
Fig.4
2V1
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Apply KVL in RHS loop
+−+=−+= 212112123
2
3
42)(22)(2 IIIIVIIV
)4(3
2
3
2212 −−−−−+−=⇒ IIV
Comparing equation (3) with equation (1) and equation (4) with equation (2)
−=
3
2
3
2
3
2
3
4
2221
1211
ZZ
ZZ Answer
2. Short Circuit Admittance Parameters or Y-Parameters:
We know for Y-parameters
( ) ( )2121 ,, VVfII =
=
2
1
2221
1211
2
1
V
V
YY
YY
I
I
)1(2121111 −−−−−+= VYVYI
)2(2221212 −−−−−+= VYVYI
From above the equations the equivalent circuit will be
Case I: Output port is short circuited, i.e. V2 = 0
Put V2 = 0 in equation (1) & (2)
admittancepointdrivinginputcircuitshort
01
111
2
===V
V
IY
V1
I1
Y11 Y12I2 V2
I2
Y22 Y21I1
(a) Equivalent Circuit for Y-Parameters
(b) Case I: V2 = 0
Network V1
I1 I2
V2=0
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admittancetransferforwardcircuitshort
01
221
2
===V
V
IY
Case II: Input port is short circuited, i.e. V1 = 0
Put V1 = 0 in equation (1) & (2)
admittancetransferreversecircuitshort
02
112
1
===V
V
IY
admittancepointdrivingoutputcircuitshort
02
222
1
===V
V
IY
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )211 ,VVfI =
( )212 ,VVfI =
Step 3: Compare the above two equations with the standard equations of y-parameters.
Example 3: Determine Z & Y-parameters for the network shown in figure 6a.
Solution: We know for z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
And for Y-parameters
)3(2121111 −−−−−+= VYVYI
)4(2221212 −−−−−+= VYVYI
Fig. 5
(c) Case II: V1 = 0
Network V1=0
I1 I2
V2
1Ω 2Ω V1
I1
V2
I2
(a)
3I2
1
2 V1
I1
V2
I2
(b)
3I2
I1-3I2
I1-2I2
Fig.6
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Apply KVL LHS
)2(2)3(1 21211 IIIIV −+−= )5(73 211 −−−−−−=⇒ IIV
Apply KVL RHS loop
)2(2 212 IIV −= )6(42 212 −−−−−−=⇒ IIV
Comparing equation (1) with equation (5) and equation (2) with equation (6)
−
−=
42
73
2221
1211
ZZ
ZZ Answer
Now rearrange equation (5) and (6) according to equation (3) & (4) respectively.
( ) ( ) 7645 ×−× )7(2
72 211 −−−−−+−=⇒ VVI
( ) ( ) 3625 ×−× )8(2
31 212 −−−−−+−=⇒ VVI
Comparing equation (7) with equation (3) and equation (8) with equation (4)
−
−=
2
31
2
72
2221
1211
YY
YY Answer
3. Transmission Parameters or T-Parameters or ABCD Parameters:
We know for T-parameters
( ) ( )2211 ,, IVfIV −=
−
=
2
2
1
1
I
V
DC
BA
I
V
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
Note: Equivalent circuit is not possible.
Case I: Output port is open circuited, i.e. I2 = 0
Put I2 = 0 in equation (1) & (2) (a) Case I: I2 = 0
Network V1
I1 I2=0
V2
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ratiotransfervoltagereversecircuitopen
02
1
2
===I
V
VA
admittancetransferreversecircuitopen
02
1
2
===I
V
IC
Case II: Input port is short circuited, i.e. V1 = 0
Put V1 = 0 in equation (1) & (2)
impedancetransferreversecircuitshort
02
1
1
=−
==V
I
VB
ratiotransfercurrentreversecircuitshort
02
1
1
=−
==V
I
ID
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )221 , IVfV =
( )221 , IVfI =
Step 3: Compare the above two equations with the standard equations of T-parameters.
4. InverseTransmission Parameters or T’-Parameters or A
’ B’ C’ D’ Parameters:
We know for T’-parameters
( ) ( )1122 ,, IVfIV −=
−
=
1
1
''
''
2
2
I
V
DC
BA
I
V
)1(1'
1
'
2 −−−−−−= IBVAV
)2(1'
1
'
2 −−−−−−= IDVCI
Note: Equivalent circuit is not possible.
Fig. 7
(b) Case II: V1 = 0
Network V1=0
I1 I2
V2
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Case I: Input port is open circuited, i.e. I1 = 0
Put I1 = 0 in equation (1) & (2)
ratiotransfervoltagereversecircuitopen'
02
1
1
===I
V
VA
admittancetransferforwardcircuitopen
01
2'
1
===I
V
IC
Case II: Input port is short circuited, i.e. V1 = 0
Put V1 = 0 in equation (1) & (2)
impedancetransferforwardcircuitshort
01
2'
1
=−
==V
I
VB
ratiotransfercurrentforwardcircuitshort
01
2'
1
=−
==V
I
ID
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )112 , IVfV =
( )112 , IVfI =
Step 3: Compare the above two equations with the standard equations of T’-parameters.
Example 4: Determine Z & T & T’-parameters for the network shown in figure 9a.
(a) Case I: I1 = 0
Network V1
I1=0 I2
V2
Fig. 8
(b) Case II: V1 = 0
Network V1=0
I1 I2
V2
1Ω 1Ω V1
I1
V2
I2
(a)
1Ω
2I12I2
V1
I1
V2
I2
(b)
2I12I2
I1+2I2 I2+2I1
2I1+2I2
I1+I2
1 Ω
1 Ω 1 Ω
Fig.9
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Solution: We know for z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
We know for T-parameters
)3(221 −−−−−−= BIAVV
)4(221 −−−−−−= DICVI
We know for T’-parameters
)5(1'
1
'
2 −−−−−−= IBVAV
)6(1'
1
'
2 −−−−−−= IDVCI
Two loop equations of fig.9b are
)7(32)(1)2(1 21121211 −−−−−+=⇒+++= IIVIIIIV
)8(23)(1)2(1 21221212 −−−−−+=⇒+++= IIVIIIIV
Comparing equation (1) & (2) with equation (7) & (8)
=
23
32
2221
1211
ZZ
ZZ Answer
Now rearrange equation (7) and (8) according to equation (3) & (4) respectively.
( ) ( ) 2837 ×−× )9(3
5
3
2221 −−−−−+=⇒ IVV
( ) ( ) 3827 ×−× )10(3
2
3
1221 −−−−−−=⇒ IVI
Comparing equation (9) &(10) with equation (3) & (4) respectively
−=
3
2
3
1
3
5
3
2
DC
BA Answer
Now rearrange equation (7) and (8) according to equation (5) & (6) respectively.
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( ) ( ) 3827 ×−× )11(3
5
3
2112 −−−−−−=⇒ IVV
From equation (8) )12(3
2
3
1112 −−−−−−=⇒ IVI
Comparing equation (11) & (12) with equation (5) & (6) respectively
−=
3
2
3
1
3
5
3
2
''
''
DC
BA Answer
5. Hybrid Parameters or h-Parameters:
We know for h-parameters
( ) ( )2121 ,, VIfIV =
=
2
1
2221
1211
2
1
V
I
hh
hh
I
V
)1(2121111 −−−−−+= VhIhV
)2(2221212 −−−−−+= VhIhI
From above the equations the equivalent circuit will be
Case I: Output port is short circuited, i.e. V2 = 0
Put V2 = 0 in equation (1) & (2)
impedancepointdrivinginputcircuitshort
01
111
2
===V
I
Vh
admittancetransferforwardcircuitshort
01
221
2
===V
V
Ih
V1
I1
h11
h12V2
(a) Equivalent Circuit for h-Parameters
V2
I2
h22 h21V1
(b) Case I: V2 = 0
Network V1
I1 I2
V2=0
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Case II: Input port is open circuited, i.e. I1 = 0
Put I1 = 0 in equation (1) & (2)
ratiotransfervoltage reversecircuit open
02
112
1
===I
V
Vh
admittancetransferforwardcircuitopen
02
222
1
===I
V
Ih
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )121 ,VIfV =
( )122 ,VIfI =
Step 3: Compare the above two equations with the standard equations of h-parameters.
6. Inverse Hybrid Parameters or g-Parameters:
We know for h-parameters
( ) ( )2121 ,, IVfVI =
=
2
1
2221
1211
2
1
I
V
gg
gg
V
I
)1(2121111 −−−−−+= IgVgI
)2(2221212 −−−−−+= IgVgV
From above the equations the equivalent circuit will be
(c) Case I: I1 = 0
Network V1
I1=0 I2
V2
Fig. 10
V1
I1
g11 g12I2
(b) Equivalent Circuit for g-Parameters
V2
I2
g22
g21V1
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Case I: Output port is open circuited, i.e. I2 = 0
Put I2 = 0 in equation (1) & (2)
admittanceinputcircuitopen
01
111
2
===I
V
Ig
gainvoltageforwardcircuitopen
01
221
2
===I
V
Vg
Case II: Input port is short circuited, i.e. V1 = 0
Put V1 = 0 in equation (1) & (2)
gaincurrentreversecircuitshort
02
112
1
===V
I
Ig
impedanceoutputcircuitshort
02
222
1
===V
I
Vg
Alternate method:
Step 1: Apply KVL/KCL in the given circuit.
Step 2: Rearrange the above equations in the following forms
( )211 , IVfI = & ( )212 , IVfV =
Step 3: Compare the above two equations with the standard equations of g-parameters.
Example 5: Determine Y & h & g-parameters for the network shown in figure 12a.
Solution: We know for Y-parameters
)1(2121111 −−−−−+= VYVYI
)2(2221212 −−−−−+= VYVYI
(b) Case I: I2 = 0
Network V1
I1 I2=0
V2
Fig. 11
(b) Case II: V1 = 0
Network V1=0
I1 I2
V2
Fig.12
1Ω 1Ω V1
I1
V2
I2
(a)
1Ω
1Ω 1Ω V1
I1
V2
I2
(b)
1Ω V1 V2
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We know for h-parameters
)3(2121111 −−−−−+= VhIhV
)4(2221212 −−−−−+= VhIhI
We know for g-parameters
)5(2121111 −−−−−+= IgVgI
)6(2221212 −−−−−+= IgVgV
Apply KCL at node V1 of fig.12b
)7(211
211211
1 −−−−−−=⇒−
+= VVIVVV
I
Apply KCL at node V2 of fig.12b
)8(2111
212221
2 −−−−−+−=⇒=−
+ VVIVVV
I
Comparing equation (1) & (2) with equation (7) & (8)
−
−=
21
12
2221
1211
YY
YY Answer
Now rearrange equation (7) and (8) according to equation (3) & (4) respectively.
From equation (7) )9(2
1
2
1211 −−−−−+= VIV
2)8()7( ×+ )10(2
3
2
1212 −−−−−+−= VII
Comparing equation (3) & 4) with equation (9) & (10)
−=
2
3
2
1
2
1
2
1
2221
1211
hh
hh Answer
Now rearrange equation (7) and (8) according to equation (5) & (6) respectively.
)8(2)7( +× )11(2
1
2
3211 −−−−−−= IVI
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From equation (8) )12(2
1
2
1212 −−−−−+= IVV
Comparing equation (5) & 6) with equation (11) & (12)
−=
2
1
2
1
2
1
2
3
2221
1211
gg
gg Answer
Inter-relationship of parameters:
Since all the parameters describe the same two port network, they are inter related. One set of
parameters may be expressed in terms of the other set.
1. Z-Parameters:
We know for z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
(a) Z-Parameters in terms of Y-Parameters:
We know for Y-parameters
)3(2121111 −−−−−+= VYVYI
)4(2221212 −−−−−+= VYVYI
Now rearrange equation (3) & (4) according to equation (1) & (2) respectively
1222 )4()3( YY ×−× )5(212
122
1 −−−−−−=⇒ IY
YI
Y
YV
Where 211222112221
1211YYYY
YY
YYY −==
1121 )4()3( YY ×−× )6(211
121
1 −−−−−+−=⇒ IY
YI
Y
YV
Compering equation (1) & (2) with equation (5) & (6) respectively
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−
−
=
y
Y
y
Y
y
Y
y
Y
ZZ
ZZ
1121
1222
2221
1211
Note: [ ] 121221
1222
1121
1222
2221
1211 1 −=
−
−=
−
−
=
Y
YY
YY
Y
y
Y
y
Y
y
Y
y
Y
ZZ
ZZ
So [ ] [ ] 1−= YZ
(b) Z-Parameters in terms of T-Parameters:
We know for T-parameters
)3(221 −−−−−−= BIAVV
)4(221 −−−−−−= DICVI
Now rearrange equation (3) & (4) according to equation (1) & (2) respectively
BD ×−× )4()3( )5(211 −−−−−+=⇒ IC
TI
C
AV
Where CBADDC
BAT −==
equ(3)From )6(1
212 −−−−−+=⇒ IC
DI
CV
Compering equation (1) & (2) with equation (5) & (6) respectively
=
C
D
C
C
T
C
A
ZZ
ZZ
12221
1211
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(c) Z-Parameters in terms of T’-Parameters:
We know for T’-parameters
)3(1'
1
'
2 −−−−−−= IBVAV
)4(1'
1
'
2 −−−−−−= IDVCI
Now rearrange equation (3) & (4) according to equation (1) & (2) respectively
equ(4)From )6(1
2'1'
'
1 −−−−−+=⇒ IC
IC
DV
'' )4()3( AC ×−× )5(2'
'
1'
'
2 −−−−−+=⇒ IC
AI
C
ATV
Where ''''''
''
' BCDADC
BAT −==
Compering equation (1) & (2) with equation (5) & (6) respectively
=
'
'
'
'
''
'
2221
1211
1
C
A
C
T
CC
D
ZZ
ZZ
(d) Z-Parameters in terms of h-Parameters:
We know for h-parameters
)3(2121111 −−−−−+= VhIhV
)4(2221212 −−−−−+= VhIhI
Now rearrange equation (3) & (4) according to equation (1) & (2) respectively
1222 )4()3( hh ×−× )5(222
121
22
1 −−−−−+=⇒ Ih
hI
h
hV
From equation (4) )6(1
2
22
1
22
212 −−−−−+−=⇒ I
hI
h
hV
Compering equation (1) & (2) with equation (5) & (6) respectively
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−
=
2222
21
22
12
22
2221
1211
1
hh
h
h
h
h
h
ZZ
ZZ
(e) Z-Parameters in terms of g-Parameters:
We know for g-parameters
)3(2121111 −−−−−+= IgVgI
)4(2221212 −−−−−+= IgVgV
Now rearrange equation (3) & (4) according to equation (1) & (2) respectively
From equation (3) )5(1
2
11
121
11
1 −−−−−−−=⇒ Ig
gI
gV
1121 )4()3( gg ×−× )6(211
1
11
212 −−−−−−+=⇒ I
g
gI
g
gV
Compering equation (1) & (2) with equation (5) & (6) respectively
−
=
1111
21
11
12
11
2221
1211
1
g
g
g
g
g
g
g
ZZ
ZZ
Example 6: Calculate values of Z-parameters, if the values of other parameters are given bellow:
(a) A = 2, B = -1, C = 3, D = -2
(b) h11 =1, h12 = -2, h21 = -3 h22 = 2
(c) Y11 = 1/3, Y12 = 2/3, Y21 = -1/3, Y22 =1/6
Solution: 1−=−= BCADT
Answer
−
−
=
=
3
2
3
1
3
1
3
2
12221
1211
C
D
C
C
T
C
A
ZZ
ZZ
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19 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
421122211 −=−= hhhhh
−−=
−−
−−
=
−
=
2/12/3
12
2
1
2
3/1
2
2
2
4
1
2222
21
22
12
22
2221
1211
hh
h
h
h
h
h
ZZ
ZZ Answer
18/521122211 =−= YYYYY
−=
−
−
=
5
6
5
6
5
12
5
3
1121
1222
2221
1211
y
Y
y
Y
y
Y
y
Y
ZZ
ZZ Answer
2. Y-Parameters: Similar to above
(a) In terms of Z-Parameters:
(b) In terms of T-Parameters:
(c) In terms of T’-Parameters:
(d) In terms of h-Parameters:
(e) In terms of g-Parameters:
3. T-Parameters: Similar to above
(a) In terms of Z-Parameters:
(b) In terms of Y-Parameters:
(c) In terms of T’-Parameters:
(d) In terms of h-Parameters:
(e) In terms of g-Parameters:
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20 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
4. T’-Parameters: Similar to above
(a) In terms of Z-Parameters:
(b) In terms of Y-Parameters:
(c) In terms of T-Parameters:
(d) In terms of h-Parameters:
(e) In terms of g-Parameters:
5. h-Parameters: Similar to above
(a) In terms of Z-Parameters:
(b) In terms of Y-Parameters:
(c) In terms of T-Parameters:
(d) In terms of T’-Parameters:
(e) In terms of g-Parameters:
6. g-Parameters: Similar to above
(a) In terms of Z-Parameters:
(b) In terms of Y-Parameters:
(c) In terms of T-Parameters:
(d) In terms of T’-Parameters:
(e) In terms of h-Parameters:
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21 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Relationships between parameters:
Note:-
1. [ ] [ ] 1−= YZ & [ ] [ ] 1−= ZY
2. [ ] [ ] 1' −≠ TT & [ ] [ ] 1' −≠ TT
3. [ ] [ ] 1−= gh & [ ] [ ] 1−= hg
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22 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
T & Π -Network:
Z-parameters of a T-network Y-parameters of a network−Π
ca ZZZ +=11 ca YYY +=11
cb ZZZ +=22 cb YYY +=22
cZZZ == 2112 cYYY −== 2112
T to Π& Π to T Transformation:
T to Π tionstransformadeltatostar⇒ and Π to T tionstransformastartodelta⇒
T to Π (Star to Delta) Π to T (Delta to Star)
b
accbba
a
Y
YYYYYYZ
1
111111
+
+
= cba
ca
a ZZZ
ZZ
Y ++=
1
a
accbba
b
Y
YYYYYYZ
1
111111
+
+
= cba
cb
b ZZZ
ZZ
Y ++=
1
c
accbba
c
Y
YYYYYYZ
1
111111
+
+
= cba
ab
c ZZZ
ZZ
Y ++=
1
Za
V1
I1
V2
I2
(a)
Zb
ZcYa V1
I1
V2
I2
(b)
Yb
Yc
Fig.13
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23 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Condition for reciprocity & reciprocal network:
A two port network is said to be reciprocal, if the ration of the excitation to response is invarient
to an interchage of the exciation and response in the network. Mathematicall, we can say from
following figure
01
2
02
1
12
if
==
=VV
I
V
I
V
0
'
10
'
212 ==
−=
−⇒
V
s
V
s
I
V
I
V
NetworkRecoprocal'2'
1
0
'
10
'
212
⇒==⇒==
IIORI
V
I
V
V
s
V
s
1. Reciprocity in terms of Z-Parameters:
We know for Z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
From fig.14a V1=Vs, I1=I1, I2=-I2’, V2=0 Put these conditions in equation (1), (2)
)3(0 21
21122211
'
2'
222121
'
221111 −−−−−−
=⇒
−=
−=
Z
ZZZZ
I
V
IZIZ
IZIZVss
From fig.14b V1=0, I1=-I1’, I2= I2, V2=Vs Put these conditions in equation (1), (2)
)4(0
12
21122211
'
1222
'
121
221
'
111 −−−−−−
=⇒
+−=
+−=
Z
ZZZZ
I
V
IZIZV
IZIZs
s
Fig. 14(V1=Vs, I1=I1, I2=-I2’, V2=0)
(a)
Network Vs
I1
'
2I V2=0
I2
Network V1=0
'
1I
I2
Vs
(V1=0, I1=-I1’, I2= I2, V2=Vs)
(b)
I1
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24 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
But for reciprocal network
21120
'
10
'
212
ZZI
V
I
V
V
s
V
s =⇒===
2. Reciprocity in terms of Y-Parameters:
We know for Y-parameters
)1(2121111 −−−−−+= VYVYI
)2(2221212 −−−−−+= VYVYI
From fig.14a V1=Vs, I1=I1, I2=-I2’, V2=0 Put these conditions in equation (1), (2)
} )3(21'2
21
'
2
111
−−−−−−=⇒=−
=
YI
VVYI
VYI
ss
s
From fig.14b V1=0, I1=-I1’, I2= I2, V2=Vs Put these conditions in equation (1), (2)
}
s
ss
VYI
YI
VVYI
222
12'
1
12
'
1 )4(
=
−−−−−−=⇒=−
But for reciprocal network
21120
'
10
'
212
YYI
V
I
V
V
s
V
s =⇒===
3. Reciprocity in terms of T-Parameters:
We know for T-parameters
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
2112Hence YY =
2112Hence ZZ =
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25 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
From fig.14a V1=Vs, I1=I1, I2=-I2’, V2=0 Put these conditions in equation (1), (2)
}'
21
'
2
'
2 )3(
DII
BI
VBIV ss
=
−−−−−=⇒=
From fig.14b V1=0, I1=-I1’, I2= I2, V2=Vs Put these conditions in equation (1), (2)
)4(0
'
12
'
1
2 −−−−−−
=⇒
−=−
−=
BCAD
B
I
V
DICVI
BIAVs
s
s
But for reciprocal network
11
0
'
10
'
212
==−⇒===
DC
BAorBCAD
I
V
I
V
V
s
V
s
4. Reciprocity in terms of T’-Parameters: Similar to above
5. Reciprocity in terms of h-Parameters:
We know for h-parameters
)1(2121111 −−−−−+= VhIhV
)2(2221212 −−−−−+= VhIhI
From fig.14a V1=Vs, I1=I1, I2=-I2’, V2=0 Put these conditions in equation (1), (2)
)3(21
11
'
2121
'
2
111−−−−−−=⇒
=−
=
h
h
I
V
IhI
IhVss
From fig.14b V1=0, I1=-I1’, I2= I2, V2=Vs Put these conditions in equation (1), (2)
1Hence''
''
=DC
BA
1Hence =DC
BA
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26 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
)4(0
12
11
'
122
'
1212
12
'
111 −−−−−=⇒
+=
+=
h
h
I
V
VhIhI
VhIhs
s
s
But for reciprocal network
2112
0
'
10
'
212
hhI
V
I
V
V
s
V
s −=⇒===
6. Reciprocity in terms of g-Parameters: Similar to above
Condition for symmetry & Symmetrical networks:
A two port network is said to be symmertical if the ports can be interchaged without changing
the port voltages and currents. Mathematicall, we can say from following figure
NetworklSymmetricaif
020102
2
01
1
1212
⇒=⇒===== I
s
I
s
III
V
I
V
I
V
I
V
1. Symmetry in terms of Z-Parameters:
We know for Z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1)
111IZVs =
2112Hence gg −=
2112Hence hh −=
Fig. 15(V1=Vs, I1=I1, I2=0, V2=V2)
(a)
Network Vs
I1 I2=0
V2 Network
V1
I1=0 I2
Vs
(V1=V1, I1=0, I2= I2, V2=Vs)
(b)
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27 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
11
0101
1
22
ZI
V
I
V
I
s
I
==⇒==
From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (2)
222IZVs =
22
0202
2
11
ZI
V
I
V
I
s
I
==⇒==
221102
2
01
1
12
if ZZI
V
I
V
II
=====
2. Symmetry in terms of Y-Parameters:
We know for Y-parameters
)1(2121111 −−−−−+= VYVYI
)2(2221212 −−−−−+= VYVYI
From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1) & (2)
212111 VYVYI s +=
222210 VYVY s +=
21122211
22
0101
1
22
YYYY
Y
I
V
I
V
I
s
I−
==⇒==
From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (1) & (2)
sVYVY 121110 +=
sVYVYI 221212 +=
21122211
11
0202
2
11
YYYY
Y
I
V
I
V
I
s
I−
==⇒==
2211Hence ZZ =
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28 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
2211
02
2
01
1
12
if YYI
V
I
V
II
=====
3. Symmetry in terms of T-Parameters:
We know for T-parameters
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1) & (2)
2AVVs =
21 CVI =
C
A
I
V
I
V
I
s
I
==⇒== 0101
1
22
From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (1) & (2)
21 BIAVV s −=
20 DICVs −=
C
D
I
V
I
V
I
s
I
==⇒== 0202
2
11
DAI
V
I
V
II
===== 02
2
01
1
12
if
4. Symmetry in terms of T’-Parameters: Similar to above
''Hence DA =
DA =Hence
2211Hence YY =
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29 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
5. Symmetry in terms of h-Parameters:
We know for h-parameters
)1(2121111 −−−−−+= VhIhV
)2(2221212 −−−−−+= VhIhI
From fig.15a V1=Vs, I1=I1, I2=0, V2=V2 Put these conditions in equation (1) & (2)
212111 VhIhVs +=
2221210 VhIh +=
22
21122211
0101
1
22
h
hhhh
I
V
I
V
I
s
I
−==⇒
==
From fig.15b V1=V1, I1=0, I2=I2, V2=Vs Put these conditions in equation (1) & (2)
sVhV 121 =
sVhI 222 =
220202
2 1
11
hI
V
I
V
I
s
I
==⇒==
11if2221
1211
21122211
02
2
01
1
12
==−⇒===
hh
hhorhhhh
I
V
I
V
II
6. Symmetry in terms of g-Parameters: Similar to above
1Hence2221
1211 =gg
gg
1Hence2221
1211 =hh
hh
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30 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Summary table for reciprocal and symmetrical two port network:-
S.No Parameters Condition of Reciprocity Condition of Symmetry
1. [ ]Z 2112 ZZ = 2211 ZZ = 2. [ ]Y
2112 YY = 2211 YY =
3. [ ]T 1=
DC
BA
DA =
4. [ ]'T 1
''
''
=DC
BA
'' DA =
5. [ ]h 2112 hh −= 1
2221
1211 =hh
hh
6. [ ]g 2112 gg −= 1
2221
1211 =gg
gg
Example 7: Calculate values of Z-parameters, and check whether given network is reciprocal
and symmetrical or not.
Solution: Take the Laplace Transformation (s-domain) of above circuit and convert it into
equivalent T-circuit. See fig.16b
Where
Ω= 2aZ
Ω+
=
=ss
Zc81
2
4
12
( ) Ω+= sZb 24
2 Ω
4F
(a)
2 Ω
4 Ω 2H ZaV1
I1
V2
I2
(b)
Zb
Zc
Fig.16
-
31 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
We know Z-parameters of a T-network
s
sZZZ ca
81
16411 +
+=+=
s
ssZZZ cb
81
16346 2
22 +++
=+=
s
ZZZ c81
22112 +
===
Hence [ ]
+++
++
++
++
=
s
ss
s
s
s
s
s
s
Z
81
16346
81
164
81
164
81
164
2 Answer
Clearly lSymmetricaNot2211 ⇒≠ ZZ
Reciprocal2112 ⇒= ZZ
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32 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Interconnections of two port networks:
1. Series connection or series-series connection:
Z-parameters for network A Z-parameters for network B
=
a
a
aa
aa
a
a
I
I
ZZ
ZZ
V
V
2
1
2221
1211
2
1
=
b
b
bb
bb
b
b
I
I
ZZ
ZZ
V
V
2
1
2221
1211
2
1
From above fig.17
ba III 111 == & ba III 222 ==
ba VVV 111 += & ba VVV 222 +=
So
+
=
+
+=
b
b
a
a
ba
ba
V
V
V
V
VV
VV
V
V
2
1
2
1
22
11
2
1
+
=
b
b
bb
bb
a
a
aa
aa
I
I
ZZ
ZZ
I
I
ZZ
ZZ
2
1
2221
1211
2
1
2221
1211
+
=
2
1
2221
1211
2
1
2221
1211
I
I
ZZ
ZZ
I
I
ZZ
ZZ
bb
bb
aa
aa
(Using ba III 111 == & ba III 222 == )
++
++=
2
1
22222121
12121111
2
1
I
I
ZZZZ
ZZZZ
V
V
aaaa
baba
Fig.17
-
33 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
So
++
++=
aaaa
baba
ZZZZ
ZZZZ
ZZ
ZZ
22222121
12121111
2221
1211
Or [ ] [ ] [ ]ba ZZZ +=
2. Parallel connection or Parallel-Parallel connection:
Y-parameters for network A Y-parameters for network B
=
a
a
aa
aa
a
a
V
V
YY
YY
I
I
2
1
2221
1211
2
1
=
b
b
bb
bb
b
b
V
V
YY
YY
I
I
2
1
2221
1211
2
1
From above fig.18
ba III 111 += & ba III 222 +=
ba VVV 111 == & ba VVV 222 ==
So
+
=
+
+=
b
b
a
a
ba
ba
I
I
I
I
II
II
I
I
2
1
2
1
22
11
2
1
+
=
b
b
bb
bb
a
a
aa
aa
V
V
YY
YY
V
V
YY
YY
2
1
2221
1211
2
1
2221
1211
+
=
2
1
2221
1211
2
1
2221
1211
V
V
YY
YY
V
V
YY
YY
bb
bb
aa
aa
Fig.18
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34 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
(Using ba VVV 111 == & ba VVV 222 == )
++
++=
2
1
22222121
12121111
2
1
V
V
YYYY
YYYY
I
I
aaaa
baba
So
++
++=
aaaa
baba
YYYY
YYYY
YY
YY
22222121
12121111
2221
1211
Or [ ] [ ] [ ]ba YYY +=
3. Cascade Connection:
T-parameters for network A T-parameters for network B
−
=
a
a
aa
aa
a
a
I
V
DC
BA
I
V
2
2
1
1
−
=
b
b
bb
bb
b
b
I
V
DC
BA
I
V
2
2
1
1
From above fig.19
aII 11 = & ba II 12 =− & 22 II b =
aVV 11 = & ba VV 12 = & 22 VV b =
Fig.19
-
35 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
So
=
−
=
=
b
b
aa
aa
a
a
aa
aa
a
a
I
V
DC
BA
I
V
DC
BA
I
V
I
V
1
1
2
2
1
1
1
1
−
=
b
b
bb
bb
aa
aa
I
V
DC
BA
DC
BA
2
2
−
=
2
2
1
1
I
V
DC
BA
DC
BA
I
V
bb
bb
aa
aa
So
=
bb
bb
aa
aa
DC
BA
DC
BA
DC
BA
Or [ ] [ ][ ]ba TTT =
Similarly we can also drive a relation for T’-parameters
[ ] [ ][ ]''' ba TTT =
4. Series-parallel connection:
Similar to above
[ ] [ ] [ ]ba hhh +=
Fig.20
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36 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
5. Parallel-series connection:
Similar to above
[ ] [ ] [ ]ba ggg +=
Overall parameters for different connections:
S.No Interconnection Overall Parameter
1. Series or Series-series [ ] [ ] [ ]ba ZZZ +=
2. Parallel or Parallel-parallel [ ] [ ] [ ]ba YYY +=
3. Cascade or Tandem [ ] [ ][ ]ba TTT =
4. Series-Parallel [ ] [ ] [ ]ba hhh +=
5. Parallel-Series [ ] [ ] [ ]ba ggg +=
Fig.21
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37 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Example 8: Determine the Y-parameters for the twin-T network shown in following figure, if all
resistances are of 1 ohm. (Fig.22)
Solution: Clearly above circuit is the parallel connection of two T-networks
We know Z-parameters of a T-network
[ ]
+
+=
cbc
cca
ZZZ
ZZZZ
So [ ] [ ]
==
21
12ba ZZ [ ] [ ] [ ]
−
−=
===⇒
−−
3
2
3
1
3
1
3
2
21
121
1
aba ZYY
[ ] [ ] [ ]
−
−=+=
3
4
3
2
3
2
3
4
ba YYY Answer
Example 9: Determine the transmission parameters for the network shown in following figure
using the concept of interconnection of four two-port networks N1, N2, N3 and N4 in cascade.
Solution:
We can directly write T-parameters of different network
=
10
211T
=
12
012
sT
=
10
213
sT
+=
11
014
sT
Fig.22
Fig.23
2 Ω
2F
2H
1F 1 Ω
N1 N2 N3 N4
-
38 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
So required T-parameters of given network
[ ] [ ][ ][ ][ ]
+
==
11
01
10
21
12
01
10
214321
s
s
sTTTTT
[ ]
++++
+++++=
141344
22838108
223
223
ssss
sssssT Answer
Example 10: Find out Z-parameters of following network.
Solution: We know for Z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
Apply KVL in loop 1-2’-1’ of fig.24b
( ) ( ) )3(132311 −−−−−−+−−= IIIZIIZV dc
Apply KVL in loop 2-1’-2’ of fig.24b
( ) ( ) )4(132322 −−−−−−+++= IIIZIIZV db
Voltage at 1-1 terminals via 1-2-1’= Voltage at 1-1 terminals via 1-2’-1’
( ) ( ) ( ) )5(13231323 −−−−−−++−=++ IIIZIIZIIZIZ dcba
Eliminate I3 from equations (1), (2) & (3) and rearrange them according to equations (1)
& (2)
( )( )
)6(211 −−−−−
+++
−+
+++
++= I
ZZZZ
ZZZZI
ZZZZ
ZZZZV
dcba
dacb
dcba
dcba
( )( ))7(212 −−−−−
+++++
+
+++−
= IZZZZ
ZZZZI
ZZZZ
ZZZZV
dcba
dbca
dcba
dacb
Za
Zb
Zc
Zd
(a) Fig.24
1
1’
2
2’
Za
Zb
Zc
Zd
(b)
V1 V2
I1 I2 I3
I1-I3
I2+I3
I1 I2+I3-I1 I2
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39 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Comparing equations (6) & (7) with equations (1) & (2)
[ ]
( )( )
( )( )
+++
++
+++
−
+++
−
+++
++
=
dcba
dbca
dcba
dacb
dcba
dacb
dcba
dcba
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
Z
For symmetrical network, i.e. cbda ZZZZ == &
[ ]
( ) ( )
( ) ( )
+−
−+
=
22
22
baab
abba
ZZZZ
ZZZZ
Z
Example 11: Find out Z-parameters of following network.
Solution:
Apply KVL at V1
)1(211 −−−−−+= VZIV
Apply KVL at V2
)2(212 −−−−−+
=Y
IIV
Rearranging these two equations, we have
( ) )3(.1 221 −−−−−−+= IZVZYV
)4(.1 221 −−−−−−= IYVI
Comparing equations (3) & (4) with standard T-parameters equations, we have
[ ]
+=
1
1
Y
ZZYT
Z
Y
Fig.25
V1 V2
I1 I2
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40 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Lattice Network: A two-port lattice network is shown in fig.31a. It can be simplified into
equivalent bridge network as shown in fig.31b. This is very common network used in filter
sections and also used as attenuator.
Z-parameters can be find out of a lattice network (See Example10)
[ ]
( )( )
( )( )
+++
++
+++
−
+++
−
+++
++
=
dcba
dbca
dcba
dacb
dcba
dacb
dcba
dcba
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
ZZZZ
Z
For symmetrical lattice network, i.e. cbda ZZZZ == &
[ ]
( ) ( )
( ) ( )
−−−=+=
−−−=−=⇒
+−
−+
=)2(
)1(
22
22
1211
1211
cb
da
baab
abba
ZZZZ
ZZZZ
ZZZZ
ZZZZ
Z
Hence if Z-parameters of a two-port network are known, we can find out its equivalent
lattice network using above equations (1) & (2)
Ladder Network: A two-port ladder network is shown in fig.27. It is mode of cascaded infinite
L-network. It is used in network synthesis.
T-parameters can be find out of a L-network (See Example11)
Za
Zb
Zc
Zd
(b)Bridge Equivalent N/W of Lattice N/W
2 2’
V2V1
1
1’
2
2’
V2V1
1
1’
Za
Zb
Zc
Zd
(a)Lattice Network Fig.26
Z
YV1
I1
Z
Y
Fig.27
Z
Y
Z
Y V2
I2
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41 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
[ ] [ ] [ ] [ ] [ ]
+=====
1
1......4321
Y
ZZYTTTTT n
So T-parameters of a ladder network
[ ] [ ][ ][ ][ ] [ ]n
nY
ZZYTTTTTT
+==
1
1.......4321
Example 12: Obtain the lattice equivalent of a symmetrical T- network as shown in Fig28a.
Solution: A symmetrical and reciprocal two port network can be realized as a symmetrical two
port lattice network.
Z-parameters of above T-network
[ ] Reciprocal&lSymmetirca32
23⇒
=Z
We know for lattice network
523
123
1211
1211
=+=+==
=−=−==
ZZZZ
ZZZZ
cb
da
Equivalent lattice network is shown in fig28b
Open circuit and short circuit Impedance:
Open circuit impedance
( ) ( )2202
2211
01
11
12
& ZI
VZZ
I
VZ
I
o
I
o ≡=≡===
Short circuit impedance
≡=
≡=
== 2202
22
1101
11
1&
1
12
YI
VZ
YI
VZ
V
s
V
s
(a)
1 Ω 1 Ω 2 Ω
1 Ω
(b) Equivalent Lattice Network
1 Ω
5 Ω
5 Ω
Fig.28
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42 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
1. Open circuit and short circuit impedances in terms of T-parameters:
We know T-parameter:
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
So )4(&)3(
02
22
01
11
12
−−−==−−−====
C
D
I
VZ
C
A
I
VZ
I
o
I
o
)6(&)5(
02
22
01
11
12
−−−==−−−====
A
B
I
VZ
D
B
I
VZ
V
s
V
s
2. T-parameters in terms of Open circuit and short circuit impedances:
From above equations (3), (4), (5) & (6)
so
o
ZZ
ZA
22
1
−±=
so
os
ZZ
ZZB
22
12 −
±=
( )soo ZZZ
C221
1
−±=
( )soo ZZZ
ZD221
20
1
−±=
Z1o 2-2’ is open circuited
Impedance at 1-1’
Z1s 2-2’ is short circuited
Impedance at 1-1’
Note:
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43 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Input and output Impedances:
Input impedance: If a load impedance ZL is connected across the output port, the impedance at
input port is known as input impedance,
1
1
I
VZin =
(i) Input impedance in terms of Z-parameters:
We know for Z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
From fig.29 )3(22 −−−−−−= LZIV
Put the value of V2 from equation (2) to equation (3)
)4(122
2122222121 −−−−−+
−=⇒−=+ I
ZZ
ZIZIIZIZ
L
L
Put the value of I2 from equation (4) to equation (1)
+
−+= 1
22
21121111 I
ZZ
ZZIZV
L
So
(ii) Input impedance in terms of T-parameters:
We know for T-parameters
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
From fig.29 )3(22 −−−−−−= LZIV
Fig.29
Network V1
I1 I2
V2 ZL
L
Lin
ZZ
ZZZZZZ
I
VZ
++−
==22
1121122211
1
1
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44 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
Put the value of V2 from equation (3) to equation (1) & equation (2)
( ) 221 BIZIAV L −−=
( ) 221 DIZICI L −−=
From above two equations
(iii) Input impedance in terms of h-parameters:
We know for Z-parameters
)1(1 212111
−−−−−+= VhIhV
)2(2221212 −−−−−+= VhIhI
From fig.29 )3(22 −−−−−−= LZIV
Put the value of I2 from equation (2) to equation (3)
( ) )4(1
1
22
2122221212 −−−−−+
−=⇒+−= I
Zh
ZhVZVhIhV
L
LL
Put the value of V2 from equation (4) to equation (1)
+−
+= 122
21121111
1I
Zh
ZhhIhV
L
L
So
Output impedance: If a load impedance ZL is connected across the input port, the impedance at
output port is known as output impedance,
2
2
I
VZop =
Fig. 30
Network V1
I1 I2
V2 ZL
DCZ
BAZ
I
VZ
L
Lin +
+==
1
1
( )L
Lin
Zh
hZhhhh
I
VZ
22
1121122211
1
1
1++−
==
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45 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
(i) Output impedance in terms of Z-parameters:
We know for Z-parameters
)1(2121111 −−−−−+= IZIZV
)2(2221212 −−−−−+= IZIZV
From fig.30 )3(11 −−−−−−= LZIV
Put the value of V1 from equation (3) to equation (1)
)4(211
2112121111 −−−−−+
−=⇒+=− I
ZZ
ZIIZIZZI
L
L
Put the value of I1 from equation (4) to equation (2)
+−
+= 111
21221212 I
ZZ
ZZIZV
L
So
(ii) Input impedance in terms of T-parameters:
We know for T-parameters
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
From fig.30 )3(11 −−−−−−= LZIV
Put the value of V1 & I1 from equation (1) & (2) to equation (3)
( ) LZDICVBIAV 2222 −−=−
So
L
Lop
ZZ
ZZhZZZ
I
VZ
+
+−==
11
2221122211
2
2
ACZ
BDZ
I
VZ
L
Lop +
+==
2
2
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46 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
(iii) Input impedance in terms of h-parameters:
We know for h-parameters
)1(1 212111
−−−−−+= VhIhV
)2(2221212 −−−−−+= VhIhI
From fig.30 )3(11 −−−−−−= LZIV
Put the value of V1 from equation (3) to equation (1)
)4(211
1212121111 −−−−−+
−=⇒+=− V
Zh
hIVhIhVI
L
L
Put the value of I1 from equation (4) to equation (2)
VhVZh
hhI
L
222
11
12212 +
+
−=
So
Image Impedances:
In a two port network, if the impedance at input port with impedance Zi2 connected at output port be Zi1 and the impedance at output port with impedance Zi1 connected
across input port be Zi2 as shown in following figure. Then these impedances Z1i & Z2i are
known as image impedances.
From fig.31a
portinputatimpedancepointDriving1
11 ==
I
VZ i
L
Lop
Zhhhhh
Zh
I
VZ
2221122211
11
2
2
+−
+==
(b)
Network V1
I1 I2
V2 Zi1
(a)
Network V1
I1 I2
V2 Zi2
Fig.31
Zi1→ ←Zi2
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47 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
From fig.31b
portoutputatimpedancepointDriving2
22 ==
I
VZ i
For symmetrical network
impedancesticCharacteri21 === Cii ZZZ
(i) Image Impedance in terms of Input and output Impedance:
ipi ZZ =1 and opi ZZ =2
(ii) Image Impedance in terms of T-Parameters:
)1(2
21 −−−−−+
+==
DCZ
BAZZZ
i
iipi
)2(1
12 −−−−−+
+==
ACZ
BDZZZ
i
iopi
Solving equations (1) & (2)
CD
ABZ i =1 and
CA
DBZ i =2
(iii)Image Impedance in terms of Open circuit and short circuit impedances:
We know
soi ZZD
B
C
A
CD
ABZ 111 =×==
And soi ZZA
B
C
D
CA
DBZ 222 =×==
Note: The image impedances don’t completely define a two port network. So we need another
parameter which we shall get from the voltage and current rations as follows:
We know for T-parameters
)1(221 −−−−−−= BIAVV
)2(221 −−−−−−= DICVI
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48 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
And reversedisIparametersTIn:Note)3( 2222 −−−−−−−= iZIV
From equation (3) 2
22
iZ
VI −= put in equation (1)
222
221 V
Z
BA
Z
VBAVV
ii
+=
+=
)4(22
1 −−−−−
+=⇒
iZ
BA
V
V
Again from equation (3) put the value of V2 to equation (2)
( ) ( ) 222221 IDCZDIZICI ii +−=−−=
( ) )5(22
1 −−−−−+−=⇒ DCZI
Ii
But AC
BDZ i =2 put in equation (4) and (5)
)6(2
1 −−−−−+=+=D
ABCDA
AC
BDBA
V
V
And )7(2
1 −−−−−−+=+=−A
ABCDDD
AC
BDC
I
I
From above equation (6) and (7)
+
+=×−
A
ABCDD
D
ABCDA
I
I
V
V
2
1
2
1
BCABCDAD ++= 2
( )2BCAD +=
So 122
11 −+=+=− ADADBCADIV
IV 1=− BCADQ
Let θcosh=AD
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49 www.eedofdit.webnode.com Prepared by: Nafees Ahmed EED, DIT, DehraDun
So θsinh1 =−AD
Therefore θθθ eIV
IV=+=− sinhcosh
22
11
Or constanttransferImageln2
1ln
22
11
22
11 =
−=−=
IV
IV
IV
IVθ
Again 222111 & IZVIZV ii −== , the image transfer constant may be written as
=
=
2
22
2
11
222
111 ln2
1ln
2
1
IZ
IZ
IIZ
IIZ
i
i
i
iθ
+
=
+
=
2
1
2
1
2
2
2
1
2
1 lnln2
1lnln
2
1
I
I
Z
Z
I
I
Z
Z
i
i
i
i
Other form of image transfer constant
AD1cosh−=θ
1sinhsinh 11 −== −− ADBCθ
o
s
o
s
Z
Z
Z
Z
AD
BC
2
21
1
111 tanhtanhtanh −−− ===θ
Let 1
1tanh
2
2
1
1
+−
=+−
=== −−
θ
θ
θθ
θθ
θe
e
ee
ee
Z
Zm
o
s
⇒ φθ jrem
me =
−+
=1
12 ( )1o1s Z&Zfromcalculatedbecanr,φ
Hence ( ) βαφπθ jnjr +=
++=2
ln2
1 n=1, 2, 3…….etc
( ) constantnattenuatioImageln2
1== rα
constantphase2
=
+=φ
πβ n