Transmission Parameters:
A network is an interconnection of basic electrical elements. A network is called an electric circuit when it has current and voltage sources connected to it. Thus an electric is a closed energized network. Figure below shows an electric network and an electrical circuit.

Two-Port Network
We can write the equations for the dependent variable in terms of the independent variables
(V1,V2) = f(I1,I2)
[V] =[Z][I]
For a two-port network
V1=Z11I1 + Z12I2
V2=Z21I1 + Z22I2
Solving
When I2 =0, the equation reduces to
V1=Z11I1 ; so we get Z11= V1/I1
V2=Z21I1 ; so we get Z21= V2/I1
When I1 =0, the equation reduces to
V1=Z12I2 ; so we get Z12= V1/I2
V2=Z22I2 ; so we get Z22= V2/I2
Equivalent Circuit of Loop equations
We know the two loop equations:
V1=Z11I1 + Z12I2
V2=Z21I1 + Z22I2
Rewriting these equations remembering that : Z12I2 and Z21I1 are current controlled voltage sources (CCVS)
V1=(Z11 – Z12)I1 + Z12(I1 + I2 )
V2= (Z21– Z12)I1 + (Z22 – Z12)I2 +Z12 (I1 + I2)

Short Circuit Admittance
The short circuit admittance is found by writing the node equations. These equations help us to write the dependent variables in terms of the independent variable as:
(I1, I2) = f(V1, V2)
I1=Y11I1 + Y12I2
I2=Y21I1 + Y22I2
Which can be written in matrix form as:
[I] = [Y][V]
Admittance Parameters
- First we short the output terminal, so V2=0 substituting we get:
I1=Y11V1 or Y11 = I1 / V1
I2=Y21V1 or Y21 = I2 / V1
- Then we short circuit the input port , so V1=0
I1= Y12V2 or Y12= I1 / V2
I2= Y22V2 or Y22 = I2 / V2
Equivalent Circuit
The equivalent circuit equations can be written by rearranging the above equation:
I1=(Y11+ Y12)V1 + Y12(V1 – V2)
I2=(Y21 – Y12)V1 + (Y22 – Y12)V2 + Y12(V2 – V1)

Ex. Determine the Y parameter of the following network

- Shorting the output port, V2=0
Therefore
V1-1/3I1 -4/3(I1+I2) =0 => V1 – 5/3I1 -4/3I2 = 0 (1)
and -1/3I2 – 4/3(I2-I1) =0 => -5/3I2 -4/3 I1 =0
solving we get I2 = -4/5I1 and I1 = -5/4I2. (2)
Substituting I2 from eqn (2) in eqn (1) we get
V1 – 5/3I1 – 4/3(-4/5I1 )= 0
Or V1 = 9/15I1 (3)
Therefore Y11 = I1/V1 = 15/9
Y11 = 5/3mho (4)
and also V1 = 9/15x(-5/4I2)
So V1= 3/4I2 (Ignoring direction)
Y21 = I2 / V1 = 4/3mho (5)
(ii) Now short circuiting at the input port, V1=0
Applying KVL again
V2 – 1/3I2 – 4/3(I2 + I1) =0 => V2 – 5/3I2 – 4/3 I1 =0 -(6)
and – 1/3 I1 – 4/3 (I1 + I2) =0 => – 5/3I1 – 4/3I2 = 0 or
I1= -4/5 I2 and I2 = -5/4I1 (7)
Substituting I1 in eqn (6) we get
V2 – 5/3I2 – 4/3 (-4/5I2) =0 V2= 9/15I2 (8)
so, Y22 = I2/V2 15/9
Or Y22 = 5/3mho
From (7) and (8) V2 = 9/15x(5/4I1) =3/4I1
So V2= 3/4I1
Y12 = I1/V2 = 4/3mho
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Example-1: 
Example-2:
Exercise: