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=  Y₁₂V_{2                                             or}                    Y₁₂= I₁ / V₂

I₂=  Y₂₂V_{2                                             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  = 9/15I_{1                                                                                                                                                   (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

Click here to view: Transmission Parameter

Example-1: 

Example-2:

Exercise: