## Transmission Parameters

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)I+Z12 (I1 + I2)

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]

• 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

## Example-2:   