ABCD Parameter of Two-Port Network

ABCD Parameter of Two-Port Network

The ABCD parameters are used to represent the mathematical model of the transmission line in terms of the two-port network. And hence, the ABCD parameter is also known as Transmission Parameter.

The input voltage and current of the two-port network can be expressed in terms of the output voltage and current of the two-port network. The transmission line is a four-terminal, linear, passive, and bilateral network. Therefore, the two-port network must be four-terminal, linear, passive, and bilateral networks.

The transmission line is the biggest part of the power system network and it is most prone to fault. Hence, a lot of research needed while designing the transmission line. The power system engineers need to understand the mathematical model of the transmission line and the ABCD parameter helps a lot to research the transmission line.

As we know according to the length of a line, the transmission lines are classified as Short, Medium, and Long Transmission lines. The mathematical model of the transmission line is used to derive the actual behavior of the transmission line. These different types of lines have different behavior in the actual power system network. And hence, the mathematical models are different for a different type of transmission line. Therefore, the ABCD parameters are also different for a different type of transmission line.

Consider a below two-port network that has four terminals (two input terminals and two output terminals).

ABCD Parameter
ABCD Parameter

The equations of ABCD Parameters give us the relation between the sending end and receiving end specifications. These equations are described below.

    \[ V_1 = A V_2 + B(I_2) \]

    \[ I_1 = C V_2 + D(I_2) \]

Matrix form of the above equations are,

    \[ \begin{bmatrix} V_1 \\ I_1 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_2 \\ I_2 \end{bmatrix} \]

In this equation, A, B, C, and D are transmission line parameters and we will derive the values for these parameters.

Calculation of ABCD Parameter

If we consider that the port-1 is sending end and port-2 is receiving end of a transmission line, then the two-port network of a transmission line is as the below figure.

ABCD Parameter for Transmission Line
ABCD Parameter for Transmission Line

For the above figure, equations of ABCD parameters are as shown in the below,

(1)   \begin{equation*} \begin{align} V_s &= A V_r + B(I_r) \\ I_s &= C V_r + D(I_r) \end{align} \end{equation*}

To calculate the value of ABCD parameters, we need to perform two cases. In the first case, we need to open the receiving end terminals and we will get A and C parameter. And in the second case, we need to short the receiving end terminals and we will get B and D parameter.

Case-1

Now, we consider the first case in which, the receiving end of the line is open-circuited. And hence the current flowing through the receiving end IR is zero.

Put this value in the equation-1,

    \[ V_s = A V_r + B(0) \]

    \[ I_s = C V_r + D(0) \]

Therefore,

    \[ A = \left \frac{V_s}{V_r} \right\vert_{I_r = 0} \]

    \[ C = \left \frac{I_s}{V_r} \right\vert_{I_r = 0} \]

The value of A gives us the ratio of sending end voltage to the receiving end (open-circuit) voltage. It is a ratio of the same quantity. So, it is unitless.

The value of C gives us the ratio of sending end current to receiving end (open-circuit) voltage. It is a ratio of current to voltage. So, the unit of C parameter is mho.

Case-2

Now, we consider the second case in which, the terminals of receiving end of a line are short-circuited. And hence the current flowing through the receiving end VR is zero.

Put this value in the equation-1,

    \[ V_s = A (0) + B(I_r) \]

    \[ I_s = C (0) + D(I_r) \]

Hence,

    \[ B = \left \frac{V_s}{I_r} \right\vert_{V_r = 0} \]

    \[ D = \left \frac{I_s}{I_r} \right\vert_{V_r = 0} \]

The value of B, gives us the ratio of sending end voltage to the receiving end (short-circuit) current. It is a ratio of voltage to current. Therefore, the unit of the B parameter is ohm.

The value of D, gives us the ratio of sending end current to the receiving end (short-circuit) current. It is a ratio of the same quantity. So, the D parameter is unitless.

Reciprocity Conditions for ABCD Parameter

A network is said to be reciprocal if the port-2 voltage is due to applied current at port-1 is the same as the port-1 voltage when the applied current is the same as port-2.

Generally, the network that consists of passive elements (resistor, inductor, and capacitor) only, that network are reciprocal networks and the network that consists of the active element (transistor or generator), that networks are not a reciprocal network.

Find the reciprocity condition of ABCD parameter, we have to follow two conditions,

Reciprocity Condition-1

    \[ V_1=V_s \quad V_2=0 \quad I_2=I_2' \]

Put these values in equation-1,

    \[ V_s = 0 + B(I_2') \]

    \[ \frac{V_s}{I_2'} = B \]

Reciprocity Condition-2

    \[ V_2=V_s \quad V_1=0 \quad I_1=-I_1' \]

Put these values in equation-1,

    \[ 0 = A V_s + B I_2 \]

    \[ A V_s = - B I_2 \]

    \[ I_2 = -\frac{A}{B} V_s \]

    \[ -I_1' = C V_s + D\left[-\frac{A}{B} V_s \right] \]

    \[ -I_1' = V_s\left( C + D\left[-\frac{A}{B} \right] \right) \]

    \[ \frac{V_s}{I_1'} = \frac{-1}{C + D\left[-\frac{A}{B} \right]} \]

    \[ \frac{V_s}{I_1'} = \frac{-B}{BC-AD} \]

Now, compare both conditions,

    \[ \frac{V_s}{I_2'} = \frac{V_s}{I_1'} \]

    \[ B = \frac{-B}{BC-AD} \]

    \[ B = \frac{B}{AD-BC} \]

    \[ AD - BC = 1 \]

Symmetry Conditions for ABCD Parameter

The electrical network is said to be symmetrical if the input impedance is equal to the output impedance. It is not necessary that, these circuits are physically symmetrical.

Find the Symmetry condition of ABCD parameter, we have to follow two conditions,

Symmetry Condition-1

    \[V_1 = V_s \quad and \quad I_2=0 \]

Put these values in equation-1,

    \[ V_s = A V_2 \]

    \[ V_2 = \frac{V_s}{A} \]

    \[ I_1 = C V_2 \]

    \[ V_2 = \frac{I_1}{C} \]

Now,

    \[ \frac{V_s}{A} = \frac{I_1}{C} \]

    \[ \frac{V_s}{I_1} = \frac{A}{C} \]

Symmetry Condition-2

[ V_2 = V_s \quad and \quad I_1 = 0\]

    \[ V_1 = A V_s + BI_2 \]

    \[ 0 = CV_s + DI_2 \]

    \[ CV_s = -DI_2\]

    \[ \frac{V_s}{I_2} = \frac{-D}{C} \]

    \[ \left| \frac{V_s}{I_2} \right| = \left| \frac{-D}{C} \right| \]

    \[ \frac{V_s}{I_2} = \frac{D}{C} \]

Impedances of both ports are same for the symmetrical network.

    \[ \frac{V_s}{I_2} = \frac{V_s}{I_1} \]

    \[ \frac{D}{C} = \frac{A}{C} \]

    \[ A = D \]

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