Question

Derive the ABCDparameters for the two networks in series, as shown in Figure 5.4.

Short answer

Therefore, the A,B,C,Dparameters are$A_1{A}_2+B_1{C}_2,A_1{B}_2+B_{1}CD,C_1{A}_2+D_1{C}_2\mathrm{and}{C}_1{B}_2+D_1{D}_2.$ .

Step-by-step solution

Given data.

Draw the given circuit

Here, $A_1,B_1,C_1\mathrm{and}{D}_1$are the parameters of network 1 and $A_2,B_2,C_2\mathrm{and}{D}_2$are the parameters of network 2.

Here, $V_s$is the sending end voltage, $V_R$is the receiving end voltage, $l_S$is the sending end current and $l_R$is the receiving end current.

Also, $l_x$is the sending end current and $V_x$is the sending end voltage of network 1.

Determine the formulas of sending end voltage and current.

Write the formula of sending end voltage in terms of transmission line parameters.

${\mathrm{V}}_{\mathrm{S}}={\mathrm{AV}}_{\mathrm{R}}+{\mathrm{Bl}}_{\mathrm{R}}$ ……. (1)

Here, $V_S$is the sending end voltage,$V_R$is the receiving end voltage, and $l_R$is the transmission line current.

Write the formula of sending end current in terms of transmission line parameters.

${\mathrm{l}}_{\mathrm{S}}={\mathrm{CV}}_{\mathrm{R}}+{\mathrm{Dl}}_{\mathrm{R}}$ ……. (2)

Here, $l_S$ is the sending end current, $V_R$ is the receiving end voltage, and $l_R$ is the receiving end current.

Determine the ABCD parameters.