The ABCD parameters of the following 2-port network are

This question was previously asked in

GATE EC 2015 Official Paper: Shift 3

Option 2 : \(\left[ {\begin{array}{*{20}{c}} {3.5 + j2}&{30.5}\\ {0.5}&{3.5 - j2} \end{array}} \right]\)

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

ABCD parameter for any two-port network given below is expressed as:

\(\begin{array}{*{20}{c}} {{V_1} = A{V_2} - B{I_2}}\\ {{I_1} = C{V_2} - D{I_2}} \end{array} \equiv \left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\)

**Calculation:**

\(\begin{array}{*{20}{c}} {A = {{\left. {\frac{{{V_1}}}{{{V_2}}}} \right|}_{{I_2} = 0}}}&{C = {{\left. {\frac{{{I_1}}}{{{V_2}}}} \right|}_{{I_2} = 0}}}\\ {B = {{\left. {\frac{{ - {V_1}}}{{{I_2}}}} \right|}_{{V_2} = 0}}}&{D = {{\left. {\frac{{ - {I_1}}}{{{I_2}}}} \right|}_{{V_2} = 0}}} \end{array}\)

**For I _{2} = 0,**

Applying voltage division across Z_{3}:

\({V_2} = \frac{2}{{7 + j4}}{V_1} \) ---- (1)

\(\frac{{{V_1}}}{{{V_2}}} = \left( {3.5 + j2} \right) = A\)

Applying Ohms Law across Z_{3}

\({V_2} = {I_1}{Z_3} \) ---- (2)

\( \frac{{{I_1}}}{{{V_2}}} = \frac{1}{{{Z_3}}} = C = 0.5\)

**For V _{2} = 0,**

Applying current division:

\(\frac{2}{{7 - j4}}{I_1} = \left( { - {I_2}} \right)\)

\( \Rightarrow - \frac{{{I_1}}}{{{I_2}}} = \frac{{7 - j4}}{2} \) ----(3)

\(-\frac{I_1}{I_2}= 3.5 - j2 = D\)

Apply KVL,

\({V_1} = \left[ {\left( {5 + j4} \right) + \left( 2 \right)} \right]{I_1} - 2{I_2} \)

\(= - \left( {7 + j4} \right)\frac{{\left( {7 - j4} \right)}}{2}{I_2} - 2{I_2}\;\)

\( - \frac{{{V_1}}}{{{I_2}}} = 30.5 = B\) ---- (4)

Form 1, 2, 3 & 4

\(\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3.5 + j2}&{30.5}\\ {0.5}&{3.5 - j2} \end{array}} \right]\)