For the two-port network shown below, the short-circuit admittance parameter matrix is

This question was previously asked in

GATE EC 2010 Official Paper

Option 1 : \(\begin{bmatrix} 4 & -2 \\\ -2 & 4 \end{bmatrix}S\)

__Concept:__

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{Y_{11}}}&{{Y_{12}}}\\ {{Y_{21}}}&{{Y_{22}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{V_2}} \end{array}} \right]\)

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

__Calculation:__

When V_{2} = 0 or short circuit port 2

Then, I_{1} = Y_{11} V_{1}

I_{2} = Y_{21} V_{1}

V_{1} and I_{1} relation can be drawn from current division rule as

\(V_1 = 0.5\times ({\frac{0.5}{0.5+0.5}})I_1\)

V_{1} = 0.25I_{1}

I_{1} = 4V_{1}

Y11 = 4 S

similarly, V1 and I2 relation can be drawn as

V_{1} = 0.5(-I_{2})

I_{2} = -2V_{1}

Y21 = -2 S

By applying the same procedure for port 1

When V1 = 0 or short circuit port 1

Then, I1 = Y12 V2

I2 = Y22 V2

V2 and I2 relation can be drawn from current division rule as

\(V_2 = 0.5\times ({\frac{0.5}{0.5+0.5}})I_2\)

V2 = 0.25I2

I2 = 4V2

Y22 = 4 S

similarly, V1 and I2 relation can be drawn as

V2 = 0.5(-I1)

I1 = -2V2

Y12 = -2 S

\( \left[ {\begin{array}{*{20}{c}} {{Y_{11}}}&{{Y_{12}}}\\ {{Y_{21}}}&{{Y_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4}&{-2}\\ {-2}&{4} \end{array}} \right]~S\)

Hence** option 1 is correct**

For the given π network

Y parameter can be calculated by

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{Y_1} + {Y_2}}&{ - {Y_2}}\\ { - {Y_2}}&{{Y_2} + {Y_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{V_2}} \end{array}} \right]\\ \)

Substituting the value of Y_{1}, Y_{2,} and Y_{3}

\(= \left[ {\begin{array}{*{20}{c}} {\left( {2 +2} \right)}&{\left( { - 2} \right)}\\ {\left( { - 2} \right)}&{\left( {2 + 2} \right)} \end{array}} \right] \\\)

\(= \left[ {\begin{array}{*{20}{c}} {4}&{ - 2}\\ { - 2}&{4} \end{array}} \right]~S\)