The circuit is given in the figure.

Solution:

$ \begin{gathered} \\ I_1(s)=a+b \\\\ \frac{V_1(s)-V_x(s)}{1}=\frac{V_x(s)}{2}+\frac{V_x(s)-V_2(s)}{2} \\\\ 2 V_1(s)-2 V_x(s)=2 V_x(s)-V_2(s) \\\\ 4 V_x(s)=2 V_1(s)+V_2(s) \\\\ V_x(s)=\frac{1}{2} V_1(s)+\frac{1}{4} V_2(s)....(1) \\\\ \text { KCL } at \text { node }(2) \\\\ I_2(s)+b=C \\\\ \therefore I_2(s)=c-b \\\\ I_2(s)=\frac{V_2(s)}{4}-\left[\frac{V_x(s)-V_2(s)}{2}\right] \\\\ =\frac{1}{4} V_2(s)+\frac{1}{2} V_2(s)-\frac{1}{2} V_x(s) \\\\ =\frac{3}{4} V_2(s)-\frac{1}{2} V_x(s)....(2) \\ \end{gathered} \\ $

put eqn (1) in (2),

$ \begin{aligned} \\ \therefore I_2(s) &=\frac{3}{4} V_2(s)-\frac{1}{2}\left[\frac{1}{2} V_1(s)+\frac{1}{4} V_2(s)\right] \\\\ &=\frac{3}{4} V_2(s)-\frac{1}{4} V_1(s)-\frac{1}{8} V_2(s) \\\\ I_2(s) &=-\frac{1}{4} V_1(s)+\frac{5}{8} V_2(s)...(3) \\ \end{aligned} \\ $

from fig.

$ \begin{aligned} \\ I_1(s) &=\frac{V_1(s)-V_x(s)}{1} \\\\ &=V_1(s)-V_x(s)...(4) \\ \end{aligned} \\ $

put equation (1) in (4)

$ \begin{aligned} \\ &I_1(s)=V_1(s)-\left[\frac{1}{2} V_1(s)+\frac{1}{4} V_2(s)\right] \\\\ &I_1(s)=\frac{1}{2} V_1(s)-\frac{1}{4} V_2(s)....(5) \\ \end{aligned} \\ $

from eqn (3) and (5) Required $Y$-parameter are $ \left[\begin{array}{ll} \ y_{11} & y_{12} \\ y_{21} & y_{22} \ \end{array}\right]=\left[\begin{array}{ll} \ 1 / 2 & -1 / 4 \\ -1 / 4 & 5 / 8 \ \end{array}\right] \ $