Question 17.9: The networks Na and Nb in Figure 17–14 are connected in para......

To get the desired y-parameters we use the fact that $[y]=[y_{a}]+[y_{b}]$ in a paralle connection. We are given $[z_{a}]$ and $[y_{b}]$, so we need to convert the z-matrix of $N_{a}$ into a y-matrix. Equation (17–17) indicates that $[y]=[z]^{-1}$, so the desired y-matrix is found to be

$[\pmb{y}] = [\pmb{z}_{a}]^{-1} +[\pmb{y}_{b}]= \begin{bmatrix} 15& 5 \\5& 10 \end{bmatrix}^{-1}+ \begin{bmatrix} 0.2& -0.2 \\-0.2& 0.7 \end{bmatrix}$

$= \begin{bmatrix} 0.08&-0.04\\-0.04 &0.12\end{bmatrix}+\begin{bmatrix}0.2& -0.2 \\-0.2& 0.7 \end{bmatrix}=\begin{bmatrix}0.28 &-0.24\\-0.24& 0.82 \end{bmatrix}S$

$\begin{bmatrix}y_{11}& y_{12}\\y_{21}& y_{22} \end{bmatrix}=[\pmb{z}]^{-1}=\frac{adj[z]}{det[z]}=\begin{bmatrix}\frac{z_{22}}{Δ_{z}}&\frac{-z_{12}}{Δ_{z}}\\\frac{-z_{21}}{Δ_{z}}&\frac{z_{11}}{Δ_{z}}\end{bmatrix}$     (17–17)

Given the y-matrix of the parallel connection, the y-parameters are seen to be $y_{11} = 0.28  S, y_{12} = y_{21} = -0.24$ S, and $y_{22}$ = 0.82 S.