Question 17.9: The networks Na and Nb in Figure 17–14 are connected in para......
The networks N_{a} and N_{b} in Figure 17–14 are connected in parallel and have two-port matrices
[\pmb{z}_{a}] = \begin{bmatrix} 15& 5 \\5& 10 \end{bmatrix} Ω and [\pmb{y}_{b}] = \begin{bmatrix} 0.2& -0.2 \\-0.2& 0.7 \end{bmatrix} S
Find the y-parameters of the parallel connection.
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.