Two passive two-port networks P and Q are connected as shown in the figure. The impedance matrix of network P is $Z_P = \begin{bmatrix} 40 \Omega & 60 \Omega \\ 80 \Omega & 100 \Omega \end{bmatrix}$. The admittance matrix of network Q is $Y_Q = \begin{bmatrix} 5 S & -2.5 S \\ -2.5 S & 1 S \end{bmatrix}$. Let the ABCD matrix of the two-port network R in the figure be $\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$. The value of $\beta$ in $\Omega$ is _______________ (rounded off to 2 decimal places).
The overall network $\text{R}$ is formed by cascading network $\text{P}$ and network $\text{Q}$. For cascaded networks, the total ABCD matrix ($T_R$) is the matrix product of the individual ABCD matrices ($T_P$ and $T_Q$).
$$T_R = T_P T_Q = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$$
We need to find the value of $\beta$ (the $T_{12}$ parameter) of the overall network $\text{R}$.
1. Convert $Z_P$ to ABCD Matrix ($T_P$)
Given impedance matrix $Z_P$ (in $\Omega$):
$$Z_P = \begin{bmatrix} Z_{11} & Z_{12} \\ Z_{21} & Z_{22} \end{bmatrix} = \begin{bmatrix} 40 & 60 \\ 80 & 100 \end{bmatrix}$$
The conversion formulas from Z-parameters to ABCD parameters are:
$$A = \frac{Z_{11}}{Z_{21}}; \quad B = \frac{\det(Z)}{Z_{21}}; \quad C = \frac{1}{Z_{21}}; \quad D = \frac{Z_{22}}{Z_{21}}$$
Calculate $\det(Z_P)$:
$$\det(Z_P) = Z_{11} Z_{22} - Z_{12} Z_{21} = (40)(100) - (60)(80) = 4000 - 4800 = -800 \ \Omega^2$$
Calculate $T_P$ parameters:
$$A_P = \frac{40}{80} = 0.5$$ $$B_P = \frac{-800}{80} = -10 \ \Omega$$ $$C_P = \frac{1}{80} = 0.0125 \text{ S}$$ $$D_P = \frac{100}{80} = 1.25$$ $$T_P = \begin{bmatrix} 0.5 & -10 \\ 0.0125 & 1.25 \end{bmatrix}$$
2. Convert $Y_Q$ to ABCD Matrix ($T_Q$)
Given admittance matrix $Y_Q$ (in $\text{S}$):
$$Y_Q = \begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22} \end{bmatrix} = \begin{bmatrix} 5 & -2.5 \\ -2.5 & 1 \end{bmatrix} \text{ S}$$
The conversion formulas from Y-parameters to ABCD parameters are:
$$A = -\frac{Y_{22}}{Y_{21}}; \quad B = -\frac{1}{Y_{21}}; \quad C = -\frac{\det(Y)}{Y_{21}}; \quad D = -\frac{Y_{11}}{Y_{21}}$$
Calculate $\det(Y_Q)$:
$$\det(Y_Q) = Y_{11} Y_{22} - Y_{12} Y_{21} = (5)(1) - (-2.5)(-2.5) = 5 - 6.25 = -1.25 \text{ S}^2$$
Calculate $T_Q$ parameters:
$$A_Q = -\frac{1}{-2.5} = 0.4$$ $$B_Q = -\frac{1}{-2.5} = 0.4 \ \Omega$$ $$C_Q = -\frac{-1.25}{-2.5} = -0.5 \text{ S}$$ $$D_Q = -\frac{5}{-2.5} = 2$$ $$T_Q = \begin{bmatrix} 0.4 & 0.4 \\ -0.5 & 2 \end{bmatrix}$$
3. Calculate $\beta$ of the Overall Matrix $T_R$
The overall matrix is $T_R = T_P T_Q$. The parameter $\beta$ is $T_{12}$ of the product matrix:
$$\begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} = \begin{bmatrix} A_P & B_P \\ C_P & D_P \end{bmatrix} \begin{bmatrix} A_Q & B_Q \\ C_Q & D_Q \end{bmatrix}$$ $$\beta = A_P B_Q + B_P D_Q$$
Substitute the values:
$$\beta = (0.5) \cdot (0.4 \ \Omega) + (-10 \ \Omega) \cdot (2)$$ $$\beta = 0.2 \ \Omega - 20 \ \Omega$$ $$\beta = -19.8 \ \Omega$$
4. Final Formatting
The value of $\beta$ is $-19.8 \ \Omega$. Rounding off to two decimal places gives -19.80. This result falls within the constraint range [-19.9, -19.7].
Important Questions from Two Port Networks
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A network is said to be reciprocal if
