Question

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).
\includegraphics[width=0.5\linewidth]{48.png}

Hint:

For cascaded two-port networks, use the relation \( [A B; C D]_R = [A B; C D]_P \times [A B; C D]_Q \).

Answer 1

Step 1: Analyze the network connection. For cascaded networks, the ABCD parameters are obtained by matrix multiplication of the ABCD matrices of individual networks \( P \) and \( Q \).
Certainly, let's convert the given equations into LaTeX format:
Given Z parameter matrix of network P is
\[Z_{P} = \begin{bmatrix} 40~\Omega & 60~\Omega
80~\Omega & 100~\Omega \end{bmatrix}\] And the admittance matrix of network Q is
\[Y_{Q} = \begin{bmatrix} 5s & -2.5s
-2.5s & 1s \end{bmatrix}\] In case of cascade connection of two-port network
Standard z-parameter equation:
\[V_{1} = Z_{11}I_{1} + Z_{12}I_{2}\] \[V_{2} = Z_{21}I_{1} + Z_{22}I_{2}\] Standard ABCD parameter equation:
\[V_{1} = AV_{2} - BI_{2}\] \[I_{1} = CV_{2} - DI_{2}\] Now according to the question
\[V_{1} = 40I_{1} + 60I_{2}\] ... (i) \[V_{2} = 80I_{1} + 100I_{2}\] ... (ii) From equation (ii)
80I_{1} = V_{2} - 100I_{2} \qquad ...(iii) I_{1} = \frac{V_{2}}{80} - \frac{100}{80}I_{2} \text{Put the value of } I_{1} \text{ in equation (i)} V_{1} = 40\left(\frac{V_{2}}{80} - \frac{100}{80}I_{2}\right) + 60I_{2} V_{1} = \frac{V_{2}}{2} - 50I_{2} + 60I_{2} V_{1} = \frac{V_{2}}{2} + 10I_{2} V_{1} = \frac{V_{2}}{2} - (-10)I_{2} \qquad ...(iv) \text{From equation (iii) and (iv)} \begin{bmatrix} A & B
C & D \end{bmatrix}_{P} = \begin{bmatrix} \frac{1}{2} & -10
\frac{1}{80} & \frac{100}{80} \end{bmatrix} \text{Now,} \text{Standard Y - parameter equation} I_{1} = Y_{11}V_{1} + Y_{12}V_{2} I_{2} = Y_{21}V_{1} + Y_{22}V_{2} \text{According to question Y - Parameter equation :} I_{1} = 5V_{1} - 2.5V_{2} \qquad ...(v) I_{2} = -2.5V_{1} + V_{2} \qquad ...(vi) \text{From equation (vi):} 2.5V_{1} = V_{2} - I_{2} V_{1} = \frac{V_{2}}{2.5} - \frac{I_{2}}{2.5} \qquad ...(vii) \text{Put the value of } V_{1} \text{ in equation 5} I_{1} = 5\left[\frac{V_{2}}{2.5} - \frac{I_{2}}{2.5}\right] - 2.5V_{2} I_{1} = \frac{5}{2.5}V_{2} - 2.5V_{2} - \frac{5}{2.5}I_{2} I_{1} = V_{2}\left(2 - \frac{5}{2.5}\right) - 2I_{2}
I_{1} = \left(-\frac{1}{2}\right)V_{2} - 2I_{2} \qquad ...(viii) \text{From equation (vii) and (viii)} \begin{bmatrix} A & B
C & D \end{bmatrix}_{Q} = \begin{bmatrix} \frac{1}{2.5} & \frac{1}{2.5}
-\frac{1}{2} & 2 \end{bmatrix} \begin{bmatrix} A & B
C & D \end{bmatrix} = \begin{bmatrix} A & B
C & D \end{bmatrix}_{P} \begin{bmatrix} A & B
C & D \end{bmatrix}_{Q} = \begin{bmatrix} \frac{1}{2} & -10
\frac{1}{80} & \frac{10}{8} \end{bmatrix} \begin{bmatrix} \frac{2}{5} & \frac{2}{5}
-\frac{1}{2} & 2 \end{bmatrix} \begin{bmatrix} A & B
C & D \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{2} \times \frac{2}{5} + 10 \times \frac{1}{2}\right) & \left(\frac{1}{2} \times \frac{2}{5} - 20\right)
\left(\frac{2}{5} \times \frac{1}{80} - \frac{1}{8} \times \frac{1}{2}\right) & \left(\frac{1}{80} \times \frac{2}{5} \times 2\right) \end{bmatrix} \begin{bmatrix} A & B
C & D \end{bmatrix} = \begin{bmatrix} \left(\frac{1}{5} + 5\right) & \left(\frac{1}{5} - 20\right)
\frac{1}{200} & \left(\frac{1}{200} - \frac{10}{4}\right) \end{bmatrix} \begin{bmatrix} A & B
C & D \end{bmatrix} = \begin{bmatrix} 5.2 & -19.8
0.005 & -2.495 \end{bmatrix} Step 2: Perform calculations. Using the impedance and admittance matrices, calculate the equivalent ABCD parameters of the two-port network \( R \). The parameter \( \beta \) is calculated as: \[ \beta = -19.90 \, \Omega \] \[\alpha = 5.2 \] \[\Gamma = 0.005 \] \[\delta = 2.49 \]