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).

 

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

Correct Answer: -19.9

Step 1: Analyze the network connection

For cascaded two-port networks, the overall ABCD parameters are obtained by multiplying the ABCD matrices of the individual networks.

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Given:

Z-parameter matrix of network P:

\[ Z_P = \begin{bmatrix} 40 & 60 \\ 80 & 100 \end{bmatrix} \; \Omega \]

Admittance matrix of network Q:

\[ Y_Q = \begin{bmatrix} 5s & -2.5s \\ -2.5s & 1s \end{bmatrix} \]

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Step 2: Convert Z-parameters of P to ABCD parameters

Standard Z-parameter equations:

\[ V_1 = Z_{11}I_1 + Z_{12}I_2 \] \[ V_2 = Z_{21}I_1 + Z_{22}I_2 \]

Substituting values:

\[ V_1 = 40I_1 + 60I_2 \quad (i) \] \[ V_2 = 80I_1 + 100I_2 \quad (ii) \]

From (ii):

\[ I_1 = \frac{V_2}{80} - \frac{100}{80}I_2 \]

Substitute in (i):

\[ V_1 = 40\left(\frac{V_2}{80} - \frac{100}{80}I_2\right) + 60I_2 \] \[ V_1 = \frac{V_2}{2} + 10I_2 \]

Comparing with standard ABCD equations:

\[ V_1 = AV_2 - BI_2,\quad I_1 = CV_2 - DI_2 \]

ABCD matrix of P:

\[ \begin{bmatrix} A & B \\ C & D \end{bmatrix}_P = \begin{bmatrix} \frac{1}{2} & -10 \\ \frac{1}{80} & \frac{100}{80} \end{bmatrix} \]

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Step 3: Convert Y-parameters of Q to ABCD parameters

Standard Y-parameter equations:

\[ I_1 = Y_{11}V_1 + Y_{12}V_2 \] \[ I_2 = Y_{21}V_1 + Y_{22}V_2 \]

Substituting values:

\[ I_1 = 5V_1 - 2.5V_2 \quad (v) \] \[ I_2 = -2.5V_1 + V_2 \quad (vi) \]

From (vi):

\[ V_1 = \frac{V_2}{2.5} - \frac{I_2}{2.5} \]

Substitute into (v):

\[ I_1 = 5\left(\frac{V_2}{2.5} - \frac{I_2}{2.5}\right) - 2.5V_2 \] \[ I_1 = -\frac{1}{2}V_2 - 2I_2 \]

Thus, ABCD matrix of Q:

\[ \begin{bmatrix} A & B \\ C & D \end{bmatrix}_Q = \begin{bmatrix} \frac{2}{5} & \frac{2}{5} \\ -\frac{1}{2} & 2 \end{bmatrix} \]

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Step 4: Find ABCD parameters of cascaded network R

\[ \begin{bmatrix} A & B \\ C & D \end{bmatrix}_R = \begin{bmatrix} A & B \\ C & D \end{bmatrix}_P \begin{bmatrix} A & B \\ C & D \end{bmatrix}_Q \]

\[ = \begin{bmatrix} \frac{1}{2} & -10 \\ \frac{1}{80} & \frac{10}{8} \end{bmatrix} \begin{bmatrix} \frac{2}{5} & \frac{2}{5} \\ -\frac{1}{2} & 2 \end{bmatrix} \]

\[ = \begin{bmatrix} 5.2 & -19.8 \\ 0.005 & -2.495 \end{bmatrix} \]

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Step 5: Final ABCD parameters

\[ \alpha = 5.2,\quad \beta = -19.9~\Omega,\quad \gamma = 0.005,\quad \delta = 2.49 \]