Question

For the two port network shown below, The impedances \( Z_{11}, Z_{12}, \) and \( Z_{22} \) (in \( \Omega \)) are A. \(6\Omega\), B. \(9\Omega\), C. \(24\Omega\). Choose the correct sequence of \(Z_{11}, Z_{12}, Z_{22}\) in answer from the options given below:

• A. A, B, C
• B. A, C, B
• C. B, A, C
• D. C, B, A

Hint:

For Pi-networks, calculating Y-parameters first is straightforward. Then, simply invert the Y-matrix to get the Z-matrix. Remember \( [Z] = [Y]^{-1} \).

Answer 1

The Correct Option is C

Step 1: Identify the network type and parameters.
The given circuit is a Pi (\(\Pi\))-network. It is often easier to first calculate the Admittance (Y) parameters for a Pi-network and then convert them to Impedance (Z) parameters.
The admittances are: \(Y_a = 1/10\) S, \(Y_b = 1/60\) S, \(Y_c = 1/30\) S.

Step 2: Calculate the Y-parameters.
For a Pi-network:
\[ Y_{11} = Y_a + Y_c = \frac{1}{10} + \frac{1}{30} = \frac{3+1}{30} = \frac{4}{30} = \frac{2}{15} \, S \] \[ Y_{22} = Y_b + Y_c = \frac{1}{60} + \frac{1}{30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20} \, S \] \[ Y_{12} = Y_{21} = -Y_c = -\frac{1}{30} \, S \] The Y-matrix is: \[ [Y] = \begin{pmatrix} \frac{2}{15} & -\frac{1}{30} \\ -\frac{1}{30} & \frac{1}{20} \end{pmatrix} \]
Step 3: Convert Y-parameters to Z-parameters.
The Z-matrix is the inverse of the Y-matrix: \( [Z] = [Y]^{-1} \).
\[ [Z] = \frac{1}{\Delta_Y} \begin{pmatrix} Y_{22} & -Y_{12} \\ -Y_{21} & Y_{11} \end{pmatrix} \] First, calculate the determinant \( \Delta_Y \): \[ \Delta_Y = (Y_{11})(Y_{22}) - (Y_{12})(Y_{21}) = \left(\frac{2}{15}\right)\left(\frac{1}{20}\right) - \left(-\frac{1}{30}\right)\left(-\frac{1}{30}\right) \] \[ \Delta_Y = \frac{2}{300} - \frac{1}{900} = \frac{6-1}{900} = \frac{5}{900} = \frac{1}{180} \] Now find the Z-matrix: \[ [Z] = \frac{1}{1/180} \begin{pmatrix} 1/20 & 1/30 \\ 1/30 & 2/15 \end{pmatrix} = 180 \begin{pmatrix} 1/20 & 1/30 \\ 1/30 & 2/15 \end{pmatrix} \] \[ [Z] = \begin{pmatrix} 180/20 & 180/30 \\ 180/30 & 180 \cdot (2/15) \end{pmatrix} = \begin{pmatrix} 9 & 6 \\ 6 & 24 \end{pmatrix} \]
Step 4: Identify the individual Z-parameters and match the sequence.
From the Z-matrix, we have:
\( Z_{11} = 9 \, \Omega \)
\( Z_{12} = 6 \, \Omega \)
\( Z_{22} = 24 \, \Omega \)
Matching these values with the given options:
\( Z_{11} = 9\Omega \rightarrow B \)
\( Z_{12} = 6\Omega \rightarrow A \)
\( Z_{22} = 24\Omega \rightarrow C \)
The required sequence for \( Z_{11}, Z_{12}, Z_{22} \) is B, A, C.