Question

For the two-port network shown in figure, The voltages and currents are given by a relation $V_1 = 2 I_1 + I_2$ and $V_2 = I_1 + I_2$. The admittance matrix $[Y]$ of this network is

• A. $\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$
• B. $\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$
• C. $\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$
• D. $\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix}$

Hint:

To quickly invert a $2 \times 2$ matrix:
Swap the diagonal elements, negate the off-diagonal elements, and divide by the determinant.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem provides the Z-parameter equations of a two-port network and asks for the corresponding admittance $[Y]$ matrix.

Step 2: Key Formula or Approach:
The relations given are in the form of impedance (Z) parameters:
\[ V_1 = Z_{11} I_1 + Z_{12} I_2 \]
\[ V_2 = Z_{21} I_1 + Z_{22} I_2 \]
The admittance matrix $[Y]$ is the inverse of the impedance matrix $[Z]$:
\[ [Y] = [Z]^{-1} \]

Step 3: Detailed Explanation:

• Identify the Z-matrix elements from the given equations:
\[ Z_{11} = 2,\quad Z_{12} = 1 \]
\[ Z_{21} = 1,\quad Z_{22} = 1 \]
\[ [Z] = \begin{bmatrix} 2 & 1 1 & 1 \end{bmatrix} \]

• Find the determinant of $[Z]$:
\[ \det(Z) = (2 \times 1) - (1 \times 1) = 2 - 1 = 1 \]

• Calculate the inverse matrix $[Z]^{-1}$:
\[ [Y] = \frac{1}{\det(Z)} \begin{bmatrix} Z_{22} & -Z_{12} -Z_{21} & Z_{11} \end{bmatrix} \]
\[ [Y] = \frac{1}{1} \begin{bmatrix} 1 & -1 \\-1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} \]

Step 4: Final Answer:
The admittance matrix $[Y]$ is $\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$, which corresponds to Option (A).