Two-Port Network Y-Matrix Analysis

This analysis focuses on a two-port network characterized by its short-circuit admittance matrix (Y-matrix). The given Y-matrix is:

Our goal is to determine whether this network is reciprocal or non-reciprocal, and whether it is passive or active.

Reciprocity Check for Two-Port Network

The condition for a two-port network to be reciprocal is that its Y-matrix must be symmetric. This means the element $y_{12}$ must be equal to the element $y_{21}$ ($y_{12} = y_{21}$).

From the provided Y-matrix:

• $$ y_{12} = - \frac{1}{2} $$
• $$ y_{21} = \frac{1}{2} $$

By comparing these values, we see that $ - \frac{1}{2} \neq \frac{1}{2} $. Since $y_{12} \neq y_{21}$, the Y-matrix is not symmetric. Consequently, the two-port network is classified as non-reciprocal.

Passivity Check for Two-Port Network

A network is considered passive if it does not generate power; it can only dissipate or store energy. For a two-port network described by its Y-matrix, passivity can be confirmed by checking the following conditions:

• The real parts of the driving-point admittances ($y_{11}$ and $y_{22}$) must be non-negative. Mathematically, $Re(y_{11}) \ge 0$ and $Re(y_{22}) \ge 0$.
• The determinant of the conjugate transpose of the Y-matrix ($det(Y^*)$) must be non-negative. That is, $det(Y^*) \ge 0$.

Driving Point Admittances Check

Let's check the first condition using the diagonal elements of the given Y-matrix:

• $y_{11} = 0$. The real part is $Re(y_{11}) = 0$. This satisfies the condition $0 \ge 0$.
• $y_{22} = 0$. The real part is $Re(y_{22}) = 0$. This also satisfies the condition $0 \ge 0$.

Since both $Re(y_{11})$ and $Re(y_{22})$ are non-negative, the first condition for passivity is met.

Determinant Check

Next, we evaluate the second condition involving the determinant of the conjugate transpose, $Y^*$. First, find the transpose of the Y-matrix, $Y^T$:

The conjugate transpose $Y^*$ is obtained by taking the complex conjugate of each element of $Y^T$. As all elements in this matrix are real numbers, their complex conjugates are themselves. Thus, $Y^* = Y^T$:

Now, we compute the determinant of $Y^*$:

$$ det(Y^*) = (y^*_{11} \times y^*_{22}) - (y^*_{12} \times y^*_{21}) $$

Substituting the values:

$$ det(Y^*) = (0 \times 0) - \left( \frac{1}{2} \times (-\frac{1}{2}) \right) $$

$$ det(Y^*) = 0 - \left( -\frac{1}{4} \right) $$

$$ det(Y^*) = \frac{1}{4} $$

The result $det(Y^*) = \frac{1}{4}$ satisfies the condition $det(Y^*) \ge 0$. This confirms the second condition for passivity.

Since both sets of conditions are met, the network is classified as passive.

Classification Summary

Based on the calculations derived from the short-circuit admittance matrix:

• The network is non-reciprocal due to the asymmetry ($y_{12} \neq y_{21}$) of the Y-matrix.
• The network is passive as it satisfies the necessary conditions related to driving-point admittances and the determinant of the conjugate transpose matrix.

Therefore, the two-port network described by the given Y-matrix is non-reciprocal and passive.