Question

How to convert z parameters to Y parameters?

Answer 1

To convert **Z-parameters** (impedance parameters) to **Y-parameters** (admittance parameters), follow these steps:

### Matrix Representation:
If you have the Z-parameter matrix:

\[
\mathbf{Z} = \begin{bmatrix}
Z_{11} & Z_{12} \\
Z_{21} & Z_{22}
\end{bmatrix}
\]

The Y-parameter matrix is the inverse of the Z-parameter matrix:

\[
\mathbf{Y} = \mathbf{Z}^{-1}
\]

Thus,

\[
\mathbf{Y} = \begin{bmatrix}
Y_{11} & Y_{12} \\
Y_{21} & Y_{22}
\end{bmatrix} = \frac{1}{\text{Det}(\mathbf{Z})} \cdot \begin{bmatrix}
Z_{22} & -Z_{12} \\
-Z_{21} & Z_{11}
\end{bmatrix}
\]

Where **Det()** is the determinant of the Z-matrix:

\[
\text{Det}(\mathbf{Z}) = Z_{11} Z_{22} - Z_{12} Z_{21}
\]

### Steps to Convert:

1. **Find the determinant of the Z-matrix:**
   \[
   \text{Det}(\mathbf{Z}) = Z_{11}Z_{22} - Z_{12}Z_{21}
   \]

2. **Calculate each Y-parameter:**
   \[
   Y_{11} = \frac{Z_{22}}{\text{Det}(\mathbf{Z})}
   \]
   \[
   Y_{12} = \frac{-Z_{12}}{\text{Det}(\mathbf{Z})}
   \]
   \[
   Y_{21} = \frac{-Z_{21}}{\text{Det}(\mathbf{Z})}
   \]
   \[
   Y_{22} = \frac{Z_{11}}{\text{Det}(\mathbf{Z})}
   \]

This method applies to a 2-port network, which is common in electrical engineering. If you have a larger matrix (for multi-port networks), the same principle of matrix inversion applies.

Answer 2

To convert Z parameters (impedance parameters) to Y parameters (admittance parameters), you need to use the following relationships between the Z and Y matrices. For a 2-port network, the Z matrix is given by:

\[ \mathbf{Z} = \begin{bmatrix}
Z_{11} & Z_{12} \\
Z_{21} & Z_{22}
\end{bmatrix} \]

And the Y matrix is:

\[ \mathbf{Y} = \begin{bmatrix}
Y_{11} & Y_{12} \\
Y_{21} & Y_{22}
\end{bmatrix} \]

The relationship between Z and Y parameters is:

\[ \mathbf{Y} = \mathbf{Z}^{-1} \]

In matrix terms, to find the Y parameters:

1. Compute the determinant of the Z matrix: \( \text{det}(\mathbf{Z}) = Z_{11}Z_{22} - Z_{12}Z_{21} \).
2. Calculate the inverse of the Z matrix:

\[ \mathbf{Z}^{-1} = \frac{1}{\text{det}(\mathbf{Z})} \begin{bmatrix}
Z_{22} & -Z_{12} \\
-Z_{21} & Z_{11}
\end{bmatrix} \]

3. This inverse matrix will be the Y matrix.

Let me know if you need an example or more details on any of these steps!