# Relationships between Parameters

Since the six sets of parameters relate the same input and output terminal variables of the same two-port network, they should be interrelated. If two sets of parameters exist, we can relate one set to the other set. Let us demonstrate the process with two examples.
Given the $z$ parameters, let us obtain the $y$ parameters. From Eq (1),
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{z}_{11} & \mathbf{z}_{12} \\\mathbf{z}_{21} & \mathbf{z}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=[\mathbf{z}]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right] \tag{1}$$
or
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=[\mathbf{z}]^{-1}\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{2}$$
Also, Eq. (3),
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{y}_{11} & \mathbf{y}_{12} \\\mathbf{y}_{21} & \mathbf{y}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]=[\mathbf{y}]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{3}$$
Comparing Eqs. (2) and (3), we see that $$[\mathbf{y}]=[\mathbf{z}]^{-1} \tag{4}$$ The adjoint of the $[\mathbf{z}]$ matrix is
$$\left[\begin{array}{cc}\mathbf{z}_{22} & -\mathbf{z}_{12} \\-\mathbf{z}_{21} & \mathbf{z}_{11}\end{array}\right]$$
and its determinant is
$$\Delta_{z}=\mathbf{z}_{11} \mathbf{z}_{22}-\mathbf{z}_{12} \mathbf{z}_{21}$$
Substituting these into Eq. (4), we get
$$\left[\begin{array}{ll}\mathbf{y}_{11} & \mathbf{y}_{12} \\\mathbf{y}_{21} & \mathbf{y}_{22}\end{array}\right]=\frac{\left[\begin{array}{cc}\mathbf{z}_{22} & -\mathbf{z}_{12} \\-\mathbf{z}_{21} & \mathbf{z}_{11}\end{array}\right]}{\Delta_{z}} \tag{5}$$
Equating terms yields
$$\mathbf{y}_{11}=\frac{\mathbf{z}_{22}}{\Delta_{z}}, \quad \mathbf{y}_{12}=-\frac{\mathbf{z}_{12}}{\Delta_{z}}, \quad \mathbf{y}_{21}=-\frac{\mathbf{z}_{21}}{\Delta_{z}}, \quad \mathbf{y}_{22}=\frac{\mathbf{z}_{11}}{\Delta_{z}}$$
As a second example, let us determine the $h$ parameters from the $z$ parameters. From Eq. (6),
$$\begin{array}{l}\mathbf{V}_{1}=\mathbf{z}_{11} \mathbf{I}_{1}+\mathbf{z}_{12} \mathbf{I}_{2} \quad (6.1) \\\mathbf{V}_{2}=\mathbf{z}_{21} \mathbf{I}_{1}+\mathbf{z}_{22} \mathbf{I}_{2} \quad (6.2) \end{array}$$
Making $\mathbf{I}_{2}$ the subject of Eq. (6.2),
$$\mathbf{I}_{2}=-\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}} \mathbf{I}_{1}+\frac{1}{\mathbf{z}_{22}} \mathbf{V}_{2} \tag{7}$$
Substituting this into Eq. (6.1),
$$\mathbf{V}_{1}=\frac{\mathbf{z}_{11} \mathbf{z}_{22}-\mathbf{z}_{12} \mathbf{z}_{21}}{\mathbf{z}_{22}} \mathbf{I}_{1}+\frac{\mathbf{z}_{12}}{\mathbf{z}_{22}} \mathbf{V}_{2} \tag{8}$$
Putting Eqs. (7) and (8) in matrix form,
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{cc}\frac{\Delta_{z}}{\mathbf{z}_{22}} & \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}} \\-\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}} & \frac{1}{\mathbf{z}_{22}}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{9}$$
From Eq. (10),
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{h}_{11} & \mathbf{h}_{12} \\\mathbf{h}_{21} & \mathbf{h}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{V}_{2}\end{array}\right] \tag{10}$$
Comparing this with Eq. (9), we obtain
$$\mathrm{h}_{11}=\frac{\Delta_{z}}{Z_{22}}, \quad \mathrm{~h}_{12}=\frac{\mathrm{l}_{12}}{\mathrm{Z}_{22}}, \quad \mathrm{~h}_{21}=-\frac{\mathrm{I}_{21}}{\mathrm{Z}_{22}}, \quad \mathrm{~h}_{22}=\frac{1}{\mathrm{Z}_{22}}$$
Example 1: Find $[\mathbf{z}]$ and $[\mathrm{g}]$ of a two-port network if $$[\mathbf{T}]=\left[\begin{array}{cc}10 & 1.5 \Omega \\2 \mathrm{~S} & 4\end{array}\right]$$
Solution: If $\mathbf{A}=10, \mathbf{B}=1.5, \mathbf{C}=2, \mathbf{D}=4$, the determinant of the matrix is $$\Delta_{T}=\mathbf{A D}-\mathbf{B C}=40-3=37$$
$$\begin{array}{cc}\mathbf{z}_{11}=\frac{\mathbf{A}}{\mathbf{C}}=\frac{10}{2}=5, & \mathbf{z}_{12}=\frac{\Delta_{T}}{\mathbf{C}}=\frac{37}{2}=18.5 \\\mathbf{z}_{21}=\frac{1}{\mathbf{C}}=\frac{1}{2}=0.5, & \mathbf{z}_{22}=\frac{\mathbf{D}}{\mathbf{C}}=\frac{4}{2}=2 \\\mathbf{g}_{11}=\frac{\mathbf{C}}{\mathbf{A}}=\frac{2}{10}=0.2, & \mathbf{g}_{12}=-\frac{\Delta_{T}}{\mathbf{A}}=-\frac{37}{10}=-3.7 \\\mathbf{g}_{21}=\frac{1}{\mathbf{A}}=\frac{1}{10}=0.1, & \mathbf{g}_{22}=\frac{\mathbf{B}}{\mathbf{A}}=\frac{1.5}{10}=0.15\end{array}$$
Thus,
$$[\mathbf{z}]=\left[\begin{array}{cc}5 & 18.5 \\0.5 & 2\end{array}\right] \Omega, \quad[\mathbf{g}]=\left[\begin{array}{cc}0.2 \mathrm{~S} & -3.7 \\0.1 & 0.15 \Omega\end{array}\right]$$
Example 2: Obtain the $y$ parameters of the op amp circuit in Fig. 1. Show that the circuit has no $z$ parameters.
Fig. 1: For Example 2.
Solution:
Since no current can enter the input terminals of the op amp, $\mathbf{I}_{1}=0$, which can be expressed in terms of $\mathbf{V}_{1}$ and $\mathbf{V}_{2}$ as $$\mathbf{I}_{1}=0 \mathbf{V}_{1}+0 \mathbf{V}_{2}$$ Where it gives $$\mathbf{y}_{11}=0=\mathbf{y}_{12}$$ Also, $$\mathbf{V}_{2}=R_{3} \mathbf{I}_{2}+\mathbf{I}_{o}\left(R_{1}+R_{2}\right)$$ where $\mathbf{I}_{o}$ is the current through $R_{1}$ and $R_{2}$. But $\mathbf{I}_{o}=\mathbf{V}_{1} / R_{1}$. Hence, $$\mathbf{V}_{2}=R_{3} \mathbf{I}_{2}+\frac{\mathbf{V}_{1}\left(R_{1}+R_{2}\right)}{R_{1}}$$ which can be written as $$\mathbf{I}_{2}=-\frac{\left(R_{1}+R_{2}\right)}{R_{1} R_{3}} \mathbf{V}_{1}+\frac{\mathbf{V}_{2}}{R_{3}}$$ Comparing this with admittance equation shows that $$\mathbf{y}_{21}=-\frac{\left(R_{1}+R_{2}\right)}{R_{1} R_{3}}, \quad \mathbf{y}_{22}=\frac{1}{R_{3}}$$ The determinant of the $[\mathbf{y}]$ matrix is $$\Delta_{y}=\mathbf{y}_{11} \mathbf{y}_{22}-\mathbf{y}_{12} \mathbf{y}_{21}=0$$ Since $\Delta_{y}=0$, the $[\mathbf{y}]$ matrix has no inverse; therefore, the $[\mathbf{z}]$ matrix does not exist according to Eq. (A). $$[y] = [z]^{-1} \tag{A}$$ Note that the circuit is not reciprocal because of the active element.