# Impedance Parameters

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Impedance and admittance parameters are commonly used in the synthesis of filters. They are also useful in the design and analysis of impedance-matching networks and power distribution networks. We discuss impedance parameters in this section and admittance parameters in the next section.
Fig. 1: The linear two-port network: (a) driven by voltage sources, (b) driven by current sources.
A two-port network may be voltage-driven as in Fig. 1(a) or current-driven as in Fig. 1(b). From either Fig. 1(a) or (b), the terminal voltages can be related to the terminal currents as $$\bbox[10px,border:1px solid grey]{\begin{array}{l}\mathbf{V}_{1}=\mathbf{z}_{11} \mathbf{I}_{1}+\mathbf{z}_{12} \mathbf{I}_{2} \\\mathbf{V}_{2}=\mathbf{z}_{21} \mathbf{I}_{1}+\mathbf{z}_{22} \mathbf{I}_{2}\end{array}} \tag{1}$$ or in matrix form as $$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{z}_{11} & \mathbf{z}_{12} \\\mathbf{z}_{21} & \mathbf{z}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=[\mathbf{z}]\left[\begin{array}{c}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right] \tag{2}$$ where the $\mathbf{z}$ terms are called the impedance parameters, or simply $z$ parameters, and have units of ohms.
The values of the parameters can be evaluated by setting $\mathbf{I}_{1}=0$ (input port open-circuited) or $\mathbf{I}_{2}=0$ (output port open-circuited). Thus, $$\begin{array}{ll}\mathbf{z}_{11}=\left.\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}\right|_{\mathbf{I}_{2}=0}, & \mathbf{z}_{12}=\left.\frac{\mathbf{V}_{1}}{\mathbf{I}_{2}}\right|_{\mathbf{I}_{1}=0} \\\mathbf{z}_{21}=\left.\frac{\mathbf{V}_{2}}{\mathbf{I}_{1}}\right|_{\mathbf{I}_{2}=0}, & \mathbf{z}_{22}=\left.\frac{\mathbf{V}_{2}}{\mathbf{I}_{2}}\right|_{\mathbf{I}_{1}=0}\end{array} \tag{3}$$ Since the $z$ parameters are obtained by open-circuiting the input or output port, they are also called the open-circuit impedance parameters. Specifically,
$\mathbf{z}_{11}=$ Open-circuit input impedance
$\mathbf{z}_{12}=$ Open-circuit transfer impedance from port 1 to port 2
$\mathbf{z}_{21}=$ Open-circuit transfer impedance from port 2 to port 1
$\mathbf{z}_{22}=$ Open-circuit output impedance
According to Eq. (3), we obtain $\mathbf{z}_{11}$ and $\mathbf{z}_{21}$ by connecting a voltage $\mathbf{V}_{1}$ (or a current source $\mathbf{I}_{1}$ ) to port 1 with port 2 open-circuited as in Fig. 2(a) and finding $\mathbf{I}_{1}$ and $\mathbf{V}_{2}$;
Fig. 2: Determination of the z parameters: (a) finding $z_{11}$ and $z_{21}$, (b) finding $z_{12}$ and $z_{22}$.
we then get $$\mathbf{z}_{11}=\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}, \quad \mathbf{z}_{21}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{1}}$$
Similarly, we obtain $\mathbf{z}_{12}$ and $\mathbf{z}_{22}$ by connecting a voltage $\mathbf{V}_{2}$ (or a current source $\mathbf{I}_{2}$ ) to port 2 with port 1 open-circuited as in Fig. 2(b) and finding $\mathbf{I}_{2}$ and $\mathbf{V}_{1}$; we then get $$\mathbf{z}_{12}=\frac{\mathbf{V}_{1}}{\mathbf{I}_{2}}, \quad \mathbf{z}_{22}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{2}}$$ The above procedure provides us with a means of calculating or measuring the $z$ parameters.
Sometimes $\mathbf{z}_{11}$ and $\mathbf{z}_{22}$ are called driving-point impedances, while $\mathbf{z}_{21}$ and $\mathbf{z}_{12}$ are called transfer impedances. A driving-point impedance is the input impedance of a two-terminal (one-port) device. Thus, $\mathbf{z}_{11}$ is the input driving-point impedance with the output port open-circuited, while $\mathbf{z}_{22}$ is the output driving-point impedance with the input port open-circuited.
When $\mathbf{z}_{11}=\mathbf{z}_{22}$, the two-port network is said to be symmetrical. This implies that the network has mirrorlike symmetry about some center line; that is, a line can be found that divides the network into two similar halves.
When the two-port network is linear and has no dependent sources, the transfer impedances are equal $\left(\mathbf{z}_{12}=\mathbf{z}_{21}\right)$, and the two-port is said to be reciprocal. This means that if the points of excitation and response are interchanged, the transfer impedances remain the same.
Fig. 3: Interchanging a voltage source at one port with an ideal ammeter at the other port produces the same reading in a reciprocal two-port.
As illustrated in Fig. 3, a two-port is reciprocal if interchanging an ideal voltage source at one port with an ideal ammeter at the other port gives the same ammeter reading. The reciprocal network yields $\mathbf{V}=\mathbf{z}_{12} \mathbf{I}$ according to Eq. (1) when connected as in Fig. 3(a), but yields $\mathbf{V}=\mathbf{z}_{21} \mathbf{I}$ when connected as in Fig. 3(b). This is possible only if $\mathbf{z}_{12}=\mathbf{z}_{21}$. Any two-port that is made entirely of resistors, capacitors, and inductors must be reciprocal. For a reciprocal network, the T-equivalent circuit in Fig. 4(a) can be used. If the network is not reciprocal, a more general equivalent network is shown in Fig. 4(b); notice that this figure follows directly from Eq. (1).
Fig. 4: (a) T-equivalent circuit (for reciprocal case only), (b) general equivalent circuit.
Fig. 5: An ideal transformer has no z parameters.
It should be mentioned that for some two-port networks, the $z$ parameters do not exist because they cannot be described by Eq. (1). As an example, consider the ideal transformer of Fig. 5. The defining equations for the two-port network are: $$\mathbf{V}_{1}=\frac{1}{n} \mathbf{V}_{2}, \quad \mathbf{I}_{1}=-n \mathbf{I}_{2}$$ Observe that it is impossible to express the voltages in terms of the currents, and vice versa, as Eq. (18.1) requires. Thus, the ideal transformer has no $z$ parameters. However, it does have hybrid parameters,
Example 1: Determine the $z$ parameters for the circuit in Fig. 6.
Fig. 6: For Example 1.
Solution:
METHOD 1 To determine $\mathbf{z}_{11}$ and $\mathbf{z}_{21}$, we apply a voltage source $\mathbf{V}_{1}$ to the input port and leave the output port open as in Fig. 7(a). Then, $$\mathbf{z}_{11}=\frac{\mathbf{V}_{1}}{\mathbf{I}_{1}}=\frac{(20+40) \mathbf{I}_{1}}{\mathbf{I}_{1}}=60 \Omega$$ that is, $\mathbf{z}_{11}$ is the input impedance at port 1. $$\mathbf{z}_{21}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{1}}=\frac{40 \mathbf{I}_{1}}{\mathbf{I}_{1}}=40 \Omega$$ To find $\mathbf{z}_{12}$ and $\mathbf{z}_{22}$, we apply a voltage source $\mathbf{V}_{2}$ to the output port and leave the input port open as in Fig. 7(b).
Fig. 7: For Example 1: (a) finding $z_{11}$ and $z_{21}$, (b) finding $z_{12}$ and $z_{22}$.
Then, $$\mathbf{z}_{12}=\frac{\mathbf{V}_{1}}{\mathbf{I}_{2}}=\frac{40 \mathbf{I}_{2}}{\mathbf{I}_{2}}=40 \Omega, \quad \mathbf{z}_{22}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{2}}=\frac{(30+40) \mathbf{I}_{2}}{\mathbf{I}_{2}}=70 \Omega$$ Thus, $$[\mathbf{z}]=\left[\begin{array}{ll}60 \Omega & 40 \Omega \\40 \Omega & 70 \Omega\end{array}\right]$$ METHOD 2 Alternatively, since there is no dependent source in the given circuit, $\mathbf{z}_{12}=\mathbf{z}_{21}$ and we can use Fig. 4(a). Comparing Fig. $6$ with Fig. 4(a), we get $$\begin{array}{l}\mathbf{z}_{12}=40 \Omega=\mathbf{z}_{21} \\\mathrm{z}_{11}-\mathrm{z}_{12}=20 \quad \Longrightarrow \quad \mathrm{z}_{11}=20+\mathrm{z}_{12}=60 \Omega \\\mathbf{z}_{22}-\mathbf{z}_{12}=30 \quad \Longrightarrow \quad \mathbf{z}_{22}=30+\mathbf{z}_{12}=70 \Omega \\\end{array}$$
Example 2: Find $\mathbf{I}_{1}$ and $\mathbf{I}_{2}$ in the circuit in Fig. $8$.
Fig. 7: For Example 2.
Solution: This is not a reciprocal network. We may use the equivalent circuit in Fig. 4(b) but we can also use Eq. (1) directly. Substituting the given $z$ parameters into Eq. (1), $$\begin{array}{l}\mathbf{V}_{1}=40 \mathbf{I}_{1}+j 20 \mathbf{I}_{2} \quad (2.1)\\ \mathbf{V}_{2}=j 30 \mathbf{I}_{1}+50 \mathbf{I}_{2} \quad (2.2) \end{array}$$ Since we are looking for $\mathbf{I}_{1}$ and $\mathbf{I}_{2}$, we substitute $$\mathbf{V}_{1}=100 \angle 0^{\circ}, \quad \mathbf{V}_{2}=-10 \mathbf{I}_{2}$$ into Eqs. (2.1) and (2.2), which become $$\begin{array}{c}100=40 \mathbf{I}_{1}+j 20 \mathbf{I}_{2} \quad (2.3)\\ -10 \mathbf{I}_{2}=j 30 \mathbf{I}_{1}+50 \mathbf{I}_{2} \quad \Longrightarrow \quad \mathbf{I}_{1}=j 2 \mathbf{I}_{2} \quad (2.4) \end{array}$$ Substituting Eq. (2.4) into Eq. (2.3) gives $$100=j 80 \mathbf{I}_{2}+j 20 \mathbf{I}_{2} \quad \Longrightarrow \quad \mathbf{I}_{2}=\frac{100}{j 100}=-j$$ From Eq. (4), $\mathbf{I}_{1}=j 2(-j)=2$. Thus, $$\mathbf{I}_{1}=2 \angle 0^{\circ} \mathrm{A}, \quad \mathbf{I}_{2}=1 \angle-90^{\circ} \mathrm{A}$$

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