# Transmission Parameters

Whatsapp
Since there are no restrictions on which terminal voltages and currents should be considered independent and which should be dependent variables, we expect to be able to generate many sets of parameters. Another set of parameters relates the variables at the input port to those at the output port. Thus,
\bbox[10px,border:1px solid grey]{\begin{aligned}\mathbf{V}_{1}=\mathbf{A} \mathbf{V}_{2}-\mathbf{B I}_{2} \\\mathbf{I}_{1}=\mathbf{C V}_{2}-\mathbf{D} \mathbf{I}_{2}\end{aligned}} \tag{1}
or
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{1}\end{array}\right]=\left[\begin{array}{ll}\mathbf{A} & \mathbf{B} \\\mathbf{C} & \mathbf{D}\end{array}\right]\left[\begin{array}{c}\mathbf{V}_{2} \\-\mathbf{I}_{2}\end{array}\right]=[\mathbf{T}]\left[\begin{array}{c}\mathbf{V}_{2} \\-\mathbf{I}_{2}\end{array}\right] \tag{2}$$
Equations (1) and (2) relate the input variables $\left(\mathbf{V}_{1}\right.$ and $\left.\mathbf{I}_{1}\right)$ to the output variables $\left(\mathbf{V}_{2}\right.$ and $\left.-\mathbf{I}_{2}\right)$. Notice that in computing the transmission parameters, $-\mathbf{I}_{2}$ is used rather than $\mathbf{I}_{2}$, because the current is considered to be leaving the network, as shown in Fig. 1, as opposed to entering the network as in Fig. ( i ).
Fig. 1: Terminal variables used to define the ABCD parameters.
Fig. ( i ): two-port network.
This is done merely for conventional reasons; when you cascade two-ports (output to input), it is most logical to think of $\mathbf{I}_{2}$ as leaving the two-port. It is also customary in the power industry to consider $\mathbf{I}_{2}$ as leaving the two-port.
The two-port parameters in Eqs. (1) and (2) provide a measure of how a circuit transmits voltage and current from a source to a load. They are useful in the analysis of transmission lines (such as cable and fiber) because they express sending-end variables $\left(\mathbf{V}_{1}\right.$ and $\left.\mathbf{I}_{1}\right)$ in terms of the receiving-end variables $\left(\mathbf{V}_{2}\right.$ and $\left.-\mathbf{I}_{2}\right)$. For this reason, they are called transmission parameters. They are also known as $\mathbf{A B C D}$ parameters. They are used in the design of telephone systems, microwave networks, and radars.
The transmission parameters are determined as
$$\begin{array}{ll}\mathbf{A}=\left.\frac{\mathbf{V}_{1}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{2}=0}, & \mathbf{B}=-\left.\frac{\mathbf{V}_{1}}{\mathbf{I}_{2}}\right|_{\mathbf{V}_{2}=0} \\\mathbf{C}=\left.\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}\right|_{\mathbf{I}_{2}=0}, & \mathbf{D}=-\left.\frac{\mathbf{I}_{1}}{\mathbf{I}_{2}}\right|_{\mathbf{V}_{2}=0}\end{array} \tag{3}$$
Thus, the transmission parameters are called, specifically,
$$\begin{array}{l}\mathbf{A}=\text { Open-circuit voltage ratio } \\\mathbf{B}=\text { Negative short-circuit transfer impedance } \\\mathbf{C}=\text { Open-circuit transfer admittance } \\\mathbf{D}=\text { Negative short-circuit current ratio }\end{array}$$
$\mathbf{A}$ and $\mathbf{D}$ are dimensionless, $\mathbf{B}$ is in ohms, and $\mathbf{C}$ is in siemens. Since the transmission parameters provide a direct relationship between input and output variables, they are very useful in cascaded networks.
Our last set of parameters may be defined by expressing the variables at the output port in terms of the variables at the input port. We obtain
$$\begin{array}{l}\mathbf{V}_{2}=\mathbf{a} \mathbf{V}_{1}-\mathbf{b} \mathbf{I}_{1} \\\mathbf{I}_{2}=\mathbf{c V}_{1}-\mathbf{d I}_{1} \\\end{array} \tag{4}$$
or
$$\left[\begin{array}{l}\mathbf{V}_{2} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{a} & \mathbf{b} \\\mathbf{c} & \mathbf{d}\end{array}\right]\left[\begin{array}{r}\mathbf{V}_{1} \\-\mathbf{I}_{1}\end{array}\right]=[\mathbf{t}]\left[\begin{array}{r}\mathbf{V}_{1} \\-\mathbf{I}_{1}\end{array}\right] \tag{5}$$
The parameters $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{d}$ are called the inverse transmission parameters. They are determined as follows:
$$\begin{array}{ll}\mathbf{a}=\left.\frac{\mathbf{V}_{2}}{\mathbf{V}_{1}}\right|_{\mathbf{I}_{1}=0}, & \mathbf{b}=-\left.\frac{\mathbf{V}_{2}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{1}=0} \\\mathbf{c}=\left.\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}\right|_{\mathbf{I}_{1}=0}, & \mathbf{d}=-\left.\frac{\mathbf{I}_{2}}{\mathbf{I}_{1}}\right|_{\mathbf{V}_{1}=0}\end{array} \tag{6}$$
From Eq. (6) and from our experience so far, it is evident that these parameters are known individually as
$\mathbf{a}=$ Open-circuit voltage gain
$\mathbf{b}=$ Negative short-circuit transfer impedance
$\mathbf{c}=$ Open-circuit transfer admittance
$\mathbf{d}=$ Negative short-circuit current gain
While a and $\mathbf{d}$ are dimensionless, $\mathbf{b}$ and $\mathbf{c}$ are in ohms and siemens, respectively. In terms of the transmission or inverse transmission parameters, a network is reciprocal if $$\mathbf{A D}-\mathbf{B C}=1, \quad \mathbf{a d}-\mathbf{b c}=1 \tag{7}$$ These relations can be proved in the same way as the transfer impedance relations for the $z$ parameters.
Example 1: Find the transmission parameters for the two-port network in Fig. $2$.
Fig. 2: For Example 1.
Solution: To determine $\mathbf{A}$ and $\mathbf{C}$, we leave the output port open as in Fig. 3(a) so that $\mathbf{I}_{2}=0$ and place a voltage source $\mathbf{V}_{1}$ at the input port.
Fig. 3: For Example 1: (a) finding A and C, (b) finding B and D.
We have
$$\mathbf{V}_{1}=(10+20) \mathbf{I}_{1}=30 \mathbf{I}_{1} \quad \text { and } \quad \mathbf{V}_{2}=20 \mathbf{I}_{1}-3 \mathbf{I}_{1}=17 \mathbf{I}_{1}$$
Thus,
$$\mathbf{A}=\frac{\mathbf{V}_{1}}{\mathbf{V}_{2}}=\frac{30 \mathbf{I}_{1}}{17 \mathbf{I}_{1}}=1.765, \quad \mathbf{C}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}=\frac{\mathbf{I}_{1}}{17 \mathbf{I}_{1}}=0.0588 \mathrm{~S}$$
To obtain $\mathbf{B}$ and $\mathbf{D}$, we short-circuit the output port so that $\mathbf{V}_{2}=0$ as shown in Fig. 3(b) and place a voltage source $\mathbf{V}_{1}$ at the input port. At node $a$ in the circuit of Fig. 3(b), KCL gives
$$\frac{\mathbf{V}_{1}-\mathbf{V}_{a}}{10}-\frac{\mathbf{V}_{a}}{20}+\mathbf{I}_{2}=0 \tag{1.1}$$
But $\mathbf{V}_{a}=3 \mathbf{I}_{1}$ and $\mathbf{I}_{1}=\left(\mathbf{V}_{1}-\mathbf{V}_{a}\right) / 10$. Combining these gives $$\mathbf{V}_{1}=13 \mathbf{I}_{1} \tag{1.2}$$ Substituting Eq. (1.2) into Eq. (1.1) and replacing the first term with $\mathbf{I}_{1}$,
$$\mathbf{I}_{1}-\frac{3 \mathbf{I}_{1}}{20}+\mathbf{I}_{2}=0 \quad \Longrightarrow \quad \frac{17}{20} \mathbf{I}_{1}=-\mathbf{I}_{2}$$
Therefore,
$$\mathbf{D}=-\frac{\mathbf{I}_{1}}{\mathbf{I}_{2}}=\frac{20}{17}=1.176, \quad \mathbf{B}=-\frac{\mathbf{V}_{1}}{\mathbf{I}_{2}}=\frac{-13 \mathbf{I}_{1}}{(-17 / 20) \mathbf{I}_{1}}=15.29 \Omega$$
Example 2: The ABCD parameters of the two-port network in Fig. $4$ are
$$\left[\begin{array}{cc}4 & 20 \Omega \\0.1 \mathrm{~S} & 2\end{array}\right]$$
The output port is connected to a variable load for maximum power transfer. Find $R_{L}$ and the maximum power transferred.
Fig. 4: For Example 2.
Solution: What we need is to find the Thevenin equivalent $\left(\mathbf{Z}_{\mathrm{Th}}\right.$ and $\left.\mathbf{V}_{\mathrm{Th}}\right)$ at the load or output port. We find $\mathbf{Z}_{\mathrm{Th}}$ using the circuit in Fig. 5(a).
Fig. 5: Solution of Example 2: (a) finding $Z_{Th}$, (b) finding $V_{Th}$, (c) finding $R_L$ for maximum power transfer.
Our goal is to get $\mathbf{Z}_{\mathrm{Th}}=\mathbf{V}_{2} / \mathbf{I}_{2}$. Substituting the given $\mathbf{A B C D}$ parameters into Eq. (1), we obtain
$$\begin{array}{c}\mathbf{V}_{1}=4 \mathbf{V}_{2}-20 \mathbf{I}_{2} \quad (2.1) \\ \mathbf{I}_{1}=0.1 \mathbf{V}_{2}-2 \mathbf{I}_{2} \quad (2.2) \end{array}$$
At the input port, $\mathbf{V}_{1}=-10 \mathbf{I}_{1}$. Substituting this into Eq. (2.1) gives $$-10 \mathbf{I}_{1}=4 \mathbf{V}_{2}-20 \mathbf{I}_{2}$$ or $$\mathbf{I}_{1}=-0.4 \mathbf{V}_{2}+2 \mathbf{I}_{2} \tag{2.3}$$ Setting the right-hand sides of Eqs. (2.2) and (2.3) equal,
$$0.1 \mathbf{V}_{2}-2 \mathbf{I}_{2}=-0.4 \mathbf{V}_{2}+2 \mathbf{I}_{2} \quad \Longrightarrow \quad 0.5 \mathbf{V}_{2}=4 \mathbf{I}_{2}$$
Hence,
$$\mathbf{Z}_{\mathrm{Th}}=\frac{\mathbf{V}_{2}}{\mathbf{I}_{2}}=\frac{4}{0.5}=8 \Omega$$
To find $\mathbf{V}_{\mathrm{Th}}$, we use the circuit in Fig. 5(b). At the output port $\mathbf{I}_{2}=0$ and at the input port $\mathbf{V}_{1}=50-10 \mathbf{I}_{1}$. Substituting these into Eqs. (2.1) and (2.2),
$$\begin{array}{c}50-10 \mathbf{I}_{1}=4 \mathbf{V}_{2} \quad (2.4)\\ \mathbf{I}_{1}=0.1 \mathbf{V}_{2}\quad (2.5) \end{array}$$
Substituting Eq. (2.5) into Eq. (2.4),
$$50-\mathbf{V}_{2}=4 \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{V}_{2}=10$$
Thus, $$\mathbf{V}_{\mathrm{Th}}=\mathbf{V}_{2}=10 \mathrm{~V}$$ The equivalent circuit is shown in Fig. 5(c). For maximum power transfer,
$$R_{L}=\mathbf{Z}_{\mathrm{Th}}=8 \Omega$$
the maximum power is
$$P=I^{2} R_{L}=\left(\frac{\mathbf{V}_{\mathrm{Th}}}{2 R_{L}}\right)^{2} R_{L}=\frac{\mathbf{V}_{\mathrm{Th}}^{2}}{4 R_{L}}=\frac{100}{4 \times 8}=3.125 \mathrm{~W}$$

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