Admittance Parameters
In the previous section we saw that impedance parameters may not exist for a two-port network. So there is a need for an alternative means of describing such a network. This need is met by the second set of parameters, which we obtain by expressing the terminal currents in terms of the terminal voltages.

In either Fig. 1(a) or (b), the terminal currents can be expressed in terms of the terminal voltages as
or in matrix form as
The $ \mathbf{y} $ terms are known as the admittance parameters (or, simply, y parameters) and have units of siemens.
The values of the parameters can be determined by setting $ \mathbf{V}_{1}=0 $ (input port short-circuited) or $ \mathbf{V}_{2}=0 $ (output port short-circuited). Thus,
Since the $ y $ parameters are obtained by short-circuiting the input or output port, they are also called the short-circuit admittance parameters. Specifically,
$ \mathbf{y}_{11}= $ Short-circuit input admittance
$ \mathbf{y}_{12}= $ Short-circuit transfer admittance from port 2 to port 1
$ \mathbf{y}_{21}= $ Short-circuit transfer admittance from port 1 to port 2
$ \mathbf{y}_{22}= $ Short-circuit output admittance Following Eq. (1), we obtain $ \mathbf{y}_{11} $ and $ \mathbf{y}_{21} $ by connecting a current $ \mathbf{I}_{1} $ to port 1 and short-circuiting port 2 as in Fig. 1(a), finding $ \mathbf{V}_{1} $ and $ \mathbf{I}_{2} $, and then calculating
Similarly, we obtain $ \mathbf{y}_{12} $ and $ \mathbf{y}_{22} $ by connecting a current source $ \mathbf{I}_{2} $ to port 2 and short-circuiting port 1 as in Fig. 1(b), finding $ \mathbf{I}_{1} $ and $ \mathbf{V}_{2} $, and then getting
This procedure provides us with a means of calculating or measuring the $ y $ parameters. The impedance and admittance parameters are collectively referred to as immittance parameters.
For a two-port network that is linear and has no dependent sources, the transfer admittances are equal $ \left(\mathbf{y}_{12}=\mathbf{y}_{21}\right) $. This can be proved in the same way as for the $ z $ parameters. A reciprocal network $ \left(\mathbf{y}_{12}=\mathbf{y}_{21}\right) $ can be modeled by the $ \Pi $-equivalent circuit in Fig. 2(a).
If the network is not reciprocal, a more general equivalent network is shown in Fig. $ 2(\mathrm{~b}) $

Fig. 1: Determination of the y parameters: (a) finding $y_{11}$ and $y_{21}$, (b) finding $y_{12}$
and $y_{22}$.
$$\begin{array}{l}\mathbf{I}_{1}=\mathbf{y}_{11} \mathbf{V}_{1}+\mathbf{y}_{12} \mathbf{V}_{2} \\\mathbf{I}_{2}=\mathbf{y}_{21} \mathbf{V}_{1}+\mathbf{y}_{22} \mathbf{V}_{2}\end{array}$$
$$\left[\begin{array}{l}\mathbf{I}_{1} \\\mathbf{I}_{2}\end{array}\right]=\left[\begin{array}{ll}\mathbf{y}_{11} & \mathbf{y}_{12} \\\mathbf{y}_{21} & \mathbf{y}_{22}\end{array}\right]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]=[\mathbf{y}]\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{V}_{2}\end{array}\right]$$
The values of the parameters can be determined by setting $ \mathbf{V}_{1}=0 $ (input port short-circuited) or $ \mathbf{V}_{2}=0 $ (output port short-circuited). Thus,
$$\begin{array}{ll}
\mathbf{y}_{11}=\left.\frac{\mathbf{I}_1}{\mathbf{V}_1}\right|_{\mathbf{V}_2=0}, & \mathbf{y}_{12}=\left.\frac{\mathbf{I}_1}{\mathbf{V}_2}\right|_{\mathbf{V}_1=0} \\
\mathbf{y}_{21}=\left.\frac{\mathbf{I}_2}{\mathbf{V}_1}\right|_{\mathbf{V}_2=0}, & \mathbf{y}_{22}=\left.\frac{\mathbf{I}_2}{\mathbf{V}_2}\right|_{\mathbf{V}_1=0}
\end{array} \tag{1}$$
$ \mathbf{y}_{12}= $ Short-circuit transfer admittance from port 2 to port 1
$ \mathbf{y}_{21}= $ Short-circuit transfer admittance from port 1 to port 2
$ \mathbf{y}_{22}= $ Short-circuit output admittance Following Eq. (1), we obtain $ \mathbf{y}_{11} $ and $ \mathbf{y}_{21} $ by connecting a current $ \mathbf{I}_{1} $ to port 1 and short-circuiting port 2 as in Fig. 1(a), finding $ \mathbf{V}_{1} $ and $ \mathbf{I}_{2} $, and then calculating
$$\mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}, \quad \mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}$$
$$\mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}, \quad \mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}$$
For a two-port network that is linear and has no dependent sources, the transfer admittances are equal $ \left(\mathbf{y}_{12}=\mathbf{y}_{21}\right) $. This can be proved in the same way as for the $ z $ parameters. A reciprocal network $ \left(\mathbf{y}_{12}=\mathbf{y}_{21}\right) $ can be modeled by the $ \Pi $-equivalent circuit in Fig. 2(a).

Fig. 2: (a) $ \Pi $-equivalent circuit (for reciprocal case only), (b) general equivalent circuit.
Example 1: Obtain the $ y $ parameters for the $ \Pi $ network shown in Fig. $ 3 $.
Solution:
METHOD I To find $ \mathbf{y}_{11} $ and $ \mathbf{y}_{21} $, short-circuit the output port and connect a current source $ \mathbf{I}_{1} $ to the input port as in Fig. 4(a). Since the $ 8\Omega $ resistor is short-circuited, the $ 2-\Omega $ resistor is in parallel with the $ 4\Omega $ resistor. Hence,
By current division,
To get $ \mathbf{y}_{12} $ and $ \mathbf{y}_{22} $, short-circuit the input port and connect a current source I $ _{2} $ to the output port as in Fig. 4(b).

The $ 4-\Omega $ resistor is short-circuited so that the $ 2\Omega $ and $ 8\Omega $ resistors are in parallel.
By current division,
METHOD 2 Alternatively, comparing Fig. $ 3 $ with Fig. 2(a),
as obtained previously.
Fig. 3: For Example 1.
METHOD I To find $ \mathbf{y}_{11} $ and $ \mathbf{y}_{21} $, short-circuit the output port and connect a current source $ \mathbf{I}_{1} $ to the input port as in Fig. 4(a). Since the $ 8\Omega $ resistor is short-circuited, the $ 2-\Omega $ resistor is in parallel with the $ 4\Omega $ resistor. Hence,
$$\mathbf{V}_{1}=\mathbf{I}_{1}(4 \| 2)=\frac{4}{3} \mathbf{I}_{1}, \quad \mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}=\frac{\mathbf{I}_{1}}{\frac{4}{3} \mathbf{I}_{1}}=0.75 \mathrm{~S}$$
$$-\mathbf{I}_{2}=\frac{4}{4+2} \mathbf{I}_{1}=\frac{2}{3} \mathbf{I}_{1}, \quad \mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}=\frac{-\frac{2}{3} \mathbf{I}_{1}}{\frac{4}{3} \mathbf{I}_{1}}=-0.5 \mathrm{~S}$$

Fig. 4: For Example 1: (a) finding
$y_{11}$ and $y_{21}$, (b) finding $y_{12}$ and $y_{22}$.
$$\mathbf{V}_{2}=\mathbf{I}_{2}(8 \| 2)=\frac{8}{5} \mathbf{I}_{2}, \quad \mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{\mathbf{I}_{2}}{\frac{8}{5} \mathbf{I}_{2}}=\frac{5}{8}=0.625 \mathrm{~S}$$
$$-\mathbf{I}_{1}=\frac{8}{8+2} \mathbf{I}_{2}=\frac{4}{5} \mathbf{I}_{2}, \quad \mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}=\frac{-\frac{4}{5} \mathbf{I}_{2}}{\frac{8}{5} \mathbf{I}_{2}}=-0.5 \mathrm{~S}$$
$$\begin{array}{c}\mathbf{y}_{12}=-\frac{1}{2} \mathrm{~S}=\mathbf{y}_{21} \\\mathbf{y}_{11}+\mathbf{y}_{12}=\frac{1}{4} \quad \Longrightarrow \quad \mathbf{y}_{11}=\frac{1}{4}-\mathbf{y}_{12}=0.75 \mathrm{~S} \\\mathbf{y}_{22}+\mathbf{y}_{12}=\frac{1}{8} \quad \Longrightarrow \quad \mathbf{y}_{22}=\frac{1}{8}-\mathbf{y}_{12}=0.625 \mathrm{~S}\end{array}$$
Example 2: Determine the $ y $ parameters for the two-port shown in Fig. $ 5 $.
Solution: We follow the same procedure as in the previous example. To get $ \mathbf{y}_{11} $ and $ \mathbf{y}_{21} $, we use the circuit in Fig. 6(a), in which port 2 is short-circuited and a current source is applied to port 1 .

At node 1 ,
But $ \mathbf{I}_{1}=\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8} $; therefore,
Hence,
and
At node 2 ,
or
Hence,
Similarly, we get $ \mathbf{y}_{12} $ and $ \mathbf{y}_{22} $ using Fig. 6(b). At node 1,
But $ \mathbf{I}_{1}=\frac{0-\mathbf{V}_{o}}{8} $; therefore,
or
Hence,
At node 2 ,
or
Thus,
Notice that $ \mathbf{y}_{12} \neq \mathbf{y}_{21} $ in this case, since the network is not reciprocal.
Fig. 5: For Example 2.

Fig. 6: Solution of Example 2: (a) finding $y_{11}$ and $y_{21}$, (b) finding $y_{12}$ and $y_{22}$.
$$\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8}=2 \mathbf{I}_{1}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-0}{4}$$
$$\begin{array}{c}0=\frac{\mathbf{V}_{1}-\mathbf{V}_{o}}{8}+\frac{3 \mathbf{V}_{o}}{4} \\0=\mathbf{V}_{1}-\mathbf{V}_{o}+6 \mathbf{V}_{o} \quad \Longrightarrow \quad \mathbf{V}_{1}=-5 \mathbf{V}_{o}\end{array}$$
$$\mathbf{I}_{1}=\frac{-5 \mathbf{V}_{o}-\mathbf{V}_{o}}{8}=-0.75 \mathbf{V}_{o}$$
$$\mathbf{y}_{11}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{1}}=\frac{-0.75 \mathbf{V}_{o}}{-5 \mathbf{V}_{o}}=0.15 \mathrm{~S}$$
$$\frac{\mathbf{V}_{o}-0}{4}+2 \mathbf{I}_{1}+\mathbf{I}_{2}=0$$
$$-\mathbf{I}_{2}=0.25 \mathbf{V}_{o}-1.5 \mathbf{V}_{o}=-1.25 \mathbf{V}_{o}$$
$$\mathbf{y}_{21}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{1}}=\frac{1.25 \mathbf{V}_{o}}{-5 \mathbf{V}_{o}}=-0.25 \mathrm{~S}$$
$$\frac{0-\mathbf{V}_{o}}{8}=2 \mathbf{I}_{1}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}$$
$$0=-\frac{\mathbf{V}_{o}}{8}+\frac{\mathbf{V}_{o}}{2}+\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}$$
$$0=-\mathbf{V}_{o}+4 \mathbf{V}_{o}+2 \mathbf{V}_{o}-2 \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{V}_{2}=2.5 \mathbf{V}_{o}$$
$$\mathbf{y}_{12}=\frac{\mathbf{I}_{1}}{\mathbf{V}_{2}}=\frac{-\mathbf{V}_{o} / 8}{2.5 \mathbf{V}_{o}}=-0.05 \mathrm{~S}$$
$$\frac{\mathbf{V}_{o}-\mathbf{V}_{2}}{4}+2 \mathbf{I}_{1}+\mathbf{I}_{2}=0$$
$$-\mathbf{I}_{2}=0.25 \mathbf{V}_{o}-\frac{1}{4}\left(2.5 \mathbf{V}_{o}\right)-\frac{2 \mathbf{V}_{o}}{8}=-0.625 \mathbf{V}_{o}$$
$$\mathbf{y}_{22}=\frac{\mathbf{I}_{2}}{\mathbf{V}_{2}}=\frac{0.625 \mathbf{V}_{o}}{2.5 \mathbf{V}_{o}}=0.25 \mathrm{~S}$$
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