Two-Port Networks

Models of two-port networks

Many complex passive and linear circuits can be modeled by a two-port network model as shown below. A two-port network is represented by four external variables: voltage $V_1$ and current $I_1$ at the input port, and voltage $V_2$ and current $I_2$ at the output port, so that the two-port network can be treated as a black box modeled by the relationships between the four variables $V_1$, $V_2$, $I_1$ and $I_2$. There exist six different ways to describe the relationships between these variables, depending on which two of the four variables are given, while the other two can always be derived.

Note: All voltages and currents below are complex variables and represented by phasors containing both magnitude and phase angle. However, for convenience the phasor notation $\dot{V}$ and $\dot{I}$ are replaced by $V$ and $I$ respectively.

• Z or impedance model: Given two currents $I_1$ and $I_2$ find voltages $V_1$ and $V_2$ by:
  
  $$\left\{ \begin{array}{l} V_1=Z_{11}I_1+Z_{12}I_2 \\ V_2=Z_{... ... ={\bf Z}\left[ \begin{array}{l} I_1 I_2\end{array} \right]$$
  
  
  Here all four parameters $Z_{11}$, $Z_{12}$, $Z_{21}$, and $Z_{22}$ represent impedance. In particular, $Z_{21}$ and $Z_{12}$ are transfer impedances, defined as the ratio of a voltage $V_1$ (or $V_2$) in one part of a network to a current $I_2$ (or $I_1$) in another part $Z_{12}=V_1/I_2$. ${\bf Z}$ is a 2 by 2 matrix containing all four parameters.

• Y or admittance model: Given two voltages $V_1$ and $V_2$, find currents $I_1$ and $I_2$ by:
  
  $$\left\{ \begin{array}{l} I_1=Y_{11}V_1+Y_{12}V_2 \\ I_2=Y_{... ... ={\bf Y}\left[ \begin{array}{l} V_1 V_2\end{array} \right]$$
  
  
  Here all four parameters $Y_{11}$, $Y_{12}$, $Y_{21}$, and $Y_{22}$ represent admittance. In particular, $Y_{21}$ and $Y_{12}$ are transfer admittances. ${\bf Y}$ is the corresponding parameter matrix.

• A or transmission model: Given $V_2$ and $I_2$, find $V_1$ and $I_1$ by:
  
  $$\left\{ \begin{array}{l} V_1=A_{11}V_2+A_{12}(-I_2) \\ I_... ...={\bf A}\left[ \begin{array}{r} V_2 -I_2\end{array} \right]$$
  
  
  Here $A_{11}$ and $A_{22}$ are dimensionless coefficients, $A_{12}$ is impedance and $A_{21}$ is admittance. A negative sign is added to the output current $I_2$ in the model, so that the direction of the current is out-ward, for easy analysis of a cascade of multiple network models.

• H or hybrid model: Given $V_2$ and $I_1$, find $V_1$ and $I_2$ by:
  
  $$\left\{ \begin{array}{l} V_1=H_{11}I_1+H_{12}V_2 \\ I_2=H... ... ={\bf H}\left[ \begin{array}{l} I_1 V_2\end{array} \right]$$
  
  
  Here $H_{12}$ and $H_{21}$ are dimensionless coefficients, $H_{11}$ is impedance and $H_{22}$ is admittance.

Generalization to nonlinear circuits

The two-port models can also be applied to a nonlinear circuit if the variations of the variables are small and therefore the nonlinear behavior of the circuit can be piece-wise linearized. Assume $z=f(x,y)$ is a nonlinear function of variables $x$ and $y$. If the variations $\triangle x$ and $\triangle y$ are small, the function can be approximated by a linear model

$$\triangle z=\triangle f(x,y)=\frac{\partial f}{\partial x}\triangle x + \frac{\partial f}{\partial y}\triangle y$$

with the linear coefficients

$$\frac{\partial f}{\partial x}=\lim_{\triangle x\rightarrow 0}... ...rrow 0} \frac{\triangle f}{\triangle y}\vert _{\triangle x=0}$$

Finding the model parameters

For each of the four types of models, the four parameters can be found from variables $V_1$, $V_2$, $I_1$ and $I_2$ of a network by the following.

• For Z-model:
  
  $$Z_{11}=\frac{V_1}{I_1}\vert _{I_2=0},\;\;\;\; Z_{12}=\frac{V... ...}\vert _{I_2=0},\;\;\;\; Z_{22}=\frac{V_2}{I_2}\vert _{I_1=0}$$
  
  

• For Y-model:
  
  $$Y_{11}=\frac{I_1}{V_1}\vert _{V_2=0},\;\;\;\; Y_{12}=\frac{I... ...V_2=0},\;\;\;\; Y_{22}=\frac{I_2}{V_2}\vert _{Y_1=0},\;\;\;\;$$
  
  

• For A-model:
  
  $$A_{11}=\frac{V_1}{V_2}\vert _{I_2=0},\;\;\;\; A_{12}=\frac{V... ...}\vert _{I_2=0},\;\;\;\; A_{22}=\frac{I_1}{I_2}\vert _{V_2=0}$$
  
  

• For H-model:
  
  $$H_{11}=\frac{V_1}{I_1}\vert _{V_2=0},\;\;\;\; H_{12}=\frac{V... ...}\vert _{V_2=0},\;\;\;\; H_{22}=\frac{I_2}{V_2}\vert _{I_1=0}$$
  
  

If we further define

$${\bf V}=[V_1, V_2]^T,\;\;\;\;\;\;{\bf I}=[I_1, I_2]^T$$

then the Z-model and Y-model above can be written in matrix form:

$${\bf V}={\bf Z} {\bf I},\;\;\;\;\;\;{\bf I}={\bf Y} {\bf V},\;\;\;\; {\bf Y}={\bf Z}^{-1}$$

Example:

Find the Z-model and Y-model of the circuit shown.

$$\left\{ \begin{array}{l} V_1=Z_{11}I_1+Z_{12}I_2 \\ V_2=Z_{21}I_1+Z_{22}I_2 \end{array} \right.$$

• First assume $I_2=0$, we get

  
  

  $$Z_{11}=V_1/I_1=j\omega L+1/j\omega C,\;\;\;\;\;Z_{21}=V_2/I_1=1/j\omega C$$
  
  

• Next assume $I_1=0$, we get
  
  $$Z_{22}=V_2/I_2=1/j\omega C,\;\;\;\;\;\;Z_{12}=V_1/I_2=1/j\omega C$$
  
  

The parameters of the Y-model can be found as the inverse of $Z$:

$$\left[\begin{array}{cc}Y_{11}&Y_{12} Y_{21}&Y_{22}\end{arra... ...a L\\ -1/j\omega L & j\omega C+1/j\omega L\end{array}\right]$$

Note:

$$\left[ \begin{array}{rr} a & b c & d \end{array} \right]^{... ...}\left[ \begin{array}{rr} d & -b -c & a \end{array} \right]$$

Combinations of two-port models

• Series connection of two 2-port networks: ${\bf Z}={\bf Z}_1+{\bf Z}_2$
• Parallel connection of two 2-port networks: ${\bf Y}={\bf Y}_1+{\bf Y}_2$
• Cascade connection of two 2-port networks: ${\bf A}={\bf A}_1 \cdot {\bf A}_2$

Example: A The circuit shown below contains a two-port network (e.g., a filter circuit, or an amplification circuit) represented by a Z-model:

$${\bf Z}=\left[\begin{array}{ll} Z_{11} & Z_{12} \\ Z_{21} &... ... 4\Omega & j3\Omega \\ j3\Omega & 2\Omega \end{array}\right]$$

The input voltage is $V_0=3\angle 0^\circ$ with an internal impedance $Z_0=5\Omega$ and the load impedance is $R_L=4\Omega$. Find the two voltages $V_1$, $V_2$ and two currents $I_1$, $I_2$.

Method 1:

• First, according the Z-model, we have
  
  $$\left\{ \begin{array}{l} V_1=Z_{11}I_1+Z_{12}I_2= 4I_1+j3I_2 \\ V_2=Z_{21}I_1+Z_{22}I_2=j3I_1+ 2I_2 \end{array} \right.$$
  
  

• Second, two more equations can be obtained from the circuit:
  
  $$\left\{ \begin{array}{l} V_1=V_0-Z_0 I_1=3-5I_1 \\ V_2=-Z_L I_2=-4 I_2 \end{array} \right.$$
  
  

• Substituting the last two equations for $V_1$ and $V_2$ into the first two, we get
  
  $$\left\{ \begin{array}{l} 9I_1+j3I_2=3 j3I_1+6I_2=0 \end{array} \right.$$
  
  

• Solving these we get
  
  $$I_1=\frac{2}{7},\;\;\;\;\;I_2=-\frac{j}{7}$$
  
  

• Then we can get the voltages
  
  $$V_1=\frac{11}{7},\;\;\;\;\;V_2=\frac{j4}{7}$$
  
  

Method 2: We can also use Thevenin's theorem to treat everything before the load impedance as an equivalent voltage source with Thevenin's voltage $V_T$ and resistance $R_T$, and the output voltage $V_2$ and current $I_2$ can be found.

• Find $Z_T=V_2/I_2$ with voltage $V_0$ short-circuit:
  • The Z-model:
    
    $$\left\{ \begin{array}{l} V_1=Z_{11}I_1+Z_{12}I_2= 4I_1+j3I_2 \\ V_2=Z_{21}I_1+Z_{22}I_2=j3I_1+ 2I_2 \end{array} \right.$$
    
    

  • Also due to the short-circuit of voltage source $V_0=0$, we have
    
    $$V_1=V_0-I_1 Z_0=0-5I_1$$
    
    

  • equating the two expressions for $V_1$, we get
    
    $$4I_1+j3I_2=-5I_1,\;\;\;\;\mbox{i.e.}\;\;\;\;I_1=-\frac{j}{3} I_2$$
    
    

  • Substituting this $I_1$ into the equation for $V_2$ above, we get
    
    $$V_2=j3I_1+2I_2=I_2+2I_2=3I_2$$
    
    

  • Find $Z_T=V_2/I_2$:
    
    $$Z_T=\frac{V_2}{I_2}=3$$
    
    

• Find open-circuit voltage $V_T$ with $I_2=0$:
  • Since the load is an open-circuit, $I_2=0$, we have
    
    $$\left\{ \begin{array}{l} V_1=Z_{11}I_1=4I_1 \\ V_2=Z_{21}I_1=j3 I_1 \end{array} \right.$$
    
    

  • Find $I_1$:
    
    $$I_1=\frac{V_0-V_1}{Z_0}=\frac{3-4I_1}{5}$$
    
    
    Solving this to get $I_1=1/3$
  • Find open-circuit voltage $V_T=V_2$:
    
    $$V_2=Z_{21}I_1=j3 \frac{1}{3}=j$$
    
    

• Find load voltage $V_2$:
  
  $$V_2=\frac{V_T}{Z_T+Z_L}\;Z_L=\frac{j}{3+4}\;4=\frac{j4}{7}$$
  
  

• Find load voltage $I_2$:
  
  $$I_2=-\frac{V_2}{Z_L}=-\frac{j4}{7}\;\frac{1}{4}=-\frac{j}{7}$$
  
  

Principle of reciprocity:

Consider the example circuit on the left above, which can be simplified as the network in the middle. The voltage source is in the branch on the left, while the current $\dot{I}_2$ is in the branch on the right, which can be found to be (current divider):

$$\dot{I}_2=\frac{V}{Z_1+Z_2 Z_3/(Z_2+Z_3)}\;\frac{Z_3}{Z_2+Z_3} =V \frac{Z_3}{Z_1Z_2+Z_1Z_3+Z_2Z_3}$$

We next interchange the positions of the voltage source and the current, so that the voltage source is in the branch on the right and the current to be found is in the branch on the left, as shown on the right of the figure above. The current can be found to be

$$\dot{I}_1=\frac{V}{Z_2+Z_1 Z_3/(Z_1+Z_3)}\frac{Z_3}{Z_1+Z_3} =V \frac{Z_3}{Z_1Z_2+Z_1Z_3+Z_2Z_3}$$

The two currents $\dot{I}_1$ and $\dot{I}_2$ are exactly the same! This result illustrates the following reciprocity principle, which can be proven in general:

In any passive (without energy sources), linear network, if a voltage $V$ applied in branch 1 causes a current $I$ in branch 2, then this voltage $V$ applied in branch 2 will cause the same current $I$ in branch 1.

This reciprocity principle can also be stated as:

In any passive, linear network, the transfer impedance $Z_{12}$ is equal to the reciprocal transfer impedance $Z_{21}$.

Based on this reciprocity principle, any complex passive linear network can be modeled by either a T-network or a $\Pi$-network:

• T-Network Model:

  From this T-model, we get
  

  $$\left\{ \begin{array}{l} V_1=(Z_1+Z_3)I_1+Z_3I_2=Z_{11}I_1+Z_... ..._2=Z_3I_1+(Z_2+Z_3)I_2=Z_{21}I_1+Z_{22}I_2 \end{array} \right.$$
  
  
  Comparing this with the Z-model, we get
  
  $$Z_{11}=Z_1+Z_3,\;\;\;\;\;Z_{22}=Z_2+Z_3,\;\;\;\;\;Z_{12}=Z_{21}=Z_3$$
  
  
  Solving these equations for $Z_1$, $Z_2$ and $Z_3$, we get
  
  $$Z_1=Z_{11}-Z_{12},\;\;\;\;Z_2=Z_{22}-Z_{21},\;\;\;\;Z_3=Z_{12}=Z_{21}$$
  
  

• $\Pi$-Network Model:

  From this $\Pi$-model, we get:
  

  $$\left\{ \begin{array}{l} I_1=Y_1V_1+Y_3(V_1-V_2)=(Y_1+Y_3)V_... ...2=-Y_3V_1+(Y_2+Y_3)V_2=Y_{21}V_1+Y_{22}V_2 \end{array} \right.$$
  
  
  Comparing this with the Y-model, we get
  
  $$Y_{11}=Y_1+Y_3,\;\;\;\;\;Y_{22}=Y_2+Y_3,\;\;\;\;\;Y_{12}=Y_{21}=-Y_3$$
  
  
  Solving these equations for $Y_1$, $Y_2$ and $Y_3$, we get
  
  $$Y_1=Y_{11}+Y_{12},\;\;\;\;Y_2=Y_{22}+Y_{21},\;\;\;\;Y_3=-Y_{12}=-Y_{21}$$
  
  

Example 1: Convert the given T-network to a $\Pi$ network.

Solution: Given $Z_1=j5$, $Z_2=-j5$, $Z_3=1$, we get its Z-model:

$$Z_{11}=Z_1+Z_3=1+j5,\;\;\;\;Z_{22}=Z_2+Z_3=1-j5,\;\;\;\;Z_{12}=Z_{21}=Z_3=1$$

The Z-model can be expressed in matrix form:

$$\left[ \begin{array}{l} V_1 V_2\end{array} \right]= \left... ...\right] \left[ \begin{array}{l} I_1 I_2\end{array} \right]$$

This Z-model can be converted into a Y-model:

$$\left[ \begin{array}{l} I_1 I_2\end{array} \right]= \left... ...\right] \left[ \begin{array}{l} V_1 V_2\end{array} \right]$$

This Y-model can be converted to a $\Pi$ network:

$$Y_1=Y_{11}+Y_{12}=-j/5,\;\;\;\;Z_2=Y_{22}+Y_{21}=j/5,\;\;\;\; Y_3=-Y_{21}=-Y_{12}=1/25$$

These admittances can be further converted into impedances:

$$Z_1=1/Y_1=5j,\;\;\;\;Z_2=1/Y_2=-j5,\;\;\;\;Z_3=1/Y_3=25$$

The same results can be obtained by Y to delta conversion.

Example 2: Consider the ideal transformer shown in the figure below. Assume $R_0=10\Omega$, $R_L=5\Omega$, and the turn ratio is $n=n_1/n_2=2$. Describe this circuit as a two-port network.

• Set up basic equations:
  
  $$\left\{ \begin{array}{l} V_1=10\;I_1+2\;V_2 I_2-V_2/5=-2\;I_1 \end{array} \right.$$
  
  

• Rearrange the equations in the form of a Z-model. The second equation is
  
  $$V_2=10\;I_1+5\;I_2$$
  
  
  Substituting into the first equation, we get
  
  $$V_1=30\;I_1+10\;I_2$$
  
  
  The Z-model is:
  
  $$\left[ \begin{array}{l} V_1 V_2 \end{array} \right]= \lef... ...right] \left[ \begin{array}{l} I_1 I_2 \end{array} \right]$$
  
  
  As $Z_{12}=Z_{21}=10$, this is a reciprocal network.

Alternatively, we can set up the equations in terms of the currents:

• 
  
  $$\left\{ \begin{array}{l} I_1=(V_1-2\;V_2)/10 I_2=-2\;I_1+V_2/5 \end{array} \right.$$
  
  

• Rearrange the equations in the form of a Y-model. The first equation is
  
  $$I_1=V_1/10-V_2/5$$
  
  

• Substituting into the second equation, we get
  
  $$I_2=-V_1/5+2V_2/5+V_2/5=-V_1/5+3V_2/5$$
  
  
  The Y-model is:
  
  $$\left[ \begin{array}{l} I_1 I_2 \end{array} \right]= \lef... ...right] \left[ \begin{array}{l} V_1 V_2 \end{array} \right]$$
  
  

Finally, we can verify that ${\bf Z}^{-1}={\bf Y}$

$${\bf Z}^{-1}=\left[ \begin{array}{rr} 30 & 10 10 & 5 \end{... ...y}{rr} 1/10 & -1/5 -1/5 & 3/5 \end{array} \right] ={\bf Y}$$