Distributed Parameters 4 – 2-Port Parameters

It’s time to readdress the definitions of parameters as they pertain to a general 2-port network which may consist of a distributed network:

Define V₁ = V(s,0) and V₂ = V(s,d)

$\begin{displaymath}\textnormal{and:}\;\;I_1\;=\;-\frac{\,d V(s,0)\,}{d x}\,\frac{1}{\,Z(s,0)\,}\qquad\qquad I_2\;=\;-\frac{\,d V(s,d)\,}{d x}\,\frac{1}{\,Z(s,d)\,}\end{displaymath}$

The 2-port transmission parameters are stated as:

$\begin{displaymath}\begin{array}{cc}y_{11}\,=\,\dfrac{I_1}{\,V_1\,}\left|_{_{V_2=0}} & y_{12}\,=\,\dfrac{I_1}{\,V_2\,}\left|_{_{V_1=0}} \\ \\y_{21}\,=\,\dfrac{I_2}{\,V_1\,}\left|_{_{V_2=0}} & y_{22}\,=\,\dfrac{I_2}{\,V_2\,}\left|_{_{V_1=0}}\end{array}\qquad\Rightarrow\qquad\begin{array}{cc}y_{11}\,=\,\dfrac{-1\;}{\,Z(0)\,}\,\dfrac{\,V'(0)\,}{\,V(0)\,}\; & \;y_{12}\,=\,\dfrac{-1\;}{\,Z(0)\,}\,\dfrac{\,V'(0)\,}{\,V(d)\,} \\ \\y_{21}\,=\,\dfrac{+1\;}{\,Z(d)\,}\,\dfrac{\,V'(d)\,}{\,V(0)\,}\; &\; y_{22}\,=\,\dfrac{+1\;}{\,Z(d)\,}\,\dfrac{\,V'(d)\,}{\,V(d)\,}\end{array}\end{displaymath}$

And it turns out that y₁₂ = y₂₁

Defining the solutions are a straightforward derivation, the voltage solutions generally result in transcendental functions having no closed forms. Solutions are then obtained from truncated infinite series forms.

A 2^nd-order differential equation has two independent solutions; the complete solution may be expressed as:

$V(s,x)\,=\,a\,V_A(s,x)\,+\,b\,V_B{s,x)$

where coefficients a and b are defined by the system boundary conditions and V_A and V_B are the independent solutions.

Normalizing the basic solution at x = 0 is given as:

$\begin{displaymath}\begin{array}{cc}V_A\,=\,1\:&\;V'_A\,=\,0\\\\V_B\,=\,0 \;&\;V'_B\,=\,1\end{array}\end{displaymath}$

V_A and V_B are both functions of s and x. Two properties are observed in the interval x between 0 and d:

$V_A\,V'_B\,-\,V'_A\,V_B\;\neq\;0\;\;\textnormal{the two solutions can not be zero simultaneously}$
$\dfrac{1}{\,Z(s,x)\,}\,(\,V_A\,V'_B\,-\,V'_A\,V_B\,)\;=\;k$

where Z(s,x) = Z(s) F(x) and k is a constant independent of x; k is found to be equal to $Z(s,0)^{-1}$ .

$\dfrac{1}{\,Z(s,x)\,}\,(\,V_A\,V'_B\,-\,V'_A\,V_B\,)\;=\;\dfrac{1}{\,Z(s,0)\,}\quad\textnormal{for all s and interval x}$

This helps provide a solution for the short-circuit admittance parameters. The basic solution set is expressed:

$V(s,x)\,=\,a\,V_A(s,x)\,+\,b\,V_B(s,x)$

$\dfrac{1}{\,Z(s,x)\,}\,[\,a\,V'_A(s,x)\,+\,b\,V'_B(s,x)\,]$

At the input:

$V(s,x)\,=\,a\,V_A(s,0)\,+\,b\,V_B(s,0)\;\;\Rightarrow\;\;a(1)\,+\,b(0)\;=\;a$

$\dfrac{1}{\,Z(s,0)\,}\,[\,a\,V'_A(s,0)\,+\,b\,V'_B(s,0)\,]\;\;\Rightarrow\;\;\dfrac{1}{\,Z(s,0)\,}\,[\,a(0)\,+\,b(1)\,]\;=\;\dfrac{b}{\,Z(s,0)\,}$

At the output, the y-parameters are expressed:

$\begin{displaymath}\begin{array}{cc}y_{11}\,=\,\dfrac{-1\;}{\,Z(s,0)\,}\,\dfrac{\,V_A(s,d)\,}{\,V_B(s,d)\,}\; & \;y_{12}\,=\,\dfrac{-1\;}{\,Z(s,0)\,}\,\dfrac{1}{\,V_B(s,d)\,} \\ \\y_{21}\,=\,\dfrac{-1\;}{\,Z(s,0)\,}\,\dfrac{1}{\,V_B(s,d)\,}\; &\; y_{22}\,=\,\dfrac{+1\;}{\,Z(s,0)\,}\,\dfrac{\,V'_B(s,d)\,}{\,V_B(s,d)\,}\end{array}\end{displaymath}$

These still only represent a general case solutions lacking specific knowledge of the distributed network.

For a uniformly distributed RC network:

$\begin{displaymath}\begin{align}V_A(s,d)\;&=\;\mathit{cosh}[\,d\,\sqrt{\,r\,c\,s\,}\,]\\\\V_B(s,d)\;&=\;\dfrac{\,\mathit{sinh}[\,\sqrt{\,r\,c\,s\,}\,]}{\,\sqrt{\,r\,c\,s\,}\,}\\\\V'_B(s,d)\;&=\;\mathit{cosh}[\,d\,\sqrt{\,r\,c\,s\,}\,]\end{align}\end{displaymath}$

with the zeroes of $V_A$ and $V'_B$ located at

$\begin{displaymath}s_z\;=\;-\frac{1}{\,d^2\,R\,C\,}\,\frac{\,(\,2\,n\,-\,1\,)^2\,\pi^2\,}{4}\end{displaymath}$

while the zeroes of $V_B$ are located at:

$\begin{displaymath}s_z\;=\;-\frac{n^2\,\pi^2}{\,d^2\,R\,C\,}\qquad\textnormal{for n = 1,2 3, ...}\end{displaymath}$

The advantage of pre-defining a basic solution set may be seen in a representative expression for the distributed RC network”

$\begin{displaymath}y_{11}\;=\;-\frac{1}{\,r(0)\,}\,\frac{\;V'_A(s,d)\,V_A(s,d)\,-\,V_A(s,d)\,V'_B(s,0)\;}{\;V_A(s,0)\,V_B(s,d)\,-\,V_A(s,d)\;\,V_B(s,0)\;}\end{displaymath}$

The other admittance parameters have similar expressions … left as an exercise for the reader.

Keeping in mind that V_A and V_B are transcendental functions, the difficulty of using this solution methodology may be appreciated and explain why this writer has not repeated the experience since grad school … and is not willing to undergo the experience again. Beyond transcribing my notes for posterity on this site.

The internet lives forever.

Hello future!

The open-circuit impedance parameters are found in a similar manner.

$\begin{displaymath}\begin{array}{cc}z_{11}\,=\,\dfrac{V_1}{\,I_1\,}\left|_{_{I_2=0}} & z_{12}\,=\,\dfrac{V_1}{\,I_2\,}\left|_{_{I_1=0}} \\ \\z_{21}\,=\,\dfrac{V_2}{\,I_1\,}\left|_{_{I_2=0}} & z_{22}\,=\,\dfrac{V_2}{\,I_2\,}\left|_{_{I_1=0}}\end{array}\qquad\Rightarrow\qquad\begin{array}{cc}z_{11}\,=\,Z(s,0)\,\dfrac{V'_B(s,d)}{\,Z(0)\,}\,\dfrac{\,V'(0)\,}{\,V(0)\,}\; & \;z_{12}\,=\,\dfrac{-1\;}{\,Z(0)\,}\,\dfrac{\,V'(0)\,}{\,V(d)\,} \\ \\z_{21}\,=\,\dfrac{+1\;}{\,Z(d)\,}\,\dfrac{\,V'(d)\,}{\,V(0)\,}\; &\; z_{22}\,=\,\dfrac{+1\;}{\,Z(d)\,}\,\dfrac{\,V'(d)\,}{\,V(d)\,}\end{array}\end{displaymath}$

With the other parameters found in a similar manner.

The functions describing the solution set generally do not have closed form, but infinite power and product series are more useful in determining pole-zero locations anyway.

The Picard Method is one means of determining a solution set. Assume:

$\begin{displaymath}\frac{\,d V(s,x)\,}{d x}\;=\;-Z(s,x)\,I\qquad\frac{\,d I(s,x)\,}{d x}\;=\;-Y(s,x)\,V\end{displaymath}$

and set-up:

$\begin{displaymath}V_x\;=\;V_1\,-\,\int_0^x\,Z\,I\,dx \qquad I_x\;=\;I_1\,-\,\int_0^x\,Y\,V\,dx\end{displaymath}$

where Z is the per unit length series impedance and Y is the per unti length shunt admittance

The approximate solutions are of the form:

$\begin{displaymath}\begin{align}V_x\;=\;V_1\,\Sum\,\xi_{2n}\;-\; I_1\,\Sum\,\xi_{2n-1}\\\\I_x\;=\;I_1\,\Sum\,\psi_{2n}\;-\; V_1\,\Sum\,\psi_{2n-1}\end{align}\end{displaymath}$

where $V_x$ and $I_x$ are the voltage and current at point x and $V_1$ and $I_1$ are the input voltage and current as defined below:

The parameters are defined:

$\begin{displaymath}\begin{align}\xi_0\;=\;1 &\qquad \psi_0\;=\;1\\\\\xi_n\;=\;\int_0^x\,Z\,\psi_{n-1}\,d x &\qquad \psi_n\;=\;\int_0^x\,Y\,\xi_{n-1}\, d x\end{align}\end{displaymath}$

$\begin{displaymath}\left[ \begin{array}{c}V_x \\ -I_x \end{array} \right]\;=\;\left[ \begin{array}{cc}\sum \, \xi_{2n} & \sum \, \xi_{2n-1} \\ \sum \, \psi_{2n-1} & \sum \, \psi_{2n}\end{array} \right]\;\left[ \begin{array}{c}V_1 \\ -I_1 \end{array} \right]\end{displaymath}$

The short-circuit admittance parameters are:

$\begin{displaymath}Y\;=\;\left[\begin{array}{cc}\dfrac{\sum \, \xi_{2n}\,d}{\,\sum \, \xi_{2n-1}\,d\,} & -\dfrac{1}{\,\sum \, \xi_{2n-1}\,d\,}\\\\-\dfrac{1}{\,\sum \, \xi_{2n-1}\,d\,} & \dfrac{\sum \, \psi_{2n}\,d}{\,\sum \, \xi_{2n-1}\,d\,}\end{array} \right]\end{displaymath}$

where

$\begin{displaymath}\begin{align}A\;&=\;\sum \, \psi_{2n}\;=\;\frac{r(0)}{\,r(d)\,}\,V'_B(s,d)\\B\;&=\;\sum \, \xi_{2n-1}\;=\;r(0)\,V_B(s,d)\\C\;&=\;\sum \, \psi_{2n-1}\;=\;\frac{\,V'_A(s,d)\,}{\,r(d)\,}\\D\;&=\;\sum \, \xi_{2n}\;=\;V_A(s,d)\end{align}\end{displaymath}$

This methodology is suitable for determining solutions for non-uniform lines.

That’s good for now. Next: An Example

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