Talk:Distributed Parameter Line Model - Revision history

Jules: Created page with "== Footnotes == === Derivation of A₁ and A₂ === Based on the boundary conditions at the receiving end of the line ( $x = 0 \,$ ), i.e. :
2020-11-22T07:40:25Z

Created page with "== Footnotes == === Derivation of A<sub>1</sub> and A<sub>2</sub> === Based on the boundary conditions at the receiving end of the line (<math>x = 0 \, </math>), i.e. : <ma..."

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== Footnotes ==

=== Derivation of A<sub>1</sub> and A<sub>2</sub> ===

Based on the boundary conditions at the receiving end of the line (<math>x = 0 \, </math>), i.e.

: <math>\boldsymbol{V}(0) = \boldsymbol{V_{r}} \, </math>
: <math>\boldsymbol{I}(0) = \boldsymbol{I_{r}} \, </math>

The voltage and current equations are as follows:

: <math> \boldsymbol{V}(0) = \boldsymbol{V_{r}} = A_{1} e^{\boldsymbol{\gamma} (0)} + A_{2} e^{-\boldsymbol{\gamma} (0)} \, </math>

: <math> \boldsymbol{I}(0) = \boldsymbol{I_{r}} = \frac{A_{1} e^{\gamma (0)} - A_{2} e^{-\gamma (0)}}{\boldsymbol{Z}_{c}} \, </math>

Therefore,

: <math> \boldsymbol{V_{r}} = A_{1} + A_{2} \, </math>

: <math> \boldsymbol{I_{r}} = \frac{A_{1} - A_{2}}{\boldsymbol{Z}_{c}} \, </math>

Substituting <math> A_{1} = \boldsymbol{I_{r}} \boldsymbol{Z}_{c} + A_{2} \, </math> into <math> \boldsymbol{V_{r}} \, </math>:

: <math> \boldsymbol{V_{r}} = \boldsymbol{I_{r}} \boldsymbol{Z}_{c} + A_{2} + A_{2} \, </math>

: <math> \Rightarrow A_{2} = \frac{\boldsymbol{V_{r}} - \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \, </math>

Solving for A<sub>1</sub>, we get:

: <math> \boldsymbol{V_{r}} = A_{1} + \frac{\boldsymbol{V_{r}} - \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \, </math>

: <math> \Rightarrow A_{1} = \frac{\boldsymbol{V_{r}} + \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \, </math>

=== Diagonality of Modal Characteristic Impedance Matrix ===

The modal characteristic impedance matrix is:

: <math> [Z_c] = [\gamma]^{-1} [T_{v}]^{-1} [Z] [T_{i}] \, </math>

If <math>[\gamma] \, </math> is diagonal, then we shall prove that <math>[Z_c] \, </math> is also diagonal. This is done by proving that <math> [T_{v}]^{-1} [Z] [T_{i}] \, </math> is diagonal.

We saw that after transformation, the first order differential equations for voltage and current are:

: <math> \frac{d \boldsymbol{V'}}{dx} = [T_{v}]^{-1} [Z] [T_{i}] \boldsymbol{I'} \, </math> ... Equ. (F1)

: <math> \frac{d \boldsymbol{I'}}{dx} = [T_{i}]^{-1} [Y] [T_{v}] \boldsymbol{V'} \, </math> ... Equ. (F2)

We also saw that the square of the modal propagation constants are the eigenvalues of [Z][Y] and [Y][Z], i.e.

: <math>[\gamma]^{2} = [\lambda] = [T_{v}]^{-1} [Z] [Y] [T_{v}] = [T_{i}]^{-1} [Y] [Z] [T_{i}] \, </math> ... Equ. (F3)

If we were to multiply the coefficients of Equations (F1) and (F2), we'd get:

: <math> ([T_{v}]^{-1} [Z] [T_{i}]) ([T_{i}]^{-1} [Y] [T_{v}]) = [T_{v}]^{-1} [Z] [Y] [T_{v}] \, </math> ... Equ. (F4a)

or

: <math> ([T_{i}]^{-1} [Y] [T_{v}]) ([T_{v}]^{-1} [Z] [T_{i}]) = [T_{i}]^{-1} [Y] [Z] [T_{i}] \, </math> ... Equ. (F4b)

We can therefore equate (F4a) and (F4b) with (F3):

: <math> ([T_{v}]^{-1} [Z] [T_{i}]) ([T_{i}]^{-1} [Y] [T_{v}]) = ([T_{i}]^{-1} [Y] [T_{v}]) ([T_{v}]^{-1} [Z] [T_{i}]) = [\gamma]^{2} \, </math> ... Equ. (F5)

We know that the modal propagation constant matrix <math>[\gamma]^{2} \, </math> is diagonal. Thus for Equation (F5) to hold, then <math> ([T_{v}]^{-1} [Z] [T_{i}]) \, </math> and <math> ([T_{i}]^{-1} [Y] [T_{v}]) \, </math> must also be diagonal. And therefore the modal characteristic impedance matrix is also diagonal.