# Talk:Distributed Parameter Line Model

## Footnotes

### Derivation of A1 and A2

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

$\boldsymbol{V}(0) = \boldsymbol{V_{r}} \,$
$\boldsymbol{I}(0) = \boldsymbol{I_{r}} \,$

The voltage and current equations are as follows:

$\boldsymbol{V}(0) = \boldsymbol{V_{r}} = A_{1} e^{\boldsymbol{\gamma} (0)} + A_{2} e^{-\boldsymbol{\gamma} (0)} \,$
$\boldsymbol{I}(0) = \boldsymbol{I_{r}} = \frac{A_{1} e^{\gamma (0)} - A_{2} e^{-\gamma (0)}}{\boldsymbol{Z}_{c}} \,$

Therefore,

$\boldsymbol{V_{r}} = A_{1} + A_{2} \,$
$\boldsymbol{I_{r}} = \frac{A_{1} - A_{2}}{\boldsymbol{Z}_{c}} \,$

Substituting $A_{1} = \boldsymbol{I_{r}} \boldsymbol{Z}_{c} + A_{2} \,$ into $\boldsymbol{V_{r}} \,$:

$\boldsymbol{V_{r}} = \boldsymbol{I_{r}} \boldsymbol{Z}_{c} + A_{2} + A_{2} \,$
$\Rightarrow A_{2} = \frac{\boldsymbol{V_{r}} - \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \,$

Solving for A1, we get:

$\boldsymbol{V_{r}} = A_{1} + \frac{\boldsymbol{V_{r}} - \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \,$
$\Rightarrow A_{1} = \frac{\boldsymbol{V_{r}} + \boldsymbol{I_{r}} \boldsymbol{Z}_{c}}{2} \,$

### Diagonality of Modal Characteristic Impedance Matrix

The modal characteristic impedance matrix is:

$[Z_c] = [\gamma]^{-1} [T_{v}]^{-1} [Z] [T_{i}] \,$

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

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

$\frac{d \boldsymbol{V'}}{dx} = [T_{v}]^{-1} [Z] [T_{i}] \boldsymbol{I'} \,$ ... Equ. (F1)
$\frac{d \boldsymbol{I'}}{dx} = [T_{i}]^{-1} [Y] [T_{v}] \boldsymbol{V'} \,$ ... Equ. (F2)

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

$[\gamma]^{2} = [\lambda] = [T_{v}]^{-1} [Z] [Y] [T_{v}] = [T_{i}]^{-1} [Y] [Z] [T_{i}] \,$ ... Equ. (F3)

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

$([T_{v}]^{-1} [Z] [T_{i}]) ([T_{i}]^{-1} [Y] [T_{v}]) = [T_{v}]^{-1} [Z] [Y] [T_{v}] \,$ ... Equ. (F4a)

or

$([T_{i}]^{-1} [Y] [T_{v}]) ([T_{v}]^{-1} [Z] [T_{i}]) = [T_{i}]^{-1} [Y] [Z] [T_{i}] \,$ ... Equ. (F4b)

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

$([T_{v}]^{-1} [Z] [T_{i}]) ([T_{i}]^{-1} [Y] [T_{v}]) = ([T_{i}]^{-1} [Y] [T_{v}]) ([T_{v}]^{-1} [Z] [T_{i}]) = [\gamma]^{2} \,$ ... Equ. (F5)

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