Talk:Distributed Parameter Line Model

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Footnotes

Derivation of A1 and A2

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 A1, 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.