# Talk:Travelling Wave Line Model

## Footnotes

### Derivation of Voltage Equation

The general solution to the transmission line wave equations are:

$I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \,$
$V(x, t) = v^{+}(x - vt) + v^{-}(x + vt) \,$

Where $v = \frac{1}{\sqrt{LC}} \,$ is the velocity of propagation (m/s).

$i^{+}(t) \,$ and $i^{-}(t) \,$ are arbitrary current functions of time.
$v^{+}(t) \,$ and $v^{-}(t) \,$ are arbitrary voltage functions of time.

The Laplace Transform of the above equations are:

$I(x, s) = I^{+}(s)e^{-\frac{sx}{v}} + I^{-}(s)e^{\frac{sx}{v}} \,$ ... Equ. (1)
$V(x, s) = V^{+}(s)e^{-\frac{sx}{v}} + V^{-}(s)e^{\frac{sx}{v}} \,$ ... Equ. (2)

Recall the Telegrapher's equation for current:

$\frac{\partial I(x,t)}{\partial x} = - C \frac{\partial V(x,t)}{\partial t} \,$

Taking the Laplace Transform of this equation, we get:

$\frac{\partial I(x,s)}{\partial x} = - sC V(x,s) \,$ ... Equ. (3)

Substituting Equations (1) and (2) into Equ. (3):

$\frac{\partial}{\partial x} \left[ I^{+}(s)e^{-\frac{sx}{v}} + I^{-}(s)e^{\frac{sx}{v}} \right] = - sC \left[ V^{+}(s)e^{-\frac{sx}{v}} + V^{-}(s)e^{\frac{sx}{v}} \right] \,$

We can differentiate the left-hand side:

$\frac{s}{v} \left[ - I^{+}(s)e^{-\frac{sx}{v}} + I^{-}(s)e^{\frac{sx}{v}} \right] = - sC \left[ V^{+}(s)e^{-\frac{sx}{v}} + V^{-}(s)e^{\frac{sx}{v}} \right] \,$

Equating the $e^{-\frac{sx}{v}}$ and $e^{\frac{sx}{v}}$ terms on both sides, we get the following pair of equations:

$- \frac{1}{v} I^{+}(s) = - C V^{+}(s) \,$
$\frac{1}{v} I^{-}(s) = - C V^{-}(s) \,$

Solving for voltage yields the following:

$V^{+}(s) = I^{+}(s) Z_{c} \,$ ... Equ. (4)
$V^{-}(s) = -I^{-}(s) Z_{c} \,$ ... Equ. (5)

Where $Z_{c} = \frac{1}{vC} = \frac{\sqrt{LC}}{C} = \sqrt{\frac{L}{C}}$ is the characteristic impedance (Ohms)

We can use the above equations to re-write the voltage equation of Equ. (2) in terms of current as follows:

$V(x, s) = Z_{c} \left[ I^{+}(s)e^{-\frac{sx}{v}} - I^{-}(s)e^{\frac{sx}{v}} \right] \,$

Finally, taking the inverse Laplace Transform of the equation above, we get:

$V(x, t) = Z_{c} \left[ i^{+}(x - vt) - i^{-}(x+vt) \right] \,$