Talk:Travelling Wave Line Model

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Footnotes

Derivation of Voltage Equation

The general solution to the transmission line wave equations are:

[math] I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \, [/math]
[math] V(x, t) = v^{+}(x - vt) + v^{-}(x + vt) \, [/math]

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

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

The Laplace Transform of the above equations are:

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

Recall the Telegrapher's equation for current:

[math] \frac{\partial I(x,t)}{\partial x} = - C \frac{\partial V(x,t)}{\partial t} \, [/math]

Taking the Laplace Transform of this equation, we get:

[math] \frac{\partial I(x,s)}{\partial x} = - sC V(x,s) \, [/math] ... Equ. (3)

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

[math] \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] \, [/math]

We can differentiate the left-hand side:

[math] \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] \, [/math]

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

[math] - \frac{1}{v} I^{+}(s) = - C V^{+}(s) \, [/math]
[math] \frac{1}{v} I^{-}(s) = - C V^{-}(s) \, [/math]

Solving for voltage yields the following:

[math] V^{+}(s) = I^{+}(s) Z_{c} \, [/math] ... Equ. (4)
[math] V^{-}(s) = -I^{-}(s) Z_{c} \, [/math] ... Equ. (5)

Where [math]Z_{c} = \frac{1}{vC} = \frac{\sqrt{LC}}{C} = \sqrt{\frac{L}{C}} [/math] 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:

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

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

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