Jules: Created page with "== Footnotes == === Derivation of Voltage Equation === The general solution to the transmission line wave equations are: : $I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \,..."$

2020-11-22T07:41:57Z

Created page with "== 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) \,..."

New page

== 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 [http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms 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>

Talk:Travelling Wave Line Model - Revision history

Jules: Created page with "== Footnotes == === Derivation of Voltage Equation === The general solution to the transmission line wave equations are: : I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \,..."

Jules: Created page with "== Footnotes == === Derivation of Voltage Equation === The general solution to the transmission line wave equations are: : $I(x, t) = i^{+}(x - vt) + i^{-}(x + vt) \,..."$