I'm reading a book (Numerical Techniques in Electromagnetics by Sadiku) & just finished the section on finite difference methods. As what I thought would be an easy exercise, I tried to apply what I'd learned to the telegraphers equations that describe the voltage, V(x, t), and current, I(x, t), on a transmission line with some realistic (meaning ugly) boundary conditions:

The equations are 2 coupled (by the boundary conditions) wave equations:

$\displaystyle

\frac{\partial^2 V}{\partial t^2} = u^2 \frac{\partial^2 V}{\partial x^2}

$

$\displaystyle

\frac{\partial^2 I}{\partial t^2} = u^2 \frac{\partial^2 I}{\partial x^2}

$

Where u is the velocity of propagation and 0 < x < L, t > 0

The boundary & initial conditions are

$\displaystyle

V(0, 0) = V_g(0) \frac{Z_0}{Z_0 + R_g}

$

$\displaystyle

I(0, 0) = V_g(0) \frac{1}{Z_0 + R_g}

$

$\displaystyle

V(0, t) = V_g(t) - R_g I(0, t)

$

$\displaystyle

V(L, t) = R_L I(L, t)

$

V(x, 0) = I(x, 0) = 0 for x>0 & everything(t<0) = 0

RL, Rg, & Z0 are real positive constants. Vg(t) is a known function of time only.

So I turn each into a difference equations using the centralized 2nd order approximation

$\displaystyle

\frac{V(i, j+1) - 2 V(i, j) + V(i, j-1)}{\Delta t^2} = u^2 \frac{V(i+1, j) - 2 V(i, j) + V(i-1, j)}{\Delta x^2}

$

same for I. I solve for V(i, j+1) & I(i, j+1). With j (time) in my outer loop & i (x) in my inner loop I start stepping across x for each time t. Everything is fine until I reach that last boundary condition. I have a chicken & egg problem. First impulse is to step V forward and calculate I or vice versa. But neither will give me the right answer. I'm going to have the same problem at x=0 after that first step.

The problem is easy if RL = Rg = 0. I can just solve the equation for V since V(L, t) = 0.

But, for RL & Rg > 0, how do I handle those two boundary conditions that relate V & I at x=0 and x=L?

I know there are other (& probably better) ways to solve this, even analytically or just intuitively. But I need the practice with FD.

Thanks,

Apchar