# Transmission Line derivation

1. Sep 24, 2011

### yayscience

Eqn. (2.1)
$$\frac{\partial v}{\partial z} = -L \frac{\partial i}{\partial t}$$
Eqn. (2.2)
$$\frac{\partial i}{\partial z} = -C \frac{\partial v}{\partial t}$$
Eqn. (2.5)
$$v(z,t)=V^+f(t-\frac{z}{v_p})+V^-f(t+\frac{z}{v_p})$$

"By substituting Equation (2.5) into Equations (2.1) and (2.2), we
determine that the current has the form:"
$$i(z,t)=\frac{V^+}{Z_0}f(t-\frac{z}{v_p})+\frac{V^-}{Z_0}f(t+\frac{z}{v_p})$$

So, I've seen this derivation about four different ways, and I know this is the result, but I can't see what moves I need to substitute 2.5 into 2.1 and 2.2 to get 2.5. Can someone please enlighten me? Thanks all!

2. Sep 25, 2011

### uart

Just some preliminaries first. If you differentiate 2.1 wrt "t" and 2.2 wrt "z" (and then repeat the other way around) you get the following second order PDE's.

$$\frac{\partial^2 v}{\partial z^2} = L C \frac{\partial^2 v}{\partial t^2}$$

$$\frac{\partial^2 i}{\partial z^2} = L C \frac{\partial^2 i}{\partial t^2}$$

These separable PDE's give the motivation for the form given in 2.5 and tell us that the phase velocity is:

$$v_p = \frac{1}{\sqrt{L C}}$$

Since the above second order PDE's tell us that i(z,t) will be in the same form as v(z,t) we can therefore write:

$$i(z,t)=I^+ \, f(t-\frac{z}{v_p})+I^- \, f(t+\frac{z}{v_p})$$

By differentiating the above wrt "t" and differentiating 2.5 wrt "z" and substituting into 2.1 we can equate coefficients to show the following:

$$I+ = \frac{V^+}{L v_p} = \frac{V^+}{Z_0}$$

$$I- = - \frac{V^-}{L v_p} = - \frac{V^-}{Z_0}$$

Hope that helps.