Jim Kelley wrote:



Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V



Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?


Depends on whether 'a' is in series or in shunt.

73, ac6xg

Apparently I should have added that distributed resistive losses in
series create distributed voltage drops with a common current through
all, but distributed resistive losses in shunt create a distribution of
Norton current sources, where the shunt current on the transmission line
or radiator at a given point is the sum of all equivalent current
sources between that point and the end of the transmission line/radiator.

73, Jim AC6XG