Discussion of Tony Wakefield's waveforms.

A length of 75Ω coaxial cable was trickle
charged up to 8 volts ( = 4 squares on the photos)
from a battery via two 1 meghom resistors.

An infinite length of coaxial cable (actually a 75 Ω
resistor) was attached to the right hand end. The voltage at the right hand end
of the capacitor immediately dropped to half (two squares), which triggered the
oscilloscope. This point is shown on all bottom traces.

The time delay from end to end of the capacitor is
about four squares. However, in (1) we see that the half amplitude output from
the charged capacitor lasted twice as long, about eight squares.

(2) shows that the drop in
signal from four squares to two squares is delayed by one square 25% along the
charged cable. The end of the signal is one square earlier here.

(3) shows that the drop in
signal from four squares to two squares is delayed by two squares 50% along the
charged cable. The end of the signal is another square earlier.

(4) shows that the drop in
signal from four squares to two squares is delayed by three squares 75% along
the charged cable. The end of the signal is another square earlier.

(5) shows that the drop in
signal from four squares to zero squares is delayed by four squares 100% along
the charged cable. The final, full drop is another square earlier.

In my
predictions , (1) thru (5) are called E1, D, C, B, A. My predictions were
correct.

The energy is reciprocating from end to end until
the switch is closed, when the rightwards travelling energy is emitted,
followed the leftwards travelling energy which must first reflect at the left
hand end. Here are
“photographs” of the state of the whole cable at intervals of time, instead of
oscilloscope pictures taken at a single point. Before the switch was closed,
the energy in the charged capacitor was already reciprocating from end to end
at the speed of light (for the dielectric). When the switch is closed, the
first portion of energy is emitted, so some of the capacitor drops from 8v
(four squares) to 4v (two squares). The back end of the 8v then retreats
towards the left. When it reaches the left hand end, the drop from 4v (two squares)
to 0v begins to travel to the right.

Ivor
Catt. 10 May 2012.