On 2014-11-01 20:44:55 +0000, Jeff Liebermann said:

On Sat, 1 Nov 2014 15:26:52 -0000, "gareth"
wrote:

***** Without the remnants of non-radiated power, there could NOT be
a standing wave!

That's quite true. Standing waves require a transmission line. If
all the RF has been radiated, and there are no "remnants" left in the
transmission line, there can be no standing waves because there is no
RF.

Think about the other boundary conditions.

If you unplug the coax cable and antenna, and then transmit into an
open circuit, there are no standing waves. All the RF power is
converted to heat in the output stage. There's no transmission line
upon which to produce standing waves and there's no antenna to
radiate. Without a transmission line or antenna, there can be no
radiation and therefore, not standing waves.

The other extreme is also true. If you have an infinitely long
lossless coaxial cable, with either an open, short, or black hole at
the far end, there are no reflections because the wave will never
quite reach the open or short to produce a reflection. Without a
reflection, there can be no standing waves.

However, this does not change the fact that standing waves do not 'use
up' any of the power fed to the aerial (in principle, increased current
intensity increases resistive losses, but this loss can be made
arbitrarily low by having a lower wire resistance). Standing waves do
not in principle use 'power' at all and certainly do not dissipate
energy that otherwise would be radiated. They require a signal to be
applied to the transmission line but, whether the power is radiated at
the other end or the signal merely meets a mismatch, say an open
circuit, the standing wave does not affect, or need to use, any of the
power that leaves the other end. Indeed they work just as well if no
power whatever is used, as in the open circuit case.



--

Percy Picacity