In all transmission lines, including coax, there are various shapes of
transverse electric and magnetic fields that can exist for the particular
transmission line geometry. For each shape, the "propagation constant" can
be calculated. Many transmission lines (at lower frequencies) have only one
shape with propagates with low attenuation. The other shapes can exist, but
their "propagation constant" is such that they decrease exponentially with
distance. The propagation constant for each shape can be calculated, and is
often a function of frequency.
When there is a discontinuity in a line, other shapes than the usual
one must exist at the point of the discontinuity. (for example, in order to
ensure that the transverse electric field is zero the surface of a
conducting shape that is part of the line discontinuity). Thus, these other
shapes exist (at a certain amplitude) at the point of discontinuity. The
amplitude of the other shapes decreases exponentially at distances away from
the discontinuity. The rate of the fall-off will depend on the particular
shape, according to its propagation constant.
Thus, the distance needed to be back to regular old TEM propagation in a
coax will depend on the particular discontinuity, and the propagation
constants of the "higher order modes" or different field shapes, of a coax
line.
I have seen examples worked out for waveguide propagation and a step
change in waveguide width. There are probably worked examples of coax
discontinuities in the literature, also.
These non-propagating shapes are usually called " evanescent modes", and
this would be a good search term to use to investigate this further.
Cliff Curry
"Ian White G/GM3SEK" wrote in message
...
Cecil Moore wrote:
Owen Duffy wrote:
Cecil, do you have some quantitative explanation / support for this?
Nope, but there were no disagreeing postings.
I am not asking whether or not field conditions (and V/I on the
conductors) immediate to the discontinuity are not Zo of either of the
lines, just where has the 2% of a wavelength come from?
As I remember it came from the spacing between conductors Vs wavelength.
The spacing between conductors is about 0.1 inches for RG-58. How many
times that value would you think it would take for a transmission line
to force its Z0 upon the signals? At 10 MHz, 2% of a wavelength (24
inches)
is about 250 times the spacing between conductors.
Maybe the electromagnetics people have a useful way to visualize it...
Deep inside the coax, the electric field lines between the inner and outer
of the coax are exactly at right-angles to the main axis. Where that is
exactly true, you have a pure TE10 mode so it's also valid to assume that
V/I is exactly equal to Zo.
Very close to the end of the coax, the electric field lines from the
center conductor start to reach out and connect with whatever is out there
beyond the end of the shield. Then you no longer have pure TE10 and can no
longer assume that V/I=Zo.
Coming at it from the other direction, the question would be: how far into
the coax must you go before the field lines become accurately at
right-angles?
We can be sure that the field lines won't suddenly snap from being
divergent to being accurately at right-angles, so what we're really asking
is: how far before the field lines are near-enough at right angles to make
V/I=Zo a good engineering approximation?
Intuitively, the diverging field lines only seem likely to occur within a
few diameters of the end of the shield. Field lines always connect with
highly conducting surfaces at right-angles, and they won't like to be
sharply bent to run along the axis of the coax.
In other words, the effect would seem to be mainly a function of shield
diameter D. Again intuitively, I can't see where wavelength would come
into it, unless D itself is a significant fraction of the wavelength
(which is normally never true, and even microwave engineers try to avoid
it).
Following this picture of diverging field lines, there should also be a
secondary effect depending on how the inner and shield of the coax are
connected to the circuit outside.
All of this suggests that it's impossible to give a single answer that
would be valid for all cases (unless you choose a number that's so big, it
can't fail to be correct... like "120 radials" :-)
However, none of this speculation is of any practical consequence. All
practical experience indicates that if a line is so short that V/I is not
quite equal to Zo, the impedance transformation along that line will be so
small that the effect of any Zo error is lost in the noise.
--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek