Roy Lewallen Inscribed thus:

Baron wrote:

Please could you elaborate on how and why a common mode current has a
different VF on a balanced line.

Sure.

First, a balanced line, whether it's twinlead or coax, doesn't have
any common mode current, by definition -- the lack of common mode is
what makes it balanced. We're talking about a physically symmetrical
line.

Whenever you have a two conductor line, you effectively have two
transmission lines, differential mode and common mode. Although you
actually have only one current on each conductor, by taking advantage
of the principle of superposition you can mathematically separate the
two currents into two *sets* or components of currents, analyze their
effects separately to gain a better understanding, and simply add the
results if you want to know the overall solution. The sum of the
common mode and differential currents are the actual conductor
currents, and the sum of the common mode and differential responses is
the actual response.

The differential or transmission line mode waves (voltage and current)
are the components which are equal and opposite on the two conductors,
so the field is strongest between the two conductors, fringing outward
in the case of ladder line. The presence of the dielectric material in
a major portion of the field slows down the waves, lowering the
velocity factor. In the case of coax, the field is entirely within the
dielectric so we can easily calculate the velocity factor if we know
the dielectric constant of the material. In the case of ladder line,
we don't know what fraction of the field is in the air and what's in
the dielectric without a very advanced computer program, so we have to
measure the velocity factor. The fraction and therefore velocity
factor changes, by the way, with frequency, a phenomenon known as
dispersion.

The common or antenna mode waves are the components that are equal and
in the same direction or polarity on the two conductors. The field is
the same as it would be if the two conductors were connected together
to make a single conductor. One conductor of the common mode
transmission line is the two conductors of the ladder line, and the
other is the Earth and/or surrounding conductors. These two common
mode transmission line conductors are usually much farther apart than
the ladder line conductors, so the common mode characteristic
impedance is higher than the differential mode impedance. The velocity
factor is usually higher, too, because the field is between the two
common mode conductors -- the ladder line and the Earth --, and almost
none of it is in the line dielectric. So its velocity factor is nearly
1. In my TDR demonstration, the common mode open end reflection
occurred before the larger differential mode reflection because of the
higher velocity factor, so it looked like a differential mode
reflection from a point short of the end. (And I helped reinforce this
mistake in order to get the audience's attention.)

Any two conductor line supports both modes and behave the same, but
coax is a little easier to understand because the differential and
common mode currents are actually physically separate -- so no
mathematical hocus-pocus is necessary. The differential currents and
waves are entirely inside the cable, and the common mode currents and
waves are outside. The velocity factor inside (differential mode) is
determined by the dielectric material, and the velocity factor of the
outside (common mode) is nearly 1.

Roy Lewallen, W7EL

Thankyou, Jim & Roy.
Your explanations were most enlightening. I just couldn't get my head
around the "how & why" the VF should be different. I have also
realised why I have sometimes seen more than one TDR reflection from a
perfectly good transmission line.

73's
--
Best Regards:
Baron.