As
the syllabus states Questions on the Advanced syllabus
will assume the candidate is familiar with basic
principles studied at Foundation and Intermediate levels
so let's start with some recapping.
FEEDER
A
feeder is something which connected the antenna to to the
transceiver ( or may be through and Low Pass Filter and
AMU and SWR meter).
The
Feeder may be coaxial feeder or Open Wire feeder and must
be appropriate for its use.
Open
wire feeder is two conductors separated by spreaders which
must be in the clear and is called BALANCED LINE.
Coaxial
feeder has a centre core, insulation, braid  connected to
EARTH  and outer insulation which may be used close to
poles, building, ground, etc and is called
UNBALANCED LINE. There should be no radiation through the
shielding braid either outwards on transmission nor
inwards on reception.
With
Open wire Feeder the wire diameter and its distance of
separation one from another gives it its
CHARACTERISTIC IMPEDANCE where as with coaxial cable the
diameter of the inner and outer conductors gives it its CHARACTERISTIC
IMPEDANCE.
The
term CHARACTERISTIC
IMPEDANCE has the symbol Z_{o }and is found in
the formula Z_{o
}=
√ L/C
Z_{o
}is a pure resistance. Losses occur in feeder cables
due to resistance of the copper wire and some of the RF
energy is lost as heat. So the amount of RF power
available at the antenna will be less than at the
transceiver with losses being greater, in the same cable
as frequencies rise.
If you are unsure of any of the above please
refer back to FLC and ILC training levels.
Syllabus Sections:
Feeders
4A3 31
Understand that the velocity factor of a feeder
is the ratio of the velocity of radio waves in the
feeder to that in free space and that the velocity
factor is always less than unity.
Velocity factor what is it?
From
the syllabus we learn that :
We know
that the velocity of radio wave in free space is the same as
the speed of light c=3
x 10^{8} m/s
From
the syllabus statement we learn that the velocity factor is
less than unity ( which is numerically 1 ) and for this to
be true the velocity of the radio waves in the feeder must
be slower.
Whilst
we have spoken about velocity factor in relation to feeders
the same goes for velocity factor in cables.
Why
is this important ?
Because
when making up a feeder and you want it to be a certain
number of multiplies of say 1/2 wave length then if you
merely cut the feeder without reference to the velocity
factor your feeder would be too long.
The
same goes for antennas. If you merely cut the half wave
dipole to the calculated length it too will be too long.
So the
physical length of feeders or antennas is always
less than its calculated electrical length
due to the velocity factor.
So if
we have a frequency of say 3.600MHz what would the length be
of a 1/4 wave stub?

4A3
31 continued Recall
that the velocity factor for coaxial feeder with a
polythene dielectric is approximately 0·67 or 2/3.
You
have to be able to remember that the velocity factor
for coaxial feeder with a polythene dielectric is
approximately 0·67 or 2/3

4A3
31 continued
Calculate physical feeder
lengths given the frequency and
velocity factor.
How
very confusing ... On the
formulae sheet which is part of
the documentation available to
down load now and will also be
given out in your examination the
Speed of light is quoted as one of
two possible abbreviations.
Either lower case c or lower case
v and eight can be used in
questions in the exam.
so
c
= 3 x 10^{8} m/s but this is exactly the same as
v = 3 x 10^{8} m/s
similarly c = f is
exactly the same as v = f
v = f or c = f
Note
that sometimes v is used for the speed of light and at
other times c.
The velocity of light = the
frequency x wavelength
As we are dealing in MHz (
millions of Hertz ) we can use the speed of light as
300m/sec
so
for a wavelength of 3.6MHz we can calculate the
wavelength
as
300 / 3.6 = wavelength =83.33 meters
so quarter wave = 83.33 / 4
= 20.83 metres
But
to correct for the length of the 1/4 wavelength we have to
take into account the end effect and this use velocity factor
now
apply the velocity factor 20.83 x 0.67 = 13.9 metres for a
1/4 wave leg of a dipole!!
This is simple mathematics
and you should be able to easily solve the problems.

Baluns
