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 Zo and is found in the formula  Z= √ L/C

Zo 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 108 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?

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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

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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 108 m/s but this is exactly the same as v = 3 x 108 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.

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