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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: 5a Feeder basics. 5a.1 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. Calculate physical feeder lengths given the frequency and velocity factor. Recall that the velocity factor for coaxial feeder with a polythene dielectric is approximately 0·67 or 2/3. 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? 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 So to calculate the wave length in metres we use the equation from the Intermediate Course click here to check back 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 we can use the speed of light as 300m/sec so 300 / 3.6 = wavelength =83.33 so quarter wave = 83.33 / 4 = 20.83 metres now apply the velocity factor 20.83 x 0.67 =13.9 metres This is simple mathematics and you should be able to easily solve the problems. Recall that feeder loss increases with increasing frequency and that lower loss feeders may be desirable at VHF, UHF and above.
The loss in the feeder is frequency dependent. As the frequency of operation increases feeder losses also increase and thus to have much the same signal coming out of the end as going is at the beginning you have to use better quality feeders  or as it is known "low loss feeders". So what is good for HF is not going to be so good at VHF and certainly not at UHF. 5a.2 Understand that a quarterwave length of feeder can be used as an impedance transformer. Apply simple examples of the formula Z_{0}^{2} = Z_{in} x_{ }Z_{out } The formulae for calculating the impedance of a matching section will be given to you in the exam, if a question on this topic comes up so the important points to know are what do the mathematical notations represent. Z_{0}^{2} = Z_{in} x_{ }Z_{out } In the formula: Z_{0} = the impedance of the quarterwave length of feeder matching line Z_{in} is the impedance of the feed line to which the antenna is to be matched Z_{out} the impedance of the antenna. Note that the Z_{0} is "squared" so to find the answer as to the impedance of the matching section you would have to use :
all that we have done is "take the square root" of both sides, so as long as you understand such manipulation the result can be easily found.
Example: What would be the matching section's impedance to match ;a 75 ohms dipole to 50 ohms coax. Z_{0} = square root of ( 75 x 50 ) = 61.3 ohms.Now it is not very easy to find such a piece of coaxial feeder so what other feed line could be better used ? Twin feeder of course, but for the exam you do not have to go into how you would achieve the physical dimensions of the matching section just be able to calculate it  as a diagram of coaxial feeder is easier to remember that it why we have used it. 5a.3 Recall the basic construction and use of waveguides.
Construction Typically waveguide is made of brass, copper, silver, aluminium, or any metal that has low resistivity. They can also be made of a combination of non ferrous metals with the exception of aluminium as this can chemically react with particularly copper. It is possible to use metals with poor conductivity characteristics, if the interior walls are properly plated. The pictures above and below show wave guides. These take the place of other feeders with which you are familiar for HF and VHF and UHF (coaxial and open wire / balanced) when using Microwave frequencies and if properly made with good joint can be low loss. The guide is usually a rectangular "tube" with flanges attached to join one piece to another and as shown below can be complicated arrangements or as shown above link to coaxial feeder via an "N" type connector.








