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Revision as of 17:12, 28 September 2006 by Alphachimpbot (talk | contribs) (BOT - updating merge tags to appear in Category:Merge by month)(diff) ← Previous revision | Latest revision (diff) | Newer revision → (diff)When radio waves from clean transmitters interact with dirty connections or corroded parts (which would then act as a diode) of a television, the rusty bolt effect, including the generation of harmonics and/or intermodulation in the signal, can occur. Rusty objects can re-radiate radio signals with harmonics and other unwanted signals; a television might then receive and attempt to interpret these signals.
If one experiences this problem, one should check both the transmitter and the television for dirty connections or corroded parts. One should also check for signs of corrosion in the cables which link the equipment to the aerials and for badly made joints. Beyond this, one might check any metal objects near the antenna for rust or corrosion. Any of these could be the source of the problem.
It is possible to cure this problem in several ways:
- Remove the corroded object. This is often the best cure because if you can eliminate the object then the interference it generates will cease entirely.
- Clean the object to allow proper electrical flow.
- Place an insulator between the two objects which are making the rust bolt. This will stop the RF current entirely.
- Lower the RF field strength, the maths below explains how this will greatly reduce the intensity of the effect.
- Get a better aerial which is more directional. It may be possible to point the aerial in such a direction that it does not pick up the unwanted signal coming from the rusty bolt.
Maths associated with the rusty bolt
The transfer characteristic of an object can be described by the following two equations.
E = Eo K1 + (Eo)
For an ideal perfect linear object K2, K3, K4, K5 etc are all zero.
For a 'rusty bolt' or a frequency mixer stage K2, K3, K4 and/or K5 etc are not zero.
Harmonic generation
If the incomming signal is a sine wave {Eo sin(ωt)} then the output will be
E = Eo K1 sin(ωt) + (Eo)
Hence by lowering the size of E the harmonics will be greatly reduced.
Mixing product generation
Second order
Ef1+f2 = k Ef1 x Ef2
Ef1-f2 = k Ef1 x Ef2
Third order
Ef1+f2+f3 = k Ef1 x Ef2 x Ef3
Ef1-f2+f3 = k Ef1 x Ef2 x Ef3
Ef1+f2-f3 = k Ef1 x Ef2 x Ef3
Ef1-f2-f3 = k Ef1 x Ef2 x Ef3
Hence the second order, third order and higher order mixing products can be greatly reduced by lowing the intensity of the original signals (f1, f2, f3, f4 ...... fn)
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