B-006
B-006
Section B-006-012
If you made a half-wavelength dipole antenna for 28.150 MHz, approximately how long would it be?
Key words: half-wavelength DIPOLE. Wavelength (lambda) in metres IN FREE SPACE is 300 divided by frequency in megahertz. A 'dipole' is approximately 95 % of one half wavelength in free space = '(300 / 2) * 0.95' divided by frequency in megahertz. In this example, the dipole must be (300 / 28.15 MHz / 2) * 0.95 = 5.06 metres. The frequency is in the 10-metre band of 28.0 to 29.7 MHz, a dipole there must necessarily be 5 metres long.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is one disadvantage of a random wire antenna?
Because the 'random wire' and 'long wire' antennas frequently originate right at the back of the antenna tuner in your station, stray RF (radio frequency) can be a problem.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is the three-dimensional radiation pattern of a half-wavelength dipole in free space?
Picture a horizontal dipole viewed from above. If you plotted radiation all around it in a single horizontal plane, the plot would look like a "number eight": peak radiation at 90 degrees (broadside) from the antenna, negligible radiation from the ends. Now, imagine that pattern in successive planes around the antenna, you get a donut shape.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is the impedance at the feed point of a half-wave dipole in free space?
Ignoring ground effects, what is the radiation pattern of a horizontal half-wave dipole installed with the ends pointing North/South?
Picture a horizontal dipole viewed from above. If you plotted radiation all around it, the plot would look like a "number eight": peak radiation at 90 degrees (broadside) from the antenna, negligible radiation from the ends.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is a major advantage of an end-fed half-wave antenna (EFHW)?
In addition to working as a half-wave antenna on its lowest frequency, the antenna may present suitable impedances at other frequencies and become a usable compromise antenna.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is a disadvantage of using an antenna equipped with traps?
An antenna with traps is a multi-band antenna (i.e., resonant at more than one frequency). If the transmitter leaks harmonic energy (multiples of the operating frequency), this harmonic energy may be more readily radiated by a multi-band antenna. For example, traps are inserted in an antenna for 80 metres to permit operation on 40 metres; if your transmitter puts out 'harmonics' while you operate on 80 m ( say, 3.5 MHz ), the second harmonic falls in the 40 m band. The antenna is also resonant at that frequency and would freely radiate the harmonics.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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What is an advantage of using a trap antenna?
The only reason why antenna traps (parallel resonant circuits) are useful is to permit operation on more than one band from the same physical antenna. Through their high impedance at resonance, traps shorten the antenna by making the antenna sections beyond them inaccessible.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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If you were to cut a half-wave dipole for 3.75 MHz, what would be its approximate length?
Wavelength (lambda) in metres IN FREE SPACE is 300 divided by frequency in megahertz. The dipole is approximately 95 % of one half wavelength in free space = '(300 / 2) * 0.95' divided by frequency in megahertz. In this example, the dipole must be cut to (300 / 3.75 MHz / 2) * 0.95 = 38 metres. [ 3.75 MHz is in the 80-metre band of 3.5 to 4.0 MHz, a DIPOLE there must be below 40 metres long ].
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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