AdvancedA-002-012-006
On VHF and above, 1/4 wavelength coaxial cavities are used to give protection from high-level signals. For a frequency of approximately 50 MHz, the diameter of such a device would be about 10 cm (4 in). What would be its approximate length?
A
Answer
Circuit design and power supplies
Type
A
1.5 metres (5 ft)
B
0.6 metres (2 ft)
C
3.7 metres (12 ft)
D
2.4 metres (8 ft)
Answer Notes
To determine the physical length of a 1/4 wavelength coaxial cavity at 50 MHz, you must first calculate the electrical length of a full wavelength. The standard formula for this is Wavelength (in metres) = 300 / Frequency (in MHz).
Using this formula, 300 divided by 50 MHz gives a full wavelength of exactly 6 metres. Because the cavity requires a 1/4 wavelength design, you simply divide this full wavelength result by 4.
Dividing 6 metres by 4 gives exactly 1.5 metres. In Imperial units, 1.5 metres is roughly equivalent to 5 feet, making the 1.5-metre (5 ft) option the correct choice.
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