Properties of RadioWaves 1
Radio Waves
The length of time it takes to generate one cycle of a radio wave is known as the period and is generally signified by the Greek letter tau (τ), and measured in microseconds (µs). (1 µs = 10-6 second).
Properties of Radio Waves 1
Figure 1.2 Sinusoidal wave - period
If, for example, the period of one cycle of a radio wave is 0.125 µs then the number of cycles produced in one second would be the reciprocal of this giving:
1 |
= |
1 |
= 8 000 000 cycles per second which are known as hertz (Hz) |
τ |
0.125 ×10-6 |
This is known as the frequency (f) of the wave; hence:
f = |
1 |
(1) |
τ |
The frequency of radio waves is expressed in hertz (Hz). Since the order of magnitude of the frequency of radio waves is very high, for convenience, the following terms are used to express the frequency:
Kilohertz (kHz) |
= |
103 |
Hz = |
1 000 Hz |
Megahertz (MHz) |
= |
106 |
Hz = |
1 000 000 Hz |
Gigahertz (GHz) |
= |
109 |
Hz = |
1 000 000 000 Hz |
So in the example above the frequency would be expressed as 8 MHz.
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1 Properties of RadioWaves
Waves Radio of Properties 1
Wavelength
The speed of radio waves (c) is the same as the speed of light (which is also EM radiation) and is approximately:
300 000 000 ms-1 (= 300 × 106 ms-1), or 162 000 nautical miles per second
Wavelength ( λ )
Figure 1.3 Sinusoidal wave - wavelength
If a radio wave travels at 300 × 106 ms-1 and the period is 0.125 µs, then the length (λ) of each wave will be:
λ = c. τ |
(2) |
300 × 106 × 0.125 × 10-6 = 37.5 m |
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This is known as the wavelength. From equation (1) this can also be stated as:
λ = |
c |
(3) |
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f |
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Giving:
λ = |
300 × 106 |
= 37.5 m |
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8 × 106 |
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Hence if the frequency is known then the wavelength can be determined and if the wavelength is known then the frequency can be calculated from:
f = |
c |
(4) |
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λ |
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Properties of RadioWaves 1
Examples:
1.If the frequency of a radio wave is 121.5 MHz calculate the wavelength.
λ = |
c |
= |
300 × 106 |
= 2.47 m |
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f |
121.5 × 106 |
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2. If the wavelength is 1515 m, what is the corresponding frequency?
f = |
c |
= |
300 × 106 |
= 198 000 Hz = 198 × 103 |
Hz = 198 kHz |
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λ |
1515 |
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For ease of calculation we can simplify the formulae:
f |
= |
300 |
MHz |
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λ (m) |
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λ |
= |
300 |
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m |
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f (MHz) |
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But we must ensure that our input arguments are correct, i.e. to calculate the frequency the wavelength must be in metres and to calculate the wavelength the frequency must be input in MHz.
Examples:
3.Determine the frequency corresponding to a wavelength of 3.2 cm.
Noting that 3.2 cm = 0.032 m the calculation becomes:
f = |
300 |
= |
9375 MHz (or 9.375 GHz) |
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0.032 |
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Properties of Radio Waves 1
4.Determine the wavelength corresponding to a frequency of 357 kHz. Noting that 375 kHz = 0.375 MHz the calculation is:
λ = |
300 |
= |
800 m |
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0.375 |
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1 Properties of RadioWaves
Waves Radio of Properties 1
Frequency Bands
The radio part of the electromagnetic spectrum extends from 3 kHz to 300 GHz. For convenience it is divided into 8 frequency bands. These are shown below with the frequencies, wavelengths and the uses made of the frequency bands in civil aviation. Note that each frequency band is related to its neighbouring band(s) by a factor of 10.
Frequency Band |
Frequencies |
Wavelengths |
Civil Aeronautical Usage |
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Very Low Frequency (VLF) |
3 – 30 kHz |
100 – 10 km |
Nil |
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Low Frequency (LF) |
30 – 300 kHz |
10 – 1 km |
NDB/ADF |
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Medium Frequency (MF) |
300 – 3000 kHz |
1000 – 100 m |
NDB/ADF, long range |
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communications |
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High Frequency (HF) |
3 – 30 MHz |
100 – 10 m |
long range |
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communications |
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Short range |
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Very High Frequency |
30 – 300 MHz |
10 – 1 m |
communication, VDF, |
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(VHF) |
VOR, ILS localizer, marker |
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beacons |
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Ultra High Frequency |
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ILS glide path, DME, SSR, |
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300 – 3000 MHz |
100 – 10 cm |
Satellite communications, |
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(UHF) |
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GNSS, long range radars |
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Super High Frequency |
3 – 30 GHz |
10 – 1 cm |
RADALT, AWR, MLS, short |
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(SHF) |
range radars |
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Extremely High |
30 – 300 GHz |
10 – 1 mm |
Nil |
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Frequency (EHF) |
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