Sky Wave and Ionospheric Propagation Questions and Answers

Answer: A

Short wavelength solar radiation strips electrons from atoms and molecules in the thin upper atmosphere, and the free electrons are what make it refract radio waves. Cosmic rays and meteors contribute a little, and meteoric ionisation is exploited in meteor scatter communication, but the sun dominates. The magnetic field shapes and complicates the ionosphere without creating it. Because the source is solar, every ionospheric property varies with time of day, season and the solar cycle.

Answer: B

Ionising radiation is strongest high up but there is little gas to ionise, while lower down there is plenty of gas but the radiation has been absorbed, so the production rate peaks at an intermediate height. Different wavelengths and different constituents peak at different heights, which gives the several layers. Gravitational separation does occur but is not the layer mechanism, and the layers are physically real. A product of two opposing trends peaking in between is the classic Chapman argument.

Answer: C

The D layer, at around seventy to ninety kilometres, is dense enough that electrons collide frequently with neutral molecules, so energy taken from the wave is lost as heat rather than re-radiated. It absorbs strongly at low frequencies and disappears at night. The E and F layers refract rather than absorb. Distinguishing a refracting layer from an absorbing one is essential, since one carries the signal and the other destroys it.

Answer: D

The amplitude of the electron's forced oscillation is inversely related to frequency, so at low frequencies it moves further, collides more with neutrals and transfers more energy to heat. Absorption therefore falls roughly as the inverse square of frequency, which is why daytime high frequency working moves to the higher bands. Penetration depth, electron density and antenna radiation are not the mechanism. A loss that falls with frequency, in a medium whose refraction also falls with frequency, sets up the central trade of high frequency operating.

Answer: A

The critical frequency is the highest frequency returned when the wave is sent straight up, and it is set directly by the peak electron density of the layer. Above it a vertical wave escapes into space. Oblique incidence raises the usable frequency well above it, which is what makes long distance working possible at all. Absorption and neutrality are separate matters. One measured number, the critical frequency, summarises the layer's refracting capability.

Answer: B

Taking the square root of the electron density and multiplying:

\[ f_{c} = 9 \times 10^{6} = 9 \times 10^{6} \]

hertz, that is 9 MHz. The square root dependence means a fourfold rise in electron density only doubles the critical frequency, which is why the usable band changes less dramatically through the solar cycle than the underlying ionisation does. Being able to move between electron density and critical frequency connects the physics to the operating practice directly.

Answer: C

For a plasma the refractive index is below one, and it falls further as the frequency approaches the plasma frequency, so the wave is refracted away from the normal and bends back towards earth. It goes to zero at the plasma frequency, which is where reflection occurs. A refractive index below unity implies a phase velocity above the speed of light, which is permissible because the group velocity remains below it. The counter-intuitive index is the whole basis of sky wave propagation.

Answer: D

The secant law states that the oblique frequency returned equals the critical frequency multiplied by the secant of the angle of incidence at the layer, so a shallow angle allows a much higher frequency than vertical incidence would. A factor of three or four is typical for a long single hop. Absorption, virtual height and skip distance are computed by other means. This one relation is what turns a modest critical frequency into a usable high frequency circuit.

Answer: A

Applying the secant law directly:

\[ f_{MUF} = 6 \times 3 = 18 \]

megahertz, that is 18 MHz. A wave at that frequency would escape if sent vertically but is returned at the shallow angle of a long path. This is why the longest hops support the highest frequencies, and why a station may hear a distant contact on a band while nearer stations are inaudible on it.

Answer: B

The MUF is the highest frequency the ionosphere will return on that particular path at that time, and above it the wave passes through into space. In practice an optimum working frequency of about eighty five per cent of it is used to allow for variability. Licensing, absorption and antenna match are separate constraints, though absorption sets the lower limit. A circuit is worked between two ionospherically determined bounds, and both move through the day.