OGRAD 01.02 – Radio Wave Propagation

Passing over obstacle with dimensions close to the wavelength

The correct statements are:

• Sky waves rely on refraction from ionospheric layers to return to the surface

• The “Skip Distance” is the range from the transmitter to the first sky wave return

Explanation of the errors:

• Statement 1 is False: Ground waves travel further over sea than over land because sea water has much higher conductivity (less resistance) than land.

• Statement 4 is False: Ground wave range decreases as frequency increases. Higher frequencies suffer greater attenuation (energy loss) when traveling along the surface.

The lower the frequency the greater the ionospheric attenuation The attenuation of an HF ground wave is worse over the land than over ice – this is also correct however in Exam mark “”” The lower the frequency the greater the ionospheric attenuation as correct.

Night, high

A change in the direction of the plane of polarization due to reflections in the ionosphere poorly framed question: here is a more accurate phrasing : What is the primary cause of the “night effect” observed in radio navigation signals, particularly from NDBs? Explanation: When sky waves reflect off the ionosphere at night, their plane of polarization can rotate due to varying ionospheric conditions. Since NDB signals are linearly polarized, any change in polarization angle reduces the effective signal received by the aircraft’s antenna, leading to signal fading, distortion, or bearing errors – this is what we refer to as the night effect. Summary: NDBs use linear polarization Ionospheric reflection at night rotates the polarization plane Aircraft receivers may not align with the changed polarization -> signal loss or distortion

79 NM

Correct Answer: The E-layer is higher by night than by day because the ionization levels are lower at night

HF we use HF

15 000 ft

Slows down and is attenuated

1500 nm

At night when ground waves and sky waves interfere with each other generally during night, sky wave and ground reflected waves MAY meet in anti-phase and fading of signal occurs.

Correct Answer: 1 and 3 are correct

Explanation:

Statement 1 is Correct: Surface waves travel significantly further over sea water because it has high conductivity, reducing attenuation compared to land.

Statement 3 is Correct: Diffraction is the mechanism that allows the surface wave to bend around the Earth’s curvature.

Statement 4 is Incorrect: VHF frequencies (e.g., 243 MHz) rely on space waves (Line of Sight), not surface waves

4

500 kHz low frequency greater ground range.

At night due to the reception of both skywave and ground waves

125 km, 1350 NM

30 MHz space wave – VHF or Higher frequency, for Less static interference. static interference is worst at low freqency. so for sky wave – HF Communication – range 3 to 30 MHz, at 03 MHz- too much static — (caused by thunderstorm/Rain and other activity).

Decrease effective range decrease.

Higher frequency and higher position of the reflecting layer

Based on the statements provided regarding the Maximum Usable Frequency (MUF):

The correct statements are:

• MUF increases as the angle of incidence becomes more oblique (shallower).

• Transmission above the MUF will result in the wave passing through the ionosphere into space.

Explanation:

1. False: MUF is always lower than the Critical Frequency.

   • Reason: The Critical Frequency is measured at vertical incidence (straight up). The MUF applies to oblique (angled) transmissions. Because the ionosphere can bend higher frequencies when they hit at a shallow angle, the MUF is typically higher than the Critical Frequency (usually 3 to 5 times higher).

2. True: MUF increases as the angle of incidence becomes more oblique.

   • Reason: Steeper angles require lower frequencies to reflect. Shallower (more oblique) angles allow higher frequencies to be reflected.

3. True: Transmission above the MUF will result in the wave passing through…

   • Reason: If the frequency is too high for the current ionospheric density and angle, the wave will not be bent sufficiently to return to Earth and will escape.

4. False: MUF is unaffected by the solar cycle.

   • Reason: The solar cycle directly impacts the density of the ionosphere. Higher solar activity increases ionization, which significantly increases the MUF.

attenuation

Explanation:

Statement 1 is Incorrect: Doubling power does not double the range due to the Inverse Square Law.

Statement 2 is Correct: VHF frequencies are too high to follow the earth’s curvature (surface wave) or reflect off the ionosphere normally; they are Space Waves limited by Line of Sight.

Statement 3 is Correct: According to the Inverse Square Law, signal strength diminishes with the square of the distance. To double the distance (2R), you need 2^2 = 4 times the power.

Statement 4 is Incorrect: VHF signals generally penetrate the ionosphere rather than refracting, which is why they are used for satellite comms but not long-range sky waves.

Night with sky and ground waves generally during night, sky wave and ground reflected waves MAY meet in anti-phase and fading of signal occurs.

Should be decreased at night In HF (High Frequency) communication, short-range sky wave propagation requires a lower frequency, as these waves reflect from lower ionospheric layers (like the E-layer). Long-range communication needs higher frequencies that can penetrate deeper and reflect from higher layers (like the F2-layer). Other options are incorrect: Frequencies increase with latitude. Lower frequencies are needed at night due to ionospheric changes Frequency is affected by both time of day and season

The bending of its propagation path as it passes through or over areas of different electrical conductivity

Range from the transmitter to the first returning sky wave

MF, greater, ground, sea, land

VLF VLF is refracted by the D-layer with minimal ionospheric penetration, resulting in less noise and static. It’s not effective for sky wave use due to limited range, so it’s mainly used for ground wave communication. HF, however, is ideal for sky wave propagation and long-range communication.

Reduction of its power by absorption, scattering or spreading

1000 nm, 300 nm, MF

A direct wave is bent around the form of the Earth

Short Logic:

Higher frequencies are harder to refract, so they penetrate deeper/higher before refraction. Lower frequencies refract easily in the lower layers.

The correct conclusion is: Statement I is False; Statement II is True.

Here is the breakdown:

• Statement I is False: It is actually the opposite. Higher frequency waves require a higher ionization density to be refracted. Lower frequency waves are bent more easily and thus require less ionization density required to be returned to Earth.

• Statement II is True: Because higher frequency waves are harder to bend (they possess more energy), they pass through the lower, less dense layers and must travel deeper (higher up) into the ionosphere to find a layer with sufficient electron density to refract them back. Lower frequencies are often refracted by the lower layers (like the E layer) and do not penetrate as far.

At night, due to the combination of sky waves and ground waves generally during night, sky wave and ground reflected waves MAY meet in anti-phase and fading of signal occurs.

The frequency required for short ranges will be less than the frequency required for long ranges In HF (High Frequency) communication, short-range sky wave propagation requires a lower frequency, as these waves reflect from lower ionospheric layers (like the E-layer). Long-range communication needs higher frequencies that can penetrate deeper and reflect from higher layers (like the F2-layer). Other options are incorrect: Frequencies increase with latitude (opposite of option 1). Lower frequencies are needed at night due to ionospheric changes (opposite of option 2). Frequency is affected by both time of day and season, making option 3 incorrect.