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Chapter 4 · 7 hours

Propagation and Radio Frequency Spectrum

IOE past exam questions

Past questions and answers

37 questions set from this chapter, 13 of them more than once. Most asked first.

  • Asked 3 times
  • 2077 Chaitra · 8 marks
  • 2076 Baisakh · 6 marks
  • 2072 Asoj · 6 marks

Explain the ionospheric wave propagation (operation of ionospheric communication) showing its different layers, with the structure of the ionosphere.

Answer

Ionospheric (sky-wave) propagation is the mode in which radio waves (mainly HF, 3–30 MHz) sent upward from the earth are bent back (refracted) by the ionised layers of the upper atmosphere and return to earth far away, giving long-distance communication of thousands of km.

Structure of the ionosphere

The ionosphere is the region from about 50 km to 400 km height where ultraviolet and X-ray radiation from the sun ionises gas molecules, producing free electrons and ions. Ionisation depends on sunlight, so it varies with time of day, season, latitude and the 11-year sunspot cycle. It is divided into layers:

 height (km)
  400 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F2 (250-400)
  300 |        /\                    day & night
  220 |~~~~~~~/~~\~~~~~~~~~~~~~~~~~~ F1 (150-250)
      |      /    \                  day only
  110 |~~~~~/~~~~~~\~~~~~~~~~~~~~~~~ E  (90-140)
   70 |::::/::::::::\::::::::::::::: D  (50-90)
      |   /  sky     \               day only
    0 |__/____________\_____________ earth
       Tx    skip      Rx
LayerHeightElectron density (per m³)PresentMain effect
D50–90 km10⁸–10¹⁰Day onlyAbsorbs LF/MF sky wave; reflects VLF
E90–140 km (≈110)≈10¹¹Mostly day (weak at night)Reflects MF and lower HF; fc ≈ 3–4 MHz
Sporadic Eₛ≈110 kmPatchy, highIrregularOccasional VHF long-distance
F1150–250 km (≈200)≈2–5 × 10¹¹Day onlySome HF reflection, absorption; fc ≈ 4–5 MHz
F2250–400 km≈10¹² (highest)Day and nightMain HF reflector; fc ≈ 5–12 MHz; longest hops

At night D and F1 disappear and F1 merges with F2; E becomes weak. So nighttime sky-wave uses mainly F2 (and residual E).

Operation of ionospheric communication

  1. Refraction, not mirror reflection: the refractive index of an ionised layer is n = √(1 − 81N/f²) (N in electrons/m³, f in Hz). As the wave goes up, N increases, n falls below 1, and the ray bends away from the normal, gradually turning back down.
  2. Condition to return: the wave returns if at some height sin θi = n, i.e. f ≤ 9√Nmax sec θi. Higher frequencies need oblique incidence; the highest frequency that returns for a given path is the MUF = fc sec θi.
  3. Critical frequency fc = 9√Nmax: the highest frequency reflected at vertical incidence.
  4. Skip distance and skip zone: for f > fc, waves at steep angles escape; the nearest point where the sky wave returns is the skip distance. Between the end of the ground wave and this point no signal is received (skip zone).
  5. Multi-hop: the returned wave reflects from earth and goes up again, so signals travel round the world in several hops (one F2 hop ≈ up to 4000 km).
  6. Virtual height: the height from which a mirror reflection would give the same delay; used in calculations.
  7. Absorption is largest in the D layer and at low frequency (∝ 1/f²), which is why MF sky wave is weak in daytime.
  8. Frequency choice: the optimum working frequency (OWF ≈ 0.85 MUF) is used; frequency is changed between day (higher) and night (lower).

Merits: very long range with low power; cheap. Demerits: fading, skip zone, dependence on sun and time, ionospheric storms, limited bandwidth.

  • Asked 3 times
  • 2080 Asoj · 3 marks
  • 2076 Bhadra · 3 marks
  • 2073 Magh · 4 marks

Write a short note on MW (medium wave) propagation.

Answer

Medium wave (MW) propagation covers the MF band, 300 kHz – 3 MHz; AM broadcasting uses 535–1605 kHz (526.5–1606.5 kHz in Nepal/ITU Region 3).

Daytime

  • Propagation is mainly by ground (surface) wave. Vertical polarisation is used (vertical mast radiators) because horizontal polarisation is quickly short-circuited by the earth.
  • The D layer strongly absorbs MW sky waves, so there is almost no sky wave.
  • Service range is about 100–300 km, depending on transmitter power and ground conductivity (larger over sea water, smaller over dry or rocky land).

Nighttime

  • The D layer disappears, so the sky wave is reflected by the E layer with little loss and reaches 1000–2000 km.
  • This gives long-distance reception but also interference between distant stations on the same channel and fading where ground wave and sky wave of similar strength meet (near-fading zone).

Other points

  • Attenuation of ground wave rises with frequency, so the lower end of the band covers more area.
  • Antennas are tall λ/4 Marconi masts with good radial ground systems; anti-fading antennas (≈ 0.53λ high) reduce high-angle radiation.
  • Signal is stable in daytime, so MW is used for local and regional AM broadcast, maritime and aeronautical beacons.
  • Asked 3 times
  • 2078 Chaitra · 3 marks
  • 2075 Bhadra · 4 marks
  • 2072 Asoj · 3 marks

Write a short note on radio frequency spectrum.

Answer

The radio frequency (RF) spectrum is the part of the electromagnetic spectrum used for radio communication, roughly 3 kHz to 300 GHz. ITU divides it into bands, each a decade in frequency, and each band has its own propagation mode and uses.

BandFrequencyWavelengthMain propagationTypical use
VLF3–30 kHz100–10 kmGround wave, earth–ionosphere waveguideSubmarine, navigation
LF30–300 kHz10–1 kmGround waveLW broadcast, beacons
MF0.3–3 MHz1000–100 mGround wave (day), sky wave (night)AM broadcast
HF3–30 MHz100–10 mSky wave (ionosphere)SW broadcast, long-distance links
VHF30–300 MHz10–1 mSpace wave (LOS)FM, TV, aircraft radio
UHF0.3–3 GHz1 m–10 cmSpace wave, troposcatterTV, mobile, GPS, Wi-Fi
SHF3–30 GHz10–1 cmLOS, satelliteMicrowave links, radar, satellite
EHF30–300 GHz10–1 mmLOS (rain/gas absorption)Radar, 5G mmWave, radio astronomy

Key trends: as frequency increases, ground-wave range falls, the ionosphere stops returning the wave (above ≈ 30 MHz), antennas become smaller and more directive, available bandwidth increases, and absorption by rain and gases rises. The spectrum is a limited natural resource, so its use is regulated by the ITU internationally and by the Nepal Telecommunications Authority (NTA) in Nepal.

  • Asked 3 times
  • 2073 Magh · 5 marks
  • 2072 Asoj · 5 marks
  • 2070 Magh · 4 marks

Explain the phenomenon of Duct propagation.

Answer

Duct propagation is a form of space-wave propagation in which a microwave signal is trapped in a thin layer of the lower atmosphere (a duct) and guided along the earth's curvature, far beyond the normal line-of-sight range, like a wave in a waveguide.

Cause: abnormal refractive index profile

In a standard atmosphere the refractive index n falls slowly with height, so rays bend down slightly (effective earth radius = 4/3 R). Sometimes n falls much faster with height, due to:

  • Temperature inversion (warm air over cold air), e.g. over cool sea, or after sunset over land;
  • Sharp fall of humidity with height (moist air near sea surface, dry air above);
  • Subsidence of air in high-pressure areas.

The modified refractive index is M = (n − 1 + h/R) × 10⁶. Normally M increases with height. Where dM/dh < 0 (n falls faster than 0.157 N-units per metre, i.e. dN/dh < −157 per km), the ray curvature becomes greater than the earth's curvature, and a duct forms.

   height          upper boundary (n jumps)
     |  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     |   /\    /\    /\    /\    /\
     |  /  \  /  \  /  \  /  \  /  \     trapped ray
     | /    \/    \/    \/    \/    \
  ===Tx==========================Rx===  earth / sea
        surface duct (few m to 100s m)

Mechanism

  1. A ray launched at a low angle bends downward more strongly than the earth curves.
  2. It returns to the earth (or the lower boundary of an elevated duct), reflects, goes up again, and is bent down again.
  3. The wave keeps zig-zagging inside the layer and travels hundreds or even thousands of km with low loss.

Types

  • Surface duct: lower boundary is the earth or sea surface (common over warm seas).
  • Elevated duct: layer lies above the ground, between two heights.

Conditions

  • Only wavelengths small compared with the duct thickness are trapped (like waveguide cut-off). A duct of thickness d supports roughly λ_max ≈ 2.5 d √(ΔM × 10⁻⁶); so mainly VHF high end, UHF and microwaves are ducted.
  • Both antennas should be inside or near the duct, and the wave launched at a small angle.

Effects

  • Very long, unexpected ranges for microwave and VHF/UHF signals (TV and radar "anomalous propagation").
  • Interference between distant stations on the same frequency; false radar echoes.
  • Not reliable for planned links, because ducts depend on weather.
  • Asked 3 times
  • 2074 Bhadra
  • 2072 Magh · 2 marks
  • 2070 Magh · 2 marks

Define virtual height of an ionospheric layer.

Answer

The virtual height of an ionospheric layer is the height at which a wave sent up from the earth would appear to be reflected if it travelled straight up and down at the speed of light and were reflected sharply like a mirror, instead of being gradually bent inside the layer.

              A  <- virtual reflection point
             / \       (h' = virtual height)
            /   \
   ~~~~~~~~/~~~~~\~~~~~~ actual reflection:
          (       )     gradual bending
   ------/---------\---- bottom of layer
        /           \
   ----T-------------R---- earth

It is measured with an ionosonde from the echo delay t: h' = c·t/2. The wave actually travels slower inside the layer and bends gradually, so the virtual height is always greater than the actual height of reflection. It is used to calculate skip distance and MUF.

  • Asked 2 times
  • 2081 Chaitra · 2+6 marks
  • 2080 Chaitra · 2+7 marks

Define the critical frequency (fcr). Derive an expression for the relation between skip distance and maximum usable frequency (fmuf) in SW propagation assuming that the Earth is flat.

Answer

Critical frequency

The critical frequency (fcr) of an ionospheric layer is the highest frequency that is reflected back to earth when the wave is sent vertically upward. Higher frequencies at vertical incidence pass through the layer. It depends on the maximum electron density of the layer:

fcr = 9 √Nmax    (fcr in Hz, Nmax in electrons/m³)

Typical values: E ≈ 3–4 MHz, F2 ≈ 5–12 MHz.

Relation between skip distance and MUF (flat earth)

Skip distance (D) is the shortest distance from the transmitter at which a sky wave of a given frequency returns to earth. The maximum usable frequency (f_MUF) is the highest frequency that returns to earth for a given distance (angle of incidence).

Assumptions: earth and ionosphere are flat; reflection takes place at virtual height h; the path is symmetric (both antennas on ground).

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Result

f_MUF = fcr √(1 + D²/4h²)
D     = 2h √[(f_MUF/fcr)² − 1]
  • For f ≤ fcr, D = 0 (no skip zone).
  • As frequency rises above fcr, the skip distance grows.

Example: h = 300 km, fcr = 8 MHz, f = 16 MHz: D = 2 × 300 × √(4 − 1) = 1039 km.

  • Asked 2 times
  • 2080 Asoj · 7 marks
  • 2072 Magh · 8 marks

Find out the relation between the critical frequency (fcr) and skip distance (D) considering that the Earth is flat for both antennas.

Answer

Critical frequency (fcr) is the highest frequency reflected by a layer at vertical incidence, fcr = 9√Nmax. Skip distance (D) is the minimum ground distance at which a sky wave of frequency f (> fcr) returns to earth. The relation follows from the refraction condition and flat-earth geometry.

Assumptions: flat earth and flat ionosphere; reflection at virtual height h; transmitter and receiver both on the ground (heights neglected); symmetric path.

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Here f_MUF is the working frequency f for which D is the skip distance. So the required relation is:

D = 2h √[(f / fcr)² − 1]       or
fcr = f / √(1 + D² / 4h²) = 2hf / √(4h² + D²)

Observations

  • If f = fcr, D = 0: the wave returns even straight overhead, so there is no skip zone.
  • For f > fcr, D increases with f; a higher frequency needs a more oblique ray to return.
  • A higher layer (bigger h) gives a larger skip distance for the same f/fcr ratio.
  • At night fcr falls (lower Nmax), so the skip distance for the same frequency increases; operators lower the frequency at night.

Example: h = 250 km, fcr = 6 MHz, f = 12 MHz: D = 2 × 250 × √(2² − 1) = 500 × 1.732 = 866 km.

  • Asked 2 times
  • 2075 Baisakh · 8 marks
  • 2074 Bhadra · 7 marks

What is skip distance? Derive the expression for skip distance (D) and its relationship with critical frequency (fcr) assuming flat earth surface for both antennas.

Answer

Skip distance (D) is the shortest distance, measured along the ground from the transmitter, at which a sky wave of a given frequency (above the critical frequency) comes back to earth after reflection from the ionosphere. Rays sent at steeper angles pass through the layer; the ray at the limiting angle lands at the skip distance. Closer points (beyond the ground-wave range) get no signal: this is the skip zone.

          escapes
             \    |   /
  ~~~~~~~~~~~~\~~~|~~/~~~~~~~~~~~~~ layer
               \ /\  \_____
   limiting ray /  \       \___
  ------------T----------------R------
              |<-- skip D --->|

Derivation (flat earth, both antennas on the ground)

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Here f_MUF is the operating frequency f for which the given distance is the skip distance, and fc = fcr. Hence:

D = 2h √[(f / fcr)² − 1]

Relationship with critical frequency

  • fcr = 9√Nmax depends on maximum electron density of the layer.
  • For f ≤ fcr: D = 0 (all rays return; no skip zone).
  • For f > fcr: D grows with the ratio f/fcr.
  • Higher fcr (daytime, high sunspot activity) gives a shorter skip for the same frequency.

Example: h = 300 km, fcr = 5 MHz, f = 10 MHz: D = 600 × √3 = 1039 km.

  • Asked 2 times
  • 2078 Chaitra · 8 marks
  • 2076 Bhadra · 7 marks

Define and derive the formula for maximum usable frequency and skip distance assuming earth is perfectly flat.

Answer

Definitions

  • Maximum usable frequency (MUF): the highest frequency that is returned to earth by the ionosphere for a given angle of incidence (that is, for a given distance between two points). Frequencies above the MUF pass through the layer.
  • Skip distance (D): the minimum distance from the transmitter at which a sky wave of a given frequency returns to earth.
  • Critical frequency (fc): the MUF for vertical incidence, fc = 9√Nmax.

Derivation (earth assumed perfectly flat)

Assumptions: flat earth and flat ionospheric layer; sharp reflection at virtual height h; transmitter and receiver at ground level; symmetric path.

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Final formulas

f_MUF = fc sec θi = fc √(1 + D² / 4h²)
D     = 2h √[(f_MUF / fc)² − 1]

Remarks

  • MUF increases with distance (larger θi), so long links can use higher frequencies.
  • The flat-earth result is good for distances up to about 500 km; for longer paths the earth's curvature limits θi, and single-hop MUF reaches about 3–4 × fc.
  • In practice the optimum working frequency OWF ≈ 0.85 MUF is used, to allow for ionospheric changes.

Example: fc = 7 MHz, h = 300 km, D = 1000 km: f_MUF = 7 × √(1 + 1000²/(4 × 300²)) = 7 × √3.778 = 13.6 MHz.

  • Asked 2 times
  • 2076 Bhadra · 6 marks
  • 2075 Bhadra · 6 marks

Explain all the layers of ionosphere and their importance to radio wave (ionospheric) communication.

Answer

The ionosphere is the ionised part of the upper atmosphere, from about 50 km to 400 km above the earth. Solar UV and X-rays knock electrons off gas molecules; these free electrons refract radio waves. The ionosphere has several layers with different heights and electron densities.

 height (km)
  400 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F2 (250-400)
  300 |        /\                    day & night
  220 |~~~~~~~/~~\~~~~~~~~~~~~~~~~~~ F1 (150-250)
      |      /    \                  day only
  110 |~~~~~/~~~~~~\~~~~~~~~~~~~~~~~ E  (90-140)
   70 |::::/::::::::\::::::::::::::: D  (50-90)
      |   /  sky     \               day only
    0 |__/____________\_____________ earth
       Tx    skip      Rx
LayerHeightElectron density (per m³)PresentMain effect
D50–90 km10⁸–10¹⁰Day onlyAbsorbs LF/MF sky wave; reflects VLF
E90–140 km (≈110)≈10¹¹Mostly day (weak at night)Reflects MF and lower HF; fc ≈ 3–4 MHz
Sporadic Eₛ≈110 kmPatchy, highIrregularOccasional VHF long-distance
F1150–250 km (≈200)≈2–5 × 10¹¹Day onlySome HF reflection, absorption; fc ≈ 4–5 MHz
F2250–400 km≈10¹² (highest)Day and nightMain HF reflector; fc ≈ 5–12 MHz; longest hops

Layers and their importance

  • D layer (50–90 km): lowest and weakest; exists only in daytime and vanishes after sunset. Its high collision rate makes it an absorbing layer: it absorbs MF and lower HF sky waves (absorption ∝ 1/f²). This is why AM (MW) stations are heard only by ground wave in daytime. It reflects VLF/LF waves (earth–ionosphere waveguide, used for navigation).
  • E layer (90–140 km): mostly a daytime layer, weak at night. Reflects MF and lower HF waves; gives single-hop ranges up to about 2000 km. At night the residual E layer gives long-distance MW reception. Sporadic E patches can reflect even VHF signals irregularly, causing long-distance TV/FM interference.
  • F1 layer (150–250 km): present only in daytime; merges with F2 at night. Reflects some HF waves but mainly adds absorption to waves going to F2.
  • F2 layer (250–400 km): highest and most strongly ionised; present day and night. It is the most important layer for long-distance HF (SW) communication: highest critical frequency (up to ≈ 12 MHz), single hop up to ≈ 4000 km, and round-the-world links by multi-hop. Its height and density vary greatly with season and sunspot cycle.

Overall importance to radio communication

  1. Makes long-distance HF communication and SW broadcasting possible with modest power, without satellites.
  2. Decides critical frequency, MUF and skip distance for each path and time.
  3. Causes absorption (D layer) and fading, so frequency must be chosen by time of day (higher by day, lower at night).
  4. Waves above ≈ 30 MHz pass through, which allows satellite and space communication at VHF and above.
  • Asked 2 times
  • 2076 Baisakh · 4 marks
  • 2070 Magh · 2 marks

Define critical frequency for ionospheric region.

Answer

The critical frequency (fc) of an ionospheric layer is the highest frequency that is reflected back to earth by that layer when the wave is sent vertically upward (normal incidence). Any frequency above fc penetrates the layer and escapes at vertical incidence.

It depends on the maximum electron density Nmax of the layer:

fc = 9 √Nmax      (fc in Hz, Nmax in electrons per m³)

This comes from the refractive index n = √(1 − 81N/f²); at vertical incidence the wave turns back where n = 0, i.e. f² = 81Nmax.

  • Typical values: E layer ≈ 3–4 MHz, F2 layer ≈ 5–12 MHz.
  • It is higher by day and in high sunspot years.
  • For oblique incidence, the highest returnable frequency is MUF = fc sec θi.

Example: Nmax = 1.24 × 10¹² /m³ gives fc = 9 × √(1.24 × 10¹²) ≈ 10 MHz.

  • Asked 2 times
  • 2070 Bhadra · 4+6 marks
  • 2069 Bhadra · 4+8 marks

With a neat diagram, explain the designation of radio waves according to the path they follow during propagation (surface, ground reflected, direct and sky waves). Also, compare the propagation characteristics for different radio bands.

Answer

Designation of radio waves by path

A radio wave from a transmitting antenna can reach the receiver by different paths. Waves are named after the path they follow.

                  ionosphere
  ~~~~~~~~~~~~~~~~~~~/\~~~~~~~~~~~~~~~~~~~
                    /  \  (4) sky wave
                   /    \
  Tx |------(1) direct wave--------->| Rx
     |  \                         /  |
     |    \ (2) ground-reflected /   |
     |      \_________________/      |
  ===|==(3) surface wave along earth=|===
       |<------ space wave = 1 + 2 ->|
       ground wave = space + surface
  1. Direct wave: travels in a straight line (line of sight) from the transmitting to the receiving antenna through the troposphere.
  2. Ground-reflected wave: reaches the receiver after one reflection from the earth's surface. Direct + ground-reflected waves together form the space wave (tropospheric wave); dominant at VHF and above.
  3. Surface wave: a vertically polarised wave guided along the earth's surface, following its curvature; it induces currents in the ground and is attenuated. Dominant at VLF, LF and MF.
  4. Sky wave: goes upward and is bent back to earth by the ionosphere; used for HF long-distance links.

Surface wave and space wave together are called the ground wave. Other modes are tropospheric scatter and duct propagation.

Propagation characteristics of radio bands

BandRangeMain modeCharacteristics
VLF3–30 kHzSurface wave; earth–ionosphere waveguideVery stable, worldwide; huge antennas; tiny bandwidth; penetrates sea water
LF30–300 kHzSurface waveLong range (≈ 1000 km), stable; low attenuation
MF0.3–3 MHzSurface wave (day), E-layer sky wave (night)Day range ≈ 100–300 km; night range long but with fading and interference
HF3–30 MHzSky wave (F2, E)Worldwide with multi-hop; skip zone, fading; depends on time, season, sunspots; ground wave only tens of km
VHF30–300 MHzSpace wave (LOS)Range ≈ horizon (≈ 50–100 km); passes through ionosphere; occasional sporadic-E and ducting
UHF0.3–3 GHzSpace wave, troposcatterLOS, multipath, building shadowing; small high-gain antennas
SHF3–30 GHzLOS, satelliteHighly directive dishes; rain attenuation above ≈ 10 GHz
EHF30–300 GHzLOSStrong rain and gaseous (O₂, H₂O) absorption; short range; very wide bandwidth

General rules: ground-wave attenuation rises with frequency; the ionosphere reflects only below about 30 MHz; above that, range is limited by the horizon, while bandwidth and antenna directivity increase.

  • Asked 2 times
  • 2070 Magh · 10 marks
  • 2070 Bhadra · 5 marks

Explain the effect of space wave propagation on the ground of plane and actual earth.

Answer

Space wave propagation is the mode in which the signal reaches the receiver by the direct wave plus the ground-reflected wave through the troposphere. It is used at VHF, UHF and microwaves. The ground affects it in two ways: by reflection (interference of the two waves) and, for real earth, by curvature.

(a) Over a plane (flat) earth

   Tx                          Rx
   |\ ------ direct r1 -------> |
 ht|  \                      / | hr
   |    \  reflected r2    /   |
 ==|======\______P______/======|==
   |<------------ d ---------->|
  • Path lengths: r₁ = √(d² + (ht − hr)²), r₂ = √(d² + (ht + hr)²).
  • For d ≫ ht, hr: path difference Δr = r₂ − r₁ ≈ 2ht·hr/d.
  • Phase difference = (2π/λ)(2ht·hr/d) + π (reflection at grazing angle gives coefficient ≈ −1).
  • Resultant field (E₀/d = field of direct wave at distance d):
E = (2E₀/d) · sin(2π ht hr / λd)
For large d (small angle):  sin x ≈ x
E ≈ 4π E₀ ht hr / (λ d²)
In practical form: E = 88 √P · ht · hr / (λ d²) V/m
  (P in W, heights, λ and d in metres)

Effects:

  • Field strength varies as 1/d² (not 1/d) at large distance, so it falls fast.
  • Near the transmitter, maxima and minima (lobes) occur as d changes, due to interference.
  • Raising either antenna increases the field directly (E ∝ ht·hr), so high masts are used.
  • Horizontal and vertical polarisation behave similarly at grazing incidence over most ground.

(b) Over the actual (curved) earth

  1. Radio horizon: curvature limits the line-of-sight range. With standard refraction (effective radius 4R/3):
d = √(2kR ht) + √(2kR hr) ≈ 4.12 (√ht + √hr) km   (h in m)
  1. Effective antenna heights: the curved ground rises between the antennas, so heights measured from the tangent plane at the reflection point are reduced:
h't = ht − d1²/(2kR),   h'r = hr − d2²/(2kR)

These reduced heights are used in the field formula, giving a weaker field than for plane earth. 3. Divergence factor: reflection from a convex surface spreads the reflected rays, so the reflected wave is weaker; the reflection coefficient is multiplied by a divergence factor D < 1. 4. Shadow zone: beyond the horizon, only weak diffracted fields exist; the field falls rapidly. 5. Tropospheric refraction: decrease of refractive index with height bends rays down and extends the horizon by about 15% (k = 4/3); abnormal conditions give super-refraction, ducting or sub-refraction. 6. Terrain, obstacles and roughness: hills and buildings block or diffract the wave; rough ground gives diffuse rather than mirror reflection (Rayleigh criterion); Fresnel-zone clearance (≥ 0.6 of first zone) is needed for a clean link.

Conclusion of comparison: flat-earth analysis fits short links; for longer links, curvature reduces effective heights, limits range to the radio horizon, and weakens the reflected wave.

  • 2081 Chaitra · 5+4 marks

Explain the surface wave propagation. Write its major advantages and disadvantages.

Answer

Surface wave propagation is the mode in which a vertically polarised wave travels along the surface of the earth, following its curvature. The earth acts as a lossy guiding boundary: the wave induces charges and currents in the ground, which keep it attached to the surface. It is the main mode for VLF, LF and MF (up to about 2–3 MHz).

     wave front tilts forward (wave tilt)
       |  /  /  /  /
       | /  /  /  /    → direction of travel
  Tx  /|/  /  /  /
 ====|==================================
  ~~~~ induced earth currents (loss) ~~~~

Explanation

  • Vertical polarisation is necessary; a horizontally polarised wave is short-circuited by the conducting ground and dies out quickly.
  • Wave tilt: energy is lost into the ground, so the wave front tilts forward. The E-field gains a horizontal component that drives power down into the earth. Tilt is larger for poor ground.
  • Attenuation: field strength is approximately
E = A · E₀ / d

where E₀/d is the unattenuated (inverse-distance) field and A (< 1) is the ground attenuation factor, found from Sommerfeld/Norton curves using the "numerical distance". A falls faster with:

  • higher frequency (loss increases rapidly with f),
  • lower ground conductivity σ and permittivity εr,
  • longer distance.
  • Diffraction: the wave bends around the earth's curvature by diffraction, more easily at long wavelengths, which is why low frequencies travel far.
  • Ground type: sea water (σ ≈ 4 S/m) gives the longest range; moist soil is medium; dry and rocky land is poor.
  • Range: VLF/LF up to 1000+ km; MF ≈ 100–300 km; HF only tens of km.

Advantages

  1. Stable, reliable signal with no fading (no ionospheric dependence).
  2. Works day and night, in all seasons.
  3. Not limited by line of sight; follows earth's curvature.
  4. VLF waves penetrate sea water, useful for submarine communication.
  5. Simple receiving arrangements; good for local and regional broadcast (AM).

Disadvantages

  1. Usable only at low frequencies (below ≈ 3 MHz), so bandwidth is very small.
  2. Requires very tall vertical antennas (λ/4 is 75 m at 1 MHz) and extensive ground systems.
  3. High transmitter power needed because of ground losses.
  4. Range falls sharply with frequency and over poor soil.
  5. Atmospheric noise is high at LF/MF.
  • 2080 Chaitra · 2+4+4 marks

What is a radio frequency spectrum? Explain characteristics of MW and SW radio propagation.

Answer

Radio frequency spectrum

The radio frequency spectrum is the range of electromagnetic frequencies used for radio communication, from about 3 kHz to 300 GHz. The ITU divides it into decade bands, each with its own propagation behaviour:

BandFrequencyMain propagationUse
VLF3–30 kHzGround wave, earth–ionosphere guideNavigation, submarine
LF30–300 kHzGround waveLW broadcast, beacons
MF0.3–3 MHzGround wave (day), sky wave (night)AM broadcast
HF3–30 MHzSky waveSW broadcast, long-distance
VHF30–300 MHzSpace wave (LOS)FM, TV, aircraft
UHF0.3–3 GHzSpace wave, troposcatterTV, mobile, GPS
SHF3–30 GHzLOS, satelliteMicrowave links, radar
EHF30–300 GHzLOS (high absorption)mmWave, radar

Characteristics of MW and SW propagation

MW (medium wave) propagation, 300 kHz – 3 MHz (AM broadcast 526.5–1606.5 kHz):

  • Daytime: mainly ground (surface) wave; the D layer absorbs the sky wave. Range ≈ 100–300 km depending on power and soil (longest over sea).
  • Nighttime: D layer disappears, the E layer reflects the sky wave, giving 1000–2000 km range, but also co-channel interference and fading where ground and sky waves mix.
  • Vertical polarisation and tall λ/4 mast antennas with radial ground systems are used.
  • Ground-wave attenuation increases with frequency, so the lower end of the band covers more area.
  • High atmospheric noise; narrow bandwidth (10 kHz channels).

SW (short wave) propagation, 3 – 30 MHz (HF):

  • Mainly sky wave via the F2 (and E) layer; ground wave dies within a few tens of km.
  • Very long range with modest power: one F2 hop up to ≈ 4000 km, worldwide by multi-hop.
  • Governed by critical frequency, MUF and skip distance; a skip zone exists between the end of ground wave and the first sky-wave return.
  • Fading due to multipath and changes in the ionosphere; diversity reception is used.
  • Frequency must change with time of day, season and sunspot cycle: higher frequencies by day, lower at night; OWF ≈ 0.85 MUF.
  • Affected by ionospheric storms and sudden ionospheric disturbances (radio blackouts).
  • Horizontal antennas such as dipoles, rhombic and log-periodic arrays are common.
  • 2080 Asoj · 2+6 marks

What is the sky wave? Explain the ionospheric wave propagation showing its different layers.

Answer

Sky wave

A sky wave is a radio wave that travels upward from the transmitting antenna and is bent (refracted) back to earth by the ionosphere, reaching receivers far beyond the horizon. It is the main mode for HF (3–30 MHz) long-distance communication.

Ionospheric wave propagation and layers

The ionosphere (≈ 50–400 km) is ionised by solar UV and X-rays. Its free electrons lower the refractive index n = √(1 − 81N/f²), so an upgoing wave bends away from the normal and gradually turns back to earth.

 height (km)
  400 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F2 (250-400)
  300 |        /\                    day & night
  220 |~~~~~~~/~~\~~~~~~~~~~~~~~~~~~ F1 (150-250)
      |      /    \                  day only
  110 |~~~~~/~~~~~~\~~~~~~~~~~~~~~~~ E  (90-140)
   70 |::::/::::::::\::::::::::::::: D  (50-90)
      |   /  sky     \               day only
    0 |__/____________\_____________ earth
       Tx    skip      Rx
LayerHeightElectron density (per m³)PresentMain effect
D50–90 km10⁸–10¹⁰Day onlyAbsorbs LF/MF sky wave; reflects VLF
E90–140 km (≈110)≈10¹¹Mostly day (weak at night)Reflects MF and lower HF; fc ≈ 3–4 MHz
Sporadic Eₛ≈110 kmPatchy, highIrregularOccasional VHF long-distance
F1150–250 km (≈200)≈2–5 × 10¹¹Day onlySome HF reflection, absorption; fc ≈ 4–5 MHz
F2250–400 km≈10¹² (highest)Day and nightMain HF reflector; fc ≈ 5–12 MHz; longest hops

How a sky-wave link works

  1. A wave of frequency f enters the layer at angle θi. It returns if f ≤ fc sec θi, where fc = 9√Nmax is the critical frequency. The limiting value is the MUF.
  2. Steep rays at frequencies above fc pass through; the first ray that returns lands at the skip distance. Between the ground-wave limit and this point lies the skip zone.
  3. The returned wave reflects from earth and may go up again (multi-hop), so worldwide coverage is possible.
  4. The D layer absorbs lower frequencies by day; at night D and F1 vanish, so lower frequencies are used and F2 carries the traffic.
  5. Signals fade due to multipath and changing ionisation; OWF ≈ 0.85 MUF is chosen for reliability.
  • 2079 Chaitra · 2+6 marks

Describe multiple HOP and explain how atmospheric ducts can be used for microwave propagation.

Answer

Multiple hop

Multiple-hop (multi-hop) propagation is sky-wave transmission in which the wave is reflected by the ionosphere, comes down, is reflected by the earth's surface, goes up again, and so on, several times before reaching a distant receiver.

  ~~~~~~/\~~~~~~~~/\~~~~~~~~/\~~~~~~ F2 layer
       /  \      /  \      /  \
      /    \    /    \    /    \
  ===Tx=====\==/======\==/======Rx===
           earth      earth
   hop 1      hop 2      hop 3
  • Single-hop range is limited by earth curvature: about 2000 km via E layer and about 4000 km via F2.
  • Longer distances (round the world) need multiple hops.
  • Each hop adds loss (ground reflection loss, D-layer absorption twice per hop), so the signal weakens and fades more; sea surface is a better reflector than land.
  • Multiple paths with different hop counts can arrive together, causing multipath fading and echo.

Use of atmospheric ducts for microwave propagation

An atmospheric duct is a layer of the lower atmosphere in which the refractive index decreases so rapidly with height (dM/dh < 0, where M = (n − 1 + h/R) × 10⁶) that rays bend down more strongly than the earth curves. The duct traps the wave between its boundaries (or between the upper boundary and the ground), guiding it like a waveguide.

   ~~~~~~~~~~~ top of duct ~~~~~~~~~~~~
    /\     /\     /\     /\     /\
   /  \   /  \   /  \   /  \   /  \
  Tx===\=/====\=/====\=/====\=/====Rx
       ground / sea surface

Causes: temperature inversion (warm, dry air above cool, moist air), especially over sea and on calm evenings; sharp fall of humidity with height.

How it helps microwaves:

  1. Microwave antennas placed inside (or near) the duct and aimed at low angles launch rays that are repeatedly bent back, so they follow the earth's curvature.
  2. The duct acts like a waveguide; only wavelengths small compared with duct thickness are trapped (λ_max ≈ 2.5 d √(ΔM × 10⁻⁶)), which suits VHF high end, UHF and microwaves (thickness from a few m to some hundred m).
  3. Ranges of several hundred to over 1000 km beyond the normal radio horizon become possible with low loss.

Limitations: ducts depend on weather and are not permanent, so they cannot be relied on for regular links. They often cause unwanted long-range interference and radar false echoes ("anomalous propagation"). Planned microwave links use them only opportunistically, mostly over sea.

  • 2079 Chaitra · 7 marks

Derive the expression between skip distance and maximum useable frequency (MUF) assuming earth's curvature.

Answer

Skip distance (D) is the minimum ground distance at which a sky wave of a given frequency returns to earth; MUF is the highest frequency returned for a given distance. Over long paths the earth's curvature must be included, because it limits the angle of incidence at the ionosphere.

Assumptions: spherical earth of radius R; thin reflecting layer at virtual height h; transmitter T and receiver at ground level; reflection point P above the mid-point; θ = half the angle subtended at the earth's centre.

               P  (reflection point, height h)
              /|\
             / | \      φ = angle of incidence at P
            /φ |  \
   ~~~~~~~~/~~~|~~~\~~~~~~~ ionosphere
       T  /    |    \  R    (on earth surface)
        \.     |     ./
          `.   |θ  .'       θ = D / (2R)
            `. | .'
               O  (earth centre)
   OT = R,  OP = R + h,  arc TR = D

Step 1: Drop a perpendicular from T onto OP. Its foot Q lies at distance R cos θ from O, and TQ = R sin θ. In triangle TQP:

QP    = (R + h) − R cos θ
tan φ = TQ / QP = R sin θ / (R + h − R cos θ)

Step 2: Secant law for the ionosphere (from n = √(1 − 81N/f²) and Snell's law): f_MUF = fc sec φ, with fc = 9√Nmax. So:

f_MUF = fc √(1 + tan² φ)

f_MUF = fc √{ 1 + [ R sin(D/2R) /
              (R + h − R cos(D/2R)) ]² }

Step 3: For D small compared with R, sin θ ≈ θ = D/2R and cos θ ≈ 1 − θ²/2:

tan φ ≈ (D/2) / (h + D²/8R)
f_MUF ≈ fc √{ 1 + [ (D/2) / (h + D²/8R) ]² }

Step 4: Skip distance for a given frequency f. Let k = √[(f/fc)² − 1] = tan φ:

D/2 = k (h + D²/8R)
(k/8R) D² − D/2 + k h = 0
D = (2R/k) [ 1 − √(1 − 2k²h/R) ]   (smaller root)

Check: as R → ∞, √(1 − x) ≈ 1 − x/2, so D → 2kh = 2h√[(f/fc)² − 1], the flat-earth result.

Step 5: Limit due to curvature. The largest angle φ occurs when the ray leaves the transmitter along the horizon (tangent at T):

sin φmax = R / (R + h)
f_MUF(max) = fc sec φmax
D_max (one hop) = 2R cos⁻¹[R / (R + h)]

For h = 300 km, R = 6370 km: sin φmax = 0.955, φmax = 72.8°, sec φmax = 3.37, D_max ≈ 3836 km. So single-hop MUF cannot exceed about 3.4 fc.

Example: fc = 8 MHz, h = 300 km, D = 2000 km (R = 6370 km): θ = 8.99°, tan φ = 2.632, f_MUF = 8 × √(1 + 2.632²) = 22.5 MHz.

  • 2078 Chaitra · 4+4 marks

Describe briefly the structure of the ionosphere and its uses for radio frequency spectrum. Write the characteristics of ground wave propagation.

Answer

Structure of the ionosphere and its use for the RF spectrum

The ionosphere is the part of the upper atmosphere, about 50–400 km high, ionised by the sun's UV and X-rays. It contains several layers:

LayerHeightElectron density (per m³)PresentMain effect
D50–90 km10⁸–10¹⁰Day onlyAbsorbs LF/MF sky wave; reflects VLF
E90–140 km (≈110)≈10¹¹Mostly day (weak at night)Reflects MF and lower HF; fc ≈ 3–4 MHz
Sporadic Eₛ≈110 kmPatchy, highIrregularOccasional VHF long-distance
F1150–250 km (≈200)≈2–5 × 10¹¹Day onlySome HF reflection, absorption; fc ≈ 4–5 MHz
F2250–400 km≈10¹² (highest)Day and nightMain HF reflector; fc ≈ 5–12 MHz; longest hops

At night D and F1 disappear and E weakens, leaving mainly F2.

Uses for the radio spectrum:

  • VLF/LF: reflected by the D layer's lower edge; the earth and ionosphere form a waveguide giving stable worldwide navigation and time signals.
  • MF: absorbed by D in daytime; reflected by E at night, so AM broadcasts travel far at night.
  • HF (3–30 MHz): refracted by E and F2 layers; makes long-distance and worldwide sky-wave communication (SW broadcast, maritime, military, amateur).
  • VHF and above: pass through the ionosphere (except occasional sporadic E), so they are used for satellite and space links, and local LOS services.

Characteristics of ground wave propagation

  1. Wave travels along the earth's surface, following its curvature by diffraction.
  2. Must be vertically polarised; horizontal polarisation is shorted by the ground.
  3. Attenuation increases rapidly with frequency, so it is useful only up to about 2–3 MHz (VLF, LF, MF).
  4. Range depends on ground conductivity and permittivity: longest over sea water, shortest over dry, rocky land.
  5. Wave tilt: the wave front leans forward, sending energy into the ground.
  6. Stable signal with no fading; independent of time of day and season.
  7. Needs tall antennas, good ground systems and high power.
  8. Ranges: VLF/LF up to 1000+ km, MF about 100–300 km.
  • 2075 Bhadra · 2+8 marks

What is Maximum Usable Frequency (MUF)? Derive the expression of MUF, critical frequency (fcr) and skip distance assuming curved earth.

Answer

Maximum Usable Frequency (MUF)

The MUF is the highest frequency that is returned to earth by the ionosphere between two given points (for a given angle of incidence). Higher frequencies pass through the layer. It is related to the critical frequency fcr (highest frequency reflected at vertical incidence) by the secant law: f_MUF = fcr sec φ. In practice the optimum working frequency OWF ≈ 0.85 MUF is used.

Derivation for curved earth

Assumptions: spherical earth of radius R; layer acting as a thin reflector at virtual height h; both antennas on the ground; reflection point P above the mid-point of the path.

First, the secant law. The refractive index of the layer is n = √(1 − 81N/f²). By Snell's law, a ray entering at angle φ returns where n = sin φ:

1 − 81Nmax/f² = sin² φ
81Nmax/f²     = cos² φ
f = 9√Nmax / cos φ = fcr sec φ,   fcr = 9√Nmax

Then the curved-earth geometry:

               P  (reflection point, height h)
              /|\
             / | \      φ = angle of incidence at P
            /φ |  \
   ~~~~~~~~/~~~|~~~\~~~~~~~ ionosphere
       T  /    |    \  R    (on earth surface)
        \.     |     ./
          `.   |θ  .'       θ = D / (2R)
            `. | .'
               O  (earth centre)
   OT = R,  OP = R + h,  arc TR = D

Step 1: Drop a perpendicular from T onto OP. Its foot Q lies at distance R cos θ from O, and TQ = R sin θ. In triangle TQP:

QP    = (R + h) − R cos θ
tan φ = TQ / QP = R sin θ / (R + h − R cos θ)

Step 2: Secant law for the ionosphere (from n = √(1 − 81N/f²) and Snell's law): f_MUF = fc sec φ, with fc = 9√Nmax. So:

f_MUF = fc √(1 + tan² φ)

f_MUF = fc √{ 1 + [ R sin(D/2R) /
              (R + h − R cos(D/2R)) ]² }

Step 3: For D small compared with R, sin θ ≈ θ = D/2R and cos θ ≈ 1 − θ²/2:

tan φ ≈ (D/2) / (h + D²/8R)
f_MUF ≈ fc √{ 1 + [ (D/2) / (h + D²/8R) ]² }

Step 4: Skip distance for a given frequency f. Let k = √[(f/fc)² − 1] = tan φ:

D/2 = k (h + D²/8R)
(k/8R) D² − D/2 + k h = 0
D = (2R/k) [ 1 − √(1 − 2k²h/R) ]   (smaller root)

Check: as R → ∞, √(1 − x) ≈ 1 − x/2, so D → 2kh = 2h√[(f/fc)² − 1], the flat-earth result.

Step 5: Limit due to curvature. The largest angle φ occurs when the ray leaves the transmitter along the horizon (tangent at T):

sin φmax = R / (R + h)
f_MUF(max) = fc sec φmax
D_max (one hop) = 2R cos⁻¹[R / (R + h)]

For h = 300 km, R = 6370 km: sin φmax = 0.955, φmax = 72.8°, sec φmax = 3.37, D_max ≈ 3836 km. So single-hop MUF cannot exceed about 3.4 fc.

Summary

fcr   = 9 √Nmax
f_MUF = fcr sec φ,
tan φ = R sin(D/2R) / (R + h − R cos(D/2R))
D     = (2R/k)[1 − √(1 − 2k²h/R)],  k = √[(f/fcr)² − 1]

Example: fcr = 8 MHz, h = 300 km, f = 16 MHz: k = √3 = 1.732; D = (2 × 6370/1.732)[1 − √(1 − 2 × 3 × 300/6370)] ≈ 1125 km (the flat-earth formula gives 1039 km).

  • 2075 Baisakh · 1+5 marks

What is ionosphere? Explain the ionosphere wave propagation showing its different layers.

Answer

Ionosphere

The ionosphere is the upper region of the atmosphere, about 50 km to 400 km above the earth, where solar ultraviolet and X-ray radiation ionises the gas, producing free electrons and ions that can bend radio waves back to earth.

Ionospheric wave propagation

In ionospheric (sky-wave) propagation, a wave (mainly HF, 3–30 MHz) is sent upward, refracted back by an ionised layer and received hundreds to thousands of km away.

 height (km)
  400 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F2 (250-400)
  300 |        /\                    day & night
  220 |~~~~~~~/~~\~~~~~~~~~~~~~~~~~~ F1 (150-250)
      |      /    \                  day only
  110 |~~~~~/~~~~~~\~~~~~~~~~~~~~~~~ E  (90-140)
   70 |::::/::::::::\::::::::::::::: D  (50-90)
      |   /  sky     \               day only
    0 |__/____________\_____________ earth
       Tx    skip      Rx

Layers

  • D (50–90 km): daytime only; absorbs MF/low HF; reflects VLF/LF.
  • E (90–140 km): reflects MF and lower HF; weak at night; sporadic-E patches.
  • F1 (150–250 km): daytime only; merges with F2 at night.
  • F2 (250–400 km): strongest, day and night; main reflector for long-distance HF.

Mechanism

  1. Refractive index n = √(1 − 81N/f²) falls as electron density N rises with height, so the ray bends away from the normal and turns back.
  2. Critical frequency fc = 9√Nmax is the highest frequency returned at vertical incidence; for oblique incidence the limit is MUF = fc sec θi.
  3. Waves above the MUF escape into space; at steep angles they escape, creating a skip zone up to the skip distance.
  4. Repeated reflections between earth and ionosphere (multi-hop) give worldwide range.
  5. The ionosphere changes with time of day, season and sunspots, so frequencies are changed accordingly and signals may fade.
  • 2074 Bhadra · 6 marks

The antenna of a TV transmitter is located at a height of 125 m above ground level. Calculate the distance up to which the LOS communication is possible if the height of receiving antenna is to be 9 m.

Answer

For space-wave (line-of-sight) communication, the maximum distance is the sum of the radio horizons of the two antennas. With standard atmospheric refraction, the effective earth radius is kR with k = 4/3.

Given: ht = 125 m, hr = 9 m, R = 6370 km, k = 4/3.

Formula:

d = √(2kR·ht) + √(2kR·hr)
With kR = 8493 km and h in metres:
d (km) = 4.12 (√ht + √hr)

Calculation:

√ht = √125 = 11.180
√hr = √9   = 3.000
d   = 4.12 × (11.180 + 3.000)
    = 4.12 × 14.180
    = 58.42 km

Separately: transmitter horizon = 4.12 × 11.180 = 46.06 km, receiver horizon = 4.12 × 3 = 12.36 km.

(If refraction is ignored, d = 3.57(√ht + √hr) = 50.62 km.)

Answer: LOS communication is possible up to about 58.4 km (taking standard 4/3 earth refraction).

  • 2074 Bhadra · 2+6 marks

What is a radio frequency spectrum? Give major propagation characteristics of VHF and UHF bands.

Answer

Radio frequency spectrum

The radio frequency spectrum is the range of electromagnetic frequencies used for radio communication, about 3 kHz to 300 GHz, divided by the ITU into bands:

BandFrequencyMain propagationUse
VLF3–30 kHzGround wave, earth–ionosphere guideNavigation, submarine
LF30–300 kHzGround waveLW broadcast, beacons
MF0.3–3 MHzGround wave (day), sky wave (night)AM broadcast
HF3–30 MHzSky waveSW broadcast, long-distance
VHF30–300 MHzSpace wave (LOS)FM, TV, aircraft
UHF0.3–3 GHzSpace wave, troposcatterTV, mobile, GPS
SHF3–30 GHzLOS, satelliteMicrowave links, radar
EHF30–300 GHzLOS (high absorption)mmWave, radar

Propagation characteristics of VHF (30–300 MHz) and UHF (300 MHz – 3 GHz)

  1. Space-wave (line-of-sight) propagation: direct wave plus ground-reflected wave; range limited to the radio horizon, d ≈ 4.12(√ht + √hr) km. High antenna masts or towers are used.
  2. Ionosphere does not return them: they pass through, so no sky-wave skip (except rare sporadic-E or F2 openings at low VHF). This makes them suitable for satellite links.
  3. Ground wave is negligible: surface-wave attenuation is very high at these frequencies.
  4. Tropospheric refraction bends the waves slightly, extending the horizon by about 15% (4/3 earth radius); abnormal conditions cause super-refraction and duct propagation with long-range interference.
  5. Tropospheric scatter at UHF gives beyond-horizon links of 100–800 km with high power and large antennas.
  6. Multipath and fading: reflections from ground, buildings and hills cause interference patterns (field varies as 1/d² far away), ghosting in TV and fast fading in mobile systems.
  7. Diffraction and shadowing: waves bend slightly over hills and buildings, less as frequency rises; UHF suffers more shadowing and building penetration loss than VHF.
  8. Small, high-gain antennas: short wavelength allows Yagi and panel antennas, giving directivity and frequency reuse.
  9. Low noise and wide bandwidth: atmospheric and man-made noise is low; enough bandwidth for FM, TV and data.

Uses: VHF – FM broadcast, TV, aircraft and marine radio; UHF – TV, mobile (GSM/4G), GPS, Wi-Fi, Bluetooth.

  • 2074 Bhadra

Define MUF (maximum usable frequency).

Answer

The maximum usable frequency (MUF) is the highest frequency that is reflected (returned) to earth by an ionospheric layer for a given angle of incidence, that is, for a given distance between the transmitter and receiver. Above the MUF the wave passes through the layer and is lost.

f_MUF = fc sec θi = fc √(1 + D²/4h²)   (flat earth)

where fc = 9√Nmax is the critical frequency, θi the angle of incidence, D the distance and h the virtual height. MUF is always greater than fc; for a single hop it can reach about 3–4 times fc. In practice the optimum working frequency OWF ≈ 0.85 MUF is used.

  • 2074 Bhadra · 4 marks

Write a short note on super refraction.

Answer

Super refraction is the condition in which radio waves in the troposphere bend downward more than normal, because the refractive index of the air decreases with height faster than in the standard atmosphere. The radio horizon then extends beyond its normal value.

Refractivity gradients (N = (n − 1) × 10⁶):

ConditiondN/dh (per km)Effect
Sub-refraction> 0Rays bend up; range reduced
Standard≈ −40k = 4/3; normal horizon
Super refraction−79 to −157Rays bend down more; range extended
Ducting (trapping)< −157Ray curvature exceeds earth's; wave trapped
            standard ray
   Tx ____------------------
     \ `--.___              super-refracted
      \       `--.___       ray (bends more)
  =====\=============`--.___Rx==== earth

Causes: temperature inversion (warm air above cool air), rapid fall of humidity with height, cool sea surface under warm dry air, clear calm nights.

Effects:

  • Effective earth radius factor k becomes greater than 4/3 (even infinite at the limit), so VHF/UHF/microwave signals travel beyond the normal horizon.
  • Unexpected long-distance reception; co-channel interference between distant TV/FM stations and microwave links.
  • Radar "anomalous propagation": ground or sea clutter appears at long range.
  • In extreme form it becomes duct propagation.
  • 2073 Magh · 5 marks

Find the maximum range of tropospheric transmission for which the transmitting antenna height is 100 ft and receiving antenna is 50 ft.

Answer

The maximum range of tropospheric (space-wave) transmission is the sum of the radio horizons of the two antennas. With standard atmospheric refraction (effective earth radius 4/3 R), the horizon in miles for a height in feet is:

d (miles) = √(2 ht) + √(2 hr)     (h in ft)

Given: ht = 100 ft, hr = 50 ft.

√(2 × 100) = √200 = 14.142 miles
√(2 × 50)  = √100 = 10.000 miles
d = 14.142 + 10.000 = 24.142 miles
d = 24.142 × 1.609 = 38.85 km

Check in metric units: ht = 30.48 m, hr = 15.24 m, d = 4.12(√30.48 + √15.24) = 4.12(5.521 + 3.904) = 38.83 km (same, within rounding).

Answer: maximum range ≈ 24.1 miles ≈ 38.9 km.

  • 2073 Bhadra · 8 marks

Briefly discuss the propagation characteristics of space wave and sky wave.

Answer

Space wave

A space wave (tropospheric wave) consists of the direct wave from the transmitting to the receiving antenna and the ground-reflected wave. It is the main mode at VHF, UHF and microwaves.

  Tx |\---------- direct ------------>| Rx
  ht |  \                          /  | hr
     |    \____ ground reflected _/   |
  ===|================================|===

Characteristics

  1. Line-of-sight: range limited by the radio horizon, d ≈ 4.12(√ht + √hr) km (h in m), typically 40–100 km; higher antennas give longer range.
  2. Field strength at large distance: E = 88√P·ht·hr/(λd²), so it falls as 1/d² and grows with antenna heights.
  3. Interference between direct and reflected waves gives lobes, multipath and fading.
  4. Affected by tropospheric refraction (4/3 earth, super-refraction, ducts), terrain, buildings and earth curvature (shadow zone).
  5. Above ≈ 10 GHz, rain and atmospheric gases absorb the signal.
  6. Not affected by the ionosphere; same performance day and night.
  7. Wide bandwidth, low noise; small high-gain antennas.
  8. Uses: FM, TV, mobile, microwave links, radar, satellite.

Sky wave

A sky wave goes upward and is refracted back to earth by the ionosphere; it is the main mode for HF (3–30 MHz).

  ~~~~~~~~~~~~~~/\~~~~~~~~~~~~~ ionosphere
               /  \
              /    \
  ========Tx=/======\=Rx======== earth
           |<-skip->|

Characteristics

  1. Long range: one hop up to ≈ 4000 km (F2), worldwide by multi-hop, with low power.
  2. Works only below the MUF (= fc sec θi); frequencies above it pass through. Critical frequency fc = 9√Nmax.
  3. Skip distance and skip zone: no reception between the end of ground wave and the first return.
  4. Fading due to multipath and changing ionisation; diversity reception needed.
  5. Depends strongly on time of day, season, latitude and sunspot cycle; frequency must be changed (OWF ≈ 0.85 MUF).
  6. D-layer absorption in daytime affects low frequencies.
  7. Disturbed by ionospheric storms and sudden ionospheric disturbances.
  8. Uses: SW broadcasting, maritime, military and amateur long-distance links.
PointSpace waveSky wave
MediumTroposphereIonosphere
BandVHF and aboveHF (3–30 MHz)
RangeLOS (tens of km)Thousands of km
StabilityStableVaries with sun, time
BandwidthLargeSmall
  • 2073 Bhadra · 2+6 marks

Write down the factors which affect the surface wave communication. Explain the major characteristics of MW and SW radio propagation.

Answer

Factors affecting surface wave communication

  1. Frequency: attenuation increases rapidly with frequency; surface wave is useful only up to about 2–3 MHz.
  2. Ground conductivity (σ): higher σ means lower loss; sea water (≈ 4 S/m) is best, dry/rocky soil is worst.
  3. Ground permittivity (εr): affects wave tilt and attenuation, especially at higher frequencies.
  4. Polarisation: only vertical polarisation propagates; horizontal is short-circuited by the ground.
  5. Distance and earth curvature: field falls as A/d (A = attenuation factor) and faster beyond the diffraction region.
  6. Terrain and obstacles: hills, forests, buildings and mixed land–sea paths add loss.
  7. Transmitter power and antenna (height, efficiency, ground system).
  8. Atmospheric noise (high at LF/MF) reduces usable range.

Characteristics of MW and SW radio propagation

MW (medium wave) propagation, 300 kHz – 3 MHz (AM broadcast 526.5–1606.5 kHz):

  • Daytime: mainly ground (surface) wave; the D layer absorbs the sky wave. Range ≈ 100–300 km depending on power and soil (longest over sea).
  • Nighttime: D layer disappears, the E layer reflects the sky wave, giving 1000–2000 km range, but also co-channel interference and fading where ground and sky waves mix.
  • Vertical polarisation and tall λ/4 mast antennas with radial ground systems are used.
  • Ground-wave attenuation increases with frequency, so the lower end of the band covers more area.
  • High atmospheric noise; narrow bandwidth (10 kHz channels).

SW (short wave) propagation, 3 – 30 MHz (HF):

  • Mainly sky wave via the F2 (and E) layer; ground wave dies within a few tens of km.
  • Very long range with modest power: one F2 hop up to ≈ 4000 km, worldwide by multi-hop.
  • Governed by critical frequency, MUF and skip distance; a skip zone exists between the end of ground wave and the first sky-wave return.
  • Fading due to multipath and changes in the ionosphere; diversity reception is used.
  • Frequency must change with time of day, season and sunspot cycle: higher frequencies by day, lower at night; OWF ≈ 0.85 MUF.
  • Affected by ionospheric storms and sudden ionospheric disturbances (radio blackouts).
  • Horizontal antennas such as dipoles, rhombic and log-periodic arrays are common.
  • 2073 Bhadra · 3 marks

Write a short note on tropospheric scatter propagation.

Answer

Tropospheric scatter (troposcatter) is beyond-the-horizon propagation in which UHF and microwave signals (about 300 MHz to 10 GHz) are scattered by small irregularities (turbulence and refractive index blobs) in the troposphere, so that a small part of the energy reaches a receiver beyond the radio horizon.

          common (scatter) volume
                 .:::::.
               /  :::::  \
   beam      /             \     beam
   Tx dish /                 \  Rx dish
   =======(==== earth bulge ====)=====
          |<-- 100 to 800 km -->|

Working: two high-gain antennas aim their beams just above the horizon so that the beams intersect in a common volume in the troposphere (height a few km). Turbulent eddies there scatter a small fraction of the energy forward to the receiver.

Features

  • Range 100–800 km, beyond LOS, without repeaters or satellites.
  • Very high path loss, so high power (kW) transmitters, large dish antennas (gain 40–50 dB) and sensitive receivers are needed.
  • Signal suffers Rayleigh fading (fast) and slow fading; diversity (space, frequency) is used.
  • Limited bandwidth because of multipath delay spread.
  • Reliable regardless of ionospheric conditions; not affected by solar disturbances.
  • Uses: military and remote links, offshore platforms, links across mountains or sea.
  • 2072 Magh · 2 marks

Explain skip zone for ionospheric region.

Answer

The skip zone (silent zone) is the region around a transmitter, between the point where the ground wave becomes too weak and the point where the first sky wave returns (the skip distance), in which no usable signal is received.

  ~~~~~~~~~~~~~~~/\~~~~~~~~~~ ionosphere
                /  \
   ground wave /    \
  Tx~~~~~~>   /      \  sky wave lands
  ==|=====|=========|==========
     ground  skip     reception
     wave    zone

It exists because, for f > fc, steep rays penetrate the ionosphere, while the ground wave at HF dies within a few tens of km. Its size grows with frequency and shrinks when fc is high (daytime).

  • 2072 Magh · 2+4 marks

Define maximum usable frequency. Derive the relation between the critical frequency and maximum usable frequency.

Answer

Maximum usable frequency

The maximum usable frequency (MUF) is the highest frequency that is returned to earth by the ionosphere for a given angle of incidence (that is, for a given distance between transmitter and receiver).

Relation between critical frequency and MUF

Critical frequency (fc) is the highest frequency reflected at vertical incidence.

                 ~~~~~~~ ionised layer ~~~~~~~
                      \  normal  /
                   θi  \   |    /
                        \  |   /
   Tx ___________________\_|__/_______ Rx

The refractive index of a layer with electron density N (per m³) at frequency f (Hz) is

n = √(1 − 81N / f²)

Using Snell's law, a ray entering from free space (n = 1) at angle θi is bent until its angle of refraction is 90° (ray horizontal), where it starts to return:

n = √(1 − 81N / f²)           (refractive index)
Snell's law at return point: sin θi = n
sin² θi = 1 − 81Nmax / f²
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi
Vertical incidence (θi = 0):  fc = 9√Nmax
Hence  f_MUF = fc sec θi      (secant law)

So f_MUF = fc sec θi = fc / cos θi. This is the secant law. Since sec θi ≥ 1, the MUF is always higher than the critical frequency, and it increases as the path becomes longer (more oblique).

For flat earth with virtual height h and distance D: sec θi = √(1 + D²/4h²), so f_MUF = fc√(1 + D²/4h²).

Example: fc = 6 MHz, θi = 60°: f_MUF = 6 × sec 60° = 6 × 2 = 12 MHz.

  • 2072 Asoj · 6 marks

Find out the line of sight distance between the transmitting antenna and receiving antenna if the transmitting antenna height is 45 m and the receiving antenna height is 25 m. [Given: the radius of Earth is 6,378 km].

Answer

The line-of-sight (LOS) distance is the sum of the distances from each antenna to the horizon. From the geometry of a tangent to a sphere of radius R:

   Tx                         Rx
   |\                         /|
 ht| \         d1   d2       / |hr
   |  `-.______  P  ______.-'  |
         earth (radius R)
 (R + h)² = R² + d²  →  d = √(2Rh + h²) ≈ √(2Rh)

Given: ht = 45 m, hr = 25 m, R = 6378 km = 6.378 × 10⁶ m.

d1 = √(2 R ht) = √(2 × 6.378×10⁶ × 45)
   = √(5.740×10⁸) = 23 959 m = 23.96 km
d2 = √(2 R hr) = √(2 × 6.378×10⁶ × 25)
   = √(3.189×10⁸) = 17 858 m = 17.86 km
d  = d1 + d2 = 23.96 + 17.86 = 41.82 km

Answer: LOS distance ≈ 41.8 km (using the given geometric earth radius).

Note: if standard atmospheric refraction is included (effective radius 4/3 × 6378 = 8504 km), d = 27.67 + 20.62 = 48.29 km.

  • 2071 Magh · 5+3 marks

Explain the different layers of ionosphere. What are the major characteristics of ionosphere?

Answer

Layers of the ionosphere

The ionosphere is the region of the upper atmosphere, about 50–400 km high, ionised by solar UV and X-ray radiation. It is divided into layers by height and electron density:

 height (km)
  400 |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F2 (250-400)
  300 |        /\                    day & night
  220 |~~~~~~~/~~\~~~~~~~~~~~~~~~~~~ F1 (150-250)
      |      /    \                  day only
  110 |~~~~~/~~~~~~\~~~~~~~~~~~~~~~~ E  (90-140)
   70 |::::/::::::::\::::::::::::::: D  (50-90)
      |   /  sky     \               day only
    0 |__/____________\_____________ earth
       Tx    skip      Rx
LayerHeightElectron density (per m³)PresentMain effect
D50–90 km10⁸–10¹⁰Day onlyAbsorbs LF/MF sky wave; reflects VLF
E90–140 km (≈110)≈10¹¹Mostly day (weak at night)Reflects MF and lower HF; fc ≈ 3–4 MHz
Sporadic Eₛ≈110 kmPatchy, highIrregularOccasional VHF long-distance
F1150–250 km (≈200)≈2–5 × 10¹¹Day onlySome HF reflection, absorption; fc ≈ 4–5 MHz
F2250–400 km≈10¹² (highest)Day and nightMain HF reflector; fc ≈ 5–12 MHz; longest hops
  • D layer: lowest; daytime only; high collision rate, so it absorbs MF and low-HF waves; reflects VLF/LF.
  • E layer: reflects MF and low HF; weak at night; sporadic-E patches can reflect VHF.
  • F1 layer: daytime only; merges into F2 at night; partly reflects and partly absorbs HF.
  • F2 layer: highest and densest; present day and night; main layer for long-distance HF; single hop up to ≈ 4000 km.

Major characteristics of the ionosphere

  1. Variable ionisation: depends on the sun, so it changes with time of day (diurnal), season, latitude and the 11-year sunspot cycle.
  2. Refractive index less than 1: n = √(1 − 81N/f²); decreases as electron density rises, so waves bend back to earth.
  3. Frequency dependent: higher frequencies penetrate further; above the critical frequency (fc = 9√Nmax) at vertical incidence, and above the MUF for oblique incidence, waves escape.
  4. Absorption: proportional to collision frequency and to 1/f², greatest in the D layer.
  5. Irregularities and disturbances: sporadic E, sudden ionospheric disturbances (solar flares) and ionospheric storms cause fading and blackouts.
  6. Virtual height is always more than actual height.
  7. Acts as a dispersive, time-varying, slightly anisotropic medium (because of earth's magnetic field, causing Faraday rotation).
  • 2071 Magh · 4+4 marks

Explain the duct propagation mechanism in radio wave propagation. Define critical frequency and MUF with necessary derivations.

Answer

Duct propagation mechanism

Duct propagation occurs when a layer of the lower atmosphere has a refractive index that falls very rapidly with height, so that VHF/UHF/microwave rays bend down more than the earth curves and get trapped in the layer (the duct), travelling far beyond the horizon.

   ~~~~~~~~~~~ top of duct ~~~~~~~~~~~
    /\     /\     /\     /\     /\
   /  \   /  \   /  \   /  \   /  \
  Tx===\=/====\=/====\=/====\=/====Rx
  • Cause: temperature inversion and sharp decrease of humidity with height (common over sea and on calm evenings).
  • Condition: modified refractive index M = (n − 1 + h/R) × 10⁶ decreases with height (dM/dh < 0, i.e. dN/dh < −157 per km).
  • The trapped wave is repeatedly bent down and reflected from the ground, like a wave in a waveguide; only wavelengths small compared with duct thickness propagate.
  • Effects: ranges of hundreds of km at microwave frequencies, but also interference and radar false echoes; not reliable because it depends on weather.

Critical frequency

The critical frequency (fc) is the highest frequency reflected by a layer at vertical incidence. From n = √(1 − 81N/f²), the wave turns back at vertical incidence where n = 0:

0 = 1 − 81Nmax / fc²
fc = 9 √Nmax     (Hz, N per m³)

Maximum usable frequency (MUF)

The MUF is the highest frequency returned to earth for a given angle of incidence θi (given distance). A ray turns back where sin θi = n (Snell's law):

n = √(1 − 81N / f²)           (refractive index)
Snell's law at return point: sin θi = n
sin² θi = 1 − 81Nmax / f²
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi
Vertical incidence (θi = 0):  fc = 9√Nmax
Hence  f_MUF = fc sec θi      (secant law)

For flat earth with virtual height h and distance D, sec θi = √(1 + D²/4h²):

f_MUF = fc √(1 + D² / 4h²)
  • 2071 Bhadra · 5+6 marks

Write down the factors which affect the space wave communication. Explain the major characteristics of MW and SW radio propagation.

Answer

Factors affecting space wave communication

Space wave = direct wave + ground-reflected wave, used at VHF and above.

  1. Antenna heights: LOS range d ≈ 4.12(√ht + √hr) km and field E ∝ ht·hr; higher antennas give longer range and stronger signal.
  2. Earth curvature: limits range to the radio horizon, reduces effective antenna heights, creates a shadow zone, and spreads (diverges) the reflected wave.
  3. Tropospheric refraction: normal decrease of refractive index gives k = 4/3 (horizon extended about 15%); abnormal profiles cause super-refraction, ducting or sub-refraction (fading or loss of signal).
  4. Ground reflection: reflection coefficient depends on ground conductivity, permittivity, polarisation and grazing angle; causes interference lobes and multipath fading.
  5. Frequency: higher frequency means more free-space path loss per given antenna size, more shadowing by obstacles, less diffraction.
  6. Obstacles and terrain: hills, buildings and trees block or diffract the wave; first Fresnel-zone clearance (≥ 0.6) is required.
  7. Atmospheric absorption and precipitation: rain, fog, O₂ and H₂O absorb above about 10 GHz.
  8. Distance: beyond the horizon the field falls rapidly; within LOS it falls as 1/d² at large d.
  9. Multipath from buildings and moving objects: causes fast fading (mobile radio).

Characteristics of MW and SW radio propagation

MW (medium wave) propagation, 300 kHz – 3 MHz (AM broadcast 526.5–1606.5 kHz):

  • Daytime: mainly ground (surface) wave; the D layer absorbs the sky wave. Range ≈ 100–300 km depending on power and soil (longest over sea).
  • Nighttime: D layer disappears, the E layer reflects the sky wave, giving 1000–2000 km range, but also co-channel interference and fading where ground and sky waves mix.
  • Vertical polarisation and tall λ/4 mast antennas with radial ground systems are used.
  • Ground-wave attenuation increases with frequency, so the lower end of the band covers more area.
  • High atmospheric noise; narrow bandwidth (10 kHz channels).

SW (short wave) propagation, 3 – 30 MHz (HF):

  • Mainly sky wave via the F2 (and E) layer; ground wave dies within a few tens of km.
  • Very long range with modest power: one F2 hop up to ≈ 4000 km, worldwide by multi-hop.
  • Governed by critical frequency, MUF and skip distance; a skip zone exists between the end of ground wave and the first sky-wave return.
  • Fading due to multipath and changes in the ionosphere; diversity reception is used.
  • Frequency must change with time of day, season and sunspot cycle: higher frequencies by day, lower at night; OWF ≈ 0.85 MUF.
  • Affected by ionospheric storms and sudden ionospheric disturbances (radio blackouts).
  • Horizontal antennas such as dipoles, rhombic and log-periodic arrays are common.
  • 2071 Bhadra · 8 marks

With a mathematical relation of refractive index of ionospheric layer derive a relation of critical frequency and maximum usable frequency (MUF) of radio waves with necessary explanation. Consider the earth is not curved.

Answer

Refractive index of an ionospheric layer

An ionised layer contains N free electrons per m³ (charge e, mass m). Under a wave field E sin ωt, each electron moves with velocity v:

m dv/dt = e E sin ωt
v = −(eE / mω) cos ωt

The electron current J = Nev lags E by 90° (an inductive current), so it is opposite to the displacement current ε₀∂E/∂t. Total current density:

J_total = ε₀ ∂E/∂t + Nev
        = ωε₀E cos ωt − (Ne²/mω) E cos ωt
        = ωε₀ [1 − Ne² / (ε₀ m ω²)] E cos ωt

So the effective relative permittivity is εr = 1 − Ne²/(ε₀mω²), and the refractive index n = √εr. Putting e = 1.602 × 10⁻¹⁹ C, m = 9.109 × 10⁻³¹ kg, ε₀ = 8.854 × 10⁻¹² F/m and ω = 2πf:

e² / (4π² ε₀ m) = 80.6 ≈ 81
n = √(1 − 81N / f²)      (N per m³, f in Hz)

Since n < 1 and falls as N rises with height, an upward ray bends away from the normal and returns to earth.

Critical frequency

At vertical incidence the wave returns where n = 0 at the level of maximum density Nmax:

1 − 81Nmax / fc² = 0
fc = 9 √Nmax

Critical frequency is the highest frequency reflected at vertical incidence.

Maximum usable frequency (flat earth)

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Explanation: fc depends only on the layer's peak electron density; the MUF depends also on the angle of incidence, so a longer path (larger θi) can use a higher frequency. With flat earth, MUF = fc sec θi = fc√(1 + D²/4h²), always ≥ fc. Frequencies above the MUF escape the layer; in practice OWF ≈ 0.85 MUF is used.

Example: Nmax = 10¹² /m³ gives fc = 9 × 10⁶ Hz = 9 MHz; for D = 600 km and h = 300 km, f_MUF = 9 × √(1 + 1) = 12.73 MHz.

  • 2070 Magh · 5 marks

Assume that reflection takes place at a height of 400 km and that the maximum density in the ionosphere corresponds to a 0.8 refractive index at 15 MHz. What will be the range (assume flat earth) for which the MUF is 20 MHz?

Answer

The MUF over a flat-earth path is the critical frequency multiplied by the secant of the angle of incidence (the secant law). So we first find the maximum electron density from the given refractive index, then the critical frequency, then the angle and the range.

Given

  • Virtual height of reflection, h = 400 km
  • At the layer maximum, n = 0.8 at f = 15 MHz
  • Required MUF = 20 MHz

Step 1: Maximum electron density

The refractive index of the ionosphere is n² = 1 − 81N/f² (N in electrons/m³, f in Hz).

81·Nmax = f²(1 − n²)
        = (15×10⁶)² × (1 − 0.8²)
        = 2.25×10¹⁴ × 0.36
        = 8.1×10¹³
Nmax    = 1.0×10¹² electrons/m³

Step 2: Critical frequency

fc = 9√Nmax = 9 × √(10¹²) = 9×10⁶ Hz = 9 MHz

Step 3: Angle of incidence for MUF = 20 MHz

MUF = fc·sec θi
sec θi = 20/9 = 2.222
cos θi = 0.45  →  θi = 63.26°
tan θi = √(1 − 0.45²)/0.45 = 1.9845

Step 4: Range (flat earth)

      T x ---------------- d ----------------- R x
        \                                    /
         \  θi                          θi  /
          \                                /
           \_________ reflection _________/
                     point, h = 400 km

For a flat earth, half the range is d/2 = h·tan θi, so:

d = 2h·tan θi = 2 × 400 × 1.9845 = 1587.6 km

Check with the combined formula MUF = fc·√(1 + (d/2h)²):

d = 2h·√((MUF/fc)² − 1) = 800 × √(4.938 − 1)
  = 800 × 1.9845 = 1587.6 km

Answer: Nmax = 10¹² /m³, fc = 9 MHz, θi = 63.26°, range d ≈ 1588 km (about 1590 km).

Questions from Old Question Collection (EX 653) (IOE EX 653 exam papers from 2069 to 2081 (20 papers)). Answers are written for this site; check them against your class notes.

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