Radio Wave Propagation in Mobile Networks

Outdoor propagation models predict path loss over irregular terrain and built-up areas. Two widely used empirical models are the Okumura model and the Hata model.

1. Okumura model

Based on extensive measurements in Tokyo. It gives median path loss as free-space loss plus correction curves.

L50(dB) = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

• LF = free space loss; Amu = median attenuation relative to free space (from curves); GAREA = gain due to environment (from curves).
• G(hte) = 20 log(hte/200), for 1000 m > hte > 30 m
• G(hre) = 10 log(hre/3) for hre ≤ 3 m; 20 log(hre/3) for 3 m < hre < 10 m
• Valid for 150–1920 MHz (extendable to 3 GHz), d = 1–100 km, hte = 30–1000 m.
• Simple and accurate for urban/suburban areas (standard deviation about 10–14 dB), but slow to respond to rapid terrain changes and needs curves, so it is hard to use in software.

2. Hata model

An empirical formula fitted to Okumura's curves, so no graphs are needed.

L50(urban) = 69.55 + 26.16 log fc − 13.82 log hte
             − a(hre) + (44.9 − 6.55 log hte) log d

• fc in MHz, d in km, heights in m.
• Small/medium city: a(hre) = (1.1 log fc − 0.7)hre − (1.56 log fc − 0.8) dB
• Large city: a(hre) = 3.2(log 11.75hre)² − 4.97 dB (fc ≥ 300 MHz)
• Suburban: L50 = L50(urban) − 2[log(fc/28)]² − 5.4
• Open rural: L50 = L50(urban) − 4.78(log fc)² + 18.33 log fc − 40.94
• Valid for fc = 150–1500 MHz, hte = 30–200 m, hre = 1–10 m, d = 1–20 km. Good for large-cell systems, not for PCS micro-cells.

Other outdoor models: Longley–Rice, Durkin's model, the PCS extension of Hata (COST-231, up to 2 GHz) and the Walfisch–Bertoni model.

Okumura model: L50 = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

Given: d = 50 km, hte = 100 m, hre = 10 m, f = 900 MHz, Amu = 43 dB, GAREA = 9 dB, EIRP = 1 kW, Gr = 10 dB.

Step 1: Free space path loss

λ  = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 10 log[(4π)²d²/λ²] = 20 log(4πd/λ)
   = 20 log(4π × 50 000 / 0.3333)
   = 125.51 dB

Step 2: Antenna height gains

G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB   (3 m < hre < 10 m)

Step 3: Mean (median) path loss

L50 = 125.51 + 43 − (−6.02) − 10.46 − 9
    = 155.07 dB

Step 4: EIRP in dBm

EIRP = 10 log(1000 W / 1 mW) = 10 log(10⁶) = 60 dBm

Step 5: Received power

Pr = EIRP − L50 + Gr
   = 60 − 155.07 + 10
   = −85.07 dBm

Answer: L50 = 155.07 dB, EIRP = 60 dBm, Pr = −85.07 dBm (about 3.1 pW).

The two-ray ground reflection model assumes the received field is the sum of a line-of-sight (LOS) wave and a wave reflected from the ground. The path difference between them decides whether they add or cancel.

  Tx
   |\  d' (LOS)
   | \______________________  Rx
ht |  \                  __/|
   |   \  d'' (GR)    __/   | hr
   |    \          __/      |
 --+-----\______/-----------+----- ground
          reflection point
   |<---------- d ---------->|

 Image method: image of Tx at depth ht below ground

Path lengths (method of images)

Replace the ground-reflected path by a straight line from the image of the transmitter (at −ht) to the receiver.

LOS:  d'  = √[d² + (ht − hr)²]
GR :  d'' = √[d² + (ht + hr)²]

Path difference

Using binomial expansion √(1 + x) ≈ 1 + x/2 for d ≫ ht + hr:

d'  = d√[1 + ((ht−hr)/d)²] ≈ d + (ht − hr)²/(2d)
d'' = d√[1 + ((ht+hr)/d)²] ≈ d + (ht + hr)²/(2d)

Δ = d'' − d'
  = [(ht + hr)² − (ht − hr)²] / (2d)
  = 4ht·hr / (2d)
Δ ≈ 2ht·hr / d

Phase difference and time delay

A path difference of one wavelength corresponds to 2π radians:

θΔ = 2πΔ/λ = Δωc/c ≈ 4π·ht·hr/(λd)

τd = Δ/c = θΔ/(2πfc)

Use in received field

With ground reflection coefficient Γ ≈ −1, the total field is

|ETOT| ≈ 2E0d0/d · sin(θΔ/2)

For large d, θΔ/2 < 0.3 rad, so sin(θΔ/2) ≈ θΔ/2 and

ETOT ≈ (2E0d0/d)(2πht·hr/(λd)) = k/d²
Pr = Pt·Gt·Gr·ht²·hr² / d⁴

So at large distance received power falls by 40 dB/decade, and is independent of frequency.

Fading is the rapid fluctuation of the amplitude, phase or multipath delay of a radio signal over a short time or short distance, caused by interference between two or more versions of the transmitted signal arriving at slightly different times (multipath).

Small-scale fading is classified independently by two mechanisms: multipath time delay spread and Doppler spread.

Based on multipath delay spread (time dispersion)

1. Flat fading

• Bs ≪ Bc and Ts ≫ στ (signal bandwidth smaller than coherence bandwidth).
• All frequency components fade together; spectral shape is kept, only gain changes.
• No ISI. Amplitude usually Rayleigh distributed. Example: narrowband voice channel.

2. Frequency selective fading

• Bs > Bc and Ts < στ.
• Different frequencies fade differently; the channel causes ISI and the received signal is distorted.
• Needs equalizers, RAKE or OFDM. A common rule: frequency selective if Ts < 10στ.

Based on Doppler spread (frequency dispersion)

3. Fast fading

• Tc < Ts and BD > Bs: the channel changes within one symbol.
• Causes frequency dispersion (time-selective fading) and signal distortion. Occurs at very low data rates or high speeds.

4. Slow fading

• Tc ≫ Ts and BD ≪ Bs: channel is nearly constant over many symbols.
• Channel changes slowly compared with the symbol rate.

The two classifications are independent, so a channel can be, for example, flat and slow fading at the same time.

Small-scale fading (or simply fading) is the rapid change of received signal amplitude and phase over a short distance (a few wavelengths) or short time, due to the constructive and destructive addition of multipath waves. Large-scale path loss is almost constant over such a distance.

Its effects are: rapid changes in signal strength, random frequency modulation due to Doppler shifts on different paths, and time dispersion (echoes) caused by multipath delays.

Types of small-scale fading

A. Based on multipath delay spread

1. Flat fading: Bs ≪ Bc, Ts ≫ στ. All frequencies fade together, no ISI, amplitude is Rayleigh distributed.
2. Frequency selective fading: Bs > Bc, Ts < στ. Parts of the spectrum fade differently, causing ISI and distortion; equalization is needed.

B. Based on Doppler spread 3. Fast fading: Ts > Tc, Bs < BD. Channel impulse response changes within a symbol; causes time-selective distortion. 4. Slow fading: Ts ≪ Tc, Bs ≫ BD. Channel is static over many symbols.

          Small-scale fading
         /                  \
  Delay spread          Doppler spread
   /        \             /        \
 Flat   Freq.-selective  Fast      Slow

Factors influencing small-scale fading

1. Multipath propagation: reflecting and scattering objects (buildings, trees, vehicles) create many waves with different amplitudes, phases and delays. Their vector sum fluctuates, and the spread of delays causes time dispersion and ISI.
2. Speed of the mobile: relative motion between BS and mobile gives each multipath wave a different Doppler shift fd = (v/λ)cos θ, producing random frequency modulation.
3. Speed of surrounding objects: moving vehicles and people change the channel with time. If they move faster than the mobile, they dominate the fading; otherwise their effect can be ignored and only the mobile's speed counts.
4. Transmission bandwidth of the signal: if the signal bandwidth is greater than the channel's coherence bandwidth, the signal is distorted (frequency selective); if much smaller, the amplitude fades but the signal is not distorted (flat).

Example: at 900 MHz (λ = 0.333 m) a car at 72 km/h (20 m/s) has maximum Doppler shift fm = 20/0.333 = 60 Hz, so the fading rate is in the tens of hertz.

Friis free space equation: Pr = Pt·Gt·Gr·λ² / [(4π)²d²]

Given: Pt = 50 W, d = 10 km, f = 900 MHz, Gt = 1, Gr = 2, R = 50 Ω (matched).

λ = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m

i) Received power

Pr = 50 × 1 × 2 × (0.3333)² / [(4π)² × (10⁴)²]
   = 7.04 × 10⁻¹⁰ W
   = 10 log(7.04×10⁻¹⁰) = −91.53 dBW = −61.53 dBm

ii) Magnitude of E-field

Effective aperture of the receiving antenna:

Ae = Gr·λ²/(4π) = 2 × 0.1111 / 12.566 = 0.01768 m²

Since Pr = |E|²·Ae / 120π:

|E| = √(Pr × 120π / Ae)
    = √(7.04×10⁻¹⁰ × 376.99 / 0.01768)
    = 3.87 × 10⁻³ V/m = 3.87 mV/m

iii) Power flux density

Pd = |E|²/120π = Pr/Ae
   = 7.04×10⁻¹⁰ / 0.01768
   = 3.98 × 10⁻⁸ W/m²

iv) RMS voltage at the receiver input

For a matched antenna, the open-circuit voltage V divides equally between antenna resistance and receiver, so Pr = V²/(4Rant):

V = √(4 × Rant × Pr) = √(4 × 50 × 7.04×10⁻¹⁰)
  = 3.75 × 10⁻⁴ V = 0.375 mV

(The voltage across the receiver input is V/2 = 0.188 mV.)

Answer: Pr = 7.04×10⁻¹⁰ W (−61.5 dBm); |E| = 3.87 mV/m; Pd = 3.98×10⁻⁸ W/m²; Vrms = 0.375 mV.

Both models are empirical, so they are valid only within the range of the measurements on which they are based.

Okumura model

• Frequency: 150 MHz to 1920 MHz (often extrapolated up to 3000 MHz).
• Distance: 1 km to 100 km between transmitter and receiver.
• BS antenna effective height (hte): 30 m to 1000 m.
• Mobile antenna height (hre): reference 3 m, correction for up to about 10 m.
• Terrain should be quasi-smooth; corrections are needed for hilly terrain, water, slope, etc.
• Environment must be classed as urban, suburban or open area; correction curves (Amu, GAREA) are read for these.

Hata model

• Frequency (fc): 150 MHz to 1500 MHz.
• BS antenna height (hte): 30 m to 200 m.
• Mobile antenna height (hre): 1 m to 10 m.
• Distance (d): 1 km to 20 km.
• Suited to large-cell (macrocell) systems; not suitable for personal communication systems with cell radius about 1 km. For 1500–2000 MHz, the COST-231 extension is used.

Coherence bandwidth (Bc)

Coherence bandwidth is the range of frequencies over which the channel can be considered "flat", i.e. it passes all spectral components with nearly equal gain and linear phase. Two sinusoids separated by less than Bc have strongly correlated amplitudes.

It is inversely proportional to the rms delay spread στ:

Bc ≈ 1/(50στ)   (correlation ≥ 0.9)
Bc ≈ 1/(5στ)    (correlation ≥ 0.5)

Significance:

• If signal bandwidth Bs < Bc: flat fading, no ISI, simple receiver.
• If Bs > Bc: frequency selective fading, causing ISI; equalizers, RAKE receivers or OFDM are needed.
• It sets the maximum symbol rate without equalization and the frequency spacing needed for frequency diversity.

Coherence time (Tc)

Coherence time is the time duration over which the channel impulse response is essentially unchanged. Two signals received within Tc have strongly correlated amplitudes. It is the time-domain dual of Doppler spread:

Tc ≈ 1/fm
Tc ≈ 9/(16π fm) ≈ 0.179/fm   (correlation 0.5)
Tc ≈ √(9/(16π)) / fm = 0.423/fm  (common rule)
fm = v/λ (maximum Doppler shift)

Significance:

• If symbol period Ts < Tc: slow fading, channel constant over a symbol.
• If Ts > Tc: fast fading, the channel changes within a symbol and the signal is distorted.
• It sets the time spacing for time diversity and interleaving depth, and how often channel estimates must be updated.

Example: v = 60 mph (26.82 m/s), fc = 1900 MHz → fm = 169.8 Hz, Tc = 0.423/169.8 ≈ 2.5 ms. A symbol rate above about 1/Tc = 400 symbols/s avoids distortion from motion.

Significance of frequency dispersion

Frequency dispersion is the spreading of the signal spectrum in the frequency domain, caused by the relative motion between the mobile and the base station (or moving objects). Each multipath wave arrives from a different angle, so it gets a different Doppler shift, and a single tone is received as a band of frequencies of width up to 2fm.

• It shows that the channel is time-variant; it is measured by Doppler spread BD and its dual, coherence time Tc ≈ 1/BD.
• If BD is comparable to the signal bandwidth (Ts > Tc), the channel causes fast fading and distortion.
• It limits how slowly data can be sent and how often the receiver must track the channel; it decides interleaver depth and time diversity design.

Derivation of Doppler shift and Doppler spread

A mobile moves at speed v from point X to Y (distance l) in time Δt, while receiving from a distant source S. The wave arrives at angle θ to the direction of motion.

                    S (far source)
                   /
                  /  waves nearly parallel
                 /
         X ----θ------------ Y    → v
           \___ Δl = l cosθ
           l = vΔt

Since the source is far, rays at X and Y are parallel. The difference in path length is:

Δl = l cosθ = vΔt cosθ

The phase change of the received signal:

Δφ = 2πΔl/λ = 2πvΔt cosθ / λ

Frequency change (Doppler shift) is the rate of change of phase:

fd = (1/2π)·(Δφ/Δt) = (v/λ)·cosθ

• θ = 0° (moving towards source): fd = +v/λ (maximum, frequency rises)
• θ = 180° (moving away): fd = −v/λ
• θ = 90°: fd = 0

The maximum Doppler shift is fm = v/λ. Since the multipath waves arrive from all angles θ (0 to 2π), the received spectrum spreads over fc − fm to fc + fm. The Doppler spread BD is the measure of this width, taken as

BD = fm = v/λ   (spectrum occupies 2fm in total)

Example: fc = 900 MHz, v = 72 km/h = 20 m/s → λ = 0.333 m, fm = 20/0.333 = 60 Hz.

Coherence bandwidth

Coherence bandwidth (Bc) is the range of frequencies over which the channel gain is nearly constant, so that two frequency components in this range are strongly correlated. It is inversely related to rms delay spread: Bc ≈ 1/(50στ) for 0.9 correlation and Bc ≈ 1/(5στ) for 0.5 correlation. If signal bandwidth exceeds Bc, the signal suffers frequency-selective fading.

Coherence time

Coherence time (Tc) is the time over which the channel impulse response stays essentially unchanged. It is inversely related to maximum Doppler shift: Tc ≈ 9/(16πfm), or Tc ≈ 0.423/fm. If the symbol period exceeds Tc, the signal suffers fast fading.

Factors influencing indoor propagation

Indoor channels differ from outdoor ones: distances are short, path loss varies a lot over small distances, and the environment changes with layout and people. The main factors are:

1. Building layout: open-plan offices, corridors and rooms produce very different multipath. Corridors can act as waveguides.
2. Construction materials: concrete, brick, metal frames, glass and wood have different reflection and penetration losses. Metal-coated glass gives high loss.
3. Building type: offices, factories, homes, stores and stadiums have different clutter and ceiling heights, so the path loss exponent varies (about 1.6–1.8 for LOS in buildings, 4–6 when obstructed).
4. Partition losses (same floor): hard partitions (fixed walls) and soft partitions (movable panels, furniture) add loss per wall, e.g. a concrete wall can add 10–15 dB.
5. Partition losses between floors (Floor Attenuation Factor): loss depends on floor material, number of floors and reinforcement; the first floor gives the largest loss (often 10–20 dB), with less extra loss for each added floor.
6. Antenna location and height: whether the antenna is near the ceiling, in a corridor or outside affects coverage.
7. Doors and windows, open or closed, and movement of people, which change the multipath with time.

A common indoor model is the log-distance model with partition terms:

PL(d) = PL(d0) + 10n log(d/d0) + Σ FAF + Σ PAF + Xσ

Okumura model: L50 = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

Given: d = 50 km, hte = 100 m, hre = 10 m, f = 900 MHz, Amu = 43 dB, GAREA = 9 dB (suburban), EIRP = 1 kW, Gr = 1 (0 dB).

Step 1: Free space path loss

λ  = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 20 log(4πd/λ)
   = 20 log(4π × 50 000 / 0.3333)
   = 125.51 dB

Step 2: Height gain factors

G(hte) = 20 log(hte/200) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(hre/3)   = 20 log(10/3)    = 10.46 dB

Step 3: Median path loss

L50 = LF + Amu − G(hte) − G(hre) − GAREA
    = 125.51 + 43 + 6.02 − 10.46 − 9
    = 155.07 dB

Step 4: Received power

EIRP = 1 kW = 10 log(1000/0.001) = 60 dBm

Pr = EIRP − L50 + Gr
   = 60 − 155.07 + 0
   = −95.07 dBm

Answer: Median path loss L50 = 155.07 dB; received power Pr = −95.07 dBm (about 3.1×10⁻¹³ W).

Coherence bandwidth

Coherence bandwidth (Bc) is the statistical range of frequencies over which the channel passes all spectral components with approximately equal gain and linear phase. It is inversely proportional to the rms delay spread:

Bc ≈ 1/(50στ)   for 0.9 frequency correlation
Bc ≈ 1/(5στ)    for 0.5 frequency correlation

Significance: if the signal bandwidth Bs < Bc, the signal sees flat fading with no ISI; if Bs > Bc, it sees frequency selective fading and ISI, so equalizers or OFDM are needed.

Coherence time

Coherence time (Tc) is the time over which the channel impulse response is essentially invariant. It is inversely proportional to maximum Doppler shift fm = v/λ:

Tc ≈ 9/(16π fm)  ≈ 0.179/fm
Tc ≈ 0.423/fm    (geometric mean rule, used in practice)

Significance: if the symbol period Ts < Tc, the channel is slow fading; if Ts > Tc, it is fast fading and the signal is distorted. It also sets the spacing needed for time diversity and interleaving.

Fading and its types

Fading is the rapid variation of the received signal amplitude and phase over short time or distance, caused by constructive and destructive addition of multipath waves and by motion.

Based on multipath time delay spread:

1. Flat fading: Bs ≪ Bc, Ts ≫ στ. All frequencies fade together, no ISI; spectral shape preserved.
2. Frequency selective fading: Bs > Bc, Ts < στ. Different frequencies fade differently; causes ISI.

Based on Doppler spread: 3. Fast fading: Ts > Tc, Bs < BD. Channel varies within a symbol; time-selective distortion. 4. Slow fading: Ts ≪ Tc, Bs ≫ BD. Channel constant over many symbols.

Large-scale fading (shadowing by hills and buildings) is the slower variation of the mean signal over hundreds of wavelengths; it is usually log-normally distributed.

Okumura vs Hata model

Delay spread calculation

The delays are taken as µs (the usual unit for such a profile; in seconds the numbers are the same but every time is in s and Bc in Hz).

ΣP = 0.01 + 0.1 + 0.1 + 1 = 1.21

Mean excess delay

τ̄ = Σ P(τk)τk / ΣP
  = (0.01×0 + 0.1×1 + 0.1×2 + 1×5) / 1.21
  = 5.3/1.21 = 4.38 µs

Second moment

τ²̄ = (0.01×0 + 0.1×1 + 0.1×4 + 1×25) / 1.21
    = 25.5/1.21 = 21.07 µs²

RMS delay spread

στ = √(τ²̄ − τ̄²) = √(21.07 − 4.38²)
   = √(21.07 − 19.19) = 1.37 µs

Maximum excess delay (10 dB) The strongest component is 0 dB at 5 µs; components within 10 dB of it start at τ = 1 µs (−10 dB), and the first arrival is at τ0 = 0. Using the textbook definition τmax = τX − τ0 with τX = 5 µs:

τmax(10 dB) = 5 − 0 = 5 µs

50% coherence bandwidth

Bc ≈ 1/(5στ) = 1/(5 × 1.374 µs) = 145.5 kHz ≈ 146 kHz

Answer: τ̄ = 4.38 µs, στ = 1.37 µs, τmax(10 dB) = 5 µs, Bc(50%) ≈ 146 kHz.

Doppler spread (BD) is the measure of spectral broadening of a received signal caused by the time rate of change of the mobile radio channel. It is the range of frequencies over which the received Doppler spectrum is non-zero. When a pure tone fc is sent, the received spectrum spreads between fc − fd and fc + fd, where fd is the Doppler shift; BD ≈ fm = v/λ.

Derivation of Doppler shift

A mobile moves with constant speed v from point X to point Y in time Δt, receiving a signal from a remote source S. The direction of arrival makes angle θ with the direction of motion.

                 S (remote source)
                /|
               / |
              /  |  rays from S are
             /   |  almost parallel
            / θ  |
    X ----------- Y ------>  v
    |<---- l --->|
    path difference Δl = l cosθ

Distance travelled:

l = v·Δt

Since S is very far, the two rays (to X and to Y) are nearly parallel, so the difference in path length is:

Δl = l cosθ = v·Δt·cosθ

Phase change due to this path difference:

Δφ = 2π·Δl/λ = (2π·v·Δt·cosθ)/λ

The apparent change in frequency is the rate of change of phase divided by 2π:

fd = (1/2π)·(Δφ/Δt)
fd = (v/λ)·cosθ

Results:

• Moving towards the source (θ = 0): fd = +v/λ, received frequency increases.
• Moving away (θ = 180°): fd = −v/λ, frequency decreases.
• Moving perpendicular (θ = 90°): fd = 0.
• Maximum Doppler shift fm = v/λ.

Example: fc = 1850 MHz, v = 60 mph = 26.82 m/s, λ = 0.162 m → fm = 26.82/0.1622 = 165.4 Hz; moving towards the source, received frequency = 1850.000165 MHz.

(i) and (ii) Transmit power

Pt(dBm) = 10 log(50 W / 1 mW) = 10 log(50 000) = 46.99 dBm
Pt(dBW) = 10 log(50 W / 1 W)  = 10 log(50)     = 16.99 dBW

Received power at 100 m

Friis equation with Gt = Gr = 1, f = 900 MHz:

λ = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m

Pr = Pt·Gt·Gr·λ² / [(4π)²d²]
   = 50 × 1 × 1 × (0.3333)² / [(4π)² × 100²]
   = 3.52 × 10⁻⁶ W = 3.52 µW

Pr(dBm) = 10 log(3.52×10⁻⁶ / 10⁻³) = −24.54 dBm

Received power at 10 km

In free space power falls as 1/d², i.e. 20 dB per decade. Using d0 = 100 m as reference:

Pr(10 km) = Pr(100 m) + 20 log(100/10 000)
          = −24.54 − 40
          = −64.54 dBm

Answer: Pt = 46.99 dBm = 16.99 dBW; Pr(100 m) = −24.5 dBm; Pr(10 km) = −64.5 dBm.

For free space propagation, the received power in dB form (Friis equation) is:

Pr(dBm) = Pt(dBm) + Gt(dBi) + Gr(dBi) − LFS(dB)
LFS = 20 log(4πd/λ)
    = 32.44 + 20 log f(MHz) + 20 log d(km)

Given: Pt = 165 W, f = 325 MHz, Gt = 12 dBi, Gr = 6 dBi, d = 15 km, no other losses.

Step 1: Transmit power in dBm

Pt = 10 log(165 / 0.001) = 10 log(165 000) = 52.17 dBm

Step 2: Free space path loss

λ   = 3×10⁸ / 325×10⁶ = 0.923 m
LFS = 20 log(4π × 15 000 / 0.923)
    = 106.20 dB
(check: 32.44 + 50.24 + 23.52 = 106.20 dB)

Step 3: Received power

Pr = 52.17 + 12 + 6 − 106.20
   = −36.03 dBm
Pr = 10^(−36.03/10) mW = 2.50 × 10⁻⁴ mW = 0.25 µW

Answer: Power delivered to the receiver ≈ −36.0 dBm, i.e. about 2.5 × 10⁻⁷ W (0.25 µW).

Path loss

Path loss is the reduction in signal power as a wave travels from transmitter to receiver. It is the difference (in dB) between effective transmitted power and received power, and may include antenna gains:

PL(dB) = 10 log(Pt/Pr)
       = −10 log[Gt·Gr·λ² / ((4π)²d²)]  (free space)

In practice PL(d) ∝ dⁿ, where n is the path loss exponent (2 in free space, 2.7–5 in urban areas).

Parameters of mobile multipath channels

These parameters, obtained from the power delay profile and Doppler spectrum, are used to classify small-scale fading.

A. Time dispersion parameters (from the power delay profile P(τ))

1. Mean excess delay τ̄: first moment of the power delay profile.

τ̄ = Σ ak²τk / Σ ak² = Σ P(τk)τk / Σ P(τk)

2. RMS delay spread στ: square root of the second central moment; the main measure of time dispersion.

στ = √(τ²̄ − (τ̄)²),  τ²̄ = Σ P(τk)τk² / Σ P(τk)

3. Maximum excess delay (X dB): time during which multipath energy falls to X dB below the maximum, τX − τ0.

B. Coherence bandwidth (Bc) Range of frequencies over which the channel is flat. Bc ≈ 1/(50στ) (0.9 correlation) or 1/(5στ) (0.5 correlation). Compared with signal bandwidth, it decides flat vs frequency selective fading.

C. Doppler spread (BD) Width of the spectrum when a pure tone is sent; BD ≈ fm = v/λ. It measures how fast the channel changes.

D. Coherence time (Tc) Time over which the channel is constant; Tc ≈ 0.423/fm. Compared with symbol period, it decides fast vs slow fading.

The Ericsson multiple breakpoint model is an empirical indoor path loss model, obtained by Ericsson Radio Systems from measurements in a multi-floor office building. Path loss versus distance is drawn as straight-line segments with four breakpoints, and the slope (path loss exponent) increases with distance.

• It assumes about 30 dB loss at d0 = 1 m, which is accurate for 900 MHz with unity-gain antennas.
• It gives both an upper and a lower bound on path loss; the actual loss lies between them, with a uniform distribution assumed in between.
• The slope is small close to the transmitter (near free space, about 20 dB/decade) and rises sharply beyond a few tens of metres as more walls and floors block the path (typical lower-bound slopes about 20, 30, 60 and 120 dB/decade, with breakpoints near 10 m, 20 m and 40 m).
• It is easy to use and suits planning of indoor coverage (e.g. office PCS and WLAN), but it is building specific.

Diffraction is the bending of radio waves around the edges of obstacles (hills, building edges) so that the signal reaches the shadow region behind them, even without line of sight. It is explained by Huygens' principle: every point on a wavefront acts as a source of secondary wavelets, which combine behind the obstacle. Diffraction loss depends on how far the obstacle blocks the Fresnel zones.

Fresnel zone geometry (knife-edge)

A transmitter and receiver are separated by an obstacle of effective height h above the LOS line, at distances d1 and d2.

                 |\  diffracted path
                 | \
        ________ h  \________
  Tx  /  α  β     |    γ     \  Rx
   o---------------+----------o    LOS line
   |<---- d1 ----->|<-- d2 -->|
   ht              obstacle   hr

Assume h ≪ d1, d2 and h ≫ λ.

Path difference

Length of the diffracted path (two straight segments over the edge):

√(d1² + h²) + √(d2² + h²)
 ≈ d1 + h²/(2d1) + d2 + h²/(2d2)   (binomial approx.)

Path difference over the direct path (d1 + d2):

Δ ≈ (h²/2)·(1/d1 + 1/d2)
Δ ≈ (h²/2)·(d1 + d2)/(d1·d2)

Phase difference

φ = 2πΔ/λ = (2π/λ)·(h²/2)·(d1 + d2)/(d1·d2)

Using small angles, the angle α ≈ h(d1 + d2)/(d1·d2). Defining the Fresnel–Kirchhoff diffraction parameter:

v = h·√[ 2(d1 + d2) / (λ·d1·d2) ]
  = α·√[ 2d1d2 / (λ(d1 + d2)) ]

the phase difference becomes:

φ = (π/2)·v²

Fresnel zones

Fresnel zones are regions where the path difference is nλ/2 (phase difference nπ). The radius of the n-th zone at the obstacle is:

rn = √[ nλ·d1·d2 / (d1 + d2) ]

• Odd zones add to the direct signal, even zones cancel it.
• If the obstacle does not block the first Fresnel zone (about 55% clearance is enough), diffraction loss is very small. As blockage increases (v increases), diffraction loss increases; at v = 0 (edge on the LOS line) the loss is 6 dB.

Hata model (urban):

L50 = 69.55 + 26.16 log fc − 13.82 log hte − a(hre)
      + (44.9 − 6.55 log hte) log d
Medium city:
a(hre) = (1.1 log fc − 0.7)hre − (1.56 log fc − 0.8) dB

Given: fc = 950 MHz, hte = 45 m, hre = 5 m, d = 10 km. All values are inside Hata's validity range (150–1500 MHz, 30–200 m, 1–10 m, 1–20 km).

Correction factor

log 950 = 2.9777
a(hre) = (1.1×2.9777 − 0.7)×5 − (1.56×2.9777 − 0.8)
       = (2.5755)×5 − 3.8452
       = 12.877 − 3.845
       = 9.03 dB

Hata path loss

26.16 log 950 = 77.90
13.82 log 45  = 22.85
44.9 − 6.55 log 45 = 44.9 − 10.83 = 34.07
log 10 = 1

L50 = 69.55 + 77.90 − 22.85 − 9.03 + 34.07×1
    = 149.64 dB

Free space path loss

LFS = 32.44 + 20 log f(MHz) + 20 log d(km)
    = 32.44 + 59.55 + 20
    = 112.00 dB

Comparison

L50 − LFS = 149.64 − 112.00 = 37.64 dB

Answer: a(hre) = 9.03 dB, Hata path loss = 149.64 dB, free space loss = 112.0 dB. The Hata loss is about 37.6 dB higher, because a real city has buildings, ground reflection, diffraction and scattering that free space ignores; also loss in Hata rises at (44.9 − 6.55 log hte) ≈ 34 dB/decade, compared with 20 dB/decade in free space.

Frequency selective fading

Frequency selective fading occurs when the channel has a constant gain and linear phase over a bandwidth smaller than the signal bandwidth, i.e. when

Bs > Bc   and   Ts < στ

Different frequency components of the signal fade by different amounts, so the received signal is distorted. In time, multiple delayed copies of each symbol overlap the next symbols, causing inter-symbol interference (ISI). It needs equalizers, RAKE receivers or OFDM. A common rule: the channel is frequency selective if Ts < 10στ.

Rayleigh distribution

In a mobile channel with many scattered paths and no dominant (LOS) path, the in-phase and quadrature components of the received signal are independent Gaussian random variables with zero mean and variance σ². The envelope r = √(I² + Q²) then follows the Rayleigh distribution:

p(r) = (r/σ²)·exp(−r²/(2σ²)),  0 ≤ r ≤ ∞
p(r) = 0,                       r < 0

• σ = rms value of received voltage before envelope detection; σ² = time-average power.
• CDF: P(R ≤ r) = 1 − exp(−r²/(2σ²))

p(r)
 |   .-.
 |  /   '.
 | /      '.
 |/         '--.____
 +-----σ--------------- r
   (peak at r = σ)

The phase is uniformly distributed over 0 to 2π. Rayleigh fading is used to model flat fading in dense urban areas. When a strong LOS path exists, the envelope follows the Ricean distribution instead.

Doppler effect in small-scale fading

The Doppler effect is the apparent change in the frequency of a received signal due to relative motion between transmitter and receiver. A mobile moving at speed v receives a wave arriving at angle θ to its direction of motion with a frequency shift

fd = (v/λ)·cosθ

In a multipath channel, waves arrive from many angles, so each has a different Doppler shift. The received spectrum spreads over fc ± fm (fm = v/λ). This Doppler spread causes random frequency modulation and makes the channel time-varying; it decides whether fading is fast or slow (through coherence time Tc ≈ 0.423/fm).

Free space path loss of the cordless phone

Given: f = 52 000 kHz = 52 MHz, d = 50 m. Assuming unity gain antennas:

λ = c/f = 3×10⁸ / 52×10⁶ = 5.769 m

PL = 20 log(4πd/λ)
   = 20 log(4π × 50 / 5.769)
   = 20 log(108.91)
   = 40.74 dB
(check: 32.44 + 20 log 52 + 20 log 0.05
       = 32.44 + 34.32 − 26.02 = 40.74 dB)

Answer: Free space path loss ≈ 40.7 dB.

Frequency selective fading

Frequency selective fading occurs when the bandwidth of the transmitted signal is greater than the coherence bandwidth of the channel (Bs > Bc), or equivalently the symbol period is less than the rms delay spread (Ts < στ). Different frequency components fade differently; in time, the received signal contains multiple delayed, attenuated copies of each symbol, causing ISI and distortion.

Relationship between coherence bandwidth and delay spread

Delay spread describes the time dispersion of the channel; coherence bandwidth describes the same effect in the frequency domain. They are inversely related because the channel's frequency response is the Fourier transform of its impulse response.

Time domain parameters (from power delay profile P(τ)):

Mean excess delay  τ̄  = Σ P(τk)τk / Σ P(τk)
Second moment      τ²̄ = Σ P(τk)τk² / Σ P(τk)
RMS delay spread   στ = √(τ²̄ − τ̄²)

Coherence bandwidth: the range of frequencies over which two frequency components have strong amplitude correlation.

Bc ≈ 1/(50στ)   (correlation above 0.9)
Bc ≈ 1/(5στ)    (correlation above 0.5)

 Time domain               Frequency domain
 P(τ)                      |H(f)|
 |  |                      |~~~~~~~~~~~  small στ
 |  | |                    |   → wide Bc (flat)
 +--+-+-- τ (short)        +------------- f

 |  |   |    |             |~\_/~\_/~\_  large στ
 +--+---+----+-- τ (long)  +------------- f
                              → narrow Bc (selective)

Why: if the channel has paths with delay difference Δτ, their phases differ by 2πfΔτ. As frequency changes by about 1/Δτ, the relative phase changes by 2π, so the combined gain goes through peaks and nulls. Larger delay spread gives faster variation with frequency, i.e. smaller Bc.

Consequences:

• Bs ≪ Bc (Ts ≫ στ): flat fading.
• Bs > Bc (Ts < στ): frequency selective fading, ISI.
• Maximum symbol rate without equalizer is limited by στ (e.g. στ/Ts ≤ 0.1).

Example: στ = 1.37 µs → Bc(0.5) = 1/(5×1.37 µs) = 146 kHz, Bc(0.9) = 14.6 kHz. A 30 kHz AMPS channel would be flat for 0.5 correlation but a 200 kHz GSM signal would need equalization.

The relation is approximate; the exact coherence bandwidth depends on the actual impulse response, so Bc and στ are linked only statistically.

Propagation models predict the average received signal strength at a given distance (large-scale models) or its rapid fluctuations over short distances (small-scale/fading models).

Basic propagation models

1. Free space model (Friis) Used when there is a clear, unobstructed LOS path, e.g. satellite and microwave links.

Pr(d) = Pt·Gt·Gr·λ² / [(4π)²d²L]
PL(dB) = 20 log(4πd/λ)

Power falls as 1/d² (20 dB/decade). Valid only in the far field (d ≥ df).

2. Two-ray ground reflection model Considers the direct path plus a ground-reflected path. For large d:

Pr = Pt·Gt·Gr·ht²·hr² / d⁴

Power falls as 1/d⁴ (40 dB/decade). More accurate than free space for long distances over flat ground.

3. Log-distance path loss model

PL(d) = PL(d0) + 10n log(d/d0)

n = 2 free space, 2.7–3.5 urban, 3–5 shadowed urban.

4. Log-normal shadowing Adds a zero-mean Gaussian random variable Xσ (dB) to log-distance loss to account for obstacles: PL(d) = PL(d0) + 10n log(d/d0) + Xσ.

Empirical outdoor models (Okumura, Hata) and indoor models (partition loss, Ericsson) build on these.

Far field distance

The far field (Fraunhofer region) begins at

df = 2D²/λ,  with df ≫ D and df ≫ λ

Given D = 1 m, f = 900 MHz:

λ  = 3×10⁸ / 900×10⁶ = 0.333 m
df = 2 × 1² / 0.333 = 6 m
Check: 6 m ≫ D (1 m) and 6 m ≫ λ (0.333 m)

Answer: Far field distance df = 6 m.

Fading is the rapid fluctuation of the amplitude, phase or delay of a received radio signal over a short time or travel distance, caused by interference between multiple versions of the transmitted signal (multipath) arriving at slightly different times.

Factors influencing small-scale fading

1. Multipath propagation: reflecting and scattering objects create several waves with different amplitudes, phases and delays. Their vector sum varies rapidly, and the spread of delays causes time dispersion and ISI.
2. Speed of the mobile: motion gives each multipath component a different Doppler shift, causing random frequency modulation. Faster movement means faster fading.
3. Speed of surrounding objects: moving vehicles, people and doors make the channel time-varying. If objects move faster than the mobile, they dominate; otherwise they can be ignored.
4. Transmission bandwidth of the signal: if signal bandwidth is larger than the coherence bandwidth, the signal is distorted (frequency selective); if smaller, only its amplitude fades (flat fading).

Derivation of Doppler shift

A mobile moves with constant speed v from X to Y (distance l) in time Δt, receiving from a distant source S at angle θ to the direction of motion.

             S (distant)
            /
           /   rays nearly parallel
          / θ
   X ------------- Y  →  v
   |<---- l ----->|
   Δl = l cosθ

Distance moved:       l  = v·Δt
Path difference:      Δl = l·cosθ = v·Δt·cosθ
Phase change:         Δφ = 2πΔl/λ = 2π·v·Δt·cosθ / λ
Doppler shift:        fd = (1/2π)(Δφ/Δt)
                      fd = (v/λ)·cosθ

• Towards the source (θ = 0): fd = +v/λ (frequency increases).
• Away from the source (θ = 180°): fd = −v/λ.
• Maximum Doppler shift: fm = v/λ.

Example: fc = 900 MHz (λ = 0.333 m), v = 72 km/h = 20 m/s → fm = 20/0.333 = 60 Hz.

Free space path loss (with unity gain antennas) is given by the Friis equation:

PL(dB) = 10 log(Pt/Pr) = 20 log(4πd/λ)

Given: f = 52 000 kHz = 52 MHz, d = 50 m (range of the phone).

Step 1: Wavelength

λ = c/f = 3×10⁸ / 52×10⁶ = 5.769 m

Step 2: Check far field

The phone's antenna is small (much less than λ), so the far field begins within a few wavelengths; 50 m is about 8.7λ, so the Friis equation can be used.

Step 3: Path loss

4πd/λ = 4π × 50 / 5.769 = 108.91

PL = 20 log(108.91) = 40.74 dB

Check using the practical form:

PL = 32.44 + 20 log f(MHz) + 20 log d(km)
   = 32.44 + 20 log 52 + 20 log 0.05
   = 32.44 + 34.32 − 26.02
   = 40.74 dB

Answer: Free space path loss at the 50 m range ≈ 40.7 dB (the received power is about 1/11 860 of the transmitted power for unity-gain antennas).

The log-distance path loss model states that the average received power falls off logarithmically with distance, whether indoors or outdoors. The average path loss is proportional to the n-th power of distance:

PL(d) ∝ (d/d0)ⁿ

PL(d) [dB] = PL(d0) + 10n log(d/d0)

• n = path loss exponent: how fast loss grows with distance.
• d0 = close-in reference distance in the far field (1 km for large cells, 100 m or 1 m for micro/indoor cells).
• PL(d0) = loss at d0, measured or found from free space: 20 log(4πd0/λ).

Log-normal shadowing: at the same distance, actual loss varies with surroundings, so a zero-mean Gaussian term Xσ (in dB, σ typically 6–12 dB) is added:

PL(d) = PL(d0) + 10n log(d/d0) + Xσ

Example: PL(d0 = 100 m) = 70 dB, n = 3.5 → at 1 km, PL = 70 + 35 × log 10 = 105 dB.

Okumura model: L50 = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

Given: L50 = 167 dB, fc = 2.1 GHz, hte = 40 m, hre = 2 m, large city, Amu = 34 dB, GAREA = 0 dB.

Step 1: Antenna height gain factors

G(hte) = 20 log(hte/200) = 20 log(40/200) = −13.98 dB
G(hre) = 10 log(hre/3)   = 10 log(2/3)    = −1.76 dB
         (hre ≤ 3 m, so 10 log is used)

Step 2: Free space loss required

LF = L50 − Amu + G(hte) + G(hre) + GAREA
   = 167 − 34 + (−13.98) + (−1.76) + 0
   = 117.26 dB

Step 3: Distance from free space loss

λ  = c/f = 3×10⁸ / 2.1×10⁹ = 0.1429 m
LF = 20 log(4πd/λ)

4πd/λ = 10^(117.26/20) = 7.294 × 10⁵
d = 7.294×10⁵ × λ/(4π)
  = 7.294×10⁵ × 0.011368
  = 8292 m

Answer: T–R separation d ≈ 8.29 km.

Note: 2.1 GHz is slightly above Okumura's basic 1920 MHz range, but the model is commonly extrapolated up to 3 GHz; Amu is taken as given.

Advantages of the two-ray model over free space model

• Free space assumes only a direct path; in real mobile links there is almost always a ground-reflected wave as well. The two-ray model includes it.
• It is accurate for predicting large-scale signal strength over distances of several km for systems with tall towers (above about 50 m), and also for LOS microcells in urban areas.
• It shows that beyond a break distance power falls as 1/d⁴ (40 dB/decade), which matches measurements better than the 1/d² of free space.
• It shows the effect of antenna heights (loss decreases as ht and hr increase) and the alternating peaks and nulls close to the transmitter.
• Path loss becomes independent of frequency at large d.

Derivation of path loss

  Tx
   |\  E_LOS (d')
ht | \_______________________ Rx
   |  \                    _/|
   |   \  E_g (d'')     _/   | hr
 --+----\____________/-------+-- ground
   |<----------- d --------->|

Path difference (method of images, image of Tx at −ht):

d'  = √[d² + (ht − hr)²],  d'' = √[d² + (ht + hr)²]
Δ = d'' − d' ≈ 2ht·hr/d   (for d ≫ ht + hr)
θΔ = 2πΔ/λ = 4π·ht·hr/(λd)

Total field. With free space field E0 at reference d0, LOS field E0d0/d', and ground reflection coefficient Γ ≈ −1 (grazing incidence):

ETOT = E_LOS + E_g
|ETOT| = (E0d0/d)·|1 − e^(−jθΔ)|
       = (E0d0/d)·√[(1 − cosθΔ)² + sin²θΔ]
       = (E0d0/d)·√(2 − 2cosθΔ)
       = 2(E0d0/d)·sin(θΔ/2)

For large d, θΔ/2 < 0.3 rad, so sin(θΔ/2) ≈ θΔ/2:

|ETOT| ≈ 2(E0d0/d)·(2π·ht·hr/(λd)) = k/d²

Received power. Since Pr ∝ |E|², and using the Friis equation at the reference:

Pr = Pt·Gt·Gr·ht²·hr² / d⁴

Path loss in dB:

PL(dB) = 40 log d − (10 log Gt + 10 log Gr
                     + 20 log ht + 20 log hr)

Thus at large distances, path loss rises 40 dB per decade of distance, decreases by 6 dB when either antenna height doubles, and does not depend on frequency.

On the reverse link the mobile transmits and the BTS receives. The largest allowed path loss is found from the link budget, and the distance is then found by inverting Okumura's model.

Link budget diagram

 Mobile Tx 15 dBm
   | −3 dB cable
   v +20 dBi antenna      EIRP = 32 dBm
   ~~~~~~~~~~~~~~ path loss L ~~~~~~~~~~~~~>
                     BTS antenna +5 dBi
                          | −3 dB cable
                          v
                  BTS Rx input = 34 − L dBm
   must be ≥ sensitivity (−80) + margin (10)

Step 1: Maximum allowed path loss

Tx power (mobile)          +15 dBm
Mobile cable loss           −3 dB
Mobile antenna gain        +20 dBi
BTS antenna gain            +5 dBi
BTS cable loss              −3 dB
-------------------------------------
Signal before path loss    +34 dBm
BTS sensitivity            −80 dBm
Link margin                 10 dB

Lmax = 34 − (−80) − 10 = 104 dB

Step 2: Okumura model

L50 = LF + Amu − G(hte) − G(hre) − GAREA
G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB

104 = LF + 43 + 6.02 − 10.46 − 9
LF  = 104 − 29.56 = 74.44 dB

Step 3: Distance from free space loss (900 MHz)

λ = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 20 log(4πd/λ)

4πd/λ = 10^(74.44/20) = 5270.5
d = 5270.5 × 0.3333 / (4π) = 139.8 m

Answer: The reverse link supports a maximum BTS–mobile distance of about 140 m.

Note: Amu = 43 dB and GAREA = 9 dB are curve values normally read for a 50 km, 900 MHz link; they are used as given. This makes the predicted range short. With the curve values for short distances, the real range would be larger.

To avoid an equalizer, the rms delay spread must be small compared with the symbol period: στ/Ts ≤ 0.1, i.e. Ts ≥ 10στ.

Step 1: Convert powers to linear

ΣP = 1 + 1 + 0.1 + 0.01 = 2.11

Step 2: Mean excess delay

τ̄ = Σ P·τ / ΣP
  = (1×0 + 1×50 + 0.1×75 + 0.01×100) / 2.11
  = 58.5 / 2.11 = 27.73 µs

Step 3: Second moment

τ²̄ = Σ P·τ² / ΣP
    = (0 + 2500 + 0.1×5625 + 0.01×10 000) / 2.11
    = 3162.5 / 2.11 = 1498.82 µs²

Step 4: RMS delay spread

στ = √(τ²̄ − τ̄²)
   = √(1498.82 − 27.73²)
   = √(1498.82 − 768.71)
   = √730.11 = 27.02 µs

Step 5: Smallest symbol period and greatest symbol rate

Ts ≥ στ/0.1 = 10 × 27.02 = 270.2 µs

Rs = 1/Ts = 1/270.2 µs = 3.70 × 10³ symbols/s

Answer: Ts(min) ≈ 270.2 µs, so the greatest symbol rate without an equalizer is about 3.70 ksymbols/s. (Coherence bandwidth for reference: Bc ≈ 1/(5στ) ≈ 7.4 kHz.)

A link is feasible if the received power at each end is at least the receiver sensitivity. Path loss is found from Okumura's model and then a link budget is done for both directions.

Given: d = 10 km, f = 2.4 GHz, Amu = 20 dB, GAREA = 13 dB, hte = 100 m, hre = 10 m.

Step 1: Free space loss

λ  = 3×10⁸ / 2.4×10⁹ = 0.125 m
LF = 20 log(4πd/λ) = 20 log(4π × 10 000 / 0.125)
   = 120.05 dB

Step 2: Height gains and median path loss

G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB

L50 = LF + Amu − G(hte) − G(hre) − GAREA
    = 120.05 + 20 + 6.02 − 10.46 − 13
    = 122.61 dB

Step 3: Link budget diagram

 Access point                          Client
 20 dBm Tx                           15 dBm Tx
 −3 dB cable                         −3 dB cable
 +5 dBi ant  <---- 122.61 dB ---->  +20 dBi ant
 Rx sens −80 dBm                    Rx sens −75 dBm

Step 4: Forward link (AP → client)

Pr = 20 − 3 + 5 − 122.61 + 20 − 3
   = −83.61 dBm
Sensitivity = −75 dBm
Margin = −83.61 − (−75) = −8.61 dB  (fails)

Step 5: Reverse link (client → AP)

Pr = 15 − 3 + 20 − 122.61 + 5 − 3
   = −88.61 dBm
Sensitivity = −80 dBm
Margin = −88.61 − (−80) = −8.61 dB  (fails)

Answer: The 10 km link is NOT feasible; both directions fall about 8.6 dB short of the receiver sensitivity (and there is no fade margin). It could be made feasible by using higher-gain antennas (e.g. a 15 dBi or higher antenna at the AP), lower-loss cables, higher transmit power within limits, or a shorter distance.

Scattering

Scattering occurs when a radio wave meets an object whose size is comparable to or smaller than the wavelength, or a rough surface (foliage, street signs, lamp posts, rough ground), and the energy is re-radiated in many directions. It explains why received signal is often stronger than reflection and diffraction models predict. A surface is treated as rough when its height variation exceeds the critical height hc = λ/(8 sin θi) (Rayleigh criterion).

Two-ray ground reflection model

  Tx
   |\  LOS path d'
ht | \_______________________ Rx
   |  \                    _/|
   |   \  reflected d''  _/  | hr
 --+----\____________/-------+-- ground
   |<----------- d --------->|

1. Field expressions. With E0 the field at reference d0, the LOS and ground-reflected fields are:

E_LOS = (E0d0/d')·cos[ωc(t − d'/c)]
E_g   = Γ·(E0d0/d'')·cos[ωc(t − d''/c)]

For grazing incidence, Γ ≈ −1.

2. Path difference (method of images):

d'  = √[(ht − hr)² + d²] ≈ d + (ht − hr)²/(2d)
d'' = √[(ht + hr)² + d²] ≈ d + (ht + hr)²/(2d)
Δ = d'' − d' ≈ 2ht·hr/d

3. Phase difference and delay:

θΔ = 2πΔ/λ = Δ·ωc/c
τd = Δ/c

4. Total field. Taking d' ≈ d'' ≈ d in the amplitudes and Γ = −1:

|ETOT| = (E0d0/d)·|1 − e^(−jθΔ)|
       = (E0d0/d)·√(2 − 2cosθΔ)
       = 2(E0d0/d)·sin(θΔ/2)

5. Large-distance approximation. When θΔ/2 < 0.3 rad (i.e. d > 20πht·hr/(3λ) ≈ 20ht·hr/λ), sin(θΔ/2) ≈ θΔ/2:

|ETOT| ≈ (2E0d0/d)·(2π·ht·hr/(λd)) ≈ k/d²   V/m

6. Received power and path loss. Since power ∝ |E|²:

Pr = Pt·Gt·Gr·ht²·hr² / d⁴

PL(dB) = 40 log d − (10 log Gt + 10 log Gr
                     + 20 log ht + 20 log hr)

So at large distances received power falls at 40 dB/decade, compared with 20 dB/decade in free space, and the path loss is independent of frequency.

Okumura model

The Okumura model is an empirical model based on extensive measurements in and around Tokyo. It is one of the most widely used models for macrocell planning in urban areas.

L50(dB) = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

• LF = free space loss.
• Amu(f,d) = median attenuation relative to free space in an urban area over quasi-smooth terrain, read from curves (reference hte = 200 m, hre = 3 m).
• G(hte) = BS antenna height gain = 20 log(hte/200), 30 m < hte < 1000 m.
• G(hre) = mobile antenna height gain = 10 log(hre/3) for hre ≤ 3 m; 20 log(hre/3) for 3 m < hre < 10 m.
• GAREA = gain due to environment (suburban, quasi-open, open), from curves.
• Valid for 150–1920 MHz (extendable to 3 GHz), d = 1–100 km.
• Merits: simple, accurate for urban/suburban cellular systems (about 10–14 dB standard deviation).
• Demerits: slow to respond to rapid terrain changes; works better in urban than rural areas; curves are hard to use in software (hence the Hata formula).

Calculation

Given: d = 50 km, hte = 100 m, hre = 10 m, f = 900 MHz, Amu = 43 dB, GAREA = 9 dB.

λ  = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 20 log(4π × 50 000 / 0.3333) = 125.51 dB

G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB

L50 = 125.51 + 43 − (−6.02) − 10.46 − 9
    = 155.07 dB

Answer: Median path loss L50 ≈ 155.07 dB.

Doppler spread

Doppler spread (BD) is the measure of spectral broadening caused by the time rate of change of the mobile channel. When a pure tone fc is sent, the received spectrum (Doppler spectrum) extends from fc − fm to fc + fm, because multipath waves arrive from different angles θ and each gets a Doppler shift fd = (v/λ)cosθ. BD is taken as the maximum Doppler shift:

BD = fm = v/λ

If the signal bandwidth is much larger than BD, Doppler spread has negligible effect (slow fading). Its time-domain dual is coherence time Tc ≈ 0.423/fm.

Coherence bandwidth

Coherence bandwidth (Bc) is the range of frequencies over which the channel can be considered flat, i.e. two frequency components have strongly correlated amplitudes. It is derived from rms delay spread στ:

Bc ≈ 1/(50στ)  (0.9 correlation)
Bc ≈ 1/(5στ)   (0.5 correlation)

Classification of fading

Based on rms delay spread (multipath time dispersion):

1. Flat fading: Ts ≫ στ (Bs ≪ Bc). All frequency components fade equally; no ISI. Common rule: flat if Ts ≥ 10στ.
2. Frequency selective fading: Ts < στ (Bs > Bc). Received signal has multiple delayed copies, causing ISI; equalization needed.

Based on coherence time (Doppler spread): 3. Fast fading: Ts > Tc (Bs < BD). Channel changes within a symbol; causes signal distortion. 4. Slow fading: Ts ≪ Tc (Bs ≫ BD). Channel stays constant over many symbols.

 Ts (symbol period)
  ^
  | Flat-slow   | Flat-fast
10στ+-----------+-----------
  | Freq.sel.-  | Freq.sel.-
  |   slow      |   fast
  +-------------+----------> Ts
               Tc

The two classifications are independent: e.g. a channel can be flat and slow fading at the same time.

The maximum cell radius is the distance at which Hata path loss equals the largest loss the link can tolerate. Antenna gains and cable losses are not given, so they are taken as 0 dB.

Step 1: Maximum allowed path loss

Pt = 50 W = 10 log(50 000 mW) = 46.99 dBm
Pr,min = −91 dBm

Lmax = Pt − Pr,min = 46.99 − (−91) = 137.99 dB

Step 2: Mobile antenna correction (medium city)

fc = 900 MHz, log 900 = 2.9542, hre = 1 m:

a(hre) = (1.1 log fc − 0.7)hre − (1.56 log fc − 0.8)
       = (3.2497 − 0.7)×1 − (4.6086 − 0.8)
       = 2.5497 − 3.8086
       = −1.26 dB

Step 3: Hata equation (urban)

hte = 30 m, log 30 = 1.4771:

L50 = 69.55 + 26.16 log fc − 13.82 log hte − a(hre)
      + (44.9 − 6.55 log hte) log d
    = 69.55 + 77.28 − 20.41 + 1.26
      + (44.9 − 9.68) log d
    = 127.68 + 35.22 log d

Step 4: Solve for d

127.68 + 35.22 log d = 137.99
log d = 10.31 / 35.22 = 0.2927
d = 10^0.2927 = 1.96 km

Step 5: Cell area

For a hexagonal cell of radius R:

A = (3√3/2)·R² = 2.598 × 1.962² = 10.0 km²

(If a circular cell is assumed, A = πR² = 12.1 km².)

Answer: Maximum cell radius ≈ 1.96 km; hexagonal cell area ≈ 10.0 km² (≈ 12.1 km² if circular). All parameters are within Hata's range (hte 30–200 m, hre 1–10 m, d 1–20 km).

Large-scale vs small-scale propagation models

Large-scale models predict the mean received signal strength at a given T–R distance; small-scale (fading) models describe the rapid fluctuations of signal strength over very short distances or times.

Propagation mechanisms in a mobile environment

1. Reflection

• Occurs when a wave hits an object with dimensions much larger than λ, e.g. the earth's surface, buildings and walls.
• Part of the wave is reflected and part transmitted, depending on the Fresnel reflection coefficient Γ, which depends on material, polarization, angle and frequency.
• Ground reflection gives the two-ray model, in which loss rises at 40 dB/decade. Reflections from many buildings cause multipath.

2. Diffraction

• Occurs when the path is blocked by a surface with sharp edges (hills, rooftops).
• Secondary wavelets (Huygens' principle) bend around the obstacle and reach the shadow region, so a signal is received even without LOS.
• Loss depends on how much of the first Fresnel zone is blocked; modelled by knife-edge diffraction with parameter v.

3. Scattering

• Occurs when the wave meets objects smaller than or comparable to λ, or rough surfaces (foliage, lamp posts, signs).
• Energy is spread in all directions, giving additional received energy not predicted by reflection and diffraction alone.
• Roughness is judged by the critical height hc = λ/(8 sin θi).

           diffraction at roof edge
  BS  ----->  _/|_____         scattering (tree)
   \          |  bldg |  ~~~>   *
    \ reflection      |        / \
     \______ground_____\_____ MS

Other effects: absorption/penetration loss through walls and foliage, and waveguiding along streets. Together these produce path loss, shadowing and multipath fading.

Using the link equation in dB with free space loss:

Pr = Pt − Lcable + Gt − LFS + Gr − Lmismatch
LFS = 20 log(4πd/λ) = 32.44 + 20 log f(MHz) + 20 log d(km)

Given: Pt = 10 W, f = 250 MHz, cable 20 m at 3 dB/100 m, Gt = 9 dBi, d = 25 km, Gr = 4 dBi, receiver feeder loss negligible, mismatch loss 0.2 dB.

Step 1: Transmit power in dBm

Pt = 10 log(10 W / 1 mW) = 40 dBm

Step 2: Transmitter cable loss

Lcable = 20 m × (3 dB / 100 m) = 0.6 dB

Step 3: Free space path loss

λ   = 3×10⁸ / 250×10⁶ = 1.2 m
LFS = 20 log(4π × 25 000 / 1.2)
    = 108.36 dB
(check: 32.44 + 47.96 + 27.96 = 108.36 dB)

Step 4: Power at the receiver

Pr = 40 − 0.6 + 9 − 108.36 + 4 − 0.2
   = −56.16 dBm
Pr = 10^(−56.16/10) mW = 2.42 × 10⁻⁶ mW = 2.42 nW

Answer: Power delivered to the receiver ≈ −56.2 dBm ≈ 2.42 × 10⁻⁹ W.

Mobile radio propagation is described by two effects that act on very different distance scales: large-scale path loss (including shadowing) and small-scale fading. The received signal is their combination.

 Pr (dBm)
   |\  /\/\  small-scale fading (fast ripples)
   | \/    \/\/\
   |   ----  \/\/\/\    local mean (shadowing)
   |       ----   \/\/\/\
   |  - - - - - - - - - -   mean path loss ~ 10n log d
   +----------------------------- log d

Large-scale path loss

• Describes the mean received power averaged over 5λ–40λ, as a function of T–R distance. It changes slowly over hundreds of metres.
• Distance-dependent loss: Pr ∝ 1/dⁿ; n = 2 in free space (Friis), 4 for two-ray at large d, 2.7–5 in urban areas.

PL(d) = PL(d0) + 10n log(d/d0) + Xσ

• Shadowing (log-normal fading): obstacles such as hills and buildings make the local mean vary randomly about the distance trend. Xσ is a zero-mean Gaussian in dB, with σ about 6–12 dB.
• Models: free space, two-ray, log-distance, Okumura, Hata, and indoor partition models.
• Used for coverage area, cell size, link budget and interference planning.

Small-scale fading

• Rapid change of amplitude and phase over a few wavelengths (fractions of a second), caused by multipath waves adding constructively and destructively. Signal may vary by 30–40 dB within half a wavelength.
• Effects: rapid amplitude changes, random frequency modulation (Doppler) and time dispersion (echoes, ISI).
• Factors: multipath, speed of the mobile, speed of surrounding objects, and signal bandwidth.
• Types: flat or frequency selective (by delay spread στ vs Ts), fast or slow (by coherence time Tc vs Ts).
• Statistics: Rayleigh envelope when there is no LOS path, Ricean when a strong LOS exists.
• Countered by diversity, equalization, interleaving and coding.

Small scale fading

Small scale fading is the rapid change in the amplitude, phase and multipath delay of a received radio signal over a short distance (a few wavelengths) or a short time (a few seconds). It happens because two or more copies of the transmitted signal (multipath waves) reach the receiver at slightly different times and with different phases, and add constructively or destructively. Large-scale path loss can be treated as constant over such a small distance.

Its three main effects are:

• Rapid changes in signal strength over a small travel distance or time interval.
• Random frequency modulation due to different Doppler shifts on different multipath signals.
• Time dispersion (echoes) caused by multipath propagation delays.

 Power (dB)
   |   /\    /\  /\     /\
   |  /  \/\/  \/  \/\_/  \   <- deep fades
   | /                     \     every ~ λ/2
   +----------------------------> distance

Factors influencing small scale fading

1. Multipath propagation – Reflecting and scattering objects (buildings, vehicles, trees) create many waves with random amplitudes, phases and arrival times. Their vector sum fluctuates strongly, and the spread of arrival times causes time dispersion and ISI.

2. Speed of the mobile – Relative motion between the base station and the mobile gives each multipath component a different Doppler shift, f_d = (v/λ) cos θ. Higher speed means a larger Doppler spread and faster fading (smaller coherence time).

3. Speed of surrounding objects – Moving cars, people and other objects in the radio channel also create time-varying Doppler shifts. If they move faster than the mobile, their effect dominates; otherwise it can be ignored.

4. Transmission bandwidth of the signal – The channel has a coherence bandwidth B_c (inversely related to its delay spread).

   • If signal bandwidth B_s < B_c, all frequency components fade together: flat fading, little distortion.
   • If B_s > B_c, different frequency components fade differently: frequency selective fading, pulse spreading and ISI.

Together, the delay spread (from multipath) and the Doppler spread (from motion) decide whether the channel is flat or frequency selective, and fast or slow fading.

Given: d = 5 km, f = 900 MHz, G_r = 2.55 dB, E₀ = 10⁻³ V/m at d₀ = 1 km, h_t = 50 m, h_r = 1.5 m.

a) Length and effective aperture of the antenna

Wavelength:

λ = c / f = (3 × 10⁸) / (900 × 10⁶) = 0.3333 m

Length of a quarter-wave monopole:

L = λ / 4 = 0.3333 / 4 = 0.0833 m  (8.33 cm)

Gain in linear form:

G = 10^(2.55/10) = 1.799

Effective aperture:

A_e = G λ² / (4π)
    = 1.799 × (0.3333)² / (4π)
    = 1.799 × 0.1111 / 12.566
    = 0.0159 m²

b) Received power using two-ray ground reflection model

First check that the far-distance approximation is valid:

20 h_t h_r / λ = 20 × 50 × 1.5 / 0.3333 = 4500 m
d = 5000 m > 4500 m  → approximation valid

For large d, the total received field is:

E_TOT(d) ≈ (2 E₀ d₀ / d) × (2π h_t h_r / (λ d))

Substitute:

2 E₀ d₀ / d        = 2 × 10⁻³ × 1000 / 5000 = 4 × 10⁻⁴ V/m
2π h_t h_r / (λ d) = 2π × 50 × 1.5 / (0.3333 × 5000)
                   = 471.24 / 1666.7 = 0.2827
E_TOT              = 4 × 10⁻⁴ × 0.2827
                   = 1.131 × 10⁻⁴ V/m = 113.1 µV/m

Received power (power density × effective aperture), with free-space impedance 120π Ω:

P_r = (|E|² / 120π) × A_e
    = ((1.131 × 10⁻⁴)² / 376.99) × 0.0159
    = (3.393 × 10⁻¹¹) × 0.0159
    = 5.40 × 10⁻¹³ W

In decibels:

P_r = 10 log(5.40 × 10⁻¹³) = −122.68 dBW = −92.68 dBm

Answer: L = 0.0833 m, A_e = 0.0159 m², E = 113.1 µV/m and P_r ≈ 5.40 × 10⁻¹³ W (−122.68 dBW, or −92.68 dBm).

Path loss vs fading

Path loss is the average drop in signal power as the distance from the transmitter grows. Fading is the quick change of the received signal around that average, caused by multipath and motion.

Time dispersion fading

Time dispersion happens because multipath copies of a symbol arrive at different delays, so the received pulse is spread out in time. It is described by the rms delay spread σ_τ in time, or the coherence bandwidth B_c ≈ 1/(5σ_τ) in frequency. Depending on how the signal compares with these, there are two types.

1. Flat fading

• Condition: B_s ≪ B_c and T_s ≫ σ_τ.
• The channel has constant gain and linear phase over the full signal bandwidth, so all spectral parts of the signal fade together.
• The shape of the spectrum is kept but the received amplitude goes up and down (deep fades). Little or no ISI.
• Amplitude usually follows a Rayleigh distribution. Called a narrowband channel.
• Example: AMPS 30 kHz channel in a typical urban area.

2. Frequency selective fading

• Condition: B_s > B_c and T_s < σ_τ (a common rule is T_s < 10σ_τ).
• Different frequency components of the signal get different gains and phases.
• Received signal contains several delayed, attenuated versions of the symbol, so the pulse is spread in time and causes inter-symbol interference (ISI).
• Called a wideband channel; needs equalizers, RAKE receivers or OFDM.

 Flat fading           Freq. selective fading
 |H(f)|                |H(f)|
 |~~~~~~~~~~~~~        | /\    /\    /\
 |  [signal]           |/  \/\/ [signal]\
 +------------- f      +------------------ f
  B_s << B_c             B_s > B_c

Rayleigh fading happens when there is no line-of-sight (LOS) path and the received signal is the sum of many scattered waves of similar strength. Rician fading happens when there is one strong dominant (LOS) component plus many weak scattered waves.

Here A is the peak amplitude of the dominant signal and I₀ is the modified Bessel function of the first kind and zero order. When A → 0 (K → 0 or −∞ dB) the Rician distribution reduces to Rayleigh.

Step 1 – Convert powers to linear values

P (dB):    0   −10    −20    −23
P (lin):   1   0.1    0.01   0.00501
τ (ns):    0   100    200    400
ΣP  =  1 + 0.1 + 0.01 + 0.00501 = 1.11501

Step 2 – Mean excess delay

τ̄ = Σ P(τ_k) τ_k / Σ P(τ_k)
  = (1×0 + 0.1×100 + 0.01×200 + 0.00501×400) / 1.11501
  = (0 + 10 + 2 + 2.005) / 1.11501
  = 14.005 / 1.11501 = 12.56 ns

Step 3 – Second moment

τ̄² (mean square) = Σ P τ² / Σ P
  = (0.1×100² + 0.01×200² + 0.00501×400²) / 1.11501
  = (1000 + 400 + 801.9) / 1.11501
  = 1974.8 ns²

Step 4 – RMS delay spread

σ_τ = √(1974.8 − 12.56²) = √(1974.8 − 157.8)
    = √1817.0 = 42.63 ns

Step 5 – Coherence bandwidth (90% correlation)

B_c(90%) ≈ 1 / (50 σ_τ) = 1 / (50 × 42.63 × 10⁻⁹)
         = 469.2 kHz

(For reference, the 50% coherence bandwidth is 1/(5σ_τ) ≈ 4.69 MHz.)

Step 6 – Decide the type of fading A channel is flat for a system when its signal bandwidth B_s is smaller than B_c.

Answer: σ_τ = 42.63 ns, B_c(90%) ≈ 469.2 kHz; the channel is flat fading for both AMPS (30 kHz) and GSM (200 kHz). GSM uses about 43% of B_c, so it is close to the limit and still uses an equalizer for more dispersive channels.

The parameters of a mobile multipath channel describe how the channel spreads the signal in time (because of multipath delays) and in frequency (because of motion). They are obtained from the power delay profile (PDP) P(τ) and the Doppler spectrum.

 P(τ) dB
  |  |
  |  |  |
  |  |  |     |
  |  |  |  |  |   |  <- threshold (e.g. −30 dB)
  +--+--+--+--+---+----> τ
     τ0 τ1 τ2 ...

1. Time dispersion parameters

• Mean excess delay (τ̄) – the first moment of the PDP: τ̄ = Σ a_k² τ_k / Σ a_k² = Σ P(τ_k) τ_k / Σ P(τ_k).
• RMS delay spread (σ_τ) – square root of the second central moment: σ_τ = √(τ̄² − (τ̄)²), where τ̄² = Σ P(τ_k) τ_k² / Σ P(τ_k). It is the most useful single measure of how much a symbol is spread. Typical values: ns indoors, µs outdoors.
• Maximum excess delay (X dB) – the time after the first arrival in which the multipath energy falls X dB below the maximum: τ_X − τ₀.

2. Coherence bandwidth (B_c)

The range of frequencies over which the channel is "flat" (passes all components with nearly equal gain and linear phase). It is inversely proportional to σ_τ:

• B_c ≈ 1/(50 σ_τ) for frequency correlation above 0.9.
• B_c ≈ 1/(5 σ_τ) for correlation above 0.5.

3. Doppler spread (B_D)

The width of the received spectrum when a pure tone f_c is sent. Due to motion, the spectrum spreads from f_c − f_m to f_c + f_m, where f_m = v/λ is the maximum Doppler shift. If the signal bandwidth is much larger than B_D, the Doppler effect is negligible.

4. Coherence time (T_c)

The time over which the channel impulse response stays nearly unchanged; it is the time-domain dual of Doppler spread.

• T_c ≈ 9/(16π f_m) (correlation 0.5)
• Common rule of thumb: T_c = √(9/(16π f_m²)) = 0.423 / f_m.

Summary

Example: σ_τ = 1 µs gives B_c(50%) ≈ 200 kHz; a 200 kHz GSM signal would then see some frequency selectivity.

Indoor propagation differs from outdoor because distances are short, the transmitter power is low, and walls, floors, furniture and people strongly affect the signal. Indoor channels are classified as line-of-sight (LOS) or obstructed (OBS). Two widely used models are described below.

1. Partition losses (same floor) model

Buildings have many internal partitions (walls, office cubicles, doors). Each partition between transmitter and receiver adds a measured loss.

• Hard partitions are part of the building structure (brick, concrete walls).
• Soft partitions are movable and do not reach the ceiling (cubicle walls, cloth office dividers).

Path loss is the free-space or reference loss plus the sum of losses of all partitions crossed:

PL(d) = PL(d₀) + 10 n log(d/d₀) + Σ PAF  (dB)

Typical partition attenuation values (Rappaport): cloth partition ≈ 1.4 dB, double plasterboard wall ≈ 3.4 dB, concrete wall ≈ 13 dB, metal-framed partition ≈ 5 to 12 dB.

2. Partition losses between floors (floor attenuation factor) model

Losses between floors depend on the building material, the floor type and the windows. The floor attenuation factor (FAF) is the extra loss for each floor between transmitter and receiver. The first floor gives the largest loss; additional floors add less.

           Floor 3   Rx
           ─────────────   FAF ≈ 6–10 dB
           Floor 2
           ─────────────   FAF ≈ 13–16 dB
           Floor 1   Tx

Typical values: about 13 dB for one floor, 18.7 dB for two floors, 24.4 dB for three floors (office building measurements).

3. Log-distance path loss model (with variants)

Many indoor measurements fit the log-normal form:

PL(dB) = PL(d₀) + 10 n log(d/d₀) + X_σ

• n = path loss exponent depending on building and surroundings (about 1.6–1.8 in an open LOS corridor, 2 to 3 in offices, up to 4–6 in obstructed paths).
• X_σ = zero-mean Gaussian random variable (dB) with standard deviation σ, which models shadowing.

The attenuation factor model (Seidel) combines both ideas:

PL(d) = PL(d₀) + 10 n_SF log(d/d₀) + FAF + Σ PAF

where n_SF is the exponent measured on the same floor.

Use

These models are used to plan Wi-Fi access points, femtocells and in-building DAS systems, for example to decide how many access points a three-storey office needs.

Given: P_t = 150 W, receiver threshold = −104 dBm, path loss at 1 m (d₀) = 15 dB, path loss exponent n = 4.

Step 1 – Transmit power in dBm

P_t = 10 log(150 W / 1 mW) = 10 log(150000)
    = 51.76 dBm

Step 2 – Maximum allowed path loss The signal can be received while P_r ≥ −104 dBm, so:

PL_max = P_t − P_r(min) = 51.76 − (−104)
       = 155.76 dB

Step 3 – Log-distance path loss model

PL(d) = PL(d₀) + 10 n log(d/d₀),   d₀ = 1 m
155.76 = 15 + 10 × 4 × log(d)
40 log(d) = 140.76
log(d) = 3.519
d = 10^3.519 = 3304 m

Answer: The radio coverage range of the base station is about 3.30 km (radius ≈ 3304 m).

Doppler spread (B_D) is the width of the received spectrum when a single pure tone of frequency f_c is transmitted over a mobile channel. Because of relative motion, each multipath wave arrives with a different Doppler shift f_d = (v/λ) cos θ, so the received spectrum spreads over f_c − f_m to f_c + f_m, where f_m = v/λ is the maximum Doppler shift. B_D measures how fast the channel changes in time.

Coherence time (T_c) is the time-domain dual of Doppler spread. It is the time duration over which the channel impulse response is nearly constant, i.e. two received signals separated by less than T_c are strongly correlated.

T_c ≈ 1 / f_m
T_c ≈ 9 / (16π f_m)      (correlation > 0.5)
T_c ≈ 0.423 / f_m        (geometric mean, common rule)

Use: If symbol period T_s < T_c (B_s > B_D), the channel is slow fading; if T_s > T_c, it is fast fading.

Example: v = 60 km/h (16.67 m/s), f = 900 MHz (λ = 0.333 m) gives f_m = 50 Hz and T_c ≈ 0.423/50 = 8.46 ms.

Given: f = 800 MHz, h_t = 30 m, h_r = 2 m, d = 10 km. Antenna gains are taken as unity (0 dB) because they are not given.

Path loss by two-ray model

For large distances (d ≫ √(h_t h_r)), the two-ray ground reflection model gives a path loss that does not depend on frequency:

PL(dB) = 40 log d − (10 log G_t + 10 log G_r
                     + 20 log h_t + 20 log h_r)

With G_t = G_r = 1:

PL = 40 log(10000) − 20 log(30) − 20 log(2)
   = 160 − 29.54 − 6.02
   = 124.44 dB

Path loss by free-space model

λ  = c / f = 3 × 10⁸ / 800 × 10⁶ = 0.375 m
PL = 20 log(4π d / λ)
   = 20 log(4π × 10000 / 0.375)
   = 20 log(335103) = 110.50 dB

(Same using PL = 32.44 + 20 log f(MHz) + 20 log d(km) = 32.44 + 58.06 + 20 = 110.50 dB.)

Comparison

Difference = 124.44 − 110.50 = 13.93 dB

Answer: Two-ray path loss = 124.44 dB, free-space path loss = 110.50 dB. The two-ray model predicts about 13.9 dB more loss, because at large distances the ground-reflected wave nearly cancels the direct wave, so power falls as 1/d⁴ instead of 1/d². The two-ray result is more realistic for a mobile link over flat ground.

Doppler spread

Doppler spread (B_D) is the range of frequencies over which the received Doppler spectrum is non-zero when a pure sinusoid is sent. Due to motion, the received spectrum occupies f_c − f_m to f_c + f_m, where f_m = v/λ is the maximum Doppler shift. Its time-domain dual is the coherence time T_c ≈ 0.423/f_m.

Types of small scale fading based on Doppler spread

Numerical (power delay profile)

From the figure: P(0) = 0 dB, P(1 µs) = −10 dB, P(2 µs) = 0 dB.

Linear powers: 1, 0.1, 1 → ΣP = 2.1

Mean excess delay

τ̄ = (1×0 + 0.1×1 + 1×2) / 2.1 = 2.1 / 2.1 = 1 µs

Second moment

τ̄² = (1×0² + 0.1×1² + 1×2²) / 2.1 = 4.1 / 2.1 = 1.952 µs²

RMS delay spread

σ_τ = √(1.952 − 1²) = √0.952 = 0.976 µs

Coherence bandwidth

B_c(90%) ≈ 1/(50 σ_τ) = 1/(50 × 0.976 µs) = 20.49 kHz
B_c(50%) ≈ 1/(5 σ_τ)  = 1/(5 × 0.976 µs)  = 204.9 kHz

Answer: τ̄ = 1 µs, σ_τ = 0.976 µs, B_c(90%) ≈ 20.5 kHz and B_c(50%) ≈ 205 kHz.

Rayleigh fading distribution

When there is no line-of-sight path, the received signal is the sum of many scattered waves with random phases. By the central limit theorem its in-phase and quadrature parts are Gaussian, so the envelope r follows the Rayleigh pdf:

p(r) = (r/σ²) exp(−r²/2σ²),   0 ≤ r ≤ ∞
     = 0,                     r < 0

• σ = rms value of the received signal before envelope detection; σ² = time-average power.
• Mean = 1.2533σ, median = 1.177σ, variance = 0.4292σ².
• Models the worst case: deep fades (20–40 dB) occur often. Typical of dense urban areas.

Ricean fading distribution

When a dominant, steady component (LOS path) exists along with random scattered waves, the envelope follows the Ricean pdf:

p(r) = (r/σ²) exp(−(r² + A²)/2σ²) I₀(Ar/σ²),
        A ≥ 0, r ≥ 0

• A = peak amplitude of the dominant signal; I₀ = modified Bessel function of first kind, zero order.
• Rician factor K = A²/(2σ²), or K(dB) = 10 log(A²/2σ²).
• As A → 0 (K → −∞ dB) the dominant path fades away and the Ricean distribution becomes Rayleigh. Large K means a nearly constant (Gaussian-like) envelope.

 p(r)   K = −∞ dB (Rayleigh)
  |   /\       K = 6 dB
  |  /  \      /\
  | /    \    /  \
  |/      \__/    \__
  +--------------------> r