Introduction to multipath propagation
Channel variability
Path loss models
Many wireless systems operate in urban areas where often a direct path (LOS, line-of-sight) between TX and RX does not exist.
Reflection/transmission (red):
Diffraction (green):
Scattering (blue):
Large scale (blue):
Medium scale (red):
Small scale (green):
The path loss (PL) is defined as
Pt = transmitted power
Pr = received power
Often expressed in dB units:
A common practice in telecommunications engineering is to express the power P with respect to a reference power P0:
P0 = 1 mW → dBm “units”
P0 = 1 W → dBW “units”
Note that dBm and dBW are actually dimensionless units (like radians).
Using dBm/dBW units in the path loss model one has:
Maxwell equations
Free space path loss model
Ray tracing models
Empirical models
Simplified path loss models
The model is valid only for values of d >> df (far-field):
Alternative formulation:
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P0= power at distance d0
d0 =reference distance (≥ df )
typically d0 = 1 m (indoor), 100 m–1 km (outdoor).
Consider an indoor wireless LAN with fc = 900 MHz, cells of radius 10 m and nondirectional antennas. Under the free-space path loss model, what transmit power is required at the access point in order for all terminals within the cell to receive a minimum power of 10 μW? How does this change if the system frequency is 5 GHz?
Answers: Pt = 1.45 W @ 900 MHz; Pt = 43.9 W @ 5 GHz.
Under the free-space path-loss model, find the transmit power required to obtain a received power of 1 dBm for a wireless system with isotropic antennas (Gl = 1) and a carrier frequency fc = 5GHz, assuming a distance d=10m. Repeat for d = 100m.
Answers: Pt = 5.53 kW @ 10 m; Pt = 553 kW @ 100 m!
More realistic values in the following example…
Consider an indoor wireless IEEE 802.11g LAN with fc = 2.4 GHz, RX sensitivity -82 dBm and TX output power 15 dBm, using nondirectional antennas. Under the free-space path loss model, find the maximum distance between TX and RX.
How does this change for an 802.11a LAN with fc = 5 GHz (assume the same RX sensitivity and TX output power)?
Answers: d = 704 m @ 2.4 GHz; d = 338 m @ 5 GHz.
Represents wavefronts as “light rays”.
Geometry determines received signal from each signal component.
Typically includes reflected rays, can also include scattered and diffracted rays.
Requires site parameters:
Scenario: LOS with one dominant reflection from the ground
For d < dc , Pr falls off ∝ 1/d2
For d > dc , Pr falls off ∝ 1/d4
dc is a natural choice for the cell radius to minimize interference to users outside the cell.
In practical cases, dc tends to be too large, which adversely affects system capacity.
Determine the critical distance for the two-ray model in an urban microcell (ht = 10 m, hr = 3 m) and an indoor microcell (ht = 3m, hr = 2m) for fc = 2 GHz.
Answer: dc = 800 m (urban); dc = 160 m (indoor).
Find the critical distance dc under the two-ray model for a large macrocell in a suburban area with the base station mounted on a tower or building (ht = 20m) , the receiver at height hr = 3m, and fc = 2 GHz. Is this a good size for cell radius in a suburban macrocell? Why or why not?
Answer: dc = 1600 m; good size for a suburban area.
Commonly used for high-level system design:
Consider a receiver with noise power -160 dBm within the signal bandwidth of interest. Assume a simplified path-loss model with d0 = 1 m, K obtained from the free-space path-loss formula with omnidirectional antennas and fc = 1 GHz, and γ = 4.
For a transmit power of Pt = 10 mW, find the maximum distance between the transmitter and receiver such that the received signal-to-noise power ratio is 20 dB.
Answer: d = 869 m
Consider an indoor wireless IEEE 802.11g LAN with fc = 2.4 GHz, RX sensitivity -82 dBm and TX output power 15 dBm, using nondirectional antennas. Assume a simplified path-loss model with d0 = 1 m, K obtained from the free-space path-loss formula with omnidirectional antennas, and γ = 3. Find the maximum distance between TX and RX.
How does this change for an 802.11a LAN with fc = 5 GHz (assume the same RX sensitivity and TX output power)?
Answers: d = 79 m @ 2.4 GHz; d = 49 m @ 5 GHz.
Okumura model:
Hata model
COST 231 model:
Multi-slope and dual-slope models.
Indoor channel models.
Scenario: large urban and suburban macrocells
Based on extensive measurements in Tokyo and its suburban areas (1966).
Not particularly accurate (difference between predicted and real values of the order of 10-14 dB).
Four steps:
a) calculate free-space path loss at the considered distance and carrier frequency
b) add median attenuation at the considered distance and carrier frequency
c) subtract the TX and RX antenna gains (see following formulas)
d) subtract the gain due to the specific environment.The values of Aμ(fc, d) and GAREA are obtained from
Okumura empirical plots (see Rappaport book )
[ht] = m, [hr] = m
Numerical approximation of the Okumura model (restricted parameter range)
Standard formula for path loss in urban areas:
PL, urban(d)dB = 69.55 + 26.16 log10(fc) – 13.82log10(ht)
– a(hr) + [44.9 – 6.55 log10(ht)] log10(d)
[ht] = m , [fc] = MHz, [d] = km
a(hr) correction factor for the mobile antenna height based on the size of the coverage area (see following slide)
For small to medium-sized cities:
For large cities:
[hr] = m, [fc] = MHz
The Hata model can be extended to suburban and rural propagation:
[fc] = MHz
[fc] = MHz
K ranges from 35.94 (countryside) to 40.94 (desert).
Find the value of the path loss using the Okumura model for d = 50 km, ht = 100m, hr = 10 m in a suburban environment. If the base station transmitter radiates an EIRP at 1 kW at a carrier frequency of 900 MHz, find the power at the receiver (assume a unity gain receiving antenna).
Repeat using the Hata model.
Note: EIRP = Pt * Gt
Answer: PL = 155.04 dB; Pr = -95.04 dBm [Okumura]; PL = 154.54 dB; Pr = -94.54 dBm [Hata]
Using the Hata model, determine the path loss of a 900 MHz cellular system operating in a large city from a base station with the height of 100 m and mobile station installed in a vehicle with antenna height of 2 m. The distance between the mobile and the base station is 4 km.
Answer: PL = 137.29 dB
Scenario: outdoor and indoor microcells:
Particular case: dual-slope model (N=2).
Source: Stanford University
The two-ray model can be approximated by the dual-slope model, with one breakpoint at the critical distance dc and attenuation slopes s1=20 dB/decade (γ1=2) and s2=40 dB/decade (γ2=4)
Source: Stanford University
The indoor radio channel is different from the outdoor radio channel:
Indoor models must also take into account attenuations across floors and partitions.
PL(d) = analytical/empirical path loss at distance d
FAF = Floor Attenuation Factor
PAF = Partition Attenuation Factor
Nf = # of floors
Np = # of partitions
Typical values of FAF and PAF vary widely with the materials and are reported in the literature (see table).
Path loss models greatly simplify Maxwell’s equations
Models vary in complexity and accuracy
Power falloff with distance is inversely proportional to d2 in free space, to d4 in two-ray model
Empirical models used in 2G simulations
Main characteristics of path loss captured by the simple model:
3. Current and emerging wireless systems
5. Shadowing
A. Goldsmith. Wireless Communications. Cambridge University Press, 2005 (chap. 2)
T. Rappaport. Wireless Communications. Principles and Practice. Prentice-Hall, 1996 (chap. 3)
Supplementary material eventually available on the website