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Giacinto Gelli » 4.Path loss models


Outline

Introduction to multipath propagation
Channel variability

  • Path loss/shadowing/multipath fading

Path loss models

  • Free-space model
  • Two-ray model
  • Simplified path loss model
  • Empirical models

Multipath propagation

Many wireless systems operate in urban areas where often a direct path (LOS, line-of-sight) between TX and RX does not exist.

Many wireless systems operate in urban areas where often a direct path (LOS, line-of-sight) between TX and RX does not exist.


Propagation mechanisms

Reflection/transmission (red):

  • the signal impinges on objects whose dimensions are larger than λ

Diffraction (green):

  • the signal impinges on objects with sharp edges

Scattering (blue):

  • the signal impinges on several objects whose dimensions are smaller than or comparable to λ

Wireless channel variability

Large scale (blue):

  • (100-1000 m)
  • path loss (includes average shadowing)

Medium scale (red):

  • (10-100 m outdoor, < 10 m indoor)
  • shadowing (due to obstructions)

Small scale (green):

  • (~ wavelength)
  • multipath fading

Path loss

The path loss (PL) is defined as

P_L = \frac{P_t}{P_r} > 1

Pt = transmitted power

Pr = received power

Often expressed in dB units:

P_L [dB] = 10 \, \log_{10} \left( \frac{P_t}{P_r} \right) > 0


The dBm and dBW “units”

A common practice in telecommunications engineering is to express the power P with respect to a reference power P0:

P [dB] = 10 \, \log_{10} \left( \frac{P}{P_0} \right)

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:

\[P_{r}[\mbox{dBm}]=P_{t}[\mbox{dBm}]-P_{L}[\mbox{dB}]\]\[P_{r}[\mbox{dBW}]=P_{t}[\mbox{dBW}]-P_{L}[\mbox{dB}]\]

Path loss models

Maxwell equations

  • Complex and impractical

Free space path loss model

  • Too simple, inadequate for most terrestrial communications

Ray tracing models

  • Require site-specific information

Empirical models

  • Don’t always generalize to other environments

Simplified path loss models

  • Main characteristic: good for high-level analysis

Free-space path loss


Free-space path loss (cont’d)


Free-space path loss (cont’d)

The model is valid only for values of d >> df (far-field):

  • df = 2 D2/λ (Fraunhofer distance)
  • D largest linear dimension of the TX antenna
  • The model is not valid for d = 0

Alternative formulation:
< style=”padding-left: 60px;”>P_r = P_0 \left( \frac{d_0}{d} \right)^2

P0= power at distance d0

d0 =reference distance (≥ df )

typically d0 = 1 m (indoor), 100 m–1 km (outdoor).

Examples

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…

Example: 802.11g/a link budget

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.

Ray tracing approximation

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:

  • Geometry
  • Dielectric properties of the objects

Two-ray model

Scenario: LOS with one dominant reflection from the ground

  • simplest among ray tracing methods
  • suitable for radio systems with relatively high antennas

Two-ray model (cont’d)


Two-ray model (cont’d)


Cell dimensioning based on dc

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.


Example

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.

Simplified path loss model

P_r  = P_t \, K \, \left( \frac{d_0}{d} \right)^{\gamma}

Commonly used for high-level system design:

  • d0 reference distance (1-10 m indoors, 10-100 m outdoors)
  • K dimensionless constant (empirically determined, sometimes assumed equal to free-space attenuation at distance d0 assuming omnidirectional antennas)
  • γ path loss exponent (empirically determined, see table)
  • free-space and two-ray models can be seen as particular cases

Example

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

Example: 802.11g/a link budget

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.

Empirical path loss models

Okumura model:

  • empirically based (site/frequency specific)
  • difficult to use (based on graphics)

Hata model

  • analytical approximation to Okumura model

COST 231 model:

  • extends Hata model to higher frequency (2 GHz)

Multi-slope and dual-slope models.
Indoor channel models.

Okumura model

Scenario: large urban and suburban macrocells

  • distance: 1-100 km
  • frequency range: 150-1920 MHz (typically extrapolated up to 3000 MHz)
  • antenna height 30-1000 m (base), 3-10 m (mobile)

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).

Okumura model (cont’d)

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 )


Okumura model (cont’d)

G(h_t) = 20 \, \log_{10} \left( \frac{h_t}{200} \right)\quad 30 \text{m} < h_t < 1000 \text{m}

G(h_r) = \left\{\begin{array}{ll}10 \, \log_{10} \left( \frac{h_r}{3} \right) & h_r \le 3 \text{m} \\ 20 \, \log_{10} \left( \frac{h_r}{3} \right) & 3 \text{m} < h_r \le 10 \text{m} \end{array}
[ht] = m, [hr] = m

Hata model

Numerical approximation of the Okumura model (restricted parameter range)

  • Distance: 1-20 km
  • Frequency range: 150-1500 MHz
  • Antenna height 30-200 m (base), 1-10 m (mobile)

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)

Hata model (cont’d)

For small to medium-sized cities:

\[G(h_t)=20\,\log_{10} \left( \frac{h_t}{200}\right)\:,\quad 30\,\mbox{m}< h_t<1000\,\mbox{m}\]\[G(h_r)=\left\{\begin{array}{ll}10\,\log_{10}\left(\frac{h_r}{3}\right)\:,&h_r\le 3 \,\mbox{m}\\\\20\,\log_{10}\left(\frac{h_r}{3}\right)\:,& 3\,\mbox{m}< h_r \le 10\,\mbox{m}\end{array}\right.\]

For large cities:

\[a(h_r) =\left\{\begin{array}{ll}8.29[\log_{10}(1.54 h_r)]^2 ñ 1.1\,\mbox{dB} & f_c\le 300\,\mbox{MHz}\\\\3.2 [ \log_{10}(11.75 h_r)]^2 ñ 4.97 \,\mbox{dB} & f_c > 300 \,\mbox{MHz} \\\end{array}\right.\]

[hr] = m, [fc] = MHz

Hata model (cont’d)

The Hata model can be extended to suburban and rural propagation:

P_{L, \text{suburban}}(d) \,\text{dB} = P_{L,\text{urban}}(d) \,\text{dB} - 2 [ \log_{10} (f_c/28)]^2 - 5.4

[fc] = MHz

P_{L, \text{rural}}(d)\,\text{dB}=P_{L,\text{urban}}(d)\,\text{dB} - 4.78 [\log_{10} (f_c]^2 +18.33\,\log_{10}(f_c)-K

[fc] = MHz

K ranges from 35.94 (countryside) to 40.94 (desert).

Example

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

COST 231 model


Multi-slope model

Scenario: outdoor and indoor microcells:

  • N-1 breakpoint distances d1, d2, ..dN-1
  • N slopes s1, s2, …sN
  • The slopes can be obtained from empirical data by linear regression

Particular case: dual-slope model (N=2).

Dual-slope model

\[P_r(d)\,\mbox{dBm}=\left\{\begin{array}{ll}P_t \,\mbox{dBm}+K-10\,\gamma_1\log_{10}(d/d_0),&d_0\led\le d_c\\\\P_t\,\mbox{dBm}+K-10\,\gamma_1\log_{10}(d_c/d_0)-10\,\gamma_2\log_{10}(d/d_c),&d>d_c\end{array}\right.\]

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)

Indoor channel models

The indoor radio channel is different from the outdoor radio channel:

  • Smaller distances
  • More channel variability

Indoor models must also take into account attenuations across floors and partitions.

Indoor channel models (cont’d)

P_r \, \text{dBm} = P_t \, \text{dBm}-P_L(d) - \sum_{i=1}^{N_f} \text{FAF}_i - \sum_{i=1}^{N_p} \text{PAF}_i

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).

Conclusions

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:

P_r  = P_t \, K \, \left( \frac{d_0}{d} \right)^{\gamma}


I materiali di supporto della lezione

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

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