path-loss and shadowing (large-scale fading) - ntuย ยท path-loss and shadowing (large-scale fading)...
TRANSCRIPT
Friis Formula
TX Antenna
EIRP=๐"๐บ"
๐
Power spatial density %&'
14๐๐+
๐ด-
ร
ร
RX Antenna
โ ๐0=๐"๐บ"๐ด-4๐๐+
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Antenna Apertureโข Antenna Aperture=Effective Area
โข Isotropic Antennaโs effective area ๐จ๐,๐๐๐ โ๐๐
๐๐
โข Isotropic Antennaโs Gain=1
โข ๐ฎ = ๐๐ ๐๐ ๐จ๐
โข Friis Formula becomes: ๐ท๐ =๐ท๐๐ฎ๐๐ฎ๐๐๐
๐๐ ๐ ๐ = ๐ท๐๐จ๐๐จ๐๐๐๐ ๐
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๐ท๐ =๐ท๐๐ฎ๐๐ด-4๐๐+
Friis Formula
โข ๐๐
๐๐ ๐ ๐ is often referred as โFree-Space Path Lossโ (FSPL)
โข Only valid when d is in the โfar-fieldโ of the transmitting antenna
โข Far-field: when ๐ > ๐ ๐, Fraunhofer distance
โข ๐ ๐ =๐๐ซ๐
๐, and it must satisfies ๐ ๐ โซ ๐ซ and ๐ ๐ โซ ๐
โข D: Largest physical linear dimension of the antennaโข ๐: Wavelength
โข We often choose a ๐ ๐ in the far-field region, and smaller than any practical distance used in the system
โข Then we have ๐ท๐ ๐ = ๐ท๐ ๐ ๐๐ ๐๐
๐
๐ท๐ =๐ท๐๐ฎ๐๐ฎ๐๐๐
๐๐ ๐ ๐
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Received Signal after Free-Space Path Loss
๐ ๐ = ๐น๐๐ ๐ฎ๐๐ฎ๐๐๐๐ โ๐๐๐ ๐ ๐
๐๐ ๐ ๐N ๐ ๐๐๐ ๐๐๐ ๐๐๐
phase difference due to propagation distance
Free-Space Path Loss
Complex envelope
Carrier (sinusoid)
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Example: Far-field Distanceโข Find the far-field distance of an antenna with
maximum dimension of 1m and operating frequency of 900 MHz (GSM 900)
โข Ans:
โข Largest dimension of antenna: D=1m
โข Operating Frequency: f=900 MHz
โข Wavelength: ๐ = ๐๐ =
๐ร๐๐๐
๐๐๐ร๐๐๐ = ๐. ๐๐
โข ๐ ๐ =๐๐๐
๐ = ๐๐.๐๐ = ๐. ๐๐ (m)
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Example: FSPL โข If a transmitter produces 50 watts of power, express the transmit
power in units of (a) dBm and (b) dBW. โข If 50 watts is applied to a unity gain antenna with a 900 MHz
carrier frequency, find the received power in dBm at a free space distance of 100 m from the antenna. What is the received power at 10 km? Assume unity gain for the receiver antenna.
โข Ans:
โข ๐๐๐๐๐๐๐ ๐๐ = ๐๐๐ ๐ฉ๐พ = ๐๐๐๐๐ฆโข Received Power at 100m
๐ท๐(๐๐๐๐) =๐๐ร๐ร๐ร ๐ร๐๐๐
๐๐๐ร๐๐๐๐
๐๐ ร๐๐๐ ๐ = ๐. ๐ร๐๐c๐ ๐พ= โ๐๐. ๐(๐ ๐ฉ๐พ)
โข Received Power at 10km
๐ท๐ ๐๐๐๐ = ๐ท๐ ๐๐๐๐๐๐๐๐๐๐๐๐
๐
= ๐. ๐ร๐๐c๐๐ ๐พ= โ๐๐. ๐(๐ ๐ฉ๐พ)
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Zuvio 1: FSPL โข If a transmitter produces 50 watts of power, express the
transmit power in units of (a) dBm and (b) dBW.
โข If 50 watts is applied to a unity gain antenna with a 900 MHz carrier frequency, find the received power in dBm at a free space distance of 100 m from the antenna. What is the received power at 10 km? Assume unity gain for the receiver antenna.
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Two-ray Model
TX Antenna
๐
๐
๐
๐ฅ ๐ฅโฒ
RX Antenna
โ"โ0
๐บj
๐บk ๐บl
๐บm
๐+c0jo ๐ก =
๐ ๐๐4๐
๐บj๐บl๐t ๐ก exp โ๐2๐๐๐๐ +
๐ ๐บk๐บm๐t ๐ก โ ๐ exp โ๐2๐ ๐ฅ + ๐ฅ|๐
๐ฅ + ๐ฅ| exp ๐2๐๐k๐ก
Delayed since x+xโ is longer. ๐ = (๐ฅ + ๐ฅ| โ ๐)/๐
R: ground reflection coefficient (phase and amplitude change) 9
Two-ray Model: Received Power
โข ๐ท๐ = ๐ท๐๐๐๐
๐ ๐ฎ๐๐ฎ๐๐ + ๐ฎ๐๐ฎ๐ ๐๐๐ c๐๐ซ๐
๐๏ฟฝ๐|
๐
โข The above is verified by empirical results.
โข ๐ซ๐ = ๐๐ (๐ + ๐| โ ๐)/๐
โข ๐ + ๐| โ ๐ = ๐๐ + ๐๐ ๐ + ๐ ๐ โ ๐๐ โ ๐๐ ๐ + ๐ ๐
10
l
x
xโ
xโ
x
Two-ray Model: Received Power
โข When ๐ โซ ๐๐ + ๐๐, ๐ซ๐ = ๐๐ ๐๏ฟฝ๐๏ฟฝc๐๐
โ ๐๐ ๐๐๐๐๐๐
โข For asymptotically large d, ๐ + ๐| โ ๐ฅ โ ๐ , ๐ โ ๐,๐๐๐๐ โ๐ฎ๐๐ฎ๐ , ๐ โ โ๐(phase is inverted after reflection)
โข ๐ฎ๐๐ฎ๐๐
+ ๐ฎ๐๐ฎ๐ ๐๐๐ c๐๐ซ๐๐๏ฟฝ๐|
๐โ ๐ฎ๐๐ฎ๐
๐
๐๐ + ๐๐๐ โ๐๐ซ๐ ๐
โข ๐ + ๐๐๐ โ๐๐ซ๐ ๐ = ๐ โ ๐๐๐ ๐ซ๐๐+ ๐๐๐๐๐ซ๐ = ๐ โ
๐๐๐จ๐ฌ ๐ซ๐ = ๐๐๐๐๐(๐ซ๐๐) โ ๐ซ๐๐
โข ๐ท๐ โ๐ ๐ฎ๐๐ฎ๐๐๐ ๐
๐๐๐ ๐๐๐๐๐๐
๐๐๐ญ =
๐ฎ๐๐ฎ๐๐๐๐๐๐ ๐
๐๐๐ญ
11Independent of ๐ now
Indoor Attenuationโข Factors which affect the indoor path-loss:
โข Wall/floor materialsโข Room/hallway/window/open area layoutsโข Obstructing objectsโ location and materialsโข Room size/floor numbers
โข Partition Loss:
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Partition type Partition Loss (dB) for 900-1300 MHz
Floor 10-20 for the first one,6-10 per floor for the next 3,A few dB per floor afterwards.
Cloth partition 1.4
Double plasterboard wall 3.4
Foil insulation 3.9
Concrete wall 13
Aluminum siding 20.4
All metal 26
Simplified Path-Loss Modelโข Back to the simplest:
โข ๐๏ฟฝ: reference distance for the antenna far field (usually 1-10m indoors and 10-100m outdoors)โข ๐พ: constant path-loss factor (antenna, average channel
attenuation), and sometimes we use
โข ๐พ: path-loss exponent
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๐0 = ๐"๐พ๐๏ฟฝ๐
๏ฟฝ
๐พ =๐
4๐๐๏ฟฝ
+
Some empirical results
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Measurements in Germany Cities
Environment Path-loss Exponent
Free-space 2
Urban area cellular radio 2.7-3.5
Shadowed urban cellular radio
3-5
In building LOS 1.6 to 1.8
Obstructed in building 4 to 6
Obstructed in factories 2 to 3
Empirical Path-Loss Modelโข Based on empirical measurementsโข over a given distanceโข in a given frequency rangeโข for a particular geographical area or building
โข Could be applicable to other environments as wellโข Less accurate in a more general environment
โข Analytical model: ๐ท๐/๐ท๐ is characterized as a function of distance.
โข Empirical Model: ๐ท๐/๐ท๐ is a function of distance including the effects of path loss, shadowing, and multipath.โข Need to average the received power measurements to remove
multipath effects ร Local Mean Attenuation (LMA) at distance d.
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Example: Okumura Model
โข Okumura Model:๐ท๐ณ ๐ ๐ ๐ฉ = ๐ณ ๐๐,๐ + ๐จ๐ ๐๐,๐ โ ๐ฎ ๐๐ โ ๐ฎ ๐๐ โ ๐ฎ๐จ๐น๐ฌ๐จโข ๐ณ ๐๐,๐ : FSPL, ๐จ๐ ๐๐,๐ : median attenuation in addition to FSPL
โข ๐บ โ" = 20 log๏ฟฝ๏ฟฝ(๏ฟฝ +๏ฟฝ๏ฟฝ) , ๐บ โ0 = ยก
10 log๏ฟฝ๏ฟฝ๏ฟฝยขยฃ , โ0 โค 3๐,
20 log๏ฟฝ๏ฟฝ๏ฟฝยขยฃ , 3๐ < โ0 < 10๐.
:antenna height gain factor.โข ๐ฎ๐จ๐น๐ฌ๐จ: gain due to the type of environment
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Example: Piecewise Linear Modelโข N segments with N-1
โbreakpointsโ
โข Applicable to both outdoor and indoor channels
โข Example โ dual-slope model:
โข ๐พ: constant path-loss factorโข ๐พ๏ฟฝ: path-loss exponent for ๐๏ฟฝ~๐kโข ๐พ+: path-loss exponent after ๐k 18
๐0 ๐ =๐"๐พ
๐๏ฟฝ๐
๏ฟฝยฉ
๐"๐พ๐k๐
๏ฟฝยฉ ๐๐k
๏ฟฝ'
๐๏ฟฝ โค ๐ โค ๐k,
๐ > ๐k.
Shadow Fadingโข Same T-R distance usually have different path loss
โข Surrounding environment is different
โข Reality: simplified Path-Loss Model represents an โaverageโ
โข How to represent the difference between the average and the actual path loss?
โข Empirical measurements have shown that
โข it is random (and so is a random variable)โข Log-normal distributed
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Log-normal distributionโข A log-normal distribution is a probability distribution
of a random variable whose logarithm is normally distributed:
โข ๐ฅ: therandomvariable(linearscale)โข ๐, ๐+:mean and variance of the distribution (in dB)
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๐ยน ๐ฅ; ๐, ๐+ =1
๐ฅ๐ 2๐exp โ
log ๐ฅ โ ๐ +
2๐+
logarithm of the random variable
normalized so that the integration of the pdf=1
Log-normal Shadowingโข Expressing the path loss in dB, we have
โข ๐ฟ๐: Describes the random shadowing effects
โข ๐ฟ๐~๐ต(๐, ๐๐)(normal distribution with zero mean and ๐๐ variance)
โข Same T-R distance, but different levels of clutter.
โข Empirical Studies show that ๐ ranges from 4 dB to 13dB in an outdoor channel
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๐ยพ ๐ ๐๐ต = ๐ยพ ๐ + ๐ร = ๐ยพ ๐๏ฟฝ + 10๐พ log๐๐๏ฟฝ
+ ๐ร
Why is it log-normal distributed?โข Attenuation of a signal when passing through an
object of depth d is approximately:
โข ๐ผ:Attenuation factor which depends on the material
โข If ๐ถ is approximately the same for all blocking objects:
โข ๐" = โ ๐ร ร : sum of all object depths
โข By central limit theorem, ๐ ๐~๐ต(๐, ๐๐)when the number of object is large (which is true).
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๐ ๐ = exp โ๐ผ๐
๐ ๐" = exp โ๐ผร๐ร ร
= exp โ๐ผ๐"
Cell Coverage Areaโข Cell coverage area: expected percentage of locations
within a cell where the received power at these locations is above a given minimum.
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Some area within the cell has received power lower than ๐&ร ร
Some area outside of the cell has received power higher than ๐&ร ร
Cell Coverage Areaโข We can boost the transmission power at the BS
โข Extra interference to the neighbor cells
โข In fact, any mobile in the cell has a nonzero probability of having its received power below ๐ท๐๐๐.
โข Since Normal distribution has infinite tailsโข Make sense in the real-world:
in a tunnel, blocked by large buildings, doesnโt matter if it is very close to the BS
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Cell Coverage Area
โข Cell coverage area is given by
โข ๐ท๐จ โ ๐ ๐ท๐ ๐ > ๐ท๐๐๐
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๐ช = ๐ฌ๐๐ ๐น๐ร ๐ ๐ท๐ ๐ > ๐ท๐๐๐๐๐๐ ๐จ ๐ ๐จ
๐๐๐๐๐๐๐๐
=๐๐ ๐น๐ร ๐ฌ ๐ ๐ท๐ ๐ > ๐ท๐๐๐๐๐๐ ๐จ ๐ ๐จ
๐๐๐๐๐๐๐๐
1 if the statement is true, 0 otherwise. (indicator function)
๐ช =๐๐ ๐น๐ร ๐ท๐จ๐ ๐จ
๐๐๐๐๐๐๐๐=
๐๐ ๐น๐ ร ร ๐ท๐จ
๐น
๐๐๐ ๐
๐๐
๐๐ ๐ฝ
Cell Coverage Area
โข Q-function:
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๐ร = ๐ ๐0ร ๐ โฅ ๐&ร ร = ๐๐&ร ร โ ๐" โ ๐ยพ ๐
๐
๐ ๐ง โ ๐ ๐ > ๐ง = ร12๐
ร
รexp โ
๐ฆ+
2 ๐๐ฆ
Log-normal distributionโs standard deviation
z
Cell Coverage Areaโข Solving the equations yield:
โข ๐ = ๐ท๐๐๐c๐ท๐ ๐น๐ , ๐ = ๐๐๐ธ๐๐๐๐๐ ๐
๐
โข If ๐ท๐๐๐ = ๐ท๐ ๐น
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๐ถ = ๐ ๐ + exp2 โ 2๐๐๐+ ๐
2 โ ๐๐๐
average received power at cell boundary (distance=R)
๐ถ =12 + exp
2๐+ ๐
2๐
Exampleโข Find the coverage area for a cell with โข a cell radius of 600mโข a base station transmission power of 20 dBmโข a minimum received power requirement of -110 dBm.
โข path loss model: ๐0(๐) = ๐"๐พmรm
๏ฟฝ, ๐พ = 3.71, ๐พ = โ31.54๐๐ต, ๐๏ฟฝ = 1, shadowing
standard deviation ๐ = 3.65dBโข Ans:
โข ๐ท๐ ๐น = ๐๐โ ๐๐.๐๐ โ ๐๐ร๐.๐๐ร๐๐๐๐๐ ๐๐๐ = โ๐๐๐.๐๐ ๐ฉ๐
โข ๐ = c๐๐๐๏ฟฝ๐๐๐.๐๐.๐๐ = ๐. ๐๐,๐ = ๐๐ร๐.๐๐ร๐.๐๐๐
๐.๐๐ = ๐.๐๐
โข ๐ช = ๐ธ ๐.๐๐ + ๐๐๐ โ๐. ๐๐ ๐ธ โ๐. ๐๐๐ = ๐.๐ (not good)
โข If we calculate C for a minimum received power requirement of -120 dBmโข C=0.988! 29
๐ถ = ๐ ๐ + exp2 โ 2๐๐๐+ ๐
2 โ ๐๐๐
๐ = ๐ท๐๐๐c๐ท๐ ๐น๐
, ๐ = ๐๐๐ธ๐๐๐๐๐ ๐๐
Example: road corners path loss
30
5 m
40 m
10 m
Radio: ChipconCC2420IEEE 802.15.4, 2.4 GHzTX pwr: 0 dBm
8 dBi peak gainomni-directionalantenna
Intel-NTU Connected Context Computing Center
โข Compare the path-loss exponent of three different locations:
1. Corner of NTU_CSIE building
2. XinHai-Keelong intersection
3. FuXing-HePing intersection
Link Measurements โPath loss around the corner building
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1 2 3
Passing-by vehicles
Occasionally Frequently Frequently
Buildings Around
No Few buildings
Some high buildings
Intersection Narrow Wide Wide
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Zuvio 2โข Out of all 4G (LTE) cellular service providers in Taiwan,
if we consider only FSPL (1) which provider would give you the best SNR? (2) whatโs the difference (in dB) of SNR comparing signals from the โbestโ and the โworstโ providers?
โข You can assume that all the other factors are the same (base station antennas and cellular phone antennas)
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