Path-loss and Shadowing (Large-scale Fading)
PROF. MICHAEL TSAI 2011/10/20
Friis Formula
TX Antenna
EIRP=
Power spatial density
41
×
×
RX Antenna
⇒ = 4
2
Antenna Aperture
• Antenna Aperture=Effective Area
• Isotropic Antenna’s effective area , ≐
• Isotropic Antenna’s Gain=1
• =
• Friis Formula becomes: = ! = !
3
= 4
Friis Formula
•
! is often referred as “Free-Space Path Loss” (FSPL)
• Only valid when d is in the “far-field” of the transmitting antenna
• Far-field: when ! > !#, Fraunhofer distance
• !# = $, and it must satisfies !# ≫ $ and !# ≫
• D: Largest physical linear dimension of the antenna
• &: Wavelength
• We often choose a !' in the far-field region, and smaller than any practical distance used in the system
• Then we have ! = !' !!
'
=
!
4
Received Signal after Free-Space Path Loss
= ( )* − ,!
! -. )* ,#/
phase difference due to propagation distance
Free-Space Path Loss
Complex envelope
Carrier (sinusoid)
5
Example: Far-field Distance
• Find the far-field distance of an antenna with
maximum dimension of 1m and operating frequency of 900 MHz (GSM 900)
• Ans:
• Largest dimension of antenna: D=1m
• Operating Frequency: f=900 MHz
• Wavelength: = #/ = 3''×1'0×1'24 = '. 00
• !# = 6 = '.00 = 4. '4 (m)
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Example: FSPL
• If a transmitter produces 50 watts of power, express the transmit power in units of (a) dBm and (b) dBW.
• 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.
• Ans:
• 1' 8-1' 9' = 1: !;< = : =>?
• Received Power at 100m
(1''A) = 9' × 1 × 1 × 0 × 1'3'' × 1'2 4
× 1'' = 0. 9 × 1'C4 <
= −9. 9 (!;<)
• Received Power at 10km
1'DA = 1''A 1'' 1''''
= 0. 9 × 1'C1' <
= −3. 9 (!;<) 7
Two-ray Model
TX Antenna
E
F
G G′
RX Antenna ℎ
ℎ
J
K
L
M
NCJO P = QR &
4
JLST P exp − X2F&
F + Q KMST P − [ exp − X2 G + G& \
G + G\ exp X2]KP
Delayed since x+x’ is longer. [ = (G + G\ − F)/_
R: ground reflection coefficient (phase and amplitude change) 8
Two-ray Model:
Received Power
• = `8a + /!)d)\)* C,bc
• The above is verified by empirical results.
• bc = () + )\ − 8)/
• ) Z )\ + 8 e Z e Z ! + e + e Z !
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l x
x’
x’
x
Two-ray Model: Received Power
• When ! ≫ e + e, bc = )d) fC8 ≈ e!e
• For asymptotically large d, ) + )\ ≈ h ≈ !, E ≈ ',
ijik ≈ /!, l ≈ −1 (phase is inverted after reflection)
• `a
8 + /!)d)\)* C,bc ≈ !`a 1 + )* −,bc
• 1 + )* −,bc = 1 − / bc + mbc = −
nop bc = m(bc ) ≈ bc
• ≈ !`a e!e qr = `!ae e qr
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Independent of & now
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K = 4ℎℎ
&
ℎ
Could be a natural choice of cell size
Indoor Attenuation
• Factors which affect the indoor path-loss:
• Wall/floor materials
• Room/hallway/window/open area layouts
• Obstructing objects’ location and materials
• Room size/floor numbers
• 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 Model
• Back to the simplest:
• s: reference distance for the antenna far field (usually 1-10m indoors and 10-100m outdoors)
• t: constant path-loss factor (antenna, average channel attenuation), and sometimes we use
• u: path-loss exponent
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= t s
v
t = &
4s
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 Model
• Based on empirical measurements
• over a given distance
• in a given frequency range
• for a particular geographical area or building
• Could be applicable to other environments as well
• Less accurate in a more general environment
• Analytical model: / is characterized as a function of distance.
• Empirical Model: / is a function of distance including the effects of path loss, shadowing, and multipath.
• Need to average the received power measurements to remove multipath effects Local Mean Attenuation (LMA) at distance d.
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Example: Okumura Model
• Okumura Model:
w ! !; = w #/, ! + x #/, ! − e − e − (y
• w #/, ! : FSPL, x #/, ! : median attenuation in addition to FSPL
• ℎ = 20 log~s(ss ) , ℎ 10 log~s , ℎ ≤ 3, 20 log~s , 3 < ℎ < 10.
:antenna height gain factor.
• (y: gain due to the type of environment
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Example: Piecewise Linear Model
• N segments with N-1
“breakpoints”
• Applicable to both
outdoor and indoor channels
• Example – dual-slope model:
• t: constant path-loss factor
• u~: path-loss exponent for s~K
• u: path-loss exponent after K 17
= t s
v
t K
v
K
v
s K,
" K.
Shadow Fading
• Same T-R distance usually have different path loss
• Surrounding environment is different
• Reality: simplified Path-Loss Model represents an
“average”
• How to represent the difference between the average and the actual path loss?
• Empirical measurements have shown that
• it is random (and so is a random variable)
• Log-normal distributed
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Log-normal distribution
• A log-normal distribution is a probability distribution of a random variable whose logarithm is normally
distributed:
• G: the random variable (linear scale)
• , : mean and variance of the distribution (in dB)
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] G; , = 1
G 2exp − log G − 2
logarithm of the random variable
normalized so that the integration of the pdf=1
Log-normal Shadowing
• Expressing the path loss in dB, we have
• : Describes the random shadowing effects
• ~(', )
(normal distribution with zero mean and variance)
• Same T-R distance, but different levels of clutter.
• Empirical Studies show that ranges from 4 dB to 13dB in an outdoor channel
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= + ¡ = s + 10u log
s + ¡
Why is it log-normal distributed?
• Attenuation of a signal when passing through an object of depth d is approximately:
• ¢: Attenuation factor which depends on the material
• If £ is approximately the same for all blocking objects:
• = ∑ ¥ ¥: sum of all object depths
• By central limit theorem, ! ~(x, ) when the number of object is large (which is true).
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¦ = exp −¢
¦ = exp −¢ § ¥
¥
= exp −¢
Path Loss, Shadowing, and Multi-Path
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Cell Coverage Area
• 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
¥¨ Some area outside of
the cell has received power higher than
¥¨
Cell Coverage Area
• We can boost the transmission power at the BS
• Extra interference to the neighbor cells
• In fact, any mobile in the cell has a nonzero
probability of having its received power below Am.
• Since Normal distribution has infinite tails
• Make sense in the real-world:
in a tunnel, blocked by large buildings, doesn’t matter if it is very close to the BS
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Cell Coverage Area
• Cell coverage area is given by
• ≐ * > Am
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© = y 1
( ª 1 > Am m ! !
/88 ``
= 1
( ª y 1 > Am m ! !
/88 ``
1 if the statement is true, 0 otherwise.
(indicator function)
© = 1
( ª !
/88 `` = 1
( ª ª (
' !
' !«
Cell Coverage Area
• Q-function:
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¬ = ® N ≥ ¥¨ = ° ¥¨ − − N
° ± ≐ > ± = ª 1 2
²
³
exp −´
2 ´
Log-normal distribution’s standard deviation
z
Cell Coverage Area
• Solving the equations yield:
• ` = AmC ( , a = 1'µ8-1'
• If Am = (
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¶ = ° · + exp 2 − 2·¸
¸ ° 2 − ·¸
¸
average received power at cell boundary (distance=R)
¶ = 1
2+ exp 2
¸ ° 2
¸
Example
• Find the coverage area for a cell with
• a cell radius of 600m
• a base station transmission power of 20 dBm
• a minimum received power requirement of -110 dBm.
• path loss model: () = t MM¹ v, u = 3.71, t = −31.54 , s = 1, shadowing standard deviation = 3.65 dB
• Ans:
• ( = ' − 01. 9 − 1' × 0. :1 × 8-1' 4'' = −11. 4 !;A
• ` = C11'd11.40.49 = 1. 4, a = 1'×0.:1×'.0
0.49 = . 1
• © = ¾ 1. 4 + )* −'. 4 ¾ −'. 2': = '. 4 (not good)
• If we calculate C for a minimum received power requirement of -120 dBm
• C=0.988!
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¶ = ° · + exp 2 − 2·¸
¸ ° 2 − ·¸
¸
` = AmC (
, a = 1'µ8-1'
Example: road corners path loss
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5 m 40 m
10 m Radio: Chipcon CC2420
IEEE 802.15.4, 2.4 GHz TX pwr: 0 dBm
8 dBi peak gain omni-directional antenna
Intel-NTU Connected Context Computing Center
• 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 –
Path 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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Doppler Effect
• Difference in path lengths bh = ! /« = ¿br nop«
• Phase change bc = bh = ¿br nop«
• Frequency change, or Doppler shift,
#! = 1
bc
br = ¿
/«
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Example
• Consider a transmitter which radiates a sinusoidal carrier frequency of 1850 MHz. For a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving
1. directly toward the transmitter.
2. directly away from the transmitter
3. in a direction which is perpendicular to the direction of arrival of the transmitted signal.
• Ans:
• Wavelength=& = ÀK
Á = ~ÃÄs×~s×~s Š= 0.162 ()
• Vehicle speed Æ 60ℎ 26.82 È 1. ]M É.Ãs.~Écos 0 160 ʱ
2. ]M É.Ãs.~Écos +160 (ʱ)
3. Since cos Ë 0, there is no Doppler shift!
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#! 1
bc
br ¿
/«
Doppler Effect
• If the car (mobile) is moving toward the direction of the arriving wave, the Doppler shift is positive
• Different Doppler shifts if different « (incoming angle)
• Multi-path: all different angles
• Many Doppler shifts Doppler spread
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