Path-loss and Shadowing (Large-scale Fading)

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)

6

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

N_CJO P = QR &

4

_J_LST P exp − X2F&

F + Q _K_MST 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

•  =  _^{  `}₈^a + ^/!_)d)\^{)* C,bc }

• The above is verified by empirical results.

• bc = () + )^\ − 8)/

• ) Z )^\ + 8  e Z e_ Z !^ + e + e_ Z !^

9

l x

x’

x’

x

Two-ray Model: Received Power

• When ! ≫ e + e, bc = ^{ )d)}_ ^fC8 ≈ ^e_!^e

• For asymptotically large d, ) + )^\ ≈ h ≈ !, E ≈ ',

i_ji_k ≈ _/!, l ≈ −1 (phase is inverted after reflection)

• ^`a

8 + ^/!_)d)\^{)* C,bc } ≈ _!^{`a } 1 + )* −,bc ^

• 1 + )* −,bc ^ = 1 − / bc ^ + m^bc =  −

nop bc = m^(^bc_ ) ≈ bc^

•  ≈ ^_!^{`a  e</sup><sub>!</sub><sup>e </sup> q<sub>r</sub> = <sup>`}_!^a_^{e e } q_r

10

Independent of & now

11

_K = 4ℎ_ℎ_

&

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:

12

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

13

_ =
_t _s



v

t = &

4_s

Some empirical results

14

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.

15

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 ^_ƒ^‚ , ℎ_ ≤ 3†, 20 log_~s ^_ƒ^‚ , 3† < ℎ_ < 10†.

:antenna height gain factor.

• _(y: gain due to the type of environment

16

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

18

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)

19

]_™ 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

• ›_œ: Describes the random shadowing effects

• ›_œ~(', œ^)

(normal distribution with zero mean and œ^ variance)

• Same T-R distance, but different levels of clutter.

• Empirical Studies show that œ ranges from 4 dB to 13dB in an outdoor channel

20

_ž  Ÿ =
_ž  +  _¡ =
_ž _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, œ^) when the number of object is large (which is true).

21

¦  = exp −¢

¦ _ = exp −¢ § _¥

¥

= exp −¢_

Path Loss, Shadowing, and Multi-Path

22

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.

23

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

24

Cell Coverage Area

• Cell coverage area is given by

• _ ≐ *   > _Am

25

© = 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:

26

_¬ = ­
® N ≥
_{ ¥¨} = °
_¥¨ −
_ −
_ž N

˜

° ± ≐ ­   > ± = ª 1 2

²

³

exp −´

2 ´

Log-normal distribution’s standard deviation

z

Cell Coverage Area

• Solving the equations yield:

• ` = ^AmC_œ ⁽ , a = ^1'µ8-_œ^{1' }

• If _Am =  (

27

¶ = ° · + 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 ^M_M^{¹ v}, u = 3.71, t = −31.54 Ÿ, _s = 1, shadowing standard deviation ˜ = 3.65 dB

• Ans:

•  ( = ' − 01. 9 − 1' × 0. :1 × 8-_1' 4'' = −11. 4 !;A

• ` = ^C11'd11.4_0.49 = 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!

28

¶ = ° · + exp 2 − 2·¸

¸ ° 2 − ·¸

¸

` = ^{AmC (}

œ , a = ^{1'µ8-1' }

œ

Example: road corners path loss

29

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

30

1 2 3 Passing-by

vehicles

Occasionally Frequently Frequently Buildings

Around

No Few

buildings

Some high buildings Intersection Narrow Wide Wide

31

Doppler Effect

• Difference in path lengths bh = ! /« = ¿br nop«

• Phase change bc = ^bh_ = ^¿br_ nop«

• Frequency change, or Doppler shift,

#_! = 1

 bc

br = ¿

 /«

32

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=& = _À^K

Á = _~ÃÄs×~s^ƒ×~s _Å = 0.162 (†)

• Vehicle speed Æ  60†­ℎ  26.82 ^_È 1. ]_M  ^É.Ã_s.~Écos 0  160 ʱ

2. ^]M  ^É.Ã_s.~Écos  +160 (ʱ)

3. Since cos ^Ë  0, there is no Doppler shift!

33

#_!  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

34