Path-loss and Shadowing(Large-scale Fading)

PROF. MICHAEL TSAI2017/10/23

Friis Formula

TX Antenna

EIRP=𝑃"𝐺"

𝑑

Power spatial density %&'

14𝜋𝑑+

𝐴-

×

×

RX Antenna

⇒ 𝑃0=𝑃"𝐺"𝐴-4𝜋𝑑+

2

Antenna Aperture• Antenna Aperture=Effective Area

• Isotropic Antenna’s effective area 𝑨𝒆,𝒊𝒔𝒐 ≐𝝀𝟐

𝟒𝝅

• Isotropic Antenna’s Gain=1

• 𝑮 = 𝟒𝝅𝝀𝟐 𝑨𝒆

• Friis Formula becomes: 𝑷𝒓 =𝑷𝒕𝑮𝒕𝑮𝒓𝝀𝟐

𝟒𝝅𝒅 𝟐 = 𝑷𝒕𝑨𝒕𝑨𝒓𝝀𝟐𝒅𝟐

3

𝑷𝒓 =𝑷𝒕𝑮𝒕𝐴-4𝜋𝑑+

Friis Formula

• 𝝀𝟐

𝟒𝝅𝒅 𝟐 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 𝒅 > 𝒅𝒇, Fraunhofer distance

• 𝒅𝒇 =𝟐𝑫𝟐

𝝀, and it must satisfies 𝒅𝒇 ≫ 𝑫 and 𝒅𝒇 ≫ 𝝀

• D: Largest physical linear dimension of the antenna• 𝜆: Wavelength

• We often choose a 𝒅𝟎 in the far-field region, and smaller than any practical distance used in the system

• Then we have 𝑷𝒓 𝒅 = 𝑷𝒓 𝒅𝟎𝒅𝟎𝒅

𝟐

𝑷𝒓 =𝑷𝒕𝑮𝒕𝑮𝒓𝝀𝟐

𝟒𝝅𝒅 𝟐

4

Received Signal after Free-Space Path Loss

𝒓 𝒕 = 𝑹𝒆𝝀 𝑮𝒕𝑮𝒓𝒆𝒙𝒑 −𝒋𝟐𝝅𝒅𝝀

𝟒𝝅𝒅 𝒈N 𝒕 𝒆𝒙𝒑 𝒋𝟐𝝅𝒇𝒄𝒕

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: 𝝀 = 𝒄𝒇 =

𝟑×𝟏𝟎𝟖

𝟗𝟎𝟎×𝟏𝟎𝟔 = 𝟎. 𝟑𝟑

• 𝒅𝒇 =𝟐𝐃𝟐

𝝀 = 𝟐𝟎.𝟑𝟑 = 𝟔. 𝟎𝟔 (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:

• 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝟓𝟎 = 𝟏𝟕𝒅𝑩𝑾 = 𝟒𝟕𝐝𝐁𝐦• Received Power at 100m

𝑷𝒓(𝟏𝟎𝟎𝒎) =𝟓𝟎×𝟏×𝟏× 𝟑×𝟏𝟎𝟖

𝟗𝟎𝟎×𝟏𝟎𝟔𝟐

𝟒𝝅×𝟏𝟎𝟎 𝟐 = 𝟑. 𝟓×𝟏𝟎c𝟔 𝑾= −𝟓𝟒. 𝟓(𝒅𝑩𝑾)

• Received Power at 10km

𝑷𝒓 𝟏𝟎𝒌𝒎 = 𝑷𝒓 𝟏𝟎𝟎𝒎𝟏𝟎𝟎𝟏𝟎𝟎𝟎𝟎

𝟐

= 𝟑. 𝟓×𝟏𝟎c𝟏𝟎 𝑾= −𝟗𝟒. 𝟓(𝒅𝑩𝑾)

7

Zuvio 1: 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.

8

Two-ray Model

TX Antenna

𝑑

𝜃

𝑙

𝑥 𝑥′

RX Antenna

ℎ"ℎ0

𝐺j

𝐺k 𝐺l

𝐺m

𝑟+c0jo 𝑡 =

𝑅𝑒𝜆4𝜋

𝐺j𝐺l𝑔t 𝑡 exp −𝑗2𝜋𝑙𝜆𝑙 +

𝑅 𝐺k𝐺m𝑔t 𝑡 − 𝜏 exp −𝑗2𝜋 𝑥 + 𝑥|𝜆

𝑥 + 𝑥| exp 𝑗2𝜋𝑓k𝑡

Delayed since x+x’ is longer. 𝜏 = (𝑥 + 𝑥| − 𝑙)/𝑐

R: ground reflection coefficient (phase and amplitude change) 9

Two-ray Model: Received Power

• 𝑷𝒓 = 𝑷𝒕𝝀𝟒𝝅

𝟐 𝑮𝒂𝑮𝒃𝒍 + 𝑮𝒄𝑮𝒅𝒆𝒙𝒑 c𝒋𝚫𝝓

𝒙�𝒙|

𝟐

• The above is verified by empirical results.

• 𝚫𝝓 = 𝟐𝝅(𝒙 + 𝒙| − 𝒍)/𝝀

• 𝒙 + 𝒙| − 𝒍 = 𝒉𝒕 + 𝒉𝒓 𝟐 + 𝒅𝟐 − 𝒉𝒕 − 𝒉𝒓 𝟐 + 𝒅𝟐

10

l

x

x’

x’

x

Two-ray Model: Received Power

• When 𝒅 ≫ 𝒉𝒕 + 𝒉𝒓, 𝚫𝝓 = 𝟐𝝅 𝒙�𝒙�c𝒍𝝀

≈ 𝟒𝝅𝒉𝒕𝒉𝒓𝝀𝒅

• For asymptotically large d, 𝒙 + 𝒙| ≈ 𝐥 ≈ 𝒅, 𝜃 ≈ 𝟎,𝐆𝐚𝐆𝐛 ≈𝑮𝒄𝑮𝒅, 𝐑 ≈ −𝟏(phase is inverted after reflection)

• 𝑮𝒂𝑮𝒃𝒍

+ 𝑮𝒄𝑮𝒅𝒆𝒙𝒑 c𝒋𝚫𝝓𝒙�𝒙|

𝟐≈ 𝑮𝒂𝑮𝒃

𝒅

𝟐𝟏 + 𝒆𝒙𝒑 −𝒋𝚫𝝓 𝟐

• 𝟏 + 𝒆𝒙𝒑 −𝒋𝚫𝝓 𝟐 = 𝟏 − 𝒄𝒐𝒔 𝚫𝝓𝟐+ 𝒔𝒊𝒏𝟐𝚫𝝓 = 𝟐 −

𝟐𝐜𝐨𝐬 𝚫𝝓 = 𝟒𝒔𝒊𝒏𝟐(𝚫𝝓𝟐) ≈ 𝚫𝝓𝟐

• 𝑷𝒓 ≈𝝀 𝑮𝒂𝑮𝒃𝟒𝝅𝒅

𝟐𝟒𝝅𝒉𝒕𝒉𝒓𝝀𝒅

𝟐𝐏𝐭 =

𝑮𝒂𝑮𝒃𝒉𝒕𝒉𝒓𝒅𝟐

𝟐𝐏𝐭

11Independent of 𝜆 now

12

𝑑k =4ℎ"ℎ0𝜆

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:

13

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:

• 𝑑�: reference distance for the antenna far field (usually 1-10m indoors and 10-100m outdoors)• 𝐾: constant path-loss factor (antenna, average channel

attenuation), and sometimes we use

• 𝛾: path-loss exponent

14

𝑃0 = 𝑃"𝐾𝑑�𝑑

�

𝐾 =𝜆

4𝜋𝑑�

+

Some empirical results

15

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: 𝑷𝒓/𝑷𝒕 is characterized as a function of distance.

• Empirical Model: 𝑷𝒓/𝑷𝒕 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 à Local Mean Attenuation (LMA) at distance d.

16

Example: Okumura Model

• Okumura Model:𝑷𝑳 𝒅 𝒅𝑩 = 𝑳 𝒇𝒄,𝒅 + 𝑨𝝁 𝒇𝒄,𝒅 − 𝑮 𝒉𝒕 − 𝑮 𝒉𝒓 − 𝑮𝑨𝑹𝑬𝑨• 𝑳 𝒇𝒄,𝒅 : FSPL, 𝑨𝝁 𝒇𝒄,𝒅 : median attenuation in addition to FSPL

• 𝐺 ℎ" = 20 log��(� +��) , 𝐺 ℎ0 = ¡

10 log���¢£ , ℎ0 ≤ 3𝑚,

20 log���¢£ , 3𝑚 < ℎ0 < 10𝑚.

:antenna height gain factor.• 𝑮𝑨𝑹𝑬𝑨: gain due to the type of environment

17

Example: Piecewise Linear Model• N segments with N-1

“breakpoints”

• Applicable to both outdoor and indoor channels

• Example – dual-slope model:

• 𝐾: constant path-loss factor• 𝛾�: path-loss exponent for 𝑑�~𝑑k• 𝛾+: path-loss exponent after 𝑑k 18

𝑃0 𝑑 =𝑃"𝐾

𝑑�𝑑

�©

𝑃"𝐾𝑑k𝑑

�© 𝑑𝑑k

�'

𝑑� ≤ 𝑑 ≤ 𝑑k,

𝑑 > 𝑑k.

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

19

Log-normal distribution• A log-normal distribution is a probability distribution

of a random variable whose logarithm is normally distributed:

• 𝑥: therandomvariable(linearscale)• 𝜇, 𝜎+:mean and variance of the distribution (in dB)

20

𝑓¹ 𝑥; 𝜇, 𝜎+ =1

𝑥𝜎 2𝜋exp −

log 𝑥 − 𝜇 +

2𝜎+

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

• 𝑿𝝈: Describes the random shadowing effects

• 𝑿𝝈~𝑵(𝟎, 𝝈𝟐)(normal distribution with zero mean and 𝝈𝟐 variance)

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

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

21

𝑃¾ 𝑑 𝑑𝐵 = 𝑃¾ 𝑑 + 𝑋Á = 𝑃¾ 𝑑� + 10𝛾 log𝑑𝑑�

+ 𝑋Á

Why is it log-normal distributed?• Attenuation of a signal when passing through an

object of depth d is approximately:

• 𝛼:Attenuation factor which depends on the material

• If 𝜶 is approximately the same for all blocking objects:

• 𝑑" = ∑ 𝑑ÅÅ : sum of all object depths

• By central limit theorem, 𝒅𝒕~𝑵(𝝁, 𝝈𝟐)when the number of object is large (which is true).

22

𝑠 𝑑 = exp −𝛼𝑑

𝑠 𝑑" = exp −𝛼Ç𝑑ÅÅ

= exp −𝛼𝑑"

Path Loss, Shadowing, and Multi-Path

23

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.

24

Some area within the cell has received power lower than 𝑃&ÅÈ

Some area outside of the cell has received power higher than 𝑃&ÅÈ

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 𝑷𝒎𝒊𝒏.

• 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

25

Cell Coverage Area

• Cell coverage area is given by

• 𝑷𝑨 ≐ 𝒑 𝑷𝒓 𝒓 > 𝑷𝒎𝒊𝒏

26

𝑪 = 𝑬𝟏𝝅𝑹𝟐Ê 𝟏 𝑷𝒓 𝒓 > 𝑷𝒎𝒊𝒏𝒊𝒏𝒅𝑨 𝒅𝑨

𝒄𝒆𝒍𝒍𝒂𝒓𝒆𝒂

=𝟏𝝅𝑹𝟐Ê 𝑬 𝟏 𝑷𝒓 𝒓 > 𝑷𝒎𝒊𝒏𝒊𝒏𝒅𝑨 𝒅𝑨

𝒄𝒆𝒍𝒍𝒂𝒓𝒆𝒂

1 if the statement is true, 0 otherwise. (indicator function)

𝑪 =𝟏𝝅𝑹𝟐Ê 𝑷𝑨𝒅𝑨

𝒄𝒆𝒍𝒍𝒂𝒓𝒆𝒂=

𝟏𝝅𝑹𝟐 Ê Ê 𝑷𝑨

𝑹

𝟎𝒓𝒅𝒓

𝟐𝝅

𝟎𝒅𝜽

Cell Coverage Area

• Q-function:

27

𝑃Ì = 𝑝 𝑃0Î 𝑟 ≥ 𝑃&ÅÈ = 𝑄𝑃&ÅÈ − 𝑃" − 𝑃¾ 𝑟

𝜎

𝑄 𝑧 ≐ 𝑝 𝑋 > 𝑧 = Ê12𝜋

Ò

Óexp −

𝑦+

2 𝑑𝑦

Log-normal distribution’s standard deviation

z

Cell Coverage Area• Solving the equations yield:

• 𝒂 = 𝑷𝒎𝒊𝒏c𝑷𝒓 𝑹𝝈 , 𝒃 = 𝟏𝟎𝜸𝒍𝒐𝒈𝟏𝟎 𝒆

𝝈

• If 𝑷𝒎𝒊𝒏 = 𝑷𝒓 𝑹

28

𝐶 = 𝑄 𝑎 + exp2 − 2𝑎𝑏𝑏+ 𝑄

2 − 𝑎𝑏𝑏

average received power at cell boundary (distance=R)

𝐶 =12 + exp

2𝑏+ 𝑄

2𝑏

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: 𝑃0(𝑑) = 𝑃"𝐾mÙm

�, 𝛾 = 3.71, 𝐾 = −31.54𝑑𝐵, 𝑑� = 1, shadowing

standard deviation 𝜎 = 3.65dB• Ans:

• 𝑷𝒓 𝑹 = 𝟐𝟎− 𝟑𝟏.𝟓𝟒 − 𝟏𝟎×𝟑.𝟕𝟏×𝒍𝒐𝒈𝟏𝟎 𝟔𝟎𝟎 = −𝟏𝟏𝟒.𝟔𝒅𝑩𝒎

• 𝒂 = c𝟏𝟏𝟎�𝟏𝟏𝟒.𝟔𝟑.𝟔𝟓 = 𝟏. 𝟐𝟔,𝒃 = 𝟏𝟎×𝟑.𝟕𝟏×𝟎.𝟒𝟑𝟒

𝟑.𝟔𝟓 = 𝟒.𝟒𝟏

• 𝑪 = 𝑸 𝟏.𝟐𝟔 + 𝒆𝒙𝒑 −𝟎. 𝟒𝟔 𝑸 −𝟎. 𝟖𝟎𝟕 = 𝟎.𝟔 (not good)

• If we calculate C for a minimum received power requirement of -120 dBm• C=0.988! 29

𝐶 = 𝑄 𝑎 + exp2 − 2𝑎𝑏𝑏+ 𝑄

2 − 𝑎𝑏𝑏

𝒂 = 𝑷𝒎𝒊𝒏c𝑷𝒓 𝑹𝝈

, 𝒃 = 𝟏𝟎𝜸𝒍𝒐𝒈𝟏𝟎 𝒆𝝈

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

• 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

31

1 2 3

Passing-by vehicles

Occasionally Frequently Frequently

Buildings Around

No Few buildings

Some high buildings

Intersection Narrow Wide Wide

32

Zuvio 2• 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) what’s the difference (in dB) of SNR comparing signals from the “best” and the “worst” providers?

• You can assume that all the other factors are the same (base station antennas and cellular phone antennas)

33