Path loss and Shadowing

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Path-loss and Shadowing (Large-scale Fading) PROF. MICHAEL TSAI 2011/10/20

Transcript of Path loss and Shadowing

Path-loss and Shadowing

(Large-scale Fading)

PROF. MICHAEL TSAI

2011/10/20

Friis Formula

TX Antenna

EIRP=�����

Power spatial density���

14��

×

×

RX Antenna

⇒ ��= ����� 4��

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Antenna Aperture

• Antenna Aperture=Effective Area

• Isotropic Antenna’s effective area ��,��� ≐ ����

• Isotropic Antenna’s Gain=1

• � = ���� ��

• Friis Formula becomes: �� = � � ������! � = � � ��

��!�

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�� = � � � 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 �� ! = �� !' !!'

�� = � � ������! �

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Received Signal after

Free-Space Path Loss

� = (�� � ���)* − ,��!

���! -. �)* ,��#/

phase difference due to propagation distance

Free-Space Path Loss

Complex envelope

Carrier (sinusoid)

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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: � = /# = 0×1'2

3''×1'4 = '. 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 × 1'2

3'' × 1'4�

�� × 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

FG G′

RX Antenna

ℎ� ℎ�

�J�K �L

�M

N�C�JO 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

• �� = � ���

� �`�a8 + �/�!�)* C,bc

)d)\�

• 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)

•�`�a8 + �/�!�)* C,bc

)d)\�≈ �`�a

!�

1 + �)* −,bc �

• 1 + �)* −,bc � = 1 − /�� bc � + ��m�bc = � −�nop bc = ���m�(bc

� ) ≈ bc�

• �� ≈ � �`�a��!

� ��e e��!

� qr = �`�ae e�!�

�qr

10Independent of & now

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�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:

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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 = &4�s

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; �, �� = 1G� 2 exp − 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 �A�m.

• 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

• �� ≐ * �� � > �A�m

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© = y 1�(� ª 1 �� � > �A�m�m!� !�

/�88`��`= 1

�(� ª y 1 �� � > �A�m�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 12

²

³

exp −´�

2�´

Log-normal distribution’s standard deviation

z

Cell Coverage Area

• Solving the equations yield:

• ` = �A�mC�� (� , a = 1'µ8�-1' �

• If �A�m = �� (

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¶ = ° · + exp 2 − 2·¸

¸�°

2 − ·¸

¸

average received power at cell boundary (distance=R)

¶ =12

+ exp2

¸�°

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 M¹M

v, u = 3.71,t = −31.54��, �s = 1, shadowing

standard deviation � = 3.65dB• 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!

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¶ = ° · + exp 2 − 2·¸

¸�°

2 − ·¸

¸

` =�A�mC�� (

�, 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 = !/��« = ¿brnop«• 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=& = KÀÁ = �×~sÂ

~ÃÄs×~sÅ = 0.162(�)• Vehicle speed Æ � 60�­ℎ � 26.82 �È1. ]M � �É.Ã�s.~É� cos 0 � 160 ʱ2. ]M � �É.Ã�s.~É� cos � +160(ʱ)3. Since cos Ë� � 0, there is no Doppler shift!

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#! � 1��bcbr � ¿� /��«

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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