path-loss and shadowing (large-scale fading)
TRANSCRIPT
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��
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 �� ! = �� !' !!'
�
�� = � � ������! �
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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 !�
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 ≈ ',
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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