PATH LOSS
Definition of Path Loss
• Path loss includes all of the lossy effects
associated with distance
Transmit
Power, Pt
Antenna Gain
Gt
Pti
Transmitter
Feeder Loss
Lt
Antenna Gain
Gr
Received
Pri
Power, Pr
Path Loss
L
Receiver
Feeder Loss
Lr
[Saunders,`99]
Motivation
• Need path loss to determine range of operation (using a
link budget)
• This module considers two cases,
• Free space
• Flat earth
Received Power
• The power appearing at the receiver input terminals is
Pt Gt Gr
Pr =
Lt LLr
• All gains G and losses L are expressed as power ratios
and the powers are in Watts
dBm and dBW
• Powers may also be expressed in
• dBm, the number of dB the power exceeds 1 milliwatt
• dBW, the number of dB the power exceeds 1 Watt.
Pr (in Watts)
Pr (in dBm) = 10 log10
10 −3Watts
Assume …
G
Definitions of Units
• dB
[Ratio]
•
• dBm
[Unit conversion]
•
• dBW
•
[Unit conversion]
dBm and dBW
dBm
dBW
Adding/Subtracting dB and dBM
•
Adding dB values is the same as multiplying with regular numbers. So if you add
10dB to a decibel value it is the same as multiplying a regular number by 10.
•
Subtracting dB values is the same as dividing with regular numbers.
•
It is OK to add dB values to an initial dBm value. This is the same as starting with
an input power level and adding amplification or subtracting attenuation from that
power level. The final answer will be your output power level in dBm.
Reference: BesserNet, http://www.bessernet.com/articles/dbm/Convert020.html
Adding/Subtracting dB and dBM
Example:
•
•
In the figure above we have an input power level of 10dBm to which we
add 20dB of amplification. The result is an output power of 30dBm.
This is the same as starting with 10mW of input power and multiplying
that by a factor of 100, giving an output power of 1,000 mW.
Reference:
BesserNet, http://www.bessernet.com/articles/dbm/Convert020.html
EIRP
• The effective isotropic radiated power (EIRP) is
Pt Gt
Pti =
Lt
• The effective isotropic received power is
Pr Lr EIRP
Pri =
=
Gr
L
Antenna Gains
• Antenna gain may be expressed in dBi or dBd
• dBi: maximum radiated power relative to an isotropic antenna
• dBd: maximum radiated power relative to a half-wave dipole
antenna
• A half-wave dipole has a peak gain of 2.15 dBi
Path Loss
• The path loss is the ratio of the EIRP to the
effective isotropic received power
Pti
L=
Pri
• Path loss is independent of system parameters
except for the antenna radiation pattern
• The pattern determines which parts of the environment
are illuminated
Free-Space Path Loss
• In the far-field of the transmit antenna, the free-space path
loss is given by
L=
(4π ) 2 d 2
• The far-field is any distance
d >>
2D
λ
λ2
d from the antenna, such that
2
, d >> D, and d >> λ
where D is the largest dimension of the antenna.
Power and Electric Field
• The peak power flux density (W/m2) in free space:
E
Pt Gt
EIRP
Pd =
=
=
2
2
4πd
Lt 4πd
η
=
E
2
120πΩ
=
E
2
2
377Ω
where |E| = envelope of the electric field in V/m
• This holds in the neighborhood (but far field) of
transmitters on towers
Effective Aperture
• Antenna gain may be expressed in terms of
effective aperture, Ae
G=
4πAe
2
λ
• For aperture antennas, such as dish
antennas, Ae = Aη , where η is the antenna
efficiency and A is the area of the aperture
• The aperture intercepts the power flux density
Pri = Pd Ae
Flat Earth (2-Ray) Model
• If there is a line-of-sight (LOS) path, then the second
strongest path is the ground bounce
Transmitter
LOS
Receiver
Ground
Bounce
Typical Relative Dimensions
• d>>ht, d>>hr for a typical mobile communications
geometry
Transmitter
LOS
Receiver
ht
Ground
Bounce
d
hr
Field Near Transmitter
• Let the field at a distance do in the neighborhood of, but
also in the far field of, the transmit antenna be E(do,t) , and
its envelope be Eo
• Assuming the transmitter is high enough,
2
Pt Gt
Eo
=
2
Lt 4πd o 120π
• The field at some other distance d>do is
& , d )#
Eo d o
E (d , t ) =
cos$$ ω c *t − ' !!
d
% + c ("
Low Grazing Angle
• At such a low (grazing) angle of incidence (θ=a
few degrees), the reflection coefficient is -1 for
horizontal polarization
Transmitter
Receiver
θ"
Γ= 1"
θ"
Field at Receiver
• The direct and bounce paths add coherently
ETOT (d , t ) = E (d !, t ) − E (d !!, t )
d !! = d1!! + d 2!!
d
ht
d1
Γ= 1"
d
d2
hr
Long Baseline Effects
• Since d is so large,
1
1 1
≈
≈
d " d "" d
6 d0 3
6 d 00 3
'
$
'
j
ω
t
−
j
ω
Eo d o ! c 45 c 12 ! Eo d o ! c 45t − c 12 $!
E (d , t ) =
Re &e
Re &e
#−
#
d0
!%
!" d 00
!%
!"
Eo d o
≈
d
Eo d o
=
d
6 d 00 3
'! jω c 64t − dc0 31
jω c 4 t − 1 $
!
Re &e 5 2 − e 5 c 2 #
!%
!"
'! jω c 64t − d 00 31 - jω c 64 d 00− d 0 31 *$!
Re &e 5 c 2 + e 5 c 2 − 1(#
+
(!
!%
,
)"
A Trick
• Pull an exponential with half the phase out to make a sine
3 d 44 − d 4 0
3 d 44 − d 4 0 $
' 3 d 44 0
j
ω
−
j
ω
c
c
1 2c .
1 2c . *
3 d 44 − d 4 0
/
/ (
!
Eo d o ! jω c 12 t − c ./ jω c 12 2 c ./ + e 2
−e 2
Re &e
e
2 j+
(#
d
2j
+
(!
!
,
)"
%
3 d 44 0
3 d 44 − d 4 0
'
$!
j
ω
t
−
j
ω
c
c
4
4
4
1 2c .
*
2 Eo d o ! 12 c ./
d
−
d
3
0
/
=
Re &e
e 2
j sin ++ ω c 1
((#
.
d
!%
, 2 2c / )!"
Field Envelope at Receiver
• Recall d
>d
• The envelope of the field is then
ETOT
2 Eo d o & , d -- − d - ) #
=
sin $$ ω c *
!!
'
d
% + 2c ( "
• Can show that d "" − d " ≈
2ht hr
, and
d
/ & d '' − d ' # ,
& d '' − d ' #
sin --ω c $
** ≈ ω c $
!
!
2
c
2
c
"+
%
"
. %
Power Received
• Making the substitutions yields
ETOT
2 Eo d o 2πht hr
=
d
λd
• The power received is
& ETOT 2 #& G λ2 #
!$ r !
Pri = Pd Ae = $
$ 120π !$% 4π !"
%
"
Flat Earth Path Loss
• Recalling
gives
2
Pt Gt
Eo
=
2
Lt 4πd o 120π
Pt Gt Gr ht2 hr2
Pri =
Lt d 4
• The flat earth path loss is therefore
d4
L= 2 2
ht hr
Path Loss (dB)
1000
100
10
10
100
1000
Path Length, d (m)
10000
Propagation path loss Lp (dB) with distance over a flat reflecting surface;
hb = 7.5 m, hm = 1.5 m, fc = 1800 M Hz.
Lp =
1
!"
λc
4πd
#2
4 sin2
"
2πhbhm
λc d
#$−1
1
6
• In reality, the earth’s surface is curved and rough, and the signal
strength typically decays with the inverse β power of the distance,
and the received power is
Ωt
dβ
where k is a constant of proportionality. Expressed in units of dBm,
the received power is
Ωp = k
Ωp
(dBm)
= 10log10 (k) + Ωt
(dBm)
− 10βlog10 (d)
• β is called the path loss exponent. Typical values of β are have been
determined by empirical measurements for a variety of areas
Terrain
Free Space
Open Area
North American
North American
North American
Japanese Urban
Suburban
Urban (Philadelphia)
Urban (Newark)
(Tokyo)
β
2
4.35
3.84
3.68
4.31
3.05
1
7
Using a Reference Power Measurement
• Suppose that a reference measurement of
received power, Po, is taken at some point in the
far field of the antenna
• Then the power taken at some more distant point
may be expressed relative to the reference
power:
& do #
Pri = Po $ !
%d "
n
Summary
• Free space path loss depends only on distance and
wavelength, and falls off as 1/d2
• Flat earth path loss
• depends also on the antenna heights, and falls off as 1/d4
• Has a pretty good fit to urban and suburban environments, even
though it is an idealization, derived only for horizontal polarization
• The power of d is called the path loss exponent
• For mobile comm, this exponent is typically between 3.5
and 4
References
• [Saunders,`99] Simon R. Saunders, Antennas
and Propagation for Wireless Communication
Systems, John Wiley and Sons, LTD, 1999.
• [Rapp, 96] T.S. Rappaport, Wireless
Communications, Prentice Hall, 1996
• [Lee, 98] W.C.Y. Lee, Mobile Communications
Engineering, McGraw-Hill, 1998