# Summary of Path Loss in Propagation by jackl17

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```									Summary of Path Loss
in Propagation

Narayan Mandayam
Understanding RF Propagation

Goals
3. Estimate network design parameters
1. Transmitters and their location
2. Transmit power
3. Antenna type
Interesting Scenarios
At which locations will
correct reception take
place?
Antenna Basics
Pdirectiona
G                    l

Pisotropic

Isotropic                     Dipole    High gain
directional

0 dBi                        2.2 dBi   14 dBi
Free Space Propagation Model
Isotropic power
PR             PT                  2
PDi                   W /m       density
4 d
2
d
PT
PT G T            Power density along
PD                       the direction of
4 d
2

PR  PD A eff             Antenna


2
PT G T                    Aeff

PR                    Aeff        G         4
4 d
strength when the transmitter
2     Also known
line-of-sight path between them
                      as Friis free
PR  PT G T G R                        space formula
  d 
Path Loss (relative measure)
2
PR           PR            
 GT G R       
PT            d 
Pt
3   f is in MHz
PR              0 . 57 * 10
 GT G R               2        d is in Km
PT                 ( df )

 PR 
    
 P   ( G T ) dB  ( G R ) dB  ( 32 . 5  20 log     10
d  20 log   10
f)
 T  dB

Path Loss represents signal attenuation
(measured on dB) between the effective
transmitted power and the receive power
(excluding antenna gains)
Path Loss (Example)
Assume that antennas are isotropic.
PR         Calculate receive power (in dBm) at free
space distance of 100m from the antenna.
Pt                          What is PR at 10Km?

50 W           PR 
    
= 47 dBm       P   ( G T ) dB  ( G R ) dB  ( 32 . 5  20 log   10
d  20 log    10
f)
 T  dB

 PR                                                                         59
    
 P   0  0  ( 32 . 5  20 log     10
0 . 1  20 log   10
900 )
 T  dB

-20 (for d = 0.1)                              20 (for d = 10)
 PR                                         PR 
    
    
 P    71 . 5 dB                           P    111 . 5 dB
 T  dB                                      T  dB

( PR ) dBm  47  71 . 5   24 . 5 dBm     ( PR ) dBm  47  111 . 5   64 . 5 dBm
Path Loss (another example)
P a th L o s s V s . D is ta n c e

2 .4 G H z     5 GHz

160

140

120

100
P a th L o s s (d B )

80

60

40

20

0
0   5   10   15   20   25   30   35    40    45       50       55    60   65   70   75   80   85   90   95   100

D is ta n c e (K m )
Path Loss (another example)
P a th L o s s V s . D is ta n c e

2.4 G H z    5 G Hz

150

140

130

120
P ath L o ss (d B )

110

100

90

80

70

60
0.01    0.1                          1                   10   100

D ista n ce (K m ) L o g S ca le

near field                          path loss in 2.4 Ghz band
Pr
r  8m              r > 8m
Pt
near field         far field
r                   r2               r3.3
Pr

path loss = 10 log (4r2/)                  r  8m

= 58.3 + 10 log (r3.3 /8)       r > 8m
Indoor Signal Measurement

Signal Strength

Channel 3     Channel 4         Channel 5   Channel 10

0
-10 1   61    121      181        241       301     361   421

-20
-30
-40

-50
-60
-70
-80
-90
Time

Signal Strength

-81
-83
-85
dBm

-87
-89
-91
-93
-95
1   51   101        151        201   251
Packet number

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