Mobile Radio Propagation

1
Mobile Radio Propagation

• Mobile radio channel is an important controlling
  factor in wireless communication systems.
• Transmission path between transmitter and
  receiver can vary in complexity.
• LOS (Line Of Sight) ? Simplest(e.g., buildings,
  mountains, foliage - trees/bushes, speed of
  mobile).

2
...Mobile Radio Propagation

• Wired channels are stationary and predictable
• Radio channels are extremely random and have
  complex models.
• Modeling of radio channels is done in statistical
  fashion based on measurements for each individual
  communication system or frequency spectrum.

3
Propagation Models

• Aim
• To predict the average received signal strength
  at a given distance from the transmitter -
  Large scale propagation models, hundreds or
  thousands of meters.
• To predict the variability of the signal
  strength, at close spatial proximity to a
  particular location - Small scale or fading
  models.

4
Free Space Propagation Model
5
Units of Received Signal Strength

• Electrical Field
• Magnitude E IEI Volts/meter
• E - Vector
• Direction E xEx yEy zEz

6
...Units of Received Signal Strength

• Power
• P ? Watts or dBm
• P (dBm) 10 log10 P (mW)
• P(mW) P(dBm)
• 10 10
• 1 0
• 10-1 -10
• 10-2 -20

7
Propagation Models

• Free Space Propagation Transmitter and receiver
  have a clear, unobstructed LOS path between them.
• Reflection From the surface of the earth and
  from buildings and walls. Usually dimensions of
  reflecting object are much greater than
  wavelength.
• Diffraction Bending of electromagnetic waves
  around sharp edges such as, sharp towers or peaks.

8
...Propagation Models

• Scattering Due to objects in the medium that
  are small compared to wavelength and the number
  of objects is many (e.g., foliage, street signs,
  lamp posts, rain, shower).

9
Free Space Propagation

• Transmitter/receiver have clear, unobstructed
  LOS path between them.
• T d R

10
...Free Space Propagation

• Pr Pt Gt Gr ?2 Valid if d gtgt 2D2 / ?
• (4?)2 d2 L
• D Max dimension of transmitting antenna.
• Power
• Pt Transmitted Power,
• Pr Received Power
• Gt Transmitter antenna gain
• Gr Receiver antenna gain
• L System loss factor ( L ? 1, transmission
  lines etc, but not due to propagation)

11
Antenna Gain

• Gain of antenna
• G 4?Ae / ?2,
• where Ae is effective aperture of antenna
• ? c / f (Hz) 3 108 / f meters

12
Electric Field - Power Relation

• In free space, the power density is given by
• Pd Pt Gt E2
• 4? d2 ?
• where, ? Intrinsic impedance of free space.

13
...Electric Field - Power Relation

• ? ?0 / ?0
• For air, ? (4? 10-7) /(8.85 10-12 )
  377 ?
• Pr Pd Ae IErI 2 Ae IErI2 ?2 Gr
• ? ? (4?)

14
Example

• If the power at a receiving antenna is Pr 7
  10 -10 W, antenna gain Gr 2 and transmitting
  frequency is 900 MHz, determine the electric
  field strength at the receiver.

15
Solution

• f 900 MHz gt
• ? (3 108) / (900 106) 0.33 m
• IEI (Pr 377 4?) / (0.33 0.33 2)1/2
• 0.0039 V/m

16
Example

• Given a transmitter produces 50W of power. If
  this power is applied to a unity gain antenna
  with 900 MHz carrier frequency, find the received
  power 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.

17
Solution

• Pr Pt Gt Gr ?2
• (4?)2 d2 L
• Pt 50 W
• Gt 1
• Gr 1
• ? (3 108) / (900 106) 0.33 m
• L 1
• d 100 m

18
Solution

• ?
• Pr 3.5 10-3 mW
• Pr (10 km) Pr (100 m) (100m/10km)2
• 3.5 10-3 mW 1/100
• 3.5 10-6 mW

19
Relating Received Power Level to Receiver Input
Voltage
20
...Received Power Level to Receiver Input
Voltage
Rant
To matched Receiver
Pr
V
R
I
Vant

• For matched receiver, R Rant
• I Vant / (2 Rant)
• Power Received V2 / R V2/ Rant Vant2/ (4
  Rant)
• Or Vant Pr (d) 4 Rant
• Vant - ?E.dl where E is the electric field V/m

21
Reflections from Ground and Buildings

• Electric Properties of Material Bodies
• Permittivity ? F/m ? Farads/m
• Permeability ? H/m ? Henries/m
• Conductivity ? S/m ? Siemens/m
• ? ?0 ?r
• ?0 Permittivity of free space 8.85 10-12
  F/m
• ?r Relative permittivity

22
Laws of Reflection at the Boundary Between Two
Dielectrics

• Er ? ? Reflection coefficient
• Et T 1 ? ?
• Transmission coefficient

Ei
Er
Ei
?i
?r
Ei
?i ?r
Et
23
Vertical Propagation (Or Parallel Polarization)
Ei
Er
Hi
Hr
?1, ?1, ?1
?i
?r
?2, ?2, ?2
? t
Et

• E-field in the plane of incidence

24
Vertical Propagation (Or Parallel Polarization)

• ? -?r sin?i (?r - cos2?i)1/2
• ?r sin?i (?r - cos2?i)1/2
• ?2 sin?t - ?1 sin?i
• ?2 sin?t ?1 sin?i

25
Horizontal Propagation (Or Perpendicular
Polarization)
Ei
Er
Hi
Hr
?1, ?1, ?1
?i
?r
?2, ?2, ?2
? t
Et

• E-field normal to the plane of incidence

26
Horizontal Propagation (Or Perpendicular
Polarization)

• ? -?r sin?i (?r - cos2?i)1/2
• ?r sin?i (?r - cos2?i)1/2
• ?2 sin?t - ?1 sin?i
• ?2 sin?t ?1 sin?i

27
Brewster Angle No Reflected Wave

• ? 0
• ð?r sin?B (?r - cos2?B)1/2
• ?r2 sin2?B ?r - cos2?B
• ?r 1 sin2?B

28
2 Cases

• sin?B (er -1)/(er2 -1)1/2 First
  medium is air e1 e0, e2 e0er
• sin?B (er2 - er )/(er2 -1)1/2 Second
  medium is air e2 e0, e1 e0er

29
Reflection from Perfect Conductor
Ei
Er
?i
?r
Et

• Parallel / Perpendicular / vert.
  polarization horiz. polarization
• ?i ?r ?i ?r
• Ei Er Ei - Er

30
Ground Reflection (2-Ray Model)
T (transmitter)
ETOT ELOS Eg
ELOS
Ei
R (receiver)
ht
ErEg
hr
?i
?0
d
31
Field Equations

• d several kms
• ht 50-100m
• ETOT ELOS Eg
• ETOT(d)
• For d gt 20hthr / ?
• Received power Pr

32
Example

• A mobile is located 5 km away from a base
  station, and uses a vertical ?/4 monopole antenna
  with a gain of 2.55dB to receive cellular radio
  signals. The electric field at 1 km from the
  transmitter is measured to be 10-3 V/m. The
  carrier frequency used is 900 MHz.
• (a) Find the length and gain of the receiving
  antenna.

33
Example

• A mobile is located 5 km away from a base
  station, and uses a vertical ?/4 monopole
  antenna with a gain of 2.55dB to receive
  cellular radio signals. The electric field at 1
  km from the transmitter is measured to be 10-3
  V/m. The carrier frequency used is 900 MHz.
• (b) Find the received power at the mobile using
  the 2-way ground model assuming the height of the
  transmitting antenna is 50 m and receiving
  antenna is 1.5 m above the ground.

34
Solution

• d0 1 km
• E0 10 -3 V/m

ht 50 m
hr 1.5 m
d 5 km
35

• (a)
• f 900 MHz
• ? (3 108) / (900 106) 0.33 m
• Length of receiving antenna,
• L ? / 4 0.33/4 0.0833 m 8.33 cm

36
(b) Gain of antenna 2.55 dB gt 1.8 Er (d)
2 10-3 1 103 2? 50 1.5 (5
103)2 0.333 113.1 10-6 V/m
37
Pr (d) I Er I2 ?2 Gr ? 4?
(113.1 10-6) 2 (0.333) 2 1.8 377
4? 5.4 10-13 W -92.68 dBm
38
Diffraction

• Diffraction allows radio signals to propagate
  around the curved surface or propagate behind
  obstructions.
• Based on Huygens principle of wave propagation.

39
Knife-edge Diffraction Geometry

• (a) T is transmitter and R is receiver, with an
  infinite knife-edge obstruction blocking the
  line-of-sight path.

40
Knife-edge Diffraction Geometry

• (b) T R are not the same height...

41
Knife-edge Diffraction Geometry

• ...If ? and ? are small and hltltd1 and d2, then h
  h are virtually identical and the geometry
  may be redrawn as in (c).

42
Knife-edge Diffraction Geometry

• (c) Equivalent where the smallest height (in
  this case hr ) is subtracted from all other
  heights.

T
?
?
ht - hr
R
hobs-hr
?
d2
d1
43
Assumptions

• h ltlt d1, d2
• h gtgt ?
• Excess path length
• ? ?
• ? ?

44
...Assumptions

• h ltlt d1, d2
• h gtgt ?
• Phase difference
• ? 2 ? ? / ?
• 2 ? h2 (d1 d2 )
• 2 ? d1 d2

45
Diffraction Parameter

• v

46
Three Cases

• Case I h gt 0
• Case II h 0
• Case III h lt 0

47
Case I h gt 0

• ? and ? are positive since h is positive.

?
h
T
R
d1
d2
48
Case II h 0

• ? and ? equal 0, since h equals 0.

d1
d2
T
R
49
Case III h lt 0

• ? and ? are negative, since h is negative.

d1
d2
T
R
h
?
50

• The electric field strength of the diffracted
  wave is given by
• Ed F(v) Eo
• where Eo is the free space field strength in the
  absence of both ground and knife edge.

51
Approximate Value of Fresnel Integral F(v)

• Gd (dB) 20 log I F(v) I

52

• v Range Gd (dB)
• v? -1 0
• -1?v? 0 20 log (0.5 0.62 v)
• 0?v?1 20 log (0.5 e-0.95v )
• 1? v? 2.4 20 log (0.4
• v?2.4 20 log (0.225 / v)

53
Example

• Compute the diffraction loss between the
  transmitter and receiver assuming
• ? 1/3 m
• d1 1 km
• d2 1 km
• h 25 m

54
Solution

• Given ? ? 1/3 m
• d1 1 km
• d2 1 km
• h 25 m
• V
• 2.74

55

• Using the table,
• Gd (dB) 20 log (0.225/2.74)
• -22 dB
• Loss 22 dB

56
Scattering

• When a radio wave impinges on a rough surface,
  the reflected energy is spread out or diffused in
  all directions.
• Ex., lampposts and foliage.
• The scattered field increases the strength of the
  signal at the receiver.

57
Radar Cross Section (RCS) Model

• RCS (Radar Cross Section)
• Power density of scattered wave in direction of
  receiver
• Power density of radio wave incident on the
  scattering object

58
Radar Cross Section (RCS) Model

• PR PT GT ?2 RCS
• (4?)3 dT 2 dR 2
• Where,
• PT Transmitted Power
• GT Gain of Transmitting antenna
• dT Distance of scattering object from
  Transmitter
• dR Distance of scattering object from Receiver

59
Practical Link Budget

• Most radio propagation models are derived using a
  combination of analytical and empirical models.
• Empirical approach is based on fitting curves or
  analytical expressions that recreate a set of
  measured data.

60
...Practical Link Budget

• Advantages of empirical models
• Takes into account all propagation factors,
  both known and unknown.
• DisadvantagesNew models need to be measured for
  different environment or frequency.

61
Log-Distance Path Model

• Over many years, some classical propagation
  models have been developed, which are used to
  predict large-scale coverage for mobile
  communication system design.

T d0 R PT PR(d0) PR(d)
62
...Log-Distance Path Model

• Path loss at d0 PT/P(d0) K(d0)n PL(d0)
• Path loss at d PT/P(d) K(d)n PL(d)
• PL(d) / PL(d0) (d/d0)n
• PL(d) dB PL(d0) dB 10n log10 (d/d0)

63
Received Power in Log-distance model

• PR(d) dbm Pt dbm PL(d) db
• n -gt path loss exponent
• d0 -gt reference distance close to transmitter
• Environment n
• Free space 2
• Urban area cellular radio 2.7 3.5
• LOS in building 1.6 1.8

64
Log-Normal Shadowing

• Log-distance path loss normal gives only the
  average value of path loss.
• Surrounding environment may be vastly different
  at two locations having the same T R separation
  d.

65
Log-Normal Shadowing

• More accurate model includes a random variable to
  account for change in environment.
• PL(d) db PL(d) X?
• PL(d0) 10n log (d / d0) X?
• X? -gt Zero mean Gaussian random variable (dB)
• ? -gt Standard deviation (dB)

66
Received Power in Log-Normal Shadowing Model

• PR(d) dbm PTdbm PL(d) db
• Values of n and ? are computed from measured
  data.
• Linear regression method which minimizes the
  difference between measured and estimated path
• Estimated over a wide range of measurement
  locations and T R separations.

67
...Received Power in Log-Normal Shadowing Model

• Probability PR (d) gt ?
• Probability PR (d) lt ?

68
Calculation of Q Function

• Q(z) Q function
• Q(-z) 1- Q(z)
• Q(0) 1/ 2
• Q(z) obtained from Appendix F,
• Table F.1, page 647

69
Calculation of Q Function
70
Example

• Four received power measurements were taken at
  the distances of 100m, 200m, 1 km and 3 km from
  a transmitter. These measured values are given in
  the following table.
• The path loss equation model for other
  measurements follows log normal shadowing model
  where d0 100 m.

71
Example

• A. Find the minimum mean square error (MMSE)
  estimate for the path loss exponent n.
• B. Calculate the standard deviation about the
  mean value.
• C. Estimate the received power at d 2 km using
  the resulting model.
• D. Predict the likelihood that the received
  signal at 2 km will be greater than 60 dBm.

72
Solution

• T-R distance Measured Power
• 100 m 0 dBm
• 200 m - 20 dBm
• 1 km - 35 dBm
• 3 km - 70 dBm

Let Pi be the average received power at
distance di Pi (d) Pi (d0) 10n log (d
/100) d d0 100m gt P0 0 dBm
73
A.

• d1 200 m, P1 -3n,
• d2 1 km, P3 -10n,
• d3 3 km, P4 -14.77n
• Mean square error J ? (P Pi)2
• (0 0)2 -20 (-3n) 2 -35 (-10n) 2
  -70 (-14.77n) 2
• 6525 2887.8n 327.153n2
• Minimum value gt dJ(n) / dn
• 654.306n 2887.8 0 ? n 4.4

74
B.

• Variance ?2 J / 4 ( P Pi)2 / 4
• (0 0) (-20 13.2)2 (-35 44)2 (-70
  64.988)2
• 4
• 152.36 / 4 38.09
• ? ? 6.17 dB

75
C.

• Pi (d 2 km)
• 0 10(4.4) log (2000/100)
• -57.24 dBm

76
D.

• Probability that the received signal will be
  greater than 60 dBm is
• _____
• PR PR(d) gt -60 dBm Q (g- PR (d)) / s
• Q (-60 57.24) / 6.17
• Q - 0.4473
• 1 Q 0.4473
• 1 0.326
• 0.674 gt 67.4

77
of Coverage Area

• Given a circular coverage area of radius R...
• In the area A, the received power PR ? ?
• The area A is defined as U(?)

R
r
Area A
78
Calculation of Coverage Area U(?)

• U (?) (1 / ? R2 ) ƒ Prob PR (R) gt ? dA
• _____
• Where Prob PR (R) gt ? Q ? - PR (R) / ?

79
Final Equation for U(?)

• The error function erf(z)

80
Alternate method

• Use Table 4.18 (page 143)

81
Example

• For the previous problem, predict the percentage
  of area with a 2 km radius cell that receives
  signals greater than 60 dBm.

82
Solution

• From solution to previous example,
• Prob PR (R) gt ? 0.674
• gt(? / n) 6.17 / 4.4
• 1.402
• From table 4.18, Fraction of total area 0.92
  gt 92

83
Other Propagation Models

• Outdoor propagation models
• Longley Rice model point-to-point communication
  systems (40MHz100MHz)
• Okumaras model widely used in urban areas
  (150 MHz 300 MHz)
• Hata model graphical path loss (150 MHz 1500
  MHz)

84
Other Propagation Models

• Indoor propagation models
• Log-distance path loss model
• Ericsson multiple breakdown model