1. MSIT 413: Wireless Technologies Week 3
Michael L. Honig
Department of EECS
Northwestern University
January 2011

2. Cellular Frequency AssignmentsB A C E D G A C B F D G A C B E F D G C B E F D G A C B E D G A C B
Will cellular layout look this regular in practice? Why not use squares instead of hexagons? F G A D
cell cluster
(contains all channels)
co-channel cells

3. Cellular Terminology
Cell cluster: group of N neighboring cells which use the complete set of available frequencies.
Cell cluster size: N
Frequency reuse factor: 1/N
Uplink or reverse link: Mobiles  Base station
Downlink or forward link: Base station  mobiles
Co-channel cells: cells which are assigned the same frequencies C D F A B G E
cell cluster
Increase in frequency reuse means less spectrum is needed to cover a region.

4. Performance Measure: Signal-to-Interference-Plus-Noise Ratio (SINR)
Expressed in dB (10 log (SINR))
Typically, the interference power dominates (ignore noise)
SINR  Signal-to-Interference Ratio (SIR or S/I)
Total interference power is the sum over all interferers:
More co-channel users  more interference
If there is no noise, can everyone increase their S/I by raising their powers?

5. Voice Capacity Defined as
Example: SE = 0.1 Erlang/MHz/km2  10 MHz needed to support 1 Erlang/km2 (Note that 1 km2 may correspond to a cell.)
To convert Erlangs to users, must estimate traffic per user, e.g., if on average, each user is active 1/10 of the time, then 1 Erlang corresponds to 10 users.
Traffic per cell depends on:
Number of channels
Grade of Service (e.g., typically 2%)
S/I requirement (determined by cluster size N)
What is a reasonable performance measure for a data service?

6. Channel Allocation
Consider GSM: 25 MHz (simplex), 200 kHz channels  125 channels (1 is used for control signaling)
How to allocate channels to cells?
Suburb
(lightly loaded)
Shopping mall
Train station

7. Channel Allocation
Consider GSM: 25 MHz (simplex), 200 kHz channels  125 channels (1 is used for control signaling)
How to allocate channels to cells?
Objective: equalize blocking probability (target is around 2%)
Fixed channel allocation assigns fixed set of channels to each cell
Drawback?
Suburb
(lightly loaded)
mall
Train station

8. Channel Allocation
Consider GSM: 25 MHz (simplex), 200 kHz channels  125 channels (1 is used for control signaling)
How to allocate channels to cells?
Objective: equalize blocking probability (target is around 2%)
Fixed channel allocation assigns fixed set of channels to each cell.
Drawback: traffic is time-varying
Dynamic channel allocation varies the number of channels per cell, depending on the load.
Suburb
(lightly loaded)
mall
Train station

9. Dynamic Channel Allocation
Channels already in use
2,6
1,3,4
new call: search for channel
1,3 5

10. Dynamic Channel Allocation
Channels already in use
2,6
3,4
new call: search for channel
1,3 4 5
Channels assigned from complete set of available channels
Each user searches for a channel with high SINR (low interference)
No frequency planning!

11. 802.11b/g Channels Channels: 1 6 11
Comparison table
Channels: 1 6 11
14 overlapping (staggered) channels (11 in the U.S.)
Center frequencies are separated by 5 MHz
Bandwidth/interference controlled by “spectral mask”
30 dB attenuation 11 MHz from center frequency
50 dB attenuation 22 MHz from center frequency

12. Dynamic Channel Allocation: 802.11b/g

13. Dynamic Channel Allocation: 802.11b/g
interference
Add new router
Channel 6

14. Dynamic Channel Allocation: 802.11b/g
Switches to channel 1
Channel 11
Channel 1
interference
Channel 6

15. Dynamic Channel Allocation: 802.11b/g
Switches to channel 6
interference
Channel 6

16. Dynamic Channel Allocation: 802.11b/g
interference
Switches to channel 1
Dynamic channel assignment becomes unstable!

17. Overlapping Channel Assignment
Interference
(less than before)
Interference
(less than before)
power
Channel 3
Channels: 1 3 6
frequency

18. FCA vs. DCA FCA DCA Low complexity Better under heavy traffic
Sensitive to changes in traffic
Variable grade of service
Higher probability of outage
Suitable for macro-cellular systems (e.g., cellular)
Low call setup delay
Requires careful frequency planning
Centralized assignment
Moderate/High complexity
Must monitor channel occupancy, traffic distribution, S/I (centralized)
Better under light/moderate traffic
Insensitive to changes in traffic
Stable grade of service
Low probability of outage (call termination)
Suitable for micro-cellular systems (e.g., cordless)
Moderate/high call setup delay
No frequency planning
Assignment can be centralized or distributed

19. Why Study Radio Propagation?
To determine coverage
Must determine path loss
Function of
Frequency
Distance
Terrain (office building, urban, hilly, rural, etc.)
Can we use the same
channels?
What determines “useful” spectrum?
Need “large-scale” models

20. Why Study Radio Propagation?
What determines “useful” spectrum? 20

21. Why Study Radio Propagation?
To enable robust communications (MODEM design)
How can we guarantee reliable communications?
What data rate can we provide?
Must determine signal statistics:
Probability of outage
Duration of outage
Received Power
Deep fades may cause an outage
time
Slow fading means slower variations in received signal strength with time.
Radio propagation has motivated the trend to CDMA and OFDM.
Need “small-scale” models

22. Will provide answers to…
What are the major causes of attenuation and fading?
Why does the achievable data rate decrease with mobility?
Why are wireless systems evolving to wider bandwidths (spread spectrum and OFDM)?
Why does the accuracy of location tracking methods increase with wider bandwidths?

23. Propagation Key Words Large-scale effects Path-loss exponent
Shadow fading
Small-scale effects
Rayleigh fading
Doppler shift and Doppler spectrum
Coherence time / fast vs slow fading
Narrowband vs wideband signals
Multipath delay spread and coherence bandwidth
Frequency-selective fading and frequency diversity

24. Propagation Mechanisms: 1. Free Space
distance d
reference distance d0=1
Note that P0 decreases with frequency.
Reference power at reference distance d0
Path loss exponent=2 P0
In dB: Pr = P0 (dB) – 20 log (d)
slope = -20 dB per decade
Pr (dB)
P0 = Gt Gr (/4)2
log (d)
antenna gains
wavelength

25. Wavelength  (meters) = c (speed of light) / frequency
Wavelength >> size of object  signal penetrates object.
Wavelength << size of object  signal is absorbed and/or reflected by object.
Large-scale effects refers to propagation over distances of many wavelengths. Small-scale effects refers to propagation over a distances of a fraction of a wavelength.
Work out dimensions in equations.
How should wavelength be related to antenna size?

26. Dipole Antenna 802.11 dipole antenna cable from transmitter
wire (radiator)

27. Radiation Pattern: Dipole Antenna
Dipole axis
Dipole axis
What happens if you are directly above the antenna?
What happens if you tilt the antenna?
Electromagnetic wave radiates out from the dipole axis.
Cross-section of doughnut pattern

28. Antenna Gain Pattern Dipole pattern (close to isotropic)
Red curve shows the antenna gain versus
angle relative to an isotropic pattern (perfect circle) in dB.
Often referred to as dBi, dB “isotropic”.
-5 dB (factor of about 1/3) relative to isotropic
pattern
Dipole pattern (close to isotropic)

29. Antenna Gain Pattern
Dipole pattern (vertical)
90 degree sector

30. Attenuation: Wireless vs. Wired
1 GHz Radio (free space)
Unshielded Twisted Pair
Path loss ~ 30 dB for the first meter + 20 dB / decade
70 dB / 100 meters (2 decades)
90 dB / 1 km (3 decades)
130 dB / 100 km!
Increases as log (distance)
Repeaters are infeasible for satellites
Path loss ~ 13 dB / 100 meters or 130 dB / 1 km
Increases linearly with distance
Requires repeaters for long distances
Short distance  Wired has less path loss.
Large distance  Wireless has less path loss.

31. Propagation Mechanisms
Incident E-M wave q
reflected wave
transmitted wave
Length of boundary >> wavelength l
2. Reflection
3. Diffraction
4. Scattering
Signal loss depends
on geometry
Hill

32. Why Use > 500 MHz?

33. Why Use > 500 MHz? There is more spectrum available above 500 MHz.
Lower frequencies require larger antennas
Antenna dimension is on the order of a wavelength = (speed of light/frequency) = 0.6 M @ 500 MHz
Path loss increases with frequency for the first meter
10’s of GHz: signals are confined locally
More than 60 GHz: attenuation is too large (oxygen absorbs signal)
Transition to HDTV: Feb 17, Prime spectrum becomes available for other radio commercial services. Why is analog TV spectrum “prime”?

34. 700 MHz Auction Broadcast TV channels 52-69 relocated in Feb. 2009.
6 MHz channels occupying 698 – 806 MHz
Different bands were auctioned separately:
“A” and “B” bands: for exclusive use (like cellular bands)
“C” band (11 MHz): must support open handsets, software apps
“D” band (5 MHz): shared with public safety (has priority)
Commenced January 24, 2008, ended in March
Analog TV broadcasters to move to channels

35. Why all the Hubbub?
This band has excellent propagation characteristics for cellular types of services (“beach-front property”).
Carriers must decide on technologies: 3G, LTE, WiMax,…
Rules for spectrum sharing can be redefined…

37. C Band Debate
Currently service providers in the U.S. do not allow any services, applications, or handsets from unauthorized 3rd party vendors.
Google asked the FCC to stipulate that whoever wins the spectrum must support open applications, open devices, open services, open networks (net neutrality for wireless).
Verizon wants to maintain “walled-garden”.
FCC stipulated open applications and devices, but not open services and networks: spectrum owner must allow devices or applications to connect to the network as long as they do not cause harm to the network
Aggressive build-out requirements:
Significant coverage requirement in four years, which continues to grow throughout the 10-year term of the license.

38. Sold to… Verizon Other winners: AT&T (B block), Qualcomm (B, E blocks)
Total revenue: $19.6 B
$9.6 B from Verizon, $6.6 B from AT&T
Implications for open access, competition?

39. D Band Rules
Winner gets to use both D band and adjacent public service band (additional 12 MHz!), but service can be preempted by public safety in emergencies.
Winner must build out public safety network: must provide service to 75% of the population in 4 years, 95% in 7 years, 99.3% in 10 years
Minimum bid: $1.3 B; estimated cost to deploy network: $10-12 B
Any takers? …

40. D Band Rules
Winner gets to use both D band and adjacent public service band (additional 12 MHz!), but service can be preempted by public safety in emergencies.
Winner must build out public safety network: must provide service to 75% of the population in 4 years, 95% in 7 years, 99.3% in 10 years
Minimum bid: $1.3 B; estimated cost to deploy network: $10-12 B
Any takers? … Nope! Highest bid was well below reserve…

41. Radio Channels Troposcatter Microwave LOS Mobile radio Indoor radio
Troposcatter enables “over-the-horizon” communications (e.g., ham radio). T T
Mobile radio
Indoor radio

42. Sinusoidal Signal Electromagnetic wave s(t) = A sin (2  f t + )
Time delay = 12,
Phase shift  = 12/50 cycle = 86.4 degrees
Amplitude A=1
s(t)
Period= 50 sec, frequency f = 1/50 cycle/sec
Time t (seconds)

43. Two Signal Paths s1(t) s2(t) Received signal r(t) = s1(t) + s2(t)
Suppose s1(t) = sin 2f t.
Then s2(t) = h s1(t - ) = h sin 2f (t - )
attenuation
(e.g., h could be ½)
delay
(e.g.,  could be 1 microsec.)

44. Sinusoid Addition (Constructive)
r(t) + =
s2(t)
Adding two sinusoids with the same frequency gives another sinusoid with the same frequency!

45. Sinusoid Addition (Destructive)
r(t) + =
s2(t)
Signal is faded.

46. Indoor Propagation Measurements
Hypothetical large indoor environment
Ceiling
Why does a have less range than 11b/g?
Normalized received power vs. distance

47. Power Attenuation In dB: Pr = P0 (dB) – 10 n log (d)
distance d
reference distance d0=1
Reference power at reference distance d0
Path loss exponent P0
slope (n=2) = -20 dB per decade
In dB: Pr = P0 (dB) – 10 n log (d)
Pr (dB)
slope = -40 (n=4)
log (d)

48. Path Loss Exponents ENVIRONMENT PATH LOSS EXPONENT, n Free space 2
Urban cellular radio
2.7 to 3.5
Shadowed urban cellular radio
3 to 5
In building line-of-site
1.6 to 1.8
Obstructed in building
4 to 6
Obstructed in factories
2 to 3

49. Large-Scale Path Loss (Scatter Plot)
Average Received Power (dBm)
Distance (meters)

50. Shadow Fading
Random variations in path loss as mobile moves around buildings, trees, etc.
Modeled as an additional random variable:
“normal” (Gaussian)
probability distribution
Pr = P0 – 10 n log d + X
“log-normal” random variable
standard deviation - 
received power in dB
For cellular:  is about 8 dB

51. Large-Scale Path Loss (Scatter Plot)
Most points are less than
 dB from the mean

52. Empirical Path Loss Models
Propagation studies must take into account:
Environment (rural, suburban, urban)
Building characteristics (high-rise, houses, shopping malls)
Vegetation density
Terrain (mountainous, hilly, flat)
Okumura’s model (based on measurements in and around Tokyo)
Median path loss = free-space loss + urban loss + antenna gains + corrections
Obtained from graphs
Additional corrections for street orientation, irregular terrain
Numerous indoor propagation studies for

53. SINR Measurements: 1xEV-DO
drive test plots

54. Link Budget How much power is required to achieve target S/I?
dBs add: Target S/I (dB) + path loss (dB) + other losses (components) (dB) - antenna gains (dB) Total Power needed at transmitter (dB)
Actual power depends on noise level.
Given 1 microwatt noise power, if the transmit power is 60 dB above the noise level then the transmit power is 1 Watt.
dB is a ratio of two powers, dBm is a ratio with mW in the denominator.

55. Example wireless channel Transmitter Receiver 40 dB attenuation
Received power
must be > -30 dBm
What is the required
Transmit power?
Recall that dBm measures the signal power relative to 1 mW (milliwatt) = Watt. To convert from S Watts to dBm, use S (dBm) = 10 log (S / 0.001)
Transmitted power (dBm) = = 10 dBm = 10 mW
What if the received signal-to-noise ratio must be 5 dB, and the noise power is -45 dBm?
P (transmitter) – channel loss (dB) = P (receiver)
Received SNR = 5 dB means P (receiver) – noise power = 5 dB

56. Urban Multipath No direct Line of Sight between mobile and base
Radio wave scatters off of buildings, cars, etc.
Severe multipath

57. Narrowband vs. Wideband
Narrowband means that the bandwidth of the transmitted signal is small (e.g., < 100 kHz for cellular). It therefore looks “almost” like a sinusoid.
Multipath changes the amplitude and phase.
Wideband means that the transmitted signal has a large bandwidth (e.g., > 1 MHz for cellular).
Multipath causes “self-interference”.
What is the bandwidth of a sinusoid? Duration?
What is the bandwidth of a short pulse?

58. Narrowband Fading
Received signal r(t) = h1 s(t - 1 ) + h2 s(t - 2) + h3 s(t - 3 ) + …
attenuation
for path 1 (random)
delay for path 1 (random)
If the transmitted signal is sinusoidal (narrowband), s(t) = sin 2f t,
then the received signal is also sinusoidal, but with a different
(random) amplitude and (random) phase:
r(t) = A sin (2f t + )
Each term in the sum is a sinusoidal type of signal.
Transmitted s(t)
Received r(t)
A,  depend on environment, location of transmitter/receiver

59. Rayleigh Fading Can show: A has a “Rayleigh” distribution
 has a “uniform” distribution
(all phase shifts are equally likely)
Probability (A < a) = 1 – e-a2/P0
where P0 is the average received power (averaged over different locations)
Ex: P0 =1, a=1: Pr(A<1) = 1 – e-1 = 0.63 (probability that signal is faded)
P0 = 1, a=0.1: Pr(A<0.1) = 1 – e-1/100 ≈ (prob that signal is severely faded)
Prob(A < a) 1
Assume received amplitude without fading is 1 (corresponds to P0=1).
1-e-a2/P0 a

60. Small-Scale Fading
a = 0.1
What should the speed of the variations depend on?

61. Small-Scale Fading Fade rate depends on Mobile speed
Speed of surrounding objects
Frequency

62. Short- vs. Long-Term Fading
Short-term fading
Long-term fading
Signal Strength (dB) T T
Time (t)
Long-term (large-scale) fading:
Distance attenuation
Shadowing (blocked Line of Sight (LOS))
Variations of signal strength over distances
on the order of a wavelength

63. Combined Fading and Attenuation
Received power
Pr (dB)
distance attenuation
Time (mobile is moving away from base)

64. Combined Fading and Attenuation
Received power
Pr (dB)
distance attenuation
shadowing
Time (mobile is moving away from base)

65. Combined Fading and Attenuation
Received power
Pr (dB)
distance attenuation
shadowing
Rayleigh fading
Time (mobile is moving away from base)

66. Example Diagnostic Measurements: 1XEV-DO
drive test measurements
drive path

67. Time Variations: Doppler Shift
Audio clip (train station)

68. Time Variations: Doppler Shift
velocity v
distance d = v t
Propagation delay = distance d / speed of light c = vt/c
transmitted
signal s(t)
delay increases
received
signal r(t)
propagation
delay
Received signal r(t) = sin 2f (t- vt/c) = sin 2(f – fv/c) t
received frequency
Doppler shift fd = -fv/c 68

69. Doppler Shift (Ex)
Mobile moving away from base  v > 0, Doppler shift < 0
Mobile moving towards base  v < 0, Doppler shift > 0
Carrier frequency f = 900 MHz, v = 60 miles/hour = meters/sec
Mobile  Base: fd = fv/c = (900 × 106) × / (3 × 108) ≈ 80 Hz
meters/sec

70. Doppler (Frequency) Shift
½ Doppler “cycle”
80 Hz Doppler cycle means the phase “turns around” 80 times per second.
in phase
out of phase
Frequency= 1/50
Frequency= 1/45

71. Doppler Shift (Ex)
Mobile moving away from base  v > 0, Doppler shift < 0
Mobile moving towards base  v < 0, Doppler shift > 0
Carrier frequency f = 900 MHz, v = 60 miles/hour = meters/sec
Mobile  Base: fd = fv/c = (900 × 106) × / (3 × 108) ≈ 80 Hz
Suppose the data rate is 9600 bits/sec, 80 Hz Doppler shift
 phase inversion every (9600/80)/2 = 60 bits!
As the data rate increases, Doppler shift becomes less significant, i.e., channel is stable over more transmitted bits.
IS-136 data rate: 48.6 kbps GSM data rate 270 kbps

72. Application of Doppler Shift: Astronomy
Doppler shift determines
relative velocity of
distant objects
(e.g., stars, galaxies…)
Observed “spectral lines”
(radiation is emitted
at discrete frequencies)
Is the cluster moving away or coming closer?
“red shift”: object
is moving away
“blue shift” object
is moving closer
sun light spectrum
spectrum of galaxy supercluster 72

73. Application of Doppler Shift: Police Radar
What happens if other objects are present?
Doppler shift can be used to compute relative speed. 73

74. Scattering: Doppler Spectrum
distance d = v t
transmitted
signal s(t)
received
signal ??
power
Received signal is the sum of all scattered waves
Doppler shift for each path depends on angle (vf cos /c )
Typically assume that the received energy is the same from all directions (uniform scattering)
freq.
frequency of s(t)

75. Scattering: Doppler Spectrum
distance d = v t
transmitted
signal s(t)
Doppler Spectrum
(shows relative strengths of Doppler shifts)
Doppler shift fd
power
power
2fd
frequency
frequency of s(t)
frequency
frequency of s(t) + Doppler shift fd

76. Scattering: Doppler Spectrum
distance d = v t
transmitted
signal s(t)
power
Doppler spectrum
frequency
power
2fd
frequency of s(t)
frequency of s(t) + Doppler shift fd

77. Rayleigh Fading
deep fade
phase shift
Received waveform
Amplitude (dB)

78. Channel Coherence Time
Coherence Time: Amplitude and phase
are nearly constant.
Rate of time variations depends on Doppler shift:
(velocity X carrier frequency)/(speed of light)
Coherence Time varies as 1/(Doppler shift).

79. Fast vs. Slow Fading transmitted bits time time received amplitude
Fast fading: channel changes
every few symbols. Coherence time
is less than roughly 100 symbols.
Slow fading: Coherence time
lasts more than a few 100 symbols.

80. Fade Rate (Ex)
fc = 900 MHz, v = 60 miles/hour  Doppler shift ≈ 80 Hz. Coherence time is roughly 1/80, or 10 msec
Data rate (voice): 10 kbps or 0.1 msec/bit  100 bits within a coherence time (fast fading)
GSM data rate: 270 kbps  about 3000 bits within a coherence time (slow fading)

81. Channel Characterizations: Time vs. Frequency
Frequency-domain description
Time-domain description
Multipath
channel
Amplitude attenuation,
Delay (phase shift)
input s(t) is a sinusoid
“narrowband” signal
The impulse response is analogous to echoes heard when clapping your hands in an auditorium.
r(t)
s(t)
Multipath
channel
time t
time t
multipath components
input s(t) is an impulse (very short pulse)
“wideband” signal
(Note: an impulse has zero duration and infinite bandwidth!)

82. Pulse Width vs. Bandwidth
Power
signal pulse
bandwidth = 1/T
Narrowband
frequency
time T
signal pulse
What happens to the signal as the bandwidth goes to zero? To infinity?
Power
bandwidth = 1/T
Wideband
time
frequency T

83. Power-Delay Profile
Received power vs. time in response to a transmitted short pulse.
delay spread 
For cellular systems (outdoors), the delay spread is typically a
few microseconds.

84. Two-Ray Impulse Response
reflection (path 2)
direct path (path 1)
s(t)
reflection is attenuated
r(t) 
time t
time t

85. Two-Ray Impulse Response
reflection (path 2)
direct path (path 1)
s(t)
reflection is attenuated
r(t) 
time t
time t
 = [(length of path 2) – (length of path 1)]/c

86. Urban Multipath
s(t)
r(t)
time t
time t
r(t)
different location
for receiver
time t
Spacing and attenuation of multipath components depend on
location and environment.

87. Delay Spread and Intersymbol Interference
s(t)
r(t)
Multipath
channel
time t
time t
Time between pulses is >> delay spread, therefore the received
pulses do not interfere.
r(t)
Intersymbol interference is analogous to distortion due to echoes when shouting in a cave.
s(t)
Multipath
channel
time t
Time between pulses is < delay spread, which causes
intersymbol interference. The rate at which symbols can be
transmitted without intersymbol interference is 1 / delay spread.

88. Coherence Bandwidth channel gain frequency f1 f2
coherence bandwidth Bc
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath.
frequency f1 f2
Is the coherence bandwidth determined by man or nature?
The channel gain is approximately constant within a coherence bandwidth Bc. Frequencies f1 and f2 fade independently if |f1 – f2 | >> Bc.

89. Coherence Bandwidth and Delay Spread
frequency
channel
gain
coherence bandwidth Bc
delay spread 
channel
gain
delay spread 
In general, the shape of the frequency response depends on the timing and position of the multipath.
coherence bandwidth Bc
frequency
Coherence bandwidth is inversely proportional to delay spread:
Bc ≈ 1/.

90. Narrowband Signal signal power channel (narrowband) gain frequency f1
coherence bandwidth Bc
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath.
frequency f1 f2
The signal power is confined within a coherence band.
Flat fading: all signal frequencies are affected the same way.

91. Wideband Signals signal power channel (wideband) gain frequency f1 f2
coherence bandwidth Bc
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath.
frequency f1 f2
What are advantages and disadvantages of frequency-selective fading?
A wideband signal spans many coherence bands.
Frequency-selective fading: different parts of the signal (in frequency) are affected differently by the channel.

92. Frequency Diversity signal power channel (wideband) gain frequency f1
coherence bandwidth Bc
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath.
frequency f1 f2
Wideband signals exploit frequency diversity. Spreading power across many coherence bands reduces the chances of severe fading.
Wideband signals are distorted by the channel fading (distortion causes intersymbol interference).
What are advantages and disadvantages of frequency-selective fading?
Analogous to investing: coherence bands represent companies/sectors; power is investing budget.

93. Coherence Bandwidth for Cellular
signal power
(wideband)
coherence bandwidth Bc
channel
gain
Frequencies far outside the coherence
bandwidth are affected differently by multipath.
frequency
What are advantages and disadvantages of frequency-selective fading? f1 f2
For the cellular band, Bc is around 100 to 300 kHz.
How does this compare with the bandwidth of cellular systems?

94. Fading Experienced by Wireless Systems
Standard Bandwidth Fade rate
AMPS kHz (NB) Fast
IS kHz Fast
GSM kHz Slow
IS-95 (CDMA) MHz (WB) Fast
3G MHz Slow to Fast
(depends on rate)
LTE up to 20 MHz Slow
> 20 MHz Slow
Bluetooth > 5 MHz (?) Slow