Title: Mobile radio propagation: Smallscale fading
1Mobile radio propagation Smallscale fading
 Fading is the rapid fluctuation of the radio
signal amplitude over a short period of time .  Fading is caused by interference between two or
more versions of transmitted signal, which arrive
at the receiver at slightly different times.  These multipath waves combine at the receiver to
give a resultant signal, which can vary in delay,
amplitude and phase.
2Multipath effects
 Rapid changes in signal strength over a small
distance or time interval.  Random frequency modulation due to varying
Doppler shift on different multipath signals.  Time dispersion (echoes) caused by multipath
propagation delay.
3Causes of fading
 In urban areas, fading occurs because the height
of mobile is lesser than the height of
surrounding structures, such as buildings and
trees.  Existence of several propagation paths between
transmitter and receiver.
4Factors influencing small signal fading
 Multipath propagations
 Speed of mobile (Doppler shift)
 Speed of surrounding objects.
 Transmission bandwidth of signal and bandwidth of
channel.
5Analysis of multipath channel
Transmitter
Spatial position
d
6Convolution model for multipath signal
A2 x(t t 2)
LOS
T
R
x(t)
A1 x(t t 1)
 Received signal y(t) A0 x(t) A1 x(t  t 1)
A2 x(t  t 2) ...
7Time varying system model for channel
 For a fixed position d, the channel between
transmitter receiver can be modulated as a
linear time varying system (LTV system).  Impulse response of the LTI system can be given
as h(d,t).  If x(t) is the transmitted signal, the received
signal can be represented as  y(d,t) x(t) h(d,t)
 denotes convolution
8System definition
 Linear Time Varying (LTV) System
h(t, t)
x(t)
y(t)
9Signal definitions
C(t)
10...Signal definitions
 j2p fct
 y(t) Re r(t) e
 j2p fct
 h(t, t) Re hb(t, t) e
11Base band equivalent channel impulse response
model
hb(t, t)
c(t)
r(t)
12Modeling of the base band impulse response model
hb(t,t)
t3
t2
t1
t0
t N2 t N1
t o t 1 t 2
13Excess delay concept
 The delay axis t, t olt t lt t n1 is divided
into equal time delay segments called excess
delay bins.  t 0 0
 t 1 ? t
 t 2 2 ? t
 t N1 (N1)? t
14Excess delay concept
 All multipath signals received within the bins
are represented by a single resolvable multipath
component having delay t i .  This model can analyze transmitted signals having
bandwidths less than 2/ ? t.
15Simplified mathematical model for baseband
impulse response
 If the channel is assumed to be time invariant,
over a small period of time or over small
distance interval  N1 j ? i
 hb(t) ? ai e ?t ti
 i 0?????
 Output r(t) c(t) hb(t)
16...Simplified mathematical model for base band
impulse response
 N1 j? i
 r(t) ? ai e ct ti
 i 0?????
 For measuring or predicting the impulse response
a probing pulse c(t) ?t is used.
17Relationship between bandwidth and received power
 Wideband signals
 N1 j ? i
 Received signal r(t) ? ai e ptti
 i 0
 Instantaneous received power amplitude
 N1
 r(t)2 ? ak2
 k 0
 gtTotal received power sum of the power of
individual multipath components.
18Average smallscale received power
 N1
 Ea,? PwB Ea,? ? ai exp j?i2
 i 0
 N1 
 ? ai2
 i 0
 Ea,? average overall possible values of ai
and ? I in a local area.  ai 2 sample average area local measurement
area, generally measured using multipath
measurement equipment.
19Relationship between bandwidth and received power
 Narrowband signals
 N1 j?i
 Received signal r(t) ? ai e
ptti  i
0  Instantaneous received power
N1 j?i r(t)2 ? ai e
2 i 0
20Instantaneous Multipath received power amplitude
 N1 j?i (t,t)
 r(t)2 ? ai e 2
 i 0
21Average power over a local area
 N1 j?i (t, t)
 Ea,? PwB Ea,? ? ai e
2  i 0
22Conclusions
 When the transmitted signal has?a wide bandwidth
gtgt bandwidth of the channel multipath structure
is completely resolved by the receiver at any
time and the received power varies very little.  When the transmitted signal has a very narrow
bandwidth (example the base band signal has a
duration greater than the excess delay of the
channel) then multipath is not resolved by the
received signal and large signal fluctuations
occur (fading).
23Example
 Assume a discrete channel impulse response is
used to model urban radio channels with excess
delays as large as 100 ?s and microcellular
channels with excess delays not larger than 4 ?s.
If the number of multipath bins is fixed at 64
find  (a) ? t
 (b) Maximum bandwidth, which the two models can
accurately represent.
24Solution
 Delays in channel ? t, 2 ? t . N ? t
 Maximum excess delay of channel
 t N N ? t 100 ? s.

 N 64
 ?t tN /N 100 ?s /64 1.5625 ?s
 Maximum bandwidth represented accurately by model
2/ ?t  1.28 MHz
25For microcellular channel
 Maximum excess delay of channel
 t N N ? t 4 ? s.

 N 64

 ? t t N /N 4 ? s /64 62.5 ns
 Maximum bandwidth represented accurately by model
2/ ? t
32 MHz
26Example
 Assume a mobile traveling at a velocity of 10m/s
receives two multipath components at a carrier
frequency of 1000MHz.  The first component is assumed to arrive at t 0
with an initial phase of 0? and a power of
70dBm.  The second component is 3dB weaker than the first
one and arrives at t 1? s, also with the
initial phase of 0?.
27...Example
 If the mobile moves directly in the direction of
arrival of the first component and directly away
from the direction of arrival of the second
component, compute the following  (a) The narrow band and wide band received power
over the interval 00.5s  (b) The average narrow band received power.
28Narrow band instantaneous power
 N1 j?i (t,t)
 r(t)2 ? ai e 2
 i
0  Now 70dBm gt 100 pw so a1 v 100 pw
 and 73dBm gt 50 pw so a2 v 50 pw
 ?i 2pd/? 2pvt/?
 ? (3108)/(100106) 0.3 m
 ?1 2p10t/0.3 209.4 t rad.
29 ?2 ?1 209.4 t rad.
 t 0
 r(t)2? v100 ? v50 2 291pw
 t 0.1
 r(t)2 v100. e j209.4 x 0.1 v50. e j209.4
x 0.1  78.2pw
 t 0.2
 r(t)2 v100. e j209.4 x 0.2 v50. e j209.4
x 0.2  81.5pw
30 t 0.3
 r(t)2 291pw
 t 0.4
 r(t)2 78.2pw
 t 0.5
 r(t)2 81.5pw
31Wideband instantaneous power
 N1
 r(t)2 ? ak2 100 50 150 pW
 k 0
32Average narrow band received power
 Ea,? PCW 2(291) 2(78.2) 2(81.5) /6
 150.233pw
 The average narrow band power and wideband power
are almost the same over 0.5s.  While the narrow band signal fades over the
observation interval, the wideband signal remains
constant.
33Smallscale multipath measurements
 Direct Pulse Measurements
 Spread Spectrum Sliding Correlator Measurement
 Swept Frequency Measurement
34Types of Small Scale Fading
Doppler Spread
Multipath time delay
Fast Fading
Slow fading
Flat fading
Frequency Selective Fading
35...Types of Small Scale Fading
 2 main propagation mechanisms
 Multipath time delay spread
 Doppler spread
 Two types of fading are independent of each other.
36Multipath terms associated with fading
 Ts Symbol period or reciprocal bandwidth
 Bs Bandwidth of transmitted signal
 Bc coherence bandwidth of channel
 Bc 1/(50??)??where ??? is rms delay spread
37...Multipath terms associated with fading
 __ _
 ?? 2 ? 2  ( ??)2
 _
 ? (? ak2 ??) / (? ak2) mean Excess delay
 __
 ? 2 (? ak2 ??2) / (? ak2)
38Fading effects due to Doppler spread
 fc frequency of pure or transmitted sinusoid
 Received signal spectrum
 fc / fd, fd
 Doppler shift
fc
S
?
V
39Doppler spread and coherence time
 Doppler frequency shift fd (v / ?) cos ? ,
 Where Wavelength ? c / fc meters
 Doppler Spread BD fm Maximum Frequency
deviation v / ?  Coherence time Tc 0.423 / fm
40Types of fading
 Flat fading
 Mobile channel has constant gain and linear phase
response.  Spectral characteristics of the transmitted
signal are maintained at receiver.  Bs ltlt Bc
 gt Ts gtgt ??
41...Types of fading
 Frequency selective fading
 Mobile channel has a constant gain and linear
phase response over a bandwidth.  Bs gt Bc
 gt Ts lt ??
 Common rule of thumb
 If Ts gt 10 ?? gt Flat fading
 If Ts lt 10 ?? gt Frequency selective fading
42How to decide flat or frequency selective fading?
 Common rule of thumb
 If Ts 10 ?? gt Flat fading
 If Ts lt 10 ?? gt Frequency selective fading
43Fast fading channel
 The channel impulse response changes rapidly
within the symbol duration.  This causes frequency dispersion due to Doppler
spreading, which leads to signal distortion.  Ts gt Tc
 Bs lt BD
44Slow fading channel
 The channel impulse response changes at a rate
much slower than the transmitted signal s(t).  Ts ltlt Tc
 Bs gtgt BD
 Velocity of mobile (or velocity of objects in
channel) and base band signaling determines slow
fading or fast fading.
45Rayleigh and Ricean distributions...
 Rayleigh fading distribution
 In mobile radio channels, the Rayleigh
distribution is commonly used to describe the
statistical time varying nature of the received
envelope of flat fading signal.
46 Pdf (Probability density function)
 p(r) (r/?2) e (r2/2?2) (0 r ?)
 0 r lt 0
 ? ? rms value of received voltage before envelope
detection.
47Cumulative distribution function (cdf)
 R
 P (R) P( r ? R) ? p(r) dr
 0
 1 e (R2/2?2)
 8
 Mean Value ER ? r p(r) dr
 0
 ???/2
 1.25336 ?
48 ?R2 ER2  ?E(R)?2
 8
 ? r2 p(r) dr  ?2p/2
 0
 0.4292 ?2
 Median value for r gt ½
 ? pr dr gt r (median) 1.77?
 rmedian
 Median value for r ? p(r ) dr gt rmedian
1.77?
0
49Ricean fading distribution
 When there is a dominant (non fading) signal
component present such as LOS propagation path,
the small scale fading envelope distribution is
Ricean.  This can be modeled as random, multipath
components arriving at different angles
superimposed on a stationary dominant signal.
50...Ricean fading distribution
 p(r) (r/?2)e (r2 A2) / (2?2) Io(Ar/?2 )
 for A ? 0, r ? 0
 p(r) 0 for r lt 0
 Io ? Modified Bessel function of first kind and
zero order
51Ricean factor
 K(dB) 10 log(A2/2?2) dB
 10 log (Deterministic signal power/ variance
of multipath)
52Level crossing and fading statistics
 Level crossing rate (LCR) is defined as the
expected rate at which the normalized Rayleigh
fading envelope, crosses a specified level in a
positive going direction.
53Simplified equation for LCR
 NR ?(2?) fm ?e?2
 fm Maximum Doppler frequency
 ? R/Rms is the value of the specified level R,
normalized to the local rms amplitude of the
fading signal
54Example
For a Rayleigh fading signal, compute the
positive going level crossing rate for ? 1,
when the maximum Doppler frequency (fm) is 20
Hz. What is the maximum velocity of the mobile
for this Doppler frequency if the carrier is 900
MHz?
55Solution
? 1 fm 20 Hz The number of zero level
crossings is NR ?2? (20) e1
18.44 Crossings/Sec Maximum velocity of mobile
fd ? 20 (3 X 108)/(900X106) 6.66 m/s
56Average fade duration
 Average period of time for which the received
signal is below a specified level R.
57Formula for Average fade duration
58Example
 Find the average fade duration for threshold
level ? 0.01, ? 0.1 and ? 1, when the
Doppler frequency is 20Hz.
59Solution
 ? ? e?2 1
 ?fm?2?
 0.01 19.9?s
 0.1 200?s
 1.0 3.43ms
60Statistical methods for multipath fading channels
 Clarks model for Flat Fading
 TwoRay Rayleigh Fading Model
 Saleh and Valenzuela Indoor statistical Model
 SIRCIM (Simulation of Indoor Radio Channels
Impulse Response Models)  SMRCIM (Simulation of Mobile Radio Channel
ImpulseResponse Models)