Title: Mobile radio propagation: Small-scale fading
1Mobile radio propagation Small-scale 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 N-2 t N-1
t o t 1 t 2
13Excess delay concept
- The delay axis t, t olt t lt t n-1 is divided
into equal time delay segments called excess
delay bins. - t 0 0
- t 1 ? t
- t 2 2 ? t
- t N-1 (N-1)? 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 - N-1 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
- N-1 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
- N-1 j ? i
- Received signal r(t) ? ai e ptti
- i 0
- Instantaneous received power amplitude
- N-1
- r(t)2 ? ak2
- k 0
- gtTotal received power sum of the power of
individual multipath components.
18Average small-scale received power
- N-1
- Ea,? PwB Ea,? ? ai exp j?i2
- i 0
- N-1 ---
- ? 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
- N-1 j?i
- Received signal r(t) ? ai e
ptti - i
0 - Instantaneous received power
N-1 j?i r(t)2 ? ai e
2 i 0
20Instantaneous Multipath received power amplitude
- N-1 j?i (t,t)
- r(t)2 ? ai e 2
- i 0
21Average power over a local area
- N-1 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 0-0.5s - (b) The average narrow band received power.
28Narrow band instantaneous power
- N-1 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
- N-1
- 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.
33Small-scale 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) e-1
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
- Two-Ray Rayleigh Fading Model
- Saleh and Valenzuela Indoor statistical Model
- SIRCIM (Simulation of Indoor Radio Channels
Impulse Response Models) - SMRCIM (Simulation of Mobile Radio Channel
Impulse-Response Models)