Transcript No Slide Title

Small-scale Mobile radio propagation



1
Small scale propagation implies signal quality
in a short distance or time range
In this small range, fading or rapid fluctuation
of the signal amplitude is observed
One cause of fading is multipath or the
process of signals reaching the receiver
through different mechanisms such as LOS,
reflection, diffration and scattering.
Multipath effects



2
Rapid changes in signal amplitude over a
small distance or time interval.
Rapid changes in signal phaseover a
small distance or time interval.
Time dispersion (echoes) caused by
multipath propagation delay.
Causes of fading


3
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.
Factors influencing small signal fading
4

Multipath propagations

Speed of mobile (Doppler shift)

Speed of surrounding objects

Bandwidths of signal and channel
Analysis of multipath channel
Transmitter
Receiver
Spatial position
d
5
Convolution model for multipath signal
T , x(t)
LOS
A2 x(t- t 2)
R, y(t)
A1 x(t- t 1)
Received signal:
6
y(t) = A0 x(t) + A1 x(t - t 1) + A2 x(t - t 2) + ...
System definition of multipath
x(t)
7
h(t)
y(t)
Baseband signal definition

Transmitted signal
c(t)- pulse
x(t) = Re { c(t) e
8
j 2  f ct
}
Received signal and system response
Received signal
j2 fct
y(t) = Re {r(t) e
}

Impulse response
j2 fct
h(t) = Re {hb(t) e
}

9
Base band equivalent channel
c(t)
10
hb(t)
r(t)
Modeling of the baseband multipath model
Mathematical model
r(t) = c(t) * hb(t, t)

hb(t,t)
t3
t2
t1
t0
11
to t1 t2
t N-2 t N-1
Excess delay concept

The delay axis t, t o<= t <= t n-1 is divided into
equal time delay segments called excess delay
bins.
t0 = 0
t1 =  t
t2 = 2  t
t N-1 = (N-1) t
12
Delay component design


13
All multipath signals received within the
bins are represented by a single
resolvable multipath component having
delay t i .
Design equation for bin width t:
Bandwidth of signal = 2/t
Final model for multipath response
r(t) =






14
N-1 j i
 ai e
c[t –ti]
i=0
c(t) – Transmitted pulse
r(t) – Received pulse
N – Number of multipaths
ai – Amplitude of multipath i
i – Amplitude of multipath i
ti – Amplitude of multipath i
Wideband multipath signals
N-1
ji
Received signal r(t) =  ai e p[t–ti]
i=0
Instantaneous received power:
N-1
|r(t)|2 =  |ak|2
k=0
=>Total received power = sum of the power of
individual multipath components.
15
Average wideband received power
16
N-1
Ea, [PwB] = Ea, [  |ai exp ji|2]
i=0
N-1 _
=  ai2
i=0
Ea, = average power
_
ai 2 = sample average signalusing multipath
measurement equipment.
Narrowband multipath signals
Received signal: r(t) =
N-1
ji
 ai e p[t–ti]
i=0
Instantaneous received power:
|r(t)|2 =
17
N-1
ji
|  ai e |2
i=0
Average narrowband received power
Ea, [PwB ]= Ea,
18
N-1
ji (t, t)
[ |  ai e
|2 ]
i=0
Conclusions


19
When the transmitted signal has a wide bandwidth >>
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).
Example
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.
20
Solution
For urban radio channel
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/  = 1.28 MHz
21
For 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
22
Example
Assume a mobile traveling at a velocity of 10m/s
receives two multipath components at a carrier
frequency of 1000 MHz.
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.
23
... 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.
24
Solution
(a) Narrow band instantaneous power
N-1
ji (t,t)
|r(t)|2 = |  ai e
|2
i=0
Now –70dBm => 100 pw so a1 = √ 100 pw
and –73dBm => 50 pw so a2 = √ 50 pw
i = 2d/ = 2vt/
 = (3*108)/(100*106) = 0.3 m
1= 2*10*t/0.3 = 209.4 t rad.
25
2 = -1 = -209.4 t rad.
t=0
|r(t)|2 = | √100 + √50 | 2 = 291pw
t = 0.1
|r(t)|2 = |√100 e j209.4 x 0.1 + √50 e -j209.4 x 0.1| 2
= 78.2pw
t = 0.2
|r(t)|2 = |√100 e j209.4 x 0.2 + √50 e -j209.4 x 0.2 | 2
= 81.5pw
26
t = 0.3
|r(t)|2 = 291pw
t = 0.4
|r(t)|2 = 78.2pw
t = 0.5
|r(t)|2 = 81.5pw
27
Wideband instantaneous power
N-1
|r(t)|2 =  |ak|2 = 100 + 50 = 150 pW
k=0
28
(b) Average 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.
29
Small-scale multipath measurements



30
Direct Pulse Measurements
Spread Spectrum Sliding Correlator Measurement
Swept Frequency Measurement
Types of Small Scale Fading
Multipath time delay
Doppler Spread
Slow fading
Flat fading
31
Frequency
Selective
Fading
Fast
Fading
Mechanisms that cause fading

2 main propagation mechanisms:
Multipath time delay spread
 Doppler spread


32
These two mechanisms are independent of
each other.
Multipath terms associated with fading
Ts = Symbol period or reciprocal bandwidth
Bs = Bandwidth of transmitted signal
Bc = Coherence bandwidth of channel
Bc= 1/(50t) where t is rms delay spread
33
Calculation of Delay Spread
__
_
t 2 = t 2 - ( t )2
Where:
_
t = ( ak2 t) / ( ak2)
__
t 2 = ( ak2 t2) / ( ak2)
34
Fading effects due to Doppler spread
fc

V
fc = frequency incident signal
Received signal spectrum = fc +/- fd
fd = Doppler shift
35
Doppler spread and coherence time
Doppler frequency shift: fd = (v / ) cos  ,
Wavelength  = c / fc
Maximum Frequency deviation = fm = v / 
Doppler Spread BD = fm
Coherence time = Tc = 0.423 / fm
36
Mathematical estimation of fading
Flat fading
Mobile
channel has constant gain and linear
phase response.
Spectral
characteristics of the transmitted signal
are maintained at receiver
Condition:
Bs << Bc
37
=> Ts >> t
Frequency selective fading


Mobile channel has a constant gain and
linear phase response over a finite
bandwidth
Condition:
Bs > Bc
=> Ts < t
38
Flat fading or frequency selective fading?
Common rule of thumb
39

If Ts ≥ 10 t => Flat fading

If Ts < 10 t => Frequency selective fading
Fast 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.

Condition:

Ts > Tc
Bs < BD
40
Slow fading channel



The channel impulse response changes at a
much slower rate than the transmitted signal
Velocity of mobile (or velocity of objects in
channel)
Condition:
Ts << Tc
Bs >> BD
41
Rayleigh and Ricean distributions


42
In mobile radio channels, the Rayleigh
distribution is commonly used to describe
the statistical time varying nature of the
received fading signal
When there is a dominant (non-fading)
signal component present such as
LOS propagation path, the small scale
fading envelope distribution is Ricean
Statistical Models

Probability density function:
p(r) =
(r/2)
2
2
–(r
+
A
)
e
2
(2 )
Io(Ar/2 ), for r  0
p(r)= 0 for r < 0
Io  Modified Bessel function
r  Received fading signal voltage
A  LOS amplitude (A=0 for Rician)
43
2  Variance of fading signal
Level crossing and fading statistics

Level crossing rate (LCR) is the rate at which
the normalized Rayleigh fading envelope
crosses a specified level in a positive going
direction.
LCR = NR = (2) fm e-2
fm = Maximum Doppler frequency
 = R/Rms = specified level R, normalized to the
rms value of fading signal
44
Average fade duration

Average period of time for which the received
signal is below a specified level R.
__
2
t = e – 1
________
fm2
45
Example
(a) For a Rayleigh fading signal, compute the
positive going level crossing rate for  = 1, when
the maximum Doppler frequency
(fm) is 20 Hz.
(b) What is the maximum velocity of the mobile
for this Doppler frequency if the carrier is
900 MHz?
46
Solution
=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
47
Example
Find the average fade duration for threshold
level  = 0.01,  = 0.1 and  = 1, when the
Doppler frequency is 20 Hz.
48
Solution
49

t = e2 – 1
fm2
0.01
0.1
1.0
19.9s
200s
3.43ms
Statistical methods for fading channels





50
Clark’s 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)