Transcript Lecture-18

TLEN 5830 Wireless Systems
Lecture Slides
01-November-2016
•
Lab-01 (Pre-Lab) Concepts
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Additional reference materials
Required Textbook:
Antennas and Propagation for Wireless Communication Systems, by Simon R.
Saunders and Alejandro Aragon-Zavala, ISBN 978-0-470-84879-1; March 2007
(2nd edition).
Optional References:
Wireless Communications and Networks, by William Stallings, ISBN 0-13040864-6, 2002 (1st edition);
Wireless Communication Networks and Systems, by Corey Beard & William
Stallings (1st edition); all material copyright 2016
Wireless Communications Principles and Practice, by Theodore S. Rappaport,
ISBN 0-13-042232-0 (2nd edition)
Acknowledgements:
Some slide content from Ray Nettleton (Former ITP faculty and Director)
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Who is this Man?
Gugliemo Marconi lived from
1874 to 1935. In 1896 he
received the first patent for a
wireless device; next year he
founded the first wireless
telegraphy company. He
demonstrated transatlantic radio
telegraphy in 1901 and received
the Nobel Prize for Physics in
1909. His invention was
credited with the saving of lives
in the Titanic disaster in 1912.
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Diffraction
Line of Secondary
Sources (Huygens)
Transmitter
d1
h
Receiver
d2
LOS Path (Blocked)
ht
hr
In the simplest case we model obstructions as knife-edged, i.e. the tip
has no width. The difference in phase between two rays combining at
the receiver is given by
f = (p/2) n2 where n = h(2 (d1+d2) / ld1d2)1/2
Where n is the Fresnel-Kirchoff Diffraction Parameter
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The Fresnel Integral
The wave arriving at the receiver is the sum of all
the wavelets from along the plane directly above
the obstruction. The field strength is therefore
given by an integral over this plane:

Er = {(1+j)/2} n exp(-jpt2/2) dt = F(n)
F(n) is the complex Fresnel integral and is
available in compilations of math tables and in
math modeling software.
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Fresnel Zones
If the direct and reflected/refracted paths differ by exactly l/2
the waves will subtract (interfere destructively). If they are
equal in magnitude they will cancel.
A Fresnel zone is the locus of all points such that the paths
differ by nl/2. The first, second and third are shown below.
Path TDR differs from path TR by 3l/2
T
R
D
These loci are ellipsoids
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Path Loss Exponent
In designing wireless systems it is necessary to
model the channel macroscopically. We can’t apply
reflection and diffraction models for every building in
a city, for every radio tower and for every possible
location for the mobile. So we model path loss with
distance as a random variable given by
Lp(dB) = k + 10nlog10(d) + Xs
Where k + 10nlog10(d) is the mean and Xs is a
Gaussian random variable in decibels (log-normal)
with zero mean and variance s. The path loss
exponent is n
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Probability: Normal Distribution
p(x)
(zero mean)
Mean
Probability Density Function:
p(x) = (2p)-1/2 exp(-x2/2s2)
Probability x is between a & b:
b
P(a<x<b) = a p(x) dx
Standard Probability x is greater than b:
Deviation


P(x>b) = b p(x) dx = Q(b)
Probability x is less than a:
a
0
x

P(x<a) = - p(x) dx
= 1 – Q(a)


Area = - p(x) dx = 1; Mean m = - x p(x) dx; Std Dev s = - x2 p(x) dx
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Coverage “Holes”
Mean is a
function of
distance
p(S)
Area under
curve = 1
s is a
function of
distance
Probability of
failed
communication
TLEN 5830 Wireless Systems
Receiver
Threshold:
doesn’t work
below here
Signal Strength S dB
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Mean and Variance
• We already know that the value of n for free space
is 2 (inverse square law)
• For urban cellphones it is typically in the range of
2.7 to 5;
• In buildings with LOS it is 1.6 to 1.8 (due to
ducting.)
• In buildings without LOS it ranges from 3 to 6.
The standard deviation of the signal also varies with
environment and distance. Outdoors in cities it has
the range 6 to 15dB. In buildings it is even more
variable particularly with construction type.
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Numerical Example
An area exhibits a path loss n = 4 and a log normal fading with s
= 6dB. Signal strength at a receiver has an average of -110dBm
and its threshold is -116dBm. What is the probability that the
receiver is below its threshold?
Here m = -110, s = 6 and T is -116 (i.e. 1 standard deviation
below the mean signal strength). The probability of being below
threshold is therefore
P(S<T ) = P(S<-116) = 1 – Q(-1) = Q(1)* = 0.1587 (from tables in
Rappaport).
* Because of symmetry, Q(-a) = 1- Q(a)
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Where to Find the Q Function
• In Microsoft Excel it is hidden in the function Normdist:
“x” is the argument
“mean” is the mean m
“standard_dev” is s
Set “cumulative” to TRUE
for the function
P(a<x) = - p(a) da
= 1 – Q(x) = Q(-x)
x
If you want to normalize against s and subtract m yourself, use
the simpler function normsdist which assumes zero mean and
unit standard deviation, and therefore only one argument x.
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The Scattering Channel
Line of Sight Necessary in
mmW systems
Scattering channel no line of sight - most
common in mobile
systems
Multipath - causes
intersymbol
interference,
mitigated by narrow
antenna beams
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Channel Impulse Response
Received
Power
Received
Power
Transmitted
Power
Multiple reflections
from objects in the
scattering field
spread the signal
Time in time. The
impulse response
changes in time if
the vehicle or
anything in the
Time
scattering field is
moving.
Time
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PDF of Received Power & Phase
The I and Q components of
the received signal are
jointly Gaussian with zero
mean and variance equal
to the received power.
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0.9
0.8
0.7
0.6
Received Pow er 0.5
0.4
0.3
0.2
1.6
0.7
0.1
-0.2
Q component
-2
2
1.6
0.8
0.4
1.2
I component
0
-1.1
-0.4
-0.8
-1.2
-1.6
-2
0
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Cartesian to Polar Conversion
Using R2 = I2 + Q2 and
Tan F = Q/I, we
observe that:
F is uniform (-p, p);
and R is Rayleigh
distributed:
1.9
R
1.6
1.3
0.7
F
1
0.6
0.7
0.5
0.4
0.1
p(r)
0.4
Q component
-0.2
p(r) = r/s2 exp(-r2/2s2),
r > 0; = 0 elsewhere.
0.3
-0.5
0.2
-0.8
0.1
-1.1
0
-1.4
0
-1.7
0.5
1
1.5
2
2.5
3
3.5
4
r
1.9
1.6
1.3
1
0.7
0.4
0.1
-0.2
-0.5
-0.8
-1.1
-1.4
-1.7
-2
-2
I component
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Typical Fading Profile
Signal Strength dB
The amount of
Rayleigh fading
does not
depend on
signal strength
Shadow fading follows the trend 1/10n
Distance from base station
Lord Rayleigh
1842-1919
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Doppler Shift
f = f0v/c cos f
Christian Doppler
1803-1853
where f is the angle between the ray and the
motion vector, f0 is the carrier frequency in Hz, v is
velocity in m/s and c is velocity of light in m/s.
Wave compressed
Frequency raised
f
Wave expanded
Frequency lowered
Motion
vector
Wave received
as transmitted
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Example
f = fo v/c cos f
suppose fo = 900MHz, f = 0,
v = 62mph = 100km/hr
100,000m/hr / 3,600sec/hr
= 27.8m/s
f = 900 x 106 x [27.8 / 3 x 108] cos 0
= 83.4 Hz
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Doppler Spread
A vehicle moving through a scattering field
encounters reflections at every possible Doppler
shift between f0v/c and –f0v/c. This increases the
received signal bandwidth by 2f0v/c.
Motion
Vector
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Doppler Spectrum
It’s clear that a car driving down a street has more scatterers in
front and behind (small f) than to the side. The power spectral
density of Doppler shift is biased toward the edges:
Sd(f ) = 1.5 / {pfm [1 – ((f-fc)/fm)2]1/2
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7
6
PSD
5
4
3
2
1
0
-1.5
-1
-0.5
0
0.5
1
1.5
Frequency
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Summary
• Path loss trend: 1/dn or -10nlog(d)
• Slow fading: Log normal (Gaussian in dB), not
frequency selective
• Fast fading: amplitude is Rayleigh, phase is uniform (p,p), frequency selective as wavelength is critical
• Time spread: multiple “echoes” from the channel
causes symbols to be widened
• Doppler spread: combination of multiple reflections
at different relative velocities causes spectrum to be
widened
• Generally speaking we can use diversity to
compensate for all but shadow fading
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The Cellular Principle
These cells use the same frequencies.
Interference is controlled through the ratio of
cell radius to cell spacing
Base A
Base B
Cell B
Radius r
Cell A
Spacing d
The principle is to pick r and d such that
the ratio of signal to (interference plus
noise) is sufficiently above the receiver’s
threshold, accounting for fading etc.
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Signal to Interference Ratio
A
(Ignore base station height; <<d)
B
mobile
d
r
The worst-case signal to interference case is shown.
SIR(dB)= K + 10nlog10(r) – [K + 10nlog10(d-r)]
= 10n {log10(r) - log10(d-r)} = 10nlog10(r/[d-r])
So system performance depends on:
a) the value of n (2 for free space, up to 4 in urban)
b) the ratio r/[d-r]
c) the receiver threshold
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Signal to Interference Ratio - Mobile
A
(Ignore base station height; <<d)
B
mobile
d
r
The worst-case signal to interference case is shown.
SIRm(dB)
= K - 10nlog10(r) – [K - 10nlog10(d-r)]
= 10n {log10(d-r) - log10(r)} = 10nlog10([d-r] / r)
So system performance depends on:
a) the value of n (2 for free space, up to 3-4 for urban)
b) the ratio [d-r] / r
c) the receiver threshold
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Example
SIRm(dB) = 10nlog10(r/[d-r]) -or- SIRm = ([d-r] / r)n
Suppose the target SIRm is 30dB, or 1000, and n = 4.
Then 30 = 10nlog10([d-r]/r); 3/4 = log10([d-r]/r);
100.75 = [d-r]/r; 5.623r = [d-r]; {5.623 + 1}r = d;
And d / r = 6.623
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Signal to Interference Ratio - Base
A
(Ignore base station height; <<d)
Mobile a
r
B
Mobile b
d
r
The worst-case signal to interference case is shown.
Note the distances for this case are the same; r and [d-r].
In practice if we bring noise back into the equation, and note
that mobiles are more power limited than base stations, it is
usually the uplink (mobile to base) that is the weaker of the two
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Frequency Re-Use
4.62km
4.62km
1km
4.62km
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Tessellation (Tiling)
• Find regular polygons and patterns of polygons that tessellate
the plane – i.e. cover the plane entirely, with no overlaps and
no gaps, by translation and rotation only.
2
3
7
1
4
6
5
Triangles and
Parallelograms
Squares and
Rectangles
Note only the triangular case requires
rotation as well as translation – the others
work by translation alone.
Hexagons and
certain clusters
of k hexagons
k = p2+q2 +pq
k, p, q integers
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Cellular Geometry
What is the re-use distance of this layout?
2
r
r
d
3
4.5 r
7
4
5
1
4
6
1
6
3
7
2
5
r
{r2 - (r/2)2}1/2
= 0.75 r
r/2
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Re-Use Distance for k=7
d
0.75 r
4.5 r
d = r {4.52 + 0.75}1/2 = 4.62 r
For n = 4, this would give us
SIR = 10nlog10([d-r] / r) = 40log10([d/r] -1) = 22.3dB
Note that this only accounts for one interference
source, e.g. if we had six-sector antennas
throughout. In the next lecture we will consider the
effect of multiple interference sources.
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