Title: Mobile Wireless Networking CS691 004 Spring 2005
1Mobile Wireless Networking CS691 004 (Spring
2005)
- Radio Propagation Models (Part I)
Xiaoyan Hong UA CS Department hxy_at_cs.ua.edu
2Outline
- Physics of radio propagation
- Two types of propagation models
- Outdoor vs. indoor radio propagation models
- Simple link budget calculations
3Physics of radio propagation
- Radio propagation affects
- Tx area, data rate, battery drain, etc
- Physics
- Free-space propagation
- Refraction
- Diffraction
- Scattering
- Reflecting
- absorption
- Coupling loss
- Federal Standard 1037C
- http//glossary.its.bldrdoc.gov/fs-1037/
4Radio Propagation two effects
- Large-Scale effects
- variation in mean received signal strength over
large T-R distances (100s or 1000s of meters) and
long time-scales - measured by averaging over 5? to 40?, i.e. 1-10m
in cellular/PCS 1-2GHz band. - models
- Small-scale effects
- fluctuations of the received signal strength
about a local mean over small travel distances
(few ?s) and short time intervals (seconds)
5Small-scale and Large-scale Fading (Indoors)
- Signal varies rapidly as T-R separation changes,
but the local average signal changes much more
slowly - variation of received signal strength with
distance from the transmitter - shadow fading (large timescale) caused by large
obstructions - fading caused by local scatterers around the
receiver - Three effects mean path loss, slow variation
about the mean, rapid variation
6path loss
- http//glossary.its.bldrdoc.gov/fs-1037/dir-026/_3
873.htm - path loss In a communication system, the
attenuation undergone by an electromagnetic wave
in transit between a transmitter and a receiver. - Note 1 Path loss may be due to many effects such
as free-space loss, refraction, reflection,
aperture-medium coupling loss, and absorption. - Note 2 Path loss is usually expressed in dB.
Synonym path attenuation.
7path loss (contd)
- path loss
- free-space loss The signal attenuation that
would result if all absorbing, diffracting,
obstructing, refracting, scattering, and
reflecting influences were sufficiently removed
so as to have no effect on propagation. - Note Free-space loss is primarily caused by
beam divergence, i.e. , signal energy spreading
over larger areas at increased distances from the
source. - coupling loss
- 1. The loss that occurs when energy is
transferred from one circuit, circuit element, or
medium to another. - Note Coupling loss is usually expressed in the
same units--such as watts or dB--as in the
originating circuit element or medium. - 2. In fiber optics, the power loss that occurs
when coupling light from one optical device or
medium to another.
8dn Path Loss Model
- Assume average power (in dB) decreases
proportional to log of distance - PL(d0) is the mean path loss in dB at close-in
reference distance d0. - D0 1m, indoor environment
- D0 100m or 1km, outdoor environment
- n is the empirical quantity. n 2 free space.
9log-normal shadowing
- Measurements show that at a given path loss has a
normal distribution d - is a zero-mean Gaussiona r.v. (in dB) with
standard deviation (in dB) - says how good the model is.
10Free Space (FS) Propagation Model
- Used when transmitter and receiver have a clear,
unobstucted, line-of-sight (LOS) path. - E.g., satellite channels, microwave LOS radio
links - FS Power at a receiver antenna at a distance d
from transmitter antenna is - Where and are antenna gains
- Lgt1 is the system loss factor not related to
propagation (e.g., losses due to filter losses,
hardware etc0 - Path loss
- Predicts Pr only for large enough d (wavelength,
linear dimension of antenna, etc).
11Free Space (FS) Propagation Model (contd)
- Large-scale models use a received power
reference point d0 - Pass loss can be expressed as path loss at
reference point d0 - Without implicit measure of PL at d0, it can be
measured as below
12Example PL calculation for FS and Urban
- Evaluate the path loss at a distance of 10km for
a radio signal with a carrier frequency of 900MHz
for free space and standard urban channels - Solution
- (a) for FS
- let d0 1km, n 2,
- 20log10(4 d0/ )
91.5 dB - PL(d) 10 n log10(d/d0)
111.5dB - (b) For standard urban
- Let d0 1km, n 4,
- PL(d) 10 n log10 (d/d0)
131.5dB - So in the urban case, there is 20dB more loss
over the 10 km path
13A summary
- dn Path Loss Model
- Free Space (FS) Propagation Model
14Propagation Mechanisms in Space with Objects
- Reflection (with Transmittance and Absorption)
- radio wave impinges on an object gtgt ? (30 cm _at_ 1
GHz) - surface of earth, walls, buildings, atmospheric
layers - if perfect (lossless) dielectric object, then
zero absorption - if perfect conductor, then 100 reflection
- reflection a function of material, polarization,
frequency, angle - Diffraction
- radio path is obstructed by an impenetrable
surface with sharp irregularities (edges) - secondary waves bend around the obstacle
- explains how RF energy can travel even without
LOS - a.k.a shadowing
- Scattering (diffusion)
- when medium has large number of objects lt ? (30
cm _at_ 1 GHz) - similar principles as diffraction, energy
reradiated in many directions - rough surfaces, small objects (e.g. foliage, lamp
posts, street signs)
15Reflection, Diffraction, and Scattering in
Real-Life
- Received signal often a sum of contributions from
different directions - random phases make the sum behave as noise
(Rayleigh Fading)
16Example Ground Reflection (2-Ray) Model
- Model found a good predictor for large-scale
signal strength over distances of several
kilometers for mobile systems with tall towers
(heights gt 50m) as well as for LOS microcell
channels - Can show (physics) that for large
- Much more rapid path loss than expected due to
free space
17Link budget Calculation using Path Loss Model
- Bit-error-rate (BER) is a function of SNR
(signal-to-noise ratio) at the receiver - SNR Pr / N
- The greater the SNR the better the reception
quality - Link budget calculations allow one to compute SNR
- requires estimate of power received from
transmitter at a receiver - Tx antenna, Rx antenna, Rx amplifier, path loss
(free space, shadowing) - also, estimate of noise and power received from
interferers - SNR (dB) Pr(d) (dBm) N
(dBm) - Where, thermal noise N KT0BF, or
- N (dBm) -174 (dBm) 10 log
B (in Hz) F (dB) - where B is the receivers bandwidth , and F is
the noise figure of the receiver - And, K 1.3810(-23) J/K is the Boltzmanns
constant, - T0 290K standard temperature.
- And, Pr(d) (dBm) Pt (dBm) Gt(dBm) Gr (dBm)
PL (d) (dB)
18Example1 Link Budget Calculation
- Received power vs. distance for FS and urban area
- Given
- cellular phone with 0.6W transmit power, Pt
0.6W 27.78dBm - unity gain antenna, 900 MHz carrier frequency, f
900MHz - What is the received power in dBm at a FS
distance 5km? - What is the Pr(5km) in a shadowed Urban area with
- FS for dlt1km, and n 4 for dgt1km?
- Solution given, d0 1km, Gt and Gr 0,
- c/f 3108 / (9108) 1/3 m,
- ((4 )2d02)/ 1.42109
91.53dB - (a) FS n 2, d 5km,
- PL(d) (dB) 10nlog10(d/d0)
105.53dB - Pr Pt (dBm) Gt Gr PL(d) (dB) 27.8 0
0 105.5 -77.7dBm - (b) Urban n 4, d 5km
- PL(d) 119.5 dB,
- Pr -91.7dBm
19Example2 Link Budget Calculation (contd)
- Maximum separation distance vs. txed power (with
fixed BW) - Given
- cellular phone with 0.6W transmit power, Pt
0.6W 27.78dBm - unity gain antenna, 900 MHz carrier frequency, f
900MHz - SNR must be at least 25 dB for proper reception
- receiver BW is B 30 KHz, and noise figure F
10 dB - What will be the maximum distance?
- Solution (previous resultsd0 1km,
91.5 dB) - N -174 dBm 10 log 30000 10 dB -119 dBm
- For SNR gt 25 dB, we must have Pr gt (-11925)
-94 dBm - This allows path loss PL(d) Pt - Pr lt 122 dB
- (a) FS n 2,
- so that 122 gt 91.5 102log(d/(1 km)), d lt
33.5 km - (b) shadowed urban with n 4,
- So that 122 gt 91.5 104log(d/(1 km)), d lt 5.8
km
20Example3 Link Budget Calculation (contd)
- Maximum BW vs. txed power (with fixed separation
distance) - Given
- cellular phone with 0.6W transmit power, Pt
0.6W 27.78dBm - unity gain antenna, 900 MHz carrier frequency, f
900MHz - SNR must be at least 25 dB for proper reception
- Max separation is 5km, and noise figure F 10 dB
- What will be the maximum BW?
- Solution
- (a) FS n 2,
- previous results at d 5km PL(d) 105.5 dB, Pr
-77.7dBm - Assume SNR must be 25dB, max noise power
- N -77.7 25 -102.7dB.
- N -174 dBm 10 log10 (B) 10 dB -102.7 dBm
, so B 1.349MHz. - (b) shadowed urban with n 4,
- At d 5km, PL(d) 119.5dB, Pr -91.7dBm
- max noise power N -91.7 25 -116.7dB
- N -174 dBm 10 log10 (B) 10 dB -116.7 dBm
, so B 53.7kHz (20 times smaller for the same
quality of reception than FS!)
21Indoor Path Loss Models (e.g. for WLANs)
- Two characteristics of indoor environments
- - small distances
- - much greater environmental variability even
for small T-R separations - e.g. doors closed vs. opens, ceiling vs. desk
mounted antennas, - walls, floors, furniture, people
moving around - Partition losses on same floor
- - wide variety of partitions.... with different
electrical physical characteristics - - hard partitions (to the ceiling) vs soft
partitions - - extensive databases of measurements
- Partition losses between floors
- - depends on construction material, number of
windows, presence of tinting... - - characterized by Floor Attenuation Factors
(FAF) - The Log-normal shadowing path loss model, found
to be valid!
22Indoor Path Loss Models (contd)
- Log-normal shadowing path loss model
- Attenuation factor model
- where is the exponent value for same
floor measurement
23Indoor Path Loss Models (contd)
24More on Large Scale Path Loss
- RF signal penetration into buildings
- - depends on building material, height,
percentage of windows, height, orientation,
transmission frequency - e.g. signal strength inside the building
increases with height (LOS to upper floor walls) - e.g. metallic tints can provide 3 to 30 dB
attenuation in a single glass pane - - n is between 3.0 and 6.2, with average of 4.5
- Ray tracing CAD tools for site specific modeling
- deterministically model indoor and outdoor
propagation using ray tracing - e.g. use building blueprints from, say, AutoCAD,
or, aerial photographs - replacing statistical models
25Reading list for this lecture
- T. S. Rappaport, K. Blankenship, and H. Xu,
"Propagation and Radio System Design Issues in
Mobile Radio Systems for the GloMo Project,"
tutorial from Virginia Tech. Available in pdf.
26Capacity of Wireless Networks
- Slides are from Sachin Adlakha
27Channel Capacity
- In 1948 Claude Shannon showed that the maximum
rate at which information can be transferred is
limited by capacity of the channel. This is
Channel coding theorem. - For Gaussian channel the capacity is given by
- C W log2(1 SNR/W) bits/sec
- Here W is the channel bandwidth
28Communication Network
- Multiple Transmitters and receivers introduce
interference and thus reduce the capacity. - How Much traffic can a wireless network carry?
29Model
- n nodes located in a region of area 1 sq.m.
- Each node can transmit at W bits per second.
- Packets are sent from source to destination in a
multihop fashion. - Packets can be buffered at intermediate nodes
while awaiting transmission.
30Arbitrary Networks
- n nodes are arbitrary located in a disk of unit
area. - Each node has arbitrary chosen destination at
some arbitrary rate. - Each node can choose arbitrary range and power
level for each transmission.
31Random Networks
- In a random scenario , n nodes are randomly
located i.e. independently and uniformly located
on surface S2 of 3D sphere of unit area or disk
of unit area. - All nodes are homogeneous i.e. all transmission
employ same nominal range or power.
32When are packets successfully received?
- Protocol Model
- Receiver R should be
- Within r of its transmitter T
- Outside footprint (1?)r of any
- other transmitter T using range r
- Physical model
- Models a situation where a minimum SIR is
required
r
(1?)r
33Transport Capacity of arbitrary network
- Define one bit-meter when one bit is transported
over distance of one meter. - Main result 1
- Under protocol model, the transport capacity of
arbitrary network is - bit meters per
second. - Main result 2
- Under physical model, is
feasible while - is not for appropriate c, c.
- Thus each node can send approx.
bit-meters/sec
34Random Network Scenario
- Define throughput capacity as the average of
number of bits per second that can be transmitted
by every node to its destination. - A throughput is of the order if
? constants c ? 0 and c lt ? ? - Main result 3
- In case of both surface of sphere and planar disk
the order of the throughput capacity is
35- Main result 4
- For the physical model, a throughput of
- bit per second
is feasible , -
- while is not for
appropriate c and c.
36- Random networks the throughput capacity is
- Implications
- Throughput decreases with n. Design networks
with smaller number of users. - For random networks, selecting appropriate power
levels is imperative for network to be connected,
while limiting interference. - Designing MAC protocols that avoid collisions and
approach the capacity as derived above
37Adding Mobility!!!
- The results derived above are for static
networks. - What if the nodes are mobile??
- Grossglauser and Tse proved that asymptotically
mobility increases the capacity of ad-hoc
networks.
38How to exploit mobility?
- Exploit diversity by giving up latency
constraints - Transfer packets from source to destination only
when they are close. - However the fraction of time the nodes are close
to each other is small. - So distribute the packets to many different nodes
as possible. - Similar to Opportunistic transmission or
multi-user diversity
39Mobility Model
- n nodes lying in the open disk of unit area
- The location of ith node at time t is given by
Xi(t) - Xi(t) is stationary and ergodic with stationary
distribution uniform on the the open disk - Trajectories of different users are independent
and identically distributed - Two cases
- No relaying of packets packets directly to
destination - Nodes can serve as relays have infinite buffer
40- Model of successful transmission is same as
physical model - Throughput definition
- Let ? be a scheduling and relaying policy
- Let Mi?(t) be number of packets that destination
d(i) receives at time t under policy ?. - The throughput ?(n) is feasible if there is a
policy ? such that for every source-destination
pair
41Mobile Nodes without relaying
- Minimum interference requirement places an upper
bound on number of successful transmission at any
time. - Main result
- For no relaying, the achievable throughput per
S-D goes to zero at least as fast as i.e.
42Mobile nodes with relaying
- Direct transmission increases interference.
- So use multi-hop i.e relay the packet to large
number of intermediate relay nodes. - Two stage algorithm
- Stage 1 Schedule packets from sources to relays
(or final destination - Stage 2 Scheduling packets from relays to final
destination
43Throughput with relaying nodes
- For the scheduling policy ?, the expected number
of feasible sender receiver pair is O(n). - Main result
- The two phased algorithm achieves a throughput
per S-D pair of O(1), i.e. there exists a
constant c gt 0 such that
44Key Facts
- This is asymptotic throughput capacity of
wireless ad-hoc networks. - Exploits long range fluctuations in fading
process due to node mobility - Key idea is that every node can get close to any
other node - The result is purely theoretical in nature i.e it
might take large delay for such capacity to
achieve. - Not applicable to real time communications.
45Other Issues
- Design MAC protocol, power assignments and
routing to achieve theoretical throughput - Recent work by Kumar et al focuses on achieving
the theoretical bounds - Also how would the capacity of mobile network
change if we place a bound on tolerable delay??
46Conclusions
- Communication capacity of wireless networks is
limited due to multiple access. - Each transmission consumes valuable resource
space thereby limiting number of simultaneous
transmissions. - Mobility provides diversity by trading off
latency to improve capacity asymptotically