Wireless Propagation

Signal Strength• Measure signal strength in

– dBW = 10*log(Power in Watts)– dBm = 10*log(Power in mW)

• 802.11 can legally transmit at 10dBm (1W).• Most 802.11 PCMCIA cards transmit at 20dBm.• Mica2 (cross bow wireless node) can transmit from –20dBm to 5dBm.

(10microW to 3mW)• Mobile phone base station: 20W, but 60 users, so 0.3W / user, but antenna has

gain=18dBi.• Mobile phone handset – 21dBm

Noise• Interference

– From other users– From other equipment

• E.g., microwave ovens 20dBm 50% duty-cycle with 16ms period.– Noise in the electronics – e.g., digital circuit noise on analogue parts.– Non-linearities in circuits.– Often modeled as white Gaussian noise, but this is not always a valid assumption.

• Thermal noise– Due to thermal agitation of electrons. Present in all electronics and transmission

media.– kT(W/hz)

• k Boltzmann’s constant = 1.38×10-23• T – temperture in Kelvin (C+273)

– kTB(W) • B bandwidth

– E.g.,• Temp = 293,=> -203dB, -173dBm /Hz• Temp 293 and 22MHz => -130dB, -100dBm

Signal to Noise Ratio (SNR)• SNR = signal power / noise power• SNR (dB) = 10*log10(signal power / noise power) • Signal strength is the transmitted power multiplied by a gain – impairments• Impairments

– The transmitter is far away.– The signal passes through rain or fog and the frequency is high.– The signal must pass through an object.– The signal reflects of an object, but not all of the energy is reflected.– The signal interferes with itself – multi-path fading– An object not directly in the way impairs the transmission.

Receiver Sensitivity• The received signal must have a strength that is larger than the receiver

sensitivity• 20dB larger would be good. (More on this later)• E.g.,

– 802.11b – Cisco Aironet 250 (the most sensitive)• 1Mbps: -94dBm; 2Mbps: -91dBm; 5.5Mbps: -89dBm; 11Mbps: -85dBm

– Mobile phone base station: -119dBm– Mobile phone hand set: -118dBm– Mica2 at 868/916MHz: -98dBm

Simple link budget• Determine if received signal is larger than the receiver sensitivity• Must account for effective transmission power

– Transmission power– Antenna gain– Losses in cable and connectors.

• Path losses– Attenuation– Ground reflection– Fading (self-interference)

• Receiver – Receiver sensitivity– Losses in cable and connectors

Antenna gain• isotropic antenna – transmits energy uniformly in all directions.• Antenna gain is the peak transmission power over any direction divided by the

power that would be achieved if an isotropic antenna is used. The units is dBi.• Sometime, the transmission power is compared to a ½ wavelength dipole. In

this case, the unit is dBD.• The ½ wavelength dipole has a gain of 2.14dB.

Horizontal direction

Vertical direction

Antenna gain• Antenna gain is increased by focusing the antenna

– The antenna does not create energy, so a higher gain in one direction must mean a lower gain in another.

– Note: antenna gain is based on the maximum gain, not the average over a region. This maximum may only be achieved only if the antenna is carefully aimed.

This antenna is narrower and results in 3dB higher gain than the dipole, hence, 3dBD or 5.14dBi

This antenna is narrower and results in 9dB higher gain than the dipole, hence, 9dBD or 11.14dBi

Antenna gainInstead of the energy going in all horizontal directions, a reflector can be placed so it only goes in one direction => another 3dB of gain, 3dBD or 5.14dBi

Further focusing on a sector results in more gain.A uniform 3 sector antenna system would give 4.77 dB more.A 10 degree “range” 15dB more.The actual gain is a bit higher since the peak is higher than the average over the “range.”

Mobile phone base stations claim a gain of 18dBi with three sector antenna system. • 4.77dB from 3 sectors – 13.33 dBi• An 11dBi antenna has a very narrow range.

Simple link budget – 802.11b receiver sensitivity

• Thermal noise: -174 dBm/Hz• Channel noise (22MHz): 73 dB• Noise factor: 5 dB• Noise power (sum of the above): -96 dBm• Receiver requirements:

– 3 dB interference margin– 0 dB is the minimum SINR

• Min receiver signal strength: -93 dBm

Simple link budget – 802.11 example• From base station

– +20dBm transmission power– +6dBi transmit antenna gain– +2.2dBi receiver antenna gain– -91dBm minimum receiver power– => 119.2 dB path losses– => 99 dB path losses if 20dB of link margin is added (to ensure the link works well.)

• From PCMCIA to base station– +0dBm transmission power– +6dBi transmit antenna gain– +2.2dBi receiver antenna gain– -91dBm minimum receiver power– => 99.4 dB path losses– => 79.2 dB path losses if 20dB of link margin is added (to ensure the link works well.)

• From PCMCIA to PCMCIA– +0dBm transmission power– +2.2dBi transmit antenna gain– +2.2dBi receiver antenna gain– -91dBm minimum receiver power– => 95.4 dB path losses– => 75.4 dB path losses if 20dB of link margin is added (to ensure the link works well.)

Simple link budget – mobile phone –downlink example

• Transmission power (base station): 20W (can be as high as 100W)• Transmission power for voice (not control): 18W• Number of users: 60• Transmission power/user: 0.3W, 300mW, 24.8dBm• Base station antenna gain (3-sectors): 18dBi• Cable loss at base station: 2dB• Effective isotropic radiated power: 40dBm (sum of the above)• Receiver:• Thermal noise: -174 dBm/Hz• Mobile station receiver noise figure (noise generated by the receiver, Johnson Noise,

ADC quantization, clock jitter): 7dB• Receiver noise density: -167 dB/Hz (-174+7)• Receiver noise: -101.2 dBm (assuming 3.84MHz bandwidth for CDMA)• Processing gain: 25dB (CDMA is spread, when unspread(demodulated) and filtered,

some of the wide band noise is removed)• Required signal strength: 7.9dB• Receiver sensitivity: -101.2 – 25 + 7.9 = -118.3• Body loss (loss due to your big head): 3dB• Maximum path loss: 40 – (-118.3) –3 = 155.3

Simple link budget – mobile phone – uplink example

• Transmission power (mobile): 0.1W (21 dBm) • Antenna gain: 0 dBi• Body loss: 3 dB• Effective isotropic radiated power: 18 dBm (sum of the above) (maximum allowabel by

FCC is 33 dBm at 1900MHz and 20dBm at 1700/2100 MHz• Receiver/base station• Thermal noise: -174 dBm/Hz• Mobile station receiver noise figure (noise generated by the receiver, Johnson Noise,

ADC quantization, clock jitter): 5dB• Receiver noise density: -169 dB/Hz (-174+5)• Receiver noise: -103.2 dBm (assuming 3.84MHz bandwidth for CDMA)• Processing gain: 25dB (CDMA is spread, when unspread(demodulated) and filtered,

some of the wide band noise is removed)• Margin for interference: 3dB (more interference on the uplink than on the downlink)• Required signal strength: 6.1dB• Receiver sensitivity: -119.0• Maximum path loss: 153.3

Required SNR• For a given bit-error probability, different modulation

schemes need a higher SNR

• Eb is the energy per bit• No is the noise/Hz

• Bit-error is given as a function of Eb / No

• Required SNR = Eb / No * Bit-rate / bandwidth

• A modulation scheme prescribes a Bit-rate / bandwidth relationship

• E.g., for 10^-6 BE probability over DBPSK requires 11 dB + 3 dB = 14 dB SNR

MICA2 link budget• http://ceng.usc.edu/~anrg/downloads/LinkModellingTutorial.pdf

Shannon Capacity• Given SNR it is possible to find the theoretical maximum bit-rate:• Effective bits/sec = B log2(1 + SNR), where B is bandwidth• E.g.,

– B = 22MHz, – Signal strength = -90dBm– N = -100dBm– => SNR = 10dB => 10– 22×106 log2(1 + 10) = 76Mbps– Of course, 802.11b can only do 1Mbps when the signal strength is at –90dBm.

Propagation• Required receiver signal strength – Transmitted signal strength is often around

– 99 dB 802.11 base station -> laptop– 79.2 dB 802.11b laptop -> base station– 75.4 dB laptop -> laptop– 155.3 Mobile phone downlink– 153.3 Mobile phone uplink.

• Where does all this energy go… – Free space propagation – not valid but a good start– Ground reflection

• 2-ray – only valid in open areas. Not valid if buildings are nearby.– Wall reflections/transmission– Diffraction– Large-scale path loss models

• Log-distance• Log-normal shadowing• Okumura• Hata• Longley-Rice

– Indoor propagation– Small-scale path loss

• Rayleigh fading• Rician Fading

Free Space Propagation• The surface area of a sphere of radius d is 4π d2, so that

the power flow per unit area w(power flux in watts/meter2) at distance d from a transmitter antenna with input accepted power pT and antenna gain GT is

• The received signal strength depends on the “size” or aperture of the receiving antenna. If the antenna has an effective area A, then the received signal strength is

PR = PT GT (A/ (4 π d2))

• Define the receiver antenna gain GR = 4 π A/λ2.• λ = c/f

• 2.4GHz=> λ= 3e8m/s/2.4e9/s = 12.5 cm• 933 MHz => λ=32 cm.

• Receiver signal strength: PR = PT GT GR (λ/4πd)2

• PR (dBm) = PT (dBm) + GT (dBi) + GR (dBi) + 10 log10((λ/4π)2)-10log10(d2)

• 2.4 GHz => 10 log10 ((λ/4π)2) = -40 dB• 933 MHz => 10 log10 ((λ/4π)2) = -32 dB

Free Space Propagation - examples• Mobile phone downlink

– λ = 12.5 cm– PR (dBm) = (PTGGL) (dBm) - 40 dB + 10 log10 (1/d2)– Or PR-PT - 40 dB = 10 log10(1/d2) – Or 155 – 40 = 10 log10 (1/d2) =– Or (155-40)/20 = log10 (1/d)– Or d = 10^ ((155-40)/20) = 562Km or Wilmington DE to Boston MA

• Mobile phone uplink– d = 10^ ((153-40)/20) = 446Km

• 802.11– PR-PT = -90dBm– d = 10^((90-40)/20) = 316 m– 11Mbps needs –85dBm– d = 10^((85-40)/20) = 177 m

• Mica2 Mote– -98 dBm sensitivity– 0 dBm transmission power– d = 10^((98-30)/20) = 2511 m

Ground reflection• Free-space propagation can not be valid since I’m pretty sure that my cell

phone does reach Boston.• You will soon see that the Motes cannot transmit 800 m.• There are many impairments that reduce the propagation. • Ground reflection (the two-ray model) – the line of sight and ground reflection

cancel out.

Ground reflection (approximate)• Approximation! When the wireless signal hits the ground, it is completely

reflected but with a phase shift of pi (neither of these is exactly true).• The total signal is the sum of line of sight and the reflected signal.• The LOS signal is = Eo/dLOS cos(2 π / λ t)• The reflected signal is -1× Eo /dGR cos(2 π / λ (t – (dGR-dLOS)))• Phasors:

– LOS = Eo/dLOS ∠0– Reflected = Eo/dGR ∠ (dGR-dLOS) 2 π / λ

• For large d dLOS = dGR

• Total energy– E = (Eo/dLOS) ( (cos ((dGR-dLOS) 2 π / λ ) – 1)2 + sin2((dGR-dLOS) 2 π / λ ) ) ½

– E = (Eo/dLOS) 2 sin((dGR-dLOS) π / λ )

Ground reflection (approximate)• dGR-dLOS

dGR = ((ht+hr)^2 + d^2)^1/2dLOS = ((ht-hr)^2 + d^2)^1/2 dGR-dLOS ≈ 2hthr/d -> 0 as d-> inf2 sin((dGR-dLOS) π / λ ) -> 0, For large d, 2 sin((dGR-dLOS) π / λ ) ≈ C/d

So total energy is 1/d^2And total power is energy squared, or K/d^4

Ground reflection (approximate)• For d > 5ht hr, Pr = (hthr)^2 / d^4 Gr GT PT• Pr – PT – 10log((hthr)^2) - log(Gr GT ) = 40 log(1/d)• Examples:• Mobile phone

– Suppose the base station is at 10m and user at 1.5 m– d = 10^((155 – 12)/40) = 3.7Km

• 802.11– Suppose the base station is at 1.5m and user at 1.5 m– d = 10^((90 – 3.5)/40) = 145m

• But this is only accurate when d is large 145m might not be large enough

Ground reflection (more accurate)• When the signal reflects off of the ground, it is partially absorbed and the phase

shift is not exactly pi.

• Polarization• Transmission line model of reflections

Polarization

The polarization could be such that the above picture is rotated by pi/2 along the axis.It could also be shifted. If a rotated and shifted

Polarization

The peak of the electric field rotates around the axis.

PolarizationIf a antenna and the electric field have orthogonal

polarization, then the antenna will not receive the signal

Polarization• When a linearly polarized electric field reflects off of a vertical or

horizontal wall, then the electric field maintains its polarization.

• In practice, there are non-horizontal and non-vertical reflectors, and antenna are not exactly polarized. In practice, a verticallypolarized signal can be received with a horizontally polarized antenna, but with a 20 dB loss.

• Theoretically, and sometimes in practice, it is possible to transmit two signals, one vertically polarized and one horizontally.

Vertically/ horizontally polarized

Vertically/ horizontally polarized

Snell's Law for Oblique Incidence

y

θ θ

θΤ

x

kT

kR

kI

kR

kT

kI

θ

θ

θΤ

yk

xk

kI = kR = ω µ0ε0

kT = ω µ0ε rε0

Boundary conditions require that all wave vectors have the same component parallel to the interface kI sinθ = k R sinθ = kT sinθT

or sinθ = εr sinθT

Graphical interpretation of Snell’s law

Transmission Line Representation for Transverse Electric (TE) Polarization

Ez+ −Hx

ZaTE = ηa cosθ

η = µ / ε

ZdTE = ηd cosθT

Ez = V + exp − jky cosθ( )+V − exp + jkycosθ( ){ }exp − jkx sinθ( )

Hx =cosθ

η V + exp − jky cosθ( )−V − exp + jky cosθ( ){ }exp − jkx sinθ( )

Hy = −sinθ

ηEz

y

θ θ

z

H

E

x

θΤ

H

E

Transmission Line Representation for Transverse Magnetic (TM) Polarization

Ex+ −Hz

TdTMd

aTMa

Z

Z

θη

εµη

θη

cos

/

cos

=

=

=

Hz = I+ exp − jkycosθ( )+ I− exp + jkycosθ( ){ }exp − jkxsinθ( )

Ex = −ηcosθ I + exp − jkycosθ( )− I− exp + jkycosθ( ){ }exp − jkxsinθ( )Ey = ηsinθHz

y

θ θ

zx

θΤ

H H

E E

Reflection from a Dielectric Half-Space• TE Polarization

• TM Polarization

ΓE =ER( )z

EI( )z

ΓE =Zd

TE − ZaTE

ZdTE + Za

TE =cosθ − εr cosθT

cosθ + εr cosθT

ΓH =HR( )z

HI( )z

ΓH =Za

TM − ZdTM

ZaTM + Zd

TM =εr cosθ − cosθT

εr cosθ + cosθT

90º

-1

ΓE

θ

90º

-1

ΓH

θθΒ

θ B = tan−1 εr( )

no phase shift

Magnitude of Reflection Coefficients at a Dielectric Half-Space

TE Polarization TM Polarization

0 15 30 45 60 75 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Ref

lect

ion

coef

ficie

nt|Γ

Ε|

Incident Angle θΙ

εr=81

εr=25εr=16

εr=9

εr=4

εr=2.56

0 15 30 45 60 75 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Ref

lect

ion

coef

ficie

nt|Γ

Η|

Incident Angle θΙ

εr=81

εr=25εr=16

εr=9

εr=4

εr=2.56

Ground reflection• See Mathcad file

Path losses• Propagation• Ground reflection• Other reflections

• We could assume that walls are perfect reflectors (|Γ|=1). But that would be poor approximation for some angles and materials. Also, this would assume that the signal is not able to propagate into buildings, which mobile phone users know is not the case.

Reflection and Transmission at Walls• Transmission line formulation• Homogeneous walls• Attenuation in walls• Inhomogeneous walls

Transmission Line Formulation for a Wall

ZdTE

ZaTE

ZaTE

ZdTE

ZaTE

ZaTE

H I

E I w

Transmission Line Method

air airwall

Z(w) ZL= Za

ZaZa Zw

Standing Wave

0- w

TransmittedIncident

Reflected

Z w( )= ZwZa cosκw w + jZw sinκw wZw cosκw w + jZa sinκw w

,

κw = kw cosθw and Z TE = η cosθ = ωµ0 κ or Z TM = ηcosθ = κ ωε0 εr

Γ =VRef

VInc

= Z (w) − Za

Z(w) + Za

and PRef

PInc

= Γ 2

For no attenuation in the wall, conservation of power gives PTrans

PInc

=1− Γ 2

Reflection at Masonry Walls(Dry Brick: er ≈ 5, e”=0)

ZdTE

ZaTE

ZaTE

0 10 20 30 40 50 60 70 80 900

0.2

0.4

0.6

0.8

1

900MHzTE

1.8GHzTE

900MHzTM

1.8GHzTM

Angle of Incidence θI (degree)

|Γ |2

κw w = nπθB

20cmE I

H I

Brewster angle

Reflection Accounting for Wall Loss

kw = ωc

εr − j ′ ′ ε

κw = kw2 − ka

2 sin2 θ = βw − jαw

ZwTE = ωµ0 κ w or Zw

TM =κ w ωε0 εr − j ′ ′ ε ( )

ZaZw, κwZ(w)

0- wz

For lossy material with κw = βw − jαw use cos and sin with complex

arguments to compute Z(w) = ZwZa cosκ ww + jZw sinκ wwZw cosκ ww + jZa sinκ ww

Or expand cos and sin in terms of exponents to write

Z(w) = Zw

Za e jβw weαw w + e− jβ wwe−αww( )+ Zw e jβ wweαww − e− jβ wwe−αww( )Zw e jβw weαw w + e− jβ wwe−αww( )+ Za e jβ wweαww − e− jβ wwe−αww( )

Note that for α ww ≥1, Z(w) ≅ Zw

The relative dielectric constant has an imaginary component

Comparison with Measured |G| 4 GHz for Reew = 4, Imew = 0.1 and l = 30 cm Landron, et al., IEEE Trans. AP, March 1996)

0 15 30 45 60 75 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Measured data

Angle of Incidence θ

TE Polarization

|Γ|

w = ∞

w = 30cm

0 15 30 45 60 75 900

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Measured data

Angle of Incidence θ

TM Polarization

|Γ|

w = ∞

w = 30cm

Transmission Loss Through Wall, cont.Then the power ratio is

PTr

PIn =2ZwZa

2

2ZwZa cosκw w + j Zw2 + Za

2( )sinκw w2

Recall that when the layer has loss, the wavenumber inside the wall is

complex kw =ωc

εr − j ′ ′ ε and the component of the wave vector

perpendicular to the surfaces is κ w = kw2 − ka

2 sin2 θ = βw − jαw

Expanding cos and sin in terms of exponents,

PTr

PIn =4Zw Za

2

2ZwZa e jβw weαw w + e− jβ wwe−αww( )+ Zw2 + Za

2( )e jβ wweαww − e− jβ wwe−αww( )2

Now the Γ might be imaginary => phase

See mathcad file

Dielectric constants

When conductivity exists, use complex dielectric constant given by

ε = εo(εr - jε") where ε" = σ/ωεo and εo ≈ 10-9/36π

Material* εr σ (mho/m) ε" at 1 GHzLime stone wall 7.5 0.03 0.54Dry marble 8.8 0.22Brick wall 4 0.02 0.36Cement 4 - 6 0.3Concrete wall 6.5 0.08 1.2Clear glass 4 - 6 0.005 - 0.1Metalized glass 5.0 2.5 45Lake water 81 0.013 0.23Sea Water 81 3.3 59Dry soil 2.5 -- --Earth 7 - 30 0.001 - 0.03 0.02 - 0.54

* Common materials are not well defined mixtures and often contain water.

Diffraction• Idea:

– The wave front is made of little sources that propagate in all directions.– If the line of sight signal is blocked, then the wave front sources results propagation

around the corner.– The received power is from the sum of these sources sources

Define excess path ∆ = h2 (d1+d2)/(2 d1d2)Phase differenceφ= 2π∆/λ

Normalize Fresnel-Kirchoff diffraction parameter

( )21

212dd

ddhλ

ν+

=

Knife edge diffraction• Path loss from transmitter to receiver is

( ) ( )2

222

2/exp2

1⎟⎟⎠

⎞⎜⎜⎝

⎛−

+== ∫

∞

dttjjFEE

o

d

ν

πν

-10 -5 0 5 10-30

-25

-20

-15

-10

-5

0

5

Rec

eive

d Si

gnal

(dB

)

v

Multiple diffractions• If there are two diffractions, there are some models. For more than 2 edges, the

models are not very good.

Large-scale Path Loss Models• Log-distance

– PL(d) = K (d/do)n

– PL(d) (dB) = PL(do) + 10 n log10(d)

Redo examples

Large-scale Path Loss Models• Log-normal shadowing

– PL(d) (dB) = PL(do) + 10 n log10(d) + X– X is a Gaussian distributed random number

• 32% chance of being outside of standard deviation.• 16% chance of signal strength being 10^(11/10) = 12 times larger/smaller than 10 n log10(d) • 2.5% chance of the signal being 158 times larger/smaller.•The fit shown is not very good.•This model is very popular.

Outdoor propagation models• Okumura

– Empirical model– Several adjustments to free-space propagation– Path Loss L(d) = Lfree space + Amu(f,d) – G(ht) – G(hr) – GArea

– A is the median attenuation relative to free-space– G(ht) = 20log(ht /200) is the base station height gain factor– G(hr) is the receiver height gain factor

• G(hr) = 10log(hr /3) for hr <3• G(hr) = 20log(hr /3) for hr >3

– Garea is the environmental correction factor

• Hata

Hata Model• Valid from 150MHz to 1500MHz• A standard formula• For urban areas the formula is:

– L50(urban,d)(dB) = 69.55 + 26.16logfc - 13.82loghte – a(hre) + (44.9 – 6.55loghte)logd

wherefc is the ferquency in MHzhte is effective transmitter antenna height in meters (30-200m)hre is effective receiver antenna height in meters (1-10m)d is T-R separation in kma(hre) is the correction factor for effective mobile antenna height which is a function of coverage area

a(hre) = (1.1logfc – 0.7)hre – (1.56logfc – 0.8) dB for a small to medium sized city

Indoor propagation models• Types of propagation

– Line of sight– Through obstructions

• Approaches– Log-normal– Site specific – attenuation factor model

• Log-normal– PL(d)[dBm] = PL(d0) + 10nlog(d/d0) + Xs

• n and s depend on the type of the building• Smaller value for s indicates the accuracy of the path loss model.

Path Loss Exponent and Standard Deviation Measured for DifferentBuildings

6.83.31300Metalworking

9.72.14000Textile/Chemical

Factory OBS

7.03.0900Indoor Street

Suburban Home

5.81.61300Metalworking

6.01.81300Paper/Cereals

7.02.14000Textile/Chemical

3.02.01300Textile/Chemical

Factory LOS

14.12.61900Office, soft partition

9.62.4900Office, soft partition

7.03.01500Office, hard partition

5.21.8914Grocery Store

8.72.2914Retail Storesσ (dB)nFrequency (MHz)Building

Site specific – attenuation factor model• PL(d) (dB) = PL(do) + 10 n log(d/do) + FAF + Σ PAF

• FAF floor attenuation factor - Losses between floors– Note that the increase in attenuation decreases as the number of floors increases.

• PAF partition attenuation factor - Losses due to passing through different types of materials.

7.231.6Through 3 Floors

5.427.5Through 2 Floors

2.916.2Through 1 Floor

Office Building 2

1.527.0Through 4 Floors

1.724.4Through 3 Floors

2.818.7Through 2 Floors

7.012.9Through 1 Floor

Office Building 1σ (dB)FAF (dB)Building

FAF

Small-scale path loss

The received signal is the phasor sum of the contributions of each reflection.

A small change in the position of the receiver or transmitter can cause a large change in the received signal strength.

See matlab file

The received signal is the sum of the contributions of each

reflection.

They are summed as phasors.

Rayleigh and Rician Fading• The inphase and quadrature parts can be modeled as independent Gaussian

random variables.

• The energy is the (X^2 + Y^2)^ ½ where X and Y are Gaussian => the energy is Rayleigh distributed.

• The power is (X^2 + Y^2) which is exponentially distributed.

• Rician – if there is a strong line-of-sight component as well as reflections. Then the signal strength has a Ricain distribution.

Summary• The signal strength depends on the environment in a complicate way. • If objects are possible obstructing, then the signal strength may be log-normal

distributed => large deviation from free-space• If the signal is narrow band, then the the signal could be completely canceled

out due to reflections and multiple paths.• Reflection, transmission, and diffraction can all be important

Path Loss

•location 1, free space loss is likely to give an accurate estimate of path loss. •location 2, a strong line-of-sight is present, but ground reflections can significantly influence path loss. The plane earth loss model appears appropriate. •location 3, plane earth loss needs to be corrected for significant diffraction losses, caused by trees cutting into the direct line of sight. •location 4, a simple diffraction model is likely to give an accurate estimate of path loss. •location 5, loss prediction fairly difficult and unreliable since multiple diffraction is involved.