- By
**feo** - Follow User

- 153 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Wireless Propagation' - feo

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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, giving effective power of 42.
- 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 = 1.3810-23
- 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.

Vertical direction

Horizontal 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 gain

Instead 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
- 22106 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 4d2, 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 pTand antenna gain GTis

- 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 d2))
- Define the receiver antenna gain GR = 4 A/2.
- = c/f
- 2.4GHz=> = 3e8m/s/2.4e9/s = 12.5 cm
- 933 MHz => =32 cm.
- Receiver signal strength: PR = PT GT GR (/4d)2
- PR (dBm) = PT (dBm) + GT (dBi) + GR (dBi) + 10 log10 ((/4d)2)
- 2.4 GHz => 10 log10 ((/4d)2) = -40 dB
- 933 MHz => 10 log10 ((/4d)2) = -32 dB

Free Space Propagation - examples

- Mobile phone downlink
- = 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 / t)
- The reflected signal is -1 Eo /dGR cos(2 / (t – (dGR-dLOS)))
- Phasors:
- LOS = Eo/dLOS0
- Reflected = Eo/dGR (dGR-dLOS) 2 /
- For large d dLOS = dGR
- Total energy
- E = (Eo/dLOS) ( (cos ((dGR-dLOS) 2 / ) – 1)2 + sin2((dGR-dLOS) 2 / ) ) ½
- E = (Eo/dLOS) 2 sin((dGR-dLOS) / )

Ground reflection (approximate)

- dGR-dLOS

dGR = ((ht+hr)^2 + d^2)^1/2

dLOS = ((ht-hr)^2 + d^2)^1/2

dGR-dLOS 2hthr/d -> 0 as d-> inf

2 sin((dGR-dLOS) / ) -> 0,

For large d, 2 sin((dGR-dLOS) / ) C/d

So total energy is 1/d^2

And 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.

Polarization

If a antenna and the electric field have orthogonal polarization, then the antenna will not receive the signal

Polarization

Vertically/ horizontally polarized

- 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 vertically polarized 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

Reflection from a Dielectric Half-Space

GE

90º

q

-1

GH

90º

qB

q

-1

- TE Polarization
- TM Polarization

no phase shift

Magnitude of Reflection Coefficients at a Dielectric Half-Space

1

1

0.9

0.9

er=81

er=81

0.8

0.8

Reflection coefficient |GH|

Reflection coefficient |GE|

er=25

0.7

0.7

er=25

er=16

er=16

0.6

0.6

er=9

er=9

0.5

0.5

0.4

0.4

er=4

er=4

0.3

0.3

er=2.56

er=2.56

0.2

0.2

0.1

0.1

Incident AngleqI

Incident AngleqI

0

0

0

15

30

45

60

75

90

0

15

30

45

60

75

90

TE Polarization

TM Polarization

Ground reflection

- See Mathcad file

Path losses

- Propagation
- Ground reflection
- Other reflections
- We could assume that walls are perfect reflectors (||=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

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

1

0.8

900MHz

20cm

TE

0.6

G 2

1.8GHz

TE

0.4

900MHz

TM

0.2

1.8GHz

TM

0

0

10

20

30

40

50

60

70

80

90

Angle of Incidence qI (degree)

B

ZaTE

ZdTE

ZaTE

Brewster angle

Reflection Accounting for Wall Loss

Za

Z(w)

Zw,kw

z

- w

0

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)

1

1

Measured data

Measured data

0.9

0.9

w =

0.8

0.8

0.7

0.7

w = 30cm

0.6

0.6

G

G

w =

0.5

0.5

0.4

0.4

0.3

0.3

w = 30cm

0.2

0.2

0.1

0.1

0

0

0

15

30

45

60

75

90

0

15

30

45

60

75

90

Angle of Incidence q

Angle of Incidence q

TE Polarization

TM Polarization

Dielectric constants

When conductivity exists, use complex dielectric constant given by

e = eo(er - je") where e" = s/weo and eo 10-9/36p

Material* ers (mho/m) e" at 1 GHz

Lime stone wall 7.5 0.03 0.54

Dry marble 8.8 0.22

Brick wall 4 0.02 0.36

Cement 4 - 6 0.3

Concrete wall 6.5 0.08 1.2

Clear glass 4 - 6 0.005 - 0.1

Metalized glass 5.0 2.5 45

Lake water 81 0.013 0.23

Sea Water 81 3.3 59

Dry 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 difference
- = 2/

Normalize Fresnel-Kirchoff diffraction parameter

Knife edge diffraction

5

0

-5

-10

Received Signal(dB)

-15

-20

-25

-30

-10

-5

0

5

10

v

- Path loss from transmitter to receiver is

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)logdwhere

fc is the ferquency in MHz

hte 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 km

a(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 Different Buildings

Site specific – attenuation factor model

- PL(d) (dB) = PL(do) + 10 n log(d/do) + FAF + 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.

Small-scale path loss

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

They are summed as phasors.

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

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.

Download Presentation

Connecting to Server..