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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.

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Presentation on theme: "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."— Presentation transcript:

1 Wireless Propagation

2 Signal Strength Measure signal strength in –dBW = 10*log(Power in Watts) –dBm = 10*log(Power in mW) can legally transmit at 10dBm (1W). Most 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

3 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  T – temperture in Kelvin (C+273) –kTB(W) B bandwidth –E.g., Temp = 293,=> -203dB, -173dBm /Hz Temp 293 and 22MHz => -130dB, -100dBm

4 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.

5 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

6 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

7 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

8 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

9 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 – dBi An 11dBi antenna has a very narrow range.

10 Simple link budget – b 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

11 Simple link budget – example From base station –+20dBm transmission power –+6dBi transmit antenna gain –+2.2dBi receiver antenna gain –-91dBm minimum receiver power –=> 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.)

12 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: 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: – = Body loss (loss due to your big head): 3dB Maximum path loss: 40 – (-118.3) –3 = 155.3

13 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: 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: Maximum path loss: 153.3

14 Required SNR For a given bit-error probability, different modulation schemes need a higher SNR E b is the energy per bit N o is the noise/Hz Bit-error is given as a function of E b / N o Required SNR = E b / N o * 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

15 MICA2 link budget

16 Shannon Capacity Given SNR it is possible to find the theoretical maximum bit-rate: Effective bits/sec = B log 2 (1 + SNR), where B is bandwidth E.g., –B = 22MHz, –Signal strength = -90dBm –N = -100dBm –=> SNR = 10dB => 10 –22  10 6 log 2 (1 + 10) = 76Mbps –Of course, b can only do 1Mbps when the signal strength is at –90dBm.

17 Propagation Required receiver signal strength – Transmitted signal strength is often around –99 dB base station -> laptop –79.2 dB b 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

18 Free Space Propagation The surface area of a sphere of radius d is 4  d 2, so that the power flow per unit area w(power flux in watts/meter 2 ) at distance d from a transmitter antenna with input accepted power p T and antenna gain G T 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 P R = P T G T (A/ (4  d 2 )) Define the receiver antenna gain G R = 4  A/ 2. = c/f 2.4GHz=> = 3e8m/s/2.4e9/s = 12.5 cm 933 MHz => =32 cm. Receiver signal strength: P R = P T G T G R ( /4  d) 2 P R (dBm) = P T (dBm) + G T (dBi) + G R (dBi) + 10 log 10 (( /4  d) 2 ) 2.4 GHz => 10 log 10 (( /4  d) 2 ) = -40 dB 933 MHz => 10 log 10 (( /4  d) 2 ) = -32 dB

19 Free Space Propagation - examples Mobile phone downlink – = 12.5 cm –PR (dBm) = (PTGGL) (dBm) - 40 dB + 10 log 10 (1/d 2 ) –Or PR-PT - 40 dB = 10 log 10 (1/d 2 ) –Or 155 – 40 = 10 log 10 (1/d 2 ) = –Or (155-40)/20 = log 10 (1/d) –Or d = 10^ ((155-40)/20) = 562Km or Wilmington DE to Boston MA Mobile phone uplink –d = 10^ ((153-40)/20) = 446Km –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

20 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.

21 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 = E o /d LOS cos(2  / t) The reflected signal is -1  E o /d GR cos(2  / (t – (d GR -d LOS ))) Phasors: –LOS = E o /d LOS  0 –Reflected = E o /d GR  (d GR -d LOS ) 2  / For large d d LOS = d GR Total energy –E = (E o /d LOS ) ( (cos ((d GR -d LOS ) 2  / ) – 1) 2 + sin 2 ((d GR -d LOS ) 2  / ) ) ½ –E = (E o /d LOS ) 2 sin((d GR -d LOS )  / )

22 Ground reflection (approximate) d GR -d LOS d GR = ((ht+hr)^2 + d^2)^1/2 d LOS = ((ht-hr)^2 + d^2)^1/2 d GR -d LOS  2hthr/d -> 0 as d-> inf 2 sin((d GR -d LOS )  / ) -> 0, For large d, 2 sin((d GR -d LOS )  / )  C/d So total energy is 1/d^2 And total power is energy squared, or K/d^4

23 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 –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

24 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

25 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

26 Polarization The peak of the electric field rotates around the axis.

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

28 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 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

29 Snell's Law for Oblique Incidence y   x    Graphical interpretation of Snell’s law

30 Transmission Line Representation for Transverse Electric (TE) Polarization E z  HxHx y  z x 

31 Transmission Line Representation for Transverse Magnetic (TM) Polarization E x  HzHz y   z x 

32 Reflection from a Dielectric Half-Space TE Polarization TM Polarization 90º EE  90º HH   no phase shift

33 Magnitude of Reflection Coefficients at a Dielectric Half-Space TE Polarization TM Polarization Reflection coefficient |    | Incident Angle    r =81  r =25  r =16  r =9  r =4  r = Reflection coefficient |    | Incident Angle    r =81  r =25  r =16  r =9  r =4  r =2.56

34 Ground reflection See Mathcad file

35 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.

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

37 Transmission Line Formulation for a Wall Z d TE Z a TE Z d TE Z a TE w

38 Transmission Line Method air wall Z(w)Z(w) ZL= ZaZL= Za ZaZa ZaZa ZwZw Standing Wave 0 - w Transmitted Incident Reflected

39 Reflection at Masonry Walls (Dry Brick: er  5, e”=0) Z d TE Z a TE MHz TE 1.8GHz TE 900MHz TM 1.8GHz TM Angle of Incidence  I (degree)   BB 20cm Brewster angle

40 Reflection Accounting for Wall Loss ZaZa Zw, wZw, w Z(w)Z(w) 0- w z The relative dielectric constant has an imaginary component

41 Comparison with Measured |G| 4 GHz for Reew = 4, Imew = 0.1 and l = 30 cm Landron, et al., IEEE Trans. AP, March 1996) Measured data Angle of Incidence  TE Polarization  w  w  30cm Measured data Angle of Incidence  TM Polarization  w  w  30cm

42 Transmission Loss Through Wall, cont. Now the  might be imaginary => phase See mathcad file

43 Dielectric constants When conductivity exists, use complex dielectric constant given by  =  o (  r - j  ") where  " =  o and  o  /36  Material*  r  mho/m)  " at 1 GHz Lime stone wall Dry marble Brick wall Cement Concrete wall Clear glass Metalized glass Lake water Sea Water Dry soil Earth * Common materials are not well defined mixtures and often contain water.

44 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  = h 2 (d 1 +d 2 )/(2 d 1 d 2 ) Phase difference  = 2  / Normalize Fresnel-Kirchoff diffraction parameter

45 Knife edge diffraction Path loss from transmitter to receiver is Received Signal(dB) v

46 Multiple diffractions If there are two diffractions, there are some models. For more than 2 edges, the models are not very good.

47 Large-scale Path Loss Models Log-distance –PL(d) = K (d/d o ) n –PL(d) (dB) = PL(d o ) + 10 n log 10 (d) Redo examples

48 Large-scale Path Loss Models Log-normal shadowing –PL(d) (dB) = PL(d o ) + 10 n log 10 (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 log 10 (d) 2.5% chance of the signal being 158 times larger/smaller. The fit shown is not very good. This model is very popular.

49 Outdoor propagation models Okumura –Empirical model –Several adjustments to free-space propagation –Path Loss L(d) = L free space + A mu (f,d) – G(h t ) – G(h r ) – G Area –A is the median attenuation relative to free-space –G(h t ) = 20log(h t /200) is the base station height gain factor –G(h r ) is the receiver height gain factor G(h r ) = 10log(h r /3) for h r <3 G(h r ) = 20log(h r /3) for h r >3 –G area is the environmental correction factor Hata

50 Hata Model Valid from 150MHz to 1500MHz A standard formula For urban areas the formula is: –L 50 (urban,d)(dB) = logf c logh te – a(h re ) + (44.9 – 6.55logh te )logd where f c is the ferquency in MHz h te is effective transmitter antenna height in meters (30-200m) h re is effective receiver antenna height in meters (1-10m) d is T-R separation in km a(h re ) is the correction factor for effective mobile antenna height which is a function of coverage area a(h re ) = (1.1logf c – 0.7)h re – (1.56logf c – 0.8) dB for a small to medium sized city

51 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.

52 Path Loss Exponent and Standard Deviation Measured for Different Buildings BuildingFrequency (MHz)n  (dB) Retail Stores Grocery Store Office, hard partition Office, soft partition Office, soft partition Factory LOS Textile/Chemical Textile/Chemical Paper/Cereals Metalworking Suburban Home Indoor Street Factory OBS Textile/Chemical Metalworking

53 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. BuildingFAF (dB)  (dB) Office Building 1 Through 1 Floor Through 2 Floors Through 3 Floors Through 4 Floors Office Building 2 Through 1 Floor Through 2 Floors Through 3 Floors

54 FAF

55 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.

56 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.

57 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

58 Path Loss location 1, free space loss is likely to give an accurate estimate of path loss.free space 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.plane earth loss location 3, plane earth loss needs to be corrected for significant diffraction losses, caused by trees cutting into the direct line of sight.diffraction location 4, a simple diffraction model is likely to give an accurate estimate of path loss.diffraction location 5, loss prediction fairly difficult and unreliable since multiple diffraction is involved.diffraction


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