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- 1. 13.4 Basic Propagation Mechanisms & Transmission Impairments(1) Reflection: propagating wave impinges on object with size >> λ• examples include ground, buildings, walls(2) Diffraction: transmission path obstructed by objects with edges• 2ndry waves are present throughout space (even behind object)• gives rise to bending around obstacle (NLOS transmission path)(3) Scattering propagating wave impinges on object with size < λ• number of obstacles per unit volume is large (dense)• examples include rough surfaces, foliage, street signs, lamp posts
- 2. 2Models are used to predict received power or path loss (reciprocal)based on refraction, reflection, scattering• Large Scale Models• Fading Modelsat high frequencies diffraction & reflections depend on• geometry of objects• EM wave’s, amplitude, phase, & polarization at point of intersection
- 3. 33.5 Reflection: EM wave in 1stmedium impinges on 2ndmedium• part of the wave is transmitted• part of the wave is reflected(1) plane-wave incident on a perfect dielectric (non-conductor)• part of energy is transmitted (refracted) into 2ndmedium• part of energy is transmitted (reflected) back into 1stmedium• assumes no loss of energy from absorption (not practically)(2) plane-wave incident on a perfect conductor• all energy is reflected back into the medium• assumes no loss of energy from absorption (not practically)
- 4. 4(3) Γ = Fersnel reflection coefficient relates Electric Field intensityof reflected & refracted waves to incident wave as a function of:• material properties,• polarization of wave• angle of incidence• signal frequencyboundary between dielectrics(reflecting surface)reflected waverefracted waveincident wave
- 5. 5(4) Polarization: EM waves are generally polarized• instantaneous electric field components are in orthogonaldirectionsin space represented as either:(i) sum of 2 spatially orthogonal components (e.g. vertical& horizontal)(ii) left-handed or right handed circularly polarized components• reflected fields from a reflecting surface can be computed usingsuperposition for any arbitrary polarizationE||E⊥
- 6. 63.5.1 Reflection from Dielectrics• assume no loss of energy from absorptionEM wave with E-field incident at ∠θi with boundary between 2dielectric media• some energy is reflected into 1stmedia at ∠θr• remaining energy is refracted into 2ndmedia at ∠θt• reflections vary with the polarization of the E-fieldplane of incidencereflecting surface= boundarybetween dielectricsθiθrθtplane of incidence = plane containing incident, reflected, & refracted rays
- 7. 7Two distinct cases are used to study arbitrary directions of polarization(1) Vertical Polarization: (Evi) E-field polarization is• parallel to the plane of incidence• normal component to reflecting surface(2) Horizontal Polarization: (Ehi) E-field polarization is• perpendicular to the plane of incidence• parallel component to reflecting surfaceplane of incidenceθiθrθtEviEhiboundary between dielectrics(reflecting surface)
- 8. 8• Ei & Hi = Incident electric and magnetic fields• Er & Hr = Reflected electric and magnetic fields• Et = Transmitted (penetrating) electric fieldHi HrEi Erθi θrθtε1,µ1, σ1ε2,µ2, σ2EtVertical Polarization: E-field inthe plane of incidenceHiHrEiErθi θrθtε1,µ1, σ1ε2,µ2, σ2EtHorizontal Polarization: E-fieldnormal to plane of incidence
- 9. 9(1) EM Parameters of Materials∀ε = permittivity (dielectric constant): measure of a materials abilityto resist current flow• µ = permeability: ratio of magnetic induction to magnetic fieldintensity• σ = conductance: ability of a material to conduct electricity,measured in Ω-1dielectric constant for perfect dielectric (e.g. perfect reflector oflossless material) given byε0 = 8.85 ×10-12F/m
- 10. 10often permittivity of a material, ε is related to relative permittivity εrε = ε0 εrlossy dielectric materials will absorb power permittivity describedwith complex dielectric constant(3.18)where ε’ =fπσ2(3.17)ε = ε0 εr -jε’highly conductive materials ∀εr & σ are generally insensitive to operating frequencyrfεεσ0<• ε0 and εr are generally constant• σ may be sensitive to operating frequency
- 11. 11Material εr σ σ/εrε0 f (Hz)Poor Ground 4 0.001 2.82 ×107 108Typical Ground 15 0.005 3.77 ×107 108Good Ground 25 0.02 9.04 ×107 108Sea Water 81 5 6.97 ×109 108Fresh Water 81 0.001 1.39 ×106 108Brick 4.44 0.001 2.54 ×1074⋅109Limestone 7.51 0.028 4.21 ×1084⋅109Glass, Corning 707 4 0.00000018 5.08 ×103 106Glass, Corning 707 4 0.000027 7.62 ×105 108Glass, Corning 707 4 0.005 1.41 ×108 1010
- 12. 12• because of superposition – only 2 orthogonal polarizations need beconsidered to solve general reflection problemMaxwell’s Equation boundary conditions used to derive (3.19-3.23)Fresnel reflection coefficients for E-field polarization at reflectingsurface boundary• Γ|| represents coefficient for || E-field polarization• Γ⊥ represents coefficient for ⊥ E-field polarization(2) Reflections, Polarized Components & Fresnel ReflectionCoefficients
- 13. 13Fersnel reflection coefficients given by(i) E-field in plane of incidence (vertical polarization)Γ|| =ititirEEθηθηθηθηsinsinsinsin1212+−= (3.19)(ii) E-field not in plane of incidence (horizontal polarization)Γ⊥ =titiirEEθηθηθηθηsinsinsinsin1212+−= (3.20)ηi = intrinsic impedance of the ithmedium• ratio of electric field to magnetic field for uniform plane wave inithmedium• given by ηi = ii εµ
- 14. 14velocity of an EM wave given by ( ) 1−µεboundary conditions at surface of incidence obey Snell’s Law( ) ( ) )90sin()90sin( 222111 θεµθεµ −=− (3.21)θi = θr (3.22)Er = Γ Ei (3.23a)Et = (1 + Γ )Ei (3.23b)Γ is either Γ|| or Γ⊥ depending on polarization• | Γ | ≈ 1 for a perfect conductor, wave is fully reflected• | Γ | ≈ 0 for a lossy material, wave is fully refracted−−= −)90sin(sin90211it θηηθ
- 15. 15• radio wave propagating in free space (1stmedium is free space)• µ1 = µ2Γ|| =iriririrθεθεθεθε22cossincossin−+−+−(3.24)Γ⊥ =iriiriθεθθεθ22cossincossin−+−−(3.25)Simplification of reflection coefficients for vertical and horizontalpolarization assuming:Elliptically Polarized Waves have both vertical & horizontal components• waves can be depolarized (broken down) into vertical & horizontalE-field components• superposition can be used to determine transmitted & reflectedwaves
- 16. 16(3) General Case of reflection or transmission• horizontal & vertical axes of spatial coordinates may not coincidewith || & ⊥ axes of propagating waves• for wave propagating out of the page define angle ∠θmeasured counter clock-wise from horizontal axesspatial horizontal axisspatial vertical axisθ⊥||orthogonal componentsof propagating wave
- 17. 17↔vertical & horizontalpolarized componentscomponents perpendicular& parallel to plane of incidenceEiH , EiV EdH , EdV• EdH , EdV = depolarized field components along the horizontal &vertical axes• EiH , EiV = horizontal & vertical polarized components of incidentwaverelationship of vertical & horizontal field components at the dielectricboundaryEdH, EdV EiH , EiV = Time Varying Components of E-field=iviHCTdvdHEERDREE(3.26)- E-field components may be represented by phasors
- 18. 18for case of reflection:• D⊥⊥ = Γ⊥• D|| || = Γ||for case of refraction (transmission):• D⊥⊥ = 1+ Γ⊥• D|| || = 1+ Γ||R = − θθθθcossinsincos, θ = angle between two sets of axesDC = ⊥⊥||||00DDR = transformation matrix that maps E-field componentsDC = depolarization matrix
- 19. 191.00.80.60.40.20.00 10 20 30 40 50 60 70 80 90|Γ|||εr=12εr=4angle of incidence (θi)Brewster Angle (θB)for εr=12vertical polarization(E-field in plane of incidence)for θi < θB: a larger dielectric constant smaller Γ|| & smaller Erfor θi > θB: a larger dielectric constant larger Γ|| & larger ErPlot of Reflection Coefficients for Parallel Polarization for εr= 12, 4
- 20. 20εr=12εr=4|Γ⊥|1.00.90.80.70.60.50.40.30 10 20 30 40 50 60 70 80 90angle of incidence (θi)horizontal polarization(E-field not in plane ofincidence)for given θi: larger dielectric constant larger Γ⊥ and larger ErPlot of Reflection Coefficients for Perpendicular Polarization for εr=12, 4
- 21. 21e.g. let medium 1 = free space & medium 2 = dielectric• if θi 0o(wave is parallel to ground)• then independent of εr, coefficients |Γ⊥| 1 and |Γ||| 1Γ|| = 1coscoscossincossin22022=−−=−+−+−= iririririririθεθεθεθεθεθεθΓ⊥ = 1coscoscossincossin22022−=−−−=−+−−= iriririiriiθεθεθεθθεθθthus, if incident wave grazes the earth• ground may be modeled as a perfect reflector with |Γ| = 1• regardless of polarization or ground dielectric properties• horizontal polarization results in 180° phase shift
- 22. 223.5.2 Brewster Angle = θB• Brewster angle only occurs for vertical (parallel) polarization• angle at which no reflection occurs in medium of origin• occurs when incident angle θi is such that Γ|| = 0 θi = θB• if 1stmedium = free space & 2ndmedium has relative permittivity εrthen (3.27) can be expressed as112−−rrεεsin(θB) = (3.28)sin(θB) =211εεε+(3.27)θB satisfies
- 23. 23e.g. 1stmedium = free spaceLet εr = 4 11614−−sin(θB) = = 0.44θB = sin-1(0.44) = 26.6oLet εr = 15 1151152−−sin(θB) = = 0.25θB = sin-1(0.25) = 14.5o
- 24. 243.6 Ground Reflection – 2 Ray ModelFree Space Propagation model is inaccurate for most mobile RFchannels2 Ray Ground Reflection model considers both LOS path & groundreflected path• based on geometric optics• reasonably accurate for predicting large scale signal strength fordistances of several km• useful for- mobile RF systems which use tall towers (> 50m)- LOS microcell channels in urban environmentsAssume• maximum LOS distances d ≈ 10km• earth is flat
- 25. 25Let E0 = free space E-field (V/m) at distance d0• Propagating Free Space E-field at distance d > d0 is given byE(d,t) = −cdtwddEccos00(3.33)• E-field’s envelope at distance d from transmitter given by|E(d,t)| = E0 d0/d(1) Determine Total Received E-field (in V/m) ETOTELOSEiEr= Egθi θ0dETOT is combination of ELOS & Eg• ELOS = E-field of LOS component• Eg = E-field of ground reflected component• θi = θr
- 26. 26d’d”ELOSEiEgθi θ0dhthrE-field for LOS and reflected wave relative to E0 given by:and ETOT = ELOS + EgELOS(d’,t) = −cdtwddEccos00(3.34)Eg(d”,t) = −cdtwddEΓ c"cos"00(3.35)assumes LOS & reflected waves arrive at the receiver with- d’ = distance of LOS wave- d” = distance of reflected wave
- 27. 27From laws of reflection in dielectrics (section 3.5.1)θi = θ0 (3.36)Eg = Γ Ei (3.37a)Et = (1+Γ) Ei (3.37b)Γ = reflection coefficient for groundEgd’d”ELOSEiθi θ0Et
- 28. 28resultant E-field is vector sum of ELOS and Eg• total E-field Envelope is given by |ETOT| = |ELOS + Eg| (3.38)• total electric field given by+−cdtwddEccos00−−cdtwddEc"cos")1( 00(3.39)ETOT(d,t) =Assumei. perfect horizontal E-field Polarizationii. perfect ground reflectioniii. small θi (grazing incidence) Γ ≈ -1 & Et ≈ 0• reflected wave & incident wave have equal magnitude• reflected wave is 180oout of phase with incident wave• transmitted wave ≈ 0
- 29. 29• path difference ∆ = d” – d’ determined from method of images( ) ( ) 2222dhhdhh rtrt +−−++∆ = (3-40)if d >> hr + ht Taylor series approximations yields (from 3-40)∆ ≈dhh rt2 (3-41)(2) Compute Phase Difference & Delay Between Two ComponentsELOSdd’d”θi θ0hthr∆hht+hrEi Eg
- 30. 30once ∆ is known we can compute• phase difference θ∆ =cwc⋅∆=∆λπ2(3-42)• time delay τd =cfc πθ2∆=∆(3-43)As d becomes large ∆ = d”-d’ becomes small• amplitudes of ELOS & Eg are nearly identical & differ only in phase"000000ddEddEddE≈≈ (3.44)if Δ = λ/n θ∆ = 2π/n0 π 2πλΔ
- 31. 31(3) Evaluate E-field when reflected path arrives at receiver( )0cos")1("cos0000ddEcddwddEc −+ −(3.45)ETOT(d,t)|t=d”/c =t = d”/creflected path arrives at receiver at− ∆1cos00cwddEc≈( )[ ]1cos00−∆θddE=( )[ ]100−∠ ∆θddE=
- 32. 32(3.46)( )( )∆∆ +−θθ 22200sin1cosddE=( ) ∆∆ +−θθ 220022001 sinddEcosddE|ETOT(d)|=== ∆2sin2 00 θddE∆−θcos2200ddE(3.47)(3.48)ETOT"00ddEddE 00θ∆Use phasor diagram to find resultant E-field from combined direct &ground reflected rays:(4) Determine exact E-field for 2-ray ground model at distance d
- 33. 33As d increases ETOT(d) decreases in oscillatory manner• local maxima 6dB > free space value• local minima ≈ -∞ dB (cancellation)• once d is large enough θΔ < π & ETOT(d) falls off asymtoticallywith increasing d-50-60-70-80-90-100-110-120-130-140101102103104mfc = 3GHzfc = 7GHzfc = 11GHzPropagation Loss ht = hr = 1, Gt = Gr = 0dB
- 34. 34if d satisfies 3.50 total E-field can be approximated as:k is a constant related to E0 ht,hr, and λraddhh rt3.022212<≈∆=∆λπλπθ(3.49)d > (3.50)λλπ rtrt hhhh 20320≈this impliesFor phase difference, θ∆ < 0.6 radians (34o) sin(0.5θ∆ ) ≈ θ∆ ∆22 00 θddE|ETOT(d)| ≈e.g. at 900MHz if ∆ < 0.03m total E-field decays with d2200 22dkdhhddE rt≈λπ(3.51)ETOT(d) ≈ V/m
- 35. 35Received Power at d is related to square of E-field by 3.2, 3.15, & 3.51Pr(d) = (3.52b)=πλππ 4120)(120)( 2220 rReGdEAdEPr(d) = 422dhhGGP rtrtt (3.52a)• received power falls off at 40dB/decade• receive power & path loss become independent of frequencyrthhif d >>
- 36. 36Path Loss for 2-ray model with antenna gains is expressed as:• for short Tx-Rx distances use (3.39) to compute total E field• evaluate (3.42) for θ∆ = π (180o) d = 4hthr/λ is where the groundappears in 1stFresnel Zone between Tx & Rx- 1stFresnel distance zone is useful parameter in microcell pathloss modelsPL(dB) = 40log d - (10logGt + 10logGr + 20log ht + 20 log hr ) (3.53)PL =1422 −=dhhGGPP rtrtrt• 3.50 must hold

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