Antennas _ Propagation Lecture Notes-ELECTROMAGNETIC WAVE by YAdocs

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                            Antennas & Propagation

                                LECTURE NOTES
                                  VOLUME V

                 ELECTROMAGNETIC WAVE
                      PROPAGATION

                            by Professor David Jenn




                                                                          (ver1.3)
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            Propagation of Electromagnetic Waves
      Radiating systems must operate in a complex changing environment that interacts with
      propagating electromagnetic waves. Commonly observed propagation effects are depicted
      below.

                                           4                            SATELLITE


                     IONOSPHERE
                                                                                    1   DIRECT
                                                    3                               2   REFLECTED
                                                                    5
                                                                                    3   TROPOSCATTER
                                                        1
                                                                                    4   IONOSPHERIC HOP
                                                    2                               5   SATELLITE RELAY
                                                                                    6   GROUND WAVE

                                                   6
                             TRANSMITTER                                   EARTH
                                                            RECEIVER


      Troposphere: lower regions of the atmosphere (less than 10 km)
      Ionosphere: upper regions of the atmosphere (50 km to 1000 km)

      Effects on waves: reflection, refraction, diffraction, attenuation, scattering, and
      depolarization.
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            Survey of Propagation Mechanisms (1)
      There are may propagation mechanisms by which signals can travel between the radar
      transmitter and receiver. Except for line-of-sight (LOS) paths, the mechanism’s
      effectiveness is generally a strong function of the frequency and transmitter-receiver
      geometry.
      1. direct path or "line of sight" (most radars; SHF links from ground to satellites)
                                                                    RX
                                      TX                            o
                                       o
                                             SURFACE

      2. direct plus earth reflections or "multipath" (UHF broadcast; ground-to-air and air-
         to-air communications)
                                      TX                            o
                                       o                            RX


                                             SURFACE

      3. ground wave (AM broadcast; Loran C navigation at short ranges)
                                      TX
                                        o                           RX
                                                                    o
                                                 SURFACE


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            Survey of Propagation Mechanisms (2)

      4. ionospheric hop (MF and HF broadcast and communications)

                                                                    F-LAYER OF
                                                                   IONOSPHERE

                            TX                                     E-LAYER OF
                             o                           o RX      IONOSPHERE
                                      SURFACE


      5. waveguide modes or "ionospheric ducting" (VLF and LF communications)
                                                                 D-LAYER OF
                                 TX                              IONOSPHERE
                                  o                    o RX
                                      SURFACE


      Note: The distinction between ionospheric hops and waveguide modes is based more on
      the mathematical models than on physical processes.




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            Survey of Propagation Mechanisms (3)

      6. tropospheric paths or "troposcatter" (microwave links; over-the-horizon (OTH)
         radar and communications)
                                                                   TROPOSPHERE

                                   TX                        o RX
                                    o
                                               SURFACE
      7. terrain diffraction

                                        TX                         o RX
                                         o
                                                  MOUNTAIN

      8. low altitude and surface ducts (radar frequencies)
                                                                       SURFACE DUCT (HIGH
                                                                      DIELECTRIC CONSTANT)

                            TX o                         o
                                        SURFACE               RX


      9. Other less significant mechanisms: meteor scatter, whistlers

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              Illustration of Propagation Phenomena




      (From Prof. C. A. Levis, Ohio State University)

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   Propagation Mechanisms by Frequency Bands
VLF and LF                  Waveguide mode between Earth and D-layer; ground wave at short
(10 to 200 kHz)             distances
LF to MF                    Transition between ground wave and mode predominance and sky
(200 kHz to 2 MHz)          wave (ionospheric hops). Sky wave especially pronounced at night.
HF                          Ionospheric hops. Very long distance communications with low power
(2 MHz to 30 MHz)           and simple antennas. The “short wave” band.
VHF                         With low power and small antennas, primarily for local use using direct
(30 MHz to 100 MHz)         or direct-plus-Earth-reflected propagation; ducting can greatly increase
                            this range. With large antennas and high power, ionospheric scatter
                            communications.
UHF                         Direct: early-warning radars, aircraft-to satellite and satellite-to-satellite
(80 MHz to 500 MHz)         communications. Direct-plus-Earth-reflected: air-to-ground
                            communications, local television. Tropospheric scattering: when large
                            highly directional antennas and high power are used.
SHF                         Direct: most radars, satellite communications. Tropospheric refraction
(500 MHz to 10 GHz)         and terrain diffraction become important in microwave links and in
                            satellite communication, at low altitudes.


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           Applications of Propagation Phenomena

  Direct               Most radars; SHF links from ground to satellites
  Direct plus Earth    UHF broadcast TV with high antennas; ground-to-air and air-to-
  reflections          ground communications
  Ground wave          Local Standard Broadcast (AM), Loran C navigation at relatively
                       short ranges
  Tropospheric paths Microwave links
  Waveguide modes      VLF and LF systems for long-range communication and navigation
                       (Earth and D-layer form the waveguide)
  Ionospheric hops     MF and HF broadcast communications (including most long-distance
  (E- and F-layers)    amateur communications)
  Tropospheric scatter UHF medium distance communications
  Ionospheric scatter  Medium distance communications in the lower VHF portion of the
                       band
  Meteor scatter       VHF long distance low data rate communications




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                   Multipath From a Flat Ground (1)
      When both a transmitter and receiver are operating near the surface of the earth,
      multipath (multiple reflections) can cause fading of the signal. We examine a single
      reflection from the ground assuming a flat earth.
                                                                                   RECEIVER

                                                        d
                                                                 θ′ = 0
                                                                                   •    •
                TRANSMITTER
                                         .
                                         A
                                         •
                                              θ=0
                                                            Ro
                                                                               C
                                                                                        D
                                                                                              hr
                                                                          R2
                                     •
                    ht              B ψ R1                       ψ
                                                                          EARTH'S SURFACE
                                                                          (FLAT)
                     ht     IMAGE              REFLECTION POINT
                                                      jφ Γ
                             •                         ρe

      The reflected wave appears to originate from an image.

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                   Multipath From a Flat Ground (2)
      Multipath parameters:
             1. Reflection coefficient, Γ = ρ e jφΓ . For low grazing angles, ψ ≈ 0 , the
                approximation Γ ≈ −1 is valid for both horizontal and vertical polarizations.
             2. Transmit antenna gain: Gt (θ A ) for the direct wave; Gt (θ B ) for the reflected wave.
             3. Receive antenna gain: Gr (θC ) for the direct wave; Gr (θ D ) for the reflected wave.
             4. Path difference: ∆R = (R1 + R2 ) − Ro
                                         4 3
                                        1 24          {
                                       REFLECTED     DIRECT

      Gain is proportional to the square of the electric field intensity. For example, if Gto is the
      gain of the transmit antenna in the direction of the maximum (θ = 0 ), then
                                                                2
                                    Gt (θ ) = Gto E t norm (θ ) ≡ Gto f t (θ ) 2
      where Et norm is the normalized electric field intensity. Similarly for the receive antenna
      with its maximum gain in the direction θ ′ = 0
                                                                2
                                  Gr (θ ′) = Gro E rnorm (θ ′) ≡ Gro f r (θ ′) 2



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                   Multipath From a Flat Ground (3)
      Total field at the receiver
                        E tot =     E ref     + Edir
                                    {           {
                                  REFLECTED    DIRECT
                                                                   ≡F
                                                            644444474444448

                                                   − jkR
                             = f t (θ A ) f r (θC )
                                                    e    o
                                                           1 + Γ f t (θ B ) f r (θ D ) e − jk∆R 
                                                     4π Ro 
                                                                 f t (θ A ) f r (θ C )          
                                                                                                 
      The quantity in the square brackets is the path-gain factor (PGF) or pattern-propagation
      factor (PPF). It relates the total field at the receiver to that of free space and takes on
      values 0 ≤ F ≤ 2 .
             • If F = 0 then the direct and reflected rays cancel (destructive interference)
             • If F = 2 the two waves add (constructive interference)
      Note that if the transmitter and receiver are at approximately the same heights, close to the
      ground, and the antennas are pointed at each other, then d >> ht ,hr and
                                                    Gt (θ A ) ≈ Gt (θ B )
                                                    Gr (θ C ) ≈ Gr (θ D )
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                   Multipath From a Flat Ground (4)

      An approximate expression for the path difference is obtained from a series expansion:

                                                              1 ( hr − ht )
                                                                            2
                                   Ro = d + ( hr − ht ) ≈ d +
                                                2                  2
                                                              2      d
                                                                   1 (ht + hr ) 2
                                   R1 + R2 = d + ( ht + hr ) ≈ d +
                                                       2                   2
                                                                   2      d
      Therefore,
                                                                       2hr ht
                                                             ∆R ≈
                                                                         d
      and
                                                               (
              | F |= 1 − e − jk 2 h r h t / d = e jkh r h t / d e − jkh r ht   /d
                                                                                                   )
                                                                                    − e jkh r h t / d = 2 sin (khr ht / d )

      The received power depends on the square of the path gain factor
                                                                         2
                                              2  kht hr      kht hr 
                            Pr ∝ | F | = 4 sin 
                                      2
                                                          ≈ 4        
                                                   d   d 
      The last approximation is based on h r , ht << d and Γ ≈ -1.

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                   Multipath From a Flat Ground (5)
      Two different forms of the argument are frequently used.
      1. Assume that the transmitter is near the ground ht ≈ 0 and use its height as a reference.
      The elevation angle is ψ where
                        h − ht ∆h hr                                 Ro
                                                                                  ∆h = hr − ht
                  tanψ = r    ≡   ≈
                           d    d   d                                     ψ
                                                                d
      2. If the transmit antenna is very close to the ground, then the reflection point is very
      near to the transmitter and ψ is also the grazing angle:

                            ∆R = b − a = 2 ht sin ψ   a
                                                           ψ
                                                           b
                                                ht     ψ                      ψ

      If the antenna is pointed at the horizon (i.e., its maximum is parallel to the ground) then
      ψ ≈ θ A.

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                   Multipath From a Flat Ground (6)
      Thus with the given restrictions the PPF can be expressed in terms of ψ
                                    | F |= 2 sin (kht tanψ )
      The PPF has minima at:        kht tanψ = nπ ( n = 0, 1, K, ∞ )
                                    2π
                                        ht tanψ = nπ
                                     λ
                                    tanψ = nλ / ht

      Maxima occur at:              kht tanψ = mπ / 2 ( m = 1, 3, 5,K , ∞ )
                                    2π              2n + 1
                                        ht tanψ =          π ( n = 0 ,1,K ,∞ )
                                     λ                2
                                             (2 n + 1)λ
                                    tanψ =
                                                 4ht

      Plots | F | are called a coverage diagram. The horizontal axis is usually distance and
      the vertical axis receiver height. (Note that because d >> hr the angle ψ is not directly
      measurable from the plot.)

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                   Multipath From a Flat Ground (7)
      Coverage diagram: Contour plots of | F | in dB for variations in hr and d normalized to
      a reference range d o . Note that when d = d o then E tot = E dir .
                                             d 
                                     | F |= 2 o  sin (kht tanψ )
                                              d 

                                                      60
                                                                                         d o = 2000 m
                            RECEIVER HEIGHT, hr (m)




                                                      50
                                                                                           ht = 100λ
                                                      40

                                                      30

                                                      20

                                                      10

                                                       0
                                                      1000   2000          3000          4000       5000
                                                                      RANGE, d (m)


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                   Multipath From a Flat Ground (8)
      Another means of displaying the received field is a height-gain curve. It is a plot of | F |
      in dB vs hr at a fixed range.
      • The constructive and destructive interference as a function of height can be identified.
      • At low frequencies the periodicity of the curve at low heights can be destroyed by the
         ground wave.
      • Usually there are many reflected wave paths between the transmitter and receiver, in
         which case the peaks and nulls are distorted.
      • This technique is often used to determine the optimum tower height for a broadcast
         radio antenna.
                                                        10


                                                        5
                                PATH GAIN FACTOR (dB)




                                                        0


                                                        -5


                                                    -10


                                                    -15


                                                    -20
                                                             0   10     20      30       40     50   60
                                                                      RECEIVER HEIGHT, hr (m)



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                                         Multipath Example
A radar antenna is mounted on a 5 m mast and tracks a point target at 4 km. The target is 2 m
above the surface and the wavelength is 0.2 m. (a) Find the location of the reflection point on
the x axis and the grazing angle ψ . (b) Write an expression for the one way path gain factor F
when a reflected wave is present. Assume a reflection coefficient of Γ ≈ −1 .
                                                   (b) The restrictions on the heights and
5m
                                                        distance are satisfied for the following
                                     2m                 formula
                       ψ                 ψ
                                                                                                    2π ( 2)( 5) 
                                                                         F = 2 sin  t r  = 2 sin 
                                                      x                               kh h
    x=0                                      x=4 km                                                             
                            Reflection
                                                                                     d            (0.2 )( 4000 
                              Point
                                                                           = ( 2 )(0.785) = 0.157
(a) Denote the location of the reflection
    point by xr and use similar triangles                               The received power varies as F 2 , thus
                 5       2
       tanψ =      =
                x r 4000 − x r                                                           ( ) = −16.1 dB
                                                                                   10 log F
                                                                                              2

           xr = 2.86 km                                                 The received power is 16.1 dB below the
               ψ = tan -1 (5 / 2860) = 0.1o                             free space value

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                            Field Intensity From the ERP
      The product Pt Gt is called the effective radiated power (ERP, or sometimes the effective
      isotropic radiated power, EIRP). We can relate the ERP to the electric field intensity as
      follows:

      • The Poynting vector for a TEM wave:
                                                                         r      2
                                                r   r r*
                                                       {
                                               W =ℜ E ×H =       }       Edir
                                                                          ηo
      • For the direct path:
                                             r    PG
                                            W = t t
                                                  4πRo2

      • Equate the two expressions: (note that ηo ≈ 120π )
                                r      2
                                Edir           Pt Gt          r      30 Pt G t Eo
                                           =           ⇒      Edir =          ≡
                                 ηo            4πRo
                                                  2                    d        d

      where Eo is called the unattenuated field intensity at unit distance.

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        Wave Reflection at the Earth’s Surface (1)
      Fresnel reflection coefficients hold when:
           1. the Earth’s surface is locally flat in the vicinity of the reflection point
           2. the surface is smooth (height of irregularities << λ )
      Traditional notation:
          1. grazing angle, ψ = 90o − θ i , and the grazing angle is usually very small (ψ < 1o )
                                                            σ                σ 
          2. complex dielectric constant, ε c = ε r ε o − j   = ε o ε r − j
                                                                                  ≡ ε (ε − jχ ) ,
                                                            ω               ε oω  o 1r 24
                                                                                         4 3
                                                                                                 εrc
                             σ
                where χ =
                           ωεo
             3. horizontal and vertical polarization reference is used
                             VERTICAL POL                       HORIZONTAL POL
              Also called   r     r                          r    r                    Also called
               transverse   E|| = EV                         E⊥ = E H                   transverse
               magnetic               n
                                      ˆ                                      n
                                                                             ˆ           electric
                                   ˆ
                                   ki                                   ˆ
                                                                        ki
                (TM) pol    ψ                                   ψ                        (TE) pol

                                      SURFACE                                SURFACE

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        Wave Reflection at the Earth’s Surface (2)
      Reflection coefficients for horizontal and vertical polarizations:

                                               (ε r − jχ ) sinψ − (ε r − jχ ) − cos 2 ψ
                                − Γ|| ≡ RV =
                                               (ε r − jχ ) sinψ + (ε r − jχ ) − cos 2 ψ

                                                 sinψ − (ε r − jχ ) − cos 2 ψ
                                  Γ⊥ ≡ RH =
                                                 sinψ + (ε r − jχ ) − cos 2 ψ

      For vertical polarization the phenomenon of total reflection can occur. This yields a
      surface guided wave called a ground wave. From Snell’s law, assuming µ r = 1 for the
      Earth,
                                                                                                   sin θ i
                            sin θ i = sin θ r = (ε r − jχ ) µ r sin θ t        ⇒       sin θ t =
                                                                              µ r =1               ε r − jχ
                                        π
      Let θ t be complex, θ t =           + jθ , where θ is real.
                                        2


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        Wave Reflection at the Earth’s Surface (3)
                     π                              π       
      Using θ t =      + jθ :         sin θ t = sin  + jθ  = cos( jθ ) = cosh θ
                     2                              2       
                                      cosθ t = − j sin( jθ ) = − j sinh θ

                                                            sin θ i
      Snell’s law becomes             sin θ t = cosh θ =
                                                            ε r − jχ
                                      cosθ t = 1 − sin 2 θ t = 1 − cosh 2 θ = sinh θ
      Reflection coefficient for vertical polarization:
                                                 jη sinh θ + ηo cosθ i
                                  Γ|| ≡ − RV =
                                                 jη sinh θ − ηo cosθ i
                        µo
      where η =                    . Note that Γ|| = 1 and therefore all of the power flow is along the
                  ε o (ε r − j χ )
      surface. The wave decays exponentially with distance into the Earth.



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        Wave Reflection at the Earth’s Surface (4)
      Example: surface wave propagating along a perfectly conducting plate

                    •   5λ plate
                    •   15 degree grazing angle
                    •   TM (vertical) polarization
                    •   the total field is plotted (incident plus scattered)
                    •   surface waves will follow curved surfaces if the radius of curvature >> λ

                                                                        INCIDENT WAVE
                                                                        (75 DEGREES OFF
                                                                        OF NORMAL)



                   CONDUCTING
                   PLATE




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                            Atmospheric Refraction (1)
      Refraction by the lower atmosphere causes waves to be bent back towards the earth’s
      surface. The ray trajectory is described by the equation: n Re sinθ = CONSTANT
      Two ways of expressing the index of refraction n (= ε r ) in the troposphere:

      1. n = 1 + χρ / ρSL + HUMIDITY TERM
                                                                       REFRACTED
                                                                   θ               θ
       Re = 6378 km = earth radius                                        RAY
                                                                                               θ
       χ ≈ 0.00029 = Gladstone-Dale
           constant
       ρ, ρ SL = mass densities at altitude
           and sea level
                                                                              Re   Re     EARTH'S
             77.6
      2. n =      ( p + 4,810 e / T )10 −6 − 1                         Re                 SURFACE
              T
      p = air pressure (millibars)
      T = temperature (K)
      e = partial pressure of
          water vapor (millibars)

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                                Atmospheric Refraction (2)
           Refraction of a wave can provide a significant level of transmission over the horizon. A
                                                                                                  ′
           bent refracted ray can be represented by a straight ray if an equivalent earth radius Re is
                                                     ′
           used. For most atmospheric conditions Re = 4 Re / 3 = 8500 km

                                                                            REFRACTED RAY
                REFRACTED                                                     BECOMES A
                   RAY                                              TX       STRAIGHT LINE           RX

 TX                                             RX                                                  hr
                                                                    ht

      ht                 LINE OF SIGHT (LOS)   hr
                            BLOCKED BY
                           EARTH'S BULGE
                                               EARTH'S          EARTH'S   EQUIVALENT EARTH
                                                                SURFACE                ′
                                                                              RADIUS, Re
                                               SURFACE

                                                                                              STANDARD
             EARTH                                                                           CONDITIONS:
           RADIUS , Re                                                                             4
                                                                                               Re ≈ Re
                                                                                                ′
                                                                                                   3




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                            Atmospheric Refraction (3)
      Distance from the transmit antenna to the horizon is Rt = (Re + ht ) − (Re ) but
                                                                         2             2
                                                                   ′           ′
       ′                         ′                        ′
      Re >> ht so that Rt ≈ 2 Re ht . Similarly Rr ≈ 2 Rehr . The radar horizon is the sum

                                              ′
                                     RRH ≈ 2 Reht + 2 Rehr′
      Example: A missile is flying 15 m             TX              Rt
                                                                                      Rr
      above the ocean towards a ground                                                                 RX
                                                      ht
      based radar. What is the approximate                                                            hr
      range that the missile can be detected
      assuming standard atmospheric               EARTH'S           ′
                                                                   Re         ′
                                                                             Re
      conditions?                                 SURFACE
                                                                                            ′
                                                                                           Re

      Using ht = 0 and hr = 15 gives a radar
      horizon of
                        ′
              R RH ≈ 2 Re hr
                      ≈ (2)(8500 × 10 3 )(15)
                      ≈ 16 km

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                            Atmospheric Refraction (4)
      Derivation of the equivalent Earth radius
                                    h                        θ (h )
                                                                                   RAY PATH
                                            TANGENT
                                                                          n (h )
                                            θo
                                                         VERTICAL



                                                                         SURFACE
      Break up the atmosphere into thin horizontal layers. Snell’s law must hold at the
      boundary between each layer, ε ( h ) sin [θ ( h ) ] = ε o sin θo
                                        h
                               h3                 M                     θ (h )       n( h3 )
                               h2                       θ2                           n( h2 )
                               h1                θ1                                  n (h1 )
                                                                      THIN LAYER IN WHICH
                                             θo                       n ≈ CONSTANT


                                                      SURFACE

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                            Atmospheric Refraction (5)
      In terms of the Earth radius,
                                 Re ε o sin θ o = ( Re + h ) ε ( R) sin [θ ( R) ]
                                 14 244 1444 24444
                                    4       3               4                3
                                     AT THE                  AT RADIUS
                                    SURFACE                   R = Re + h

      Using the grazing angle, and assuming that ε (h ) varies linearly with h
                                                               d            
                      Re ε o cosψ o = ( Re + h )  ε o + h           ε ( h )  cos[ ( h )]
                                                                                  ψ
                                                              dh            
      Expand and rearrange
                                                  d                                 d
     Re ε o {cosψ o − cos[ ( h ) ]} =  ε o + Re
                          ψ                            ε ( h )  h cos[ψ ( h )] + h 2      ε ( h ) cos[ ( h )]
                                                                                                      ψ
                                                 dh                                 dh
      If h << Re then the last term can be dropped, and since ψ is small, cosψ ≈ 1 + ψ 2 / 2
                                                                              
                               [ψ ( h )]2 ≈ ψ o + 2h 1 + Re d ε ( h ) 
                                              2
                                                  Re         ε o dh           
      The second term is due to the inhomogenity of the index of refraction with altitude.

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                             Atmospheric Refraction (6)

      Define a constant κ such that
                                                                                                   −1
                                                                                              
                 [ψ (h )]2   ≈ψo +
                               2    2h   2 2h
                                       =ψo +               where       κ = 1 +
                                                                                Re d
                                                                                        ε ( h) 
                                   κRe        ′
                                             Re                                 ε o dh        
        ′                                                        ′
      Re = κRe is the effective (equivalent) Earth radius. If Re is used as the Earth radius then
      rays can be drawn as straight lines. This is the radius that would produce the same
      geometrical relationship between the source of the ray and the receiver near the Earth’s
      surface, assuming a constant index of refraction. The restrictions on the model are:
            1.               Ray paths are nearly horizontal
            2.                ε (h ) versus h is linear over the range of heights considered
      Under standard (normal) atmospheric conditions, κ ≈ 4 / 3 . That is, the radius of the Earth
      is approximately Re =   6378 km = 8500 km . This is commonly referred to as “the four-
                              4
                         ′  
                             3
      thirds Earth approximation.”

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                                 Fresnel Zones (1)
      For the direct path phase to differ from the reflected path phase by an integer multiple of
      180 o the paths must differ by integer multiples of λ / 2
                                      ∆R = nλ / 2 ( n = 0,1,K )
      The collection of points at which reflection would produce an excess path length of nλ / 2
      is called the nth Fresnel zone. In three dimensions the surfaces are ellipsoids centered on
      the direct path between the transmitter and receiver

                LOCUS OF REFLECTION                               DIRECT
                POINTS (SURFACES OF                              PATH (LOS)
                   REVOLUTION)
                                      n=2                                     RECEIVER

                                      n =1
                TRANSMITTER
                                                                              hr
                            ht

                                                                 REFLECTING SURFACE


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                                  Fresnel Zones (2)

      A slice of the vertical plane gives the following geometry

                                                d
                                                          dr                 RX
                                    dt
                            TX
                                                                        hr
                                                         R2
                             ht           R1
                                                                  nth FRESNEL ZONE


                                                      REFLECTION POINT

      For the reflection coefficient Γ = ρ e jπ = − ρ :
          • If n is even the two paths are out of phase and the received signal is a minimum
          • If n is odd the two paths are in phase and the received signal is a maximum
      Because the LOS is nearly horizontal Ro ≈ d and therefore Ro = d t + d r ≈ d . For the
      nth Fresnel zone R1 + R2 = d + nλ / 2 .


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                                       Fresnel Zones (3)
      The radius of the nth Fresnel zone is
                                                             nλ d t d r
                                                    Fn =
                                                                d
      or, if the distances are in miles, then
                                                                nd t d r
                                                 Fn = 72.1               (feet)
                                                                f GHz d
      Transmission path design: the objective is to find transmitter and receiver locations and
      heights that give signal maxima. In general:
             1. reflection points should not lie on even Fresnel zones
             2. the LOS should clear all obstacles by 0.6 F1, which essentially gives free space
                transmission
      The significance of 0.6 F1 is illustrated by examining two canonical problems:
                (1) knife edge diffraction and
                (2) smooth sphere diffraction.

      Conversions: 0.0254 m = 1 in; 12 in = 1 ft; 3.3 ft = 1 m; 5280 ft = 1 mi; 1 km = 0.62 mi

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                                         Diffraction (1)
      Knife edge diffraction
                            l = CLEARANCE DISTANCE
                                                          l>0                    l = 0, SHADOW
                                                          l<0                      BOUNDARY


                              ht       SHARP
                                      OBSTACLE
                                                                                 hr

                                                  d
      Smooth sphere diffraction
                   l = CLEARANCE DISTANCE
                                                                l>0              l = 0, SHADOW
                                                                                   BOUNDARY
                                                                      l<0
                                                 BULGE                      hr
                               ht
                                                                                       SMOOTH
                                                                                      CONDUCTOR




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                                           Diffraction (2)
                    E tot
      A plot of           shows that at 0.6 F1 the free space (direct path) value is obtained.
                    Edir


                                                          SHADOW BOUNDARY


                         r       0
                        E
                       r
                       Edir
                                                                         FREE SPACE
                       in dB                                            FIELD VALUE

                                -5
                                -6


                                     l<0                 l>0

                               -10         0    0.6 F1
                                               CLEARANCE DISTANCE, l > 0
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                            Path Clearance Example
Consider a 30 mile point-to-point communication link over the ocean. The frequency of
operation is 5 GHz and the antennas are at the same height. Find the lowest height that
provides the same field strength as in free space. Assume standard atmospheric conditions.

       The geometry is shown below (distorted          The maximum bulge occurs at the midpoint.
       scale). The bulge factor (in feet) is given                     d ≈ dt + dr
                             dd
       approximately by b = t r , where d t                                 (15)(15)
                             1.5κ                                   bmax =               = 112.5 ft
       and d r are in miles.                                               (1.5)( 4 / 3)
                                                                                     nd t d r
 TX                         0.6F1              RX                      Fn = 72.1              ft
                                                                                     f GHz d
                 d
      ht                b   bmax          hr                        0.6 F1 = 53 ft
                  dt         dr                        Compute the minimum antenna height:
                        ′
                       Re                                              h = bmax + 0.6 F1
                                                                         = 112.5 + 53 = 165 ft


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                            Example of Link Design (1)




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                            Example of Link Design (2)




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                     Antennas Over a Spherical Earth
      When the transmitter to receiver distance becomes too large the flat Earth approximation
      is no longer accurate. The curvature of the surface causes:
      1. divergence of the power in the reflected wave in the interference region
      2. diffracted wave in the shadow region (note that this is not the same as a ground wave)
                                                    ′
      The distance to the horizon is d t = RRH ≈ 2 Re ht or, if ht is in feet, d t ≈ 2ht miles.
      The maximum LOS distance between the transmit and receive antennas is
                              d max = d t + d r ≈ 2 ht + 2hr (miles)


                                  TANGENT RAY                      INTERFERENCE
                              (SHADOW BOUNDARY)                       REGION

                                                          hr     DIFFRACTION
                                             dr                     REGION
                       ht       dt


                                                             SMOOTH
                                R′
                                 e                          CONDUCTOR
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                    Interference Region Formulas (1)
      Interference region formulas

                                                  Ro
                                                            R2
                                     R1                                 hr
                                         ψ                  ψ
                            ht
                                             dt            dr


                                                                   SMOOTH
                                           ′
                                          Re                      CONDUCTOR


      The path-gain factor is given by

                                          F = 1 + ρ e jφ Γ e − jk∆R D

      where D is the divergence factor (power) and ∆R = R1 + R2 − Ro .

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                     Interference Region Formulas (2)
      Approximate formulas1 for the interference region:
                                                                                              1
                                                                                 φ − k∆R   2
                                    F = (1 + Γ D ) − 4 Γ
                                        
                                                                          D sin 2 Γ
                                                   2
                                                                                    
                                                                                                  
                                                                                                   
                                                                                          2
      where
                                                                                                                           −1
                  2h1h2                         h1 + h2                                2
                                                                                 4 S1S 2 T       
             ∆R =        J ( S , T ) , tanψ =           K ( S , T ) , D = 1 +                    (power)
                                                                           S (1 − S 2 )(1 + T ) 
                    d                              d                                 2

                    d1                   d2
             S1 =          , S2 =                 where h1 is the smallest of either ht or hr
                      ′
                   2 Re h1                 ′
                                        2 Re h2
                              d         S T + S2
             S=                        = 1       , T = h1 / h2 (< 1 since h1 < h2 )
                                   ′
                         ′ h1 + 2 Reh2
                      2 Re                1+ T
                                                                             (1 − S1 ) + T 2 (1 − S 2 )
                                                                                   2                2
             J ( S , T ) = (1 −   S1 )(1 −
                                   2           2
                                              S2 ) ,   and K ( S , T ) =
                                                                                        1+ T2
      1
       D. E. Kerr, Propagation of Short Radio Waves, Radiation Laboratory Series, McGraw-Hill, 1951 (the formulas have been reprinted in many
      other books including R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985).

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                    Interference Region Formulas (3)
      The distances can be computed from d = d1 + d 2 and

                                d       Φ+π           −1  2 Re (h1 − h2 ) d 
                                                                ′
                            d1 = + p cos     , Φ = cos  
                                                                               ,
                                                                               
                                2        3                       p 3
                                                                               
      and
                                                                         1/ 2
                                       2                   d2
                                 p=         ′
                                          Re ( h1 + h2 ) +   
                                        3
                                                            4
                                                              
      Another form for the phase difference is
                                             2kh1h2
                                     k∆R =          (1 − S1 )(1 − S 2 ) = νζπ
                                                          2         2
                                               d
      where
                                   3
                                4 h1 / 2     3
                                            h1 / 2             h2 / h1
                            ν=            =        ,    ζ =            (1 − S1 )(1 − S 2 ) ,
                                                                             2         2
                                        ′
                               λ 2 Re 1030λ                   d / d RH

                    ′
      and d RH = 2 Reh1 (distance to the radio horizon).


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                     Diffraction Region Formulas (1)
                DIRECT RAY                            SHADOW
                TO HORIZON                           BOUNDARY


                                                                           hr     DIFFRACTED
                            ht                   d
                                                                                     RAYS


                                       R′
                                        e


      Approximate formulas for the diffraction region (frequencies > 100 MHz):

                                         F = V1 ( X )U1 ( Z1 )U1 ( Z 2 )

      where U 1 is available from tables or curves, Z i = hi / H ( i = 1,2 ), X = d / L , and
                                                 1                                   1
                                          ( R′ )
                                              2 3
                                                                                ′
                                                                                Re  3
      V1 ( X ) = 2 π X e − 2.02 X , L = 2 e  = 28.41λ1/ 3 (km), H =       = 47.55λ2 / 3 (m)
                                                                                  2
                                          4k                         2k 
                                                 

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                     Diffraction Region Formulas (2)
      A plot of U1 ( Z )




      Fig. 6.29 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985 (axis labels corrected)

                                                                                                                        41
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                                 Surface Waves (1)
      At low frequencies (1 kHz to about 3 MHz) the interface between air and the ground acts
      like an efficient waveguide at low frequencies for vertical polarization. Collectively the
      space wave (direct and Earth reflected) and surface wave are called the ground wave.
                                             SURFACE WAVE


                            ht                       d                 hr

                                     R′
                                      e


      The power density at the receiver is the free space value times an attenuation factor
                                                                   2
                                                Pr = Pdir 2 As
      where the factor of 2 is by convention. Most estimates of As are based calculations for a
      surface wave along a flat interface. Approximations for a flat surface are good for
      d ≤ 50 /( f MHz )1 / 3 miles. Beyond this distance the received signal attenuates more
      quickly.

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                                     Surface Waves (2)
                                                           kd
      Define a two parameters:           p=                                (numerical distance)
                                                + (σ / ωεo )
                                                 2   εr
                                                      2                2

                                               ε ε ω 
                                   b = tan −1  r o 
                                                σ 
                                         1.8 × 10 4σ
      A convenient formula is σ / ωεo =              . The attenuation factor for the ground wave
                                            f MHz
                               2 + 0.3 p
      is approximately As =                  − p / 2 e − 0.6 p sin b ( b ≤ 90 o )
                            2 + p + 0.6 p 2
      Example: A CB link operates at 27 MHz with low gain antennas near the ground. Find
      the received power at the maximum flat Earth distance. The following parameters hold:
      Pt = 5 W; Gt = Gr = 1; ε r = 12 and σ = 5 × 10 − 3 S/m. The maximum flat Earth range is
       d max = 50 /( 27)1 / 3 = 16.5 miles.
                            πd / λ                          16.5                                d
              p=                      = 0.25 d / λ = 0.0225      (1000) ≈ 601 →                   = 4p
                     12 + (90 / 27)
                       2            2                       0.62                                λ
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                                    Surface Waves (3)
      Check b to see if formula applies (otherwise use the chart on the next page)

                                                                 −12
                                     −1  (12 )(8.85 × 10
                                                                )( 2π )( 27 × 10 6 ) 
                                                                                       = 74.5o
                            b = tan 
                                                            5 × 10 − 3                
                                                                                     

      Attenuation constant
                                       2 + 0.3 p
                             As =                       − p / 2 e −0.6 p sin b ≈ 8.33 × 10 −4
                                    2 + p + 0.6 p 2

      The received power for the ground wave is


                            Pr = Pdir 2 As
                                             2
                                                 =
                                                     Pt Gt Aer
                                                                  2 As
                                                                         2
                                                                             =
                                                                                      (
                                                                                 Pt (1) λ2 / 4π   ) 2 As 2
                                                 4π d 2                4π d 2
                                 (5)(8.33 × 10 − 4 ) 2
                               =                       = 1.52 × 10 −14 W
                                  (4π )( 4) 2 (601) 2
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                                   Surface Waves (4)

                                                FLAT EARTH




      Fig. 6.35 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985

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                                   Ground Waves (5)
                            SPHERICAL EARTH ( ε r = 15 and σ = 10 − 2 S/m)




      Fig. 6.36 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985
                                                                                                             46
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                            Urban Propagation (1)
      Urban propagation is a unique and relatively new area of study. It is important in the
      design of cellular and mobile communication systems. A complete theoretical treatment of
      propagation in an urban environment is practically intractable. Many combinations of
      propagation mechanisms are possible, each with different paths. The details of the
      environment change from city to city and from block to block within a city. Statistical
      models are very effective in predicting propagation in this situation.
      In an urban or suburban environment there is rarely a direct path between the transmitting
      and receiving antennas. However there usually are multiple reflection and diffraction
      paths between a transmitter and receiver.
                                          BASE     • Reflections from objects close to the
                                        STATION
                                       ANTENNA
                                                     mobile antenna will cause multiple signals
                                                     to add and cancel as the mobile unit
                                                     moves. Almost complete cancellation can
                                                     occur resulting in “deep fades.” These
                                                     small-scale (on the order of tens of
         MOBILE
                                                     wavelengths) variations in the signal are
        ANTENNA
                                                     predicted by Rayleigh statistics.


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                            Urban Propagation (2)
      • On a larger scale (hundreds to thousands of wavelengths) the signal behavior, when
        measured in dB, has been found to be normally distributed (hence referred to a
        lognormal distribution). The genesis of the lognormal variation is the multiplicative
        nature of shadowing and diffraction of signals along rooftops and undulating terrain.
      • The Hata model is used most often for predicting path loss in various types of urban
        conditions. It is a set of empirically derived formulas that include correction factors for
        antenna heights and terrain.

      Path loss is the 1 / r 2 spreading loss in signal between two isotropic antennas. From the
      Friis equation, with Gt = G r = 4πAe / λ2 = 1
                                                                   2
                                         Pr (1)(1) λ2  1 
                                    Ls =    =          =      
                                         Pt   (4π r ) 
                                                     2    2 kr 

      Note that path loss is not a true loss of energy as in the case of attenuation. Path loss as
      defined here will occur even if the medium between the antennas is lossless. It arises
      because the transmitted signal propagates as a spherical wave and hence power is flowing
      in directions other than towards the receiver.

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                                       Urban Propagation (3)
      Hata model parameters * : d = transmit/receive distance (1 ≤ d ≤ 20 km)
                                f = frequency in MHz (100 ≤ f ≤ 1500 MHz)
                               hb = base antenna height ( 30 ≤ hb ≤ 200 m)
                               hm = mobile antenna height (1 ≤ hm ≤ 10 m)
      The median path loss is
        Lmed = 69.55 + 26.16 log( f ) − 13.82 log( hb ) + [44.9 − 6.55 log( hb ) ]log( d ) + a ( hm )
      In a medium city: a ( hm ) = [0.7 − 1.1 log( f )]hm + 1.56 log( f ) − 0.8
                                  1.1 − 8.29 log 2 (1.54hm ), f ≤ 200 MHz
      In a large city: a ( hm ) = 
                                  4.97 − 3.2 log (11.75hm ), f ≥ 400 MHz
                                                    2

                                     − 2 log 2 ( f / 28) − 5.4, suburban areas
      Correction factors: Lcor = 
                                     − 4.78 log ( f ) + 18.33 log( f ) − 40.94, open areas
                                                   2

      The total path loss is: Ls = Lmed − Lcor
       * Note: Modified formulas have been derived to extend the range of all parameters.




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                        Urban Propagation Simulation
      Urban propagation modeling using the wireless toolset Urbana (from Demaco/SAIC). Ray
      tracing (geometrical optics) is used along with the geometrical theory of diffraction (GTD)
      Closeup showing antenna placement
      (below)




       Carrier to Interference (C/I) ratio (right)



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                                                 Measured Data
                                        Two different antenna heights
                                                                                       Measured data in
                            h = 2.7 m                                  h = 1.6 m          an urban
                                              f = 3.35 GHz                              environment




                                             f = 8.45 GHz
                                                                                          Three different
                                                                                           frequencies




                                              f = 15.75 GHz

                                                                                         From Masui, “Microwave Path Loss
                                                                                         Modeling in Urban LOS Environ-
                                                                                         ments,” IEEE Journ. on Selected Areas
                                                                                         in Comms., Vol 20, No. 6, Aug. 2002.




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             Attenuation Due to Rain and Gases (1)

    Sources of signal attenuation in the atmosphere include rain, fog, water vapor and other
    gases. Most loss is due to absorption of energy by the molecules in the atmosphere. Dust,
    snow, and rain can also cause a loss in signal by scattering energy out of the beam.




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             Attenuation Due to Rain and Gases (2)




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             Attenuation Due to Rain and Gases (3)

      There is no complete, comprehensive macroscopic theoretical model to predict loss. A
      wide range of empirical formulas exist based on measured data. A typical model:

                    A = aR b , attenuation in dB/km
                    R is the rain rate in mm/hr
                    a = Ga f GHz E a
                    b = Gb f GHz Eb

      where the constants are determined from the following table:

                            Ga = 6.39 × 10 − 5     E a = 2.03        f GHz < 2.9
                               = 4.21 × 10 − 5         = 2.42           2.9 ≤ f GHz < 54
                               = 4.09 × 10 − 2         = 0.699           54 ≤ f GHz < 180
                             Gb = 0.851 Eb = 0.158                 f GHz < 8.5
                                = 1.41     = −0.0779                  8.5 ≤ f GHz < 25
                                = 2.63     = −0.272                    25 ≤ f GHz < 164

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            Ionospheric Radiowave Propagation (1)
    The ionosphere refers to the upper regions of the atmosphere (90 to 1000 km). This region is
    highly ionized, that is, it has a high density of free electrons (negative charges) and positively
    charged ions. The charges have several important effects on EM propagation:

      1. Variations in the electron density ( N e ) cause waves to bend back towards Earth, but
         only if specific frequency and angle criteria are satisfied. Some examples are shown
         below. Multiple skips are common thereby making global communication possible.

                                                                            N e max
                                  4
                                                                         IONOSPHERE
                                  3

                                  2


                              1
                       TX             SKIP DISTANCE


                                                                    EARTH’S SURFACE


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            Ionospheric Radiowave Propagation (2)
      2. The Earth’s magnetic field causes the ionosphere to behave like an anisotropic medium.
         Wave propagation is characterized by two polarizations (“ordinary” and “extra-
         ordinary” waves). The propagation constants of the two waves are different. An
         arbitrarily polarized wave can be decomposed into these two polarizations upon entering
         the ionosphere and recombined on exiting. The recombined wave polarization will be
         different that the incident wave polarization. This effect is called Faraday rotation.

    The electron density distribution has the general characteristics shown on the next page. The
    detailed features vary with
                    •   location on Earth,
                    •   time of day,
                    •   time of year, and
                    •   sunspot activity.

    The regions around peaks in the density are referred to as layers. The F layer often splits
    into the F1 and F2 layers.



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                  Electron Density of the Ionosphere




                                                       (Note unit is per cubic centimeter)

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                            The Earth’s Magnetosphere




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            Ionospheric Radiowave Propagation (3)

      Relative dielectric constant of an ionized gas (assume electrons only):
                                                     ω2p
                                         εr =1 −
                                                 ω (ω − jν )
      where: ν = collision frequency (collisions per second)
                         N ee 2
                    ωp =        , plasma frequency (radians per second)
                         mε o
                    N e = electron density ( / m3 )
                    e = 1.59 × 10 −19 C, electron charge
                    m = 9.0 × 10 − 31 kg, electron mass
      For the special case of no collions, ν = 0 and the corresponding propagation constant is

                                                                       ω2
                                                                        p
                                       k c = ω µoε rε o = ko 1 −
                                                                       ω2
      where ko = ω µoε o .

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            Ionospheric Radiowave Propagation (4)
      Consider three cases:
             1. ω > ω p : k c is real and e − jk c z = e − j k c z is a propagating wave
             2. ω < ω p : k c is imaginary and e − jk c z = e − k c z is an evanescent wave
             3. ω = ω p : k c = 0 and this value of ω is called the critical frequency, ωc

      At the critical frequency the wave is reflected. Note that ωc depends on altitude because
      the electron density is a function of altitude. For electrons, the highest frequency at
                                                         which a reflection occurs is
                                                                                         ωc
                                           REFLECTION                             fc =      ≈ 9 N e max
                    ω = ωc ⇒ ε r = 0          POINT                                      2π
                                                                Reflection at normal incidence requires
                  IONOSPHERE       h′                           the greatest N e .

                                  TX           EARTH’S          1
                                                                  The critical frequency is where the propagation constant is zero.
                                               SURFACE
                                                                Neglecting the Earth’s magnetic field, this occurs at the plasma
                                                                frequency, and hence the two terms are often used
                                                                interchangeably.

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            Ionospheric Radiowave Propagation (5)

      At oblique incidence, at a point of the ionosphere where the critical frequency is f c , the
      ionosphere can reflect waves of higher frequencies than the critical one. When the wave
      is incident from a non-normal direction, the reflection appears to occur at a virtual
      reflection point, h ′ , that depends on the frequency and angle of incidence.



                                 VIRTUAL
                                 HEIGHT

                                                                  IONOSPHERE

                                                h′

                                                                        EARTH’S
                                                                        SURFACE
                            TX
                                            SKIP DISTANCE




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            Ionospheric Radiowave Propagation (6)
      To predict the bending of the ray we use a layered approximation to the ionosphere just as
      we did for the troposphere.

                                                       M
                    ALTITUDE




                                                                      ε r ( z3 )       LAYERED
                               z3                               ψ3                   IONOSPHERE
                               z2             ψ2                      ε r (z2 )    APPROXIMATION
                                                                      ε r ( z1 )
                               z1        ψ1
                                    ψi                                εr = 1


      Snell’s law applies at each layer boundary
                                          sin ψ i = sin (ψ 1 ) ε r ( z1 ) = L

      The ray is turned back when ψ ( z ) = π / 2 , or sinψ i = ε r ( z )


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            Ionospheric Radiowave Propagation (7)

      Note that:
      1. For constant ψ i , N e must increase with frequency if the ray is to return to Earth
         (because ε r decreases with ω ).
      2. Similarly, for a given maximum N e ( N e max ), the maximum value of ψ i that results in
         the ray returning to Earth increases with increasing ω .
      There is an upper limit on frequency that will result in the wave being returned back to
      Earth. Given N e max the required relationship between ψ i and f can be obtained

                                  sin ψ i = ε r ( z)
                                                ω2 p
                                sin ψ i = 1 − 2
                                    2
                                                ω
                                                81N e max
                            1 − cos 2 ψ i = 1 −
                                                     f2
                                             f 2 cos 2 ψ i                 81N e max
                                 N e max =                 ⇒     f max =
                                                  81                       cos2 ψ i

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            Ionospheric Radiowave Propagation (8)
      Examples:

         1. ψ i = 45o , N e max = 2 × 1010 / m 3 : f max = (81)(2 × 1010 ) /(0 .707) 2 = 1.8 MHz
         2. ψ i = 60o , N e max = 2 × 1010 / m 3 : f max = (81)( 2 × 1010 ) /(0.5) 2 = 2.5 MHz

      The value of f that makes ε r = 0 for a given value of N e max is the critical frequency
      defined earlier:
                                              f c = 9 N e max

      Use the N e max expression from previous page and solve for f

                                       f = 9 N e max secψ i = f c secψ i

      This is called the secant law or Martyn’s law. When secψ i has its maximum value, the
      frequency is called the maximum usable frequency (MUF). A typical value is less than 40
      MHz. It can drop as low as 25 MHz during periods of low solar activity. The optimum
      usable frequency (OUF) is 50% to 80% of the MUF.


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                            Maximum Usable Frequency
    The maximum usable frequency (MUF) in wintertime for different skip distances. The MUF
    is lower in the summertime.




             Fig. 6.43 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985

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           Ionospheric Radiowave Propagation (9)

   Multiple hops allow for very long range communication links (transcontinental). Using a
   simple flat Earth model, the virtual height ( h ′ ), incidence angle (ψ i ), and skip distance (d )
                            d
   are related by tanψ i =      . This implies that the wave is launched well above the horizon.
                           2h ′
   However, if a spherical Earth model is used and the wave is launched on the horizon then
              ′
   d = 2 2 Re h′ .
    EFFECTIVE SPECULAR
     REFLECTION POINT

                                     IONOSPHERE
                                                                                          IONOSPHERE


                                h′
                   ψi                                         TX
                                                                      EARTH’S
                                                                      SURFACE
                            d

                Single ionospheric hop                              Multiple ionospheric hops
                    (flat Earth)                                         (curved Earth)
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          Ionospheric Radiowave Propagation (10)

    Approximate virtual heights for layers of the ionosphere

                                            Layer    Range for h ′ (km)
                                             F2      250 to 400 (day)
                                             F1      200 to 250 (day)
                                              F        300 (night)
                                              E             110

    Example: Based on geometry, a rule of thumb for the maximum incidence angle on the
    ionosphere is about 74 o . The MUF is

                                            MUF = f c sec(74 o ) = 3.6 f c

    For N e max = 1012 / m 3 , f c ≈ 9 MHz and the MUF = 32.4 MHz. For reflection from the F2
    layer, h ′ ≈ 300 km. The maximum skip distance will be about

                                         ′
                            d max ≈ 2 2 Re h ′ = 2 2(8500 × 10 3 )(300 × 10 3 ) = 4516 km

                                                                                                            67
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            Ionospheric Radiowave Propagation (7)
                                                                                             ′
                                                                                 1 + h ′ / Re − cosθ     1
      For a curved Earth, using the law of sines for a triangle                                      =
                                                                                          sin θ        tanψ i
                                                                               where
                                                                                            d
                                                                                    θ=
                                      ψi                                                  2 Re ′
                            R/2                         R/2
                                           h′                                  and the launch angle (antenna
                                                                               pointing angle above the horizon)
                            ∆
                                                                               is
                                                                                     ∆ = φ − 90 o = 90 o − θ −ψ i
                                d/2
                      φ                         LAUNCH ANGLE:                  The great circle path via the
                                                    o                     o
                                            ∆ = 90 −θ −ψ i = φ − 90            reflection point is R, which can be
                                                                               obtained from
                       ′
                      Re
                                                                                                 ′
                                                                                             2 Re sin θ
                                                                                       R=
                                      θ                                                        sin ψ i


                                                                                                                            68
Naval Postgraduate School            Department of Electrical & Computer Engineering                Monterey, California



            Ionospheric Radiowave Propagation (8)

      Example: Ohio to Europe skip (4200 miles = 6760 km). Can it be done in one hop?
      To estimate the hop, assume that the antenna is pointed on the horizon. The virtual height
      required for the total distance is
                    d / 2 = Reθ → θ = d / (2 Re ) = 0.3976 rad = 22.8 degrees
                               ′                 ′
                       ′                 ′          ′          ′
                    ( Re + h ′) cos θ = Re → h ′ = Re /cosθ − Re = 720 km
      This is above the F layer and therefore two skips must be used. Each skip will be half of
      the total distance:. Repeating the calculation for d / 2 = 1690 km gives
                    θ = d / (2 Re ) = 0.1988 rad = 11.39 degrees
                                ′
                           ′            ′
                    h ′ = Re /cosθ − Re = 171 km
      This value lies somewhere in the F layer. We will use 300 km (a more typical value) in
      computing the launch angle. That is, still keep d / 2 = 1690 km and θ = 11.39 degrees, but
      point the antenna above the horizon to the virtual reflection point at 300 km
                                                                             −1
                    tanψ i = sin(11.39 o ) 1 +     − cos(11.39 o )                   → ψ i = 74.4 o
                                                300
                                            8500
                                                                  
                                                                   
                                                                                                                      69
Naval Postgraduate School            Department of Electrical & Computer Engineering       Monterey, California



            Ionospheric Radiowave Propagation (9)

      The actual launch angle required (the angle that the antenna beam should be pointed above
      the horizon) is

                      launch angle, ∆ = 90 o − θ − ψ i = 90 o − 11.39 o − 74.4 o = 4.21o

      The electron density at this height (see chart, p.3) is N e max ≈ 5× 1011 / m 3 which
      corresponds to the critical frequency
                                             f c ≈ 9 N e max = 6.36 MHz
      and a MUF of

                                       MUF ≈ 6.36 sec 74.4 o = 23.7 MHz
      Operation in the international short wave 16-m band would work. This example is
      oversimplified in that more detailed knowledge of the state of the ionosphere would be
      necessary: time of day, time of year, time within the solar cycle, etc. These data are
      available from published charts.


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          Ionospheric Radiowave Propagation (10)
      Generally, to predict the received signal a modified Friis equation is used:
                                                  Pt Gt G r
                                         Pr =                     Lx Lα
                                                (4πR / λ )    2

      where the losses, in dB, are negative:
                     L x = Lpol + Lrefl − Giono
                     Lrefl = reflection loss if there are multiple hops
                     Lpol = polarization loss due to Faraday rotation and earth reflections
                     Giono = gain due to focussing by the curvature of the ionosphere
                     Lα = absorption loss
                     R = great circle path via the virtual reflection point

      Example: For Pt = 30 dBW, f = 10 MHz, Gt = G r = 10 dB, d = 2000 km, h ′ = 300 km,
      L x = 9.5 dB and Lα = 30 dB (data obtained from charts).

      From geometry compute: ψ i = 70.3o , R = 2117.8 km, and thus Pr = −108.5 dBw


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             Ducts and Nonstandard Refraction (1)
      Ducts in the atmosphere are caused by index of refraction rates of decrease with height
      over short distances that cause rays to bend back towards the surface.
                                    TOP OF DUCT




                                                  EARTH’S
                                                  SURFACE
                            TX


      • The formation of ducts is due primarily to water vapor, and therefore they tend to occur
        over bodies of water (but not land-locked bodies of water)
      • They can occur at the surface or up to 5000 ft (elevated ducts)
      • Thickness ranges from a meter to several hundred meters
      • The trade wind belts have a more or less permanent duct of about 1 to 5 m thickness
      • Efficient propagation occurs for UHF frequencies and above if both the transmitter and
        receiver are located in the duct
      • If the transmitter and receiver are not in the duct, significant loss can occur before
        coupling into the duct
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              Ducts and Nonstandard Refraction (2)
       Because variations in the index of refraction are so small, a quantity called the refractivity
       is used
                               N ( h) = [n( h) − 1]10 6    n (h ) = ε r ( h)

       In the normal (standard) atmosphere the gradient of the vertical refractive index is linear
       with height, dN / dh ≈ −39 N units/km. If dN / dh < −157 then rays will return to the
       surface. Rays in the three Earth models are shown below.
                               True Earth                     Equivalent Earth     Flat Earth
                                                           (Standard Atmosphere)




                                   Re                              ′
                                                                  Re                        ∞

From Radiowave Propagation, Lucien Boithias, McGraw-Hill



                                                                                                                73
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              Ducts and Nonstandard Refraction (3)
       Another quantity used to solve ducting problems is the modified refractivity
                                                                     ′
                                       M (h ) = N ( h ) + 10 6 (h / Re )
       In terms of M, the condition for ducting is dM /dh = dN / dh + 157 . Other values of dN / dh
       (or dM / dh ) lead to several types of refraction as summarized in the following figure and
       table. They are:
       1. Super refraction: The index of refraction decrease is more rapid than normal and the ray
       curves downward at a greater rate
       2. Substandard refraction (subrefraction): The index of refraction decreases less rapidly
       than normal and there is less downward curvature than normal




From Radiowave Propagation, Lucien Boithias, McGraw-Hill
                                                                                                               74
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             Ducts and Nonstandard Refraction (4)

                                     Summary of refractivity and ducting conditions


                                         Ray                   Atmospheric           Virtual     Horizontally
                     dN / dh           Curvature      κ         Refraction           Earth      Launched Ray
                       >0                 up         <1                           more convex
                        0                none        1            below              actual
                     dN                                           normal                           moves
                 0>       > −39                      >1                                            away
                      dh
                       -39                           4/3          normal               less        from
                                                                                     convex        Earth
                       dN                           > 4/3
               − 39 >      > − 157       down                     above
                       dh
                                                                  normal              plane      parallel to
                       -157                                                                        Earth
                                                             super-refraction       concave     draws closer
                      < -157                                                                       to Earth




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