4 radio wave propagation over the earth

Radio Wave PropagationOver the Earth

SOLO HERMELIN

Updated: 27.04.10

1

http://www.solohermelin.com

SOLORadio Wave Propagation over the Earth

Table of Content

The Physics of Radio Waves

Effects of Earth Atmosphere

Atmospheric Absorption due to Gases and Water Vapor – Clear Air

2

Sun, Background and Atmosphere

Rain Attenuation over Electromagnetic Spectrum

Attenuation due to Fog over Electromagnetic Spectrum

Troposphere (from Earth Surface up to about 15 km (~ 50 kft(

Atmospheric Refraction Effects on Target Location

Ray Tracking in Troposphere

Linear Model of Refractive Index

Spherical Earth Effects – Radar Horizon Line of Sight (No Atmosphere(

Multipath PropagationMultipath Propagation for a Flath Earth

Multipath Propagation for a Spherical Earth

SOLORadio Wave Propagation over the Earth

Table of Content (continue(

3

Multipath PropagationGround ReflectionDivergence Reflection from a Spherical Earth SurfaceGround Surface RoughnessPattern Propagation FactorGeneric Radar Equation

Radar Equation with Multipath(Radar Equation with Multipath for a Distant Target (ψ -> 0

Ionosphere (from 40 – 50 km out to several Earth radii(

Ionosphere LayersIonosphere Mechanism of RefractionIonosphere Discovery History

References

SOLOThe Physics of Radio Waves

Radio Waves are Electro-Magnetic (EM) Waves, Oscillating Electric and MagneticFields.

The Macroscopic properties of the Electro-Magnetic Field is defined by

Magnetic Field Intensity H [ ]1−⋅mA

Electric Displacement D [ ]2−⋅⋅ msA

Electric Field Intensity E [ ]1−⋅mV

Magnetic InductionB [ ]2−⋅⋅ msV

The relations between those quantities and the sources were derived by James Clerk Maxwell in 1861

James Clerk Maxwell(1831-1879)

1. Ampère’s Circuit Law (A) eJt

DH

+

∂∂=×∇

2. Faraday’s Induction Law (F) t

BE

∂∂−=×∇

3. Gauss’ Law – Electric (GE) eD ρ=⋅∇

4. Gauss’ Law – Magnetic (GM) 0=⋅∇ B

André-Marie Ampère1775-1836

Michael Faraday1791-1867

Karl Friederich Gauss1777-1855

Maxwell’s Equations:

Electric Current Density eJ

[ ]2−⋅mA

Free Electric Charge Distributioneρ [ ]3−⋅⋅ msA

zz

yy

xx

111:∂∂+

∂∂+

∂∂=∇

4

SOLO Waves

2 2

2 2 2

10

d s d s

d x v d t− =Wave Equation

Regressive wave Progressive waverun this

-30 -20 -10

0.6

1.0.8

0.40.2

In the same way for a3-D wave

( ) ( )2 2 2 2 2

22 2 2 2 2 2 2

1 1, , , , , , 0

d s d s d s d s ds x y z t s x y z t

d x d y d z v d t v d t+ + − = ∇ − =

−=

v

xtfs

+=

v

xts ϕ

−=

−=

y

y

v

xtf

yd

d

td

sd

v

xtf

yd

d

vxd

sd

2

2

2

2

2

2

22

2

&1

+=

+=

z

z

v

xt

zd

d

td

sd

v

xt

zd

d

vxd

sd

ϕ

ϕ

2

2

2

2

2

2

22

2

&1

5

EM Wave

Equations

SOLO

ELECTROMGNETIC WAVE EQUATIONS

For Homogeneous, Linear and Isotropic Medium

where are constant scalars, we have

t

E

t

DH

t

t

H

t

BE

ED

HB

∂∂=

∂∂=×∇

∂∂

∂∂−=

∂∂−=×∇×∇

=

=

εµ

µ

ε

µ

Since we have also

( )( ) ( )

=⋅∇=

∇−⋅∇∇=×∇×∇

=∂∂+×∇×∇

0&

0

2

2

2

DED

EEE

t

EE

ε

µε

t

DH

∂∂=×∇

t

BE

∂∂−=×∇

For Source-lessMedium

02

22 =

∂∂−∇

t

EE

µε

Define

meme KK

c

KKv ===

∆

00

11

εµµε

where ( )smc /103

1036

1104

11 8

9700

×=

××

==−−

∆

ππεµ

is the velocity of light in free space.

22

20

HH

tµε ∂∇ − =

∂

same way

The Physics of Radio Waves

Return to Table of contents

6

tt ∂∂×∇=∇×

∂∂

µε ,

ED

ε=

HB

µ=

SOLO

Properties of Electro-Magnetic Waves

http://www.radartutorial.eu

Given a monochromatic (sinusoidal) E-M wave ( )0 0sin 2 sin

: /

xE E f t E t k x

c

k cω

π ω

ω

= − = − ÷ =

Period T,Frequency f = 1/T

Wavelength λ = c T =c/f c – speed of light


Run This7

SOLO


8

( ) ( ){ }( ) ( ){ }

wavenumbercfck

eHxktHH

eEzktEE

fc

zktjy

zktjx

λππω

ωω

λ

ω

ω

/22:

Imsin

Imsin

00

00

=

−

−

===

=−=

=−=

Phasor and time-harmonic/ instantaneous signal representation

HES

×=Poynting Vector Vector Power

ohmsZH

E3770 == Free space Impedance

SOLO


Frequency Band Designation9

POLARIZATION

SOLO

Electromagnetic wave in free space is transverse ; i.e. the Electric and Magnetic Intensitiesare perpendicular to each other and oscillate perpendicular to the direction of propagation.

A Planar wave (in which the Electric Intensity propagates remaining in a plane – containing the propagation direction) is said to be Linearly Polarized or Plane-Polarized.

If EM wave composed of two plane waves of equal amplitude but differing in phase by 90° then the EM wave is said to be Circular Polarized.

If EM wave is composed of two plane waves of different amplitudes and/or the difference in phase is different than 0,90,180,270° then the light is aid to be Elliptically Polarized.

If the direction of the Electric Intensity vector changes randomly from time to time we say that the EM wave is Unpolarized.

E


See “Polarization” presentation for more details10

POLARIZATION

SOLO

A Planar wave (in which the Electric Intensity propagates remaining in a plane – containing the propagation direction) is said to be Linearly Polarized or Plane-Polarized.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://www.enzim.hu/~szia/cddemo/edemo0.htm (Andras Szilagyi)

Linear Polarization or Plane-Polarization

( ) yyzktj

y eAE 1∧

+−= δω


Run This11

POLARIZATION

SOLO

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

If EM wave is composed of two plane waves of equal amplitude but differing in phase by 90° then the light is said to be Circular Polarized.

http://www.optics.arizona.edu/jcwyant/JoseDiaz/Polarization-Circular.htm

( ) ( ) yx xx zktjzktj eAeAE 11 2/∧

++−∧

+− += πδωδω


Run This

12

POLARIZATION

SOLO Properties of Electro-Magnetic Waves


13

SOLO Interaction of Electromagnetic Waves with Material

• Reflection

• Refraction

• Diffraction

- the re-radiation (scattering) of EM waves from the surface of material

- the bending of EM waves at the interface of two materials

-the bending of EM waves through an aperture in, or around an edge, of a material

• Absorption- the absorption of EM energy is due to the interaction with the material

Stimulated Emission& Absorption

Run This


14

SOLO Absorption and Emission

The absorption of a photon of frequency ν by a medium corresponds to the destruction of the photon; by conservation of energy the absorbing medium must be excited to alevel with energy h ν1 > h ν0 .

Stimulated Emission& Absorption Photon emission corresponds to the creation of a photon of

frequency ν; by conservation of energy, the emitting medium must be de-excited from an excited state to a state of lower energy than the excited state h ν = h ν2 - h ν1.

Phenomenologically, absorption and emission in gas phase media composed of atoms, diatomic molecules, and even larger molecules are restricted to discrete frequencies corresponding to the difference in the energy levels in the atoms. Continuous frequencies regimes arise only when the absorbed electromagnetic frequency is sufficiently high to ionize the atoms or molecules.

Run This


15

SOLO Reflection and Refraction at a Boundary Interface

When an electromagnetic wave of frequency ω=2πf is traveling through matter, the electrons in the medium oscillate with the oscillation frequency of the electromagnetic wave. The oscillations of the electrons can be described in terms of a polarization of the matter at the incident electromagnetic wave. Those oscillations modify the electric field in the material. They become the source of secondary electromagnetic wave which combines with the incident field to form the total field.

The ability of matter to oscillate with the electromagnetic wave of frequency ω is embodied in the material property known as the index of refraction at frequency ω, n (ω).

16

SOLO Refraction at a Boundary Interface

• If EM wavefronts are incident to a material surface at an angle, then the wavefronts will bend as they propagate through the material interface. This is called refraction.

• Refraction is due to change in speed of the EM waves when it passes from one material to another.

Index of refraction: n = c / v

Snell’s Law: n1 sin θ1 = n2 sin θ2

Run This

Willebrord van Roijen Snell1580-1626

17

Radio Wave Propagation over the EarthSOLO

Effects of Earth Atmosphere

The Principal Earth Atmosphere Effects are:

Refraction

Absorption

Thermal Noise Generation

The Minor Earth Atmosphere Effects are:

Scintillation (fluctuation of signal strength( resulting for refractive-index irregularities

Rotation of the Plane of Polarization by the Atmosphere

The two major regions of Earth’s Atmosphere that affects propagation of radio waves are

Troposphere (from Earth Surface up to about 15 km (~ 50 kft(


18

SOLO

• The index of refraction, n, decreases with altitude.

• Therefore, the path of a horizontally propagating EM wave will gradually bend towards the earth.

• This allows a radar to detect objects “over the horizon”.

Atmospheric Effects

19

Radio Wave Propagation over the Earth

20

SOLO Radio Wave Propagation over the Earth

SOLO

Atmospheric Effects• Atmospheric Absorption

- increases with frequency, range, and concentration of atmospheric particles (fog, rain drops, snow, smoke,…(

• Atmospheric Refraction

- occurs at land/sea boundaries, in condition of high humidity, and at night when a thermal profile inversion exists, especially at low frequencies.

• Atmospheric Turbulence

- in general at high frequencies (optical, MMW or sub-MMW(, and is strongly dependent on the refraction index (or temperature( variations, and strong winds.

21


SOLO

Atmospheric EffectsAtmospheric Absorption due to Gases and Water Vapor – Clear Air

22


SOLO Sun, Background and Atmosphere (continue – 1)

Atmosphere

Atmosphere affects electromagnetic radiation by

( ) ( )3.2

11

==

RkmRR ττ

• Absorption • Scattering • Emission • Turbulence

Atmospheric Windows:

Window # 2: 1.5 μm ≤ λ < 1.8 μm

Window # 4 (MWIR): 3 μm ≤ λ < 5 μm

Window # 5 (LWIR): 8 μm ≤ λ < 14 μm

For fast computations we may use the transmittance equation:

R in kilometers.

Window # 1: 0.2 μm ≤ λ < 1.4 μmincludes VIS: 0.4 μm ≤ λ < 0.7 μm

Window # 3 (SWIR): 2.0 μm ≤ λ < 2.5 μm

23

SOLO

Sun, Background and Atmosphere (continue – 2)

24


SOLO

Sun, Background and Atmosphere (continue – 3)Atmosphere Absorption over Electromagnetic Spectrum

25


SOLO

Rain Attenuation over Electromagnetic Spectrum

FREQUENCY GHz

ON

E-W

AY

AT

TE

NU

AT

ION

-Db

/KIL

OM

ET

ER

WAVELENGTH


26


SOLO

Rain Attenuation over Electromagnetic Spectrum O

NE

-WA

Y A

TT

EN

UA

TIO

N-D

b/K

ILO

ME

TE

R

WAVELENGTH

a. Drizzle – 0.25 mm/hrb. Light rain – 1 mm/hrc. Moderate rain – 4 mm/hrd. Heavy rain – 16 mm/hre. Excesive rain – 40 mm/hr

Example 4 mm/hr rain at 1o GHz (3 cm wavelength(

gives about 0.1 db/km attenuation

27


SOLO

Attenuation due to Fog over Electromagnetic Spectrum O

NE

-WA

Y A

TT

EN

UA

TIO

N-D

b/K

ILO

ME

TE

R

FREQUENCY

Increase in Fog Density (liquid water content) yield higher attenuation values

Example 100’ Visibility Fog at 10 GHz gives about 0.05 db/km

attenuation

28


SOLO

Attenuation in Radar Detection Range due to Attenuation along Propagation Path

AT

TE

NU

AT

ION

-RA

NG

E N

AU

TIC

AL

MIL

ES

FREE SPACE RANGE, NAUTICAL MILES

The net Effect of rain or fog (Atmospheric Attenuation) is a Reduction in Freespace Performance

Example Freespace system of 1000 nmi in 1db/km attenuation

gives a reduction of 20 nmi

29



Troposphere (from Earth Surface up to about 15 km (~ 50 kft)

The Refractive Effect of the Troposphere is related to its dielectric constant, which is determined by the pressure, temperature and water vapor content.

The Absorption and Noise-emissive Effects are related to the same quantities andto oxygen content.

30



Atmospheric Refraction Effects on Target Location

Because of Atmospheric Refraction a Ray Bending occurs and this provides

• Range Error

• Target Elevation Angle

31



Ray Tracking in Troposphere

The relationship expressing the angular bending of a ray of light can be determined from Geometric Optics using the Fermat Principle of Less Time which states that the path chosen by a ray joining two points is that which can be traveled in the Least Possible Time.

ncv /= - Velocity of light in atmosphere

c - Velocity of light in vacuum

n - refraction index of the atmosphere

∫∫ ==2

1

2

1

1minminmin

s

s

s

s

sdncv

sdt

32



Ray Tracking in Troposphere (continue – 1)

∫=2

1

1minmin

s

s

sdnc

t

In applying Fermat Princ iple we must minimize ∫=2

1

s

s

sdnI δδ

From the figure we can see that

( )[ ] rdrddrsd

drrdsd2/122

2222

/1 φ

φ

+=→

+=

[ ] ( ) rddrdrfrdrnIs

s

s

s

/:,12

1

2

1

2/122 φφφδφδδ ==+= ∫∫

Therefore

hrr += 0where

r0 – Earth Radiush - Height above Earth

Ray Path Geometry

33




[ ] ( ) rddrdrfrdrnIs

s

s

s

/:,min1minmin2

1

2

1

2/122 φφφδφδδθθθ

==+= ∫∫

The condition for the minimum is given by the Euler-LagrangeEquation:

0=∂∂−

∂∂

φφff

rd

d

Since f is not a function of φ the Euler-Lagrange Equation becomes:

( ) 01

22/122

2

=

+=

∂∂

φφ

φ

r

rn

rd

df

rd

d

The integration of this equation gives:

From the Figure we can see that:

( ) θφφ

cos1

2/122=

+

r

r constKrn ==θcosFinal Result

( ) constKr

rn ==+

2/122

2

1 φφ

34



Ray Tracking in Troposphere – Second Method

From the Figure we have curves at constant heights h , h + dh above the Earth Surface, representing shells of constant refraction n and n + dn, respectively. A ray ab enters the lower shell at an angle θ with respect to the line Oa (connecting the center of the spherical Earth O with the point a) on the shell a. Due to atmospheric refraction, the ray bends and reaches the upper shell at an angle θ+dθ. Normals erected in the points of interception a and b of the ray with the two shells intersect at O’. This

radius of curvature is ρ.

From the Figure the length of the path ab is: ( )θφρψρ dddab −==

From the Figure in the triangle Oab, if we approximate that ab with a straight line, we can use the Sinus Theorem in a Triangle Oab to calculate:

( ) ( )hr

abd

hdhrab

d

+≅→

+++=

00

cos90sinsin θφθφ

1cos

0

−+

=

hr

dab θρ

θρ

The radius of the ray curvature

35



Ray Tracking in Troposphere – Second Method (continue – 2)

Let use Snell’s Law to compute the relation between θ and θ + d θ when the ray path passes from n to n + d n shell. Since 90ₒ -θ and 90ₒ -θ –dθ are the angles between the ray path and the

normal to the shell, we have: ( ) ( ) ( )θθθ dndnn −−+=− 90sin90sin

Using the Sine Law in the triangle Oab we obtain:

( ) θθθ coscos0

0

hr

dhhrd

+++=+

( ) ( ) ( ) ( ) constddhhrndnhrn =++++=+ θθθ coscos 00Therefore


hrrconstKrn +=== 0cosθWe recovered

1cos

0

−+

=

hr

dab θρ

θρ

36





( ) ( ) ( ) ( ) constddhhrndnhrn =++++=+ θθθ coscos 00

Start with

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) θθθθθ

θθθθθθθθθ

ddhhrnhrnddhnhrn

ddhhrndhhrnddhhrn

ddhhrndnddhhrndn

sincoscoscos

sincoscos

sincoscos

000

000

00

++−++++≅++−+++++≅

−+++≅++++

Developing the right side of this equation we obtain

From those two equations we obtain:( ) ( ) 0sincoscos 00 =++−++ θθθθ ddhhrnhrnddhn

( )( )

+

+≅++++=

<<

hr

dh

n

ndctg

dhhrn

dhnhrndctgd

hdh

00

0 θθθ

θθρ

ρ

θρ

θρctg

hr

hrdh

nnd

hr

dab

1cos

1cos

0

0

0

−+

+

+=

−+

=Using

From Figure θsin

hdab = θ

θθρ

ρ

θ sin

cos

1cossin

0

0

−+

+

+==

hr

hrdh

nnd

hdab

37





We obtained

θθ

θρ

ρ

θ sin

cos

1cossin

0

0

−+

+

+==

hr

hrdh

nnd

hdab

from which

θρθρθρcoscoscos

00 hr

dh

n

ndhd

hr

dh

++=−

+

θρ

coshdndn

−=

The refractive index of air is very near to unity (n ≈ 1). Furthermore only rays that are close to horizontal (θ ≈0) are of interest in radar.

Therefore, the Radius of the Ray Curvature, ρ, is given by:

hdnd−

≅ 1ρ38




2

1cossin

−=→=

sd

hd

sd

hd θθ

We assume that the atmosphere is stratified and its properties vary only as a function of height, and therefore is spherically symmetrical. The geometry of the ray in the Earth Atmosphere is described in Figure on the upper right.

From the Figure on the lower right we have:

( ) ( ) [ ]2

0

00

2

000

coscos

2

/1

cos1

cos1

cos1sin

0000

+

−

=

−

=−

==+==

rhhn

n

hd

rr

hnn

hdhdhdsd

hrrnrrn

θθθθ

θθ

Ray-Path Geometry in an Athmosphere Spherically Symmetricwith Respect to Earth Center

39




dsndtcRd ==Using the relation:

( ) [ ]2

0

00

/1

cos1

+

−

=

rhhn

n

hdsd

θ

we can obtain the Radar Range R as a function of initial angle θ0 and final height h1:

( ) ( ) ( )

( ) [ ]

∫∫

+

−

==11

02

0

000

01

/1

cos1

,hh

rhhn

n

hdhnsdhnhR

θθ

( ) ( ) ( ) [ ]( ) [ ]

∫∫−

+

+==11

02

00

0

02

000

01

1cos

/1

/1

cos

1,

hh

n

rhhn

hdrhhn

nsdhnhR

θ

θθ

or:

40




( ) ( ) hceenhn −−+= 11 0

Exponential Model of Refractive Index

A commonly assumed Model of the Refractive Index is

Ns ce (h in km) ce (h in feet) hp ft

200 0.1184 3.609x10-5 10,000

250 0.1256 3.829x10-5 5,000

301 0.1396 4.256x10-5 1,000

313 0.1439 4.385x10-5 700

350 0.1593 4.857x10-5 0

400 0.1867 5.691x10-5 0

450 0.2233 6.805x10-5 0

Define the Earth Surface Refractivity Ns( ) 6

0 10x1−= nN s

The dependence of Ns on Atmosphere Pressure P (millibars), partial water vapor pressure e (millibars) and the Temperature T in degrees Kelvin is

+=

T

eP

TN s

48106.77 41


Troposphere (from Earth Surface up to about 15 km (~ 50 kft)Ray Tracking in Troposphere (continue – 6)

( ) ( ) ( )

( ) [ ]

∫∫

+

−

==11

02

0

000

01

/1

cos1

,hh

rhhn

n

hdhnsdhnhR

θθ

( ) ( ) hceenhn −−+= 11 0

Exponential Model of Refractive Index

Chart of Radar Range as function of target height h1 and elevation angle θ0

42


Troposphere (from Earth Surface up to about 15 km (~ 50 kft)Ray Tracking in Troposphere (continue – 6)

( ) ( ) ( )

( ) [ ]

∫∫

+

−

==11

02

0

000

01

/1

cos1

,hh

Radar

rhhn

n

hdhnsdhnhR

θθ

( ) ( ) hceenhn −−+= 11 0

1-Way Tropospheric Range Error for Standard Atmosphere (0% Humidity)

The range error due to Ray Bending/Refraction can be significant

Exampleθ0 = 2 Degrees and Height = 70,000 ftRange Error = 140 ft 43




By integration we obtain:

∫−

=−r

r

rnrn

r

rd

0

1cos

2

000

0

θ

φφ

( ) θφφ

cos1

2/122=

+

r

r000 coscos θθ rnrn =

( ) 0002/122

2

cos1

θφ

φrn

r

rn =+

( ) ( )222000

242 1cos φθφ rrnrn +=

11cos

2

2

000

2 =

−

φ

θ

rn

rnr

44




∫−

=−r

r

rnrn

r

rd

0

1cos

2

000

0

θ

φφ

From the Figure the Diffraction Angle Error is θ0 - θt

θt can be obtained from the Sinuses Law

( )Rhr

t φθ sin90sin

0

=++

+= − φθ sincos 01

R

hrt

45




Tropospheric Refraction Errors for a Standard Atmosphere with 100% Humidity46




Tropospheric Refraction Errors for a Standard Atmosphere with 0% Humidity

Exampleθ0 = 2 Degrees,h = 1,000 nmi

θ0 –θt =0. 28 Degrees,

47



Linear Model of Refractive Index

Schelleng, Burrows, Ferrell (SBF) Model of Refractive Index is a simpler Linear Model

( ) hknhn −≈ 0

H [km] N N0-[h/4r0] x 106

0 319 = N0 319

1 277 279

3 216 201

10 92

20 20

50 0.2

( )6

6

10x1

10x1−+=

−=

Nn

nN

( ) ( ) kmRadiusEarthrrhnhn 63704/ 000 ==−≈

~

48



Linear Model of Refractive Index (continue – 1)


( ) ( )00 4/ rhnhn −≈

( ) θθ coscos 0000 hrnrn +=

00

0

0 coscos θθn

n

hr

r

+=

0

0 1

rh

n

<<≈

0

0

0

00

20

00

00

0 cos

34

34

cos4/4/

cos4/1

1cos θθθθ

hr

r

rhhhr

r

rhhr

r

+≅

−−+=

−+≅

The SBF Model of Refractive Index is equivalent to plotting the rays as straight lights in a plot in which the Earth Radius is greater by a factor of 4/3.

49




We found that the Radius of the Ray Curvature, ρ, is:

hdnd−

≅ 1ρ

The curvature of the Earth of radius r0 is the reciprocal of this radius and the relative curvature of the considered ray to that of the earth is the difference between the two curvatures, or

hd

nd

rr+=−

00

111

ρ

Equating this to the relative curvature of an imaginary earth having such a radius kr0 that the curvature of a ray propagating in the atmosphere is zero, we obtain

00

11

rkhd

nd

r=+

hdnd

rk

01

1

+=

For: ( ) ( )00 4/ rhnhn −≈04

1

rhd

nd −= 3

4=k


We recovered the SBF Model result. 50





The SBF model is reasonable realistic up to about 3 km (approximately 10,000 ft) but becomes unrealistic (predicts to much ray bending) at higher altitudes. It is apparent that the SBF model is physically incorrect at altitudes for which it results in n < 1, that is for h > (n0-1)/k, or about

8 km, because n cannot be less than 1 in an un-ionized, unbounded medium.

51




Computation of the Target Height

( ) ( ) ( ) e

A

e

A

e ahaRRhah

ee

−++−++= θ

90cos2 1

221

If we use the SBF Model with Earth Radius ae = 4/3 r0 ,the ray paths are straight lines and

we can compute the target height using the Cosine Law in the triangle ADC

Define Ae = ae +h1

( ) θsin2222eee ARRAah ++=+

( ) θsin22221 eee ARRAAhh ++=+− ( ) ( ) θsin22 22

11 ee ARRAhhhh +=+−−

( ) eeeeeeee AARRAAhaARRAAhh +++=++−++=+− θθ sin22sin22 221

221

122

2

sin2

sin2h

AARRA

ARRh

eee

e ++++

+=θ

θ 52




Computation of the Target Height (continue – 1)

122

2

sin2

sin2h

AARRA

ARRh

eee

e ++++

+=θ

θ

( ) eee

e

eee

e

AARA

ARRh

AARA

ARRhh

++++≈

++++≈

/sin1

sin2

/sin21

sin2 2

1

2

1 θθ

θθ

Let use the fact that R << Ae and h1 << ae

( ) ( )e

e

e

e

ee

ee

e

A

A

RARR

hA

A

RARR

h

AR

A

ARRh

2

sin4

12sin1

2

2

sin1sin2

2sin

12

sin22

222

1

2

1

2

1

θθθθ

θθ

−+−

+=

−+

+≈

+

++=

ea

RRhh

2

cossin

22

1

θθ ++≈

r0 = 6,370 km, so that ae = 8,493 km=4,586 nmi

h, h1 [ft]

R [nmi]

θθ 221 cos6625.0sin076,6 Rhh ++≈

53

SOLO

Spherical Earth Effects – Radar Horizon Line of Sight (No Atmosphere)

For a Antenna at a height h1 above a Spherical Earth of Radius r0 = 4587 nm the range to the horizon Rh1, is:

( ) 102

1102

02

101 2201

hrhhrrhrRrh

h

<<

≅+=−+=

[ ] [ ]fthnmRh 11 23.1≅or

The range from the Antenna to a Target at the horizon Rh, is:

( ) ( )

2010

,2

2202

110

20

220

20

210

2222021

hrhrhhrhhr

rhrrhrRrhh

h

+≅+++=

−++−+=<<

[ ] [ ] [ ]( )fthfthnmRh 2123.1 +≅or

54


SOLO

Earth Curvature Nomograph

[ ] [ ] [ ]( )fthfthnmRh 2123.1 +≅

Example 1:

fth 2001 =fth 7502 =nmRh 51=

fth 2001 =

fth 7502 =nmRh 51=

Example 2:

fth 2001 =fth 02 =nmRh 17=

nmRh 171 =

Spherical Earth Effects – Radar Horizon Line of Sight (No Atmosphere)

55




Geometry of a Specular Reflection from a Flat Earth

Multipath Propagation

Multipath refers to the existence of more that one ray path by which electromagnetic waves can travel from the radar to the target, and vice versa.

In the usual multipath situationthere are two paths:• Direct path• Indirect path via reflection from earth surface.

In the unusual multipath situation there are more than two paths due to earth non-uniformity.

56





Multipath Propagation

There are four possible ray paths:

1.Direct transmission to target and direct target to receiver2.Direct transmitted path plus reflected received path3.Reflected transmitted path plus direct received path4.Reflected transmitted path plus reflected received path

Multipath is a problem of interference of multiple waves, that have different phases and amplitudes, therefore is a Phasor Addition.

57




Multipath Propagation for a Flath Earth

Target Elevation, ( )[ ] ( ) RhhRhh rtrtd //sin 1 −≈−= −θ

Grazing Angle, ( )[ ] ( ) RhhRhh rtrt //sin 1 +≈+= −ψ

Range to Reflection Point

( )( )rtr

Rhh

r hhhRhxrt

+≈=+≈

/tan//tan

0

ψψ

From the Figure ( ) rd RRR θθ coscos 21 +=

−=

−+=∆ 1

cos

cos121

r

dRR

RRRR

θθ

rtrtrtr

rdrt

d R

hh

RR

hh

R

h

R

h

R

hh θθθθ ≈+≈++===≈−=

2121

sinsin

( ) ( ) ( ) ( )R

hh

R

hh

R

hhRRRRRR rtrtrt

drrdrdr

dr

d

r 2

222

11

2

11coscos1

cos

cos2

2

2

22222

1

1

1

=

−−+=−≈

−+−+−≈−≈

−=∆

<<

<<

<<

θθθθθθθθ θ

θ

θ

Path Length DifferenceBetween Image and Direct Path ΔR = R1+R2-Rd

Phase Difference between direct and indirect pathsλ - wavelength

R∆=∆λπϕ 2 58


Multipath Propagation for a Spherical Earth

The Refraction by the Atmosphere Effect can be taken in consideration approximately by using an Effective Earth Radius ae 03

4rae =

The basic problem is to find the reflection point B. This point is such that the reflected rays R1 and R2 make the same Grazing Angle ψ with the Horizon Plane at the reflection point.

The Direct Path from antenna A to Target D has a range (using Cosine Law in the triangle ADC)

( ) ( ) ( ) ( )( ) ( ) ( ) ( )[ ]2/sin21222

cos2

221

22

2121

2

212

22

1

φ

φ

−++−++++=

++−+++=

hahahhhhaa

hahahahaR

eeee

eeeed

( ) ( ) ( ) ( )2/sin4 221

212 φhahahhR eed +++−=

( )( ) ( )

( )( ) e

d

ahhee

d

ahh

hhR

haha

hhRe /14

sin221

212

21

,21

212

21

21 ++−−≈

++−−=

<<

<<

−φ

φ

Assuming that the Target position is known (Rd, h1, h2, ae are known) we obtain

59


Multipath Propagation for a Spherical Earth (continue – 1)

To find the other parameters we must find ϕ 1 and use ϕ 2= -ϕ ϕ 1. Then we obtain (using Cosine Law in the triangles ABC and DBC)

( )( ) ( )

( )( ) e

d

ahhee

d

ahh

hhR

haha

hhRe /14

sin221

212

21

,21

212

21

21 ++−−≈

++−−=

<<

<<

−φ

φ

We found

( ) ( )( ) ( )[ ] 2,12/sin21222

cos2

222

22

=−+−++=

+−++=

ihaahhaa

haahaaR

iieeiiee

iieeieei

φ

φ

( ) ( ) 2,12/sin4 22 =++= ihaahR iieeii φ

and( ) ( ) 2,190cos2

sin

222 =+−+=+−

iaRRaha

i

ieiieie

ψ

ψ

2,122

2sin

22

=−≈−+=<<

ia

R

R

h

aR

Rhha

e

i

i

iah

ei

iiiei

ei

ψ

If ϕ 1 was properly chosen then ψ1=ψ2=ψ. 60



We found

ee a

R

R

h

a

R

R

h

22sin 2

2

21

1

1 −=−=ψ

2,1cos1

=≈==<<

iRRaG iiiei

ψψφ

From the Figure

( ) 2121 GGaaG ee +=+== φφφ

We have

ee a

GG

GG

h

a

G

G

h

221

1

21

1

1 −−−

=−

( )1

211

22

1

211

2

22

2

2

GGa

GGGGha

Ga

Gha

e

e

e

e

−−+−=−

( )[ ] 02232 112122

13

1 =++−+− GhaGhhaGGGG ee

( )[ ] 0/2/232 112122

13

1 =++−+− ee ahahh φφφφφφ

or

ϕ 1 Computation

This is a cubic equation in ϕ 1 and has the solution

( )

( )

−=Ψ

++=

+Ψ+=

−

e

ee

ap

hh

a

hh

ap

p

3121

2

21

1

2cos

23

2

3cos

2

φ

φ

πφφ

61



To find the transmitted and received power we need to compute the angles θd and θr, to obtain the antenna gains for the direct and indirect paths. For θd computation let use Cosine Law in the triangle ACD:

( ) ( ) ( ) ( )

d

deddee haRRhahaθ

θsin

122

12

2 90cos2−

++−++=+

( ) ( )( )1

221

22

2sin

haR

Rhaha

ed

deed +

−+−+=θ

( )( ) e

d

d

ahh

ed

ded a

R

R

hh

haR

Rhhhha e

22

2sin 12

,

1

221

2212

21

−−≈+

−−+−=<<

θ

For θr computation let use Cosine Law in the triangle ABC:

( ) ( ) ( )

r

reee haRRhaaθ

θsin

112

12

12 90cos2

−

++−++=

( )( )11

21

21

2

2sin

haR

Rhaa

e

eer +

−+−=θ

( ) e

ah

e

er a

R

R

h

haR

Rhha e

22

2sin 1

1

1

11

21

211

1

−≈+

−+=<<

θ 62



Since the values of R1 + R2 and Rd are big numbers and the difference ΔR is small, a direct computation is numerically problematic.

Instead let use Cosine Law in the triangle ABD:

( )( )

( ) ( ) ψψ

ψψ

221

221

221

22

21

2cos

212

22

12

sin4sin212

2180cos2

RRRRRRRR

RRRRRd

−+=−++=

−−+=−

Path Length DifferenceBetween Image and Direct Path ΔR = R1+R2-Rd

( ) ( ) ( )dR

dd RRRRRRRRRRR ++−+=−+=∆

212122

212

21 sin4

ψ

dRRR

RRR

++=∆

21

221 sin4 ψ

Phase Difference between direct and indirect pathsλ - wavelength

R∆=∆λπϕ 2

63


Ground Reflection

Smooth Surface Reflection Coefficient (Fresnel Coefficient)

( )( )

( )( )gg

ggjhh

gg

ggjvv

h

v

e

e

ψεψψεψ

ρ

ψεψεψεψε

ρ

φ

φ

cossin

cossin

cossin

cossin

−+

−−==Γ

−+

−−==Γ

−

−

The Smooth Surface reflection coefficient depends on the frequency, on the surface dielectric coefficient ε=ε’- j ε’’, and on the radar grazing angle ψg. The vertical polarizationand the horizontal polarization coefficient are:

ej

j

σλεεεε60'

"'

−=−=

ε – complex dielectric constant of the surfaceε’ - relative dielectric constantσe – conductive of the surface materialλ - wavelength

64


Radar Coefficient Magnitude Radar Coefficient Phase

Ground Reflection

Smooth Surface Reflection Coefficient (Fresnel Coefficient)

( )( )

( )( )gg

ggjhh

gg

ggjvv

h

v

e

e

ψεψψεψ

ρ

ψεψεψεψε

ρ

φ

φ

cossin

cossin

cossin

cossin

−+

−−==Γ

−+

−−==Γ

−

−

The Smooth Surface reflection coefficient depends on the frequency, on the surface dielectric coefficient ε=ε’- j ε’’, and on the radar grazing angle ψg. The vertical polarizationand the horizontal polarization coefficient are:

( ) ( )( ) ( )

010

901

190

=Γ−=−==Γ

=Γ−=+−−=

+−==Γ

gvgh

gvgh

ψψ

ψεεεε

εεψ

Brewster Angle

For ψ = 0 the surface behaves like a mirror,without attenuation but a change of π in phase.

65


Material ε’ σe (mho/mm)

Good soil (wet) 25 0.02

Average soil 15 0.005

Poor soil (dry) 3 0.001

Snow, ice 3 0.001

Fresh water (h=1m) (h=0.03m)

8165

0.715

Salt water (h=1m) (h=0.03m)

7560

515

Electrical Properties of Typical Surfaces

ej

j

σλεεεε60'

"'

−=−=

ε’ - relative dielectric constantσe – conductive of the surface materialλ - wavelength

Smooth Surface Reflection Coefficient

66

Ground Reflection


Ground Reflection

Smooth Surface Reflection Coefficient

Magnitude and Phase of the Reflection Coefficient, Vertical Polarization, at a Number of Frequencies

Magnitude of the Reflection Coefficient, Horizontal Polarization, at a Number of Frequencies

( )( )

( )( )gg

ggjhh

gg

ggjvv

h

v

e

e

ψεψψεψ

ρ

ψεψε

ψεψερ

φ

φ

cossin

cossin

cossin

cossin

−+

−−==Γ

−+

−−==Γ

−

−

σλεεεε60'

"'

j

j

−=−=

ε’ - relative dielectric constantσe – conductive of the surface materialλ - wavelengthψg – grazing angle

67


Divergence Reflection from a Spherical Earth Surface

The overall reflection coefficient is also affected by the round Earth Divergence Factor D. When a electromagnetic wave is reflected by a round Earth it diverges because of surface curvature.

Due to divergence the reflected energy is spread and the power density is reduced by the Divergence Factor D < 1.For a given Radar and Target geometry the Divergence Factor is given by:

( ) ( ) ( )eee

e

ahahGaGG

GaD

/1/1sincos/2

cossin

2121 +++=

ψψψψ

Since ψ is a small angle for the cases when the divergence is effective, and since h1, h2 << ae the Divergence Factor is given by:

ψsin2

1

1

21

GaGG

D

e

+≈

68


SOLO

• Surface Diffraction

- increases at lower frequency, range, and higher surface roughness

69

Ground Surface Roughness



SOLO

• Path difference of the two rays is Δr =OA+OB = 2 h sin ψ

• Similarly the phase difference (Δφ) is simply k Δr or 4 π h sin ψ / λ

• By arbitrarily setting the phase difference to be less than π /2 we obtain the Rayleigh criteria for “rough surface”

Other criteria such as phase difference less than π /4 or π /8 are considered more realistic.

2

sin4 πλ

ψπ >hψ

λsin8

>h smoothי roughΔφ

π /20

70

Ground Surface Roughness (continue - 1)



SOLO

2

sin4 πλ

ψπφ<

>=∆ h

smoothי rough Δφ

π /20

2sin4

2

−

= λψσπ

ρh

es

σh – r.m.s. of the Gaussian Distribution of a surface height irregularityψ - grazing angleλ - wavelength

This defines the Coefficient of Returned Field Intensity in the Specular Direction

In the Figure on the right we see the results published by Beckmann and Spizzichino (1963) and the fitting result of aGaussian Distribution from which they derived the Coefficient of Surface Roughness as:

71

Ground Surface Roughness (continue = 2)

Radio Wave Propagation over the EarthTroposphere (from Earth Surface up to about 15 km (~ 50 kft)

SOLO

Ground Surface Roughness (continue - 3)

• Transition from Specular to Diffuse as function of Surface Roughness σh

- increases at lower frequency, range, and higher surface roughness

2sin2

2

−

= λψσπ

ρh

es

72

Radar Wave Propagation over the Earth


Radar Wave Propagation over the EarthSOLO



Pattern Propagation Factor

The phase difference of two waves is affected by two factors:1.The path length difference ΔR2.The phase change at reflection

Multipath is a problem of interference of multiple waves, that have different phases and amplitudes, therefore is a Phasor Addition.

The amplitude difference of two waves is affected by three factors:1.Each wave is subjected to the power inverse square-low reduction 2. Power loss at reflection3. Nonuformity of Antenna gain (different gains for direct and indirect paths)

73


Troposphere (from Earth Surface up to about 15 km (~ 50 kft)Pattern Propagation Factor (continue – 1)

Let compute the received Field Intensities

( ) ( ) ( ) ( ) ( ) ( )[ ]( ) ( )

( )( )

2

220

222220

21

2121

1

2

+=

++=

−+

+++

RRRkj

d

rRkjdTGT

RRkjr

RRRkjrd

RkjdTGTr

ef

fefE

efeffefEE

θθρθσ

θρθθρθσ

σTGT – Target Radar Cross Section (RCS)

– electric intensity in the direction of the beam-maximum direction0E

74

There are four possible ray paths:1.Direct transmission to target and direct target to receiver ADA2.Direct transmitted path plus reflected received path ADBA3.Reflected transmitted path plus direct received path ABDA4.Reflected transmitted path plus reflected received path ABDBA

R1+R2-R =ΔR – path length difference between indirect and direct paths

ρ – Ground Reflection Coefficient

f (θd) – is the ratio of electric intensities in the direct path to the intensity in the beam-maximum direction

f (θr) – is the ratio of electric intensities in the direction of the reflected ray to the intensity in the beam-maximum direction



Pattern Propagation Factor (continue – 2)

The received Field Intensities

75

The received Field Power is ( ) ( )( )

42

1

242

0

21~ Rkj

d

rRkjdTGTrr e

f

fefEEP ∆+=

θθρθσ

or

( ) ( )( )

420

44

0 1 FGPef

ffPP TRTGT

Rkj

d

rdTGTr σ

θθρθσ =+= ∆

( ) ( )( )

2

220 1

+= ∆Rkj

d

rRkjdTGTr e

f

fefEE

θθρθσ

ΔR=R1+R2-R – path length difference between indirect and direct paths

ρ – Ground Reflection Coefficient

Γ – is the specular (Fresnel) reflection coefficientϕρ j

vh

vh e−=Γ

D – Diffraction Factor

ρs – Surface Roughness Factor

Dsvh ρρ Γ=

P0 – Transmitted Power

GTR = |f(θd)|2 – Antenna Transmitting Gain( )

( )Rkj

d

r ef

fF ∆+=

θθρ

1 – Pattern Propagation Factor



Pattern Propagation Factor (continue – 3)

F – the Pattern Propagation Factor is defined as

( )( )

( ) ϕββλπα

θ

θρθ α +−+∆=

Γ+= −

drj

d

rsvh

d Ref

fDfF

21

( ) ( ) ( ) ( ) Rjjr

js

vh

jd

Rj

rd eefeDefeffF rd∆−−−−∆−

Γ+=+= λπ

βϕβλπ

θρθθρθ22

:

76

77

Generic Radar Equation

[ ]2

2 2

Pr Pr Re

1 1 1 1

4 4 4TGT

Xmtr TR RCVRRCVR TGT TGT

TR TR TR TGT RCVR TGT RCVR RCVR

Transmitter opagation opagation ceiverXMTR TGT TGT RCVR

P G GP A G W

L R L R L Lσ

λπ π π→ →

→ →

=

the Power [W/m2] received at the Receiver isπ

λ4

2RCVR

RCVR

GA =

( ) [ ]WLLRR

GGPP

RCVRTRRCVRTR

RRTGTRCVRTRTR

RCVR

RCVRRCVRTGTTRTGTTR

223

1000/1000/2

4

1010

πσλ αα →→ −−

=

Using

or

SOLO

1000/

1000/

10

10RCVRRCVRTGT

TRTGTTR

RRCVRTGT

RTGTTR

L

L→

→

−→

−→

=

=α

α

α – atmospheric attenuation in db/km



Radar Equation with Multipath

( )( )

( ) ϕββλπα

θ

θρθ α +−+∆=

Γ+= −

drj

d

rsvh

d Ref

fDfF

21

[ ]2

2 2

Pr Pr Re

1 1 1 1

4 4 4TGT

Xmtr TR RCVRRCVR TGT TGT

TR TR TR TGT RCVR TGT RCVR RCVR

Transmitter opagation opagation ceiverXMTR TGT TGT RCVR

P G GP A G W

L R L R L Lσ

λπ π π→ →

→ →

=

In our case RTR = R ≈ R1+R2=RRCVR , and we must add the Pattern Propagation Factor F

( ) [ ] RCVRRCVRTGTTGTTRTRTGTRCVRTRTR

RCVR LLLLLWFLR

GGPP →→== :

4

4

43

2

πσλ

( ) ( )

( )( ) drd

rsvh

dd

f

fDx

RxRxfRxjRxfF

ββϕθ

θρ

λπ

λπθ

λπ

λπθ

−+=ΨΓ

=

Ψ+∆+

Ψ+∆−=

Ψ+∆−

Ψ+∆−=

::

2sin

2cos1

2sin

2cos1

2

22

2

444

78

F – Pattern Propagation Factor



Radar Equation with Multipath

( )

( )( ) drd

rsvh

d

f

fDx

RxxfF

ββϕθ

θρλπθ

−+=ΨΓ

=

Ψ+∆−+=

::

2cos21

2

244

( ) [ ] RCVRRCVRTGTTGTTRTRTGTRCVRTRTR

RCVR LLLLLWFLR

GGPP →→== :

4

4

43

2

πσλ

79



(Radar Equation with Multipath for a Distant Target (ψ -> 0

( )( )

4sin2

2

1

24 sin2sin2

2cos21

:1:

=

+∆−+=

=−+=Ψ=Γ

=

=∆

= λψππ

λπ

πββϕθ

θρ

ψr

hR

x

drd

rsvh

hRxxF

f

fDx

r

For a Distant Target (ψ -> 0) we have

ψsin2 rhR =∆( ) ( ) 010 =Γ−=−==Γ ψψ vh

( ) 10

0

sin22

2

====

−

ψ

λψσπ

ψρh

es

1≈D

Assume a uniform Antenna Gain at Transmission and Reception f (θd) = f (θr) =1

80



(Radar Equation with Multipath for a Distant Target (ψ -> 0

44 sin2

sin16

=

λψπ rhF

For a Distant Target (ψ -> 0) we have

( ) [ ]WFLR

GGPP TGTRCVRTRTRRCVR

4

43

2

4 πσλ=

We can see that:

,1,0sin2 == nn

hr πλ

ψπ

( ) ,1,02

12sin2 =+= nn

hr πλ

ψπ

|F| is minimum (zero) for

|F| is maximum for

,1,02

sin 1 =

= − n

h

n

r

λψ

( ),1,0

2

12sin 1 =

+= − nh

n

r

λψ 81


( )

( )( ) drd

rs

d

f

fDx

RxxfF

ββϕθ

θρρλπθ

−+=Ψ=

Ψ+∆−+=

::

2cos21

2

244

( ) [ ]WFLR


4

43

2

4 πσλ=

82


( )

( )( ) drd

rs

d

f

fDx

RxxfF

ββϕθ

θρλπθ

−+=ΨΓ

=

Ψ+∆−+=

::

2cos21

2

244

( ) [ ]WFLR


4

43

2

4 πσλ=

83

AN/SPS-49 Very Long-Range Air Surveillance Radar



The Ionosphere is a region of very low gas density compared to the troposphere.

Its effect on radio wave propagation is due to the presence of free electrons,caused primarily by the ionizing action of the sun’s ultraviolet rays and X rays.

The region of the ionosphere that significantly affects radio waves extends up to above 1,000 km.

The effects observed include - Refraction, - Absorption,

- Noise Emission and - Polarization Rotation.

All those effects, for a given condition of the Ionosphere,decrease with increasing frequency, and they are completely unimportant in the

microwave region, above 1,000 MHz. 84

SOLO

Ionosphere Layers

• Ionospheric Layers are independently produced by absorption of Solar Radiation by specific molecules of atmospheric constituent

• D region - 60-90 km, dynamics dominated by neutral atmosphere - Ions are NO+ produced by UV radiation at 121.5 nm

• E region - 90 – 130 km, peak near 110 km - Ions are O2

+ and NO+ produced by UV radiation in the 100 – 150 km range, and solar X-rays in the 1 – 10 nm range

• F region (F1 and F2( - Above 130 km (typically caps at 2000 km( F1 peaks at 170 km, F2 peaks near 250 km - Ions are mainly O+ from photons in the 17 – 91 nm range

85

The Ionosphere

86

SOLO Earth Magnetic Field

87

The density of electrons (how many electrons there are per every cubic centimeter( is represented by the varying colors. Bands of high density that appear at high latitudes during the storm but disappear rapidly as it subsides are due to the high velocity particles smashing into the atoms in the atmosphere and knocking electrons free. These same high velocity particles produce the auroral lights. We can use these maps and the varying colors to find the lowest frequency that is detectable from the ground. The lowest frequency detectable, known as the critical frequency, is related to the density of electrons by the equation: f = 9x10^-3 x sqrt(N( MHz. In this equation f is the critical frequency and N is the electron density, sqrt means to take the square root of the electron density. In the maps above the electron density ranges from 33300 electrons/cm^3 (dark blue( to 249750 electrons/cm^3 (green( to 552780 electrons/cm^3 (red(.

Below is an animation comparing the ionospheric conditions during a typical day with that of a day containing an ionospheric storm. An ionospheric storm is caused by a coronal mass ejection from the sun that strikes the Earth's atmosphere. These mass ejections contain large amounts of particles that smash into the ionosphere and knock electrons loose from atoms. As discussed above the loose electrons reflect radio waves from astronomical sources back into space. The addition of loose electrons as a result of a mass ejection makes observations and communications difficult. The dark blue and purple areas are the areas where the number of loose electrons is low. In these areas there are few electrons to reflect radio waves and thus lower frequency waves are able to reach the ground. As can be seen from the animations the night time and early morning hours are best for observations due to the fact that the sun is not in the sky and its ultraviolet light is not reaching the atmosphere at this time

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Ionosphere Mechanism of Refraction

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When a radio wave reaches the ionosphere, the electric field in the wave forces the electrons in the ionosphere into oscillation at the same frequency as the radio wave. Some of the radio-frequency energy is given up to this resonant oscillation. The oscillating electrons will then either be lost to recombination or will re-radiate the original wave energy. Total refraction can occur when the collision frequency of the ionosphere is less than the radio frequency, and if the electron density in the ionosphere is great enough

The critical frequency is the limiting frequency at or below which a radio wave is reflected by an ionospheric layer at vertical incidence. If the transmitted frequency is higher than the plasma frequency of the ionosphere, then the electrons cannot respond fast enough, and they are not able to re-radiate the signal. It is calculated as shown below

Nfcritical310x9 −=

where N = electron density per cm3 and fcritical is in MHz.

The Maximum Usable Frequency (MUF( is defined as the upper frequency limit that can be used for transmission between two points at a specified time.

αsincritical

MUF

ff =

where α is the angle of the wave relative to the horizonThe cutoff frequency is the frequency below which a radio wave fails to penetrate a layer of the ionosphere at the incidence angle required for transmission between two specified points by refraction from the layer.

Typical MUF15 – 40 MHz (daytime( 3 – 14 MHz (night(

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Daily Propagation Effects

• Shortly after sunrise, the D and E layers are formed and the F layer splits into two parts. – The D layer acts as a selective absorber, attenuating low frequency

signals, making frequencies below 5 or 6 MHz useless during the day for DX work.

• The E and F1 layers increase steadily in intensity from sunrise to noon and then decreases thereafter. – Short skip propagation via the E or F1 layers when the local time at the

ionospheric refraction point is approximately noon. – The MUF’s for the E and F1 layers are about 5 and 10 MHz

respectively. • The F2 layer is sufficiently ionized to HF radio waves and return them to

earth. – For MUF’s is above 5 - 6 MHz, long distance communications are

possible. – MUF’s falls below 5 MHz, the frequencies that can be returned by the F

layer are completely attenuated by the D layer.

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Effects of Sunspots

• During a sunspot minimum, the chromosphere is very quiet and its UV emissions are very low.

• F2 MUF’s decrease, rarely rising to 20 MHz • Most long distance communications must be carried out on the lower HF

bands. • During periods of high sunspot activity:

– The best daytime bands are 12 and 10m. – At night, the best bands are 20, 17 and 15m.

• At the low end of the solar cycle, – The best daytime bands are 30 and 20m.– After dark, 40m will open for at least the early part of the evening. – In the early morning hours, only 80m will support worldwide

communications

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Ionosphere Discovery History

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Guglielmo Marconi1874- 1937

1909 Physics Nobel Price

At 12 December 1901, using a 122 m kit-supported antenna for reception, Marconi received at Signal Hill in St John’s, Newfoundland (now part of Canada( signals transmitted by transmitter at Poldhu, Cornwall (about 3,500 km(. The transmission consisted of the three dots of the Morse code letter S, whose signals were difficult to distinguish from the atmospheric noise.

In February, 1902, Marconi sailed on S.S. Philadelphia west from Great Britain, and recorded signals daily from the Poldhu station.

The test results produced coherer-tape reception up to 2,496 km, and audio reception up to 3,378 km.The reception was from longer distances at night then at day (up to 1,125 km(

In the Receiver, Marconi used a Coherer.The Coherer was a glas tube filled about half full with sharply cut metal filings, often part silver and part nickel. Silver electrodes make contact with the metal particles on both sides. When a signal is receved, at the antenna, the filings tend to cling to each other reducing the resistence of the Coherer.

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In 1902, short after Marconi’s remarkable transatlantic radio communication, Kennelly and Heaviside postulated independently an atmospheric conducting layer (E layer or Heaviside layer( which would work as a reflector of radio waves and allow distance transmissions.

Oliver Heaviside(1850 – 1925(

William Henry Eccles(1875 – 1966(

Arthur Edwin Kennelly(1861 – 1939(

Also William Henry Eccles, an assistant of Gugliemo Marconi, supported the Kennelly-Heraviside theory and suggested in 1912 that solar radiation was responsible for the observed differences in radio wave propagation during the day and night.

A two-way transatlantic communication between USA and France was accomplished for the first time on November 17, 1923.

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Two research groups, one in America (Breit & Tuve(and one in England (Appleton & Barnett( , were practically simultaneously able to prove decisively the existence of the Kennelly – Heaviside layer and to measure the altitude.

Gregory Breit(1899 – 1981(

Merle Antony Tuve(1901 – 1982(

In America, (1925, 1926( G. Breit and M.A. Tuve devised a technique for determining the height of the reflecting region. Radio waves travel with the speed of light, thus the height of the reflecting region can be calculated if one measure the time taken by the transmitted radio wave back to earth.

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In England (1926( Appleton and Barnett applied two different methods based on continuous transmission. In the first method, the elevation angle of the signal arriving at the receiver was measured. When the distance between transmitter and the receiver was known, the altitude reflecting layer could be calculated. With the second method the receiver was close to the transmitter and changes in the interference pattern of the ground wave and a nearly vertically reflected wave were observed when the transmitting frequency was slowly varied.

Miles Aylmer FultonBarnett(1901 - (

Edward Appleton(1892 - (

1947 Nobel Prize PhysicsIn the course of his investigation, Appleton discovered another reflecting layer at the height of roughly 200 to 400 km. Later Appleton called this layer F-layer and Kenneelly-Heaviside as E-layer. Subsequently a new layer was discovered at altitude between 50 to 90 km, and was called D-layer.The D-layers disappears during the dark hours. The F-layer splits into two different regions, namely F1 and F2. The F1 region, which exists only in daytime, has a peak density around 200km. In the F2 the peak density occurs at about 300 km.

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References:

SOLORadar Wave Propagation over the Earth

Blake, L., V., “Radar Range-Performance Analysis”, LexingtonBooks, 1980, Chapter 5, “Effects of the Earth’s Atmosphere”

Rohan, P., “Surveillance Radar Performance Predictions”, Peter Peregrinus Ltd, 1983, Chapter 5, “Radar Wave Propagation”

Berkowitz, R., S., Ed., “Modern Radar – Analysis, Evaluation and System Design”, John Wiley & Sons, 1965, Part V, Chapter 1, Millman, G.,H., “Atmospheric Effects on Radio Wave Propagation”

Schleher, D.,C., “Introduction Electronic Warfare”, Artech House, 1986, Chapter 6, “Radar and ECM Performance Analysis”

Skolnik, M., I., “Introduction to Radar System”, McGraw-Hill, 3th Ed., 1980,Ch. 12, “Propagation of Radio Waves”

Mahafza, B.,R.,“Radar Systems Analysis and Design Using MATLAB”, Chapman &Hall/CRC, 2000, Ch. 8, “Radio Waves Propagation”

Kerr, D.,E., Ed., “Propagation of Short Radio Waves”, Peter Peregrinus, 1987

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TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

The Rate of Vector interference is proportional to 2 π Δ R/ λ which is a function of wavelength (frequency(

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REFLECTION & REFRACTION SOLO

Introduction

4 radio wave propagation over the earth

Science

electromagnetic em waves

bending of em waves

planar wave

electromagnetic field

plane waves of equal

soloradio wave propagation

radio wave propagationover

waythe physics of radio