Ionosphere(v1.6.2)

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

EC3630 Radiowave Propagation

IONOSPHERIC WAVE PROPAGATION

by Professor David Jenn

EARTHS SURFACE

IONOSPHERE

(version 1.6.2)

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Ionospheric Radiowave Propagation (1)

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

1. Variations in the electron density ( eN ) cause waves to bend back towards Earth, but

only if specific frequency and angle criteria are satisfied. Some examples are shownbelow. Multiple skips are common thereby making global communication possible.

EARTHS SURFACE

1

2

3

4

TX

IONOSPHERE

SKIP DISTANCE

maxeN

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Ionospheric Radiowave Propagation (2)2. The Earths magnetic field causes the ionosphere to behave like an anisotropic medium.

Wave propagation is characterized by two polarizations (denoted as ordinary andextraordinary waves). The propagation constants of the two waves are different. Anarbitrarily polarized wave can be decomposed into these two polarizations uponentering the ionosphere and recombined on exiting. The recombined wave polarizationangle will be different that the incident wave polarization angle. This effect is calledFaraday 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 splitsinto the 1F and 2F layers.

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Electron Density of the Ionosphere

(From Rohan. Note unit is per cubic centimeter.)

Day

Night

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The Earths Magnetosphere

(From George Parks, Physics of Space Plasmas, Addison-Wesley)

Interaction of the sun andEarth magnetic fields

Interaction of the sun andEarth magnetic fields

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Barometric Law (1)Ionized layers are generated by solar radiation, mainly ultraviolet, which interacts withneutral molecules to produce electron-ion pairs. As a first step in predicting the ionizedlayers it is necessary to know the distribution of molecules with height.

A

Altitude

Column of

Atmosphere

dhh +

h p

dpp +

Consider a differential volume of air, as shown in the figure.The pressurep decreases with altitude, so dp is negative.There is a net buoyant force on the volume of cross sectionalareaA

AdpdppAApFb =+= )( .

This force must be balanced by the gravitational force of theweight of the gas in the volume

gAdhgdVFg ==

where is the mass density of the gas (mass per unit volume)and g is the gravitational constant. Ifm is the mean molecular

massgAdhmNgdVFg ==

Equating the last two equations gives gdhmNdp = . N is thenumber of molecules per unit volume.

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Barometric Law (2)Pressure and temperature are related by the ideal gas law: TBp Nk= (kB is Boltzmansconstant and T is temperature). Use the ideal gas law to substitute forNin dp:

TB

dp mgdhp k =

then integrate from starting height oh to a final height h

( )ln

( ) To

h

ho B

p h mgdh

p h k=

Finally, assuming that all of the quantities are approximately constant over the heights of

concern, and defining a new constant1

,TB

mg

H k= gives the barometric law for pressure vs.

height:

( ) ( )exp ( )exp ( )expT

o o

h ho

o o oh hB

h hmg dhp h p h dh p h p hk H H

= = =

His called the local scale height and thus the exponent, when normalized byH, is in Hunits. The reference height is arbitrary. The pressure, mass density, and number density

are seen to vary exponentially with height difference measured inHunits.

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Chapmans Theory (1)Chapman was the first to quantify the formation of ionized regions due to flux from thesun. In general we expect ion production to vary as follows:

1.Small at high altitudes solar flux is high, but the number of molecules availablefor ionization is small

2.Small at low altitudes the number of molecules available is large, but solar fluxis low due to attenuation at higher altitudes

3.High at intermediate altitudes sufficient number of molecules and fluxChapmans derivation proceeds along the following steps:

1.Find the amount of flux penetrating to an altitude h.2.By differentiation, find the decrease in flux with incremental height dh.3.The decrease in flux represents absorbed energy. From the result of step 2 an

expression for ion production per unit volume at height h is derived.4.The barometric law is used to find the optical depth at height h (optical depth is a

measure of the opacity of the atmosphere above h).5.The maximum height of electron-ion production is set as the reference height.6.Assuming equilibrium (no vertical wind) laws of the variation of the electron

density are derived based on the rates of electron production and loss.7.A scaling law is developed so that ionization can be represented by a single

variable rather than two (altitude and solar zenith angle).

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Chapmans Theory (2)Only the final results are presented here. Sdenotes the solar flux and is the zenith angle.

The electron density at height h and zenith angle is given by

( ) = zoe ezNhN sec1

21exp),(

where Hhhz o /)( = and oN is the electrondensity at the reference height oh . This equation

specifies a recombination type of layer (as

opposed to an attachment type layer1). It can berewritten (by a substitution of variables andrescaling) as

)secln,0(cos),( = zNzN ee

The value for any can be obtained from theo0= curve.

A

Column ofAtmosphere

dhh +

h

dSS+

S

1There are several reactions which can remove electrons from the ionosphere. The two most important classes of recombination are (1)the electron attaches itself to a positive ion to form a neutral molecule, and (2) it attaches itself to a neutral molecule to form a negativeion.

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Chapmans Theory (3)

It is found that the E and F1 regions behave closely to what is predicted by Chapmansequation.

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Dielectric Constant of a Plasma (1)

When an electric field is applied, charges in the ionosphere are accelerated. The mass ofthe ions (positive charges) is much greater than the electrons, and therefore the motion ofthe ions can be neglected in comparison with that of the electrons. The polarization vector

isdeNpNP eerrr

==

where =eN electron density3

m/ ,19

1059.1 =e C, electron charge, and dr

is the

average displacement vector between the positive and negative charges. The relative

dielectric constant is

E

P

E

PE

E

D

oo

o

or r

r

r

rr

r

r

+=

+== 1

The propagation constant is

njkc

jjk orc ===

where rn = is the index of refraction. Both the dielectric constant and index of

refraction can be complex ( rrr j = and njnn = ).

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Assume ax polarized electric field. The electrons must follow the electric field, and

therefore, xxd =r

. The equation of motion of an electron istj

xeeExmxm =+

&&& where 31100.9 =m kg, electron mass, and = collision frequency. The solution of the

equation of motion is of the form tjaex = . Substituting this back into the equation givestj

xtjtj

eeEaejmeam =+ )()( 2

with( ) /12 jm

eEa x

=. The polarization vector is

( ) /1

2

2

jm

xeEeNP

tjxe

=

r

Now define a plasma frequency,o

ep

m

eN

2= and constants

2

=

pX and

=Z so

that

( )jZ

XxeEP tjxo

=

1

r

.

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The ratio of Pr

to Er

gives the complex dielectric constant, which is equal to the square ofthe index of refraction

( ) ( )

jjZXnj prrr

=

===

2

2 11

1

Separating into real and imaginary terms gives an equivalent conductivity

orr j

= where

( )222

1

+

=m

eN

o

er and ( )22

2

+=

m

eNe

For the special case of no collisions, 0= , and the corresponding propagation constant

is real

2

2

1

pooroc kk ==

with oook = .

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Consider three cases:

1. p > : ck is real andzkjzjk cc ee

= is a propagating wave

2. p< : ck is imaginary and zkzjk cc ee = is an evanescent wave3. p = : 0=ck and this value of is called the critical frequency

1, c

At the critical frequency the wave is reflected. Note that c depends on altitude because

the electron density is a function of altitude. For electrons, the highest frequency at

EARTHS

SURFACE

TX

IONOSPHERE h

REFLECTION

POINT0== rc

which a reflection occurs is

max92

ec

c Nf =

Reflection at normal incidence requires

the greatest eN .

1The critical frequency is where the propagation constant iszero. Neglecting the Earths magnetic field, this occurs atthe plasma frequency, and hence the two terms are oftenused interchangeably.

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If the waves frequency and angle of incidence on the ionosphere are chosen correctly,the wave will curve back to the surface, allowing for very long distance communication.For oblique incidence, at a point of the ionosphere where the critical frequency is cf ,

the ionosphere can reflect waves of higher frequencies than the critical one. When thewave is incident from a non-normal direction, the reflection appears to occur at a virtualreflection point, h, that depends on the frequency and angle of incidence.

EARTHSSURFACE

TX

IONOSPHERE

h

VIRTUAL

HEIGHT

SKIP DISTANCE

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Ionospheric Radiowave Propagation (2)To predict the bending of the ray we use a layered approximation to the ionosphere just aswe did for the troposphere.

1z2

z3z

1

i

2

3

A

LTITUDE

LAYERED

IONOSPHERE

APPROXIMATION

)( 1zr

)( 2zr

)( 3zr

M

1=r

Snells law applies at each layer boundary( ) L== )(sinsin 11 zri

The ray is turned back when 2/)( =z , or )(sin zri =

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Note that:

1. For constant i , a higher eN is required for higher frequencies if the ray is to return

to Earth (because r decreases with ).2. Similarly, for a given maximum eN ( maxeN ), the maximum value of i that results

in the ray returning to Earth increases with increasing .

There is an upper limit on frequency that will result in the wave being returned back toEarth. Given

maxe

N the required relationship betweeni

andfcan be obtained

i

eie

ei

pi

ri

Nf

fN

f

N

z

2max

max

22

max

2max2

2

22

cos

81

81

cos

811cos1

1sin

)(sin

==

=

=

=

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Examples:

1. o45=i ,310

max m/102 =eN : 8.1)707.0/()102)(81(210

max ==f MHz

2. o60=i , 310max m/102 =eN : 5.2)5.0/()102)(81( 210max ==f MHz

The value offthat makes 0=r for a given value of maxeN is the critical frequencydefined earlier:

max9 ec Nf =

Use the maxeN expression from the previous page and solve forf

icie fNf secsec9 max ==

This is called the secant law or Martyns law. When isec has its maximum value

(maximum angle of incidence on the ionosphere), the frequency is called the maximum

usable frequency (MUF). A typical value is less than 40 MHz. It can drop as low as 25MHz during periods of low solar activity. The optimum usable frequency1 (OUF) is 50%to 80% of the MUF. Data for the MUF is available. This is only correct for horizontallystratified media.

1Sometimes called the optimum working frequency (OWF).

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Maximum Usable Frequency

Shown below is the MUF in wintertime for different skip distances. The MUF is lower inthe summertime.

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

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Multiple hops allow for very long range communication links (transcontinental). Using asimple flat Earth model, the virtual height (h), incidence angle ( i ), and skip distance

(d) are related by h

di = 2tan . This implies that the wave is launched well above the

horizon. However, if a spherical Earth model is used and the wave is launched on the

horizon then hRd e = 221.

EFFECTIVE SPECULAR

REFLECTION POINT

IONOSPHERE

i

h

d Single ionospheric hop

(flat Earth)

EARTHS

SURFACE

TX

IONOSPHERE

Multiple ionospheric hops

(curved Earth)

1Whether to use Re or Re depends on the amount of path through the troposphere where refraction is significant. In the remaining

notes Re is used.

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Approximate virtual heights for layers of the ionosphere

Layer Range for h (km)

2F 250 to 400 (day)

1F 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 o74 . The MUF is

cc ff 6.3)74sec(MUF ==o

For

312

max m/10=eN , 9cf MHz and the 4.32MUF = MHz. For reflection from the 2Flayer, 300h km. The maximum skip distance will be about

4516)10300)(108500(222233

max == hRd e km

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For a curved Earth, using the law of sines for a trianglei

eRh

tan

1

sin

cos/1=

+

R/2

h

d/ 2

eR

i

LAUNCH ANGLE:

90o

i = 90o

=

R/2

where

eR

d

=

2

and the launch angle (antennapointing angle above the horizon)

is

i ==oo 9090

The great circle path via thereflection point isR, which can be

obtained from

i

eRR

sin

sin2 =

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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

( ) 3976.02/2/ === ee RdRd rad = 22.8 degrees720/coscos)( ===+ eeee RRhRhR 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 16902/ =d km gives( ) 1988.02/ == eRd rad = 11.39 degrees

711/cos == ee RRh 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 16902/ =d km and = 11.39 degrees, butpoint the antenna above the horizon to the virtual reflection point at 300 km

ooo 4.74)39.11cos(8500

3001)39.11sin(tan

1

=

+=

ii

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

launch angle, = ooooo 21.44.7439.119090 == i

The electron density at this height (see chart, p.3) is 311max m/105eN whichcorresponds to the critical frequency

= max9 ec Nf 6.36 MHzand a MUF of

7.234.74sec36.6MUF = o 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 benecessary: time of day, time of year, time within the solar cycle, etc. These data areavailable from published charts.

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Generally, to predict the received signal a modified Friis equation is used:

( )

LL

R

GGPP x

rttr

2/4

=

where the losses, in dB, are negative:

ionoreflpol GLLLx +=

=reflL reflection loss if there are multiple hops

=polL polarization loss due to Faraday rotation and earth reflections=ionoG gain due to focussing by the curvature of the ionosphere=L absorption loss

R = great circle path via the virtual reflection point

Example: For 30=tP dBW,f= 10 MHz, 10== rt GG dB, d= 2000 km, 300=h km,5.9=xL dB and 30=L dB (data obtained from charts).

From geometry compute: o3.70=i ,R = 2117.8 km, and thus 5.108=rP dBw

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Charges in Magnetic Fields (1)

The motion of charged particles in a magnetic field can be derived from the Lorentzequation. For a charge q of mass m moving at velocity u

m

duF qu B m

dt= =

rr rr

Decomposing ur

into longitudinal and transverse components with respect to Br

gives

( ) ( )L T L Td

q u u B m u udt

+ = +rr r r r

Let the magnetic field be uniform and in thez direction, so thatoB B z=

r, L zu u z=

r, and T x yu u x u y= +

r

and therefore the vector equation can be written as two scalar equations

/x

y o

duqu B m

dt= , /

y

x o

duqu B m

dt=

The classical gyrofrequency is / /H o o oqB m q H m = = . To solve the two equations definea new variable x yu ju = +

0Hd

jdt

+ =

LurT

ur u

r

BrLur

Tur u

r

Br

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The solution is

( ) ( )cos sinHT T H H j t

u e u t j t

= = + r

Therefore,( )

( )

cos

sin

x T H

y T H

z L

dxu u t

dt

dyu u t

dt

dzu udt

= =

= =

= =

Integrating with respect to time

( )

( )

sin

cos

To H

H

To H

H

o L

ux x t

uy y t

z z u t

= +

= +

=

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The Larmor radius is defined as /T Hcr u = . The previous equations describe a circle ofradius cr with the constants ox , oy , oz and that depend on the initial conditions. Note that

the particle motion has the following characteristics:

1.The particle rotates around the magnetic field line in a circle of radius cr at frequency

H

2.The center of the circle (the particles guiding center) travels along the field line withvelocity Lu

rso that the particle traces out a helix, as shown in the figure

oB B z=r

Lur

Tur

ur

Guiding

Centercrr

rr

O

x

y

BroB B z=

r

Lur

Tur

ur

Guiding

Centercrr

rr

O

x

y

oB B z=r

Lur

Tur

ur

Guiding

Centercrr

rr

O

x

y

Br

Br

N l P d S h l D f El i l & C E i i M C lif i

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The Earths magnetic field is inhomogeneous. The magnetic field is strongest at the poleswhere the flux lines converge. The force on the charge varies as the gradient of the field

m

du

F M B m dt= =

rr r r

2

2

cq r BMm

=r

is the magnitude of the magnetic dipole moment. If momentum (energy) is

conserved along the path,1 2 2

2( ) constantL Hm u u+ = . Conservation of momentum implies:

Diverging flux lines

1.As Br increases H increases and Tu increases

2.As Br

increases cr decreases and Tu increases

The reverse is true when Br

decreases. Thus, Tu

increases as the flux lines converge, which requires thatLu decrease so that momentum is conserved. At some

point all energy is in the transverse component, andlongitudinal motion stops. This is a turning point wherethe particle reverses direction.

N l P t d t S h l D t t f El t i l & C t E i i M t C lif i

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Earth Magnetic Field (1)

The Earths magnetic field is comprised of three main components:1.The main field due to the outer core. The molten core consists of rotating liquid metal

that give rise to a dynamo effect.

2.Fields that arise in the Earths crust and upper mantle.3.Atmospheric magnetic field

due to currents flowing in theatmosphere and in the oceans.

The first component dominatesand is approximately a dipolefield with the magnetic dipolepointing south.

Typical values:

At poles, 40.6 to 0.7 10 TB = r

Near equator: 40.25 10 TB = r

2 4(1 Wb/m = 1 T =10 G)

From: www.ngdc.noaa.gov/seg/WMM/back.shtml

Naval Postgraduate School Department of Electrical & Computer Engineering Monterey California

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Earth Magnetic Field (2)

The Earths main magnetic field reverses on the time scale of every several hundredthousand years. It is believed that the main field is currently in the process of reversal. Theevidence includes the weakening of the magnetic field, relatively rapid migration of the

poles (below left) and an increase in anomalies. Below is a simulation result showing theEarths magnetic field during the transition period.

Simulation of the Earths magnetic field, G.A. Glatzmaier and P.H. Roberts, "A three-dimensional convective dynamo solution with rotating and finitely conducting innercore and mantle," Phys. Earth Planet. Inter., 91, 63-75 (1995).

(http://science.nasa.gov/headlines/y2003/29dec_magneticfield.htm?list770900)

Naval Postgraduate School Department of Electrical & Computer Engineering Monterey California

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Magneto-ionic Medium (1)

Thus far the Earths magnetic field has been ignored. An electron with velocity ur

moving

in a medium with a magnetic field ooo HBrr

= experiences a force om BueFrrr

= .

Assume a plane wave that is propagating in thez direction ( zeHE ~,rr

). From

Maxwells equations

[ ]( )

( )( )

=+=

+=

+=

+==

)3(0

)2(/

)1(/

zzzo

yyoyx

xxoxy

o

DjPEj

EPEjH

EPEjH

PEjDjH

rrrr

=

=

=

=

)6(0

)5(

)4(

zo

yox

xoy

o

Hj

HjE

HjE

HjE

rr

These equations show that the plane wave will be transverse only with respect to the B

r

, D

r

and Hr vectors, which is different from the isotropic case. The characteristic impedance ofthe medium is

/// oxyyx jHEHE === .


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Solve (5) for yH and use the result in (1):

43421

2

122

n

E

P

xo

xoo

+=

. A similar process

starting with equation (4) would lead toyo

y

E

Pn

+= 12 . Equating this to the expression for

2n above gives the polarization ratio of the wave,

y

x

y

x

E

E

P

PR =pol .

Rewrite the equation of motion for an average electron when both electric and magneticfields are present

( )oBrEermrmr

&rr

&r&&r +=+

There is also a term in the parenthesis for the magnetic field of the wave, but it turns out tobe negligible. We now define the longitudinal and transverse components, which areparallel and perpendicular to the direction of propagation, respectively.


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g p f p g g y, f


Definition of the coordinate system:y lies in the plane defined by )( ooo HB = andz

)transverse(sin

)allongitudin(cos

oyT

ozL

HHH

HHH

==

==

Note the direction ofx is chosen to forma right-handed system (RHS). Anycombination of propagation direction

and magnetic field can be handled withthis convention, as long as they are notparallel. If they are parallel anycombination ofx andy that forms aRHS is acceptable.

x

y

Direction of

propagation

oBr

Lies in the plane

defined byz andBoEarths magneticfield vector

LBr

TB

r

Determinedfromz andy

z

In this coordinate system the equation of motion can be written as three scalar equations,which for thex component is

( )ToLox HzHyEexmxm &&&&& +=+


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g p f p g g y f


For time-harmonic displacements, phasors can be used and the derivatives in time give amultiplicative factor j

( )ToLox HzjHyjEemxjmx +=+2

The polarization is caused by the electron displacements inx,y, andz

e

z

e

y

e

x

eN

Pz

eN

Py

eN

Px

=

=

= ,,

Substituting:e

Tzo

e

Lyox

e

x

e

x

eNHPj

eNHPjeE

eNmPj

eNmP +=

2

Multiply by mo/ and define the longitudinal and transverse gyro-frequencies1:

cos/ BLoL meH =

sin/ BToT meH =

1In general, the gyro-frequency (also called the cyclotron frequency) of an electron in a magnetic field is defined as meB

oB/ or

mHeooH

/ .


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For convenience we define the frequency ratios2

=

pX ,

LLY = ,

TTY = , and

=Z .

Recall that the plasma frequency iso

ep

m

eN

2

= . With this notation

zTyLxxo PjYPjYjZPXE += )1(

Going back to they andz components of the equation of motion, and treating them thesame, gives two more equations

xTzzo

xLyyo

PjYjZPXE

PjYjZPXE

=

+=

)1(

)1(

These are the equations needed to find n in terms of the components of Er , and Pr .Arbitrarily we choose to solve for yP and yE , eliminating zzxx EPEP ,,, .


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With several substitutions and a fair amount of algebra, two equations are obtained:

2polpolpol

pol2

polpol

)1(

)1/()1(

RPjYjZRPEXR

jZXPRYPjYjZRPEXR

yLyyo

yTyLyyo

+=

+=

Equate the right hand sides: 2polpol2 )1/( RjYjZXRYjY LTL =+

Solving this quadratic equation gives:

+

=

2/12

2

42

pol)1(4)1(2

LTT

L

YjZX

Y

jZX

Y

Y

jR m

Not all plane waves can pass through the medium. To remain plane waves they must haveone of two polarizations that are characteristic of the medium. If this equation is satisfied,both equations at the top of the page give the same solution:

pol1/

RjYjZ

XEP

L

oyy

=


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Use this ratio to find the index of refraction: [ ]pol2 1/1 RjYjZXn L= and with polR from the quadratic equation yields the Appleton-Hartree formula

2/1

2

2

42

22

)1(4)1(21

1)(

+

===

LTT

r

YjZX

Y

jZX

YjZ

Xnjnn

Physical interpretation: The Earths magnetic field causes the ionosphere to be anisotropic.

There are two modes of propagation, each with a particular polarization, that dependsentirely on the properties of the medium. The phase velocities of the two modes, from thetwo values ofn above, are different, and when they recombine they have different phaserelationships. The solution with the positive sign is called the ordinary wave; that with thenegative sign is the extraordinary wave. When the frequency is greater than about 1 MHz,

which is always the case for systems that transmit through the ionosphere, the wave can beconsidered solely longitudinal ( TH , T and TY are zero) and therefore,

)(1

22

jn

L

p

is sufficiently accurate for most calculations.


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Anisotropic Media

In an isotropic medium the phase velocity at any given point is independent of thedirection of propagation. For an isotropic medium, the direction of phase propagation (i.e.,movement of the equiphase planes) is radial with velocity pu . The energy propagation,

which is generally taken as the group velocity,1 gu is also radial. In an anisotropic medium

the directions of the equiphase planes and energy are different.

Isotropic Medium Anisotropic Medium

1This is the velocity of propagation of a packet of frequencies centered about a carrier frequency. For simple amplitude modulated

(AM) waveforms, it is the velocity of propagation of the envelope, whereas the phase velocity is the velocity of propagation of thecarrier.

Noncircular

Wavefront

Pr

Point

Source

Energy

Direction

(Radial)

Wavefront

Normal

oBr

CircularWavefront

P

r

Point

Source

Noncircular

Wavefront

Pr

Point

Source

Energy

Direction

(Radial)

Wavefront

Normal

oBr

oBr

CircularWavefront

P

r

Point

Source

CircularWavefront

P

r

Point

Source


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Phase and Group Velocities

The phase velocity is/ /pu c n = =

and the phase index of refraction (our normal index of refraction) is related to it by

pn u c = . A group index of refraction, gn , can be defined analogous to the phase index of

refraction:

g gn u c=

where /gu d d = . It is always true that2

cuu gp = . The group index can be expressed

as:

( )

2/

2

2

g g

p

g

d d cn c u c

d d

n ud c

d c fd

nd

dnn n f

df

= = =

=

=

= +


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Total Electron Content

If we neglect the magnetic field and consider the case where 0Z= and 1X

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Phase Delay and Time Delay (1)

Let the distance between point P1 and P2 be l . As a check, for a homogeneous medium,

r on nc c

= = = = l l l l

A typical value for a vertical path at 100 MHz is 217 melectrons/103 , which gives

( )( )

( )( )( )( )

217 19

31 12 8 6

3 10 1.6 1012526 rad

4 9 10 8.85 10 3 10 100 10

The time delay between a transmitter at P1 and receiver at P2 is:

=2

1)(

11 P

Pd ds

snct

With the same assumptions onZandX, the (group) time delay is

2

1

2

1 1 40.31

2

P

P

Td

Nt X ds

c c cf

+ = +

l


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Phase Delay and Time Delay (2)

The phase rotation of the ordinary wave is found to be2

o2 4 2

2 4

To

oT T L

N ek

m cY Y Y

= +

+

l

A handy identity is2

51.06 10o

e

m c

= . For the extraordinary wave

++

+=

cm

e

YYY

Nk

oLTT

To

2

242

x

42

l

Faraday rotation is the rotation of the phase angle of a linearly polarized plane wave:

( )4 22 2

4 2o x 2 2

41 1 14

2 4 41

T LT T T L

o oT L

Y Ye eN N Y Y

m c m cY Y

+ = = +

where the approximation assumes that the wave frequency is much larger than the gyrofrequency. With a magnetic field present the equations are more complex.


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Faraday Rotation Example

Find the Faraday rotation of a 135 MHz plane wave propagating a distance of 200 kmthrough the ionosphere. The electron density is constant over the path and equal to

10105 /m3. The gyro-frequency is 6108 Hz. Assume collisions can be neglected and

that the propagation is longitudinal.

The TEC is ( ) 161030

1010510200)( == == eeT NdssNN ll

/m2. For longitudinal

propagation and no collisions 0==ZYT and /BLY =

( )( )( )( )( )

radians75.3

1013522

1082101006.1

24

1

26

6165

2

BT

o

Ncm

e

x xy y

inEr

outEr

z

200 km

3.75 rad


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GPS Error Due to Ionospheric Time Delay

An increase in phase velocity is accompanied by a decrease in group velocity:

( )

( )

1

2

12

11

21

p

g

u c X

u c X

u c X

=

== +

Differential relationships

1 1 11 1

2 2ds c X dt dt X ds

c

= =

m

The path length error due to ionospheric time delay error is

{

2 2 2

1 1 1

1 1 1

22 2 2

1

40.31( )

t t P

Tt t P

u dt cX dt X X ds N f

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GPS Example

GPS uses two frequencies (L1 and L2 in the table on the previous page). If one canestimate the TEC then it is possible to correct the position errors due to the ionospherictime delay. Given that the one-way time delay for a path from the ground to the satellite

differs by 15 ns for the two frequencies, estimate the total electron contect of the path.

The change in time delay due to the ionosphere is given by the term

2

40.3 Td

Nt

cf

For the two frequencies:

( )( )

2 1

9

2 22 1

19 917 3

2 2 7 192 1

40.3 1 115 10

15 10 1 1 15 104.2 10 / m

40.3 1.34 10 2.6 10

Td d

T

Nt t

c f f

cN

f f

= = =

Applying this correction in practice is difficult because of other ambiguities that exist.


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Absorption

Once the index of refraction is determined for the medium, the attenuation constant can beobtained from

n

c

jj

=+=

For simplicity, the effect of the magnetic field is ignored in this discussion ( 0== LT YY )so that

+

+==

22

22

111)(

Z

XZj

Z

Xnjnn

yielding2

12

1

Z

XZ

ncn

c +== . In general, is not constant along a path through the

ionosphere, because the electron density, and henceXand n are changing along the path.The collision frequency, if not zero, can also change with location in the ionosphere.Thus, the loss due to absorption from 1P to 2P should be computed from

=2

1

)(expattn

P

P

dssL

where s is the distance along the path.


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Example

Find the attenuation (in dB/km) in the ionosphere forf = 30 MHz, 710= /s and810=eN /m

3.

Using the formula from the previous page with 0531.010302

106

7

=

=

=

Z

and 62

6

82

109.810302

1034.56 =

=

=

pX ,

( )11

22

2

+

=

m

eN

o

er . Substituting

gives:

=+

= 72

1048.112

1

Z

XZ

ec r

0013.0)log(20

1000 = e dB/km.

Alternatively, we could use the basic equation for attenuation

+= 11

2

2

ro

roo

where( )22

2

+

=m

eNe .

For 1110=eN the attenuation increases to 1.3 dB/km.


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Ionospheric Soundings and Ionograms

An ionosonde is an instrument used to measure the variation of critical frequency withheight. The measurements are used to determine ( )eN h . A train of short RF pulses istransmitted vertically and the time delay to reception is measured. The range to the

reflection point can be computed. The frequency of the pulse train vf is swept to obtain aplot ofh versusf, which is called an ionogram. The three segments of the curves betweenspikes represent reflections from the E, F1, and F2 layers.

From: www.ukssdc.ac.uk/ionosondes/ionogram_interpretation.html


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Transmission Curves (1)

Two paths through the ionosphere are shown below, one for vertical incidence at

frequencyfv and the other for oblique incidence at frequencyfob. We know that

c

f f< , a reflection always occurs;

cf f> , reflection only if cos i cf f < .

From the Secant law, / secob v if f = . For a flat earth, tan /(2 )i d h = , and since

[ ]22 2sec 1 tan 1 /(2 )i i d h = + = + , we can solve for h:

[ ]

[ ]

( )

2

2

2

sec 1 /(2 )

/ 1 /(2 )

2 / 1

i

ob v

ob v

d h

f f d h

dh

f f

= +

= +

=

i

vf f=

obf f=

h

/ 2d / 2d

i

vf f=

obf f=

h

/ 2d / 2d


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Transmission Curves (2)

Plot a family of curves ofh versus vf with obf treated as a parameter. Then overlaythem on an ionogram (heavy line in the figure). Together they determine what paths asignal radiated at frequency obf may take over distance d(neglecting loss and magnetic

effects). They are called transmission curves.Intersections of transmission curveswith the ionogram represent validreflection heights for the given

frequency.

In this example, at frequenciesabove about 13 MHz there are nopoints common to the transmissioncurves and ionogram. This is the

MUF.

(Curves shown are for d= 800 km,after K. Davies)

8 MHz 10 12 14

E

F1

F2

13

8 MHz 10 12 14

E

F1

F2

8 MHz 10 12 14

E

F1

F2

13


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Time Delay Contours

Nightside transionospheric time delay at 30MHz for receiver on a LEO platform at 1000km.

Upper left: Receiver field-of-view showingtime delay for ground-based transmitterswhose signals range in elevation angle from90 (zenith propagation) to 5 (definesboundary of circular area).

Upper right: GAIM vertical TEC, orbit path,and receiver location on nightside.Lower left: Bisecting cut through image inupper left from E to W.Lower right: Corresponding bisecting cut fromS to N.

(http://www.cpi.com/splash/signalprop.html)


I h i R B di

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Ionospheric Ray Bending

The paths (red curves) taken by signalsthrough two ionospheres under daytimeconditions with different altitude

distributions of the ionization. Theionospheres are displayed in color and thefrequency and elevation angle consideredfor the illustration are 30 MHz and 5,respectively. The blue curves show thecorresponding straight-line paths that

appear with curvature due to the use of aflat-Earth projection for the displays. Thetotal electron content along the straight-line paths (TECSL) is the same for bothcases (100 TEC units). The displaysillustrate different degrees of ray bending

due to different distributions of theionization along the ray paths butconstrained to the same TECSL values.

(http://www.cpi.com/splash/signalprop.html)


S i ill i (1)

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Scintillation (1)

Scintillation refers to rapid fluctuations in amplitude, phase, or angle of arrival of waves. Itis produced by small-scale refractive index variations in the troposphere and ionosphere. Anoptical effect of scintillation is the twinkling of stars.

Tropospheric scintillation is produced in the first few kilomenters of the surface and dependson season, local climate, frequency, and elevation angle. Typical peak-to-peak losses are:

Loss (dB) Conditions

1 Clear, summer0.25 to 1 Winter

2 to 6 Some clouds

The fading rate ranges from 0.5 Hz to 10 Hz. Scintillation increases with frequency and is

worst at low elevations (

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Scintillation (2)

A simple model to explain scintillation is the dielectric layers or ellipsoids model.

Plane input

wavefront

Distorted output

wavefront

Time varying inhomogeneous

dielectric region

Plane input

wavefront

Distorted output

wavefront

Time varying inhomogeneous

dielectric region

Ellipsoids with slightly different relativedielectric constants are changing shape as afunction of time. The output wavefront isdistorted due to atomospheric multipath.

Wavefront distortion gives rise to phase

errors and angle of arrival errors.

Multipath causes signal fading.


S i till ti (3)

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Scintillation (3)

Ionospheric scintillation is due to localized changes in the electron density eN , and hence

r . The effect decreases as21/f and usually can be neglected above 4 GHz, except when

severe magnetic storms are occurring. Characteristics are:

3.At 4 GHz causes 0.5 to 10 dB fading 0.1 % of the time

4.Worst at the poles and in an equatorial belt 25 o from the equator5.Worst 1 2 hours after sunset6.Independent of elevation angle

Ionospheric irregularities can take on a wide range of sizes but are typically around 1 km inextent. They tend to align with the Earths magnetic field lines and form striations orplumes.

Br

Plumes

Electron density

irregularities

Earth

Br

Field aligned striations

Br

Plumes

Electron density

irregularities

Earth

Br

Plumes

Electron density

irregularitiesBr

Plumes

Electron density

irregularities

Earth

Br


Br



S i till ti I d

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Scintillation Index

A measure of scintillation isthe scintillation index

22

4 2P PS

P

=

P = signal power

P = average power

(usually over 1 minute)

Significance:

4 0.6S < : weak

4 0.6S > : strong

4 1.0S = : saturation

http://gps.ece.cornell.edu/scintplot.html

Current information available atNational Weather Service Space Weather Prediction Centerhttp://www.swpc.noaa.gov/index.html


I h i Di t b

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Ionospheric Disturbances

Sudden disturbances are usually due to solar flares, with the following effects:

1.phase anomolies: change in h over short paths of a few hundred km (D layer)2. frequency deviations: change in frequency of a reflected signal (E and F regions)

3. cosmic noise absorption: sudden decrease in cosmic noise4. increase in TEC (E and F regions)

The time extent of the effects is from several minutes to several hours.

Magnetic storms are disturbances of the geomagnetic field that last from a few hours to

several days.

Factors that influence magnetic stormbehavior include:

polar heatingthermal expansionelectromagnetic driftneutral air dragglobal atmospheric circulationatmospheric composition

Consequences of magnetic storms:

power outages

increase in atmospheric drag on satellitesaffects the discharging of satellitesinduces large voltages in big conductors(e.g., Alaskan pipeline)

adverse effects on radio communications

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