lecture 3 and 4

7/24/2019 lecture 3 and 4

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Satellite Communications

Red Sea UniversityEngineering Faculty Page 1

Course Title

Satellites Communications

By: Lecturer. Elmustafa Sayed Ali AhmedElectrical and Electronics engineering Dept.

E-mail: [email protected]


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Course outlines

Lecture 1: Introduction To satellites.

Lecture 2: Keplers laws.

Lecture 3: Orbits and satellite coverage areas.

Lecture 4:Radio Link Analysis.

Lecture 5: Earth stations segment.

Lecture 6: satellite station segment.

Lecture 7: Satellites systems Modulation and accessing.

Lecture 8: Impairments of the Satellite link Communication.

Lecture 9: Interference in Satellite systems.

Lecture 10: Noise and rain effect in satellite link.

Lecture 11: Link budget of satellite system.

Lecture 12:VSAT.

Lecture 13: Advanced Topics.


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Lecture 3

Orbits and satellite coverage areas

Orbit definition and Terms

Although an infinite number of orbits are possible, only a very limited number of

these are of use for satellite communications. Some of the terms used describing anorbit are:

Apogee: The point farthest from the earth, ra.

Perigee: The point of closest approach to the earth, rp.

Line of AP sides: The line joining the Perigee and apogee through the center

of the earth.

Ascending node: The point where the orbit crosses the equatorial plane going

from south to north.

Descending node: The point where the orbit crosses the equatorial plane

going from north to south.

Line of node s: The line joining the ascending and descending nodes through

the center of earth.

Inclination: The angle between the orbital plane and the earthsequatorial

plane measured counter clock wise at the ascending node.


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Prograde orbi t: An orbit in which the satellite moves in the same direction as

the earths rotation. The inclination of a prograde orbit always lies between 0

and 90.

Retrograde orbi t: An orbit in which the satellite moves in direction counter to

the earths rotation. The inclination of a retrograde orbit always lies between

90 and 180.

Types of Orbits

There are three main orbits of satellites;

Polar orbit

The polar orbiting satellite follows an orbit that is close to the earth and passes

over or over close to the inclination is close to 90. The average height 800 to

1000 km above the earth. Used mainly for earth observation and surveillance

(Weather, Pollution, Monitoring, and the like) and for search and rescue work.

Inclined highly elliptical orbit (HEO)

This orbit used where communications is desired to regions of high latitude.

The orbital velocity is least at the apogee, and hence by placing the apogee

above the high latitude regions the satellite remains visible for longer period

from these regions. The height of this orbit from earth is above 36000 km.

The Geo-stationary Orbit (GEO)

The geostationary orbit is the orbit in which a satellite appears stationary

relative to the earth. This is the most widely used of all orbits, for the very

practical reason that an earth station antenna pointed at a geostationary

satellite automatically follows it. The geostationary orbit lies in the equatorial

plane, meaning the inclination is zero. Another requirement is that the satellite

must orbit the earth in the same direction as the earth spins and at the same


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speed, since the earth rotates at constant speed, the orbital speed must also be

constant. The height of this orbit from earth is about 36000 km.

Low earth orbits (LEO)

Most satellites, the International Space Station, the Space Shuttle, and the

Hubble Space Telescope are all in Low Earth Orbit (commonly called

"LEO"). This orbit is almost identical to our previous baseball orbiting

example, except that it is high enough to miss all the mountains and also high

enough that atmospheric drag won't bring it right back home again. The height

of this orbit from earth between 500 to 2000 km.

medium earth orbits (MEO)

Mid-earth orbit is also known as Semi-synchronous. Satellites said to be

semi-synchronous have a period of 1/2 a sidereal day. Thus, they orbit the

earth two times per day. Geo positioning and navigation satellites, such as

GPS and GLONASS, use this type orbit. The height of this orbit from earth is

ranged from 8000 km to 20000 km.


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Apogee and perigee heights

Apogee and perigee height are required to find the shape of the orbit from the

geometry of the ellipse.

ra = a (1+e)

rp = a (1-e)

Ha = ra-R

Hp = rp-R

Example

Calculate the perigee and apogee heights for the orbital parameters given by;

Eccentricity e= 0.0011501, semi major axis a=7192.3 km, R= 6371 km.

Solution

ra= a (1+e)

rb= a(1-e)

ra= 7192.3 (1+.0011501) =7200.6 km

rp= 7192.3 (1+.0011501)= 7184.1 km

Ha = ra-R

Hp = rp-R

Ha = 829.6 km

Hp= 813.1 km


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Inclined orbits Calculations

The study of general situation of satellite in inclined orbit is very complicated,

that because the location of the earth station is always given in terms of local

geographic coordinates which rotate with the earth while the position of

satellite is varying in space. Rectangular coordinates are generally used for

calculation of satellite positions and velocities in space while the earth station

quantities of interest may be the azimuth and elevation angles and range. Since

the satellites in this orbit are not geostationary, the required look angle and

range will change with time.

The Geostationary Orbit Calculations

Satellites in this orbit are appear to be stationary with respect to the earth, and

there are three conditions are required for an orbit to be a geostationary;

Satellite must travel eastward at the same rotational speed as the earth.

orbit must be circular

The inclination of the orbit must be zero.

Movement north and south can be avoided only when the inclination orbit

remains to zero. Means that the orbit lies in the earth equatorial plan. As

mentioned before Kepler third law used for radius of orbit calculations.

aGEO= 42164 km.

aE= 6378 km.

Geostationary height is hGEO= aGEO- aE

= 421646378

= 35786 km


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Antenna Look Angles

The user must be able to determine the azimuth and elevation angles of the

ground station antenna. For large commercially operated ground stations, the

look angle settings will be controlled by a computer while the owner of home

satellite receiving system will probably have to make such adjustments

manually. The antenna look angle with the geostationary orbit is simple

because the satellite in this orbit is stationary with respect to the earth and no

tracking required. The antenna used in home reception is quite broad and no

track necessary. This allow the antenna to be fixed in position.

To maximize transmission and reception, the direction of maximum gain of

the earth station antenna must point directly to the satellite. And two align the

antenna two angels as mentioned before are required;

Azimuth, angle measured from the true north Az

Elevation, angle measured from the local horizontal plane EL.

The earth constant are;

Mean earth radius R=6378Km

Radius of geostationary orbit Ageo= 42164Km

There are three information peaces required to determine the look angles for

geostationary orbit.

Earth station latitude denoted by E

Earth station longitude denoted by E

Longitude of sub satellite point denoted by SS


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Note:

-

Latitude north will be taken as positive angle, and latitude south taken as

negative angle.

-

Longitude east of the Greenwich meridian taken as positive angle, and

longitude west taken as negative angle.

the calculations of antenna look angle depends on ; ES is a position of the

earth station , SS is sub satellite point , S is a satellite ,and d is a range from

the earth station to the satellite. Angle is an angle to be determined. See

figure below.

From figure assume that SS is sub satellite point lie on the equator from

geostationary satellite. Angle a is measured from North Pole to the sub

satellite point. And since sub satellite point is on equator a= 90 degree.


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Angle B is the difference in longitude between the earth station and sub

satellite longitudes;

B= E-SS

The southern latitude are assigned negative values, the angle C is given by;

C= 90- E

Angle A can be found by the application of certain of Napiers rules. Forquadrant triangle these result in A being obtained from;

tan A= ||

The azimuth can be determined once angle A is known. Four situations must

be considered, these are shown in figures below, for these situations the

azimuth is given by;

Figure a: E< 0 and B0 Az= 360-A

Figure c: E> 0 and B 0 and B>0 Az= 180-A


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To find the range and elevation, it is first find the side angle of the quadrant

triangle. FromNapiers rules;

cos b = cos Ecos B

The mean radius of the earth is taken as:

R= 6371 Km

Application of the cosine rule to the plane triangle gives the range d as;

d= + 2

Application of the sine rule to the plane triangles gives the elevation:

cos EL =

sin


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Example:

An earth station at latitude 20 S and longitude 30 W is working into a

geostationary satellite situated at longitude 30 E. determine the look angles,

range and elevation angle.

Solution;

R= 6871 Km , Ageo= 42164 Km

S= 30 , E= -30 , E= -20

B= E- S= -30-30 = -60 degree

tan A= ||

= ()

= 5.064

A= tan -1(5.064) = 78.8 degree

Where E< 0 and B


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Limits of Visibility

There will be east and west limits on the geostationary arc visible from any

given earth station. The geographic coordinates of the earth station and the

antenna elevation will set the limits.

For any given earth station, the curvature of the earth will set th limits to

farthest satellite that can be seen east or west of the earth station longitude. At

the limit set by the earth curve the elevation will be zero. In practice the nose

picked up from the earth by the antenna at zero elevation is excessive, so that

an angle of elevation of 5 degree is generally assumed as being the usable

minimum degree. From the triangle plane below;

sin S =

sin95


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b= 180- sigma- S

Sigma= 90 + El min ; El min= 5 degree

b= 18095S

= 85-S

The longitude difference angle B can be determined from;

cos B =

Example;

The coordinates for an earth station are 43 south 30 east. Calculates the limit of

visibility.

Solution

R = 6371 Km , Ageo= 42164 Km

E= 30 , E= -43

S = sin -1(

sin95) = 8.657 degree.

b= 858.657 = 76.343 degree.

B= cos -1(

) = 71 degree.

The limit of visibility;

E- B = 30-71 = -41 degree.

E+ B = 30+71 = 101 degree.


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Lecture 4

Radio Link Analysis

An important issue arises in the design of satellite communication system is

the link budget analysis. As its name implies, a link budget, or more

specifically link power budget, is the totaling of all the gains and losses

incurred in operating a communication link. In particular, the balance sheet

constituting the link budget provides a detailed ac-counting of three broadly

defined items:

Apportionment of the resources available to the transmitter and the

receiver.

Sources responsible for the loss of signal power.

Sources of noise.

- Free Space Propagation Model

In a radio communication system, the propagation of the modulated signal is

accomplished by means of a transmitting antenna, the function of which is

twofold:

To convert the electrical modulated signal into an electromagnetic field. In

this capacity, the transmitting antenna acts as an "impedance-transforming"

transducer, matching the impedance of the antenna to that of free space.

To radiate the electromagnetic energy in desired directions.

At the receiver, we have a receiving antenna whose function is the opposite

of that of the transmitting antenna: It converts the electro-magnetic field into

an electrical signal from which the modulated signal is extracted. In addition,

the receiving antenna may be required to suppress radiation originating from

directions where it is not wanted.


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Typically, the receiver is located in the far field of the transmitting antenna,

in which case, for all practical purposes, we may view the transmitting antenna

as a fictitious volume less emitter or point source. A complete description of

the far field of the point source requires knowledge of the electromagnetic

field as a function of both time and space.

Consider an isotropic source radiating a total power denoted by Pt, measured

in watts. The radiated power passes uniformly through a sphere of surface area

4d2, where d is the distance (in meters) from the source. Hence, the power

density, denoted by (d), at any point on the surface of the sphere is given by

(d)=Pt/4d2watts/m2.

The equation states that the power density varies inversely as the square of the

distance from a point source. This statement is the familiar inverse-square law

that governs the propagation of electromagnetic waves in free space.

Multiplying the power density (d) by the square of the distance d at which

it is measured, we get a quantity called radiation intensity denoted by ,

=d2(d)

Whereas the power density (d) is measured in watts per square meter, the

radiation intensity is measured in watts per unit solid angle (watts per

steradian).

In the case of a typical transmitting or receiving radio antenna, the radiationintensity is a function of the spherical coordinates and defined in figure

below. Thus, in general,


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We may express the radiation intensity as (, ). The power radiated inside

an infinitesimal solid angle d is given by (, ) d where

d= sindd steradians

The total power radiated is therefore;

P =(, ) d watts

Which is a mathematical statement of the power theorem. The power theorem

states that if the radiation-intensity pattern (, ) is known for all values of

angle pair (, ) then the total power radiated is given by the integral of (,

) over a solid angle of 4 steradians. Then the average power radiated per

unit solid angle is

Pav= 1/4(, ) d

= P/4watts/steradian

Which represents the radiation intensity that is produced by an isotropic

source radiating the same total power P.


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-

Directive Gain, Directivity, and Power Gain

The ability of an antenna to concentrate the radiated power in a given direction

as in the case of the transmitting antenna or conversely, to effectively absorb

the incident power from that direction as in the case of the receiving antenna,

is specified in terms of its directive gain or directivity. For a direction

specified by the angle pair (, ), the directive gain of an antenna, denoted by

g (, ) is defined as the ratio of the radiation intensity in that direction to the

average radiated power, as shown by

g (, ) = (, )/Pav

= (, )/P/4

The directivity of an antenna, denoted by D, is defined as the ratio of the

maximum radiation intensity from the antenna to the radiation intensity from

an isotropic source. Thus, whereas the directive gain of the antenna is a

function of the anglepair (, ), the directivity D is a constant that has been

maximized for a particular direction.

A quantity called power gain does involve the radiation efficiency of theantenna. Specifically, the power gain of an antenna, denoted by G, is defined

as the ratio of the maximum radiation intensity from the antenna to the

radiation intensity from a loss less isotropic source, under the constraint that

the same input power is applied to both antennas. Using radiation to denote

the radiation efficiency factor of the antenna, the power gain G might be

related to the directivity D as

G = radiationD

Thus, the power gain of an antenna over a loss less isotropic source equals the

directivity if the antenna is 100 percent efficient


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(radiation = l) but it is less than the directivity if any losses are present in the

antenna (radiation < 1). We will assume that the antenna is 100 percent

efficient and therefore refer only to the power gain of the antenna. The power

gain of an antenna is the result of concentrating the power density in a

restricted region smaller than 4 steradians, illustrated in figure below.

Effective radiated power referenced to an isotropic source (Effective

Isotropic Radiated Power (EIRP)); the EIRP is defined as the product

of the transmitted power, Pt, and the power gain of the transmitting

antenna, Gt, as shown by

EIRP = PtGtwatts

Antenna beam width, representing a "planar" measure of the antenna's

solid angle of view; the beam width, in degrees or radians, the angle

that subtends the two points on the main lobe of the field-power pattern

at which the peak field power is reduced by 3 dBs. The higher the power

gain of the antenna, the narrower is the antenna beam width.

Another matter of interest discernible from the figure is the side lobes

of the field-power pattern. Unfortunately, every physical antenna has

side lobes, which are responsible for absorbing unwanted interfering

radiations.


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Effective Aperture

The term is the effective aperture of the antenna, which is defined as the ratio

of the power available at the antenna terminals to the power per unit area of

the appropriately polarized incident electromagnetic wave. The effective

aperture, denoted by A, is defined in terms of the antenna's power gain G as

A = ( 2/ 4) G

Where is the wavelength of the carrier. Where the wavelength and frequency

f are reciprocally related as

= c/f

Where c is the speed of light (approximately equal to 3 X l08 m/s).

The term effective aperture has particular significance in the context of

reflector antennas and electromagnetic horns that are characterized by a well-

defined aperture. For these antennas, the ratio of the antenna's effective

aperture to its physical aperture is a direct measure of the antenna's aperture

efficiency, radiation, in radiating power to a desired direction or absorbing

power from that direction. Nominal values for the efficiency radiation or

reflector antennas lie in the range or 45 to 75 percent.

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Friis Free-Space Equation

With this introductory material on antennas at hand, we are now ready to

formulate the basic propagation equation for a radio communication link.

Consider a transmitting antenna with an EIRP; we may express the power

density of the transmitting antenna as EIRP/4d2, where d is the distancebetween the receiving and transmitting antennas, the power Pr absorbed by

the receiving antenna is the product of this power density and the antenna's

effective area denoted by Ar, is shown by

Pr= (EIRP/4d2) Ar


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= (PtGtAr) / 4d2 watts

According to the reciprocity principle, we may express the effective area of

the receiving antenna as

Ar= (2/ 4) Gr

Where Gr, is the power gain of the receiving antenna. Substituting this

formula for Ar, we may express the received signal power in the equivalent

form

Pr= PtGtGr( / 4d)2

The above equation is called the Friis free-space equation.

- Path Loss

The path loss, PL, representing signal "attenuation" in decibels across the

entire communication link, is defined as the difference between the

transmitted signal power Pt, and received signal power Pr, as shown by

[PL] = 10 log10(Pt/ Pr)

[PL] = 10 log10(Pt/(Pt GtGr(/4d)2))

[PL] =10 log10(GtGr) + 10 log10(4d / )2

The minus sign associated with the first term signifies the fact that this term

represents a "gain." The second term, due to the collection of terms (4d / )

2 is called the free-space loss, denoted by Lfree space. Note that increasing the

distance d separating the receiving antenna from the transmitting antenna

causes the free-space loss to increase, which, in turn, compels us to operate

the radio communication link at lower frequencies so as to maintain the path

loss at a manageable level.

The Friis free-space equation enables us to calculate the path loss PL for

specified values of power gains Gt and Gr , the carrier wavelength , and

distance d.


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Lfree space = (4d/)2 = (4df/c)2= (4df/3108)2

[FSL] = L[dB] = 10 log10(4df/3108)2

[FSL] = L[dB] = 32.4 +20 log10d [km] + 20 log10f [MHz]

[PL] =10 log10(Gt)10 log10(Gr) + 10 log10Lfree space

[PL] =GtdBGrdB + L dB

From equation Pr= PtGtGr( / 4d)2

[Pr] = [EIRP] + [Gr] + [FSL]

Example

In satellite communication system, freespace conditions is to be assumed.

The satellite is a geo-stationary satellite (h = 36000km) above earth, the frequency

used 4GHz, the transmitting antenna gain is 15 dB and the receiving antenna gain is

45dB. Calculate:

(i)

The free- space transmission loss.

(ii) The received power, when the transmitted power is 200 watt.

(iii) The received power, when is rains in the path (Lrain= 40 dB) for the same

transmitted power.

Solution

1-

FSL= 32.4+ 20log d+20log F

= 32.4+20 log 36000+20log 4000

= 195.57 dB

2- Pr= PtGtGr (

)

[Pr/Pt]dB= [Gt] + [Gr]-[FSL]

=15+45-195.57 = -136 dB

Pr/Pt= 103.= 2.5*10-14

Pr = Pt*2.5*10-14= 200*2.5*10-14= 5P watt


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Red Sea University Engineering Faculty Page 23

3- [Pr/Pt]dB= [Gt] + [Gr]-[FSL]

= 15+45-[all Loss]

=15+45- [196+40] = -176dB

Pr/Pt= 10.= 2.5*10-18

Pr = Pt*2.5*10-18= 200*2.5*10-18= 5*10-4 P watt

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Feeder losses

Losses will occurs in the connection between the receive antenna and reviver

proper. These will be denoted by RFL, or [RFL] dB, for receiver feeder losses.

-

Antenna Misalignment Losses [AML]

When satellite link is established, the idea situation is to have the earth station

and satellite antenna aligned for maximum gain, as shown in figures below.

There are the possible sources of off axis loss, one at the satellite and one at

the earth station.

The off axis losses at the earth station is referred to as the antenna pointing

losses. To minimize this losses antenna tracking system is required for a large

diameter antenna to minimize the pointing error.

lecture 3 and 4

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