Satellite Communication Principles

Satellite Communication Principles Abdullah Al-Ahmadi College of Engineering Majmaah University URL: http://faculty.mu.edu.sa/aalahmadi; Electronic address: a.alahmadi@mu.edu.sa 1

Part I Introduction A. Key Dates in Satellite Communication History 1957 USSR Sputnik-I, the first man-made satellite, a battery-driven radio beacon. 1958 US Explorer-I, the first US satellite. 1958 US Score satellite launched and broadcast the President Eisenhower s State of the Union address. 1962 Bell Labs launched Telestar I and II, which are considered the fist classic telecommunications satellite. They had 6389 MHz uplinks and 4169 MHz downlinks and solar batteries (longer lifetime... early satellites like Sputnik and Explorer had purely battery-driven systems and died less than a year after launch). Mid-1980s GPS came online publicly. Operated by the US Air Force, it was the first globally available positioning system. Developed in the 70s with the initial impetus to guide ICBMs. 2000 LEO telephony systems (Iridium, Globalstar, Orbcomm). B. Types of Earth Orbits There are 4 types of earth orbital altitudes: Low Earth Orbit (LEO) is any orbit with an altitude less than 1500 km. These are cheapest to launch, but zip across the sky very quickly. Due to low launch and replacement costs, many communications satellites use networks of LEO satellites to trunk data and mobile telephony from one point on the earth to another. 2

This orbit is also used for surveying and weather satellites. Medium Earth Orbit (MEO) is any satellite orbit above 1500 km and less than 36,000 km. These orbits are less populated because they are more expensive to launch into. Earth coverage is better here, however, so some satellite systems that require multiple simultaneous satellites for operation (GPS) operate here. Geosynchronous Earth Orbit (GEO) is the designation for any satellite that is 36,000 km. This orbit is again highly populated because of the unique functionality a stationary satellite provides for broadcasting information to fixed dishes on earth. High Earth Orbit (HEO) is any orbit above the GEO 36,000 km altitude. Extremely rare for anything other than scientific spacecraft or de-orbited GEO satellites. 3

Part II Orbital Mechanics I. CIRCULAR ORBITS To achieve a stable orbit, a spacecraft must be first beyond the earth s atmosphere (400, 000 ft 122 km). Newton s laws of motion are: ( ) 1 s = ut + at 2 (1) 2 v 2 = u 2 + 2at (2) v = u + at (3) F = ma (4) where s is the distance traveled from time t = 0. u is the initial velocity of the object at time t = 0. v is the final velocity of the object at time t. a is the acceleration of the object. m is the mass of the object. F is the force acting on the object. Equations 4 states that the force acting on a body is equal to the mass of the body multiplied by the resulting acceleration of the body. Alternatively, the resulting acceleration is the ratio of the force acting on the body to the mass of the body. a = F m (5) The force has a unit of Newton N: 4

Figure 2.1 (p. 18) Forces acting on a satellite in a stable orbit around the earth (from Fig. 3.4 of reference 1). Gravitational force is inversely proportional to the square of the distance between the centers of gravity of the satellite and the planet the satellite is orbiting, in this case the earth. The gravitational force inward (F IN, the centripetal force) is directed toward the center of gravity of the earth. The kinetic energy of the satellite (F OUT, the centrifugal force) is directed diametrically opposite to the gravitational force. Kinetic energy is proportional to the square of the velocity of the satellite. When these inward and outward forces are balanced, the satellite moves around the earth in a free fall trajectory: the satellite s orbit. For a description of the units, please see the text. Figure 1: Forces acting on a satellite in a stable point around the earth. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. A Newton is the force required to accelerate a mass of 1 kg with an acceleration of 1 m/s 2. Thus, the lighter the mass of the body, the higher the acceleration. In a stable orbit, two main forces are acting on the satellite as shown in Figure 1: 1. A centrifugal force F OUT : Due to the kinetic energy of the satellite. Tries to push the satellite to a higher orbit. 2. A centripetal force F IN : Due to the the gravitational attraction of the planet about which the satellite is orbiting. Attempts to pull the satellite toward the earth. If these two forces are equal, the satellite will remain in a stable orbit. The acceleration a, due to centripetal force at distance r from the center of the earth is: a = µ/r 2 km/s 2 (6) 5

The constant µ is the product of the universal gravitational constant G and the mass of the earth M E. µ = GM E = 3.986004418 10 5 km 3 /s 2 (7) The constant µ is called Kepler s constant. G = 6.672 10 11 Nm 2 /kg 2 M e = 5.98 10 24 kg The mean earth radius is 6378.137 km. From Equation 4, the centripetal force acting on the satellite, F IN is given by: F IN = m ( µ/r 2) (8) = m ( GM E /r 2) The acceleration a, due to centrifugal force is given by: a = v 2 /r Hence, the centrifugal force F OUT is: F OUT = m ( v 2 /r ) (9) If the forces on the satellite are balanced, F IN = F OUT, then using Equations 8 and 9: m µ/r 2 = m v 2 /r Hence, the velocity v of a satellite in a circular orbit is given by: v = (µ/r) 1/2 In circular orbits, the distance traveled by a satellite in one orbit around a planet is equal to the circumference of that orbit, 2πr. 6

Since time is equal distance divided by velocity, then the period of the satellite s orbit, T, is: T = 2πr v = (2πr) / [(µ/r) 1/2] T = ( 2πr 3/2) / ( µ 1/2) Example: Calculate the orbital period and velocity for a satellite in the GEO at 35, 786.03 km above the surface of the earth. Solution: Orbital velocity: v = (µ/r) 1/2 ( 3.986004418 10 5 = 35, 786.03 + 6378.137 = 3.0747 km/s ) 1/2 Orbital period: T = ( 2πr 3/2) / ( µ 1/2) 2π [35, 786.03 + 6378.137]3/2 = [3.986004418 10 5 ] 1/2 = 86164.08246 seconds = 23 hours 56 minutes 4.08 seconds II. KEPLER S LAWS OF PLANETARY MOTION Jahannes Kepler (1571-1630) developed three laws of planetary motion. Kepler s three laws are: 1. The orbit of any smaller body about a larger body is always an ellipse, with the center of mass of the larger body as one of the foci. 2. The orbit of the smaller body sweeps out equal areas in equal time. Figure 2 7

Figure 2.5 (p. 22) Illustration of Kepler s second law of planetary motion. A satellite is in orbit about the planet earth, E. The orbit is an ellipse with a relatively high eccentricity, that is, it is far from being circular. The figure shows two shaded portions of the elliptical plane in which the orbit moves, one is close to the earth and encloses the perigee while the other is far from the earth and encloses the apogee. The perigee is the point of closest approach to the earth while the apogee is the point in the orbit that is furthest from the earth. While close to perigee, the satellite moves in the orbit between times t 1 and t 2 and sweeps out an area denoted by A 12. While close to apogee, the satellite moves in the orbit between times t 3 and sweeps out an area denoted by A 34. If t 1 t 2 = t 3 t 4, then A 12 = A 34. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 2: Illustration of Kepler s second law of planetary motion 3. The square of the period of revolution of the smaller body about the larger body equals a constant multiplied by the third power of the semimajor axis of the orbital ellipse. That is: T 2 = 4π2 a 3 µ where T is the orbital period, a is the semimajor axis of the orbital ellipse and µ is Kepler s constant. If the orbit is circular, then a becomes distance r. III. DESCRIBING THE ORBIT OS A SATELLITE Referring to Figure 3: The semimajor axis of the ellipse is denoted by a, and the semiminor axis, by b. The relationship between the semimajor axes a and the semiminor axis b is as follows: b = a (1 e 2 ) 8

Figure 2.6 (p. 24) The orbit as it appears in the orbital plane, The point O is the center of the earth and the point C is the center of the ellipse. The two centers do not coincide unless the eccentricity, e, of the ellipse is zero (i.e., the ellipse becomes a circle and a = b). The dimensions a and b are the semimajor and semiminor axes of the orbital ellipse, respectively. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 3: Orbit of a satellite The eccentricity e is given by: e = a2 b 2 a where e is the eccentricity and it has values between 0 and 1 0 e 1 The Perigee is the distance between the earth and the spacecraft at the closest distance to the planet and it is equal to r b = (1 e) a. The Apogee is the distance between the earth and the spacecraft at the furthest distance to the planet and it is equal to r a = (1 + e) a. Example: A satellite is in an elliptical orbit with a perigee of 1000 km and apogee of 4000 km. Using a mean earth radius of 6378.14 km, find the period of the orbit in hours, minutes, and seconds. Solution: 9

For a semimajor axis length a, earth radius r e, perigee height h p, and apogee h a : 2a = 2r e + h p + h a = 2 (6378.14) + 1000 + 4000 = 17756.28 km a = 8878.14 km Using Kepler s third law: T 2 = 4π2 a 3 µ = 4π2 (8878.14) 3 3.986004418 10 5 = 6.9308728 10 7 s 2 T = 8325.1864 s = 2 hours 18 minutes 45.19 seconds IV. LOOK ANGLE DETERMINATION Earth surface was divided into a grid-like structure of orthogonal lines: Latitude and Longitude. Latitude is the angular distance measured in degrees north or south the equator. Longitude is the angular distance measured in degrees from a given reference longitudinal line. There are 360 of longitude measured from 0 at the Greenwich. There are 90 of latitude. Latitude +90 is the north pole. Latitude 90 is the south pole. When GEO satellites are registered in Geneva, their subsatellite location is given in degrees east to avoid confusion. 10

subsatellite point would see the satellite at zenith (i.e., at an elevation angle of 90 ). The pointing direction from the satellite to the subsatellite point is the nadir direction from the satellite. If the beam from the satellite antenna is to be pointed at a location on the earth that is not at the subsatellite point, the pointing direction is defined by the angle away from nadir. In general, two off-nadir angles are given: the number of degrees north (or south) from nadir. In general, two off-nadir angles are given: the number of degrees north (or south) from nadir; and the number of degrees east (or west) from nadir. East, west, north, and south directions are those defined by the geography of the earth. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 4: The subsatellite point Thus, INTELSAT primary location is at 335.5 E (not 24.5 W). Since all U.S satellites are located west of Greenwich, it has become accepted to be described in terms of degrees W. The coordinates to which an earth station antenna must be pointed to communicate with a satellite are called the Look Angles. In order to determine the look angle, we first have to calculate the subsatellite point. The subsatellite point is the location on the surface of the earth that lies directly between the satellite and the center of the earth. Figure 4 It is the Nadir pointing direction from the satellite. To an observer at the subsatellite point: The satellite will appear to be directly overhead in the Zenith direction. A. Elevation Angle Calculation. From Figure 5: 11

r s is the vector from the center of the earth to the satellite. r e is the vector from the center of the earth to the earth station. d is the vector from the earth station to the satellite. The central angle γ measured between r e and r s is the angle between the earth station and the satellite. ψ is the angle measured from r e to d. The central angle γ is given by: cos (γ) = cos (L e ) cos (l s l e ) For a satellite to be visible from the earth station, then γ should be: ( ) γ cos 1 re r ( s ) 6378.137 γ cos 1 42164.17 γ 81.3 This called a visibility test. Then the elevation angle El can be calculated by: [( ) ] El = tan 1 rs cos γ / sin γ r e γ = tan 1 [(6.6107345 cos γ) / sin γ] γ B. Azimuth Angle Calculation. First we have to calculate an intermediate angle α by: [ ] tan α = tan 1 (ls l e ) sin (L e ) Having the intermediate angle α, the azimuth look angle Az can be found from: Case 1: Earth station in the Northern Hemisphere with: 1. Satellite to the SE of the earth station: Az = 180 α 12

Figure 2.12 (p. 33) The geometry of elevation angle calculation. The plane of the paper is the plane defined by the center of the earth, the satellite, and the earth station. The central angle is γ. The elevation angle EI is measured upward from the local horizontal at the earth station. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 5: The geometry of elevation 2. Satellite to the SW of the earth station: Az = 180 + α Case 2: Earth station in the Southern Hemisphere with: 1. Satellite to the NE of the earth station: Az = α 2. Satellite to the NW of the earth station: Az = 360 α Example: An earth station is situated at Majmaah University (25.89 N,45.35 E), needs to calculate the look angle to ARABSAT 5B (26 E). Solution: 1. Find the central angle γ cos γ = cos (L e ) cos (l s l e ) = cos (26.89) cos (26 25.89) = 0.8988 γ = 26 The central angle γ is less than 81.3, so the satellite is visible from the earth station. 13

2. Find the elevation angle El El = tan 1 [(6.6107345 cos γ) / sin γ] γ = tan 1 [(6.6107345 cos (26)) / sin (26)] 26 = 59.611 3. Find the intermediate angle α [ ] tan α = tan 1 (ls l e ) sin (L e ) [ ] tan (26 45.35) = tan 1 sin (25.89) = 38.81 4. Find the azimuth angle Since the satellite is SW of the earth station then: Az = 180 + α = 180 + 38.81 = 218.81 14

Part III Orbital Perturbation Sources of Orbital Perturbations: 1. Other celestial bodies. (a) Moon: because of its closeness to the earth. (b) Sun: because of its size. 2. Longitudinal variations: (a) Earth oblateness. (b) Mass concentration (Mascons): (252 E, 75 ) and (162, 348 ) Areas with higher and lower density. Orbital Forcing: Cycles of orbital perturbation on Earth s orbit around the sun. Largely due to Jupiter, although other planets contribute slightly. Figure 2.14 (p. 40) Influences long-term climate (Milankovitch cycles). Main results of orbital forcing: Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. 15

1. Eccentricity variations. Eccentricity will vary slightly over time due to the influence of other planets. The variation is between e=0.005 and 0.058, which occurs over 413,000 year cycles. (Currently at e=0.012). The planet receives more insolation (level of sunlight and energy from the sun) at perihelion than at aphelion. As eccentricity increases, the perihelion portion of the orbit becomes warmer and the aphelion becomes cooler. 2. Precession variations. The earth s off-axis tilt of 23 itself spins around slowly over time. The earth experiences precession in 26,000 year cycles. 3. Obliquity variations. Earth rotation axis wobbles between 21.5 and 24.5 (Currently at 23.44 ). Occurs in 41000 year cycles. As obliquity increases: Seasonal variations become more severe across the planet as higher latitudes receive more insolation in the summer and less in the winter. 4. Inclination variations. Earth s orbital inclination wobbles in and out of the sun s equatorial plane. The earth passes through the plane of the solar system twice a year. During this phase, observed meteor activity always increases. 16

Part IV Launches and Launch Vehicles A satellite cannot be placed into a stable orbit unless two parameters are correct: 1. The velocity vector. 2. The orbital height. A GEO satellite must be in an orbit at: A height of 35786.03 km above the surface of the earth (42164.17 km from the center of the earth). Inclination of zero degrees. An ellipticity of zero. A velocity of 3074.7 m/s tangential to the earth in the plane of the orbit. I. PLACING SATELLITES INTO GEOSTATIONARY ORBIT A. Geostationary Transfer Orbit and AKM Place the spacecraft with the final rocket stage still attached into low earth orbit. After a couple of orbits, the final stage is reignited and the spacecraft is launched into GTO (Geostationary Transfer Orbit). The GTO has a perigee that is the original LEO altitude and an apogee that is the GEO altitude. Figure 6 Since the rocket motors fires at apogee, it is commonly refereed to as the apogee kick motor (AKM). The AKM is used in: 1. Circularize the orbit at GEO. 2. Remove any inclination error. 17

igure 2.19 (p. 48) stration of the GTO/AKM approach to ostationary orbit (not to scale). The mbined spacecraft and final rocket ge are placed into low earth orbit (LEO) und the earth. After careful orbit termination measurements, the final ge is ignited in LEO and the spacecraft erted into a transfer orbit that lies tween the LEO and the geostationary it altitude: the so-called geostationary nsfer orbit or GTO. Again, after more reful orbit determination, the apogee k motor (AKM) is fired on the satellite d the orbit is both circularized at ostationary altitude and the inclination uced to close to zero. The satellite is n in GEO. Figure 6: GTO/AKM approach Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. B. Geostationary Transfer Orbit with Slow Orbit Raising. The thrusters are used to raise the orbit from GTO to GEO over a number of burns. Figure 7. The satellite has two power level of thrusters. One for more powerful orbit raising maneuvers. One for on-orbit (low thrust) maneuvers. C. Direct Insertion to GEO The final stages of the rocket are used to place the satellite directly into GEO rather than using the satellite own propulsion. 18

re 2.20 (p. 49) ion of slow orbit raising to geostationary t to scale). The combined spacecraft l rocket stage are placed into low earth O) around the earth. As before (see.19), the spacecraft is injected into GTO his case, once the satellite is ejected final rocket stage, it deploys many of ents that it will later use in GEO (solar etc.) and stabilizes its attitude using s and momentum wheels, rather than pin-stabilized. The higher power s are then used around the apogee to e perigee of the orbit until the orbit is at the GEO altitude. At the same time rbit is being raised, the thruster firings esigned gradually to reduce the on to close to zero. Figure 7: Slow Orbit Raising Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. 19

Part V Orbital Effects in Communications Systems Performance I. DOPPLER SHIFT The frequency of a moving radio transmitter varies with the transmitter s velocity. The true transmitter frequency f T is the frequency that the transmitter would send when at rest. The received frequency f R is higher than f T when the transmitter is moving toward the receiver. And lower than f T when the transmitter is moving away from the receiver. The relationship between the transmitted and received frequencies is: f R f T f T = f f T = V T v p or f = V T f T /c = V T /λ where V T is the component of the transmitter velocity directed toward the receiver. v p = c the phase velocity of light (2.9979 10 8 3 10 8 m/s). λ is the wavelength of the transmitted signal. If the transmitter is moving away from the receiver, then V T is negative. This change in frequency is called the Doppler Shift, The Doppler effect or Doppler. 20

For GEO satellites, the effect is negligible. Example: A LEO satellite is in a circular orbit with an altitude, h, of 1000 km. A transmitter on the satellite has a frequency of 2.65 GHz. Find: 1. The velocity of the satellite in orbit. 2. The component velocity toward an observer at an earth station as the satellite appears over the horizon, for an observer who is in the plane of the satellite orbit. 3. The Doppler shift of the received signal at the earth station. 4. The Doppler shift for this signal when it is received by the same observer if it carries a Ku-band transmitter at 20 GHz Answer: 1. The period of the satellite is: T 2 = ( 4π 2 a 3) /µ = ( 4π 2 (6378 + 1000) 3) /3.986004418 10 5 = 3.977754 10 7 s 2 T = 6306.94 s The circumference of the orbit is 2πr = 46357.3 km so the velocity of the satellite in the orbit is: v s = 46357.3/6306.94 = 7.35 km/s 2. The component of velocity toward an observer in the plane of the orbit as the satellite appears over the horizon is given by v r = v s cos θ, where θ is the angle between the satellite velocity vector and the direction of the observer. cos θ = r e / (r e + h) = 6378/7378 = 0.8645 21

hence v r = v s cos θ = 6.354 km/s = 6354 m/s 3. The Doppler shift in the received signal is: f = V T /λ = 6354/0.1132 = 56130 Hz = 56.130 khz 4. The Doppler shift at the receiver is: f = V T /λ = 6354/0.015 = 423.6 khz II. RANGE VARIATION The position of a satellite with respect to the earth exhibits a cyclic daily variations. The variation in position will lead to a variation in range between the satellite and the receiver. If time division multiple access (TDMA) is being used; careful attention must be paid to the timing of the frames within the TDMA busts. so that the individual user frames arrive at the satellite in the correct sequence at the correct time. 22

Figure 2.21 (p. 52) Eclipse geometry. (Source: Spilker, Digital Communications by Satellite, Prentice Hall, 1977.) During the equinox periods around the March 21 and September 3, the geostationary plane is in the shadow of the earth on the far side of the earth from the sun. As the satellite moves around the geostationary orbit, it will pass through the shadow and undergo an eclipse period. The length of the eclipse period will vary from a few minutes to over an hour (see Figure 2.22), depending on how close the plane of the geostationary orbit is with respect to the center of the shadow thrown by the earth. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 8: Solar eclipse III. SOLAR ECLIPSE A satellite is eclipse when the earth prevents sunlight from reaching it; that is, when the satellite is in the shadow of the earth. Figure 8. For GEO satellites, eclipses occur during two periods that being 23 days before the equinoxes (about March 21 and about September 23) and ends 23 days after the equinoxes periods. Figure 9 During full eclipse, a satellite receives no power from its solar array and must operate entirely from its batteries. IV. SUN TRANSIT OUTAGE During the equinox periods, the orbit of the satellite will also pass directly in front of the sun on the sunlit side of the earth. Figure 10 The sun is a hot microwave source with an equivalent temperature of 6000 to 10000 K at the frequencies used by communication satellites (4 to 50 GHz). The earth station antenna will therefore receive not only the signal from the satellite but also the noise temperature transmitted by the sun. 23

Figure 2.22 (p. 52) Dates and duration of eclipses. (Source: Martin, Communications Satellite Systems, Prentice Hall 1978.) Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. Figure 9: Dates and duration of eclipses. The added noise temperature will cause the fade margin of the receiver to be exceeded and an outage will occur. 24

ure 2.23 (p. 53) matic of sun outage conditions. During quinox periods, not only does the earth s ow cause eclipse periods to occur for tationary satellites, during the sunlit n of the orbit, there will be periods when un appears to be directly behind the ite. At the frequencies used by unications satellites (4 to 50 GHz), the ppears as a hot noise source. The tive temperature of the sun at these encies is on the order of 10,000 K. The se temperature observed by the earth n antenna will depend on whether the width partially, or completely, encloses un. Figure 10: Sun outage. Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt Copyright 2003 John Wiley & Sons. Inc. All rights reserved. 25