2 Brief history of satellite communication NameDate of launchnote SPUTNIK IOctober 4, 1957 the world's first orbital spacecraft. Nov 1957, Sputnik 2 and a dog escape earth and enter outerspace SCOREDecember 18, 1958 The first communication satellite which broadcasted a Christmas message for 12 days until the batteries failed Echo 1August 12, 1960 a passive reflector satellite, the technology was soon abandoned April 12, 1961 First man in space Telstar1962 First telecommunication satellite, first real-time active Intelsat1964-1979 geosynchronous earth orbit,open to use by all nations Inmarsat1979 used in international shipping ACTS1993 spot beams, on-board storage and processing, and all digital transmission DirecTV1994 begins Direct Broadcast to Home IridiumMotorola was supposed to provide mobile telephone service
3 16-2 SATELLITE NETWORKS A satellite network is a combination of nodes, some of which are satellites, that provides communication from one point on the Earth to another. A node in the network can be a satellite, an Earth station, or an end-user terminal or telephone.
4 Figure Satellite orbits
5 Table 1 Satellite frequency bands Sky UK, Eutelsat 28A; Ku band
6 What is the period of the Moon, according to Keplers law? Example 16.1 Here C is a constant approximately equal to 1/100. The period is in seconds and the distance in kilometers. The Moon is located approximately 384,000 km above the Earth. The radius of the Earth is 6378 km. Applying the formula, we get.
7 According to Keplers law, what is the period of a satellite that is located at an orbit approximately 35,786 km above the Earth? Example 16.2 Solution This means that a satellite located at 35,786 km has a period of 24 h, which is the same as the rotation period of the Earth. A satellite like this is said to be stationary to the Earth. The orbit, as we will see, is called a geosynchronous orbit.
16.8 Figure 16.14 Satellite categories Low Earth Orbit (LEO) Medium Earth Orbit (MEO) Geosynchronous Orbit (GEO) GEO: EXACTLY 22 238 miles MEO: typically around 8000 miles HEO: var. LEO: typically between 500 and 1000 miles
9 Figure 16.15 Satellite orbit altitudes
Geosynchronous Orbit (GEO) Satellite Systems Advantages: large area coverage, stay where they are at 35,786km (22,000miles) above the Earth satellite rotation is synchronous to earth three satellites can cover the whole globe low system complexity Disadvantages: long propagation delay (~125 msec) high transmission power is required
11 Figure 16.16 Satellites in geostationary orbit
Medium Earth Orbit (MEO) Satellite Systems Advantages: slightly longer propagation delays (~40 msec) slightly higher transmission power required more expensive than LEOs but cheaper than GEOs Disadvantages: coverage spot greater than a LEO, but still less than a GEO still the need to be in rotation to preserve their low altitude 6-8 hours to circle the earth. multiple MEO satellites are still needed to cover a region continuously handovers and satellite tracking are still needed, hence, high complexity
13 Global Position System (GPS) Operated by the US Department of Defense. Orbiting at an altitude about 18,000km Consists of 24 satellites in 6 orbits; 32 by Dec 2012 At any time, about 9 (>4) satellites are visible from any point on Earth A GPS receiver has an almanac that tell the current position of each satellite
14 Figure Trilateration If we now our distance from three points, we know exactly where we are. (three circles meet at one signal point)
Application of GPS 15 Military forces Navigation Clock synchronization, CDMA cellular system
Low Earth Orbit (LEO) Satellite Systems Advantages: short propagation delays (10-15 msec) low transmission power required low price for satellite and equipment Disadvantages: small coverage spot they have to be in rotation to preserve their low altitude (90 mins period) a network of at least 6 LEO satellites is required to cover a region continuously high system complexity due the need for handovers and satellite tracking
Low Earth Orbit (LEO) Satellite Systems 16.17 LEO satellites have polar orbits Altitude is between 500-2000 km Rotation period of 90-120 min. An LEO system has a cellular type of access Footprint has a diameter of 8000 km. Delay < 20 ms, accept for telephony Work together as a network, connected through intersatellite links (ISLs)
18 Figure LEO satellite system
19 Little LEO, under 1GHz, for low date rate message Big LEO: between 1-3 GHz, Globalstar and Iridium system Broadband LEO provide communication similar to fibre optic network. Teledesic Three categories of LEO
20 Figure Iridium constellation The Iridium system has 66 (planning was 77) satellites in six LEO orbits, each at an altitude of 750 km. Iridium is designed to provide direct worldwide voice and data communication using handheld terminals, a service similar to cellular telephony but on a global scale ( including poles, oceans and airways).
22 Figure Teledesic Teledesic has 288 satellites in 12 LEO orbits, each at an altitude of 1350 km. Internet in the sky. Teledesic officially suspended its satellite construction work on October 1, 2002.
24 Use Keplers formula to find the period and altitude for an Iridium satellite and Globalstar satellite. Iridium satellites are orbiting at 750 km above the earth surface. Globalstar satellites are orbiting at 1400 km above the earth surface. The radius of the earth 6378 km
25 Iridium satellites are orbiting at 750 km above the earth surface. Considering the radius of the earth 6378 km, the radius of the orbit is then (750 km + 6378 km) = 7128 km. Using the Kepler formula, we have Period = (1/100) (distance) 1.5 = (1/100) (7128) 1.5 = 6017 s = 1.67 hours
26 Globalstar satellites are orbiting at 1400 km above the earth surface. Considering the radius of the earth, the radius of the orbit is then (1400 km + 6378 km) = 7778 km. Using the Kepler formula, we have Period = (1/100) (distance) 1.5 = (1/100) (7778) 1.5 = 6860 s = 1.9 hours
27 The space shutter is an example of a LEO satellite. Sometimes, it orbits at an altitude of 250 km. a.Using a mean earth radius of 6378km, calculate the period of the shuttle orbit. b.Determine the linear velocity of the shutter along this orbit. a.a = 6378 + 250 = 6628 km T = 1/100 a 1.5 = 5396 sec = 1.5 hours b. The linear velocity is the circumference divided by the period (2πa)/T = (41645)/(5396) = 7.72 km/s