Wireless WANs: Satellite Networks Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Brief history of satellite communication Name Date of launch note SPUTNIK I October 4, 1957 the world's first orbital spacecraft. Nov 1957, Sputnik 2 and a dog escape earth and enter outerspace SCORE December 18, 1958 The first communication satellite which broadcasted a Christmas message for 12 days until the batteries failed Echo 1 August 12, 1960 a passive reflector satellite, the technology was soon abandoned April 12, 1961 First man in space Telstar 1962 First telecommunication satellite, first real-time active Intelsat 1964-1979 geosynchronous earth orbit ,open to use by all nations Inmarsat 1979 used in international shipping ACTS 1993 spot beams, on-board storage and processing, and all digital transmission DirecTV 1994 begins Direct Broadcast to Home Iridium Motorola was supposed to provide mobile telephone service Spot beams subdivide a satellite's footprint which allows the satellite to use its portion of the spectrum more efficiently On-board storage and processing allows for inter-satellite communication and the caching of information until a spot beam finds its target All-digital transmission allows a satellite to incorporate error codes into its signal which helps to overcome rain fade.

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.

Figure Satellite orbits

Table 1 Satellite frequency bands Sky UK, Eutelsat 28A; Ku band

Example 16.1 What is the period of the Moon, according to Kepler’s law? 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.

Example 16.2 According to Kepler’s law, what is the period of a satellite that is located at an orbit approximately 35,786 km above the Earth? 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.

Medium Earth Orbit (MEO) Geosynchronous Orbit (GEO) Figure 16.14 Satellite categories GEO: EXACTLY 22 238 miles MEO: typically around 8000 miles HEO: var. LEO: typically between 500 and 1000 miles Low Earth Orbit (LEO) Medium Earth Orbit (MEO) Geosynchronous Orbit (GEO)

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

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

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

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 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 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)

Figure LEO satellite system

Three categories of LEO 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

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).

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Figure Teledesic Internet in the sky. Teledesic officially suspended its satellite construction work on October 1, 2002. Teledesic was a company founded in the 1990s to build a commercial broadband satellite constellation for Internet services. Using low-earth orbiting satellites small antennas could be used to provide uplinks of as much as 100 Mbit/second and downlinks of up to 720 Mbit/second. The original 1995 proposal was extremely ambitious, costing over US$9 billion originally planning 840 active satellites with in-orbit spares at an altitude of 700 km.[1] In 1997 the scheme was scaled back to 288 active satellites at 1400 km[2] and was later scaled back further in complexity and number of satellites as the projected market demand continued to decrease. The commercial failure of the similar Iridium and Globalstar ventures (composed of 66 and 48 operational satellites, respectively) and other systems, along with bankruptcy protection filings, were primary factors in halting the project, and Teledesic officially suspended its satellite construction work on October 1, 2002.[3] Teledesic was notable for gaining early funding from Microsoft (investing US$30 million for an 8.5% stake), Craig McCaw, Bill Gates, Paul Allen and Saudi prince Alwaleed bin Talal, and for achieving allocation on the Ka-band frequency spectrum for non-geostationary services. Teledesic's merger with ICO Global Communications led to McCaw's companies taking control of ICO, which has successfully launched one test satellite. Teledesic has 288 satellites in 12 LEO orbits, each at an altitude of 1350 km.

Use Kepler’s 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

Iridium satellites are orbiting at 750 km above the earth surface 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

Globalstar satellites are orbiting at 1400 km above the earth surface 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

The space shutter is an example of a LEO satellite The space shutter is an example of a LEO satellite. Sometimes, it orbits at an altitude of 250 km. Using a mean earth radius of 6378km, calculate the period of the shuttle orbit. Determine the linear velocity of the shutter along this orbit. a = 6378 + 250 = 6628 km T = 1/100 a1.5 = 5396 sec = 1.5 hours b. The linear velocity is the circumference divided by the period (2πa)/T = (41645)/(5396) = 7.72 km/s