1. Wireless WANs: Satellite Networks
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2. 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
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.

3. 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. 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.

7. 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.

8. Medium Earth Orbit (MEO) Geosynchronous Orbit (GEO)
Figure Satellite categories
GEO: EXACTLY 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)

9. Figure 16.15 Satellite orbit altitudes

10. 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

12. 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)

15. Application of GPS Military forces Navigation
Clock synchronization, CDMA cellular system

16. 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

17. Low Earth Orbit (LEO) Satellite Systems
LEO satellites have polar orbits
Altitude is between km
Rotation period of 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. 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

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
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.

24. 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

25. 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 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
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 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
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 = = 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