Mobile Communications Exercise: Satellite Systems and Wireless LANs. Georg von Zengen, IBR, TU Braunschweig,

Mobile Communications

Exercise: Satellite Systems and

Wireless LANs

Exercise: Satellite Systems and Wireless LANs

N°1

Please define the terms “inclination” and “elevation” using the following two figures. How do these parameters influence the usefulness of a satellite?

Exercise: Satellite Systems and Wireless LANs

Inclination

Exercise: Satellite Systems and Wireless LANs

Elevation

Exercise: Satellite Systems and Wireless LANs

N°2

a) Please name the typical orbits of satellite systems and fill in the

respective distance to the earth surface.

b) Please calculate the minimum propagation delay (earth  satellite)

through satellite systems of the three orbit types. Please assume a wave propagation speed of .

c) Please name advantages and disadvantages of the orbits as well as

examples of techniques using such orbits.

3·108m

Exercise: Satellite Systems and Wireless LANs

N°2

a) Please name the typical orbits of satellite systems and fill in the

respective distance to the earth surface.

Orbit Distance to the earth surface

GEO: Geostationary Orbit 35,768 km

MEO: Medium Earth Orbit 5,000 – 20,000km

Exercise: Satellite Systems and Wireless LANs

N°2

b) Please calculate the minimum propagation delay (earth  satellite)

through satellite systems of the three orbit types. Please assume a wave propagation speed of . 3·108m

s

Orbit Distance to the earth surface Propagation Delay GEO 35768 km MEO 5000 – 20000 km LEO 500 – 1500 km t = 35768×10 3 m 3×108 m s = 35768 3×105 s=0.119s=119ms t = 5×10 6 m 3×108 m s = 5 3×10 -2 s=0.017s=17ms t = 5×10 5 m 3×108 m s = 5 3×10 -3 s=0.0017s=1.7ms

Exercise: Satellite Systems and Wireless LANs

N°2

c) Please name advantages and disadvantages of the orbits as well as

examples of techniques using such orbits.

GEO (TV Broadcast services, telephony, internet access)

+ fixed antenna positions, large footprint - high transmit power, high latency

- bad elevation in areas above 60° latitude

MEO (GPS, telephony)

• medium latency, medium footprint, medium visibility

• handover may be necessary, moving satellites LEO (Telephony)

+ low latency, small footprint: better frequency reuse - short visibility, handovers necessary

Exercise: Satellite Systems and Wireless LANs

N°3

a) How are priorities of different frame types modeled in IEEE 802.11?

How many priority level are defined in the standard?

Exercise: Satellite Systems and Wireless LANs

N°3

a) How are priorities of different frame types modeled in IEEE

802.11b? How many priority level are defined in the standard?

Different Inter Frame Spaces (IFS) model priority. Shorter IFS mean higher priority.

• Short IFS: Before ACK, CTS (10 μs)

• Point Coordination Function IFS (30 μs)

• Distributed Coordination Function IFS (50 μs)

• Extended IFS: After receiving erroneous frame

Exercise: Satellite Systems and Wireless LANs

N°3

b) How is fairness among stations implemented in IEEE 802.11?

Fairness means, that all stations will eventually be able to access the medium.

IEEE 802.11 models fairness by suspending and resuming the

Exercise: Satellite Systems and Wireless LANs

N°4

Please explain the Hidden- and Exposed Terminal Problem and outline how IEEE 802.11 (WLAN) and IEEE 802.15.1 (Bluetooth) solve these problems.

Exercise: Satellite Systems and Wireless LANs

Hidden Terminal

• In wireless networking, collisions occur at the receiver

A B C

Exercise: Satellite Systems and Wireless LANs

Exposed Terminal

• Station B cannot transmit, since the medium is sensed busy

A B C

DATA DATA

Exercise: Satellite Systems and Wireless LANs

Hidden Terminal with RTS/CTS

A B C RTS A to B CTS A CTS A DATA RTS C to B CTS C CTS C DATA

Exercise: Satellite Systems and Wireless LANs

Exposed Terminal with RTS/CTS

A B C RTS C to D CTS B CTS C DATA RTS B to A DATA RTS C to D

Exercise: Satellite Systems and Wireless LANs

N°4

Please explain the Hidden- and Exposed Terminal Problem and outline how IEEE 802.11 (WLAN) and IEEE 802.15.1 (Bluetooth) solve these problems.

• Bluetooth is a master/slave system and does not suffer from the Hidden- and Exposed Terminal Problem. All nodes that can see the master may be part of the network; all others are not.

Exercise: Satellite Systems and Wireless LANs

N°5

Assuming a wireless IEEE 802.11b-compliant network with 4 stations that are within the same area, please outline how media access using DFWMAC-DCF CSMA/CA is handled. Please ignore Acknowledgement frames assume that the MAC layer of each node receives a frame from the upper layer at the points in time designated by the arrows. The medium has been busy before t₀ and nodes chose their respective backoff timer to be their station number plus 3 in slots.

Exercise: Satellite Systems and Wireless LANs

Data Frame (5 Slots) Time [Slots] Back off (3 + n Slots) Station 4 Suspend Backoff Timer Suspend Backoff Timer Suspend Backoff Timer

Exercise: Satellite Systems and Wireless LANs

N°6

The throughput of IEEE 802.11 primarily limited by MAC mechanisms and not by the physical layer capabilities. Please calculate the maximum real-world upper-layer throughput of IEEE 802.11b at 11 Mbit/s for a frame payload size of 500, 1500 and 2312 bytes. Please assume that each packet uses the backoff timer and also requires a subsequent acknowledgement frame.

Exercise: Satellite Systems and Wireless LANs

Backoff Timer PLCP Header Payload Data PLCP Header ACK

t

DIFS MAC Header SIFS

Element Data Bitrate Time [μs]

DIFS 50

Backoff Timer 310

PLCP Header 192 bit 1 Mbit/s

MAC Header 272 bit 11 Mbit/s

Payload Data 500 / 1500 /

2312 byte 11 Mbit/s

SIFS 10

ACK Frame 112 bit 1 Mbit/s

192bit 106bit s =192×10-6s=192ms 272bit 11×106bit s = 272 11 ×10 -6 s=24.7ms 112bit 106bit =112×10-6s=112ms 500×8bit 11×106bit s = 4 11×10 -3 s=363.6ms

Exercise: Satellite Systems and Wireless LANs

Payload length Time per packet [μs] Maximum Throughput

500 byte 1254.3 1500 byte 1981.6 2312 byte 2572.2 500 byte 1254.3ms=3.189MBit / s 1500 byte 1981.6ms=6.056MBit / s 2312 byte 2572.2ms=7.191MBit / s

Goodput

=

bit

time

G

₅₀₀

=

500

×

8

×

10

-

6

1254.3

×

10

-

6

=

500

×

8

1254.3

=

3.189MBit / s