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1. 1
NETWORKS
LECTURE - 10

2. 2
Today’s topic
• Data Link Layer Functions – making sense
of the data
• Framing
• Flow Control
• Error Control
• Link (Media) access

3. Accessing Links – Media
Access protocols
Two types of “links”:
 point-to-point
– PPP for dial-up access
– point-to-point link between Ethernet switch
and host
 broadcast (shared wire or medium)
– traditional Ethernet
– upstream HFC
– 802.11 wireless LAN

4. MAC Protocols: a taxonomy
Three broad classes:
 Channel Partitioning
– divide channel into smaller “pieces” (time
slots, frequency, code)
– allocate piece to node for exclusive use
 Random Access
– channel not divided, allow collisions
– “recover” from collision
– Round-robin – Taking turns

5. Channel partitioning /
Multiplexing
Multiplexing
Analog Digital
FDM WDM TDM

6. Channel partitioning /
Multiplexing
 FDM (Frequency Division
Multiplexing): frequency subdivided.
 WDM (Wavelength Division
Multiplexing) : used on optical
channels
 TDM (Time Division Multiplexing):
channel divided into N time slots, one
per user

7. Dividing a link into channels

8. FDM

9. FDM is an analog multiplexing
technique that combines signals.
Gaps may be provided between
adjacent channels – Guard bands
FDM …

10. FDM multiplexing
process

11. FDM demultiplexing example

12. Example 1
Assume that a voice channel occupies a
bandwidth of 4 KHz. We need to combine three
voice channels into a link with a bandwidth of 12
KHz, from 20 to 32 KHz. Show the configuration
using the frequency domain without the use of
guard bands.
Solution
Shift (modulate) each of the three voice channels
to a different bandwidth.

13. Example 1

14. Example 2
Five channels, each with a 100-KHz bandwidth,
are to be multiplexed together. What is the
minimum bandwidth of the link if there is a need
for a guard band of 10 KHz between the channels
to prevent interference?
Solution
For five channels, we need at least four guard bands.
This means that the required bandwidth is at least
5 x 100 + 4 x 10 = 540 KHz.

15. Example 2

16. Example 3
Four data channels (digital), each transmitting at
1 Mbps, use a satellite channel of 1 MHz. Design
an appropriate configuration using FDM
Solution
The satellite channel is analog. We divide it into four
channels, each channel having a 250-KHz bandwidth.
Each digital channel of 1 Mbps is modulated such that
each 4 bits are modulated to 1 Hz. One solution is 16-
QAM (Quadrature Amplitude modulation).
16 Quadrature Amplitude Modulation. This is a modulation
technique in which the carrier can exist in one of sixteen
different states. As such, each state can represent four bits – 0000
through to 1111,

17. Example 3

18. FDM hierarchy

19. FDMA - Frequency division
multiple access
 channel spectrum divided into
frequency bands
 each station assigned fixed frequency
band
 unused transmission time in frequency
bands go idle
 example: 6-station LAN, 1,3,4 have pkt,
frequency bands 2,5,6 idle

20. WDM
Wave Division Multiplexing

21. WDM is an analog multiplexing
technique to combine
optical signals.

22. WDM

23. Prisms in WDM multiplexing
and demultiplexing

24. TDM
TDM is a digital multiplexing
technique to combine data.

25. TDMA Time division multiple access
access to channel in "rounds"
 each station gets fixed length slot
(length = pkt trans time) in each round
 unused slots go idle
 example: 6-station LAN, 1,3,4 have pkt,
slots 2,5,6 idle

26. TDM

27. TDM frames

28. Interleaving

29. TDM …
In a TDM, the data rate of the link is n
times faster, and the unit duration is n
times shorter.

30. Example 5
Four 1-Kbps connections are multiplexed
together. A unit is 1 bit. Find (1) the
duration of 1 bit before multiplexing, (2)
the transmission rate of the link, (3) the
duration of a time slot, and (4) the
duration of a frame?

31. Solution
We can answer the questions as follows:
1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms).
2. The rate of the link is 4 Kbps.
3. The duration of each time slot 1/4Kbps or 250 ms.
4. The duration of a frame 1 ms.

32. Example 6
Four channels are multiplexed using TDM. If each
channel sends 100 bytes/s and we multiplex 1 byte
per channel, show the frame traveling on the link,
the size of the frame, the duration of a frame, the
frame rate, and the bit rate for the link.
Solution

33. Example 6 - solution

34. Example 7
A multiplexer combines four 100-Kbps channels
using a time slot of 2 bits. Show the output with
four arbitrary inputs. What is the frame rate?
What is the frame duration? What is the bit rate?
What is the bit duration?
Solution
Data rate = 400 Kbps
Frame size = 8 bits
Frame rate = 50 K frames per second
Frame duration = 1/50K = 20 ms

35. Example 7 - solution

36. Example 8
Two channels, one with a bit rate of 100
Kbps and another with a bit rate of 200
Kbps, are to be multiplexed. How this can
be achieved? What is the frame rate?
What is the frame duration? What is the
bit rate of the link?

37. Solution
We can allocate one slot to the first
channel and two slots to the second
channel. Each frame carries 3 bits. The
frame rate is 100,000 frames per second
because it carries 1 bit from the first
channel. The frame duration is 1/100,000
s, or 10 ms. The bit rate is 100,000
frames/s x 3 bits/frame, or 300 Kbps.

38. Framing bits –
Synchronization

39. TDM Standard – T1 line

40. T-1 Standard frame structure

41. TDM hierarchy

42. DS and T lines rates
Service Line
Rate
(Mbps)
Voice
Channels
DS-1 T-1 1.544 24
DS-2 T-2 6.312 96
DS-3 T-3 44.736 672
DS-4 T-4 274.176 4032

43. E line rates
E Line
Rate
(Mbps)
Voice
Channels
E-1 2.048 30
E-2 8.448 120
E-3 34.368 480
E-4 139.264 1920

44. Summary
• Multiplexing techniques
• FDM
• WDM
• TDM
– Framing
– T1 – as an example

45. Questions
1. What is FDM ?
2. What is a guard band ?
3. What is the idea behind TDM ?
4. How would you determine the
data rate required for a TDM
system ?
5. What is the data rate of a T1
system ?

46. Keywords
Multiplexing
Multiple access
FDM
TDM
WDM
Frame rate
T1 & E1 standards

47. Thank you !

48. 48
Acknowledgement
Computer Networking: A Top Down Approach Featuring the
Internet,
2nd edition.
Jim Kurose, Keith Ross
Addison-Wesley, July 2002.
Computer Networks: A systems Approach, 4th edition,
David Peterson, Davie, Morgan Kauffman, 2004.
Thanks to the authors of the following books
for making the slides for their books freely
available. The slides for this course have
been adapted from these sources.

49. Summary
 Error handling
–Detection and ARQ
 RS232
 MAC
–CSMA/CD – the idea