1. CS412 Introduction to Computer Networking & Telecommunication
Public Switched
Telephone Network
Chi-Cheng Lin, Winona State University

2. Topics
Structure of Telephone System
Modem
Multiplexing
Switching

3. Structure of the Telephone System
(a) Fully-interconnected network.
(b) Centralized switch.
(c) Two-level hierarchy.

4. Structure of the Telephone System
A typical circuit route for a medium-distance call.

5. Major Components of the Telephone System
Local loops
Analog twisted pairs going to houses and businesses
Trunks
Digital fiber optics connecting the switching offices
Switching offices
Where calls are moved from one trunk to another

6. Digital Transmission Why digital? Low error rate
Mix signals from different sources (multimedia)
Cheaper
Maintenance is easier

7. Modem
The use of both analog and digital transmissions for a computer to computer call. Conversion is done by the modems and codecs.

8. Modem How can we transmit digital data over analog local loop?
Modulator-demodulator
A device that accepts a serial stream of bits as input and produces a modulated (analog) carrier signal as output (or vice versa)
Modulation
Superimpose information signals on to the carrier signal at transmitting end

9. Figure 5.19 Modulation/demodulation

10. Modulation Techniques
Basic modulation techniques
Amplitude modulation (AM)
(a.k.a. amplitude shift keying (ASK))
Problem: vulnerable to noise
Frequency modulation (FM or FSK)
Problem: limited by physical capacity of carrier
Phase modulation (PM or PSK)
Problem: Hard to distinguish small phase shift
Combination of modulation techniques

11. Digital Signal AM FM PM

12. Modulation Techniques
Constellation patterns
Diagrams showing legal combinations of amplitude and phase
Each high-speed modem standard has its own
Baud rate
Number of signal units transmitted per sec
Number of sample per second
One symbol is sent during each baud

13. Figure ASK

14. Figure FSK

15. Figure 5.9 PSK constellation

16. Figure PSK

17. Figure 5.11 The 4-PSK characteristics

18. Figure The 4-PSK method

19. Figure 5.12 The 8-PSK characteristics

20. Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?
Solution
For PSK the baud rate is the same as the bandwidth, which means the baud rate is But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

21. Figure 5.14 The 4-QAM and 8-QAM constellations

22. Figure 5.15 Time domain for an 8-QAM signal

23. Figure 5.16 16-QAM constellations

24. More Constellation Diagrams
(a) QPSK. (b) QAM (c) QAM-64.

25. Figure Bit and baud

26. Table 5.1 Bit and baud rate comparison ASK, FSK, 2-PSK Bit 1 N
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
ASK, FSK, 2-PSK
Bit 1 N
4-PSK, 4-QAM
Dibit 2 2N
8-PSK, 8-QAM
Tribit 3 3N
16-QAM
Quadbit 4 4N
32-QAM
Pentabit 5 5N
64-QAM
Hexabit 6 6N
128-QAM
Septabit 7 7N
256-QAM
Octabit 8 8N

27. Example 10
A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?
Solution
The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud

28. Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus,
(1000)(4) = 4000 bps

29. Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud

30. Telephone Modems A telephone line has a bandwidth of Modem standards
3000 Hz (3300 – 300) for voice
2400 Hz (3000 – 600) for data
Modem standards
V.32: 9,600 bps
V.32bis: 14,400 bps
V.34bis: 28,800 ~ 33,600 bps
V.90: download up to 56kbps (56K modem)
V.92: adjustable speed, call waiting, etc.

31. Figure 5.18 Telephone line bandwidth

32. Figure 5.20 The V.32 constellation and bandwidth

33. Figure 5.21 The V.32bis constellation and bandwidth

34. Trellis Coded Modulation
(b)
1 parity per symbol to reduce error
Examples, 2400 baud
(a) V.32 for 9600 bps.
(b) V32 bis for 14,400 bps.

35. Multiplexing Multiplexing Why multiplexing?
Set of techniques allowing simultaneous transmission of multiple signals across a single data link
Dividing total available bandwidth over a link into multiple channels
Why multiplexing?

36. Figure FDM

37. Multiplexing Frequency Division Multiplexing (FDM)
Dividing bandwidth of a link into separate channels
Wavelength Division Multiplexing (WDM)
Used over fiber optics
Similar to FDM
Time Division Multiplexing (TDM)
Combining signals from low speed channels to share time on a high-speed link
Synchronous
Asynchronous (Statistical)

38. FDM

39. Figures 6.4 & 6.5 FDM process and demultiplexing

40. Wavelength division multiplexing.
WDM
Wavelength division multiplexing.

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

42. Figure TDM frames

43. 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?
Solution
We can answer the questions as follows:
1. The duration of 1 bit is 1/1 Kbps, or s (1 ms).
2. The rate of the link is 4 Kbps.
3. The duration of each time slot 1/4 ms or 250 ms.
4. The duration of a frame 1 ms.

44. Figure Interleaving

45. 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
The multiplexer is shown in Figure 6.15.

46. Figure Example 6

47. 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
Figure 6.16 shows the output for four arbitrary inputs.

48. Figure Example 7

49. Figure Framing bits

50. Example 8
We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link.
Solution
See next slide.

51. Solution (continued) We can answer the questions as follows:
1. The data rate of each source is 2000 bps = 2 Kbps.
2. The duration of a character is 1/250 s, or 4 ms.
3. The link needs to send 250 frames per second.
4. The duration of each frame is 1/250 s, or 4 ms.
5. Each frame is 4 x = 33 bits.
6. The data rate of the link is 250 x 33, or 8250 bps.

52. Example 9
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?
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.

53. Analog to Digital Encoding
Figure 5-16
Analog to Digital Encoding
WCB/McGraw-Hill
 The McGraw-Hill Companies, Inc., 1998

54. Figure PAM

55. Figure 4.19 Quantized PAM signal

56. Figure 4.20 Quantizing by using sign and magnitude

57. Figure PCM

58. Figure 4.22 From analog signal to PCM digital code
-80 48
127

59. Figure 4.23 Nyquist theorem=

60. Example 4
What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)?
Solution
The sampling rate must be twice the highest frequency in the signal:
Sampling rate = 2 x (11,000) = 22,000 samples/s

61. Example 6
We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0 to 4000 Hz.
Sampling rate = 4000 x 2 = 8000 samples/s
Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps

62. TDM - Example Why is T1 line 1.544 Mbps?
193bits/(12510-6sec) = Mbps

63. Figure 6.19 T-1 line for multiplexing telephone lines

64. Figure 6.20 T-1 frame structure

65. TDM - Example
Multiplexing T1 streams onto higher carriers

66. Table 6.1 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
4032
The capacity of each digital channel is 64 Kbps.

67. Switching Switch Switching
Device creating connections between devices linked to it
Switching
Forwarding data from a switch to another device

68. Switching Techniques Techniques Circuit switching Message switching
Packet switching
Virtual circuit switching (later …)
End-to-end path has to be set up BEFORE any data can be sent
Data follow the same path
No danger of congestion (only in path setup phase)

69. Switching Techniques Message switching No physical path established
Store-and-forward
No limit on block size
 Problems
Routers must have disks to buffer long blocks
A single block may tie up a link for minutes
 Cure: ??

70. Switching Techniques Packet switching Similar to message switching:
Store-and-forward
Tight upper limit on block size
 allowing packets to be buffered in router main memory
No single block can tie up a link for too long
Shorter delay and higher throughput

71. Circuit Switching Vs. Packet Switching

72. Propagation delay: time to send to 1 bit from src to dest
# hops=3
Circuit switching
Message switching
Packet switching

73. Circuit Switching Vs. Packet Switching