CS412 Introduction to Computer Networking & Telecommunication Public Switched Telephone Network

Chi-Cheng Lin, Winona State University

Topics

 Structure of Telephone System  Modem  Multiplexing  Switching 2

Structure of the Telephone System

(a) (b) (c) Fully-interconnected network.

Centralized switch.

Two-level hierarchy.

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Structure of the Telephone System

 A typical circuit route for a medium distance call.

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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 5

Digital Transmission

 Why digital?

 Low error rate  Mix signals from different sources (multimedia)  Cheaper  Maintenance is easier 6

Modem

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

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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 8

Figure 5.19

Modulation/demodulation

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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 10

Digital Signal AM FM PM

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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 12

Figure 5.3

ASK

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Figure 5.6

FSK

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Figure 5.9

PSK constellation

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Figure 5.8

PSK

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Figure 5.11

The 4-PSK characteristics

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Figure 5.10

The 4-PSK method

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Figure 5.12

The 8-PSK characteristics

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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 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

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Figure 5.14

The 4-QAM and 8-QAM constellations

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Figure 5.15

Time domain for an 8-QAM signal

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Figure 5.16

16-QAM constellations

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More Constellation Diagrams

(a) QPSK.

(b) QAM-16.

(c) QAM-64.

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Figure 5.17

Bit and baud

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Table 5.1 Bit and baud rate comparison

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

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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 2 3 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud 27

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 log 2 16 = 4. Thus, (1000)(4) = 4000 bps 28

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 log 2 64 = 6. Thus, 72000 / 6 = 12,000 baud 29

Telephone Modems

  A telephone line has a bandwidth of  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.

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Figure 5.18

Telephone line bandwidth

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Figure 5.20

The V.32 constellation and bandwidth

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Figure 5.21

The V.32bis constellation and bandwidth

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Trellis Coded Modulation

  (a) (b) 1 parity per symbol to reduce error Examples, 2400 baud  (a) V.32 for 9600 bps.

 (b) V32 bis for 14,400 bps.

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Multiplexing

 

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?

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Figure 6.3

FDM

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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) 37

FDM

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Figures 6.4 & 6.5

FDM process and demultiplexing

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WDM

Wavelength division multiplexing.

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Figure 6.12

TDM

TDM

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

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Figure 6.13

TDM frames

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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 0.001 s (1 ms).

2. The rate of the link is 4 Kbps.

3. The duration of each time slot 1/4 ms or 250

m

s. 4. The duration of a frame 1 ms.

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Figure 6.14

Interleaving

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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. 45

Figure 6.15

Example 6

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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.

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Figure 6.16

Example 7

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Figure 6.17

Framing bits

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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.

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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 8 + 1 = 33 bits.

6. The data rate of the link is 250 x 33, or 8250 bps.

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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. 52

Figure 5-16

Analog to Digital Encoding

WCB/McGraw-Hill



Figure 4.18

PAM

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Figure 4.19

Quantized PAM signal

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Figure 4.20

Quantizing by using sign and magnitude

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Figure 4.21

PCM

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Figure 4.22

From analog signal to PCM digital code

48 127 -80 58

Figure 4.23

Nyquist theorem

= = = 59

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 60

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 61

TDM - Example

 Why is T1 line 1.544 Mbps?

 193bits/(125  10 -6 sec) = 1.544 Mbps 62

Figure 6.19

T-1 line for multiplexing telephone lines

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Figure 6.20

T-1 frame structure

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TDM - Example

 Multiplexing T1 streams onto higher carriers 65

Table 6.1 DS and T lines rates

Service DS-1 Line T-1 Rate (Mbps) 1.544

Voice Channels 24 DS-2 T-2 6.312

96 DS-3 DS-4 T-3 T-4 44.736

274.176

The capacity of each digital channel is 64 Kbps.

672 4032

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Switching

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

Switching Techniques

  Techniques  Circuit switching  Message switching  Packet switching  Virtual circuit switching (later …) Circuit switching  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) 68

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: ??

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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 70

Circuit Switching Vs. Packet Switching

Circuit switching Packet switching

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

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Circuit Switching Vs. Packet Switching

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