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

Topics Structure of Telephone System Modem Multiplexing Switching

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

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

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

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

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

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

Figure 5.19 Modulation/demodulation

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

Digital Signal AM FM PM

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

Figure 5.3 ASK

Figure 5.6 FSK

Figure 5.9 PSK constellation

Figure 5.8 PSK

Figure 5.11 The 4-PSK characteristics

Figure 5.10 The 4-PSK method

Figure 5.12 The 8-PSK characteristics

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.

Figure 5.14 The 4-QAM and 8-QAM constellations

Figure 5.15 Time domain for an 8-QAM signal

Figure 5.16 16-QAM constellations

More Constellation Diagrams (a) QPSK. (b) QAM-16. (c) QAM-64.

Figure 5.17 Bit and baud

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

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

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

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

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.

Figure 5.18 Telephone line bandwidth

Figure 5.20 The V.32 constellation and bandwidth

Figure 5.21 The V.32bis constellation and bandwidth

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.

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?

Figure 6.3 FDM

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)

FDM

Figures 6.4 & 6.5 FDM process and demultiplexing

Wavelength division multiplexing. WDM Wavelength division multiplexing.

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.

Figure 6.13 TDM frames

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

Figure 6.14 Interleaving

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.

Figure 6.15 Example 6

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.

Figure 6.16 Example 7

Figure 6.17 Framing bits

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.

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.

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.

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

Figure 4.18 PAM

Figure 4.19 Quantized PAM signal

Figure 4.20 Quantizing by using sign and magnitude

Figure 4.21 PCM

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

Figure 4.23 Nyquist theorem =

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

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

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

Figure 6.19 T-1 line for multiplexing telephone lines

Figure 6.20 T-1 frame structure

TDM - Example Multiplexing T1 streams onto higher carriers

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-4 274.176 4032 The capacity of each digital channel is 64 Kbps.

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

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)

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: ??

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

Circuit Switching Vs. Packet Switching

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

Circuit Switching Vs. Packet Switching