Solution Manual for Data Communications and Networking

Chapter 1 Introduction Solutions to Review Questions and Exercises Review Questions 1. The five components of a data communication system are the sender, receiver, transmission medium, message, and protocol. 2. The advantages of distributed processing are security, access to distributed databases, collaborative processing, and faster problem solving. 3. The three criteria are performance, reliability, and security. 4. Advantages of a multipoint over a point-to-point configuration (type of connection) include ease of installation and low cost. 5. Line configurations (or types of connections) are point-to-point and multipoint. 6. We can divide line configuration in two broad categories: a. Point-to-point: mesh, star, and ring. b. Multipoint: bus 7. In half-duplex transmission, only one entity can send at a time; in a full-duplex transmission, both entities can send at the same time. 8. We give an advantage for each of four network topologies: a. Mesh: secure b. Bus: easy installation c. Star: robust d. Ring: easy fault isolation 9. The number of cables for each type of network is: a. Mesh: n (n - 1) / 2 b. Star: n c. Ring: n - 1 d. Bus: one backbone and n drop lines 10. The general factors are size, distances (covered by the network), structure, and ownership. 11. An internet is an interconnection of networks. The Internet is the name of a specific worldwide network 12. A protocol defines what is communicated, in what way and when. This provides accurate and timely transfer of information between different devices on a network. 13. Standards are needed to create and maintain an open and competitive market for manufacturers, to coordinate protocol rules, and thus guarantee compatibility of data communication technologies. Exercises 14. Unicode uses 32 bits to represent a symbol or a character. We can define 232 different symbols or characters. 15. With 16 bits, we can represent up to 216 different colors. 16. a. Cable links: n (n - 1) / 2 = (6 ¥ 5) / 2 = 15 b. Number of ports: (n - 1) = 5 ports needed per device 17. a. Mesh topology: If one connection fails, the other connections will still be working. b. Star topology: The other devices will still be able to send data through the hub; there will be no access to the device which has the failed connection to the hub. c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line fails, only the corresponding device cannot operate. d. Ring Topology: The failed connection may disable the whole network unless it is a dual ring or there is a by-pass mechanism. 18. This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13. 19. Theoretically, in a ring topology, unplugging one station, interrupts the ring. However, most ring networks use a mechanism that bypasses the station; the ring can continue its operation. 20. In a bus topology, no station is in the path of the signal. Unplugging a station has no effect on the operation of the rest of the network. 21. See Figure 1.1 Figure 1.1 Solution to Exercise 21 22. See Figure 1.2. Figure 1.2 Solution to Exercise 22 23. a. E-mail is not an interactive application. Even if it is delivered immediately, it may stay in the mail-box of the receiver for a while. It is not sensitive to delay. b. We normally do not expect a file to be copied immediately. It is not very sensitive to delay. c. Surfing the Internet is the an application very sensitive to delay. We except to get access to the site we are searching. 24. In this case, the communication is only between a caller and the callee. A dedicated line is established between them. The connection is point-to-point. 25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communi... Chapter 2 Network Models Solutions to Review Questions and Exercises Review Questions 1. The Internet model, as discussed in this chapter, include physical, data link, network, transport, and application layers. 2. The network support layers are the physical, data link, and network layers. 3. The application layer supports the user. 4. The transport layer is responsible for process-to-process delivery of the entire message, whereas the network layer oversees host-to-host delivery of individual packets. 5. Peer-to-peer processes are processes on two or more devices communicating at a same layer 6. Each layer calls upon the services of the layer just below it using interfaces between each pair of adjacent layers. 7. Headers and trailers are control data added at the beginning and the end of each data unit at each layer of the sender and re... 8. The physical layer is responsible for transmitting a bit stream over a physical medium. It is concerned with a. physical characteristics of the media b. representation of bits c. type of encoding d. synchronization of bits e. transmission rate and mode f. the way devices are connected with each other and to the links 9. The data link layer is responsible for a. framing data bits b. providing the physical addresses of the sender/receiver c. data rate control d. detection and correction of damaged and lost frames 10. The network layer is concerned with delivery of a packet across multiple networks; therefore its responsibilities include a. providing host-to-host addressing b. routing 11. The transport layer oversees the process-to-process delivery of the entire message. It is responsible for a. dividing the message into manageable segments b. reassembling it at the destination c. flow and error control 12. The physical address is the local address of a node; it is used by the data link layer to deliver data from one node to anot... 13. The application layer services include file transfer, remote access, shared database management, and mail services. 14. The application, presentation, and session layers of the OSI model are represented by the application layer in the Internet model. The lowest four layers of OSI correspond to the Internet model layers. Exercises 15. The International Standards Organization, or the International Organization of Standards, (ISO) is a multinational body dedi... 16. a. Route determination: network layer b. Flow control: data link and transport layers c. Interface to transmission media: physical layer d. Access for the end user: application layer 17. a. Reliable process-to-process delivery: transport layer b. Route selection: network layer c. Defining frames: data link layer d. Providing user services: application layer e. Transmission of bits across the medium: physical layer 18. a. Communication with user’s application program: application layer b. Error correction and retransmission: data link and transport layers c. Mechanical, electrical, and functional interface: physical layer d. Responsibility for carrying frames between adjacent nodes: data link layer 19. a. Format and code conversion services: presentation layer b. Establishing, managing, and terminating sessions: session layer c. Ensuring reliable transmission of data: data link and transport layers d. Log-in and log-out procedures: session layer e. Providing independence from different data representation: presentation layer 20. See Figure 2.1. Figure 2.1 Solution to Exercise 20 21. See Figure 2.2. Figure 2.2 Solution to Exercise 21 22. If the corrupted destination address does not match any station address in the network, the packet is lost. If the corrupted... 23. Before using the destination address in an intermediate or the destination node, the packet goes through error checking that... 24. Most protocols issue a special error message that is sent back to the source in this case. 25. The errors between the nodes can be detected by the data link layer control, but the error at the node (between input port and output port) of the node cannot be detected by the data link layer. Chapter 3 Data and Signals Solutions to Review Questions and Exercises Review Questions 1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 2. The amplitude of a signal measures the value of the signal at any point. The frequency of a signal refers to the number of periods in one second. The phase describes the position of the waveform relative to time zero. 3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal. 4. Three types of transmission impairment are attenuation, distortion, and noise. 5. Baseband transmission means sending a digital or an analog signal without modulation using a low-pass channel. Broadband transmission means modulating a digital or an analog signal using a band-pass channel. 6. A low-pass channel has a bandwidth starting from zero; a band-pass channel has a bandwidth that does not start from zero. 7. The Nyquist theorem defines the maximum bit rate of a noiseless channel. 8. The Shannon capacity determines the theoretical maximum bit rate of a noisy channel. 9. Optical signals have very high frequencies. A high frequency means a short wave length because the wave length is inversely proportional to the frequency (l = v/f), where v is the propagation speed in the media. 10. A signal is periodic if its frequency domain plot is discrete; a signal is nonperiodic if its frequency domain plot is continuous. 11. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal. 12. An alarm system is normally periodic. Its frequency domain plot is therefore discrete. 13. This is baseband transmission because no modulation is involved. 14. This is baseband transmission because no modulation is involved. 15. This is broadband transmission because it involves modulation. Exercises 16. a. T = 1 / f = 1 / (24 Hz) = 0.0417 s = 41.7 ¥ 10-3 s = 41.7 ms b. T = 1 / f = 1 / (8 MHz) = 0.000000125 = 0.125 ¥ 10-6 s = 0.125 ms c. T = 1 / f = 1 / (140 KHz) = 0.00000714 s = 7.14 ¥ 10-6 s = 7.14 ms 17. a. f = 1 / T = 1 / (5 s) = 0.2 Hz b. f = 1 / T = 1 / (12 ms) =83333 Hz = 83.333 ¥ 103 Hz = 83.333 KHz c. f = 1 / T = 1 / (220 ns) = 4550000 Hz = 4.55¥ 106 Hz = 4.55 MHz 18. a. 90 degrees (p/2 radian) b. 0 degrees (0 radian) c. 90 degrees (p/2 radian) 19. See Figure 3.1 Figure 3.1 Solution to Exercise 19 20. We know the lowest frequency, 100. We know the bandwidth is 2000. The highest frequency must be 100 + 2000 = 2100 Hz. See Figure 3.2 Figure 3.2 Solution to Exercise 20 21. Each signal is a simple signal in this case. The bandwidth of a simple signal is zero. So the bandwidth of both signals are the same. 22. a. bit rate = 1/ (bit duration) = 1 / (0.001 s) = 1000 bps = 1 Kbps b. bit rate = 1/ (bit duration) = 1 / (2 ms) = 500 bps c. bit rate = 1/(bit duration) = 1 / (20 ms/10) = 1 / (2 ms) = 500 Kbps 23. a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms c. ((100,000 ¥ 8) / 1000) s = 800 s 24. There are 8 bits in 16 ns. Bit rate is 8 / (16 ¥ 10-9) = 0.5 ¥ 10-9 = 500 Mbps 25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz 26. The bandwidth is 5 ¥ 5 = 25 Hz. 27. The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.3. Figure 3.3 Solution to Exercise 27 28. The signal is nonperiodic, so the frequency domain is made of a continuous spectrum of frequencies as shown in Figure 3.4. Figure 3.4 Solution to Exercise 28 29. Using the first harmonic, data rate = 2 ¥ 6 MHz = 12 Mbps 30. dB = 10 log10 (90 / 100) = -0.46 dB 31. -10 = 10 log10 (P2 / 5) Æ log10 (P2 / 5) = -1 Æ (P2 / 5) = 10-1 Æ P2 = 0.5 W 32. The total gain is 3 ¥ 4 = 12 dB. The signal is amplified by a factor 101.2 = 15.85. 33. 100,000 bits / 5 Kbps = 20 s 34. 480 s ¥ 300,000 km/s = 144,000,000 km 35. 1 mm ¥ 1000 = 1000 mm = 1 mm 36. We have 4,000 log2 (1 + 1,000) a 40 Kbps 37. We have 4,000 log2 (1 + 10 / 0.005) = 43,866 bps 38. The file contains 2,000,000 ¥ 8 = 16,000,000 bits. With a 56-Kbps channel, it takes 16,000,000/56,000 = 289 s. With a 1-Mbps channel, it takes 16 s. 39. To represent 1024 colors, we need log21024 = 10 (see Appendix C) bits. The total number of bits are, therefore, 1200 ¥ 1000 ¥ 10 = 12,000,000 bits 40. We have SNR = (200 mW) / (10 ¥ 2 ¥ mW) = 10,000 SNRdB = 10 log10 SNR = 40 41. We have SNR= (signal power)/(noise power). SNR = [(signal voltage)2] / [(noise voltage)2] = [(signal voltage) / (noise voltage)]2 = 202 = 400 SNRdB = 10 log10 SNR a 26.02 42. We can approximately calculate the capacity as a. C = B ¥ (SNRdB /3) = 20 KHz ¥ (40 /3) = 267 Kbps b. C = B ¥ (SNRdB /3) = 200 KHz ¥ (4 /3) = 267 Kbps c. C = B ¥ (SNRdB /3) = 1 MHz ¥ (20 /3) = 6.67 Mbps 43. a. The data rate is doubled (C2 = 2 ¥ C1). b. When the SNR is doubled, the data rate increases slightly. We can say that, approximately, (C2 = C1 + 1). 44. We can use the approximate formula C = B ¥ (SNRdB /3) or SNRdB = (3 ¥ C) /B SNRdB = 3 ¥ 100 Kbps / 4 KHz = 75 SNR = 10 SNRdB/10 = 107.5 a 31,622,776 45. We have transmission time = (packet length)/(bandwidth) = (8,000,000 bits) / (200,000 bps) = 40 s 46. We have (bit length) = (propagation speed) ¥ (bit duration) a. Bit length = (2 ¥108 m) ¥ [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. b. Bit length = (2 ¥108 m) ¥ [(1 / (10 Mbps)] = 20 m. This means a bit occupies 20 meters on a transmission medium. c. Bit length = (2 ¥108 m) ¥ [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2 meters on a transmission medium. 47. a. Number of bits = bandwidth ¥ delay = 1 Mbps ¥ 2 ms = 2000 bits b. Number of bits = bandwidth ¥ delay = 10 Mbps ¥ 2 ms = 20,000 bits c. Number of bits = bandwidth ¥ delay = 100 Mbps ¥ 2 ms = 200,000 bits 48. We have Latency = processing time + queuing time + transmission time + propagation time Latency = 0.000010 + 0.000020 + 1 + 0.01 = 1.01000030 s Chapter 4 Digital Transmission Solutions to Review Questions and Exercises Review Questions 1. The three different techniques described in this chapter are line coding, block coding, and scrambling. 2. A data element is the smallest entity that can represent a piece of information (a bit). A signal element is the shortest uni... 3. The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps). The signal rate is the number of signal elements sent in 1s. The unit is the baud. 4. In decoding a digital signal, the incoming signal power is evaluated against the baseline (a running average of the received ... 5. When the voltage level in a digital signal is constant for a while, the spectrum creates very low frequencies, called DC components, that present problems for a system that cannot pass low frequencies. 6. A self-synchronizing digital signal includes timing information in the data being transmitted. This can be achieved if there are transitions in the signal that alert the receiver to the beginning, middle, or end of the pulse. 7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition coding. 8. Block coding provides redundancy to ensure synchronization and to provide inherent error detecting. In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. 9. Scrambling, as discussed in this chapter, is a technique that substitutes long zero- level pulses with a combination of other levels without increasing the number of bits. 10. Both PCM and DM use sampling to convert an analog signal to a digital signal. PCM finds the value of the signal amplitude for each sample; DM finds the change between two consecutive samples. 11. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time. 12. We mentioned synchronous, asynchronous, and isochronous. In both synchronous and asynchronous transmissions, a bit stream is... Exercises 13. We use the formula s = c ¥ N ¥ (1/r) for each case. We let c = 1/2. a. r = 1 Æ s = (1/2) ¥ (1 Mbps) ¥ 1/1 = 500 kbaud b. r = 1/2 Æ s = (1/2) ¥ (1 Mbps) ¥ 1/(1/2) = 1 Mbaud c. r = 2 Æ s = (1/2) ¥ (1 Mbps) ¥ 1/2 = 250 Kbaud d. r = 4/3 Æ s = (1/2) ¥ (1 Mbps) ¥ 1/(4/3) = 375 Kbaud 14. The number of bits is calculated as (0.2 /100) ¥ (1 Mbps) = 2000 bits 15. See Figure 4.1. Bandwidth is proportional to (3/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-L scheme. Figure 4.1 Solution to Exercise 15 16. See Figure 4.2. Bandwidth is proportional to (4.25/8)N which is within the range in Table 4.1 (B = 0 to N) for the NRZ-I scheme. Figure 4.2 Solution to Exercise 16 17. See Figure 4.3. Bandwidth is proportional to (12.5 / 8) N which is within the range in Table 4.1 (B = N to B = 2N) for the Manchester scheme. Figure 4.3 Solution to Exercise 17 18. See Figure 4.4. B is proportional to (12/8) N which is within the range in Table 4.1 (B = N to 2N) for the differential Manchester scheme. Figure 4.4 Solution to Exercise 18 19. See Figure 4.5. B is proportional to (5.25 / 16) N which is inside range in Table 4.1 (B = 0 to N/2) for 2B/1Q. Figure 4.5 Solution to Exercise 19 20. See Figure 4.6. B is proportional to (5.25/8) ¥ N which is inside the range in Table 4.1 (B = 0 to N/2) for MLT-3. Figure 4.6 Solution to Exercise 20 21. The data stream can be found as a. NRZ-I: 10011001. b. Differential Manchester: 11000100. c. AMI: 01110001. 22. The data rate is 100 Kbps. For each case, we first need to calculate the value f / N. We then use Figure 4.6 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 = 0 Æ P = 1.0 b. f /N = 50/100 = 1/2 Æ P = 0.5 c. f /N = 100/100 = 1 Æ P = 0.0 d. f /N = 150/100 = 1.5 Æ P = 0.2 23. The data rate is 100 Kbps. For each case, we first need to calculate the value f/N. We then use Figure 4.8 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 = 0 Æ P = 0.0 b. f /N = 50/100 = 1/2 Æ P = 0.3 c. f /N = 100/100 = 1 Æ P = 0.4 d. f /N = 150/100 = 1.5 Æ P = 0.0 24. a. The output stream is 01010 11110 11110 11110 11110 01001. b. The maximum length of consecutive 0s in the input stream is 21. c. The maximum length of consecutive 0s in the output stream is 2. 25. In 5B/6B, we have 25 = 32 data sequences and 26 = 64 code sequences. The number of unused code sequences is 64 - 32 = 32. In 3B/4B, we have 23 = 8 data sequences and 24 = 16 code sequences. The number of unused code sequences is 16 - 8 = 8. 26. See Figure 4.7. Since we specified that the last non-zero signal is positive, the first bit in our sequence is positive. Figure 4.7 Solution to Exercise 26 27. a. In a low-pass signal, the minimum frequency 0. Therefore, we have fmax = 0 + 200 = 200 KHz. Æ fs = 2 ¥ 200,000 = 400,000 samples/s b. In a bandpass signal, the maximum frequency is equal to the minimum frequency plus the bandwidth. Therefore, we have fmax = 100 + 200 = 300 KHz. Æ fs = 2 ¥ 300,000 = 600,000 samples /s 28. a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say fmax = 0 + 200 = 200 KHz Æ fs = 2 ¥ 200,000 = 400,000 samples/s nb = log21024 = 10 bits/sample N = 400 KHz ¥ 10 = 4 Mbps b. The value of nb = 10. We can easily calculate the value of SNRdB SNRdB = 6.02 ¥ nb + 1.76 = 61.96 c. The value of nb = 10. The minimum bandwidth can be calculated as BPCM = nb ¥ Banalog = 10 ¥ 200 KHz = 2 MHz 29. The maximum data rate can be calculated as Nmax = 2 ¥ B ¥ nb = 2 ¥ 200 KHz ¥ log24 = 800 kbps 30. We can first calculate the sampling rate (fs) and the number of bits per sample (nb) fmax = 0 + 4 = 4 KHz Æ fs = 2 ¥ 4 = 8000 sample/s Æ nb = 30000 / 8000 = 3.75 SNRdB = 6.02 ¥ nb + 1.72 = 25.8 31. We can calculate the data rate for each scheme: a. NRZ Æ N = 2 ¥ B = 2 ¥ 1 MHz = 2 Mbps b. Manchester Æ N = 1 ¥ B = 1 ¥ 1 MHz = 1 Mbps c. MLT-3 Æ N = 3 ¥ B = 3 ¥ 1 MHz = 3 Mbps d. 2B1Q Æ N = 4 ¥ B = 4 ¥ 1 MHz = 4 Mbps 32. a. For synchronous transmission, we have 1000 ¥ 8 = 8000 bits. b. For asynchronous transmission, we have 1000 ¥ 10 = 10000 bits. Note that we assume only one stop bit and one start bit. Some systems send more start bits. c. For case a, the redundancy is 0%. For case b, we send 2000 extra for 8000 required bits. The redundancy is 25%. Chapter 5 Analog Transmission Solutions to Review Questions and Exercises Review Questions 1. Normally, analog transmission refers to the transmission of analog signals using a band-pass channel. Baseband digital or analog signals are converted to a complex analog signal with a range of frequencies suitable for the channel. 2. A carrier is a single-frequency signal that has one of its characteristics (amplitude, frequency, or phase) changed to represent the baseband signal. 3. The process of changing one of the characteristics of an analog signal based on the information in digital data is called dig... 4. a. ASK changes the amplitude of the carrier. b. FSK changes the frequency of the carrier. c. PSK changes the phase of the carrier. d. QAM changes both the amplitude and the phase of the carrier. 5. We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency. 6. A constellation diagram can help us define the amplitude and phase of a signal element, particularly when we are using two ca... 7. The two components of a signal are called I and Q. The I component, called in- phase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis. 8. The process of changing one of the characteristics of an analog signal to represent the instantaneous amplitude of a baseband... 9. a. AM changes the amplitude of the carrier b. FM changes the frequency of the carrier c. PM changes the phase of the carrier 10. We can say that the most susceptible technique is AM because the amplitude is more affected by noise than the phase or frequency. Exercises 11. We use the formula S = (1/r) ¥ N, but first we need to calculate the value of r for each case. 12. We use the formula N = r ¥ S, but first we need to calculate the value of r for each case. 13. We use the formula r = log2L to calculate the value of r for each case. 14. See Figure 5.1. Figure 5.1 Solution to Exercise 14 a. We have two signal elements with peak amplitudes 1 and 3. The phase of both signal elements are the same, which we assume to be 0 degrees. b. We have two signal elements with the same peak amplitude of 2. However, there must be 180 degrees difference between the two phases. We assume one phase to be 0 and the other 180 degrees. c. We have four signal elements with the same peak amplitude of 3. However, there must be 90 degrees difference between each pha... d. We have four phases, which we select to be the same as the previous case. For each phase, however, we have two amplitudes, 1 ... 15. See Figure 5.2 Figure 5.2 Solution to Exercise 15 a. This is ASK. There are two peak amplitudes both with the same phase (0 degrees). The values of the peak amplitudes are A1 = 2 (the distance between the first dot and the origin) and A2= 3 (the distance between the second dot and the origin). b. This is BPSK, There is only one peak amplitude (3). The distance between each dot and the origin is 3. However, we have two phases, 0 and 180 degrees. c. This can be either QPSK (one amplitude, four phases) or 4-QAM (one amplitude and four phases). The amplitude is the distance between a point and the origin, which is (22 + 22)1/2 = 2.83. d. This is also BPSK. The peak amplitude is 2, but this time the phases are 90 and 270 degrees. 16. The number of points define the number of levels, L. The number of bits per baud is the value of r. Therefore, we use the formula r = log2L for each case. 17. We use the formula B = (1 + d) ¥ (1/r) ¥ N, but first we need to calculate the value of r for each case. 18. We use the formula N = [1/(1 + d)] ¥ r ¥ B, but first we need to calculate the value of r for each case. 19. First, we calculate the bandwidth for each channel = (1 MHz) / 10 = 100 KHz. We then find the value of r for each channel: B = (1 + d) ¥ (1/r) ¥ (N) Æ r = N / B Æ r = (1 Mbps/100 KHz) = 10 20. We can use the formula: N = [1/(1 + d)] ¥ r ¥ B = 1 ¥ 6 ¥ 6 MHz = 36 Mbps 21. 22. We calculate the number of channels, not the number of coexisting stations. Chapter 6 Bandwidth Utilization: Solutions to Review Questions and Exercises Review Questions 1. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link. 2. We discussed frequency-division multiplexing (FDM), wave-division multiplexing (WDM), and time-division multiplexing (TDM). 3. In multiplexing, the word link refers to the physical path. The word channel refers to the portion of a link that carries a transmission between a given pair of lines. One link can have many (n) channels. 4. FDM and WDM are used to combine analog signals; the bandwidth is shared. TDM is used to combine digital signals; the time is shared. 5. To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed analog signals from lo... 6. To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed digital signals from l... 7. WDM is common for multiplexing optical signals because it allows the multiplexing of signals with a very high frequency. 8. In multilevel TDM, some lower-rate lines are combined to make a new line with the same data rate as the other lines. Multiple... 9. In synchronous TDM, each input has a reserved slot in the output frame. This can be inefficient if some input lines have no d... 10. In spread spectrum, we spread the bandwidth of a signal into a larger bandwidth. Spread spectrum techniques add redundancy; ... 11. The frequency hopping spread spectrum (FHSS) technique uses M different carrier frequencies that are modulated by the source... 12. The direct sequence spread spectrum (DSSS) technique expands the bandwidth of the original signal. It replaces each data bit with n bits using a spreading code. Exercises 13. To multiplex 10 voice channels, we need nine guard bands. The required bandwidth is then B = (4 KHz) ¥ 10 + (500 Hz) ¥ 9 = 44.5 KHz 14. The bandwidth allocated to each voice channel is 20 KHz / 100 = 200 Hz. As we saw in the previous chapters, each digitized v... 15. a. Group level: overhead = 48 KHz - (12 ¥ 4 KHz) = 0 Hz. b. Supergroup level: overhead = 240 KHz - (5 ¥ 48 KHz) = 0 Hz. c. Master group: overhead = 2520 KHz - (10 ¥ 240 KHz) = 120 KHz. d. Jumbo Group: overhead = 16.984 MHz - (6 ¥ 2.52 MHz) = 1.864 MHz. 16. a. Each output frame carries 1 bit from each source plus one extra bit for synchronization. Frame size = 20 ¥ 1 + 1 = 21 bits. b. Each frame carries 1 bit from each source. Frame rate = 100,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /100,000 = 10 ms. d. Data rate = (100,000 frames/s) ¥ (21 bits/frame) = 2.1 Mbps e. In each frame 20 bits out of 21 are useful. Efficiency = 20/21= 95% 17. a. Each output frame carries 2 bits from each source plus one extra bit for synchronization. Frame size = 20 ¥ 2 + 1 = 41 bits. b. Each frame carries 2 bit from each source. Frame rate = 100,000/2 = 50,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /50,000 = 20 ms. d. Data rate = (50,000 frames/s) ¥ (41 bits/frame) = 2.05 Mbps. The output data rate here is slightly less than the one in Exercise 16. e. In each frame 40 bits out of 41 are useful. Efficiency = 40/41= 97.5%. Efficiency is better than the one in Exercise 16. 18. a. Frame size = 6 ¥ (8 + 4) = 72 bits. b. We can assume that we have only 6 input lines. Each frame needs to carry one character from each of these lines. This means that the frame rate is 500 frames/s. c. Frame duration = 1 /(frame rate) = 1 /500 = 2 ms. d. Data rate = (500 frames/s) ¥ (72 bits/frame) = 36 kbps. 19. We combine six 200-kbps sources into three 400-kbps. Now we have seven 400- kbps channel. a. Each output frame carries 1 bit from each of the seven 400-kbps line. Frame size = 7 ¥ 1 = 7 bits. b. Each frame carries 1 bit from each 400-kbps source. Frame rate = 400,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /400,000 = 2.5 ms. d. Output data rate = (400,000 frames/s) ¥ (7 bits/frame) = 2.8 Mbps. We can also calculate the output data rate as the sum of input data rate because there is no synchronizing bits. Output data rate = 6 ¥ 200 + 4 ¥ 400 = 2.8 Mbps. 20. a. The frame carries 4 bits from each of the first two sources and 3 bits from each of the second two sources. Frame size = 4 ¥ 2 + 3 ¥ 2 = 14 bits. b. Each frame carries 4 bit from each 200-kbps source or 3 bits from each 150 kbps. Frame rate = 200,000 / 4 = 150,000 /3 = 50,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /50,000 = 20 ms. d. Output data rate = (50,000 frames/s) ¥ (14 bits/frame) = 700 kbps. We can also calculate the output data rate as the sum of input data rates because there are no synchronization bits. Output data rate = 2 ¥ 200 + 2 ¥ 150 = 700 kbps. 21. We need to add extra bits to the second source to make both rates = 190 kbps. Now we have two sources, each of 190 Kbps. a. The frame carries 1 bit from each source. Frame size = 1 + 1 = 2 bits. b. Each frame carries 1 bit from each 190-kbps source. Frame rate = 190,000 frames/s. c. Frame duration = 1 /(frame rate) = 1 /190,000 = 5.3 ms. d. Output data rate = (190,000 frames/s) ¥ (2 bits/frame) = 380 kbps. Here the output bit rate is greater than the sum of the input rates (370 kbps) because of extra bits added to the second source. 22. a. T-1 line sends 8000 frames/s. Frame duration = 1/8000 = 125 ms. b. Each frame carries one extra bit. Overhead = 8000 ¥ 1 = 8 kbps 23. See Figure 6.1. Figure 6.1 Solution to Exercise 23 24. See Figure 6.2. Figure 6.2 Solution to Exercise 24 25. See Figure 6.3. Figure 6.3 Solution to Exercise 25 26. a. DS-1 overhead = 1.544 Mbps - (24 ¥ 64 kbps) = 8 kbps. b. DS-2 overhead = 6.312 Mbps - (4 ¥ 1.544 Mbps) = 136 kbps. c. DS-3 overhead = 44.376 Mbps - (7 ¥ 6.312 Mbps) = 192 kbps. d. DS-4 overhead = 274.176 Mbps - (6 ¥ 44.376 Mbps) = 7.92 Mbp