TEST BANK FOR Data Communications and Networking By Behrouz Forouzan (Solution Manual)

Exam (elaborations) TEST BANK FOR Data Communications and Networking By Behrouz Forouzan (Solution Manual) CHAPTER 1 Introduction Solutions to Odd-Numbered Review Questions and Exercises Review Questions 1. The five components of a data communication system are the sender, receiver, transmission medium, message, and protocol. 3. The three criteria are performance, reliability, and security. 5. Line configurations (or types of connections) are point-to-point and multipoint. 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. 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 11. An internet is an interconnection of networks. The Internet is the name of a specific worldwide 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 15. With 16 bits, we can represent up to 216 different colors. 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. 2 d. Ring Topology: The failed connection may disable the whole network unless it is a dual ring or there is a by-pass mechanism. 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. 21. See Figure 1.1 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. 25. The telephone network was originally designed for voice communication; the Internet was originally designed for data communication. The two networks are similar in the fact that both are made of interconnections of small networks. The telephone network, as we will see in future chapters, is mostly a circuit-switched network; the Internet is mostly a packet-switched network. Figure 1.1 Solution to Exercise 21 Station Station Station Repeat er Station Station Station Repeat er Station Station Station Repeater Hub 1 CHAPTER 2 Network Models Solutions to Odd-Numbered Review Questions and Exercises Review Questions 1. The Internet model, as discussed in this chapter, include physical, data link, network, transport, and application layers. 3. The application layer supports the user. 5. Peer-to-peer processes are processes on two or more devices communicating at a same layer 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 removed at the corresponding layers of the receiver. They provide source and destination addresses, synchronization points, information for error detection, etc. 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 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 13. The application layer services include file transfer, remote access, shared database management, and mail services. Exercises 15. The International Standards Organization, or the International Organization of Standards, (ISO) is a multinational body dedicated to worldwide agreement on international standards. An ISO standard that covers all aspects of network communications is the Open Systems Interconnection (OSI) model. 2 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 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 21. See Figure 2.1. 23. Before using the destination address in an intermediate or the destination node, the packet goes through error checking that may help the node find the corruption (with a high probability) and discard the packet. Normally the upper layer protocol will inform the source to resend the packet. 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. Figure 2.1 Solution to Exercise 21 B/42 C/82 A/40 Sender Sender LAN1 LAN2 R1 D/80 42 40 A D i j Data T2 80 82 A D i j Data T2 1 CHAPTER 3 Data and Signals Solutions to Odd-Numbered Review Questions and Exercises Review Questions 1. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 3. Using Fourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal. 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. 7. The Nyquist theorem defines the maximum bit rate of a noiseless 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 (λ = v/f), where v is the propagation speed in the media. 11. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal. 13. This is baseband transmission because no modulation is involved. 15. This is broadband transmission because it involves modulation. Exercises 17. a. f = 1 / T = 1 / (5 s) = 0.2 Hz b. f = 1 / T = 1 / (12 μs) =83333 Hz = 83.333 × 103 Hz = 83.333 KHz c. f = 1 / T = 1 / (220 ns) = Hz = 4.55× 106 Hz = 4.55 MHz 19. See Figure 3.1 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. 23. a. (10 / 1000) s = 0.01 s b. (8 / 1000) s = 0. 008 s = 8 ms 2 c. ((100,000 × 8) / 1000) s = 800 s 25. The signal makes 8 cycles in 4 ms. The frequency is 8 /(4 ms) = 2 KHz 27. The signal is periodic, so the frequency domain is made of discrete frequencies. as shown in Figure 3.2. 29. Using the first harmonic, data rate = 2 × 6 MHz = 12 Mbps Using three harmonics, data rate = (2 × 6 MHz) /3 = 4 Mbps Using five harmonics, data rate = (2 × 6 MHz) /5 = 2.4 Mbps 31. –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W 33. 100,000 bits / 5 Kbps = 20 s 35. 1 μm × 1000 = 1000 μm = 1 mm 37. We have 4,000 log2 (1 + 10 / 0.005) = 43,866 bps 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 41. We have SNR= (signal power)/(noise power). However, power is proportional to the square of voltage. This means we have Figure 3.1 Solution to Exercise 19 Figure 3.2 Solution to Exercise 27 Frequency domain Bandwidth = 200 − 0 = 200 Amplitude 10 volts Frequency 30 KHz 10 KHz ... 3 SNR = [(signal voltage)2] / [(noise voltage)2] = [(signal voltage) / (noise voltage)]2 = 202 = 400 We then have SNRdB = 10 log10 SNR ≈ 26.02 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). 45. We have transmission time = (packet length)/(bandwidth) = (8,000,000 bits) / (200,000 bps) = 40 s 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 4 1 CHAPTER 4 Digital Transmission Solutions to Odd-Numbered Review Questions and Exercises Review Questions 1. The three different techniques described in this chapter are line coding, block coding, and scrambling. 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. 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. 7. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition coding. 9. Scrambling, as discussed in this chapter, is a technique that substitutes long zerolevel pulses with a combination of other levels without increasing the number of bits. 11. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time. Exercises 13. We use the formula s = c × N × (1/r) for each case. We let c = 1/2. a. r = 1 → s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud b. r = 1/2 → s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud c. r = 2 → s = (1/2) × (1 Mbps) × 1/2 = 250 Kbaud d. r = 4/3 → s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud 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. 17. See Figure 4.2. 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. 2 19. See Figure 4.3. B is proportional to (5.25 / 16) N which is inside range in Table 4.1 (B = 0 to N/2) for 2B/1Q. 21. The data stream can be found as a. NRZ-I: . b. Differential Manchester: . c. AMI: . 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 → P = 0.0 b. f /N = 50/100 = 1/2 → P = 0.3 c. f /N = 100/100 = 1 → P = 0.4 d. f /N = 150/100 = 1.5 → P = 0.0 Figure 4.1 Solution to Exercise 15 Figure 4.2 Solution to Exercise 17 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 Case a Case b Case c Case d Average Number of Changes = (0 + 0 + 8 + 4) / 4 = 3 for N = 8 B (3 / 8) N 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 Case a Case b Case c Case d Average Number of Changes = (15 + 15+ 8 + 12) / 4 = 12.5 for N = 8 B (12.5 / 8) N 3 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. 27 a. In a low-pass signal, the minimum frequency 0. Therefore, we have fmax = 0 + 200 = 200 KHz. → 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. → fs = 2 × 300,000 = 600,000 samples /s 29. The maximum data rate can be calculated as Nmax = 2 × B × nb = 2 × 200 KHz × log24 = 800 kbps 31. We can calculate the data rate for each scheme: Figure 4.3 Solution to Exercise 19 a. NRZ → N = 2 × B = 2 × 1 MHz = 2 Mbps b. Manchester → N = 1 × B = 1 × 1 MHz = 1 Mbps c. MLT-3 → N = 3 × B = 3 × 1 MHz = 3 Mbps d. 2B1Q → N = 4 × B = 4 × 1 MHz = 4 Mbps 11 11 11 11 11 11 11 11 10 +3 +1 −3 −1 +3 +1 −3 −1 +3 +1 −3 −1 11 00 11 +3 +1 −3 −1 Case a Case b Case c Case d Average Number of Changes = (0 + 7 + 7 + 7) / 4 = 5.25 for N = 16 B (5.25 / 8) N 4 1 CHAPTER 5 Analog Transmission Solutions to Odd-Numbered 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. 3. The process of changing one of the characteristics of an analog signal based on the information in digital data is called digital-to-analog conversion. It is also called modulation of a digital signal. The baseband digital signal representing the digital data modulates the carrier to create a broadband analog signal. 5. We can say that the most susceptible technique is ASK because the amplitude is more affected by noise than the phase or frequency. 7. The two components of a signal are called I and Q. The I component, called inphase, is shown on the horizontal axis; the Q component, called quadrature, is shown on the vertical axis. 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 Exercises 11. We use the formula S = (1/r) × N, but first we need to calculate the value of r for each case. a. r = log22 = 1 → S = (1/1) × (2000 bps) = 2000 baud b. r = log22 = 1 → S = (1/1) × (4000 bps) = 4000 baud c. r = log24 = 2 → S = (1/2) × (6000 bps) = 3000 baud d. r = log264 = 6 → S = (1/6) × (36,000 bps) = 6000 baud 2 13. We use the formula r = log2L to calculate the value of r for each case. 15. See Figure 5.1 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. 17. We use the formula B = (1 + d) × (1/r) × N, but first we need to calculate the value of r for each case. a. log24 = 2 b. log28 = 3 c. log24 = 2 d. log2128 = 7 Figure 5.1 Solution to Exercise 15 a. r = 1 → B= (1 + 1) × (1/1) × (4000 bps) = 8000 Hz b. r = 1 → B = (1 + 1) × (1/1) × (4000 bps) + 4 KHz = 8000 Hz c. r = 2 → B = (1 + 1) × (1/2) × (4000 bps) = 2000 Hz d. r = 4 → B = (1 + 1) × (1/4) × (4000 bps) = 1000 Hz 2 3 –3 3 –2 –2 2 2 –2 2 a. b. I I I Q Q Q c. d. I Q 3 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) → r = N / B → r = (1 Mbps/100 KHz) = 10 We can then calculate the number of levels: L = 2r = 210 = 1024. This means that that we need a 1024-QAM technique to achieve this data rate. 21. a. BAM = 2 × B = 2 × 5 = 10 KHz b. BFM = 2 × (1 + β) × B = 2 × (1 + 5) × 5 = 60 KHz c.BPM = 2 × (1 + β) × B = 2 × (1 + 1) × 5 = 20 KHz 4 1 CHAPTER 6 Bandwidth Utilization: Solutions to Odd-Numbered 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. 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. 5. To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed analog signals from lower-bandwidth lines onto higher-bandwidth lines. The analog hierarchy uses voice channels (4 KHz), groups (48 KHz), supergroups (240 KHz), master groups (2.4 MHz), and jumbo groups (15.12 MHz). 7. WDM is common for multiplexing optical signals because it allows the multiplexing of signals with a very high frequency. 9. In synchronous TDM, each input has a reserved slot in the output frame. This can be inefficient if some input lines have no data to send. In statistical TDM, slots are dynamically allocated to improve bandwidth efficiency. Only when an input line has a slot’s worth of data to send is it given a slot in the output frame. 11. The frequency hopping spread spectrum (FHSS) technique uses M different carrier frequencies that are modulated by the source signal. At one moment, the signal modulates one carrier frequency; at the next moment, the signal modulates another carrier frequency. 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 15. a. Group level: overhead = 48 KHz − (12 × 4 KHz) = 0 Hz. b. Supergroup level: overhead = 240 KHz − (5 × 48 KHz) = 0 Hz. 2 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. 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 μs. 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. 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 μs. 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. 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 μs. 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. 23. See Figure 6.1. 25. See Figure 6.2. Figure 6.1 Solution to Exercise 23 O L E L Y I E B H H TDM 3 27. The number of hops = 100 KHz/4 KHz = 25. So we need log225 = 4.64 ≈ 5 bits 29. Random numbers are 11, 13, 10, 6, 12, 3, 8, 9 as calculated below: Figure 6.2 Solution to Exercise 25 N1 = 11 N2 =(5 +7 × 11) mod 17 − 1 = 13 N3 =(5 +7 × 13) mod 17 − 1 = 10 N4 =(5 +7 × 10) mod 17 − 1 = 6 N5 =(5 +7 × 6) mod 17 − 1 = 12 N6 =(5 +7 × 12) mod 17 − 1 = 3 N7 =(5 +7 × 3) mod 17 − 1 = 8 N8 =(5 +7 × 8) mod 17 − 1 = 9 000 111 TDM 4 1 CHAPTER 7 Transmission Media Solutions to Odd-Numbered Review Questions and Exercises Review Questions 1. The transmission media is located beneath the physical layer and controlled by the physical layer. 3