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Solutions to Forouzan 4th Edition of Data Communications and Networking: Chapter 3

CHAPTER 3 : Data and Signals

Solutions to Review Questions

1. 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.

2. This is baseband transmission because no modulation is involved.

3. The frequency domain of a voice signal is normally continuous because voice is a nonperiodic signal.

4. Optical signals have very high frequencies. A high frequency means a short wave length because the wave length is inversely proportional to the frequency ( = v/f), where v is the propagation speed in the media.

5. Using Fourier analysis. Fourier series gives the frequency domain of a periodic signal; Fourier analysis gives the frequency domain of a nonperiodic signal.

6. Three types of transmission impairment are attenuation, distortion, and noise.

7. This is broadband transmission because it involves modulation.

8. A low-pass channel has a bandwidth starting from zero; a band-pass channel has a
bandwidth that does not start from zero.

9. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T.

10. 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.

11. 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.

12. A signal is periodic if its frequency domain plot is discrete; a signal is nonperiodic if its frequency domain plot is continuous.

13. This is baseband transmission because no modulation is involved.

14. The Nyquist theorem defines the maximum bit rate of a noiseless channel.

15. An alarm system is normally periodic. Its frequency domain plot is therefore discrete.

Exercises

23. 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

35. The bandwidth is 5 × 5 = 25 Hz.

36. a. 90 degrees (π/2 radian)
b. 0 degrees (0 radian)
c. 90 degrees (π/2 radian)

47. 480 s × 300,000 km/s = 144,000,000 km

Solutions to Forouzan 4th Edition of Data Communications and Networking: Chapter 6

CHAPTER 6 : Bandwidth Utilization

Solutions to Review Questions

1. We discussed frequency-division multiplexing (FDM), wave-division multiplexing (WDM), and time-division multiplexing (TDM).

2. In multilevel TDM, some lower-rate lines are combined to make a new line with the same data rate as the other lines. Multiple slot TDM, on the other hand, uses multiple slots for higher data rate lines to make them compatible with the lower data rate line. Pulse stuffing TDM is used when the data rates of some lines are not an integral multiple of other lines.

3. 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.

4. 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. 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. 

6. 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.

7. Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link.

8. In spread spectrum, we spread the bandwidth of a signal into a larger bandwidth. Spread spectrum techniques add redundancy; they spread the original spectrum needed for each station. The expanded bandwidth allows the source to wrap its message in a protective envelope for a more secure transmission. We discussed frequency hopping spread spectrum (FHSS) and direct sequence spread spectrum (DSSS).

9. FDM and WDM are used to combine analog signals; the bandwidth is shared. TDM is used to combine digital signals; the time is shared.

10. To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed analog signals from lower-bandwidth lines onto higherbandwidth 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).

11. To maximize the efficiency of their infrastructure, telephone companies have traditionally multiplexed digital signals from lower data rate lines onto higher data rate lines. The digital hierarchy uses DS-0 (64 Kbps), DS-1 (1.544 Mbps), DS-2 (6.312 Mbps), DS-3 (44.376 Mbps), and DS-4 (274.176 Mbps).

12. WDM is common for multiplexing optical signals because it allows the multiplexing of signals with a very high frequency.

Exercises

18.
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 Mbps.

22.
a. T-1 line sends 8000 frames/s. Frame duration = 1/8000 = 125 μs.
b. Each frame carries one extra bit. Overhead = 8000 × 1 = 8 kbps

28.
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.

Solutions to Forouzan 4th Edition of Data Communications and Networking: Chapter 4

CHAPTER 4 : Digital Transmission

Solutions to Review Questions

1. In parallel transmission we send data several bits at a time. In serial transmission we send data one bit at a time.

2. We mentioned synchronous, asynchronous, and isochronous. In both synchronous and asynchronous transmissions, a bit stream is divided into independent frames. In synchronous transmission, the bytes inside each frame are synchronized; in asynchronous transmission, the bytes inside each frame are also independent. In isochronous transmission, there is no independency at all. All bits in the whole stream must be synchronized.

3. A data element is the smallest entity that can represent a piece of information (a bit). A signal element is the shortest unit of a digital signal. Data elements are what we need to send; signal elements are what we can send. Data elements are being carried; signal elements are the carriers.

4. 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.

5. In decoding a digital signal, the incoming signal power is evaluated against the baseline (a running average of the received signal power). A long string of 0s or 1s can cause baseline wandering (a drift in the baseline) and make it difficult for the receiver to decode correctly.

6. 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.

7. 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.

8. 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.

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.

10. The three different techniques described in this chapter are line coding, block coding, and scrambling.

11. In this chapter, we introduced unipolar, polar, bipolar, multilevel, and multitransition coding.

12. 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.

Exercises

13. The data stream can be found as
a. NRZ-I: 10011001.
b. Differential Manchester: 11000100.
c. AMI: 01110001.

19. 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 →___ P = 1.0
b. f /N = 50/100 = ½ →___ P = 0.5
c. f /N = 100/100 = 1 →___ P = 0.0
a. f /N = 0/100 = 0 →___ P = 0.0
b. f /N = 50/100 = ½ →___ P = 0.3
c. f /N = 100/100 = 1 →___ P = 0.4
d. f /N = 150/100 = 1.5 →___ P = 0.0

20. See Figure 4.7. Since we specified that the last non-zero signal is positive, the first bit in our sequence is positive.

22. 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.

23. 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.

27. We use the formula s = c × N × (1/r) for each case. We let c = ½.
a. r = 1 →___ s = (1/2) × (1 Mbps) × 1/1 = 500 kbaud
b. r = ½ →___ s = (1/2) × (1 Mbps) × 1/(1/2) = 1 Mbaud
c. r = 2 →___ s = (1/2) × (1 Mbps) × ½ = 250 Kbaud
d. r = 4/3 →___ s = (1/2) × (1 Mbps) × 1/(4/3) = 375 Kbaud

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 4.3. 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.

27. 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

Solutions to Forouzan 4th Edition of Data Communications and Networking: Chapter 1

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CHAPTER 1 : Introduction
Solutions to Review Questions

1. 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

2. 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.

3. The general factors are size, distances (covered by the network), structure, and ownership.

4. The five components of a data communication system are the sender, receiver, transmission medium, message, and protocol.

5. Advantages of a multi-point over a point-to-point configuration (type of connection) include ease of installation and low cost.

6. 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.

7. The advantages of distributed processing are security, access to distributed databases, collaborative processing, and faster problem solving.

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. An internet is an interconnection of networks. The Internet is the name of a specific worldwide network

10. Line configurations (or types of connections) are point-to-point and multipoint.

11. The three criteria are performance, reliability, and security.

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. We can divide line configuration in two broad categories:
a. Point-to-point: mesh, star, and ring.
b. Multipoint: bus

Exercises

15. 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.

16. 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.

17. 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.

20.
A. Surfing the Internet is the an application very sensitive to delay. We except to get access to the site we are searching.
B. We normally do not expect a file to be copied immediately. It is not very sensitive to delay.
C. 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.

21. 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.

22. A. Mesh topology: If one connection fails, the other connections will still be working.

24. Unicode uses 32 bits to represent a symbol or a character. We can define 232 different symbols or characters.

25. This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13.

Solutions to Forouzan 4th Edition of Data Communications and Networking: Chapter 23

CHAPTER 23 : Process-to-Process Delivery

Solutions to Selected Review Questions

1. Since the length of a datagram must be contained in a 2 byte field, the maximum size of a UDP datagram is 65,535 bytes (header plus data). However, given that the IP layer must also store the total length of the packet in a 2 byte field, the maximum length would be 20 bytes less than this, or 65,515 bytes, to leave room for the IP header. The implementation may impose a smaller limit than this.

2. The minimum size of a UDP datagram is 8 bytes at the transport layer and 28 bytes at the IP layer. This size datagram would contain no data–only an IP header with no options and a UDP header. The implementation may require padding.

3. a. The FIN bit is set. This is a FIN segment request to terminate the connection.
b. None of the control bits are set. The segment is part of a data transmission without piggybacked acknowledgment.
c. The ACK and the FIN bits are set. This is a FIN+ACK in response to a received FIN segment.

4. The largest amount of process data that can be encapsulated in a UDP datagram is 65,507 bytes. (65,535 minus 8 bytes for the UDP header minus 20 bytes for the IP header). The implementation may impose a smaller limit than this.

5. Reliability is not of primary importance in applications such as echo, daytime, BOOTP, TFTP and SNMP. In custom software, reliability can be built into the client/server applications to provide a more reliable, low overhead service.

6. Port addresses do not need to be universally unique as long as each IP address/port address pair uniquely identify a particular process running on a particular host. A good example would be a network consisting of 50 hosts, each running echo server software. Each server uses the well known port number 7, but the IP address, together with the port number of 7, uniquely identify a particular server program on a particular host. Port addresses are shorter than IP addresses because their domain, a single system, is smaller than the domain of IP addresses, all systems on the Internet.

7. Ephemeral is defined as short-lived or transitory. Ephemeral port numbers are only used for the duration of a single communication between client and server, so they are indeed short-lived.

8. The maximum size of the TCP header is 60 bytes. The minimum size of the TCP header is 20 bytes.

9. IP and UDP are both connectionless and unreliable protocols. The main difference in their reliability is that IP only calculates a checksum for the IP header and not for the data while UDP calculates a checksum for the entire datagram.

10. The smallest amount of process data that can be encapsulated in a UDP datagram is 0 bytes.

11. See Table 23.1. Table 23.1 Answer to the Question 11.

12. UDP is preferred because each user datagram can be used for each chunk of data. However, a better solution is SCTP.

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