Following Contains the Solutions of the Multiple Choices from the Web on the Publisher Site and Also Solutions to Selected Book Questions

Following contains the solutions of the multiple choices from the web on the publisher site and also solutions to selected book questions.

You need to understand/analyze the answers to a multiple choice question- not merely read/ memorize the correct answers. The values in a multiple choice may be changed resulting in a different calculated answer.

Chapter 4 Web Multiple Choices Answers (check on the web and content of the book to avoid any typo in the following answers.)

1. ______signal repeats a pattern over and over again. ans: c. A periodic 2. ______signal has no repeating pattern. ans: d. An aperiodic 3. The ______wave is the simplest analog signal. ans: a. sine 4. The sine wave is an example of ______signal. ans: a. an analog 5. The amplitude of a signal can be measured in ______. ans: d. any of the above 6. On a time-domain plot, the ______of a signal is the vertical value from a point on the curve to the x-axis. ans: a. amplitude 9. A simple sine wave completes one cycle in one microsecond. Its frequency is ______. ans: a. 1 MHz 10. The period of a signal is usually expressed in ______. ans: c. seconds 11. The frequency of a signal is usually expressed in ______. ans: a. Hz 12. The ______of a signal is usually expressed in Hz. ans: b. frequency 13. The ______of a signal is usually expressed in seconds. ans: c. period 14. The frequency of a signal is inversely related to its ______. ans: b. period 15. The ______of a signal is its number of cycles per second. ans: b. frequency 16. The ______of a signal is the time it needs to complete one cycle. ans: d. period 17. The value of a simple sine wave at time zero is its maximum positive value. The phase shift is therefore ______degrees. ans: b. 90 18. The value of a simple sine wave at time zero is zero. The next value is negative. The phase shift is therefore ______degrees. ans: c. 180 19. The value of a simple sine wave at time zero is zero. The next value is positive. The phase shift is therefore ______degrees. ans: a. 0 20. A nanosecond is ______as long as a microsecond. ans: c. 0.001 21. A picosecond is ______as long as a nanosecond. ans: c. 0.001 22. Ten thousand milliseconds equal ______. ans: b. ten seconds 23. One thousand picoseconds equal ______. ans: d. one nanosecond 24. A signal with a frequency of 10 MHz has more cycles per second than a signal with a frequency of ______. ans: a. 10 KHz 25. A signal with a period of 1 microsecond has a higher frequency than a signal with a period of ______. ans: a. one millisecond 26. A signal with a period of 1 microsecond has a lower frequency than a signal with a period of ______. ans: d. b or c 27. The equivalent of 20 MHz is ______. ans: a. 20 x 106 Hz 28. A signal with a frequency of 1 GHz has more cycles per second than a signal with a frequency of ______. ans: c. one MHz 29. A sine wave has a frequency of 10 Hz. Its period is ______. ans: c. 0.1 second 30. A sine wave completes one cycle in 20 seconds. Its frequency is ______. ans: c. 0.05 Hz 31. A signal has a constant value of 10 volts. Its frequency is ______Hz. ans: a. zero 32. A simple sine wave is offset one half cycle at time zero. This is a phase shift of ______degrees. ans: d. 180 33. A simple sine wave completes one cycle in ______degrees. ans: d. 360 34. A phase shift of 180 degrees is the same as a phase shift of ______of a cycle. ans: c. one half 35. In a time-domain plot, signal amplitude is plotted against ______. ans: c. time 36. A time-domain plot shows signal ______with respect to time. ans: a. amplitude 37. In a frequency-domain plot, the signal amplitude of a simple sine wave is plotted against ______. ans: c. frequency 38. In a frequency-domain plot of a composite signal consisting of twelve sine waves (all of different frequencies and amplitudes), there are ______vertical bars. ans: c. 12 39. A signal with constant amplitude of ten volts has a frequency of ______. ans: a. 0 40. The ______of a signal is the collection of all its component frequencies. ans: b. frequency spectrum 41. The ______of a signal is the width of its frequency spectrum. ans: a. bandwidth 42. A signal is decomposed into two sine waves, one with a frequency of 10 Hz, the other with a frequency of 90 Hz. The bandwidth of the signal is ______Hz. ans: d. 80

1 43. A signal is decomposed into three sine waves with frequencies of 10, 20, and 30 Hz. The bandwidth of the signal is ______Hz. ans: b. 20 44. The bandwidth of a signal is 10 KHz. The frequency of the sine wave with the highest frequency is 11 KHz. The frequency of the sine wave with the lowest frequency is ______Hz. ans: a. 1 45. The ______is the time required to send one bit. ans: a. bit interval 46. The ______is the number of bits sent in one second. ans: b. bit rate 47. A bit interval of 0.1 seconds means a bit rate of ______bps. ans: c. 10 48. A digital signal has a bit rate of 200 bps. The bit interval is ______seconds. ans: a. 0.005 49. A bit interval of 10 milliseconds means a bit rate of ______bps. ans: d. 100 50. A digital signal has a bit rate of 50 Kbps. The bit interval is ______microseconds. ans: b. 20

Chapter 5 Web Multiple Choices Answers (check on the web and content of the book to avoid any typo in the following answers.)

1. In ______encoding one amplitude represents a 1 bit and zero amplitude represents a 0 bit (or vice versa). ans: a. unipolar 2. In ______encoding positive and negative amplitudes represent the bits. ans: b. polar 3. In ______encoding positive, negative, and zero amplitudes represent the bits. ans: c. bipolar 4. A digital signal has its 0 bits represented by 0 volts and its 1 bit represented by 5 volts. This is ______encoding. ans: a. unipolar 5. A digital signal has its 0 bits represented by 0 volts and its 1 bit represented by -5 volts or 5 volts. This is ______encoding. ans: c. bipolar 6. A digital signal has its 0 bits represented by -5 volts and its 1 bit represented by 5. This is ______encoding. ans: b. polar 7. The DC component is a serious problem for ______encoding. ans: a. unipolar 8. Unipolar encoding has a DC component because the average ______of the signal is nonzero. ans: a. amplitude 9. NRZ-L is a ______encoding method. ans: b. polar 10. NRZ-I is a ______encoding method. ans: b. polar 11. RZ is a ______encoding method. ans: b. polar 12. Manchester encoding is a ______encoding method. ans: b. polar 13. Differential Manchester encoding is a ______encoding method. ans: b. polar 14. AMI is a ______encoding method. ans: c. bipolar 17. ______encoding is superior to ______encoding because the problem of the DC component is alleviated. ans: c. Polar; unipolar 18. Ethernet LANs use ______encoding. ans: b. Manchester 19. Token Ring LANs use ______encoding. ans: c. differential Manchester 20. In ______encoding the transition between a positive and a negative voltage represents a 1 bit. ans: a. NRZ-I 21. In ______encoding halfway through each bit interval, the signal returns to zero. ans: c. RZ 22. RZ encoding requires ______signal change(s) to encode one bit. ans: c. two 23. Manchester and differential Manchester encoding are both types of ______encoding. ans: c. biphase 24. Which of the following is not a type of bipolar encoding? ans: b. RZ 30. ______is an attempt to synchronize long strings of 0s. ans: d. a and b 37. In ______conversion we are representing analog information as a series of 0s and 1s. ans: b. analog-to-digital 38. In ______, an analog signal is sampled at equal intervals, with the resulting pulses still analog in value. ans: c. PAM 39. In ______, the first step after PAM is quantization of the analog pulses. ans: d. PCM 40. The ______sampling rate is based on the Nyquist theorem. ans: b. PAM 41. A sampling rate of ______million samples per second is needed for a signal with components ranging from 10MHz to 100 MHz. ans: d. 200 42. The process of changing one of the characteristics of a carrier analog signal based on the information in a digital signal is called ______conversion. ans: c. digital-to-analog 2 43. In ______the frequency of the carrier signal is varied based on the information in a digital signal. ans: c. FSK 44. In ______the amplitude of the carrier signal is varied based on the information in a digital signal. ans: a. ASK 45. In ______the phase of the carrier signal is varied based on the information in a digital signal. ans: b. PSK 46. In ______the phase and amplitude of the carrier signal is varied based on the information in a digital signal. ans: d. QAM 47. Most modern modems use ______for digital to analog modulation. ans: d. QAM 48. ______rate is the number of bits per second; ______rate is the number of signal units per second. ans: b. Bit; baud 49. ______rate is always less than or equal to ______rate. ans: a. Baud; bit 50. If the bit rate is 1200 bps and there are 4 bits for each signal element, then the baud rate is ______. ans: d. 300 51. If the baud rate is 1200 and there are 4 bits for each signal element, then the bit rate is ______. ans: a. 4800 52. An ASK modulated signal has a bit rate of 2000 bps; the baud rate is ______. ans: a. 2000 53. A 2-PSK modulated signal has a bit rate of 2000 bps; the baud rate is ______. ans: a. 2000 54. A 4-PSK modulated signal has a bit rate of 2000 bps; the baud rate is ______. ans: b. 1000 55. An 8-PSK modulated signal has a baud rate of 2000; the bit rate is ______bps. ans: b. 6000 56. An 8-QAM modulated signal has a baud rate of 2000; the bit rate is ______bps. ans: b. 6000 57. A 32-QAM modulated signal has a baud rate of 2000; the bit rate is ______bps. ans: b. 10000 58. A 128-QAM modulated signal has a baud rate of 2000; there are ______bits per baud. ans: c. 7 59. A 256-QAM modulated signal has a bit rate of 8000; there are ______bits per baud. ans: c. 8 60. OOK is a type of ______modulation. ans: a. ASK 61. The modulation technique most affected by noise is ______. ans: a. ASK 62. For ______, the minimum bandwidth required for transmission is equal to the baud rate. ans: d. a and b 63. The minimum bandwidth for an ASK modulated signal with a baud rate of 5000 is ______Hz. ans: c. 5000 64. On a 16-QAM-constellation diagram, each constellation point represents a ______. ans: c. quadbit 65. In FM the ______of the information signal modulates the frequency of the carrier signal. ans: a. amplitude 66. In AM the ______of the information signal modulates the amplitude of the carrier signal. ans: a. amplitude 67. In PM the ______of the information signal modulates the phase of the carrier signal. ans: a. amplitude 68. In which type of modulation can the bit rate be four times the baud rate? ans: c. PSK 69. In which type of modulation can the bit rate be three times the baud rate? ans: d. None of the above 70. In which type of modulation can the bit rate be half the baud rate? ans: d. None of the above 71. In ______modulation, the bit rate is 8 times the baud rate. ans: c. 256-QAM 72. In ______modulation, the baud rate is 1/4 times the bit rate. ans: d. None of the above 73. In a dibit modulation, the number of points in the constellation is ______. ans: b. 4 74. In a tribit modulation, the number of points in the constellation is ______. ans: c. 8 75. A 4-PSK constellation is a ______modulation. ans: a. dibit 76. An 8-QAM constellation is a ______modulation. ans: b. tribit 77. The number of points in the constellation of an 8-PSK modulation is ______the number of points for an 8-QAM. ans: c. equal to 78. If the baud rate for modulation scheme A is two times the baud rate for modulation scheme B, the required bandwidth for scheme A is ______the one for scheme B. ans: a. more than 79. If the bit rate for modulation scheme A is two times the bit rate for modulation scheme B, the required bandwidth for scheme A is ______the one for scheme B. ans: d. we cannot tell

3 ANSWERS TO BOOK QUESTIONS CHAPTER 4 solutions for selected book questions. Review Questions 1. A sine wave has three characteristics: the amplitude, the period or frequency and the phase. The amplitude is the value of the signal at any point on the wave; it is the distance from a given point on the wave to the horizontal axis. The period is the time a signal needs to complete one cycle and the frequency gives the number of periods in one second. The phase indicates the status of the first cycle and describes the position of the waveform at time zero. 2. The spectrum of a signal is the set of sine waves that constitute the signal. 3. Information can be in the form of data, voice, pictures, etc. To transmit information a transformation into electromagnetic signals is necessary. 4. Analog information: singing a song, flow of time 5. Digital information: number of pages in a book, time measurement with a digital watch 6. Analog signals have an infinite range of values, while digital signals have a limited number of values. 7. Periodic signals consist of a continuously repeated pattern, whereas aperiodic signals have no repetition pattern. 8. Analog data is a set of specific points of data and all possible points in between. Digital data is a set of specific points of data with no points in between. 9. Digital signal. 10. Frequency and period are the inverse of each other. T = 1/ f and f = 1/T. 11. Seconds, milliseconds, microseconds, nanoseconds, and picoseconds. 12. Hertz, kilohertz, megahertz, gigahertz, and terahertz. 13. A high frequency signal changes value in a short period of time; there are many changes in a short time. A low frequency signal has less changes within a certain time; the signal changes slowly. 14. The amplitude of a signal measures the value of the signal at any point. 15. The frequency of a signal refers to the number of periods in one second. 16. The phase describes the position of the waveform relative to time zero. 17. The vertical axis of both plots represents the amplitude. In the time-domain plot the horizontal axis represents the time and in the frequency-domain plot, the frequency. 18. A simple periodic signal is a sine wave. A composite signal is a collection of sine waves. 19. Frequency-domain. 20. Time-domain. 21. Time-domain. 22. The bandwidth of a signal is the highest frequency minus the lowest frequency. 23. Fourier analysis (Appendix D). 24. The bit interval is the time needed to send one bit; its counterpart in analog signals is the period. 25. Bit rate refers to the number of bit intervals per second. It is equivalent to the frequency in analog signals.

4.2 MULTIPLE CHOICE QUESTIONS 26. b 27. c 28. a 29. a 30. a 31. b 32. a 33. d 34. d 35. c 36. d 37. b 38. a 39. b 40. d 41. b 42. a 43. c 44. a 45. b

EXERCISES 46. a. 1 Hz = 10 -3 KHz 47. b. 1 MHz = 10 3 KHz a. 10 KHz c. 1 GHz = 10 6 KHz b. 25.34 MHz d. 1 THz = 10 9 KHz c. 108 x 10 6 KHz d. 2.456764 MHz n 7 49. 10 n => 10 e.g. 10 7 is 10 a. 4.17 x 10 –2 s, 41.7 ms, 4.17 x 10 4us, 4.17 x 10 7 ns, 4.17 x 10 10 ps b. 1.25 x 10 –7 s, 1.25 x 10 –4 ms, 0.125 us, 1.25 x 10 2 ns, 1.25 x 10 5 ps c. 7.14 x 10 –6 s, 7.14 x 10 –3 ms, 7.14 us, 7.14 x 10 3 ns, 7.14 x 10 6 ps d. 8.33 x 10 –14 s, 8.33 x 10 –11 ms, 8.33 x 10 –8 ms, 8.33 x 10 –5 ns, 8.33 x 10 –2 ps 50. a. 0.2 Hz, 2 x 10 –4 KHz, 2 x 10 –7 MHz, 2 x 10 –10 GHz, 2 x 10 –13 THz b. 8.33 x 10 4 Hz, 83.3 KHz, 8.33 x 10 –2 MHz, 8.33 x 10 –5 GHz, 8.33 x 10 –8 THz c. 4.55 x 10 6 Hz, 4.55x 10 3 KHz, 4.55 MHz, 4.55 x 10 –3 GHz, 4.55 x 10 –6 THz d. 1.23 x 10 10 Hz, 1.23 x 10 7 KHz, 1.23 x 10 4 MHz, 12.3 GHz, 1.23 x 10 –2 THz

51. a. 90 degrees 52. 53. 54. See Figure 4.1 b. 0 degrees a. 360 or 0 degrees a. 1/8 cycle 55. See Figure 4.2. c. 90 degrees b. 180 degrees b. 1/4 cycle 56. See Figure 4.3. d. 180 degrees c. 270 degrees c. 1/6 cycle 57. See Figure 4.4 d. 120 degrees d. 1 cycle 4

58. See Figure 4.5 59. See Figure 4.6 60. See Figure 4.7 61. The bandwidth of a signal is the width of its frequency spectrum. In both cases, the frequency spectrum is not applicable, therefore the question can not be answered on this basis. 62. a. 1 Kbps b. 500 bps c. 500 Kbps d. 4 Tbps (4 ´ 10 12 bps) 63. a. 0.01 s b. 5 us c. 0.2 us d. 1 ns 64. a. 0.01 s b. 8 ms c. 800 s 65. 500 Mbps 66. 2 MHz 67. See Figure 4.8 68. 2 MHz. See Figure 4.9. 69. 25 Hz 70. 0 Hz 71. See Figure 4.10 72. See Figure 4.11

6 CHAPTER 5 solutions for selected book questions. 5.1 REVIEW QUESTIONS 1. In this text, modulation is transformation of a digital or analog signal into another analog signal; encoding is the conversion of streams of bits into a digital signal. 2. Digital-to-digital encoding is the conversion of digital information into a digital signal. 3. Analog-to-digital conversion is the transformation an analog signal into a digital signal using sampling. 4. Digital-to-analog conversion is the modulation of a digital signal into an analog signal. 5. Analog-to-analog conversion is the modulation of an analog signal into another analog signal. 6. Amplitude modulation is more susceptible to noise. 7. QAM combines both ASK and FSK and provides many combinations of amplitude and phase. Each combination can represent more than one bit. 8. Unipolar encoding uses only one voltage level. Polar encoding uses two levels, positive for 1, negative for 0 (or vice versa), and bipolar encoding uses two alternating levels for bit 1 and zero voltage for bit 0 (or vice versa). 9. The DC component is the constant portion of a signal. 10. A synchronization problem can occur when a data stream includes a long series of 1s and 0s. A timer may have trouble determining the beginning and end of each bit. 11. In NRZ-L the signal depends on the state of the bit: a positive voltage is usually a 0, and the negative a 1. In NRZ-I the signal is inverted when a 1 is encountered. 12. Manchester encoding uses the inversion at the middle of a bit interval for both synchronization and bit representation. In differential Manchester encoding the transition at the middle of a bit is used only for synchronization, while the inversion or its absence at the beginning of the bit shows the bit representation. 13. The major disadvantage of NRZ encoding is the lack of a synchronization method for long streams of 0s or 1s. Both RZ and biphase encoding feature a signal change at the middle of each bit that is used for synchronization. 14. Both methods convert digital data into digital signals. In RZ, a 1 bit is represented by positive-to-zero, and 0 by negative-to-zero, whereas in bipolar AMI a 0 is represented by a zero voltage, while 1 is represented by alternating positive and negative values. 17. a. PAM b. Quantization c. Binary encoding d. Digital-to-digital encoding 18. The higher the number of samples taken the more accurate the digital reproduction of an analog signal. 19. The higher the number of bits allotted for each sample the more precise the digital representation of the signal will be. 20. ASK, FSK, PSK, and QAM. 21. Bit rate is the number of bits transmitted during one second, whereas baud rate is the number of signal units, which can represent more than one bit, transmitted per second. In ASK both the bit and baud rates are the same. In PSK and QAM the baud rate is less than or equal to the bit rate of the signal. 22. Modulation is the process of modification of one or more characteristics of a carrier signal by an analog signal that needs to be transmitted. 23. The carrier signal is a high-frequency signal that is modulated by the information signal. 24. ASK: the bandwidth is almost equal to the baud rate. 25. FSK: the bandwidth is almost equal to the baud rate plus the frequency shift. 26. PSK: the bandwidth is almost equal to the baud rate. 27. Amplitude and phase of each signal unit; number of bits per baud. 28. QAM: the bandwidth is almost equal to the baud rate. 29. QAM is a combination of PSK and ASK. 30. PSK is based on phase shift and therefore is less susceptible to noise. 31. AM is used for analog-to-analog conversion, ASK for digital-to-analog. 32. FM is used for analog-to-analog conversion, FSK for digital-to-analog. 33. The bandwidth of an AM carrier signal is twice the bandwidth of the modulating signal, whereas the bandwidth of an FM signal is 10 times the bandwidth of the modulating signal.

5.2 MULTIPLE CHOICE QUESTIONS 34. b 35. a 36. d 37. c 38. a 39. b 40. d 41. c 42. d 43. d 44. c 45. d 46. c 47. a 48. b 49. d 50. a 51. c 52. a 53. a 54. d 55. b 56. b 57. b 58. a 59. b 60. c 61. c 62. b 63. d 64. d 65. c 66. b 67. b 68. d 69. d

7 EXERCISES

71. See Figure 5.1 95. a. 0.91 x 127 = 116 => 01110100

75. 00100100 b. –0.25 x 127 = –32 => 10100000

76. 11001001 c. 0.56 x 127 = 71 => 01000111 77. 00101101 96. It is ASK (2 amplitudes, 1 phase) with 1 bit per baud. See 78. 01110011 Figure 5.6. 79. 00011100 97. It is ASK (2 amplitudes, 1 phase) with 1 bit per baud. See 80. 10010010 Figure 5.7 81. 10001001 98. It is PSK (1 amplitude, 2 phases) with 1 bit per baud. See 82. 01110110 Figure 5.8 83. 10100000000010 99. It is PSK (1 amplitude, 2 phases) with 1 bit per baud. See 84. 00100000100100 Figure 5.9 85. 100. It is 4-QAM (2 amplitudes, 4 phases) with 2 bits per baud. a. 1 level (plus one zero voltage) See Figure 5.10 b. 2 levels 101. ASK c. 2 levels 102. PSK d. 2 levels (plus zero voltage for half of each bit 103. QAM interval) 104. QAM e. 2 levels 105. No, 12 is not a power of 2. f. 2 levels 106. No, 18 is not a power of 2.

88. 1/8000 = 0.125 ms 107. The number of points in a constellation is a power of 2.

89. 8000 samples/sec 108. Three bits per baud 90. Two bits per sample: bit rate = 8,000 x 2 = 109. 16,000. a. BW = 4 x 2 = 8 KHz b. BW = 8 x 2 = 16 KHz 91. a. 2000 bps b. 4000 bps c. BW = (3,000 – 2,000) x 2 = 2,000 Hz = 2 KHz c. 6000 bps d. 3000 bps 110. e. 2000 bps f. 2000 bps a. BW = 12 x 10 = 120 KHz g. 1500 bps h. 6000 bps b. BW = 8 x 10 = 80 KHz c. BW = 1,000 x 10 = 10 KHz 92. a. 1000 baud b. 2000 baud c. 1500 baud d. 6000 baud

93. a. 1000 bps b. 1000 bps c. 3000 bps d. 4000 bps 94. See Figure 5.5.