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
b. 1 MHz = 10 3 KHz
c. 1 GHz = 10 6 KHz
d. 1 THz = 10 9 KHz
47.
a. 10 KHz
b. 25.34 MHz
c. 108 x 10 6 KHz
d. 2.456764 MHz
n
10 n => 10 e.g. 10 7
49.
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
b. 0 degrees
c. 90 degrees
d. 180 degrees
52.
a. 360 or 0 degrees
b. 180 degrees
c. 270 degrees
d. 120 degrees
53.
a. 1/8 cycle
b. 1/4 cycle
c. 1/6 cycle
d. 1 cycle
54.
55.
56.
57.
See
See
See
See
is 107
Figure
Figure
Figure
Figure
4.1
4.2.
4.3.
4.4
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
5
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
75. 00100100
76. 11001001
77. 00101101
78. 01110011
79. 00011100
80. 10010010
81. 10001001
82. 01110110
83. 10100000000010
84. 00100000100100
85.
a. 1 level (plus one zero voltage)
b. 2 levels
c. 2 levels
d. 2 levels (plus zero voltage for half of each bit
interval)
e. 2 levels
f. 2 levels
88. 1/8000 = 0.125 ms
89. 8000 samples/sec
90. Two bits per sample: bit rate = 8,000 x 2 =
16,000.
91.
a. 2000
c. 6000 bps
e. 2000 bps
g. 1500 bps
bps
b. 4000 bps
d. 3000 bps
f. 2000 bps
h. 6000 bps
92. a. 1000 baud
c. 1500 baud
b. 2000 baud
d. 6000 baud
93. a. 1000 bps
c. 3000 bps
94. See Figure 5.5.
b. 1000 bps
d. 4000 bps
95. a. 0.91 x 127 = 116
=> 01110100
b. –0.25 x 127 = –32
=> 10100000
c. 0.56 x 127 = 71
=> 01000111
96. It is ASK (2 amplitudes, 1 phase) with 1 bit per baud. See
Figure 5.6.
97. It is ASK (2 amplitudes, 1 phase) with 1 bit per baud. See
Figure 5.7
98. It is PSK (1 amplitude, 2 phases) with 1 bit per baud. See
Figure 5.8
99. It is PSK (1 amplitude, 2 phases) with 1 bit per baud. See
Figure 5.9
100. It is 4-QAM (2 amplitudes, 4 phases) with 2 bits per baud.
See Figure 5.10
101. ASK
102. PSK
103. QAM
104. QAM
105. No, 12 is not a power of 2.
106. No, 18 is not a power of 2.
107. The number of points in a constellation is a power of 2.
108. Three bits per baud
109.
a. BW = 4 x 2 = 8 KHz
b. BW = 8 x 2 = 16 KHz
c. BW = (3,000 – 2,000) x 2 = 2,000 Hz = 2 KHz
110.
a. BW = 12 x 10 = 120 KHz
b. BW = 8 x 10 = 80 KHz
c. BW = 1,000 x 10 = 10 KHz
8
9