1. Unit 2

2. Overview
Bandwidth
Shannon’s theorem

3. Overview Bandwidth Shannon’s theorem Communication systems
Analog Modulation AM FM
Digital Modulation
ASK
FSK

4. Bandwidth
we will discuss more deeply what signal bandwidth is, what the meaning of channel bandwidth to a communications engineer is, and what the limitations on information rate are.
Signal bandwidth:
We can divide signals into two categories: The pure tone signal (the sinusoidal wave, consisting of one frequency component), and complex signals that are composed of several components, or sinusoids of various frequencies.
T=1x10-3 s
f=1/1x10-3 =1000Hz=1 kHz
t (ms) 1
Pure signal

5. Approaching a 150 Hz square wave
The bandwidth of a signal composed of components of various frequencies (complex signal) is the difference between its highest and lowest frequency components, and is expressed in Hertz (Hz) - the same as frequency.
For example, a square wave may be constructed by adding sine waves of various frequencies:
The resulting wave resembles a square wave. If more sine waves of other frequencies were added, the resulting waveform would more closely resemble a square wave
Since the resulting wave contains 2 frequency components, its bandwidth is around =300 Hz.
Pure tone
150 Hz sine wave
Pure tone
450 Hz sine wave
Approaching a 150 Hz square wave
(ms)

6. Male voice
Since voice signals are also composed of several components (pure tones) of various frequencies, the bandwidth of a voice signal is taken to be the difference between the highest and lowest frequencies which are 3000 Hz and (close to) 0 Hz
Although other frequency components above 3000 Hz exist, (they are more prominent in the male voice), an acceptable degradation of voice quality is achieved by disregarding the higher frequency components, accepting the 3kHz bandwidth as a standard for voice communications
3000 Hz frequency component
Female voice
3000 Hz frequency component

7. channel bandwidth:
The bandwidth of a channel (medium) is defined to be the range of frequencies that the medium can support. Bandwidth is measured in Hz
With each transmission medium, there is a frequency range of electromagnetic waves that can be transmitted:
Twisted pair cable: 0 to 109 Hz (Bandwidth : 109 Hz)
Coax cable: 0 to 1010 Hz (Bandwidth : 1010 Hz)
Optical fiber: 1014 to 1016 Hz (Bandwidth : = 9.9x1015 Hz)
Optical fibers have the highest bandwidth (they can support electromagnetic waves with very high frequencies, such as light waves)
The bandwidth of the channel dictates the information carrying capacity of the channel
This is calculated using Shannon’s channel capacity formula
Increasing bandwidth

8. Shannon’s Theorem (Shannon’s Limit for Information Capacity)
Claude Shannon at Bell Labs figured out how much information a channel could theoretically carry:
I = B log2 (1 + S/N)
Where I is Information Capacity in bits per second (bps)
B is the channel bandwidth in Hz
S/N is Signal-to-Noise ratio (SNR: unitless…don’t make into decibel: dB)
Note that the log is base 2!

9. Signal-to-Noise Ratio
S/N is normally measured in dB (decibel). It is a relationship between the signal we want versus the noise that we do not want, which is in the medium.
It can be thought of as a fractional relationship (that is, before we take the logarithm):
1000W of signal power versus 20W of noise power is either:
1000/20=50 (unitless!)
or: about 17 dB ==> 10 log /20 = dB

10. Communication systems
Digital
Analog
The block diagram on the top shows the blocks common to all communication systems

11. We recall the components of a communication system:
Input transducer: The device that converts a physical signal from source to an electrical, mechanical or electromagnetic signal more suitable for communicating
Transmitter: The device that sends the transduced signal
Transmission channel: The physical medium on which the signal is carried
Receiver: The device that recovers the transmitted signal from the channel
Output transducer: The device that converts the received signal back into a useful quantity

12. Analog Modulation
The purpose of a communication system is to transmit information signals (baseband signals) through a communication channel
The term baseband is used to designate the band of frequencies representing the original signal as delivered by the input transducer
For example, the voice signal from a microphone is a baseband signal, and contains frequencies in the range of Hz
The “hello” wave is a baseband signal:

13. Since this baseband signal must be transmitted through a communication channel (such as air or cable) using electromagnetic waves, a procedure is needed to shift the range of baseband frequencies to other frequency ranges suitable for transmission; and, a corresponding shift back to the original frequency range after reception. This is called the process of modulation and demodulation
Remember the radio spectrum:
For example, an AM radio system transmits electromagnetic waves with frequencies of around a few hundred kHz (MF band)
The FM radio system operates with frequencies in the range of MHz (VHF band)
AM radio
FM radio/TV

14. Since the baseband signal contains frequencies in the audio frequency range (3 kHz), some form of frequency-band shifting must be employed for the radio system to operate properly
This process is accomplished by a device called a modulator
The transmitter block in any communications system contains the modulator device
The receiver block in any communications system contains the demodulator device
The modulator modulates a carrier wave (the electromagnetic wave) which has a frequency that is selected from an appropriate band in the radio spectrum
For example, the frequency of a carrier wave for FM can be chosen from the VHF band of the radio spectrum
For AM, the frequency of the carrier wave may be chosen to be around a few hundred kHz (from the MF band of the radio spectrum)
The demodulator extracts the original baseband signal from the received modulated signal
In Summary:
Modulation is the process of impressing a low-frequency information signal (baseband signal) onto a higher frequency carrier signal

15. Basic analog communications system
Baseband signal (electrical signal)
EM waves (modulated signal)
Transmitter
Input transducer
Transmission Channel
Modulator
EM waves (modulated signal)
Carrier
Baseband signal (electrical signal)
Receiver
Output transducer
Demodulator

16. Types of Analog Modulation
Amplitude Modulation (AM)
Amplitude modulation is the process of varying the amplitude of a carrier wave in proportion to the amplitude of a baseband signal. The frequency of the carrier remains constant
Frequency Modulation (FM)
Frequency modulation is the process of varying the frequency of a carrier wave in proportion to the amplitude of a baseband signal. The amplitude of the carrier remains constant
Phase Modulation (PM)
Another form of analog modulation technique which we will not discuss

17. Amplitude Modulation Carrier wave Baseband signal Modulated wave
Amplitude varying-frequency constant

18. Frequency Modulation Carrier wave Baseband signal Modulated wave
Large amplitude: high frequency
Baseband signal
Small amplitude: low frequency
Modulated wave
Frequency varying-amplitude constant

19. AM vs. FM AM requires a simple circuit, and is very easy to generate.
It is simple to tune, and is used in almost all short wave broadcasting.
The area of coverage of AM is greater than FM (longer wavelengths (lower frequencies) are utilized-remember property of HF waves?)
However, it is quite inefficient, and is susceptible to static and other forms of electrical noise.
The main advantage of FM is its audio quality and immunity to noise. Most forms of static and electrical noise are naturally AM, and an FM receiver will not respond to AM signals.
The audio quality of a FM signal increases as the frequency deviation increases (deviation from the center frequency), which is why FM broadcast stations use such large deviation.
The main disadvantage of FM is the larger bandwidth it requires

20. Digital Modulation
The previous section presented analog communication systems that transmit information in analog form using Amplitude or Frequency modulation
Digital communication systems also employ modulation techniques, some of which include:
Amplitude Shift Keying
Frequency Shift Keying
Phase Shift Keying

21. Basic digital communications system
Transmitter
EM waves (modulated signal)
Digital signal
Analog signal
Error correction coding
A/D converter
Transmission Channel
Input transducer
Modulator
Carrier
EM waves (modulated signal)
Receiver
analog signal
digital signal
Error detection/ correction
Output transducer
D/A converter
Demodulator

22. Some Types of Digital Modulation
Amplitude Shift Keying (ASK)
The most basic (binary) form of ASK involves the process of switching the carrier either on or off, in correspondence to a sequence of digital pulses that constitute the information signal. One binary digit is represented by the presence of a carrier, the other binary digit is represented by the absence of a carrier. Frequency remains fixed
Frequency Shift Keying (FSK)
The most basic (binary) form of FSK involves the process of varying the frequency of a carrier wave by choosing one of two frequencies (binary FSK) in correspondence to a sequence of digital pulses that constitute the information signal. Two binary digits are represented by two frequencies around the carrier frequency. Amplitude remains fixed
Phase Shift Keying (PSK)
Another form of digital modulation technique which we will discuss

23. Amplitude Shift Keying
Digital information
Carrier wave
ASK modulated signal
Amplitude varying-frequency constant
Carrier present
Carrier absent

24. Frequency Shift Keying
Digital information
Carrier 1 (frequency #1)
Carrier 2 (frequency #2)
FSK modulated signal
Frequency varying-amplitude constant

25. Topics discussed in this section:
DATA RATE LIMITS
A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
Topics discussed in this section:
Noiseless Channel: Nyquist Bit Rate
Noisy Channel: Shannon Capacity
Using Both Limits

26. Note
Increasing the levels of a signal increases the probability of an error occurring, in other words it reduces the reliability of the system. Why??

27. Capacity of a System
The bit rate of a system increases with an increase in the number of signal levels we use to denote a symbol.
A symbol can consist of a single bit or “n” bits.
The number of signal levels = 2n.
As the number of levels goes up, the spacing between level decreases -> increasing the probability of an error occurring in the presence of transmission impairments.

28. Nyquist Theorem
Nyquist gives the upper bound for the bit rate of a transmission system by calculating the bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth of the system (assuming 2 symbols/per cycle and first harmonic).
Nyquist theorem states that for a noiseless channel:
C = 2 B log22n
C= capacity in bps
B = bandwidth in Hz

29. Example 3.33
Does the Nyquist theorem bit rate agree with the intuitive bit rate described in baseband transmission?
Solution
They match when we have only two levels. We said, in baseband transmission, the bit rate is 2 times the bandwidth if we use only the first harmonic in the worst case. However, the Nyquist formula is more general than what we derived intuitively; it can be applied to baseband transmission and modulation. Also, it can be applied when we have two or more levels of signals.

30. Example 3.34
Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

31. Example 3.35
Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as

32. Example 3.36
We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?
Solution
We can use the Nyquist formula as shown:
Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.

33. Shannon’s Theorem
Shannon’s theorem gives the capacity of a system in the presence of noise.
C = B log2(1 + SNR)

34. Example 3.37
Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as
This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.

35. Example 3.38
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of The signal-to-noise ratio is usually For this channel the capacity is calculated as
This means that the highest bit rate for a telephone line is kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.

36. Example 3.39
The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as

37. Example 3.40
For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be simplified to
For example, we can calculate the theoretical capacity of the previous example as

38. Example 3.41
We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?
Solution
First, we use the Shannon formula to find the upper limit.

39. Example 3.41 (continued)
The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels.

40. Note
The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.

41. 5.1 Modulation of Digital Data
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Quadrature Amplitude Modulation
Bit/Baud Comparison

42. Figure 5.1 Digital-to-analog modulation

43. Figure 5.2 Types of digital-to-analog modulation

44. Note:
Bit rate is the number of bits per second. Baud rate is the number of signal units per second. Baud rate is less than or equal to the bit rate.

45. Example 1
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate
Solution
Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bps

46. Example 2
The bit rate of a signal is If each signal unit carries 6 bits, what is the baud rate?
Solution
Baud rate = 3000 / 6 = 500 baud/s

47. Figure ASK

48. Figure 5.4 Relationship between baud rate and bandwidth in ASK

49. Example 3
Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex.
Solution
In ASK the baud rate and bit rate are the same. The baud rate is therefore An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.

50. Example 4
Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate?
Solution
In ASK the baud rate is the same as the bandwidth, which means the baud rate is But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.

51. Example 5
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the bandwidths in each direction. Assume there is no gap between the bands in the two directions.
Solution
For full-duplex ASK, the bandwidth for each direction is
BW = / 2 = 5000 Hz
The carrier frequencies can be chosen at the middle of each band (see Fig. 5.5).
fc (forward) = /2 = 3500 Hz
fc (backward) = – 5000/2 = 8500 Hz

52. Figure 5.5 Solution to Example 5

53. Figure FSK

54. Figure 5.7 Relationship between baud rate and bandwidth in FSK

55. Example 6
Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz.
Solution
For FSK
BW = baud rate + fc1 - fc0
BW = bit rate + fc1 - fc0 = = 5000 Hz

56. Example 7
Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode.
Solution
Because the transmission is full duplex, only 6000 Hz is allocated for each direction.
BW = baud rate + fc1 - fc0
Baud rate = BW - (fc1 - fc0 ) = = 4000
But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

57. Figure PSK

58. Figure 5.9 PSK constellation

59. Figure The 4-PSK method

60. Figure 5.11 The 4-PSK characteristics

61. Figure 5.12 The 8-PSK characteristics

62. Figure 5.13 Relationship between baud rate and bandwidth in PSK

63. Example 8
Find the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode.
Solution
For PSK the baud rate is the same as the bandwidth, which means the baud rate is But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

64. Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?
Solution
For PSK the baud rate is the same as the bandwidth, which means the baud rate is But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

65. Note:
Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved.

66. Figure 5.14 The 4-QAM and 8-QAM constellations

67. Figure 5.15 Time domain for an 8-QAM signal

68. Figure 5.16 16-QAM constellations

69. Figure Bit and baud

70. Table 5.1 Bit and baud rate comparison
Modulation
Units
Bits/Baud
Baud rate
Bit Rate
ASK, FSK, 2-PSK
Bit 1 N
4-PSK, 4-QAM
Dibit 2 2N
8-PSK, 8-QAM
Tribit 3 3N
16-QAM
Quadbit 4 4N
32-QAM
Pentabit 5 5N
64-QAM
Hexabit 6 6N
128-QAM
Septabit 7 7N
256-QAM
Octabit 8 8N

71. Example 10
A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?
Solution
The constellation indicates 8-PSK with the points 45 degrees apart. Since 23 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud

72. Example 11
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution
A 16-QAM signal has 4 bits per signal unit since
log216 = 4.
Thus,
(1000)(4) = 4000 bps

73. Example 12
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution
A 64-QAM signal has 6 bits per signal unit since
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud

74. 5.2 Telephone Modems
Modem Standards

75. Note:
A telephone line has a bandwidth of almost 2400 Hz for data transmission.

76. Figure 5.18 Telephone line bandwidth

77. Modem stands for modulator/demodulator.
Note:
Modem stands for modulator/demodulator.

78. Figure 5.19 Modulation/demodulation

79. Figure 5.20 The V.32 constellation and bandwidth

80. Figure 5.21 The V.32bis constellation and bandwidth

81. Figure 5.22 Traditional modems

82. Figure K modems

83. Topics discussed in this section:
5-1 DIGITAL-TO-ANALOG CONVERSION
Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.
Topics discussed in this section:
Aspects of Digital-to-Analog Conversion Amplitude Shift Keying Frequency Shift Keying
Phase Shift Keying
Quadrature Amplitude Modulation

84. Figure 5.1 Digital-to-analog conversion

85. Figure 5.2 Types of digital-to-analog conversion

86. Note
Bit rate is the number of bits per second. Baud rate is the number of signal
elements per second.
In the analog transmission of digital data, the baud rate is less than or equal to the bit rate.

87. Example 5.1
An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate.
Solution
In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from

88. Example 5.2
An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need?
Solution
In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.

89. Figure 5.3 Binary amplitude shift keying

90. Figure 5.4 Implementation of binary ASK

91. Example 5.3
We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?
Solution
The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).

92. Example 5.4
In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction.

93. Figure 5.5 Bandwidth of full-duplex ASK used in Example 5.4

94. Figure 5.6 Binary frequency shift keying

95. Example 5.5
We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1?
Solution
This problem is similar to Example 5.3, but we are modulating by using FSK. The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means

96. Figure 5.7 Bandwidth of MFSK used in Example 5.6

97. Example 5.6
We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.
Solution
We can have L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000 Mbaud. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1000 = Figure 5.8 shows the allocation of frequencies and bandwidth.

98. Figure 5.8 Bandwidth of MFSK used in Example 5.6

99. Figure 5.9 Binary phase shift keying

100. Figure 5.10 Implementation of BASK

101. Figure 5.11 QPSK and its implementation

102. Example 5.7
Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0.
Solution
For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz.

103. Figure 5.12 Concept of a constellation diagram

104. Figure 5.13 shows the three constellation diagrams.
Example 5.8
Show the constellation diagrams for an ASK (OOK), BPSK, and QPSK signals.
Solution
Figure 5.13 shows the three constellation diagrams.

105. Figure 5.13 Three constellation diagrams

106. Quadrature amplitude modulation is a combination of ASK and PSK.
Note
Quadrature amplitude modulation is a combination of ASK and PSK.

107. Figure 5.14 Constellation diagrams for some QAMs

108. Figure 5.20 Phase modulation

109. Note
The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B.