Lecture 3.ppt

4.2
4-1 DIGITAL-TO-DIGITAL CONVERSION
In this section, we see how we can represent digital
data by using digital signals. The conversion involves
three techniques: line coding, block coding, and
scrambling. Line coding is always needed; block
coding and scrambling may or may not be needed.
Line Coding
Block Coding
Topics discussed in this section:

4.3
Figure 4.1 Line coding and decoding

4.4
Figure 4.2 Signal element versus data element

4.5
A signal is carrying data in which one data element is
encoded as one signal element ( r = 1). If the bit rate is
100 kbps, what is the average value of the baud rate if c is
between 0 and 1?
Solution
We assume that the average value of c is 1/2 . The baud
rate is then
Example 4.1

4.6
Although the actual bandwidth of a
digital signal is infinite, the effective
bandwidth is finite.
Note

4.7
The maximum data rate of a channel (see Chapter 3) is
Nmax = 2 × B × log2 L (defined by the Nyquist formula).
Does this agree with the previous formula for Nmax?
Solution
A signal with L levels actually can carry log2L bits per
level. If each level corresponds to one signal element and
we assume the average case (c = 1/2), then we have
Example 4.2

4.8
Figure 4.3 Effect of lack of synchronization

4.9
In a digital transmission, the receiver clock is 0.1 percent
faster than the sender clock. How many extra bits per
second does the receiver receive if the data rate is
1 kbps? How many if the data rate is 1 Mbps?
Solution
At 1 kbps, the receiver receives 1001 bps instead of 1000
bps.
Example 4.3
At 1 Mbps, the receiver receives 1,001,000 bps instead of
1,000,000 bps.

4.10
Block coding is normally referred to as
mB/nB coding;
it replaces each m-bit group with an
n-bit group.
Note

4.11
Figure 4.14 Block coding concept

4.12
4-2 ANALOG-TO-DIGITAL CONVERSION
We have seen in Chapter 3 that a digital signal is
superior to an analog signal. The tendency today is to
change an analog signal to digital data. In this section
we describe two techniques, pulse code modulation
and delta modulation.
Pulse Code Modulation (PCM)
Delta Modulation (DM)

4.13
Figure 4.21 Components of PCM encoder

4.14
Figure 4.22 Three different sampling methods for PCM

4.15
According to the Nyquist theorem, the
sampling rate must be
at least 2 times the highest frequency
contained in the signal.
Note

4.16
Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals

4.17
For an intuitive example of the Nyquist theorem, let us
sample a simple sine wave at three sampling rates: fs = 4f
(2 times the Nyquist rate), fs = 2f (Nyquist rate), and
fs = f (one-half the Nyquist rate). Figure 4.24 shows the
sampling and the subsequent recovery of the signal.
It can be seen that sampling at the Nyquist rate can create
a good approximation of the original sine wave (part a).
Oversampling in part b can also create the same
approximation, but it is redundant and unnecessary.
Sampling below the Nyquist rate (part c) does not produce
a signal that looks like the original sine wave.
Example 4.6

4.18
Figure 4.24 Recovery of a sampled sine wave for different sampling rates

4.19
Consider the revolution of a hand of a clock. The second
hand of a clock has a period of 60 s. According to the
Nyquist theorem, we need to sample the hand every 30 s
(Ts = T or fs = 2f ). In Figure 4.25a, the sample points, in
order, are 12, 6, 12, 6, 12, and 6. The receiver of the
samples cannot tell if the clock is moving forward or
backward. In part b, we sample at double the Nyquist rate
(every 15 s). The sample points are 12, 3, 6, 9, and 12.
The clock is moving forward. In part c, we sample below
the Nyquist rate (Ts = T or fs = f ). The sample points are
12, 9, 6, 3, and 12. Although the clock is moving forward,
the receiver thinks that the clock is moving backward.
Example 4.7

4.20
Figure 4.25 Sampling of a clock with only one hand

4.21
An example related to Example 4.7 is the seemingly
backward rotation of the wheels of a forward-moving car
in a movie. This can be explained by under-sampling. A
movie is filmed at 24 frames per second. If a wheel is
rotating more than 12 times per second, the under-
sampling creates the impression of a backward rotation.
Example 4.8

4.22
Telephone companies digitize voice by assuming a
maximum frequency of 4000 Hz. The sampling rate
therefore is 8000 samples per second.
Example 4.9

4.23
A complex low-pass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
Solution
The bandwidth of a low-pass signal is between 0 and f,
where f is the maximum frequency in the signal.
Therefore, we can sample this signal at 2 times the
highest frequency (200 kHz). The sampling rate is
therefore 400,000 samples per second.
Example 4.10

4.24
A complex bandpass signal has a bandwidth of 200 kHz.
What is the minimum sampling rate for this signal?
Solution
We cannot find the minimum sampling rate in this case
because we do not know where the bandwidth starts or
ends. We do not know the maximum frequency in the
signal.
Example 4.11

4.25
Figure 4.26 Quantization and encoding of a sampled signal

4.26
What is the SNRdB in the example of Figure 4.26?
Solution
We can use the formula to find the quantization. We have
eight levels and 3 bits per sample, so
SNRdB = 6.02(3) + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
Example 4.12

4.27
A telephone subscriber line must have an SNRdB above
40. What is the minimum number of bits per sample?
Solution
We can calculate the number of bits as
Example 4.13
Telephone companies usually assign 7 or 8 bits per
sample.

4.28
We want to digitize the human voice. What is the bit rate,
assuming 8 bits per sample?
Solution
The human voice normally contains frequencies from 0
to 4000 Hz. So the sampling rate and bit rate are
calculated as follows:
Example 4.14

4.29
Figure 4.27 Components of a PCM decoder

4.30
We have a low-pass analog signal of 4 kHz. If we send the
analog signal, we need a channel with a minimum
bandwidth of 4 kHz. If we digitize the signal and send 8
bits per sample, we need a channel with a minimum
bandwidth of 8 × 4 kHz = 32 kHz.
Example 4.15

4.31
Figure 4.28 The process of delta modulation

4.32
Figure 4.29 Delta modulation components

4.33
Figure 4.30 Delta demodulation components

4.34
4-3 TRANSMISSION MODES
The transmission of binary data across a link can be
accomplished in either parallel or serial mode. In
parallel mode, multiple bits are sent with each clock
tick. In serial mode, 1 bit is sent with each clock tick.
While there is only one way to send parallel data, there
are three subclasses of serial transmission:
asynchronous, synchronous, and isochronous.
Parallel Transmission
Serial Transmission

4.35
Figure 4.31 Data transmission and modes

4.36
Figure 4.32 Parallel transmission

4.37
Figure 4.33 Serial transmission

4.38
In asynchronous transmission, we send
1 start bit (0) at the beginning and 1 or
more stop bits (1s) at the end of each
byte. There may be a gap between
each byte.
Note

4.39
Asynchronous here means
“asynchronous at the byte level,”
but the bits are still synchronized;
their durations are the same.
Note

4.40
Figure 4.34 Asynchronous transmission

4.41
In synchronous transmission, we send
bits one after another without start or
stop bits or gaps. It is the responsibility
of the receiver to group the bits.
Note

4.42
Figure 4.35 Synchronous transmission

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.
Aspects of Digital-to-Analog Conversion
Amplitude Shift Keying
Frequency Shift Keying
Phase Shift Keying
Quadrature Amplitude Modulation

Figure 5.1 Digital-to-analog conversion

Figure 5.2 Types of digital-to-analog conversion

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.
Note

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
Example 5.1

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.

Figure 5.3 Binary amplitude shift keying

Figure 5.4 Implementation of binary ASK

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).

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.

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

Figure 5.6 Binary frequency shift keying

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

Figure 5.7 Bandwidth of MFSK used in Example 5.6

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 = 8000. Figure 5.8 shows the allocation of
frequencies and bandwidth.

Figure 5.8 Bandwidth of MFSK used in Example 5.6

Figure 5.9 Binary phase shift keying

Figure 5.10 Implementation of BASK

Figure 5.11 QPSK and its implementation

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.

Figure 5.12 Concept of a constellation diagram

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

Figure 5.13 Three constellation diagrams

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

Figure 5.14 Constellation diagrams for some QAMs

5-2 ANALOG AND DIGITAL
Analog-to-analog conversion is the representation of
analog information by an analog signal. One may ask
why we need to modulate an analog signal; it is
already analog. Modulation is needed if the medium is
bandpass in nature or if only a bandpass channel is
available to us.
Amplitude Modulation
Frequency Modulation
Phase Modulation

Figure 5.15 Types of analog-to-analog modulation

Figure 5.16 Amplitude modulation

The total bandwidth required for AM
can be determined
from the bandwidth of the audio
signal: BAM = 2B.
Note

Figure 5.17 AM band allocation

The total bandwidth required for FM can
be determined from the bandwidth
of the audio signal: BFM = 2(1 + β)B.
Note

Figure 5.18 Frequency modulation

Figure 5.19 FM band allocation

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

6.78
Chapter 6
Bandwidth Utilization:
Multiplexing and
Spreading
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6.79
Bandwidth utilization is the wise use of
available bandwidth to achieve
specific goals.
Efficiency can be achieved by
multiplexing; privacy and anti-jamming
can be achieved by spreading.
Note

6.80
6-1 MULTIPLEXING
Whenever the bandwidth of a medium linking two
devices is greater than the bandwidth needs of the
devices, the link can be shared. Multiplexing is the set
of techniques that allows the simultaneous
transmission of multiple signals across a single data
link. As data and telecommunications use increases, so
does traffic.
Frequency-Division Multiplexing
Wavelength-Division Multiplexing
Synchronous Time-Division Multiplexing
Statistical Time-Division Multiplexing

6.81
Figure 6.1 Dividing a link into channels

6.82
Figure 6.2 Categories of multiplexing

6.83
Figure 6.3 Frequency-division multiplexing

6.84
FDM is an analog multiplexing technique
that combines analog signals.
Note

6.86
Figure 6.5 FDM demultiplexing example

6.87
Assume that a voice channel occupies a bandwidth of 4
kHz. We need to combine three voice channels into a link
with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the
configuration, using the frequency domain. Assume there
are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a
different bandwidth, as shown in Figure 6.6. We use the
20- to 24-kHz bandwidth for the first channel, the 24- to
28-kHz bandwidth for the second channel, and the 28- to
32-kHz bandwidth for the third one. Then we combine
them as shown in Figure 6.6.
Example 6.1

6.89
Five channels, each with a 100-kHz bandwidth, are to be
multiplexed together. What is the minimum bandwidth of
the link if there is a need for a guard band of 10 kHz
between the channels to prevent interference?
Solution
For five channels, we need at least four guard bands.
This means that the required bandwidth is at least
5 × 100 + 4 × 10 = 540 kHz,
as shown in Figure 6.7.
Example 6.2

6.91
Four data channels (digital), each transmitting at 1
Mbps, use a satellite channel of 1 MHz. Design an
appropriate configuration, using FDM.
Solution
The satellite channel is analog. We divide it into four
channels, each channel having a 250-kHz bandwidth.
Each digital channel of 1 Mbps is modulated such that
each 4 bits is modulated to 1 Hz. One solution is 16-QAM
modulation. Figure 6.8 shows one possible configuration.
Example 6.3

6.93
Figure 6.9 Analog hierarchy

6.94
The Advanced Mobile Phone System (AMPS) uses two
bands. The first band of 824 to 849 MHz is used for
sending, and 869 to 894 MHz is used for receiving.
Each user has a bandwidth of 30 kHz in each direction.
How many people can use their cellular phones
simultaneously?
Solution
Each band is 25 MHz. If we divide 25 MHz by 30 kHz, we
get 833.33. In reality, the band is divided into 832
channels. Of these, 42 channels are used for control,
which means only 790 channels are available for cellular
phone users.
Example 6.4

6.95
Figure 6.10 Wavelength-division multiplexing

6.96
WDM is an analog multiplexing
technique to combine optical signals.
Note

6.97
Figure 6.11 Prisms in wavelength-division multiplexing and demultiplexing

6.99
TDM is a digital multiplexing technique
for combining several low-rate
channels into one high-rate one.
Note

6.100
Figure 6.13 Synchronous time-division multiplexing

6.101
In synchronous TDM, the data rate
of the link is n times faster, and the unit
duration is n times shorter.
Note

6.102
In Figure 6.13, the data rate for each input connection is
3 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit),
what is the duration of (a) each input slot, (b) each output
slot, and (c) each frame?
Solution
We can answer the questions as follows:
a. The data rate of each input connection is 1 kbps. This
means that the bit duration is 1/1000 s or 1 ms. The
duration of the input time slot is 1 ms (same as bit
duration).
Example 6.5

6.103
b. The duration of each output time slot is one-third of
the input time slot. This means that the duration of the
output time slot is 1/3 ms.
c. Each frame carries three output time slots. So the
duration of a frame is 3 × 1/3 ms, or 1 ms. The
duration of a frame is the same as the duration of an
input unit.
Example 6.5 (continued)

6.104
Figure 6.14 shows synchronous TDM with a data stream
for each input and one data stream for the output. The
unit of data is 1 bit. Find (a) the input bit duration, (b)
the output bit duration, (c) the output bit rate, and (d) the
output frame rate.
Solution
a. The input bit duration is the inverse of the bit rate:
1/1 Mbps = 1 μs.
b. The output bit duration is one-fourth of the input bit
duration, or ¼ μs.
Example 6.6

6.105
c. The output bit rate is the inverse of the output bit
duration or 1/(4μs) or 4 Mbps. This can also be
deduced from the fact that the output rate is 4 times as
fast as any input rate; so the output rate = 4 × 1 Mbps
= 4 Mbps.
d. The frame rate is always the same as any input rate. So
the frame rate is 1,000,000 frames per second.
Because we are sending 4 bits in each frame, we can
verify the result of the previous question by
multiplying the frame rate by the number of bits per
frame.

6.107
Four 1-kbps connections are multiplexed together. A unit
is 1 bit. Find (a) the duration of 1 bit before multiplexing,
(b) the transmission rate of the link, (c) the duration of a
time slot, and (d) the duration of a frame.
Solution
a. The duration of 1 bit before multiplexing is 1 / 1 kbps,
or 0.001 s (1 ms).
b. The rate of the link is 4 times the rate of a connection,
or 4 kbps.
Example 6.7

6.108
c. The duration of each time slot is one-fourth of the
duration of each bit before multiplexing, or 1/4 ms or
250 μs. Note that we can also calculate this from the
data rate of the link, 4 kbps. The bit duration is the
inverse of the data rate, or 1/4 kbps or 250 μs.
d. The duration of a frame is always the same as the
duration of a unit before multiplexing, or 1 ms. We
can also calculate this in another way. Each frame in
this case has four time slots. So the duration of a
frame is 4 times 250 μs, or 1 ms.

6.109
Figure 6.15 Interleaving

6.110
Four channels are multiplexed using TDM. If each
channel sends 100 bytes /s and we multiplex 1 byte per
channel, show the frame traveling on the link, the size of
the frame, the duration of a frame, the frame rate, and
the bit rate for the link.
Solution
The multiplexer is shown in Figure 6.16. Each frame
carries 1 byte from each channel; the size of each frame,
therefore, is 4 bytes, or 32 bits. Because each channel is
sending 100 bytes/s and a frame carries 1 byte from each
channel, the frame rate must be 100 frames per second.
The bit rate is 100 × 32, or 3200 bps.
Example 6.8

6.112
A multiplexer combines four 100-kbps channels using a
time slot of 2 bits. Show the output with four arbitrary
inputs. What is the frame rate? What is the frame
duration? What is the bit rate? What is the bit duration?
Solution
Figure 6.17 shows the output for four arbitrary inputs.
The link carries 50,000 frames per second. The frame
duration is therefore 1/50,000 s or 20 μs. The frame rate
is 50,000 frames per second, and each frame carries 8
bits; the bit rate is 50,000 × 8 = 400,000 bits or 400 kbps.
The bit duration is 1/400,000 s, or 2.5 μs.
Example 6.9

6.115
Figure 6.19 Multilevel multiplexing

6.116
Figure 6.20 Multiple-slot multiplexing

6.117
Figure 6.21 Pulse stuffing

6.118
Figure 6.22 Framing bits

6.119
We have four sources, each creating 250 characters per
second. If the interleaved unit is a character and 1
synchronizing bit is added to each frame, find (a) the data
rate of each source, (b) the duration of each character in
each source, (c) the frame rate, (d) the duration of each
frame, (e) the number of bits in each frame, and (f) the
data rate of the link.
Solution
a. The data rate of each source is 250 × 8 = 2000 bps = 2
kbps.
Example 6.10

6.120
b. Each source sends 250 characters per second;
therefore, the duration of a character is 1/250 s, or
4 ms.
c. Each frame has one character from each source,
which means the link needs to send 250 frames per
second to keep the transmission rate of each source.
d. The duration of each frame is 1/250 s, or 4 ms. Note
that the duration of each frame is the same as the
duration of each character coming from each source.
e. Each frame carries 4 characters and 1 extra
synchronizing bit. This means that each frame is
4 × 8 + 1 = 33 bits.

6.121
Two channels, one with a bit rate of 100 kbps and
another with a bit rate of 200 kbps, are to be multiplexed.
How this can be achieved? What is the frame rate? What
is the frame duration? What is the bit rate of the link?
Solution
We can allocate one slot to the first channel and two slots
to the second channel. Each frame carries 3 bits. The
frame rate is 100,000 frames per second because it carries
1 bit from the first channel. The bit rate is 100,000
frames/s × 3 bits per frame, or 300 kbps.
Example 6.11

6.122
Figure 6.23 Digital hierarchy

6.123
Table 6.1 DS and T line rates

6.124
Figure 6.24 T-1 line for multiplexing telephone lines

6.125
Figure 6.25 T-1 frame structure

6.127
Figure 6.26 TDM slot comparison

Lecture 3.ppt

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