Ch03

3.1
Chapter 3
Data and Signals

3.2
To be transmitted, data must be
transformed to electromagnetic signals.
Note

3.3
3-1 ANALOG AND DIGITAL3-1 ANALOG AND DIGITAL
Data can beData can be analoganalog oror digitaldigital. The term. The term analog dataanalog data refersrefers
to information that is continuous;to information that is continuous; digital datadigital data refers torefers to
information that has discrete states. Analog data take oninformation that has discrete states. Analog data take on
continuous values. Digital data take on discrete values.continuous values. Digital data take on discrete values.
Analog and Digital Data
Analog and Digital Signals
Periodic and Nonperiodic Signals
Topics discussed in this section:Topics discussed in this section:

3.4
Note
Data can be analog or digital.
Analog data are continuous and take
continuous values.
Digital data have discrete states and
take discrete values.

3.5
Signals can be analog or digital.
Analog signals can have an infinite
number of values in a range; digital
signals can have only a limited
number of values.
Note

3.6
Figure 3.1 Comparison of analog and digital signals

3.7
In data communications, we commonly
use periodic analog signals and
nonperiodic digital signals.
Note

3.8
3-2 PERIODIC ANALOG SIGNALS3-2 PERIODIC ANALOG SIGNALS
Periodic analog signals can be classified asPeriodic analog signals can be classified as simplesimple oror
compositecomposite. A simple periodic analog signal, a. A simple periodic analog signal, a sine wavesine wave,,
cannot be decomposed into simpler signals. A compositecannot be decomposed into simpler signals. A composite
periodic analog signal is composed of multiple sineperiodic analog signal is composed of multiple sine
waves.waves.
Sine Wave
Wavelength
Time and Frequency Domain
Composite Signals
Bandwidth

3.10
Figure 3.3 Two signals with the same phase and frequency,
but different amplitudes

3.11
Frequency and period are the inverse of
each other.
Note

3.12
Figure 3.4 Two signals with the same amplitude and phase,
but different frequencies

3.13
Table 3.1 Units of period and frequency

3.14
The power we use at home has a frequency of 60 Hz.
The period of this sine wave can be determined as
follows:
Example 3.3

3.15
Express a period of 100 ms in microseconds.
Example 3.4
Solution
From Table 3.1 we find the equivalents of 1 ms (1 ms is
10−3
s) and 1 s (1 s is 106
μs). We make the following
substitutions:.

3.16
The period of a signal is 100 ms. What is its frequency in
kilohertz?
Example 3.5
Solution
First we change 100 ms to seconds, and then we
calculate the frequency from the period (1 Hz = 10−3
kHz).

3.17
Frequency is the rate of change with
respect to time.
Change in a short span of time
means high frequency.
Change over a long span of
time means low frequency.
Note

3.18
If a signal does not change at all, its
frequency is zero.
If a signal changes instantaneously, its
frequency is infinite.
Note

3.19
Phase describes the position of the
waveform relative to time 0.
Note

3.20
Figure 3.5 Three sine waves with the same amplitude and frequency,
but different phases

3.21
A sine wave is offset 1/6 cycle with respect to time 0.
What is its phase in degrees and radians?
Example 3.6
Solution
We know that 1 complete cycle is 360°. Therefore, 1/6
cycle is

3.22
Figure 3.6 Wavelength and period

3.23
Figure 3.7 The time-domain and frequency-domain plots of a sine wave

3.24
A complete sine wave in the time
domain can be represented by one
single spike in the frequency domain.
Note

3.25
The frequency domain is more compact and
useful when we are dealing with more than one
sine wave. For example, Figure 3.8 shows three
sine waves, each with different amplitude and
frequency. All can be represented by three
spikes in the frequency domain.
Example 3.7

3.26
Figure 3.8 The time domain and frequency domain of three sine waves

3.27
A single-frequency sine wave is not
useful in data communications;
we need to send a composite signal, a
signal made of many simple sine waves.
Note

3.28
According to Fourier analysis, any
composite signal is a combination of
simple sine waves with different
frequencies, amplitudes, and phases.
Note

3.29
If the composite signal is periodic, the
decomposition gives a series of signals
with discrete frequencies;
if the composite signal is nonperiodic,
the decomposition gives a combination
of sine waves with continuous
frequencies.
Note

3.30
Figure 3.9 shows a periodic composite signal with
frequency f. This type of signal is not typical of those
found in data communications. We can consider it to be
three alarm systems, each with a different frequency.
The analysis of this signal can give us a good
understanding of how to decompose signals.
Example 3.8

3.31
Figure 3.9 A composite periodic signal

3.32
Figure 3.10 Decomposition of a composite periodic signal in the time and
frequency domains

3.33
Figure 3.11 shows a nonperiodic composite signal. It
can be the signal created by a microphone or a telephone
set when a word or two is pronounced. In this case, the
composite signal cannot be periodic, because that
implies that we are repeating the same word or words
with exactly the same tone.
Example 3.9

3.34
Figure 3.11 The time and frequency domains of a nonperiodic signal

3.35
The bandwidth of a composite signal is
the difference between the
highest and the lowest frequencies
contained in that signal.
Note

3.36
Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

3.37
If a periodic signal is decomposed into five sine waves
with frequencies of 100, 300, 500, 700, and 900 Hz, what
is its bandwidth? Draw the spectrum, assuming all
components have a maximum amplitude of 10 V.
Solution
Let fh be the highest frequency, fl the lowest frequency,
and B the bandwidth. Then
Example 3.10
The spectrum has only five spikes, at 100, 300, 500, 700,
and 900 Hz (see Figure 3.13).

3.38
Figure 3.13 The bandwidth for Example 3.10

3.39
A periodic signal has a bandwidth of 20 Hz. The highest
frequency is 60 Hz. What is the lowest frequency? Draw
the spectrum if the signal contains all frequencies of the
same amplitude.
Solution
Let fh be the highest frequency, fl the lowest frequency,
and B the bandwidth. Then
Example 3.11
The spectrum contains all integer frequencies. We show
this by a series of spikes (see Figure 3.14).

3.40

3.41
A nonperiodic composite signal has a bandwidth of 200
kHz, with a middle frequency of 140 kHz and peak
amplitude of 20 V. The two extreme frequencies have an
amplitude of 0. Draw the frequency domain of the
signal.
Solution
The lowest frequency must be at 40 kHz and the highest
at 240 kHz. Figure 3.15 shows the frequency domain
and the bandwidth.
Example 3.12

3.42

3.43
An example of a nonperiodic composite signal is the
signal propagated by an AM radio station. In the United
States, each AM radio station is assigned a 10-kHz
bandwidth. The total bandwidth dedicated to AM radio
ranges from 530 to 1700 kHz. We will show the rationale
behind this 10-kHz bandwidth in Chapter 5.
Example 3.13

3.44
Another example of a nonperiodic composite signal is
the signal propagated by an FM radio station. In the
United States, each FM radio station is assigned a 200-
kHz bandwidth. The total bandwidth dedicated to FM
radio ranges from 88 to 108 MHz. We will show the
rationale behind this 200-kHz bandwidth in Chapter 5.
Example 3.14

3.45
Another example of a nonperiodic composite signal is
the signal received by an old-fashioned analog black-
and-white TV. A TV screen is made up of pixels. If we
assume a resolution of 525 × 700, we have 367,500
pixels per screen. If we scan the screen 30 times per
second, this is 367,500 × 30 = 11,025,000 pixels per
second. The worst-case scenario is alternating black and
white pixels. We can send 2 pixels per cycle. Therefore,
we need 11,025,000 / 2 = 5,512,500 cycles per second, or
Hz. The bandwidth needed is 5.5125 MHz.
Example 3.15

3.46
3-3 DIGITAL SIGNALS3-3 DIGITAL SIGNALS
In addition to being represented by an analog signal,In addition to being represented by an analog signal,
information can also be represented by ainformation can also be represented by a digital signaldigital signal..
For example, a 1 can be encoded as a positive voltageFor example, a 1 can be encoded as a positive voltage
and a 0 as zero voltage. A digital signal can have moreand a 0 as zero voltage. A digital signal can have more
than two levels. In this case, we can send more than 1 bitthan two levels. In this case, we can send more than 1 bit
for each level.for each level.
Bit Rate
Bit Length
Digital Signal as a Composite Analog Signal
Application Layer

3.47
Figure 3.16 Two digital signals: one with two signal levels and the other
with four signal levels

3.48
A digital signal has eight levels. How many bits are
needed per level? We calculate the number of bits from
the formula
Example 3.16
Each signal level is represented by 3 bits.

3.49
Assume we need to download text documents at the rate
of 100 pages per minute. What is the required bit rate of
the channel?
Solution
A page is an average of 24 lines with 80 characters in
each line. If we assume that one character requires 8
bits, the bit rate is
Example 3.18

3.50
A digitized voice channel, as we will see in Chapter 4, is
made by digitizing a 4-kHz bandwidth analog voice
signal. We need to sample the signal at twice the highest
frequency (two samples per hertz). We assume that each
sample requires 8 bits. What is the required bit rate?
Solution
The bit rate can be calculated as
Example 3.19

3.51
Figure 3.17 The time and frequency domains of periodic and nonperiodic
digital signals

3.52
Figure 3.18 Baseband transmission

3.53
A digital signal is a composite analog
signal with an infinite bandwidth.
Note

3.54
Figure 3.19 Bandwidths of two low-pass channels

3.55
Figure 3.20 Baseband transmission using a dedicated medium

3.56
Baseband transmission of a digital
signal that preserves the shape of the
digital signal is possible only if we have
a low-pass channel with an infinite or
very wide bandwidth.
Note

3.57
In baseband transmission, the required bandwidth is proportional to
the bit rate;
if we need to send bits faster, we need more bandwidth.
Note
In baseband transmission, the required
bandwidth is proportional to the bit rate;
if we need to send bits faster, we need
more bandwidth.

3.58
Table 3.2 Bandwidth requirements

3.59
What is the required bandwidth of a low-pass channel if
we need to send 1 Mbps by using baseband transmission?
Solution
The answer depends on the accuracy desired.
a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.
b. A better solution is to use the first and the third
harmonics with B = 3 × 500 kHz = 1.5 MHz.
c. Still a better solution is to use the first, third, and fifth
harmonics with B = 5 × 500 kHz = 2.5 MHz.
Example 3.22

3.60
We have a low-pass channel with bandwidth 100 kHz.
What is the maximum bit rate of this
channel?
Solution
The maximum bit rate can be achieved if we use the first
harmonic. The bit rate is 2 times the available bandwidth,
or 200 kbps.
Example 3.22

3.61
Figure 3.23 Bandwidth of a bandpass channel

3.62
If the available channel is a bandpass
channel, we cannot send the digital
signal directly to the channel;
we need to convert the digital signal to
an analog signal before transmission.
Note

3.63
Figure 3.24 Modulation of a digital signal for transmission on a bandpass
channel

3.64
An example of broadband transmission using
modulation is the sending of computer data through a
telephone subscriber line, the line connecting a resident
to the central telephone office. These lines are designed
to carry voice with a limited bandwidth. The channel is
considered a bandpass channel. We convert the digital
signal from the computer to an analog signal, and send
the analog signal. We can install two converters to
change the digital signal to analog and vice versa at the
receiving end. The converter, in this case, is called a
modem which we discuss in detail in Chapter 5.
Example 3.24

3.65
A second example is the digital cellular telephone. For
better reception, digital cellular phones convert the
analog voice signal to a digital signal (see Chapter 16).
Although the bandwidth allocated to a company
providing digital cellular phone service is very wide, we
still cannot send the digital signal without conversion.
The reason is that we only have a bandpass channel
available between caller and callee. We need to convert
the digitized voice to a composite analog signal before
sending.
Example 3.25

3.66
3-4 TRANSMISSION IMPAIRMENT3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are notSignals travel through transmission media, which are not
perfect. The imperfection causes signal impairment. Thisperfect. The imperfection causes signal impairment. This
means that the signal at the beginning of the medium ismeans that the signal at the beginning of the medium is
not the same as the signal at the end of the medium.not the same as the signal at the end of the medium.
What is sent is not what is received. Three causes ofWhat is sent is not what is received. Three causes of
impairment areimpairment are attenuationattenuation,, distortiondistortion, and, and noisenoise..
Attenuation
Distortion
Noise

3.67
Figure 3.25 Causes of impairment

3.69
Suppose a signal travels through a transmission medium
and its power is reduced to one-half. This means that P2
is (1/2)P1. In this case, the attenuation (loss of power)
can be calculated as
Example 3.26
A loss of 3 dB (–3 dB) is equivalent to losing one-half
the power.

3.70
A signal travels through an amplifier, and its power is
increased 10 times. This means that P2 = 10P1 . In this
case, the amplification (gain of power) can be calculated
as
Example 3.27

3.71
One reason that engineers use the decibel to measure the
changes in the strength of a signal is that decibel
numbers can be added (or subtracted) when we are
measuring several points (cascading) instead of just two.
In Figure 3.27 a signal travels from point 1 to point 4. In
this case, the decibel value can be calculated as
Example 3.28

3.72
Figure 3.27 Decibels for Example 3.28

3.75
3-5 DATA RATE LIMITS3-5 DATA RATE LIMITS
A very important consideration in data communicationsA very important consideration in data communications
is how fast we can send data, in bits per second, over ais how fast we can send data, in bits per second, over a
channel. Data rate depends on three factors:channel. Data rate depends on three factors:
1.1. The bandwidth availableThe bandwidth available
2.2. The level of the signals we useThe level of the signals we use
33. The quality of the channel (the level of noise). The quality of the channel (the level of noise)
Noiseless Channel: Nyquist Bit Rate
Noisy Channel: Shannon Capacity
Using Both Limits

3.76
Increasing the levels of a signal may
reduce the reliability of the system.
Note

3.77
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.
Example 3.33

3.78
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
Example 3.34

3.79
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
Example 3.35

3.80
The Shannon capacity gives us the
upper limit; the Nyquist formula tells us
how many signal levels we need.
Note

3.81
3-6 PERFORMANCE3-6 PERFORMANCE
One important issue in networking is theOne important issue in networking is the performanceperformance ofof
the network—how good is it? We discuss quality ofthe network—how good is it? We discuss quality of
service, an overall measurement of networkservice, an overall measurement of network
performance, in greater detail in Chapter 24. In thisperformance, in greater detail in Chapter 24. In this
section, we introduce terms that we need for futuresection, we introduce terms that we need for future
chapters.chapters.
Bandwidth
Throughput
Latency (Delay)
Bandwidth-Delay Product

3.82
In networking, we use the term
bandwidth in two contexts.
❏ The first, bandwidth in hertz, refers to
the range of frequencies in a
composite signal or the range of
frequencies that a channel can pass.
❏ The second, bandwidth in bits per
second, refers to the speed of bit
transmission in a channel or link.
Note

3.83
The bandwidth of a subscriber line is 4 kHz for voice or
data. The bandwidth of this line for data transmission
can be up to 56,000 bps using a sophisticated modem to
change the digital signal to analog.
Example 3.42

3.84
If the telephone company improves the quality of the line
and increases the bandwidth to 8 kHz, we can send
112,000 bps by using the same technology as mentioned
in Example 3.42.
Example 3.43

3.85
A network with bandwidth of 10 Mbps can pass only an
average of 12,000 frames per minute with each frame
carrying an average of 10,000 bits. What is the
throughput of this network?
Solution
We can calculate the throughput as
Example 3.44
The throughput is almost one-fifth of the bandwidth in
this case.

3.86
What is the propagation time if the distance between the
two points is 12,000 km? Assume the propagation speed
to be 2.4 × 108 m/s in cable.
Solution
We can calculate the propagation time as
Example 3.45
The example shows that a bit can go over the Atlantic
Ocean in only 50 ms if there is a direct cable between the
source and the destination.

3.87
What are the propagation time and the transmission
time for a 2.5-kbyte message (an e-mail) if the
bandwidth of the network is 1 Gbps? Assume that the
distance between the sender and the receiver is 12,000
km and that light travels at 2.4 × 108 m/s.
Solution
We can calculate the propagation and transmission time
as shown on the next slide:
Example 3.46

3.88
Note that in this case, because the message is short and
the bandwidth is high, the dominant factor is the
propagation time, not the transmission time. The
transmission time can be ignored.
Example 3.46 (continued)

3.89
The bandwidth-delay product defines
the number of bits that can fill the link.
Note

3.90
Figure 3.33 Concept of bandwidth-delay product

Ch03

More Related Content

What's hot (20)

Similar to Ch03 (20)

More from Aman Jaiswal (13)

Recently uploaded (20)

Ch03