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PART 2
Physical Layer
We start the discussion of the Internet model with the bottom-most layer, the physical
layer. It is the layer that actually interacts with the transmission media, the physical
part of the network that connects network components together. This layer is involved
in physically carrying information from one node in the network to the next. Figure 1
shows the position of the physical layer in the 5-layer Internet model.
Figure 1 Position of the physical layer
Data link layer
Bit-signal
transformation
Bit
synchronization
Circuit
switching
Bit-rate
control
Multiplexing
High-speed
access
Telephone
network
Physical layer
The physical layer has complex tasks to perform. One major task is to provide
services for the data link layer. The data in the data link layer consists of 0s and 1s
organized into frames that are ready to be sent across the transmission medium. This
stream of 0s and 1s must first be converted into another entity: signals. One of the
services provided by the physical layer is to create a signal that represents this stream
of bits.
The physical layer must also take care of the physical network, the transmission
medium. The transmission medium is a passive entity; it has no internal program or
logic for control like other layers. The transmission medium must be controlled by the
physical layer. The physical layer decides on the directions of data flow. The physical
layer decides on the number of logical channels for transporting data coming from different sources.
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46 PART 2 PHYSICAL LAYER
Services
The physical layer transfers a stream of bits (in the form of a signal) from the sender to
the receiver. The transfer is node-to-node, from one node to the next. The physical layers of the two adjacent nodes provide a logical pipe through which the bits can travel.
Figure 2 shows the general services offered by the physical layer.
Figure 2 Physical layer services
Duties of
physical layer
Bit-signal
transformation
Bit-rate
control
Bit
synchronization
Multiplexing
Circuit
switching
Bit-to-Signal Transformation
The logical pipe under the physical layer is the transmission media (cable or air). Since a
transmission medium cannot carry bits, we need to represent the bits by a signal, electromagnetic energy that can propagate through a medium.
Bit-Rate Control
Although the transmission medium determines the upper limit of the data rate, the
physical layer is the controller. The design of the physical-layer hardware and software
determine the data rate.
Bit-Synchronization
The timing of the bit transfer is crucial in data communications. The physical layer
governs the synchronization of the bits by providing clocking mechanisms that control
both the sender and the receiver.
Multiplexing
Multiplexing is the process of dividing a link, the physical medium, into logical channels for better efficiency. The physical layer, using different techniques, can do this.
Although the medium itself is not actually changed, the result is several channels
instead of one. Multiplexing defined in this section of the text is needed to understand
access methods in later chapters.
Switching
Switching in data communications can be done in several layers. We have circuitswitching, packet-switching, and message switching. Circuit switching, a method that
allows two nodes to have a dedicated link, is mostly a function of the physical layer.
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PART 2 PHYSICAL LAYER 47
Packet switching is discussed in Chapter 18 as a data-link-layer issue and in Chapter 19
as a network-layer issue.
Transmission Media
The physical layer is dependent on the transmission media to carry its bits (in signal
form). Although the transmission media are not actually a part of the physical layer, the
media are controlled by this layer. Media can be guided and unguided. Twisted-pair
cable, coaxial cable, and fiber-optic cables are discussed in the guided media section.
Radio and microwave communication are included in the unguided section.
Networks and Technologies
To connect the issues discussed in Chapters 3 to 7, we have included several examples
of networks and technologies which provide services at the physical layer.
Telephone Network
Most of the networks today have their beginnings in the telephone network. Telephone
networks have been around for some time and provide voice communication around
the world. When the need for data communication started, the telephone network was
the foundation. Data were transformed into analog signals and sent over the same networks that sent voice. We discuss the telephone network as a prelude to other specific
data networks and also as a good example of a network with the physical layer issues
covered in this part of the book. We also give a brief historical background of the telephone network to understand the reasons for some recent developments such as
LATAs.
High Speed Access
Accessing the Internet requires a physical connection between the user and a company
known as the Internet service provider. We introduce modems, an Internet access
method that many users find too slow. We present two alternative technologies. DSL
technology provides a faster physical connection, again using the existing telephone
line. The cable TV network allows the use of some channels previously assigned for
video broadcasting for data transfer to and from the Internet.
Chapters
Part two of the book covers seven chapters. Chapters 3 introduces the concepts and
characteristics of signals as the vehicle for carrying data. Chapters 4 and 5 show how
we can change bits into digital or analog signals. Chapter 6 is about multiplexing, an
important issue due to the improvements in transmission media bandwidth. Although
transmission media are located below the physical layer, it is controlled by the physical layer. We have included the discussion of media in Chapter 7. Chapter 8 discusses
switching, a topic that can be related to several layers. We have, however, discussed
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48 PART 2 PHYSICAL LAYER
only circuit switching, which is mostly a physical-layer issue. To show an application
of circuit switching, we have introduced the telephone network and several topics
related to this network. The main purpose of most networks today is to access the
Internet. Chapter 9 introduces several technologies that allow the user to access the
Internet.
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CHAPTER 3
Signals
One of the major concerns of the physical layer lies in moving data in the form of electromagnetic signals across a transmission medium. Whether you are collecting numerical statistics from another computer, sending animated pictures from a design
workstation, or causing a bell to ring at a distant control center, you are working with
the transmission of data across network connections.
Generally, the data usable to a person or application are not in a form that can be transmitted over a network. For example, you cannot roll up a photograph, insert it into a wire,
and transmit it across town. You can, however, transmit an encoded description of the photograph. Instead of sending the actual photograph, you can use an encoder to create a stream
of 1s and 0s that tells the receiving device how to reconstruct the image of the photograph.
But even 1s and 0s cannot be sent as such across network links. They must be further converted to a form that transmission media can accept. Transmission media work
by conducting energy along a physical path. So a data stream of 1s and 0s must be
turned into energy in the form of electromagnetic signals.
To be transmitted, data must be transformed to electromagnetic signals.
3.1 ANALOG AND DIGITAL
Both data and the signals that represent them can take either analog or digital form.
Analog and Digital Data
Data can be analog or digital. An example of analog data is the human voice. When
someone speaks, an analog wave is created in the air. This can be captured by a microphone and converted to an analog signal or sampled later and converted to a digital signal.
An example of digital data is data stored in the memory of a computer in the form of
0s and 1s. It can be converted to a digital signal when it is transferred from one position
to another inside or outside the computer or modulated into an analog signal and then
sent through a transmission medium to another computer.
Analog and Digital Signals
Like the data they represent, signals can be either analog or digital. An analog signal has
infinitely many levels of intensity over a period of time. As the wave moves from value A
49
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50 CHAPTER 3 SIGNALS
to value B, it passes through and includes an infinite number of values along its path. A
digital signal, on the other hand, can have only a limited number of defined values, often
as simple as 1 and 0.
The simplest way to show signals is by plotting them on a pair of perpendicular axes.
The vertical axis represents the value or strength of a signal. The horizontal axis represents
the passage of time. Figure 3.1 illustrates an analog and a digital signal. The curve representing the analog signal is passing through an infinite number of points. The vertical lines of the
digital signal, however, demonstrate the sudden jump the signal makes from value to value.
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.
Figure 3.1 Comparison of analog and digital signals
Value
Value
Time
a. Analog signal
Time
b. Digital signal
Periodic and Aperiodic Signals
Both analog and digital signals can take one of two forms: periodic and aperiodic
(nonperiodic).
A periodic signal completes a pattern within a measurable time frame, called a
period, and repeats that pattern over subsequent identical periods. The completion of
one full pattern is called a cycle. An aperiodic signal changes without exhibiting a pattern
or cycle that repeats over time.
Both analog and digital signals can be periodic or aperiodic. In data communication, however, we commonly use periodic analog signals and aperiodic digital signals
to send data from one point to another.
In data communication, we commonly use periodic analog signals and aperiodic digital
signals.
3.2 ANALOG SIGNALS
Analog signals can be classified as simple or composite. A simple analog signal, a sine
wave, cannot be decomposed into simpler signals. A composite analog signal is composed of multiple sine waves.
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SECTION 3.2 ANALOG SIGNALS 51
Sine Wave
The sine wave is the most fundamental form of a periodic analog signal. Visualized as
a simple oscillating curve, its change over the course of a cycle is smooth and consistent, a continuous, rolling flow. Figure 3.2 shows a sine wave. Each cycle consists of a
single arc above the time axis followed by a single arc below it.
Figure 3.2 A sine wave
Value
•••
Time
We can mathematically describe a sine wave as
s(t) = Asin(2πft + φ)
where s is the instantaneous amplitude, A the peak amplitude, f the frequency, and φ
the phase. These three characteristics fully describe a sine wave.
Peak Amplitude
The peak amplitude of a signal represents the absolute value of its highest intensity,
proportional to the energy it carries. For electric signals, peak amplitude is normally
measured in volts (see Fig. 3.3).
Figure 3.3 Amplitude
Amplitude
Peak amplitude
•••
Time
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52 CHAPTER 3 SIGNALS
Period and Frequency
Period refers to the amount of time, in seconds, a signal needs to complete one cycle.
Frequency refers to the number of periods in one second. Note that period and frequency
are just one characteristic defined in two ways. Period is the inverse of frequency, and
frequency is the inverse of period, as shown in the following formulas.
1
f = --T
1
T = --f
and
Frequency and period are inverses of each other.
Figure 3.4 shows the concept of period and frequency.
Figure 3.4 Period and frequency
Amplitude
Frequency 6 Hz
Six periods in 1 s
1s
•••
Time
T
Period 16 s
Period is formally expressed in seconds. Frequency is formally expressed in
hertz (Hz), as shown in Table 3.1.
Table 3.1 Units of period and frequency
Unit
Seconds (s)
Equivalent
1s
Unit
hertz (Hz)
Equivalent
1 Hz
Milliseconds (ms)
10–3 s
kilohertz (KHz)
103 Hz
Microseconds (µs)
10–6 s
megahertz (MHz)
106 Hz
Nanoseconds (ns)
10–9 s
gigahertz (GHz)
109 Hz
Picoseconds (ps)
10–12 s
terahertz (THz)
1012 Hz
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SECTION 3.2 ANALOG SIGNALS 53
Example 1
Express a period of 100 ms in microseconds, and express the corresponding frequency in
kilohertz.
Solution
Let us first express 100 ms in microseconds. From Table 3.1 we find the equivalent of 1 ms (1 ms
is 10−3 s) and 1 s (1 s is 106 µs). We make the following substitutions:
100 ms = 100 × 10−3 s = 100 × 10−3 × 106 µs = 105 µs
Now we use the inverse relationship to find the frequency, changing hertz to kilohertz (1 Hz is
10−3 KHz).
1
100 ms = 100 × 10−3 s = 10−1 s
f = ------------ Hz = 10 × 10−3 KHz = 10−2 KHz
−1
10
More About Frequency
We know already that frequency is the relationship of a signal to time and that the frequency of a wave is the number of cycles it completes per second. But another way to
look at frequency is as a measurement of the rate of change. Electromagnetic signals
are oscillating waveforms; that is, they fluctuate continuously and predictably above
and below a mean energy level. A 40-Hz signal has one-half the frequency of an 80-Hz
signal; it completes one cycle in twice the time of the 80-Hz signal, so each cycle also
takes twice as long to change from its lowest to its highest voltage levels. Frequency,
therefore, though described in cycles per second (hertz), is a general measurement of the
rate of change of a signal with respect to time.
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.
If the value of a signal changes over a very short span of time, its frequency is
high. If it changes over a long span of time, its frequency is low.
Two Extremes
What if a signal does not change at all? What if it maintains a constant voltage level for the
entire time it is active? In such a case, its frequency is zero. Conceptually, this idea is a simple one. If a signal does not change at all, it never completes a cycle, so its frequency is 0 Hz.
But what if a signal changes instantaneously? What if it jumps from one level to
another in no time? Then its frequency is infinite. In other words, when a signal
changes instantaneously, its period is zero; since frequency is the inverse of period, in
this case, the frequency is 1/0, or infinite (unbounded).
If a signal does not change at all, its frequency is zero. If a signal changes instantaneously, its frequency is infinite.
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54 CHAPTER 3 SIGNALS
Phase
The term phase describes the position of the waveform relative to time zero. If we
think of the wave as something that can be shifted backward or forward along the time
axis, phase describes the amount of that shift. It indicates the status of the first cycle.
Phase describes the position of the waveform relative to time zero.
Phase is measured in degrees or radians [360° is 2π rad; 1° is 2π/360 rad, and 1 rad
is 360/(2π)]. A phase shift of 360° corresponds to a shift of a complete period; a phase
shift of 180° corresponds to a shift of one-half of a period; and a phase shift of 90° corresponds to a shift of one-quarter of a period (see Fig. 3.5).
Figure 3.5 Relationships between different phases
Amplitude
Amplitude
Time
Amplitude
Time
14 cycle
a. 0
Time
12 cycle
b. 90
c. 180
Example 2
A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees
and radians?
Solution
We know that one complete cycle is 360°. Therefore, 1/6 cycle is
1
2π
π
--- × 360 = 60° = 60 × --------- rad = --- rad = 1.046 rad
6
360
3
Examples of Sine Waves
A visual comparison of signals with different characteristics can give a better understanding of these characteristics. Figure 3.6 shows three sine waves with different peak
amplitudes, frequencies, and phases.
Time and Frequency Domains
A sine wave is comprehensively defined by its amplitude, frequency, and phase. We
have been showing a sine wave by using what is called a time-domain plot. The
time-domain plot shows changes in signal amplitude with respect to time (it is an
amplitude versus time plot). Phase and frequency are not explicitly measured on a
time-domain plot.
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SECTION 3.2 ANALOG SIGNALS 55
Figure 3.6 Sine wave examples
Amplitude
A5 f 4 0
s(t) 5 sin (24t 0)
5
•••
Time
1s
Amplitude
A 10 f 8 0
s(t) 10 sin (28t 0)
10
•••
Time
1s
Amplitude
A 5 f 2 /4
s(t) 5 sin (22t /4)
5
•••
Time
1s
To show the relationship between amplitude and frequency, we can use what is
called a frequency-domain plot. Figure 3.7 compares the time domain (instantaneous
amplitude with respect to time) and the frequency domain (peak amplitude with respect
to frequency).
The figure shows three signals with varying frequencies. Compare the models
within each pair to see which sort of data each is best suited to convey. All three
signals have a peak amplitude of 5 volts (V). The frequency of the first signal is 0; we
show it in the frequency domain with a spike at frequency 0 and a height of 5 (its
amplitude). The second signal has a frequency of 8, so we show it in the frequency
domain with a spike of height 5 and a frequency of 8. Finally, the third is shown with a
frequency of 16 at the same height. Note that in the frequency domain we can show two
characteristics of a signal with only one spike; the position is the frequency, and the
height is the peak amplitude. The phase of a signal cannot be shown in the frequency
domain; we need another domain that we won’t discuss in this book.
An analog signal is best represented in the frequency domain.
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56 CHAPTER 3 SIGNALS
Figure 3.7 Time and frequency domains
Time
domain
Frequency
domain
5
5
•••
1s
0
Time
Frequency
a. A signal with frequency 0
5
5
•••
8
Time
Frequency
1s
b. A signal with frequency 8
5
5
•••
Time
16
Frequency
1s
c. A signal with frequency 16
Composite Signals
So far, we have focused attention on simple signals (sine waves). Although a simple sine
wave signal is very useful for some purposes, it is useless for data communications. We can
send a single sine wave to carry electric energy from one place to another. For example, the
power company sends a single sine wave with a frequency of 60 Hz to distribute electric
energy to our houses and businesses. We can use a single sine wave to send an alarm to a
security center when a burglar opens a door or a window in our house. In the first case, the
sine wave is carrying energy; in the second, the presence of the signal infers danger.
If we used one single sine wave to convey a conversation over the phone, we
would always hear a buzz; it would make no sense and carry no information. If we sent
one single sine wave to convey data, we would always be sending alternating 1s and 0s,
which does not have any communication value.
A single-frequency sine wave is not useful in data communications; we need to change
one or more of its characteristics to make it useful.
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SECTION 3.2 ANALOG SIGNALS 57
If we want to use a sine wave for communication, we need to change one or more of its
characteristics. For example, when the data to be sent are a 1 bit, we can send a maximum
amplitude; when it is a 0 bit, we can send a minimum amplitude. However, we need to keep
in mind that when we change one or more characteristics of a sine wave, it is no longer a
simple sine wave. Instead, it is a composite signal made of many simple sine waves. A
mere change in the amplitude, frequency, or phase creates a new set of frequencies. Intuitively, change is related to frequency; more change means creating more new frequencies.
When we change one or more characteristics of a single-frequency signal, it becomes a
composite signal made of many frequencies.
Fourier Analysis
In the early 1900s, the French mathematician Jean-Baptiste Fourier showed that any
composite signal is a sum of a set of sine waves of different frequencies, phases, and
amplitudes. In other words, we can write a composite signal as
s(t) = A1 sin (2πf1t + φ1) + A2 sin (2πf2t + φ2) + A3 sin (2πf3t + φ3) + . . .
According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes.
For example, let us consider the square wave of Figure 3.8 with a peak amplitude
of A and a frequency of f (period T ). According to Fourier analysis, we can prove that
this signal can be decomposed into a series of sine waves as shown below.
Figure 3.8 Square wave
A
T
2T
3T
Time
4A
4A
4A
s(t) = ------- sin 2πft + ------- sin [2π(3f )t] + ------- sin [2π(5f )t] + . . .
π
3π
5π
In other words, we have a series of sine waves with frequencies f, 3f, 5f, 7f, . . . and
amplitudes 4A/π, 4A/3π, 4A/5π, 4A/7π, and so on. The term with frequency f is dominant and is called the fundamental frequency. The term with frequency 3f is called
the third harmonic, the term with frequency 5f is the fifth harmonic, and so on. To
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58 CHAPTER 3 SIGNALS
recreate the complete square wave signal requires all the odd harmonics up to infinity.
For example, if the square wave has a frequency of 5000, the components have
frequencies 5000, 15,000, 25,000, and so on. Figure 3.9 shows three of the harmonics.
Figure 3.9 Three harmonics
Amplitude
A
•••
Time
Of course, if we add these three harmonics, we do not get a square wave—we get
something which is close, but not exact, as shown in Figure 3.10. If we need something
closer to a square wave, we need to add more harmonics.
Figure 3.10 Adding first three harmonics
•••
Time
Frequency Spectrum
The description of a signal using the frequency domain and containing all its components is called the frequency spectrum of that signal. For example, Figure 3.11 shows
the frequency spectrum of a square wave and the frequency spectrum of a signal which
is very close to a square wave (only three harmonics).
Composite Signal and Transmission Medium
A signal needs to pass through a transmission medium (cable or air). However, each
medium has its own characteristics. One of the characteristics of a medium is related to
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SECTION 3.2 ANALOG SIGNALS 59
Figure 3.11 Frequency spectrum comparison
•••
3f
f
5f
7f
9f
11f
13f
nf
a. Frequency spectrum of a square wave
f
3f
5f
b. Frequency spectrum of an approximation with only three harmonics
frequency. A medium may pass some frequencies and may block or weaken others. This
means that when we send a composite signal, containing many frequencies, at one end
of a transmission medium, we may not receive the same signal at the other end. To
maintain the integrity of the signal, the medium needs to pass every frequency (and also
preserve the amplitude and phase, as we shall see later).
What we must realize is that no transmission medium is perfect. Each medium
passes some frequencies, weakens others, and blocks still others. This means that when
we send our square wave signal through a medium, we get something at the other end
which is not a square wave at all. Figure 3.12 shows the concept.
Figure 3.12 Signal corruption
Transmission medium
Input signal
Output signal
Bandwidth
The range of frequencies that a medium can pass is called its bandwidth. Because no
medium can pass or block all frequencies, the bandwidth normally refers to the range
of frequencies that a medium can pass without losing one-half of the power contained
in that signal. The bandwidth is a range and is normally referred to as the difference
between two numbers. For example, if a medium can pass frequencies between 1000
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60 CHAPTER 3 SIGNALS
and 5000 without losing most of the power contained in this range, its bandwidth is
5000 – 1000, or 4000.
The bandwidth is a property of a medium: It is the difference between the highest and
the lowest frequencies that the medium can satisfactorily pass.
If the bandwidth of a medium does not match the spectrum of a signal, some of
the frequencies are lost. For example, the square wave signal in Figure 3.8 has a
spectrum that expands to infinity. No transmission medium has such a bandwidth.
This means that passing a square wave through any medium will always deform the
signal. As another example, voice normally has a spectrum of 300 to 3300 Hz (a
bandwidth of 3000 Hz). If we use a transmission line with a bandwidth of 1000
(between 1500 and 2500 Hz), we lose some frequencies in our voice; it may not even
be recognizable.
Sometimes people use the term bandwidth with regard to a signal. For example,
they say, “This signal has a bandwidth of 1000 Hz.” In this case, what they mean
is that the signal has a spectrum with significant frequencies that span 1000 Hz. In
other words, they mean, “We need a medium with a bandwidth of 1000 Hz if we
want to send this signal without losing a significant part of it.” We can say that today,
people use the term bandwidth for media and signals interchangeably, but it was not
always so.
In this book, we use the term bandwidth to refer to the property of a medium or the
width of a single spectrum.
Figure 3.13 shows the concept of bandwidth. The figure depicts the range of frequencies a medium can pass and the relative amplitude of the frequencies passed. Note
that the media may pass some frequencies above the 5000 and below 1000, but according to the criteria we mentioned before, the amplitudes of those frequencies are less
than those in the middle.
Figure 3.13 Bandwidth
Amplitude
1000
5000
Bandwidth 5000 1000 4000 Hz
Frequency
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SECTION 3.2 ANALOG SIGNALS 61
Example 3
If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700,
and 900 Hz, what is the 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
B = fh − fl = 900 − 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Fig. 3.14).
Figure 3.14 Example 3
Amplitude
10 V
100
300
500
700
900
Frequency
Bandwidth 900 100 800 Hz
Example 4
A 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 integral frequencies of the same amplitude.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
B = fh − fl
20 = 60 − fl
fl = 60 − 20 = 40 Hz
The spectrum contains all integral frequencies. We show this by a series of spikes (see Fig. 3.15).
Figure 3.15 Example 4
40 41 42
58 59 60
Bandwidth 60 40 20 Hz
fl 40 Hz
fh 60 Hz
Frequency
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62 CHAPTER 3 SIGNALS
Example 5
A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A
medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal
faithfully pass through this medium?
Solution
The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the
range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz;
the signal is totally lost.
3.3 DIGITAL SIGNALS
In addition to being represented by an analog signal, data can be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage
(see Fig. 3.16).
Figure 3.16 A digital signal
Amplitude
1
0
1
1
0
0
0
1
•••
Time
Bit Interval and Bit Rate
Most digital signals are aperiodic, and thus period or frequency is not appropriate.
Two new terms—bit interval (instead of period) and bit rate (instead of frequency)—
are used to describe digital signals. The bit interval is the time required to send one
single bit. The bit rate is the number of bit intervals per second. This means that the
bit rate is the number of bits sent in 1 s, usually expressed in bits per second (bps).
See Figure 3.17.
Example 6
A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)?
Solution
The bit interval is the inverse of the bit rate.
1
1
Bit interval = ---------------- = ------------ = 0.000500 s = 0.000500 × 106 µs = 500 µs
bit rate 2000
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SECTION 3.3 DIGITAL SIGNALS 63
Figure 3.17 Bit rate and bit interval
Amplitude
1
1 s 8 bit intervals
Bit rate 8 bps
0
1
1
0
0
0
1
•••
Time
Bit interval
Digital Signal as a Composite Analog Signal
It should be clear so far that a digital signal, with all its sudden changes, is actually a
composite signal having an infinite number of frequencies. In other words, the bandwidth of a digital signal is infinite.
A digital signal is a composite signal with an infinite bandwidth.
Digital Signal Through a Wide-Bandwidth Medium
If a medium has a wide bandwidth, we can send a digital signal through it. Of course,
some of the frequencies are blocked by the medium, but still enough frequencies are
passed to preserve a decent signal shape. We will see that we can use a dedicated
medium such as a coaxial cable to send a digital signal through a local area network.
Digital Signal Through a Band-Limited Medium
Can we send digital data through a band-limited medium? The answer to this question
is definitely yes. We send data by using band-limited telephone lines to the Internet
every day. But what is the minimum required bandwidth B in hertz if we want to send
n bps? In other words, what is the relationship between the number of bits per second
and the required bandwidth? We will formally answer this question when we discuss
the Nyquist theorem and the Shannon capacity. In this section, we take an intuitive
approach to understand the foundation of data transmission.
Using Only One Harmonic
To simplify the discussion, imagine that our computer creates just 6 bps. We make this
unrealistic assumption just to be able to show things graphically. Every second, 6 bits
are produced by the computer. One second, we may have 111111, another second
001010, another second 101010, and so on. We use an encoding method that uses a
positive value to represent 1 and a negative value to represent 0. Figure 3.18 shows two
signals.
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64 CHAPTER 3 SIGNALS
Figure 3.18 Digital versus analog
Digital
Analog
Amplitude
1
Amplitude
1
1
1
1
1
1
1
1
1
1
1
•••
•••
Time
Time
1s
1s
a. Best case, bit rate 6, f 0
Amplitude
Amplitude
1
0
1
0
1
0
1
0
1
0
•••
1
0
•••
Time
Time
1s
1s
b. Worst case, bit rate 6, f 3
Let us see if we can simulate any of these patterns by using a single-frequency signal. The best case is 111111 or 000000. We can simulate this case by sending a signal
of frequency zero. The worst case is definitely 101010 or 010101. These are the worst
cases, because there are more changes in this pattern than in any other pattern; with
each succeeding bit there is a change. More change means higher frequency. However,
we can simulate this digital signal by using a single-frequency analog signal with a frequency of 3 Hz, one-half of the bit rate. So we have
Best case:
Worst case:
bit rate = 6
bit rate = 6
frequency = 0
frequency = 3
We can say that all other cases are between the best and the worst cases. We can
simulate other cases with a single frequency of 1 or 2 Hz (using the appropriate phase).
In other words, if we need to simulate this digital signal of data rate 6 bps, sometimes we need to send a signal of frequency 0, sometimes 1, sometimes 2, and sometimes 3 Hz. We need our medium to be able to pass frequencies of 0 to 3 Hz. Our
medium needs to have a bandwidth of 3 Hz (3 − 0).
If we generalize this simple example, we come to a very simple relationship
between the bit rate and bandwidth. To send n bps through an analog channel using the
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SECTION 3.3 DIGITAL SIGNALS 65
above approximation, we need a bandwidth B such that
n
B = --2
Using More Harmonics
The above discussion was based on one harmonic. For each pattern we send a singlefrequency signal using a frequency of 0 to 3 Hz. However, in many situations, a onefrequency signal is not very appropriate; the analog signal may look very different from
the digital signal. The receiver may not recognize the signal correctly.
To improve the shape of the signal for better communication, particularly for high
data rates, we need to add some harmonics. It is clear from our previous discussions that
we need to add some odd harmonics. If we add the third harmonic to each case, we need
B = n/2 + 3n/2 = 4n/2 Hz; if we add third and fifth harmonics, we need B = n/2 + 3n/2 +
5n/2 = 9n/2 Hz; and so on. In other words, we have
n
B >= --2
or
n <= 2B
Table 3.2 shows how much bandwidth we need to send 1000 bps using this method.
Table 3.2 Bandwidth requirements
Bit Rate
Harmonic 1
Harmonics
1, 3
Harmonics
1, 3, 5
Harmonics
1, 3, 5, 7
n = 1 Kbps
B = 500 Hz
B = 2 KHz
B = 4.5 KHz
B = 8 KHz
n = 10 Kbps
B = 5 KHz
B = 20 KHz
B = 45 KHz
B = 80 KHz
n = 100 Kbps
B = 50 KHz
B = 200 KHz
B = 450 KHz
B = 800 KHz
We want to emphasize the following: In this method as well as others, the required
bandwidth is proportional to the bit rate. If we double the bit rate, we need to double
the bandwidth.
The bit rate and the bandwidth are proportional to each other.
Digital versus Analog Bandwidth
The above discussion on the proportionality of bandwidth and bit rate leads to the idea of digital bandwidth. If we are sending analog data through a medium, we are concerned with analog bandwidth (expressed in hertz); if we are sending digital data through a medium, we are
concerned with digital bandwidth (in bits per second). Analog bandwidth is the range of frequencies that a medium can pass. Digital bandwidth is the maximum bit rate that a medium
can pass. They represent the same property of a medium, but in different scales and units.
The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per
second.
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66 CHAPTER 3 SIGNALS
Higher Bit Rate
Some readers may be puzzled by the above discussion, especially if they consider data
transmission over telephone lines. Telephone lines have a bandwidth of 3 to 4 KHz
for the regular user; we know that sometimes we send more than 30,000 bps using a
traditional modem. According to the above discussion, we should not be able to send
more than 8000 bps. How can this be reconciled? It’s so because we are using a modem
with modulation techniques that allow the representation of multiple bits in one single
period of an analog signal. We discuss these techniques in Chapter 5.
3.4 ANALOG VERSUS DIGITAL
Finally we come to this question: Should we use analog or digital signals? It really
depends on the situation and on the available bandwidth.
Low-pass versus Band-pass
A channel or a link is either low-pass or band-pass. A low-pass channel has a bandwidth with frequencies between 0 and f. The lower limit is 0, the upper limit can be any
frequency (including infinity). On the other hand, a band-pass channel has a bandwidth
with frequencies between f1 and f2. Figure 3.19 shows the bandwidth of a low-pass
channel and a band-pass channel.
Figure 3.19 Low-pass and band-pass
Amplitude
0
Low-pass channel
Amplitude
f1
f2
Band-pass channel
f1
Frequency
Frequency
Digital Transmission
A digital signal theoretically needs a bandwidth between 0 and infinity. The lower limit (0)
is fixed; the upper limit (infinity) can be relaxed if we lower our standards by accepting
a limited number of harmonics. This means a bandwidth between 0 and f for a low-pass
signal.
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SECTION 3.5 DATA RATE LIMITS 67
We have a low-pass channel only if the medium is dedicated to two devices (pointto-point) or shared between several devices in time (not in frequency). For example, in
a wired local area network, a cable can be shared between stations. We can transmit
data digitally in this system.
Digital transmission needs a low-pass channel.
Analog Transmission
An analog signal normally has a narrower bandwidth than a digital signal with frequencies between f1 and f2. In other words, an analog signal requires a band-pass channel. In
addition, the bandwidth of an analog signal can always be shifted. For example, we can
always shift a signal with a bandwidth from f1 to f2 to a signal with a bandwidth from f3
to f4 as long as the width of the bandwidth remains the same.
A band-pass channel is more available than a low-pass channel. The bandwidth of
a medium can be divided into several band-pass channels to carry several analog transmissions. For example, in analog cellular telephony, a limited bandwidth is divided
between many telephone users. Each user has a bandwidth between 0 to 30 KHz, with
each signal shifted appropriately.
Analog transmission can use a band-pass channel.
This is not to say that an analog transmission cannot use a low-pass channel; it just
means that it can use the more available band-pass channel. A low-pass channel is a special
case of a band-pass channel with f1 = 0.
3.5 DATA RATE LIMITS
A very important question 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 levels of signals we can use
3. The quality of the channel (the level of the noise).
Two theoretical formulas were developed to calculate the data rate: one by Nyquist for
a noiseless channel, another by Shannon for a noisy channel.
Noiseless Channel: Nyquist Bit Rate
For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum
bit rate
BitRate = 2 × Bandwidth × log2 L
In this formula, Bandwidth is the bandwidth of the channel, L is the number of signal
levels used to represent data, and BitRate is the bit rate in bits per second.
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68 CHAPTER 3 SIGNALS
Example 7
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:
BitRate = 2 × 3000 × log2 2 = 6000 bps
Example 8
Consider the same noiseless channel, transmitting a signal with four signal levels (for each level,
we send two bits). The maximum bit rate can be calculated as:
BitRate = 2 × 3000 × log2 4 = 12,000 bps
Noisy Channel: Shannon Capacity
In reality, we cannot have a noiseless channel; the channel is always noisy. In 1944,
Claude Shannon introduced a formula, called the Shannon capacity, to determine the
theoretical highest data rate for a noisy channel:
Capacity = Bandwidth × log2 (1 + SNR)
In this formula, Bandwidth is the bandwidth of the channel, SNR is the signal-to-noise
ratio, and Capacity is the capacity of the channel in bits per second. The signal-to-noise
ratio is the statistical ratio of the power of the signal to the power of the noise. Note that
in the Shannon formula there is no indication of the signal level, which means that no
matter how many levels we use, we cannot achieve a data rate higher than the capacity
of the channel. In other words, the formula defines a characteristic of the channel, not the
method of transmission.
Example 9
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 is calculated as
C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B × 0 = 0
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.
Example 10
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps
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SECTION 3.6 TRANSMISSION IMPAIRMENT 69
This means that the highest bit rate for a telephone line is 34,860 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.
Using Both Limits
In practice, we need to use both methods to find what bandwidth of what signal level we
need. Let us show this by an example.
Example 11
We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?
Solution
First, we use the Shannon formula to find our upper limit.
C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps
Although the Shannon formula gives us 6 Mbps, this is the upper limit. For better performance we choose something lower, for example 4 Mbps. Then we use the Nyquist formula to
find the number of signal levels.
4 Mbps = 2 × 1 MHz × log2 L
L=4
3.6 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not perfect. The imperfections cause
impairment in the signal. This means that the signal at the beginning and end of the
medium are not the same. What is sent is not what is received. Three types of impairment usually occur: attenuation, distortion, and noise (see Fig. 3.20).
Figure 3.20 Impairment types
Impairment
Attenuation
Distortion
Noise
Attenuation
Attenuation means loss of energy. When a signal, simple or composite, travels
through a medium, it loses some of its energy so that it can overcome the resistance of
the medium. That is why a wire carrying electrical signals gets warm, if not hot, after a
while. Some of the electrical energy in the signal is converted to heat. To compensate
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70 CHAPTER 3 SIGNALS
for this loss, amplifiers are used to amplify the signal. Figure 3.21 shows the effect of
attenuation and amplification.
Figure 3.21 Attenuation
Original
Attenuated
Amplified
Amplifier
Point 1
Transmission medium
Point 2
Point 3
Decibel
To show that a signal has lost or gained strength, engineers use the concept of the decibel. The decibel (dB) measures the relative strengths of two signals or a signal at two
different points. Note that the decibel is negative if a signal is attenuated and positive if
a signal is amplified.
dB = 10 log10 (P2 /P1)
where P1 and P2 are the powers of a signal at points 1 and 2, respectively.
Example 12
Imagine a signal travels through a transmission medium and its power is reduced to half. This
means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as
10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB
Engineers know that –3 dB or a loss of 3 dB is equivalent to losing half the power.
Example 13
Imagine a signal travels through an amplifier and its power is increased ten times. This means
that P2 = 10 × P1. In this case, the amplification (gain of power) can be calculated as
10 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB
Example 14
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 talking about several points instead of
just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified.
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SECTION 3.6 TRANSMISSION IMPAIRMENT 71
Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the
signal just by adding the decibel measurements between each set of points.
Figure 3.22 Example 14
1 dB
⫺3 dB
⫺3 dB
7 dB
Amplifier
Point 1
Transmission
medium
Point 2
Point 3
Transmission
medium
Point 4
In this case, the decibel can be calculated as
dB = –3 + 7 – 3 = +1
which means that the signal has gained power.
Distortion
Distortion means that the signal changes its form or shape. Distortion occurs in a composite signal, made of different frequencies. Each signal component has its own propagation
speed (see the next section) through a medium and, therefore, its own delay in arriving at
the final destination. Figure 3.23 shows the effect of distortion on a composite signal.
Figure 3.23 Distortion
Transmission
medium
Composite signal
sent
Point 1
Components,
in phase
Point 2
Components,
out of phase
Composite signal
received
Noise
Noise is another problem. Several types of noise such as thermal noise, induced noise,
crosstalk, and impulse noise may corrupt the signal. Thermal noise is the random
motion of electrons in a wire which creates an extra signal not originally sent by the
transmitter. Induced noise comes from sources such as motors and appliances. These
devices act as a sending antenna and the transmission medium acts as the receiving antenna.
Crosstalk is the effect of one wire on the other. One wire acts as a sending antenna and
the other as the receiving antenna. Impulse noise is a spike (a signal with high energy in
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72 CHAPTER 3 SIGNALS
a very short period of time) that comes from power lines, lightning, and so on. Figure 3.24
shows the effect of noise on a signal.
Figure 3.24 Noise
Transmitted
Point 1
Noise
Received
Transmission medium
Point 2
3.7 MORE ABOUT SIGNALS
Four other measurements used in data communications are throughput, propagation
speed, propagation time, and wavelength. We discuss these in this section before closing
the chapter.
Throughput
The throughput is the measurement of how fast data can pass through an entity (such
as a point or a network). In other words, if we consider this entity as a wall through
which bits pass, throughput is the number of bits that can pass this wall in one second.
Figure 3.25 shows the concept.
Direction
Im
ag
in
ar
y
w
al
l
Figure 3.25 Throughput
11110011111000010011
Transmission medium
111100101010101010101111100010
Transmission medium
Throughput is the number of bits passing
through this wall in a second.
Propagation Speed
Propagation speed measures the distance a signal or a bit can travel through a medium
in one second. The propagation speed of electromagnetic signals depends on the
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SECTION 3.7 MORE ABOUT SIGNALS 73
medium and on the frequency of the signal. For example, in a vacuum, light is propagated with a speed of 3 × 108 m/s. It is lower in air. It is much lower in a cable.
Propagation Time
Propagation time measures the time required for a signal (or a bit) to travel from one
point of the transmission medium to another. The propagation time is calculated by dividing the distance by the propagation speed. Figure 3.26 shows the concept.
Propagation time = Distance/Propagation speed
Figure 3.26 Propagation time
Propagation time ⫽ t2 ⫺ t1 ⫽ dd/Propagation speed
Distance ⫽ d
111000010011
At time t1
111000010011
At time t2
Wavelength
Wavelength is another characteristic of a signal traveling through a transmission
medium. Wavelength binds the period or the frequency of a simple sine wave to the
propagation speed of the medium. In other words, while the frequency of a signal is
independent of the medium, the wavelength depends on both the frequency and the
medium. Although wavelength can be associated with electrical signals, it is customary
to use wavelengths when talking about the transmission of light in an optical fiber. The
wavelength is the distance a simple signal can travel in one period (see Fig. 3.27).
Figure 3.27 Wavelength
At time t
Wavelength
Transmission medium
Direction of
propagation
Transmission medium
At time t ⫹ T
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74 CHAPTER 3 SIGNALS
Wavelength can be calculated given the propagation speed and the period of the
signal
Wavelength = Propagation speed × Period
However, since period and frequency are related to each other, we can also say
Wavelength = Propagation speed × (1/Frequency) = Propagation speed/Frequency
If we represent wavelength by λ, propagation speed by c (speed of light), and frequency
by f, we get
λ = c/f
The wavelength is normally measured in micrometers (microns) instead of meters. For
example, the wavelength of red light (frequency = 4 × 1014) in air is
λ = c/f = (3 × 108)/(4 × 1014) = 0.75 × 10–6 m = 0.75 µm
In a coaxial or fiber-optic cable, however, the wavelength is lower (0.5 µm) because the
propagation speed in the cable is less than in the air.
3.8 KEY TERMS
analog data
analog signal
aperiodic signal
attenuation
band-pass channel
bandwidth
bit interval
bit rate
bits per second (bps)
composite signal
cycle
decibel (dB)
digital
digital data
digital signal
distortion
Fourier analysis
frequency
frequency-domain plot
frequency spectrum
fundamental frequency
harmonics
hertz (Hz)
low-pass channel
noise
Nyquist bit rate
peak amplitude
period
periodic signal
phase
propagation speed
propagation time
Shannon capacity
signal
signal-to-noise ratio (SNR)
sine wave
throughput
time-domain plot
wavelength
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SECTION 3.10 PRACTICE SET 75
3.9 SUMMARY
❏ Data must be transformed into electromagnetic signals prior to transmission across
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
❏
a network.
Data and signals can be either analog or digital.
A signal is periodic if it consists of a continuously repeating pattern.
Each sine wave can be characterized by its amplitude, frequency, and phase.
Frequency and period are inverses of each other.
A time-domain graph plots amplitude as a function of time.
A frequency-domain graph plots each sine wave’s peak amplitude against its
frequency.
By using Fourier analysis, any composite signal can be represented as a combination
of simple sine waves.
The spectrum of a signal consists of the sine waves that make up the signal.
The bandwidth of a signal is the range of frequencies the signal occupies. Bandwidth
is determined by finding the difference between the highest and lowest frequency
components.
Bit rate (number of bits per second) and bit interval (duration of 1 bit) are terms
used to describe digital signals.
A digital signal is a composite signal with an infinite bandwidth.
Bit rate and bandwidth are proportional to each other.
The Nyquist formula determines the theoretical data rate for a noiseless channel.
The Shannon capacity determines the theoretical maximum data rate for a noisy channel.
Attenuation, distortion, and noise can impair a signal.
Attenuation is the loss of a signal’s energy due to the resistance of the medium.
The decibel measures the relative strength of two signals or a signal at two
different points.
Distortion is the alteration of a signal due to the differing propagation speeds of
each of the frequencies that make up a signal.
Noise is the external energy that corrupts a signal.
We can evaluate transmission media by throughput, propagation speed, and
propagation time.
The wavelength of a frequency is defined as the propagation speed divided by the
frequency.
3.10 PRACTICE SET
Review Questions
1. Describe the three characteristics of a sine wave.
2. What is the spectrum of a signal?
3. Contrast an analog signal with a digital signal.
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76 CHAPTER 3 SIGNALS
4. A signal has been received that only has values of –1, 0, and 1. Is this an analog or
a digital signal?
5. What is the relationship between period and frequency?
6. What are the units of period and frequency?
7. What does the amplitude of a signal measure?
8. What does the frequency of a signal measure?
9. What does the phase of a signal measure?
10. Which type of plot shows the amplitude of a signal at a given time?
11. How can a composite signal be decomposed into its individual frequencies?
12. What is a bit interval, and what is its counterpart in an analog signal?
13. What is bit rate, and what is its counterpart in an analog signal?
14. Name three types of transmission impairment.
15. What does a decibel measure?
16. What is the relationship between propagation speed and propagation time?
17. What is the wavelength of a signal and how is it calculated?
18. What does the Shannon capacity have to do with communications?
Multiple-Choice Questions
19. Before data can be transmitted, they must be transformed to ________.
a. Periodic signals
b. Electromagnetic signals
c. Aperiodic signals
d. Low-frequency sine waves
20. A periodic signal completes one cycle in 0.001 s. What is the frequency?
a. 1 Hz
b. 100 Hz
c. 1 KHz
d. 1 MHz
21. Which of the following can be determined from a frequency-domain graph of a
signal?
a. Frequency
b. Phase
c. Power
d. All the above
22. Which of the following can be determined from a frequency-domain graph of a signal?
a. Bandwidth
b. Phase
c. Power
d. All the above
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SECTION 3.10 PRACTICE SET 77
23. In a frequency-domain plot, the vertical axis measures the ________.
a. Peak amplitude
b. Frequency
c. Phase
d. Slope
24. In a frequency-domain plot, the horizontal axis measures the ________.
a. Peak amplitude
b. Frequency
c. Phase
d. Slope
25. In a time-domain plot, the vertical axis is a measure of ________.
a. Amplitude
b. Frequency
c. Phase
d. Time
26. In a time-domain plot, the horizontal axis is a measure of ________.
a. Signal amplitude
b. Frequency
c. Phase
d. Time
27. If the bandwidth of a signal is 5 KHz and the lowest frequency is 52 KHz, what is
the highest frequency?
a. 5 KHz
b. 10 KHz
c. 47 KHz
d. 57 KHz
28. What is the bandwidth of a signal that ranges from 40 KHz to 4 MHz?
a. 36 MHz
b. 360 KHz
c. 3.96 MHz
d. 396 KHz
29. When one of the components of a signal has a frequency of zero, the average
amplitude of the signal ________.
a. Is greater than zero
b. Is less than zero
c. Is zero
d. (a) or (b)
30. A periodic signal can always be decomposed into ________.
a. Exactly an odd number of sine waves
b. A set of sine waves
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78 CHAPTER 3 SIGNALS
31.
32.
33.
34.
35.
36.
37.
c. A set of sine waves, one of which must have a phase of 0°
d. None of the above
As frequency increases, the period ________.
a. Decreases
b. Increases
c. Remains the same
d. Doubles
Given two sine waves A and B, if the frequency of A is twice that of B, then the
period of B is ________ that of A.
a. One-half
b. Twice
c. The same as
d. Indeterminate from
A sine wave is ________.
a. Periodic and continuous
b. Aperiodic and continuous
c. Periodic and discrete
d. Aperiodic and discrete
If the maximum amplitude of a sine wave is 2 V, the minimum amplitude is
________ V.
a. 2
b. 1
c. –2
d. Between –2 and 2
A signal is measured at two different points. The power is P1 at the first point and
P2 at the second point. The dB is 0. This means ________.
a. P2 is zero
b. P2 equals P1
c. P2 is much larger than P1
d. P2 is much smaller than P1
________ is a type of transmission impairment in which the signal loses strength
due to the resistance of the transmission medium.
a. Attenuation
b. Distortion
c. Noise
d. Decibel
________ is a type of transmission impairment in which the signal loses strength
due to the different propagation speeds of each frequency that makes up the signal.
a. Attenuation
b. Distortion
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SECTION 3.10 PRACTICE SET 79
38.
39.
40.
41.
42.
43.
44.
c. Noise
d. Decibel
________ is a type of transmission impairment in which an outside source such as
crosstalk corrupts a signal.
a. Attenuation
b. Distortion
c. Noise
d. Decibel
The ________ has units of meters/second or kilometers/second.
a. Throughput
b. Propagation speed
c. Propagation time
d. (b) or (c)
________ has units of bits/second.
a. Throughput
b. Propagation speed
c. Propagation time
d. (b) or (c)
The ________ has units of seconds.
a. Throughput
b. Propagation speed
c. Propagation time
d. (b) or (c)
When propagation speed is multiplied by propagation time, we get the ________.
a. Throughput
b. Wavelength of the signal
c. Distortion factor
d. Distance a signal or bit has traveled
Propagation time is ________ proportional to distance and ________ proportional to
propagation speed.
a. Inversely; directly
b. Directly; inversely
c. Inversely; inversely
d. Directly; directly
Wavelength is ________ proportional to propagation speed and ________ proportional
to period.
a. Inversely; directly
b. Directly; inversely
c. Inversely; inversely
d. Directly; directly
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80 CHAPTER 3 SIGNALS
45. The wavelength of a signal depends on the ________.
a. Frequencies of the signal
b. Medium
c. Phase of the signal
d. (a) and (b)
46. The wavelength of green light in air is ________ the wavelength of green light in
fiber-optic cable.
a. Less than
b. Greater than
c. Equal to
d. None of the above
47. Using the Shannon formula to calculate the data rate for a given channel, if C = B,
then ________.
a. The signal is less than the noise
b. The signal is greater than the noise
c. The signal is equal to the noise
d. Not enough information is given to answer the question
Exercises
48. Given the frequencies listed below, calculate the corresponding periods. Express
the result in seconds, milliseconds, microseconds, nanoseconds, and picoseconds.
a. 24 Hz
b. 8 MHz
c. 140 KHz
d. 12 THz
49. Given the following periods, calculate the corresponding frequencies. Express the
frequencies in hertz, kilohertz, megahertz, gigahertz, and terahertz.
a. 5 s
b. 12 µs
c. 220 ns
d. 81 ps
50. What is the phase shift for the following?
a. A sine wave with the maximum amplitude at time zero
b. A sine wave with maximum amplitude after 1/4 cycle
c. A sine wave with zero amplitude after 3/4 cycle and increasing
d. A sine wave with minimum amplitude after 1/4 cycle
51. Show the phase shift in degrees corresponding to each of the following delays in
cycles.
a. 1 cycle
b. 1/2 cycle
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SECTION 3.10 PRACTICE SET 81
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
c. 3/4 cycle
d. 1/3 cycle
Show the delay in cycles corresponding to each of the following.
a. 45°
b. 90°
c. 60°
d. 360°
Draw the time-domain plot of a sine wave (for only 1 s) with a maximum amplitude of 15 V, a frequency of 5, and a phase of 270°.
Draw two sine waves on the same time-domain plot. The characteristics of each
signal are as follows:
signal A: amplitude 40, frequency 9, phase 0;
signal B: amplitude 10, frequency 9, phase 90.
Draw two periods of a sine wave with a phase shift of 90°. On the same diagram,
draw a sine wave with the same amplitude and frequency but with a 90° phase shift
from the first.
What is the bandwidth of a signal that can be decomposed into four sine waves
with frequencies at 0, 20, 50, and 200 Hz? All maximum amplitudes are the same.
Draw the frequency spectrum.
A periodic composite signal with a bandwidth of 2000 Hz is composed of two
sine waves. The first one has a frequency of 100 Hz with a maximum amplitude
of 20 V; the second one has a maximum amplitude of 5 V. Draw the frequency
spectrum.
Show how a sine wave can change its phase by drawing two periods of an arbitrary
sine wave with phase shift of 0° followed by the two periods of the same signal
with a phase shift of 90°.
Imagine we have a sine wave called A. Show the negative of A. In other words,
show the signal −A. Can we relate the negation of a signal to the phase shift? How
many degrees?
Which signal has a higher bandwidth, a signal that changes 100 times per second
or a signal that changes 200 times per second?
What is the bit rate for each of the following signals?
a. A signal in which 1 bit lasts 0.001 s
b. A signal in which 1 bit lasts 2 ms
c. A signal in which 10 bits last 20 µs
d. A signal in which 1000 bits last 250 ps
What is the duration of 1 bit for each of the following signals?
a. A signal with a bit rate of 100 bps
b. A signal with a bit rate of 200 Kbps
c. A signal with a bit rate of 5 Mbps
d. A signal with a bit rate of 1 Gbps
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82 CHAPTER 3 SIGNALS
63. A device is sending out data at the rate of 1000 bps.
a. How long does it take to send out 10 bits?
b. How long does it take to send out a single character (8 bits)?
c. How long does it take to send a file of 100,000 characters?
64. What is the bit rate for the signal in Figure 3.28?
Figure 3.28 Exercise 64
16 ns
•••
Time
65. What is the frequency of the signal in Figure 3.29?
Figure 3.29 Exercise 65
4 ms
Time
66. Draw the time-domain representation (for the first 1/100 s) of the signal shown in
Figure 3.30.
Figure 3.30 Exercise 66
10 V
2500 Hz
67.
68.
69.
70.
Frequency
Draw the frequency-domain representation of the signal shown in Figure 3.31.
What is the bandwidth of the composite signal shown in Figure 3.32?
What is the bandwidth of the signal shown in Figure 3.33?
A composite signal contains frequencies from 10 to 30 KHz, each with an amplitude of 10 V. Draw the frequency spectrum.
71. A composite signal contains frequencies from 10 to 30 KHz. The amplitude is zero
for the lowest and the highest signals and is 30 V for the 20-KHz signal. Assuming
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SECTION 3.10 PRACTICE SET 83
Figure 3.31 Exercise 67
7 ␮s
5
Time
Figure 3.32 Exercise 68
180
5
5
5
5
Frequency
5
Figure 3.33 Exercise 69
Frequency
280
that the amplitudes change gradually from the minimum to the maximum, draw the
frequency spectrum.
72. Two signals have the same frequencies. However, whenever the first signal is at its
maximum amplitude, the second signal has an amplitude of zero. What is the phase
shift between the two signals?
73. What is the mathematical representation of a signal with an amplitude of 10 V, a
frequency of 2500 Hz, a phase of 30°?
74. Show the frequency domain of the following signal:
s(t) = 8 + 3 sin 100πt + 5 sin 200πt
75. What is the period of the following signal?
s(t) = 4 sin 628t
76. A cosine wave is a sine wave with a 90° phase shift. Show the equivalent of the following signal in sine format.
s(t) = cos (2πft + π)
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84 CHAPTER 3 SIGNALS
77. A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one
channel, what are the data rates if we use one harmonic, three harmonics, and five
harmonics?
78. A signal travels from point A to point B. At point A, the signal power is 100 W. At
point B, the power is 90 W. What is the attenuation in decibel?
79. The attenuation of a signal is –10 dB. What is the final signal power if it was originally 5 W?
80. A signal has passed through three cascaded amplifiers, each with a 4 dB gain.
What is the total gain? How much is the signal amplified?
81. If the throughput at the connection between a device and the transmission medium
is 5 Kbps, how long does it take to send 100,000 bits out of this device?
82. The light of the sun takes approximately eight minutes to reach the earth. What is
the distance between the sun and the earth?
83. A signal has a wavelength of 1 µm in air. How far can the front of the wave travel
during five periods?
84. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is
the maximum data rate supported by this line?
85. We measure the performance of a telephone line (4 KHz of bandwidth). When the
signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this
telephone line?