Data and Signals.ppt

3.1
Chapter 3
Data and Signals
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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

3.3
3-1 ANALOG AND DIGITAL
Data can be analog or digital. The term analog data refers
to information that is continuous; digital data refers to
information that has discrete states. Analog data take on
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:

3.4
Analog and Digital Data
 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
Analog and Digital Signals
• 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.

3.6
Figure 3.1 Comparison of analog and digital signals

3.7
3-2 PERIODIC ANALOG SIGNALS
In data communications, we commonly use periodic
analog signals and nonperiodic digital signals.
Periodic analog signals can be classified as simple or
composite. A simple periodic analog signal, a sine wave,
cannot be decomposed into simpler signals. A composite
periodic analog signal is composed of multiple sine
waves.
 Sine Wave
 Wavelength
 Time and Frequency Domain
 Composite Signals
 Bandwidth

3.9
Figure 3.3 Two signals with the same phase and frequency,
but different amplitudes

3.10
Frequency and period are the inverse of
each other.
Note

3.11
Figure 3.4 Two signals with the same amplitude and phase,
but different frequencies

3.12
Table 3.1 Units of period and frequency

3.13
The power we use at home has a frequency of 60 Hz.
The period of this sine wave can be determined as
follows:
Example 3.1

3.14
The period of a signal is 100 ms. What is its frequency in
kilohertz?
Example 3.2
Solution
First we change 100 ms to seconds, and then we
calculate the frequency from the period (1 Hz = 10−3
kHz).

3.15
Frequency
• 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.

3.16
If a signal does not change at all, its
frequency is zero.
If a signal changes instantaneously, its
frequency is infinite.
Note

3.17
Phase describes the position of the
waveform relative to time 0.
Note

3.18
Figure 3.5 Three sine waves with the same amplitude and frequency,
but different phases

3.19
Figure 3.6 Wavelength and period

3.20
Figure 3.7 The time-domain and frequency-domain plots of a sine wave

3.21
A complete sine wave in the time
domain can be represented by one
single spike in the frequency domain.
Note

3.22
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

3.23
Figure 3.8 The time domain and frequency domain of three sine waves

3.24
Signals and Communication
 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.
 According to Fourier analysis, any
composite signal is a combination of
simple sine waves with different
frequencies, amplitudes, and phases.

3.25
Composite Signals and
Periodicity
 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.

3.26
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 3.4

3.27
Figure 3.9 A composite periodic signal

3.28
Figure 3.10 Decomposition of a composite periodic signal in the time and
frequency domains

3.29
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 3.5

3.30
Figure 3.11 The time and frequency domains of a nonperiodic signal

3.31
Bandwidth and Signal
Frequency
 The bandwidth of a composite signal is
the difference between the highest and the
lowest frequencies contained in that
signal.

3.32
Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

3.33
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 3.6
The spectrum has only five spikes, at 100, 300, 500, 700,
and 900 Hz (see Figure 3.13).

3.34
Figure 3.13 The bandwidth for Example 3.6

3.35
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 3.7
The spectrum contains all integer frequencies. We show
this by a series of spikes (see Figure 3.14).

3.36

3.37
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 3.8

3.38

3.39
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 3.9

3.40
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 3.10

3.41
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 3.11

3.42
3-3 DIGITAL SIGNALS
In addition to being represented by an analog signal,
information can also be represented by a digital signal.
For example, a 1 can be encoded as a positive voltage
and a 0 as zero voltage. A digital signal can have more
than two levels. In this case, we can send more than 1 bit
for each level.
 Bit Rate
 Bit Length
 Digital Signal as a Composite Analog Signal
 Application Layer

3.43
Figure 3.16 Two digital signals: one with two signal levels and the other
with four signal levels

3.44
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.

3.45
Assume we need to download text documents at the rate
of 100 pages per sec. 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 (ascii), the bit rate is
Example 3.18

3.46
What is the bit rate for high-definition TV (HDTV)?
Solution
HDTV uses digital signals to broadcast high quality
video signals. The HDTV screen is normally a ratio of
16 : 9. There are 1920 by 1080 pixels per screen, and the
screen is renewed 30 times per second. Twenty-four bits
represents one color pixel.
Example 3.20
The TV stations reduce this rate to 20 to 40 Mbps
through compression.

3.47
Figure 3.17 The time and frequency domains of periodic and nonperiodic
digital signals

3.48
Figure 3.18 Baseband transmission

3.49
Figure 3.23 Bandwidth of a bandpass channel

3.50
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

3.51
Figure 3.24 Modulation of a digital signal for transmission on a bandpass
channel

3.52
3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not
perfect. The imperfection causes signal impairment. This
means that the signal at the beginning of the medium is
not the same as the signal at the end of the medium.
What is sent is not what is received. Three causes of
impairment are attenuation, distortion, and noise.
 Attenuation
 Distortion
 Noise

3.53
Figure 3.25 Causes of impairment

3.54
Attenuation
 Means loss of energy -> weaker signal
 When a signal travels through a medium it
loses energy overcoming the resistance of
the medium
 Amplifiers are used to compensate for this
loss of energy by amplifying the signal.

3.55
Measurement of Attenuation
 To show the loss or gain of energy the unit
“decibel” is used.
dB = 10log10P2/P1
P1 - input signal
P2 - output signal

3.57
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.

3.58
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

3.59
Figure 3.27 Decibels for Example 3.28

3.60
The loss in a cable is usually defined in decibels per
kilometer (dB/km). If the signal at the beginning of a
cable with −0.3 dB/km has a power of 2 mW, what is the
power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as
Example 3.30

3.61
Distortion
 Means that the signal changes its form or shape
 Distortion occurs in composite signals
 Each frequency component has its own
propagation speed traveling through a medium.
 The different components therefore arrive with
different delays at the receiver.
 That means that the signals have different phases
at the receiver than they did at the source.

3.63
Noise
 There are different types of noise
 Thermal - random noise of electrons in the wire
creates an extra signal
 Induced - from motors and appliances, devices
act are transmitter antenna and medium as
receiving antenna.
 Crosstalk - same as above but between two
wires.
 Impulse - Spikes that result from power lines,
lighning, etc.

3.65
Signal to Noise Ratio (SNR)
 To measure the quality of a system the SNR
is often used. It indicates the strength of the
signal wrt the noise power in the system.
 It is the ratio between two powers.
 It is usually given in dB and referred to as
SNRdB.

3.66
The power of a signal is 10 mW and the power of the
noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as
follows:
Example 3.31

3.67
The values of SNR and SNRdB for a noiseless channel
are
Example 3.32
We can never achieve this ratio in real life; it is an ideal.

3.68
Figure 3.30 Two cases of SNR: a high SNR and a low SNR

3.69
3-5 DATA RATE LIMITS
A very important consideration in data communications
is how fast we can send data, in bits per second, over a
channel. Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
 Noiseless Channel: Nyquist Bit Rate
 Noisy Channel: Shannon Capacity
 Using Both Limits

3.70
Increasing the levels of a signal
increases the probability of an error
occurring, in other words it reduces the
reliability of the system. Why??
Note

3.71
Capacity of a System
 The bit rate of a system increases with an increase
in the number of signal levels we use to denote a
symbol.
 A symbol can consist of a single bit or “n” bits.
 The number of signal levels = 2n.
 As the number of levels goes up, the spacing
between level decreases -> increasing the
probability of an error occurring in the presence of
transmission impairments.

3.72
Nyquist Theorem
 Nyquist gives the upper bound for the bit rate of a
transmission system by calculating the bit rate
directly from the number of bits in a symbol (or
signal levels) and the bandwidth of the system
(assuming 2 symbols/per cycle and first
harmonic).
 Nyquist theorem states that for a noiseless
channel:
C = 2 B log22n
C= capacity in bps
B = bandwidth in Hz

3.73
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

3.74
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

3.75
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

3.76
We need to send 265 kbps over a noiseless channel with
a bandwidth of 20 kHz. How many signal levels do we
need?
Solution
We can use the Nyquist formula as shown:
Example 3.36
Since this result is not a power of 2, we need to either
increase the number of levels or reduce the bit rate. If we
have 128 levels, the bit rate is 280 kbps. If we have 64
levels, the bit rate is 240 kbps.

3.77
Shannon’s Theorem
 Shannon’s theorem gives the capacity of a
system in the presence of noise.
C = B log2(1 + SNR)

3.78
Consider an extremely noisy channel in which the value
of the signal-to-noise ratio is almost zero. In other
words, the noise is so strong that the signal is faint. For
this channel the capacity C is calculated as
Example 3.37
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.

3.79
We can calculate the theoretical highest bit rate of a
regular telephone line. A telephone line normally has a
bandwidth of 3000. The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as
Example 3.38
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.

3.80
The signal-to-noise ratio is often given in decibels.
Assume that SNRdB = 36 and the channel bandwidth is 2
MHz. The theoretical channel capacity can be calculated
as
Example 3.39

3.81
For practical purposes, when the SNR is very high, we
can assume that SNR + 1 is almost the same as SNR. In
these cases, the theoretical channel capacity can be
simplified to
Example 3.40
For example, we can calculate the theoretical capacity of
the previous example as

3.82
We have a channel with a 1-MHz bandwidth. The SNR
for this channel is 63. What are the appropriate bit rate
and signal level?
Solution
First, we use the Shannon formula to find the upper
limit.
Example 3.41

3.83
The Shannon formula gives us 6 Mbps, the upper limit.
For better performance we choose something lower, 4
Mbps, for example. Then we use the Nyquist formula to
find the number of signal levels.
Example 3.41 (continued)

3.84
The Shannon capacity gives us the
upper limit; the Nyquist formula tells us
how many signal levels we need.
Note

3.85
3-6 PERFORMANCE
One important issue in networking is the performance of
the network—how good is it? We discuss quality of
service, an overall measurement of network performance,
in greater detail in Chapter 24. In this section, we
introduce terms that we need for future chapters.
 Bandwidth - capacity of the system
 Throughput - no. of bits that can be
pushed through
 Latency (Delay) - delay incurred by a bit
from start to finish
 Bandwidth-Delay Product

3.86
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. Often referred to as
Capacity.
Note

3.87
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

3.88
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

3.89
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.

3.90
Propagation & Transmission delay
 Propagation speed - speed at which a bit
travels though the medium from source to
destination.
 Transmission speed - the speed at which all
the bits in a message arrive at the
destination. (difference in arrival time of
first and last bit)

3.91
Propagation and Transmission Delay
 Propagation Delay = Distance/Propagation speed
 Transmission Delay = Message size/bandwidth bps
 Latency = Propagation delay + Transmission delay +
Queueing time + Processing time

3.92
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.

3.93
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

3.94
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.

3.95
What are the propagation time and the transmission
time for a 5-Mbyte message (an image) if the bandwidth
of the network is 1 Mbps? 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
times as shown on the next slide.
Example 3.47

3.96
Note that in this case, because the message is very long
and the bandwidth is not very high, the dominant factor
is the transmission time, not the propagation time. The
propagation time can be ignored.

3.97
Figure 3.31 Filling the link with bits for case 1

3.98
We can think about the link between two points as a
pipe. The cross section of the pipe represents the
bandwidth, and the length of the pipe represents the
delay. We can say the volume of the pipe defines the
bandwidth-delay product, as shown in Figure 3.33.
Example 3.48

3.99
Figure 3.32 Filling the link with bits in case 2

3.100
The bandwidth-delay product defines
the number of bits that can fill the link.
Note

3.101
Figure 3.33 Concept of bandwidth-delay product

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