4123702
Data Communications System
By
Ajarn Preecha Pangsuban
Part 2 – Physical Layer
 Interacts with transmission media
 Creates a signal representing 0s and 1s
 Physical movement of data
 Determines direction of data flow
 Decides on the number of logical channels for
transporting data coming from different source
Coming up…
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Position of the physical Layer
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Physical layer services
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Transmission Media
 Guided
Twisted-pair
Coaxial cable
Fiber-optic



 Unguided
Radio
Microwave


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Networks and Technologies
 Telephone network – circuit-switched network
 High speed access
Modems
DSL
Cable



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Part 2 Chapters
Chapter 3
Signals
Chapter 4
Digital Transmission
Chapter 5
Analog Transmission
Chapter 6
Multiplexing
Chapter 7
Transmission Media
Chapter 8
Circuit Switching and Telephone Network
Chapter 9
High Speed Digital Access
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Chapter 3: Signals
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Note:
To be transmitted, data must be
transformed to electromagnetic
signals.
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Analog and Digital Signals
 Signals can be analog or digital form
 Analog signals can have an infinite number of values in a
range
 digital signals can have only a limited number of values.
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Periodic and Aperiodic Signals
 Periodic – completes a pattern within a
measurable time frame, called a period


One full pattern is a cycle
Analog signals
 Aperiodic – changes without exhibiting a pattern
Digital signals

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Periodic
Signals
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Aperiodic Signals
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Note:
In data communication, we commonly
use periodic analog signals and
aperiodic digital signals.
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Analog Signals
 Sine wave – most fundamental form of a periodic analog
signal
 Fully described by: Amplitude, Frequency and Phase
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Analog Signaling
 More susceptible to noise and less precise than a
digital signal
 Benefit - because they are more variable than
digital signals, they can convey greater subtleties
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Amplitude
 Absolute value of a signal’s highest intensity
 Normally measured in volts
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Period and Frequency
 Period - amount of time to complete one cycle, expressed
in seconds (s)
 Frequency – number of periods in one second, inverse of
period
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Frequency
 Rate of change with respect to time, expressed in hertz (Hz)
 Change in a short span of time means high frequency
 Change over a long span of time means low frequency
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Phase
 Position of the waveform relative to time zero
 Measured in degrees or radians
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Table 3.1 Units of periods and frequencies
Unit
Seconds (s)
Equivalent
1s
Unit
hertz (Hz)
Equivalent
1 Hz
Milliseconds (ms)
10–3 s
kilohertz (KHz)
103 Hz
Microseconds (ms)
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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Example Sine Waves s(t) = A sin(2ft +)
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Example 1
Express a period of 100 ms in microseconds, and express
the corresponding frequency in kilohertz.
Solution
From Table 3.1 we find the equivalent of 1 ms.We make the
following substitutions:
100 ms = 100  10-3 s = 100  10-3  106 ms = 105 ms
Now we use the inverse relationship to find the frequency,
changing hertz to kilohertz
100 ms = 100  10-3 s = 10-1 s
f = 1/10-1 Hz = 10  10-3 KHz = 10-2 KHz
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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 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2 /360 rad = 1.046 rad
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Time and Frequency Domains
 Time-domain plot – displays changes in signal
amplitude with respect to time
 Frequency-domain plot – relationship between
amplitude and frequency

Best represents an analog signal
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Figure 3.6 Sine wave examples
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Figure 3.6
Sine wave examples (cont.)
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Figure 3.6
Sine wave examples (cont.)
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Figure 3.7 Time and frequency domains
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Figure 3.7 Time and frequency domains (cont.)
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Figure 3.7 Time and frequency domains (cont.)
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Composite Signals
 Composed of many simple sine waves of differing
frequencies
 Fourier – showed any composite signal is a sum of a set
of sine waves of different frequencies, phases, and
amplitudes (Harmonics)

Fourier analysis
sin(2kft)
s(t )  A  
 k odd ,k 1
k
4

 Harmonics – components of digital signal, each having a
different frequencies, phases, and amplitudes
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Figure 3.8 Square wave
4A
4A
4A
s(t ) 
sin 2ft 
sin[ 2 (3 f )t ] 
sin[ 2 (5 f )t ]  ...

3
5
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Figure 3.9 Three harmonics
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Figure 3.10 Adding first three harmonics
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Frequency Spectrum
 Description of a signal using the frequency
domain and containing all of its components
 Dependent on medium used
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Figure 3.11 Frequency spectrum comparison
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Composite Signals and Transmission Medium
 A medium’s characteristics may affect the signal
 Some frequencies may be weakened or blocked
 Signal corruption – when square wave is sent through a
medium, other end which is not square wave at all
Figure 3.12 Signal corruption
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Bandwidth
 Range of frequencies that a medium can pass
without losing one-half of the power contained in
the signal
 Difference between the highest and the lowest
frequencies that the medium can satisfactorily
pass.
 In this book, we use the term bandwidth to refer to
the property of a medium or the width of a single
spectrum.
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Figure 3.13 Bandwidth
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Frequency Spectrum
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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
B = fh - fl = 900 - 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700,
and 900 (see Figure 13.4 )
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Figure 3.14 Example 3
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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
B = fh - fl
20 = 60 - fl
fl = 60 - 20 = 40 Hz
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Figure 3.15 Example 4
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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.
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Quiz
 In this picture, what is the bandwidth?
when f=1MHz
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Digital Signals
 Use binary (0s and 1s) to encode information
 Usually aperiodic; period and frequency are not
appropriate
 Less affected by interference (noise); fewer errors
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Figure 3.16 A digital signal
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Bit Interval and Bit Rate
 Describe digital signals by
Bit interval – time required to send one single bit
Bit rate – number of bit intervals per second, usually
expressed as bits per second (bps)


Figure 3.17 Bit rate and bit interval
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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.
Bit interval = 1/ 2000 s = 0.000500 s
= 0.000500 x 106 ms = 500 ms
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Digital signal As a Composite Analog
Signal
 A composite analog signal having an infinite
number of frequency
 The bandwidth of a digital signal is infinite
A digital signal is a composite signal
with an infinite bandwidth.
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Digital signal through a wide
bandwidth medium
 Some of frequencies are blocked by medium
 But still enough are passed to preserve a decent
signal shape
 Such as a coaxial cable to send a digital
through a LAN
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Digital signal through a band-limited
medium
 Can send digital data through a band-limited medium ?
Yes / No
 Such as the Internet via telephone line
 The relationship between bite rate (n) and the required
bandwidth (B)
n

B



Using only one harmonic
2



Using more harmonic
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n

 B   or n  2 B
2

54
Using only one harmonic (Digital versus Analog)
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Using more harmonic
 Third harmonic
n 3n 4n
B 

2 2
2
 Fifth harmonic
n 3n 5n 9n
B  

2 2
2
2
Bit Rate
Harmonic
1
Harmonics
1, 3
Harmonics
1, 3, 5
Harmonics
1, 3, 5, 7
1 Kbps
500 Hz
2 KHz
4.5 KHz
8 KHz
10 Kbps
5 KHz
20 KHz
45 KHz
80 KHz
100 Kbps
50 KHz
200 KHz
450 KHz
800 KHz
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Note:
The bit rate and the bandwidth are
proportional to each other.
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Digital versus Analog Bandwidth
 Analog bandwidth – range of frequencies a
medium can pass; expressed in hertz
 Digital bandwidth – maximum bit rate that a
medium can pass; expressed in bits per second
(bps)
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Low-pass versus Band-pass Channels
 Channel or a link is either
low-pass or band pass
 Low-pass –frequencies
between 0 and f (infinity)


Dedicated connections;
alternating
communications
Digital transmissions
 Band-pass – frequencies
between f1 and f2


Shared connections
Analog transmissions
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Data Rate Limits
 Dependent on three factors
Bandwidth available
Levels of signals we can use
Quality of the channel (level of noise)



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Noiseless Channel: Nyquist Bit Rate
 Defines the theoretical maximum bit rate
Bit Rate = 2  Bandwidth  log2 L
L is the number of signal levels used to represent data
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
Bit Rate = 2  3000  log2 2 = 6000 bps
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Noisy Channel: Shannon Capacity
 Determine the theoretical highest data rate for a
noisy channel
Capacity = Bandwidth  log2 (1 + SNR)
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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Transmission Impairment
 Imperfections cause impairment
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Attenuation
 Loss of energy
 Amplifiers are used to strengthen
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Decibel
 Measures the relative strength of two signals or a
signal at two different points
dB = 10 log 10 (P2/P1)
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
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Distortion
 Signal changes form or shape
 Each component has its own propagation speed,
therefore its own delay in arriving
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Noise
 Corruption caused by
Thermal noise – random motion of electrons,
creating an extra signal
Induced noise – outside sources such as motors and
appliances
Crosstalk – effect of one wire on another
Impulse noise – a spike for a short period from
power lines, lightning, etc.




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Noise
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More About Signals
 Throughput - how fast data can pass through an
entity (such as a point or network)
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More About Signals
 Propagation speed – distance a signal or bit can
travel through a medium in one second depend on


Medium
Frequency
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More About Signals
 Propagation time – time required for a signal (or bit)
to travel from one point of the transmission to
another
Propagation time = Distance/Propagation speed
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More About Signals (cont)
 Wavelength – distance a simple signal can travel in one
period

Depends on both the frequency and the medium
Wavelength = Propagation speed x Period
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Next…
 How do we transform data into signals to send
them to our destination?


Encoding – converting to digital
Modulation – converting to analog
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Credits
 All figures obtained from publisher-provided
instructor downloads
Data Communications and Networking, 3rd edition by
Behrouz A. Forouzan. McGraw Hill Publishing, 2004
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