Ch3 physical layer.ppt

4123702
Data Communications System
By
Ajarn Preecha Pangsuban

4123702 Data Communications System @YRU 2
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…

Position of the physical Layer

Physical layer services

Transmission Media
 Guided

Twisted-pair

Coaxial cable

Fiber-optic
 Unguided

Radio

Microwave

Networks and Technologies
 Telephone network – circuit-switched network
 High speed access

Modems

DSL

Cable

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

Chapter 3: Signals

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

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.

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

Periodic
Signals

Aperiodic Signals

In data communication, we commonly
use periodic analog signals and
aperiodic digital signals.
Note:Note:

Analog Signals
 Sine wave – most fundamental form of a periodic analog
signal
 Fully described by: Amplitude, Frequency and Phase

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

Amplitude
 Absolute value of a signal’s highest intensity
 Normally measured in volts

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

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

Phase
 Position of the waveform relative to time zero
 Measured in degrees or radians

Table 3.1 Units of periods and frequencies
Unit Equivalent Unit Equivalent
Seconds (s) 1 s hertz (Hz) 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

Example Sine Waves s(t) = A sin(2πft +Φ)

Example 1Example 1
Express a period of 100 ms in microseconds, and express
the corresponding frequency in kilohertz.
SolutionSolution
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
µs = 105
µs
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

Example 2Example 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 radπ

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

Figure 3.6 Sine wave examples

Figure 3.6 Sine wave examples (cont.)

Figure 3.7 Time and frequency domains

Figure 3.7 Time and frequency domains (cont.)

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
 Harmonics – components of digital signal, each having a
different frequencies, phases, and amplitudes
∑
∞
=
××=
1,
)2sin(4
)(
koddk k
kft
Ats
π
π

Figure 3.8 Square wave
...])5(2sin[
5
4
])3(2sin[
3
4
2sin
4
)( +++= tf
A
tf
A
ft
A
ts π
π
π
π
π
π

Figure 3.9 Three harmonics

Figure 3.10 Adding first three harmonics

Frequency Spectrum
 Description of a signal using the frequency
domain and containing all of its components
 Dependent on medium used

Figure 3.11 Frequency spectrum comparison

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

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.

Figure 3.13 Bandwidth

Frequency Spectrum

Example 3Example 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 )

Figure 3.14 Example 3

Example 4Example 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

Figure 3.15 Example 4

Example 5Example 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 haveThe answer is definitely no. Although the signal can have
the same bandwidth (1000 Hz), the range does notthe same bandwidth (1000 Hz), the range does not
overlap. The medium can only pass the frequenciesoverlap. The medium can only pass the frequencies
between 3000 and 4000 Hz; the signal is totally lost.between 3000 and 4000 Hz; the signal is totally lost.

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

Figure 3.16 A digital signal

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

Example 6Example 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
µs = 500 µs

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 signalA digital signal is a composite signal
with an infinite bandwidth.with an infinite bandwidth.

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

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)

Using only one harmonic

Using more harmonic






=
2
n
B
Bnor
n
B 2
2
<=





>=

Using only one harmonic (Digital versus Analog)

Using more harmonic
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
 Third harmonic
 Fifth harmonic
2
4
2
3
2
nnn
B =+=
2
9
2
5
2
3
2
nnnn
B =++=

The bit rate and the bandwidth areThe bit rate and the bandwidth are
proportional to each other.proportional to each other.
Note:Note:

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)

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

Data Rate Limits
 Dependent on three factors

Bandwidth available

Levels of signals we can use

Quality of the channel (level of noise)

Noiseless Channel: Nyquist Bit Rate
 Defines the theoretical maximum bit rate
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
Bit Rate = 2 × Bandwidth × log2 L

Noisy Channel: Shannon Capacity
 Determine the theoretical highest data rate for a
noisy channel
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
Capacity = Bandwidth × log2 (1 + SNR)

Transmission Impairment
 Imperfections cause impairment

Attenuation
 Loss of energy
 Amplifiers are used to strengthen

Decibel
 Measures the relative strength of two signals or a
signal at two different points
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
dB = 10 log 10 (P2/P1)

Distortion
 Signal changes form or shape
 Each component has its own propagation speed,
therefore its own delay in arriving

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.

Noise

More About Signals
 Throughput - how fast data can pass through an
entity (such as a point or network)

More About Signals
 Propagation speed – distance a signal or bit can
travel through a medium in one second depend on

Medium

Frequency

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

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

Credits
 All figures obtained from publisher-provided
instructor downloads
Data Communications and Networking, 3rd edition by
Behrouz A. Forouzan. McGraw Hill Publishing, 2004

Ch3 physical layer.ppt

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