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Transcript
Computer Communication & Networks
Lecture # 06
Physical Layer: Analog Transmission
Nadeem Majeed Choudhary
[email protected]
Physical Layer
Physical Layer Topics to Cover
Signals
Digital Transmission
Analog Transmission
Multiplexing
Transmission Media
Reasons for Conversions

Digital data  Digital Signal


Analog data  Digital Signal


Allows the use of digital transmission and switching equipment
Digital data  Analog Signal



Easy and simple to implement
Allows us of the public telephone system
Allows use of optical fiber
Analog Data  Analog Signal


Easy
Telephone system was primarily analog
Data Encoding Techniques
Analog Encoding of Digital Information

Phase shift keying



Two binary numbers (0,1) represented by phase shift of the
carrier wave
More efficient and noise resistant than FSK
Used up to 9600 on voice grade lines
Data Encoding Techniques
Analog Encoding of Digital Information






Techniques can be combined
Common to combine phase shift and amplitude shift
Can get 56kps on a voice grade line
With some techniques called multilevel signaling each
signal represents more than one bit
Baud rate = signal changes per second
Bit rate = bits per second
DIGITAL-TO-ANALOG CONVERSION

Digital-to-analog conversion is the process of
changing one of the characteristics of an analog
signal based on the information in digital data.
Data Rate Vs Signal Rate

Baud rate determines the bandwidth required
to send the signal
S = N * 1/r
Where S is signal rate, N is data rate and
r is number of data elements carried in one
signal element.

Note
Bit rate is the number of bits per second. Baud rate is the
number of signal elements per second.
In the analog transmission of digital data, the baud rate is
less than or equal to the bit rate.
Example
An analog signal carries 4 bits per signal element. If
1000 signal elements are sent per second, find the bit
rate.
Solution
In this case, r = 4, S = 1000, and N is unknown. We can
find the value of N from
Example
An analog signal has a bit rate of 8000 bps and a baud
rate of 1000 baud. How many data elements are
carried by each signal element? How many signal
elements do we need?
Solution
In this example, S = 1000, N = 8000, and r and L are
unknown. We find first the value of r and then the value
of L.
Carrier Signal

In analog signal, the sending device produces a highfrequency signal that acts as a basis for the
information signal. This base signal is called the
carrier signal or carrier frequency.

Receiving device is tuned to the frequency of the
carrier signal that it expects from the sender.

Digital information then modulates the carrier signal
by modifying one or more of its characteristics
(amplitude, frequency, or phase). This kind of
modification is called modulation (or shift keying, and
the information signal is called the modulating signal.
Data Encoding Techniques
Analog Encoding of Digital Information (cont)

Amplitude shift keying




Two binary numbers (0,1) represented by two different amplitudes of the
carrier wave
Rather inefficient
Used up to 1200 bps on voice grade lines
Used to transmit digital data over optical fiber
Binary Amplitude Shift Keying
Bandwidth

Bandwidth of a signal is the total range of
frequencies occupied by that signal.

Bandwidth of the signal is centered at carrier
frequency fc.
Data Encoding Techniques
Analog Encoding of Digital Information (cont)

Frequency shift keying




Two binary numbers (0,1) represented by two different frequencies of
the carrier wave
Less susceptible to error than ASK
Used up to 1200bps on voice grade lines
Commonly used for high frequency ( 4 to 30mhz) radio
Binary Frequency Shift Keying
Binary Phase Shift Keying
PSK Constellation
4 PSK Method
4 PSK Constellation
8 PSK Constellation
4 QAM & 8 QAM
Analog-to-Analog Modulation
Analog-to-Analog

Analog-to-analog conversion is the
representation of analog information by an
analog signal. One may ask why we need to
modulate an analog signal; it is already
analog. Modulation is needed if the medium
is bandpass in nature or if only a bandpass
channel is available to us.
Amplitude Modulation
Frequency Modulation
Phase Modulation
Bandwidth Utilization
Bandwidth utilization is the wise use of available bandwidth to achieve
specific goals.
Efficiency can be achieved by multiplexing; privacy and anti-jamming
can be achieved by spreading.
Multiplexing

Whenever the bandwidth of a medium linking
two devices is greater than the bandwidth
needs of the devices, the link can be shared.
Multiplexing is the set of techniques that
allows the simultaneous transmission of
multiple signals across a single data link. As
data and telecommunications use increases,
so does traffic.
Contd..
FDM is an analog multiplexing technique that combines analog
signals.
Example
Assume that a voice channel occupies a bandwidth of 4
kHz. We need to combine three voice channels into a link
with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the
configuration, using the frequency domain. Assume there
are no guard bands.
Solution
We shift (modulate) each of the three voice channels to a
different bandwidth. We use the 20- to 24-kHz bandwidth
for the first channel, the 24- to 28-kHz bandwidth for the
second channel, and the 28- to 32-kHz bandwidth for the
third one.
6.33
Time Division Multiplexing
TDM is a digital multiplexing technique for combining several lowrate channels into one high-rate one.
Spread Spectrum

Spread Spectrum is a means of transmission in which
the data sequence occupies a bandwidth in excess of
the minimum bandwidth necessary to send it.





Effectively the signal is mapped to a higher dimension signal
space
Signal spreading is done before transmission by using a
spreading sequence. The same sequence is used at the
receiver to retrieve the signal
Spread Spectrum is most effective against interference
(intentional or non-intentional) with fixed energy.
Main commercial applications in wireless and GPS.
Two techniques to spread the spectrum : FHSS, DSSS
DSSS



In direct sequence spread spectrum we use a
code of n bits to represent each bit.
Barker sequence is used as spreading code
in which n= 11
So, if signal is N then rate of spread signal is
11N
FHSS


Pseudorandom code generator creates a kbit pattern for every hopping period.
Frequency table uses that pattern to find the
frequency to be used and passes it to
frequency synthesizer.
Carrier signal is created by frequency
synthesizer.
Readings

Chapter 5 (B.A Forouzan)

Section 5.1, 5.2
Q&A