Transcript Physical Layer - University of Delhi

Theoretical basis for data
communication
Transmission of data
Data must be transformed to electromagnetic
signals to be transmitted.
Data : Analog or Digital
• Analog data : human voice, chirping of
birds etc , converted to
– Analog or digital signals
Digital : data stored in computer memory,
converted to
– Analog or digital signals
Examples
• Analog data as analog signal : Human voice from
our houses to the telephone exchange.
• Analog data as digital signal : most of the systems
today : Say Human voice, images sent on digital
lines .. New telephone system (digital exchanges)
• Digital data as analog signal : computer data sent
over internet using analog line .. Say telephone
line ( say our house to the exchange)
• Digital data as digital signal : say from one digital
exchange to another
Signals : Analog or digital
• Analog signal has infinitely many levels of
intensity (infinitely many values,
continuous values) over a period of time.
• Digital signal has only a limited number of
defined values(discrete values) say, 0,1.
Figure 3.1
Comparison of analog and digital signals
Figure 3.2
A sine wave
Figure 3.3
Amplitude
Figure 3.4
Period and frequency
• If a signal does not change at all, its
frequency is zero.
• If it changes instantaneously, its frequency
is infinite.
• An analog signal is best represented in the
frequency domain.
Figure 3.7
Time and frequency domains
Figure 3.7
Time and frequency domains (continued)
Figure 3.7
Time and frequency domains (continued)
Single-frequency sine wave isnot
useful for data communication
• A single sine wave can carry electric energy
from one place to another. For eg., the
power company sends a single sine wave
with a frequency of say 60Hz to distribute
electric energy to our houses.
Contd..
• If a single sine wave was used to convey
conversation over the phone, we would
always hear just a buzz.
• If we sent one sine wave to transfer data, we
would always be sending alternating 0’s and
1’s, which does not have any
communication value.
Composite Signals
• If we want to use sine wave for communication,
we need to change one or more of its
characteristics. For eg., to send 1 bit, we send a
maximum amplitude, and to send 0, the minimum
amplitude.
• When we change one or more characteristics of a
single-frequency signal, it becomes a composite
signal made up of many frequenies.
Figure 3.9
Three harmonics
Figure 3.10
Adding first three harmonics
Fourier Analysis
• In early 1900s, French Mathematician JeanBaptiste Fourier showed that any composite
signal can be represented as a combination
of simple sine waves with different
frequencies, phases and amplitudes.
• More is the number of components included
better is the approximation
• For eg., let us consider the square wave …
Time-Voltage graph
• Time on x-axis in msec, Voltage on y-axis
• The first trace in the above figure is the
sum of 2 sine waves with amplitudes
chosen to approximate a 3 Hz square
wave (time base is msec). One sine
wave has a frequency of 3 Hz and the
other has a frequency of 9 Hz. The
second trace starts with the first but
adds a 15 Hz sine wave and a 21 Hz
sine wave. It is clearly a better
approximation.
Figure 3.8
Square wave
• It can be shown (ref Kreyzsig) that this
signal consists of a series of sine waves
with frequencies f, 3f, 5f, 7f, … and
amplitudes 4A/pi, 4A/3Pi, 4A/5Pi,
4A/7Pi,… where f is the fundamental
frequency(1/T, T the time period) and A the
maximum amplitude. The term with
frequency f, 3f .. are called the first
harmonic, 3rd harmonic,… respectively.
Frequency spectrum of a signal
• The description of a signal using the
frequency domain and containing all its
components is called the frequency
spectrum of the signal.
Figure 3.11 Frequency spectrum comparison
Composite Signal and
Transmission Medium
• A signal needs to pass thru a transmission
medium. A transmission medium may pass
some frequencies, may block few and
weaken others.
• This means when a composite signal,
containing many frequencies, is passed thru
a transmission medium, we may not receive
the same signal at the other end.
Figure 3.12
Signal corruption
Bandwidth of a channel
• The range of frequencies that a medium can
pass without loosing one-half of the power
contained in that signal is called its
bandwidth.
Figure 3.13
Bandwidth
Representing data as Digital
Signals
• 1 can be encoded as a positive voltage say 5 volts,
0 as zero voltage (or negative voltage say –5 volts)
• Most digital signals are aperiodic. Thus we use
• Bit interval (instead of period) : time required to
send one bit = 1/ bit rate.
• Bit rate (instead of frequency) :number of bits per
second.
Figure 3.17
Bit rate and bit interval
Digital signal as Composite
Signal
• Digital signal is nothing but a composite
analog signal with an infinite bandwidth.
• A digital signal theoretically needs a
bandwidth between 0 and infinity. The
lower limit 0 is fixed. The upper limit may
be compromised.
Relationship b/w bit rate and
reqd. channel b/w (informal)
• Imagine that our computer creates 6bps
• In 1 second, the data created may be 111111, no
change in the value, best case
• In another, 101010, maximum change in the
values, worst case
• In another, 001010, change in between the above
two cases
• We have already shown .. More the changes
higher are the frequency components
Figure 3.18
Digital versus analog
Using single harmonic – just to
get the intuition
• The signal 111111 (or 00000 ) can be
simulated by sending a single-frequency
signal with frequency 0.
• The signal 101010 (010101) can be
simulated by sending a single-frequency
signal with frequency 3 Hz. (3 signals or
sine waves per second)
• All other cases are between the best and the worst
cases. We can simulate other cases with a single
frequency of 1 0r 2 Hz (using appropriate phase).
• I.e. to simulate the digital signal at data rate 6bps,
sometimes we need to send a signal of frequency
0, sometimes 1,sometimes 2 and sometimes 3. We
need that our medium should be able to pass
frequencies of 0-3 Hz.
Generalizing the example above
•
•
•
•
Bit rate = n bps
Best case ---- frequency 0 Hz
Worst case ----- frequency n/2 Hz
Hence B (bandwidth) = n/2
Using more harmonics
• However, as said earlier, one harmonic does not
approximate the digital signal nicely and more
harmonics are required to approximate the digital
signal.
• As shown earlier, such a signal consists of odd
harmonics
• When we add 3rd harmonic to the worst case, we
need B = n/2 + 3n/2 = 4n/2
• When we add 5th harmonic to the worst case, we
need B = n/2 + 3n/2 + 5n/2= 9n/2 and so on.
• In other words, B >= n/2 or n <= 2B
Relationship b/w bit rate and
reqd. channel b/w (informal)
• Hence we conclude that bit rate and the
bandwidth of a channel are proportional to
each other.
Analog vs Digital
• Low-pass channel : has a bandwidth with
frequencies between 0 and f (f could be anything
including infinity).
• Band-pass channel : has a bandwidth with
frequencies between f1 (>=0) and f2
• A band-pass channel is more easily available than
a low-pass channel.
Figure 3.19
Low-pass and band-pass
Digital Rate limits
• Data rate depends on 3 factors:
– The bandwidth available
– Number of levels of signals
– Quality of the channel (noise level)
Figure 3.18
Digital versus analog
Noiseless Channel: Nyquist Bit
rate
• b = 2 B log L (log is to base 2)
b : bit rate
B : Bandwidth
L : number of levels
Noisy channel : Shannon
Capacity
• C = B log (1 + SNR)
C = capacity of the channel in bps
B = Bandwidth
SNR = signal to noise ratio
Digital vs Analog contd…
• Digital signal needs a low-pass channel
• Analog signal can use a band-pass channel.
• Moreover, bandwidth of a signal can always be
shifted ( a property required for FDM – The
bandwidth of a medium can be divided into
several band-pass channels to carry several analog
transmissions at the same time.)
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
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 signalto-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
Using both the limits
• In practice we use both the limits to
determine, given the channel bandwidth,
what should be the number of levels a
signal should have.
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
Then we use the Nyquist formula to find the
number of signal levels.
4 Mbps = 2  1 MHz  log2 L  L = 4
I Acknowledge
Help from the following site
http://www.mhhe.com/engcs/compsci/forouzan/
In preparing this lecture.