Title: Module 3
1Module 3
- Data Communication Part I
2- Textbook sections
- LG 3.1 Digital Representation of Information
- LG3.2 Why Digital communication?
- LG 3.3 Characterization of Communication Channels
- LG 3.4 Fundamental Limits in Digital Transmission
- LG 3.5 Line Coding
- LG 3.6 Modems and Digital Modulation
- Topics
- Data and Signals
- Data
- Signals
- Line Coding
- Modems
- Baseband and Broadband
- Synchronous and Asynchronous Communication
- Advantages of Digital Communication
- Fundamental Limits in Digital Transmission
- Nyquist Rate
- Shannons channel coding theorem
31. Data and Signals
- Data Entities that convey meaning
- Analog data are functions of time and occupy a
limited frequency spectrum - Digital data
- Examples pressure, temperature
- Signals Electric or electromagnetic encoding of
data - Analog signal time-varying electromagnetic
signal occupying a limited frequency spectrum - Digital signal Using a different voltage level
for each of the two binary digits. - Representation
- Frequency domain representations of data and
signals - Time domain representations of data and signals
41. Data and Signals
5Time domain and frequency domain representations
of signals
Time domain
Frequency domain
61. Data and Signals
- Line Coding
- The method used for converting a binary
information sequence into a digital signal in a
digital communications system - Mapping from data bits to voltage pulses
- Types of digital-to-digital encoding
- Unipolar The voltage pulses all have the same
algebraic sign, that is, all positive or
negative. Only one level of value. - Polar signaling One logic state is represented
by a positive voltage level, and the other by a
negative voltage level. Two levels (positive and
negative) of amplitude. - Nonreturn to zero (NRZ) The voltage level is
constant during a bit interval. There is no
transition. (no return to a zero voltage level) - Differential encoding The signal is decoded by
comparing the polarity of adjacent voltage pulses
rather than the absolute values of a voltage
pulse. - Biphase At least one transition exists per bit
interval. - Bipolar Three levels of amplitude (positive,
zero, and negative)
7Types of digital-to-digital encoding
8(No Transcript)
91. Data and Signals
- Applications of Line Coding
- NRZ encoding RS232 based protocols
- Manchester encoding Ethernet networks
- Differential Manchester encoding token-ring
networks - NRZ-Inverted encoding Fiber Distributed Data
Interface (FDDI)
101. Data and Signals
11LG Figure 3.25 Line coding methods
0
1
0
1
1
1
0
0
1
Unipolar NRZ
Polar NRZ
NRZ-Inverted (Differential Encoding)
Bipolar Encoding
Manchester Encoding
Differential Manchester Encoding
121. Data and Signals
- Line coding considerations
- Synchronization (more detail next page)
- A signal change for each bit can assure
synchronization. The receiver can use these
changes to build up, update, and synchronize its
clock. - Bandwidth
- Signal changes needed to encode one bit.
- Direct Current (DC) component (less important)
- Related to using transformers
- Problem associated with NRZ coding
131. Data and Signals
- FDDI 4B/5B
- Address the inefficiency of the Manchester
encoding without suffering from the problem of
having extended durations of high or low signals. - The idea of 4B/5B is to insert extra bits into
the bit stream so as to break up long sequences
of 0s or 1s. Specially, every 4 bits of actual
data are encoded in a 5 bit code that is then
transmitted to the receiver hence the name
4B/5B. - The 5 bits are selected in such a way that each
symbol has no more than one leading 0 and no more
than two trailing 0s. Thus when sent
back-to-back, no pair of 5 bit codes results in
more than three consecutive 0s being transmitted. - The resulting 5 bit codes are then transmitted
using the NRZI encoding, which explains why the
code is only concerned about consecutive 0s -
NRZI already solves the problem of consecutive
1s. - The 4B/5B results in 80 efficiency.
141. Data and Signals
4-bit Data Symbol -gt 5-bit Code Bits -gt encoding
each code bit using NRZ-I
151. Data and Signals
- Synchronization Consideration
- Problem of unvarying signal
- When a signal is unvarying, the receiver cannot
determine the beginning and ending of each bit. - Take unipolar coding for example. A long
uninterrupted series of 1s or 0s can cause
synchronization problem. - Problem of Using Timers
- Whenever there is no signal change to indicate
the start of the next bit in a sequence, the
receiver has to rely on a timer. Given an
expected bit rate of 1000 bps, if the receiver
detects a positive voltage lasting 0.005 seconds,
it reads one 1 per 0.001 seconds, or five 1s.
However, five 1s can be stretched to 0.006
second, causing an extra 1 bits to be read by the
receiver. That one extra bit in the data stream
causes everything after it to be decoded
erroneously. - Problem of Having a Separate Clock Line
- A solution developed to control the
synchronization of unipolar transmission is to
use a separate, parallel line that carries a
clock pulse and allows the receiving device to
resynchronize its timer to that of the signals.
But doubling the number of lines used for
transmission increase the cost.
161. Data and Signals
- Modem
- A communications device that enables a computer
to transmit information over a standard telephone
line. Because a computer is digital, it works
with discrete electrical signals representing
binary 1 and binary 0. A telephone is analog and
carries a signal that can have any of a large
number of variations. Modems are needed to
convert digital signals to analog, and vice
versa. When transmitting, modems
impose(modulate) a computers digital signals on
a continuous carries frequency on the telephone.
When receiving, modems sift out (demodulate) the
information from the carries and transfer it in
digital form to the computer. - Basic function of the modulation is to produce a
signal that contains the information sequence and
that occupies frequencies in the range passed by
the channel.
17LG Figure 3.28 Amplitude, frequency, and phase
modulation techniques
Information
1
(a)
Amplitude Shift Keying
t
-1
1
Frequency Shift Keying
(b)
t
-1
1
Phase Shift Keying
(c)
t
-1
18LG Figure 3.29 Modulating a signal
(a) Information
A
(b) Baseband Signal Xi(t)
t
2T
6T
T
4T
5T
3T
0
-A
t
2A
(d) 2Yi(t) cos(2?fct)
t
-2A
19LG Figure 3.30 Modulator and demodulator
(a) Modulate cos(2?fct) by multiplying it by Ak
for (k-1)T lt t ltkT
x
Ak
Yi(t) Ak cos(2?fct)
cos(2?fct)
(b) Demodulate (recover) Ak by multiplying by
2cos(2?fct) and lowpass filtering
Lowpass Filter with cutoff W Hz
x
Yi(t) Akcos(2?fct)
Xi(t)
2cos(2?fct)
2Ak cos2(2?fct) Ak 1 cos(2??fct)
201. Data and Signals
- Baseband Technology
- A network technology that uses a small part of
the electromagnetic spectrum and sends one signal
at a time over the underlying medium. - Uses digital signaling. Digital signals are
inserted on the line as voltage pulses. - The entire frequency spectrum of the medium is
used to form the signal hence frequency-division
multiplexing (FDM) cannot be used - Transmission is bidirectional. That is, a signal
inserted at any point on the medium propagates in
both directions to the end. - Most LAN use baseband signaling (e.g., Ethernet
and FDDI)
211. Data and Signals
- Broadband Technology
- A network technology that uses a large part of
the electromagnetic spectrum to achieve higher
throughput rates. - Broadband refers to the use of analog signaling
- Usually broadband system employ frequency
division multiplexing (FDM) to allow multiple,
independent communications to proceed
simultaneously over a single underlying medium. - Much greater distances are possible with
broadband compared to baseband. This is because
the analog signals that carry the digital data
can propagate greater distance before the noise
and attenuation damage the data.
221. Data and Signals
- Asynchronous Transmission
- Bits are sent one character at a time. (A
character is in general 8 bits in length) - Timing or synchronization must only be maintained
within each character. The receiver has the
opportunity to resynchronize at the beginning of
each new character. - Start-stop technique
- Idle state When no character is being
transmitted the line between transmitter and
receiver is in an idle state. The definition
of idle is by convention, but typically is
equivalent to the signaling element for binary 1. - Start bit The beginning of a character is
signaled by a start bit with a value of binary 0. - Data bits
- Stop bit The last bit of the character is
followed by a stop bit, which is a binary 1. A
minimum length for the stop bit is specified and
this is usually 1, 1.5 or 2 times the duration of
an ordinary bit. No maximum value is specified,
Since the stop bit is the same as the idle state.
231. Data and Signals
- Example of an Asynchronous Transmission
- Given the facts that a steady stream of
characters is sent, and the interval between
characters is uniform and equal to the stop bit. - Note
- Start bit 0
- Stop bit1
- ASCII character
- A 1000001
- B 0100001
- C 1100001
- The bit pattern of sending ASCII characters ABC
(without parity bit) is - 010000011001000011011000011...1111
- start bit(0) A (1000001) stop bit (1) start
bit (0) B(0100001) stop bit (1) start bit
(0) C(1100001)) stop bit (1)
24Asynchronous Communication
251. Data and Signals
- Synchronous Transmission
- Blocks of characters or bits are transmitted
without start and stop codes. - To prevent timing drift between transmitter and
receiver, their clocks must somehow be
synchronized - One possibility is to provide a separate clock
line between transmitter and receiver - Another possibility is to embed the clocking
information in the data signal. For example for
digital signals, this can be achieved with
bi-phase encoding. - Another level of synchronization is required, to
allow the receiver to determine the beginning and
end of a block of data. - With character-oriented transmission, the frame
begins with one or more synchronization
characters. The synchronization character,
usually called SYN, is a unique bit pattern that
signals the receiver that this is the beginning
of a block - With bit-oriented transmission, a special bit
pattern signals the beginning of a block In
bit-oriented transmission, this preamble is eight
bits long and is referred to as a flag.
26Synchronous Communication
272. Advantages of Digital Transmission
- Digital transmission systems can operate with
lower signal levels or with greater distances
between repeaters than analog systems can. This
factors translates into lower overall system cost
and was the original motivation for the
introduction of digital transmission.
28LG Figure 3.5 General Transmission system
Transmitter
Receiver
Communication channel
A transmission system make use of a physical
transmission medium or channel that allows the
propagation of energy in the form of pulse or
variations in voltage, current or light intensity
292. Advantages of Digital Transmission
- Analog transmission
- Transmit a waveform, which is a function that
varies continuously with time - Digital transmission
- Transmit a given symbol that is selected from
some finite set of possibilities - For example, in binary digital transmission, the
objective is to transmit either a 0 or a 1.
30Figure 3.6 Analog versus digital transmission
- (a) Analog transmission all details must be
reproduced accurately
Received
Sent
- e.g. AM, FM, TV transmission
(b) Digital transmission only discrete levels
need to be reproduced
Received
Sent
- e.g digital telephone, CD Audio
31Figure 3.7 Typical long-distance link
Note To transmit over long distances, it is
necessary to introduce repeaters periodically to
regenerate the signal.
Transmission segment
Destination
Source
Repeater
Repeater
322. Advantages of Digital Transmission
- Impact of distance on analog transmission
- Attenuation
- Different frequency components of the signal are
attenuated differently - In general, high frequency components are
attenuated more than low-frequency components. - Delay
- Different frequency components of a signal are
delayed by different amounts - Noise
- Analog repeater
- Attempts to eliminate the distortion caused by
attenuation and delay by using equalizers
33LG Figure 3.8 An analog repeater
Recovered signal residual noise
Attenuated distorted signal noise
Amp.
Equalizer
Repeater
34LG Figure 3.9 A digital repeater
- Note
- Purpose is to determine with high probability
the original binary stream - Use an equalizer to compensate for the
distortion - The repeater does not need to completely
regenerate the original shapes of the
transmitted signal. It only needs to determine
whether the original pulse was positive or
negative.
Decision Circuit. Signal Regenerator
Amplifier Equalizer
Timing Recovery
353. Fundamental Limits in Digital Transmission
- Using electric current to send bits and using
electric voltage to encode bits - The following figure illustrates how positive
and negative voltage can be used to transmit bits
across a wire. In this example, the sender
applies a negative voltage to send a 1 bit or a
positive voltage to send a 0 bits.
- Baud rate
- Bit rate is the number of bits per second. Baud
rate is the number of signal units (pulses) per
second that are required to represent those bits. - Baud rate can be either less than, equal to, or
greater than the bit rate. - Each pulse can encode one or more bits of
information (See multi-level pulses later)
363. Fundamental Limits in Digital Transmission
- Limitations of real hardware The following
figure illustrates the voltage emitted by a real
device as it transmits a bit. In practice,
voltage are often worse than this example.
373. Fundamental Limits in Digital Transmission
- Hardware Bandwidth and the Transmission of Bits
- Each transmission system has a limited bandwidth,
which is the maximum rate that the hardware can
change a signal. If a sender attempts to
transmit changes faster than the bandwidth, the
hardware will not be able to keep up because it
will not have sufficient time to complete one
change before the sender attempts to make another
. Thus some of the changes will be lost.
383. Fundamental Limits in Digital Transmission
- Nyquist RateThe maximum data rate of a noiseless
channel - Nyquist derived an equation expressing the
maximum data rate for a finite bandwidth
noiseless channel - If an arbitrary signal has been run through a
low-pass filter of bandwidth W, the filtered
signal can be completed reconstructed by making
only 2W (exact) samples per second. - Nyquist rate is based on the sampling theorem,
which states If a signal f(t) is sampled at
regular intervals of time and at a rate higher
than twice the highest significant signal
frequency, then the samples contain all the
information of the original signal. The function
f(t) may be reconstructed from these samples by
the use of a low-pass filter. - Sampling the line faster than 2W times per second
is pointless because the higher frequency
components that such sampling could recover have
already been filtered out. - A proof of the sampling theorem is based on the
Fourier series and can be found in Data and
Computer Communications by William Stallings
393. Fundamental Limits in Digital Transmission
- Nyquist Rate
- For multilevel transmission pulses that can take
on M 2m amplitude levels, the maximum bit rate
is 2Wm bits/seconds - R 2W pulses/second m bits/pulse
- 2Wm bits/second
- where
- R is bit rate in bits/seconds
- W is the bandwidth in Hz
- 2m M, M is the number of levels
403. Fundamental Limits in Digital Transmission
- Multi-levels Pulse
- Two-levels pulse
- Binary information can be send by a pulse with
amplitude A for a 1 bit and A for a 0 bit - Bit rate is 2W bps
- Four-levels pulse
- With pulse taking on amplitude from the set
-A,-A/3,A/3,A to transmit the pairs of bits
00,01,10,11, then each pulse convey two bits of
information. (Each increment of amplitude is
equal to 2A divided by 3) - Bit rate is 4W bps.
- Eight-levels pulse
- With pulse taking on amplitude from the set
-A,-5A/7,-3A/7,A/7,A/7,3A/7,5A/7,A to
transmit the bits 000,001,010,011,100,101,110,111
, then each pulse convey three bits of
information. (Each increment of amplitude is
equal to 2A divided by 7) - Bit rate is 6W bps.
413. Fundamental Limits in Digital Transmission
- Nyquist Rate
- In the absence of noise, the bit rate can be
increased without limit by increasing the number
of signal levels M. - Each increase in number of signal levels
- requires a reduction in the spacing between
levels - Increases the probability that noise will convert
the transmitted signal levels into other signal
levels. - The presence of noise limits the reliability with
which the receiver can correctly determine the
information that was transmitted - Increase the signal amplitude can decrease the
effect of noise
42LG Figure 3.12 Signal-to-noise ratio
signal noise
signal
noise
High SNR
t
t
t
noise
signal noise
signal
Low SNR
t
t
t
Average Signal Power
SNR
Average Noise Power
SNR (dB) 10 log10 SNR
433. Fundamental Limits in Digital Transmission
- Channel Capacity
- For a given bandwidth (W) and signal-to-noise
ration (SNR), information can be transmitted up
to C bits/seconds, with arbitrarily small
probability of error by using sufficiently
complex encoding systems - It is not possible to transmit at a rate higher
than C bits/seconds by any encoding system
without a definite probability of error. In
other words, it is not possible to transmit
faster than C and have error free communication - One of the most famous of all results of
information theory is Shannons channel coding
theorem. For a given channel there exists a code
that will permit the error-free transmission
across the channel at a rate R, provided R is
less than C, the channel capacity is given by the
following formula - C Wlog2(1SNR) bits/seconds
- Note In equation above, SNR is not in decibel
(dB)
443. Fundamental Limits in Digital Transmission
- Channel Capacity
- Example
- When SNR approaches zero, then
- C Wlog2(1SNR) bits/seconds W log21 W0
0 - When SNR is 1, then
- C Wlog2(1SNR) bits/seconds W log22 W1
W - Conversion between SNR and SNR(dB)
- SNR(dB) 10Log10 SNR
- SNR 10(SNR(dB)/10)
- Equation x 2y ? log2x log22y y
- Find minimum SNR for a particular C
- C Wlog2(1SNR) gt C/W log2(1SNR) gt
- 1SNR 2(C/W) gt SNR 2(C/W) -1
453. Fundamental Limits in Digital Transmission
- Channel Capacity
- Example For a channel of 3000 Hz bandwidth, and
a signal-to-noise (SNR) of 30 dB, the Shannons
channel capacity is - SNR 10(SNR(dB)/10) 10(30/10) 103 1000
- C Wlog2(1SNR) bits/seconds
- 3000 log2(11000) 3000
(log101001)/(log102) - 3000 (3.0004)/(0.30103) 29901.3
bits/seconds - C is less than 30 K bits/seconds