Transcript CSC 335 Data Communications and Networking I

CSC 335
Data Communications
and
Networking
Lecture 3: Signal Encoding and
Conversion
Dr. Cheer-Sun Yang
Motivation
How is information coded in a format suitable
for transmission?
• What are the available communication
services and devices today?
• How are bits encoded into electric signals?
• How are analog signals and digital signals
converted?
Communication Services and
Devices
• Telephone System – switching technique
and routing methods are the two main
design issues.
• Integrated Services Digital Network
• Cellular Phones – the sender and receiver
can move
• Fax Machines
• Computers
Data Encoding
• ASCII (American Standard Code for
Information Interchange)
• EBCDIC (Extended Binary Coded Decimal
Interchange Code)
• Others – Baudot, morse, BCD
Electric Current and Data Bits
The simplest electronic communication systems
use a small electric current to encode data.
Positive voltage – represents 0 (or 1)
Negative voltage – represents 1 (or 0)
A waveform diagram can be used to illustrate how
data bits are represented and transmitted.
Electric Current and Data Bits
A waveform diagram provides a visual
representation of how an electrical signal varies
over time. For example, the diagram shows that a
longer time elapsed between the transmission of
the fourth and the fifth bits than between others.
Digital Encoding Schemes Using
Digital Signals
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Nonreturn to Zero-Level (NRZ-L)
Nonreturn to Zero Inverted (NRZI)
Manchester
Differential Manchester
Nonreturn to Zero-Level (NRZL)
• Two different voltages for 0 and 1 bits
• Voltage constant during bit interval
– no transition I.e. no return to zero voltage
• e.g. Absence of voltage for zero, constant
positive voltage for one
• More often, negative voltage for one value
and positive for the other
• This is NRZ-L
Nonreturn to Zero Inverted
• Nonreturn to zero inverted on ones
• Constant voltage pulse for duration of bit
• Data encoded as presence or absence of
signal transition at beginning of bit time
• Transition (low to high or high to low)
denotes a binary 1
• No transition denotes binary 0
• An example of differential encoding
NRZ
NRZ pros and cons
• Pros
– Easy to engineer
– Make good use of bandwidth
• Cons
– dc component
– Lack of synchronization capability and
hard to synchronize timing of sender and
receiver.
• Used for magnetic recording
• Not often used for signal transmission
Differential Encoding
• Data represented by changes rather than
levels
• More reliable detection of transition rather
than level
• In complex transmission layouts it is easy to
lose sense of polarity
Manchester
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Transition in middle of each bit period
Transition serves as clock and data
Low to high represents one
High to low represents zero
Used by IEEE 802.3
Advantages of Manchester
• Synchronization: Because there is a
predictable transition during each bit
time, the receiver can synchronize on that
transition.
• Error detection: Noise on the line would
have to invert both the signal before and
after to cause an undetected error.
How are bits encoded into digital
signals?
• Exercise with a neighbor now.
• Draw a waveform diagram depicting the
message “Hi” using NRZL, NRZI, and
Manchester encoding schemes.
– Assume that the bit representation of “H”
is
0 1 0 0 1 0 0 0 = 0X48
– Assume that the bit representation of “i”
is
0 1 1 0 1 0 0 1 = 0X69
Limitation
• Digital signals cannot be used to transmit
across a long distance.
• During transmitting digital signals, it is
susceptible to interference easily.
• Digital encoding schemes are widely used
in recording.
• Instead, analog signals are used to transmit
even digital data bits. How?
Analog Signals
• Digital computers are incompatible with
analog transmission media such as phone
lines.
• How can one use analog signals to represent
digital data bits?
• We need to convert digital data to analog
signal at the sender side and convert analog
data back to digital data at the receiver side.
Fundamental of Communications
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Fourier Series Approximation
Nyquist Theorem
Shannon’s Theorem
Modulation and Demodulation
We can add
sines
together
to make
new
functions...
g1(t)=sin(2f t)
g2(t)=1/3sin(2( 3f )t)
g3(t)= g1(t) + g2(t)
Fourier Transform


1
s(t )  a0   ai cos2it / P    bi sin 2it / P 
2
n 1
n 1
Jean B. Fourier found that
any periodic function can be expressed as an
infinite sum of sine function.
Baud Rate vs. Bit Rate
• Transmission speed can be measured in bits
per second(bps).
• Technically, transmission is rated in baud,
the number of changes in the signal per
second that the hardware generates.
• Using RS-232 standard to communicate, bit
rate rate = baud rate.
• In general, bit rate rate = N * baud rate,
where N is the number of signals in a string.
Baud Rate vs. Bit Rate
• Sender sends the bit string, by b1 b2 … bn.
• The transmitter alternately analyzes each string
and transmits a signal component uniquely
determined by the bit values. Once the component
is sent, the transmitter gets another bit string and
repeats this process.
• The different signal components make up the
actual transmitted signal. The frequency with
which the components change is the baud rate.
• At the receiving end, the process is reversed.The
receiver alternately samples the incoming signal
and generates a bit string.
Baud Rate vs. Bit Rate
• Consequently, the bit rate depends on two
things: the frequency with which a
component can change (baud rate) and n,
the number of bits in the string. That is why
the formula:(signal may have up to 2n
different amplitudes)
bit rate = n * baud rate
Nyquist Sampling Theorem
• Due to Harry Nyquist (1920)
• Nyquist showed that if ƒ is the maximum
frequency the medium can transmit, the receiver
can completely reconstruct by sampling it 2ƒ
times per second on a perfectly noiseless channel.
• In other words, the receiver can reconstruct the
signal by sampling it at intervals of 1/(2ƒ) second.
• For example, if the max frequency is 4000 Hz, the
receiver needs to sample the signal 8000 times per
second or using 2ƒ as the baud rate.
• Bit rate = 2ƒ * n. (See Table 2.9)
Any Limit on Bit Rate?
• The formula Bit rate = 2ƒ * n seems to
imply that there is no upper bound for the
data rate given the maximum frequency.
Unfortunately, this is not true for two
reasons.
How about real hardware?
• First, if we used amplitude to represent data
bits, each time we separate the amplitude
into smaller ranges to represent more data
bits, the receiver must be more sophisticated
(and more expensive) to be able to detect
smaller differences. If the differences
become too small, we eventually exceed the
ability of a device to detect them.
How about real hardware?
• Second, many channels are actually subject
to noise.
Limitation on Real Hardware
Signal-to-Noise Ratio
• S/N is known as the signal-to-noise ratio in
decibles.
• Because S is usually much larger than N,
the ration is often scaled down
logarithmically and the unit is measured in
bels and 1 dB = 0.1 bel.
• So when we refer to signal-to-noise ration,
we should be careful about units.
Signal-to-Noise Ratio
• Electrical engineers uses S/N to indicate the
quality of sound. The higher the ration is,
the better the quality is.
• B = log 10 (S/N) bels, where B is the quality
rate measured in bels.
• S/N is known as the signal-to-noise ratio
measured in decibles.
• If B=2.5 bels, then S = ___________N?
Shannon’s Theorem
• Bit rate = Bandwidth * log 2 (1+S/N) bps.
• According to this result, a bit rate around
35,000 bps is an upper limit for
conventional modems.
Shannon’s Theorem and
56KModem
• According to this result, a bit rate around
35,000 bps is an upper limit for
conventional modems.
• However, a 56kbps modem can achieve the
high rates when used to connect with an
ISP. As such, it takes advantage of the fact
that there is no analog-to-digital conversion
at the ISP site.
Example of Shannon’s Theorem
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Bandwidth = 3000 Hz,
Quality rate = 35 dB or 3.5 bels,
What is the bit rate?
Please work with your neighbors now.
Hint: You must find S/N first.
Motivation on Modulation and
Demodulation
If either analog or digital signals were used
exclusively, communications would be
simplified. However, this is impossible
especially attempting to send signals across
a long distance. Digital signals cannot be
transmitted far without being converted to
analog signals. Because telephone system is
an analog device, computer signals must be
converted to analog signals.
The Waveform of a Carrier
The wave form of an analog signal
carrier oscillates continuously even
when no signal is being sent.
Carrier
• Researchers found that a continuous,
oscillating signal will propagate farther than
other signals.
• Instead of transmitting an electric current
that only changes when the value of a bit
changes, long-distance communication
systems send a continuously oscillating
signal, usually a sine wave, called a carrier.
Data Modulation
• To send data, a transmitter modifies the
carrier slightly.
• Collectively, such modifications are called
modulation.
• The technique was originated for
transmitting radio or TV signals.
• Generally speaking, modulation is the
process to transform a digital signal into an
analog signal.
Data Demodulation
• At the receiving end, the analog signal
is transformed back to digital signals.
• The process is called demodulation.
• The device to perform modulation and
demodulation is called a modem. We
will talk about modem later.
Example of Data Modulation
The digital signal ’01’ is sent. The carrier
is reduced to 2/3 full strength to encode a
1 bit and 1/3 strength to encode a 0 bit.
Modulation Techniques
• Amplitude shift keying (ASK)
• Frequency shift keying (FSK)
• Phase shift keying (PK)
Modulation Techniques
Modulation Techniques
This modulation technique is called
Amplitude Shift keying (ASK) technique.
Amplitude Shift Keying
• Values represented by different amplitudes
of carrier
• Usually, one amplitude is zero
– i.e. presence and absence of carrier is used
• Susceptible to sudden gain changes
• Inefficient
Example of ASK
Bit Values
00
01
10
11
Amplitude
A1
A2
A3
A4
Amplitude Shifting Keying (four amplitudes),
two bits per baud
Phase Shift Keying
• Nyquist Theorem suggests that the number
of bits sent per cycle can be increased if the
encoding scheme permits multiple bits to be
encoded in a single cycle of the carrier.
• ASK and FSK work well but require at least
one cycle of a carrier wave to send a single
bit.
• PSK changes the timing of the carrier wave
abruptly to encode data. Such change is
called a phase shift.
Example of Phase Shift
Phase Shift Keying
Arrows indicate points at which the carrier abruptly
jumps to a new position in the cycle. For different
code, the phase shift is different.
Frequency Shift Keying
QAM
• Any of the simple techniques can be used
with any number of different signals.
• More signals means a greater bit rate with a
given baud rate.
• The problem is that a higher bit rate
requires more signals and reduces the
differences among them and makes the
receiver’s job more difficult.
QAM(cont’d)
• Another approach is to use a combination of
frequencies, amplitudes, or phase shifts,
which allows us to use a larger group of
legitimate signals while maintaining larger
differences among them.
• One technique is Quadrature Amplitude
Modulation (QAM), in which a group of
bits is assigned a signal defined by its
amplitude and phase shift.
Signal Associations for QAM
Two amplitudes and four phases are used to send
three bits per baud.
Performance of Digital to Analog
Modulation Schemes
• Bandwidth
– ASK and PSK bandwidth directly related to bit
rate
– FSK bandwidth related to data rate for lower
frequencies, but to offset of modulated
frequency from carrier at high frequencies
• In the presence of noise, bit error rate of PSK and
QPSK are about 3dB superior to ASK and FSK
Analog-to-Digital Conversion
• Usually, it is the reverse of what we have
just discussed. A modem examines the
incoming signals for amplitude,
frequencies, and phase shifts and generates
digital signals. This works for signals
having constant characteristics.
• What about analog signals whose
characteristics change continually such as a
voice signal?
Pulse Code Modulation
• One way of making the signal truly digital
is to assign amplitudes from a predefined
set to the sample signals.
• This process is called PCM.
The pulse amplitude is divided into eight values or 23 values.
Accuracy of PCM
1. The sampling frequency
2. The number of amplitudes chosen: in Fig
2.47, the resulted signal becomes
distorted.
Modem
• Modem = modulator + demodulator
• A modem converts digital signals to analog
signals before sending them across a phone
line.
• Another modem converts analog signals back
to digital signals before passing them to a
receiver.
Illustration of Dial-up Modem
Modems
• Intelligent Modems (Hayes Compatible)
– A user can enter commands such as continuing dialing,
beeping when disconnected, etc.
– Hayes Modem allows a user to enter AT command to
request for connection.
– ATDT5551234: AT represents AT command; D stands
for dial; T stands for tone dialing.
• Cable Modems – connects to cable TV
carrier from a PC and a TV.
• Null Modems – used for connecting two
local PC’s together. (will be discussed again
in next chapter)
3Com Cable Modem
Summary of Modem
A pair of modem is required for long-distance
communication across a leased line; each
modem contains separate circuitry to send and
receive digital data. To send data, a modem
emits a continuous carrier wave, which it then
modulates according to the values of the bits
being transferred. To receive data, a modem
detects modulation in the incoming carrier, and
uses it to recreate the data bits.
Assignment
• Read Chapter 2
• Exercises