Transcript - Snistnote

UNIT – II
Physical Layer and Media
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Data and Signals
Digital Transmission
Analog Transmission
Bandwidth Utilization: Multiplexing and Spreading
3-1
Data and Signals
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Analog and Digital
Periodic Analog Signals
Digital Signals
Transmission Impairment
Data Rate Limits
Performance
3-2
Analog and Digital
• To be transmitted, data must be transformed to
electromagnetic signals
• Data can be analog or digital. Analog data are
continuous and take continuous values. Digital
data have discrete states and take on discrete
values.
• Signals can be analog or digital. Analog signals
can have an infinite number of values in a range;
digital signals can have only a limited number of
values.
3-3
Analog and Digital Signals
3-4
Periodic and Nonperiodic Signals
• In data communication,
we commonly use
periodic analog signals
and nonperiodic digital
signals
3-5
Periodic Analog Signals
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Periodic analog signals can be classified as simple or composite.
A simple periodic analog signal, a sine wave, cannot be decomposed into
simpler signals.
A composite periodic analog signal is composed of multiple sine waves
Sine wave is described by
– Amplitude
– Period (frequency)
– phase
3-6
Amplitude
3-7
Period and Frequency
• Frequency and period are the inverse of each
3-8
Units of Period and Frequency
3-9
Example 3.5
• Express a period of 100 ms in microseconds, and express
the corresponding frequency in kilohertz
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 ms = 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
3-10
More About Frequency
• Another way to look frequency
– Frequency is a measurement of the rate of changes
– Change in a short span of time means high frequency
– Change over a long span of time means low frequency
• Two extremes
– No change at all  zero frequency
– Instantaneous changes  infinite frequency
3-11
Phase
• Phase describes the position of the waveform relative to
time zero
3-12
Sine Wave Examples
3-13
Example 3.6
• A sine wave is offset one-sixth of a cycle with respect to
time zero. What is its phase in degrees and radians?
We know that one complete cycle is 360 degrees.
Therefore, 1/6 cycle is
(1/6) 360 = 60 degrees = 60 x 2π /360 rad = 1.046 rad
3-14
Wavelength
• Another characteristic of a signal traveling through a transmission
medium
• Binds the period or the frequency of a simple sine wave to the
propagation speed of the medium
• Wavelength = propagation speed x period
= propagation speed/frequency
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Time and Frequency Domains
• A complete sine wave in the time domain can be
represented by one single spike in the frequency domain
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Example 3.7
• Time domain and frequency domain of three sine waves
with frequencies 0, 8, 16
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Composite Signals
• A single-frequency sine wave is not useful in data
communications; we need to send a composite signal, a
signal made of many simple sine waves
• When we change one or more characteristics of a singlefrequency signal, it becomes a composite signal made of
many frequencies
• According to Fourier analysis, any composite signal is a
combination of simple sine waves with different
frequencies, phases, and amplitudes
• If the composite signal is periodic, the decomposition gives
a series of signals with discrete frequencies; if the
composite signal is nonperiodic, the decomposition gives a
combination of sine waves with continuous frequencies.
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Composite Periodic Signal
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Composite Nonperiodic Signal
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Bandwidth
• The bandwidth of a composite signal is the difference between the
highest and the lowest frequencies contained in that signal
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Signal Corruption
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Example 3.11
• 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
B = fh - fl, 20 = 60 – fl, fl = 60 - 20 = 40 Hz
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Digital Signals
3-24
Bit Rate and Bit Interval
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Example 3.18
• Assume we need to download text documents at the rate of 100 pages
per minute. What is the required bit rate of the channel?
Solution
• A page is an average of 24 lines with 80 characters in each line. If we
assume that one character requires 8 bits, the bit rate is
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Digital Signal as a Composite Analog Signal
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Transmission of Digital Signals
• A digital signal is a composite analog signal with an
infinite bandwidth
• Baseband transmission: Sending a digital signal without
changing into an analog signal
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Low-Pass Channel with Wide Bandwidth
• Baseband transmission of a digital signal that preserves the shape of
the digital signal is possible only if we have a low-pass channel with
infinite or very wide bandwidth
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Low-Pass Channel with Limited Bandwidth
• Rough approximation
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Low-Pass Channel with Limited Bandwidth
• Better approximation
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Bandwidth Requirement
• In baseband transmission, the required bandwidth is proportional to the
bit rate; if we need to send bits faster, we need more bandwidth
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Broadband Transmission (Using Modulation)
• Modulation allows us to use a bandpass channel
• If the available channel is a bandpass channel, we cannot
send the digital signal directly to the channel; we need to
convert the digital signal to an analog signal before
transmission.
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Modulation for Bandpass Channel
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Transmission Impairment
3-35
Attenuation
• Loss of energy to overcome the resistance of the medium: heat
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Decibel
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Example 3.26: Suppose a signal travels through a transmission medium and its
power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the
attenuation (loss of power) can be calculated as
Example 3.28
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Distortion
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The signal changes its form or shape
Each signal component in a composite signal has its own propagation speed
Differences in delay may cause a difference in phase
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Noise
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Several types of noises, such as thermal noise, induced noise, crosstalk, and
impulse noise, may corrupt the signal
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Signal-to-Noise Ratio (SNR)
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To find the theoretical bit rate limit
SNR = average signal power/average noise power
SNRdB = 10 log10 SNR
• Example 3.31: The power of a signal is 10 mW and the power of the
noise is 1 μW; what are the values of SNR and SNRdB ?
Solution:
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Two Cases of SNRs
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Data Rate Limits
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Data rate depends on three factors:
– Bandwidth available
– Level of the signals we use
– Quality of the channel (the noise level)
Noiseless channel: Nyquist Bit Rate
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Bit rate = 2 * Bandwidth * log2L
Increasing the levels may cause the reliability of the system
Noisy channel: Shannon Capacity
–
Capacity = Bandwidth * log2(1 + SNR)
3-42
Nyquist Bit Rate: Examples
• 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
• Consider the same noiseless channel, transmitting a signal
with four signal levels (for each level, we send two bits).
The maximum bit rate can be calculated as:
Bit Rate = 2 x 3000 x log2 4 = 12,000 bps
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Shannon Capacity: Examples
• Consider an extremely noisy channel in which the value of the signalto-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
• 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
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Using Both Limits
• The Shannon capacity gives us the upper limit; the Nyquist formula
tells us how many signal levels we need.
• Example: 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?
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
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Performance
• Bandwidth (in two contexts)
– Bandwidth in hertz, refers to the range of frequencies in a composite
signal or the range of frequencies that a channel can pass.
– Bandwidth in bits per second, refers to the speed of bit transmission in a
channel or link.
• Throughput
– Measurement of how fast we can actually send data through a network
• Latency (Delay)
– Define how long it takes for an entire message to completely arrive at the
destination from the time the first bit is sent out from the source
– Latency = propagation time + transmission time + queuing time +
processing delay
– Propagation time = Distance/Propagation speed
– Transmission time = Message size/Bandwidth
• Jitter
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Bandwidth-Delay Product
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The bandwidth-delay product defines the number of bits that can fill the link
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Bandwidth-Delay Product
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Bandwidth-delay product concept
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Digital-to-Digital Conversion
• Involves three techniques:
– Line coding (always needed), block coding, and
scrambling
• Line coding: the process of converting digital data to
digital signals
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Signal Element and Data Element
• Data elements are what we need to send; signal elements are what we
can send
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Data Rate Versus Signal Rate
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Data rate defines the number of data elements (bits) sent in 1s: bps
Signal rate is the number of signal elements sent in 1s: baud
Data rate = bit rate, signal rate = pulse rate, modulation rate, baud rate
S = c x N x 1/r, where N is the date rate; c is the case factor, S is the
number of signal elements; r is the number of data elements carried by
each signal element
Although the actual bandwidth of a digital signal is infinite, the effective
bandwidth is finite
The bandwidth is proportional to the signal rate (baud rate)
The minimum bandwidth: Bmin = c x N x 1/r
The maximum data rate: Nmax = 1/c x B x r
4-51
Design Consideration for Line Coding
Scheme
• Baseline wandering
– Long string of 0s and 1s can cause a drift in the
baseline
• DC components
– DC or low frequencies cannot pass a transformer or
telephone line (below 200 Hz)
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Self-synchronization
Built-in error detection
Immunity to noise and interference
Complexity
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Lack of Synchronization
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Line Coding Schemes
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Unipolar Scheme
• One polarity: one level of signal voltage
• Unipolar NRZ (None-Return-to-Zero) is simple, but
– DC component : Cannot travel through microwave or transformer
– Synchronization : Consecutive 0’s and 1’s are hard to be synchronized 
Separate line for a clock pulse
– Normalized power is double that for polar NRZ
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Polar Scheme
• Two polarity: two levels of voltage
• Problem of DC component is alleviated (NRZ,RZ)
or eliminated (Biphaze)
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Polar NRZ
• NRZ-L (Non Return to Zero-Level)
– Level of the voltage determines the value of the bit
• NRZ-I (Non Return to Zero-Invert)
– Inversion or the lack of inversion determines the value of the bit
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Polar NRZ: NRZ-L and NRZ-I
• Baseline wandering problem
– Both, but NRZ-L is twice severe
• Synchronization Problem
– Both, but NRZ-L is more serious
• NRZ-L and NRZ-I both have an average signal
rate of N/2 Bd
• Both have a DC component problem
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RZ
• Provides synchronization for consecutive 0s/1s
• Signal changes during each bit
• Three values (+, -, 0) are used
– Bit 1: positive-to-zero transition, bit 0: negative-to-zero transition
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Biphase
• Combination of RZ and NRZ-L ideas
• Signal transition at the middle of the bit is used for
synchronization
• Manchester
– Used for Ethernet LAN
– Bit 1: negative-to-positive transition
– Bit 0: positive-to-negative transition
• Differential Manchester
– Used for Token-ring LAN
– Bit 1: no transition at the beginning of a bit
– Bit 0: transition at the beginning of a bit
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Polar Biphase
• Minimum bandwidth is 2 times that of NRZ
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Bipolar Scheme
• Three levels of voltage, called “multilevel binary”
• Bit 0: zero voltage, bit 1: alternating +1/-1
– (Note) In RZ, zero voltage has no meaning
• AMI (Alternate Mark Inversion) and pseudoternary
– Alternative to NRZ with the same signal rate and no DC
component problem
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Multilevel Scheme
• To increase the number of bits per baud by encoding a
pattern of m data elements into a pattern of n signal
elements
• In mBnL schemes, a pattern of m data elements is encoded
as a pattern of n signal elements in which 2m ≤ Ln
• 2B1Q (two binary, one quaternary)
• 8B6T (eight binary, six ternary)
• 4D-PAM5 (four-dimensional five-level pulse amplitude
modulation)
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2B1Q: for DSL
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8B6T
• Used with 100Base-4T cable
• Encode a pattern of 8 bits as a pattern of 6 (three-levels) signal
elements
• 222 redundant signal element = 36(478 among 729) - 28(256)
• The average signal rate is theoretically, Save = 1/2 x N x 6/8; in practice
the minimum bandwidth is very close to 6N/8
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4D-PAM5: for Gigabit LAN
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Multiline Transmission: MLT-3
• The signal rate for MLT-3 is one-fourth the bit rate
• MLT-3 when we need to send 100Mbps on a copper wire that cannot
support more than 32MHz
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Summary of Line Coding Schemes
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Block Coding
• Block coding is normally referred to as mB/nB coding; it
replaces each m-bit group with an n-bit group
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4B/5B
• Solve the synchronization problem of NRZ-I
• 20% increase the signal rate of NRZ-I (Biphase scheme has the signal
rate of 2 times that of NRZ-I
• Still DC component problem
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4B/5B Mapping Codes
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8B/10B
• 210 – 28 = 768 redundant groups used for disparity
checking and error detection
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Scrambling
• Biphase : not suitable for long distance communication due
to its wide bandwidth requirement
• Combination of block coding and NRZ: not suitable for
long distance encoding due to its DC component problem
• Bipolar AMI: synchronization problem  Scrambling
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B8ZS
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Commonly used in North America
Updated version of AMI with synchronization
Substitutes eight consecutive zeros with 000VB0VB
V denotes “violation”, B denotes “bipolar”
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HDB3
• High-density bipolar 3-zero
• Commonly used outside of North America
• HDB3 substitutes four consecutive zeros with 000V or B00V depending
on the number of nonzero pulses after the last substitution
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Sampling: Analog-to-Digital Conversion
• Analog information (e.g., voice)  digital signal
(e.g., 10001011…)
• Codec(Coder/Decoder): A/D converter
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PCM
• Pulse Code Modulation
• Three processes
– The analog signal is sampled
– The sampled signal is quantized
– The quantized values are encoded as streams of bits
• Sampling: PAM (Pulse amplitude Modulation)
– According to the Nyquist theorem, the sampling rate
must be at least 2 times the highest frequency contained
in the signal.
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Components of PCM Encoder
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Different Sampling Methods for PCM
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Nyquist Sampling Rate
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Sampling Rate
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Quantization
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Quantization
• Quantization level (L)
• Quantization error : depending on L (or nb )
– SNRdB = 6.02nb + 1.76 dB
• Nonuniform quantization:
– Companding and expanding
– Effectively reduce the SNRdB
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Original Signal Recovery: PCM Decoder
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PCM Bandwidth
• The min. bandwidth of a line-encoded signal
– Bmin = c x N x 1/r = c x nb x fs x 1/r
= c x nb x 2 x Banalog x 1/r
= nb x Banalog where 1/r = 1, c = 1/2
• Max. data rate of a channel
– Nmax = 2 x B x log2L bps
• Min. required bandwidth
– Bmin = N/(2 x log2L) Hz
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Delta Modulation
• To reduce the complexity of PCM
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Delta Modulation Components
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Delta Demodulation Components
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Transmission Modes
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Parallel Transmission
• Use n wires to send n bits at one time synchronously
• Advantage: speed
• Disadvantage: cost  Limited to short distances
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Serial Transmission
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On communication channel
Advantage: reduced cost
Parallel/serial converter is required
Three ways: asynchronous, synchronous, or isochronous
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Asynchronous Transmission
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Use start bit (0) and stop bits (1s)
A gap between two bytes: idle state or stop bits
It means asynchronous at byte level
Must still be synchronized at bit level
Good for low-speed communications (terminal)
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Synchronous Transmission
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Bit stream is combined into “frames”
Special sequence of 1/0 between frames: No gap
Timing is important in midstream
Byte synchronization in the data link layer
Advantage: speed  high-speed transmission
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Analog Transmission
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Digital-to-Analog Conversion
Analog-to-Analog Conversion
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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
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Types of Digital-to-Analog Modulation
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Aspects of D/A Conversion
• Data element versus signal element
• Data rate (bit rate) versus signal rate (baud rate)
– S = N x 1/r baud
S (signal rate), N (data rate),
r (number of data element in one signal element)
– Bit rate: bits per second (in bps)
– Baud rate: signal elements per second (in baud)
– Bit rate  baud rate
• Carrier signal (carrier frequency)
– High-frequency signal used to modulate the information
– Modulated signal: information modulated by the carrier signal
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Examples 5.2
•
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
S = 1000, N = 8000, and r and L are unknown. We find
first the value of r and then the value of L.
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ASK : Binary ASK
• BASK or OOK (on-off keying)
• Bandwidth for ASK: B = (1 + d) x S
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Implementation of Binary ASK
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Full-duplex ASK: Example
• In data communications, we normally use full-duplex links
with communication in both directions. We need to divide
the bandwidth into two with two carrier frequencies. In this
example, the available bandwidth for each direction is now
50 kHz, which leaves us with a data rate of 25 kbps in each
direction.
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FSK: Binary FSK
• Bandwidth for ASK: B = (1 + d) x S + 2Δf
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BFSK: Example
• We have an available bandwidth of 100 kHz which spans
from 200 to 300 kHz. What should be the carrier frequency
and the bit rate if we modulated our data by using FSK
with d = 1?
The midpoint of the band is at 250 kHz. We choose 2Δf to
be 50 kHz; this means
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Implementation of Binary FSK
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Multilevel FSK
• The frequencies need to be 2Δf apart. Min. value 2Δf needs to be S.
• B = (1 + d) x S + (L – 1) 2Δf  B = L x S with d = 0
• Example: We need to send data 3 bits at a time at a bit rate of 3 Mbps.
The carrier frequency is 10 MHz. Calculate the number of levels
(different frequencies), the baud rate, and the bandwidth
L = 23 = 8. The baud rate is S = 3 MHz/3 = 1000 Mbaud. This means
that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The
bandwidth is B = 8 × 1000 = 8000.
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PSK: Binary PSK
• Bandwidth : the same as BASK, B = (1 + d) x S
• Less than that for BFSK
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Implementation of Binary PSK
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Quadrature PSK
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Constellation Diagram
• Define the amplitude and phase of a signal element
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Constellation Diagram: Examples
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QAM
• Quadrature amplitude modulation
• Combination of ASK and PSK
• Bandwidth : the same as that required for ASK and PSK
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Analog-to-Analog Modulation
• Analog-to-analog conversion is the representation of analog
information by an analog signal
• Modulation is needed if the medium is bandpass in nature or if only a
bandpass channel is available to us
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Amplitude Modulation
• The total bandwidth required for AM can be determined
from the bandwidth of the audio signal: BAM = 2B.
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AM Band Allocation
• Bandwidth of an audio signal (speech and music) is
usually 5 kHz
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Frequency Modulation
• The total bandwidth required for FM can be determined
from the bandwidth of the audio signal: BFM = 2(1 + β)B.
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FM Band Allocation
• Bandwidth of an audio signal (speech and music)
broadcast in stereo is almost 15 kH
• FCC allows 200 kHz for each station (β =4 with some
extra guard band)
• Separated by at least 200 kHz
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Phase Modulation
• The total bandwidth required for PM can be determined
from the bandwidth and maximum amplitude of the
modulating signal: BPM = 2(1 + β)B.
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Chapter 6. Bandwidth Utilization:
Multiplexing and Spreading
1. Multiplexing
2. Spread Spectrum
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Bandwidth Utilization
• Bandwidth utilization is the wise use of available bandwidth to achieve
specific goals.
• Two categories: multiplexing and spreading
• Efficiency can be achieved by multiplexing
• Privacy and anti-jamming can be achieved by spreading.
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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.
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Categories of Multiplexing
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Frequency Division Multiplexing
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FDM is an analog multiplexing technique that combines analog signals
Signals modulate different carrier frequencies
Modulated signals are combined into a composite signal
Channel - Bandwidth range to accommodate a modulated signal
Channels can be separated by strips of unused bandwidth (guard band)
to prevent overlapping
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FDM Process
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FDM Demultiplexing Example
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FDM: Example 1
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FDM: Example 2
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FDM: Example 3
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Analog Hierarchy
• Hierarchical system used by AT&T
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Wave Division Multiplexing
• Analog multiplexing technique to combine optical
signals
• Conceptually the same as FDM
• Light signals transmitted through fiber optic
channels
• Combining different signals of different
frequencies (wavelengths)
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Prisms in WDM
• Combining and splitting of light sources are easily
handled by a prism
• Prism bends a light beam based on the incidence
angle and the frequency
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Time Division Multiplexing
• Digital multiplexing technique for combining several low-rate
channels into one high-rate one
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TDM: Time Slots and Frames
• In synchronous TDM, the data rate of the link is n times faster, and the
unit duration is n times shorter
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TDM: Example 1
• Four 1-Kbps connections are multiplexed together.
A unit is 1 bit. Find (a) the duration of 1 bit before
multiplexing, (b) the transmission rate of the link,
(c) the duration of a time slot, and (d) the duration
of a frame?
a) The duration of 1 bit is 1/1 Kbps, or 0.001 s (1
ms).
b) The rate of the link is 4 Kbps.
c) The duration of each time slot 1/4 ms or 250 μs.
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d) The duration of a frame 1 ms.
Interleaving
• Interleaving can be done by bit, by byte, or by any other data unit
• The interleaved unit is of the same size in a given system
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TDM: Example 2
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TDM: Example 3
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Empty Slots
• Synchronous TDM is not efficient in many cases
• Statistical TDM can improve the efficiency by removing the empty slot
from the frame
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Data Rate Management
• To handle a disparity in the input data rates
• Multilevel multiplexing, multiple-slot allocation and pulse stuffing
• Multilevel multiplexing
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Data Rate Management
• Multiple-slot allocation / Pulse stuffing
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Frame Synchronizing
• Synchronization between the multiplexing and demultiplexing is a
major issue in TDM
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TDM: Example 4
• We have four sources, each creating 250 characters per second. If the
interleaved unit is a character and 1 synchronizing bit is added to each
frame, find (a) the data rate of each source, (b) the duration of each
character in each source, (c) the frame rate, (d) the duration of each
frame, (e) the number of bits in each frame, and (f) the data rate of the
link.
1. The data rate of each source is 2000 bps = 2 Kbps.
2. The duration of a character is 1/250 s, or 4 ms.
3. The link needs to send 250 frames per second.
4. The duration of each frame is 1/250 s, or 4 ms.
5. Each frame is 4 x 8 + 1 = 33 bits.
6. The data rate of the link is 250 x 33, or 8250 bps
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Digital Hierarchy
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DS and T Line Rates
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T-1 Line for Multiplexing Telephone Lines
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T-1 Frame Structure
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E Line Rates
• European use a version of T lines called E lines
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Statistical TDM
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Statistical TDM
• Addressing is required in Statistical TDM
• Slot size: the ratio of the data size to address size must be
reasonable to make transmission efficient
• No synchronization bit: no need for frame-level sync.
• Bandwidth: normally less than the sum of the capacities of
each channel
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Spread Spectrum
• Combine signals from different sources to fit into a larger
bandwidth to prevent eavesdropping and jamming by
adding redundancy
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FHSS
• Frequency Hopping Spread Spectrum (FHSS)
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Frequency Selection in FHSS
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Frequency Cycles
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Bandwidth Sharing
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DSSS
• Direct Sequence Spread Spectrum (DSSS)
• Replace each data bit with n bits using a spreading code
• Each bit is assigned a code of n bits called chips
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DSSS Example
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