3-1 UNIT – II Physical Layer and Media Chapter 3 Data and Signals Chapter 4 Digital Transmission...

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UNIT – II Physical Layer and Media

Chapter 3 Data and SignalsChapter 4 Digital TransmissionChapter 5 Analog TransmissionChapter 6 Bandwidth Utilization: Multiplexing and Spreading

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Data and Signals

Analog and Digital Periodic Analog Signals Digital Signals Transmission Impairment Data Rate Limits Performance

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Analog and Digital

• To be transmitted, data must be transformed to To be transmitted, data must be transformed to electromagnetic signals electromagnetic signals

• Data can be analog or digital. Analog data are Data can be analog or digital. Analog data are continuous and take continuous values. Digital continuous and take continuous values. Digital data have discrete states and take on discrete data have discrete states and take on discrete values.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.

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Analog and Digital Signals

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Periodic and Nonperiodic Signals

• In data communication, we commonly use periodic analog signals and nonperiodic digital signals

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Periodic Analog Signals• Periodic analog signals can be classified as simple or composite. Periodic analog signals can be classified as simple or composite. • A simple periodic analog signal, a sine wave, cannot be decomposed into A simple periodic analog signal, a sine wave, cannot be decomposed into

simpler signals. simpler signals. • A composite periodic analog signal is composed of multiple sine wavesA composite periodic analog signal is composed of multiple sine waves • Sine wave is described by

– Amplitude– Period (frequency)– phase

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Amplitude

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Period and Frequency

• Frequency and period are the inverse of each

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Units of Period and Frequency

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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

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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

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Phase

• Phase describes the position of the waveform relative to time zero

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Sine Wave Examples

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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

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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 single-When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of frequency signal, it becomes a composite signal made of many frequenciesmany 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

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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

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Attenuation

• Loss of energy to overcome the resistance of the medium: heat

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Decibel

• 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

• 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

• 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)

• 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

• 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– Bit rate = 2 * Bandwidth * log2L

– Increasing the levels may cause the reliability of the system

• Noisy channel: Shannon Capacity– Capacity = Bandwidth * log2(1 + SNR)

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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 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

• 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

• The bandwidth-delay product defines the number of bits that can fill the link

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Bandwidth-Delay Product

• Bandwidth-delay product concept

Data Communications, Kwangwoon University 4-49

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


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

• 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


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)

• Self-synchronization• Built-in error detection• Immunity to noise and interference• Complexity


Lack of Synchronization


Line Coding Schemes


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


Polar Scheme

• Two polarity: two levels of voltage• Problem of DC component is alleviated (NRZ,RZ)

or eliminated (Biphaze)


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


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


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


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


Polar Biphase

• Minimum bandwidth is 2 times that of NRZ


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


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)


2B1Q: for DSL


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


4D-PAM5: for Gigabit LAN


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


Summary of Line Coding Schemes


Block Coding

• Block coding is normally referred to as mB/nB coding; it replaces each m-bit group with an n-bit group


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


4B/5B Mapping Codes


8B/10B

• 210 – 28 = 768 redundant groups used for disparity checking and error detection


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


B8ZS

• Commonly used in North America• Updated version of AMI with synchronization• Substitutes eight consecutive zeros with 000VB0VB • V denotes “violation”, B denotes “bipolar”


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


Sampling: Analog-to-Digital Conversion

• Analog information (e.g., voice) digital signal (e.g., 10001011…)

• Codec(Coder/Decoder): A/D converter


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.


Components of PCM Encoder


Different Sampling Methods for PCM


Nyquist Sampling Rate


Sampling Rate


Quantization


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


Original Signal Recovery: PCM Decoder


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


Delta Modulation

• To reduce the complexity of PCM


Delta Modulation Components


Delta Demodulation Components


Transmission Modes


Parallel Transmission

• Use n wires to send n bits at one time synchronously• Advantage: speed• Disadvantage: cost Limited to short distances


Serial Transmission

• On communication channel• Advantage: reduced cost• Parallel/serial converter is required• Three ways: asynchronous, synchronous, or isochronous


Asynchronous Transmission

• 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)


Synchronous Transmission

• 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


Analog Transmission

• Digital-to-Analog Conversion

• Analog-to-Analog Conversion


Digital-to-Analog Conversion

• Digital-to-analog conversion is the process of changing Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the one of the characteristics of an analog signal based on the information in digital datainformation in digital data


Types of Digital-to-Analog Modulation


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


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?

SolutionS = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L.


ASK : Binary ASK

• BASK or OOK (on-off keying)BASK or OOK (on-off keying)

• Bandwidth for ASK: B = (1 + d) x SBandwidth for ASK: B = (1 + d) x S


Implementation of Binary ASK


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.


FSK: Binary FSK

• Bandwidth for ASK: B = (1 + d) x S + 2Bandwidth for ASK: B = (1 + d) x S + 2Δf


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


Implementation of Binary FSK


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.


PSK: Binary PSK

• Bandwidth : the same as BASK, B = (1 + d) x SBandwidth : the same as BASK, B = (1 + d) x S

• Less than that for BFSKLess than that for BFSK


Implementation of Binary PSK


Quadrature PSK


Constellation Diagram

• Define the amplitude and phase of a signal elementDefine the amplitude and phase of a signal element


Constellation Diagram: Examples


QAM

• Quadrature amplitude modulation

• Combination of ASK and PSK

• Bandwidth : the same as that required for ASK and PSK


Analog-to-Analog Modulation

• Analog-to-analog conversion is the representation of analog Analog-to-analog conversion is the representation of analog information by an analog signalinformation by an analog signal

• Modulation is needed if the medium is bandpass in nature or if only a Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to usbandpass channel is available to us


Amplitude Modulation

• The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B.


AM Band Allocation

• Bandwidth of an audio signal (speech and music) is usually 5 kHz


Frequency Modulation

• The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B.


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


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 Whenever the bandwidth of a medium linking two devices is greater than the bandwidth needs of the devices, the link can be shared. than the bandwidth needs of the devices, the link can be shared.

• Multiplexing is the set of techniques that allows the simultaneous Multiplexing is the set of techniques that allows the simultaneous transmission of multiple signals across a single data link.transmission of multiple signals across a single data link.

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Categories of Multiplexing

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Frequency Division Multiplexing

• 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.

d) The duration of a frame 1 ms.

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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 Combine signals from different sources to fit into a larger bandwidth to prevent eavesdropping and jamming by bandwidth to prevent eavesdropping and jamming by adding redundancyadding redundancy

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FHSS

• Frequency Hopping Spread Spectrum (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)Direct Sequence Spread Spectrum (DSSS)• Replace each data bit with n bits using a spreading codeReplace each data bit with n bits using a spreading code• Each bit is assigned a code of n bits called chipsEach bit is assigned a code of n bits called chips

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DSSS Example

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3-1 UNIT – II Physical Layer and Media Chapter 3 Data and Signals Chapter 4 Digital Transmission...

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