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Chapter 3: Signals

Analog and Digital Signals

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

Analog and DigitalAnalog and DigitalAnalog and Digital DataAnalog and Digital SignalsPeriodic and Aperiodic Signal

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DataDataData can be

Analog infinite number of values in a range

Digital limited number of defined values

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Analog Signals Sine wave : most fundamental form of a periodic analog signal

Amplitude Absolute value of a signal’s highest intensity, Normally in

volts Frequency

number of periods in one second, inverse of period Change in a short span of time means high frequencyhigh frequency

Phase Position of the waveform relative to time zero (degrees or

radians )

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Time and Frequency DomainsTime and Frequency Domains Time-domain plot

displays changes in signal amplitude with respect to time

Frequency-domain plot compares time domain and frequency domain

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Digital SignalsDigital Signals Use binary (0s and 1s) to encode information Less affected by interference (noise) Fewer errors Describe digital signals by

Bit intervalBit interval time required to send one bit

Bit rateBit rate number of bit intervals per sec (bps)

Analog bandwidthAnalog bandwidth range of frequencies a medium can pass (hertz)

Digital bandwidthDigital bandwidth maximum bit rate that a medium can pass (bps)

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Data Rate LimitsData Rate Limits How to determine the maximum bit rate (bps) How to determine the maximum bit rate (bps)

over a channel?over a channel? Data rate depends on 3 factors

Bandwidth available Levels of signals we can use Quality of the channel (level of noise)

Two theoretical formulas were developed to calculate the data rate Nyquist for a noiseless channel Shannon for noisy channel

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Noiseless ChannelNoiseless Channel Nyquist Bit RateNyquist Bit Rate

Defines the theoretical maximum bit rate

BitBit Rate = 2 Rate = 2 BandwidthBandwidth log log22 LL

L L is the number of signal levels used to represent data is the number of signal levels used to represent data

Example

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

BitBit Rate = 2 Rate = 2 3000 3000 log log22 2 = 6000 bps 2 = 6000 bps

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Noisy ChannelNoisy ChannelShannon CapacityShannon Capacity

Determine the theoretical highest data rate for a noisy channel C = B C = B loglog22 (1 + SNR) (1 + SNR)

Example

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. then Channel capacity Channel capacity = 3000 log= 3000 log22 (1 + 3162) (1 + 3162)

= 3000 log= 3000 log22 (3163) (3163)

= 3000 = 3000 11.62 11.62 = 34,860 bps= 34,860 bps

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Transmission ImpairmentTransmission Impairment Imperfections cause impairment, which means that a signal at

the beginning and the end of the medium are not the same Three types of impairmentsThree types of impairments

1) Attenuation1) Attenuation Loss of energy, Amplifiers are used to strengthen To show that a signal has lost or gained strength, engineers use

the concept of decibel (db) The Decibel measures the relative strength of two signals or a

signal at two different points Example

A signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. Calculate the attenuation (loss of power)?

attenuation = 10 log10 (P2/P1) = 10 log10 (P2/P1)

= 10 log10 (0.5P1/P1) = 10 log10 (0.5P1/P1)

= 10 log10 (0.5) = 10 log10 (0.5) = 10(–0.3) = –3 dB = 10(–0.3) = –3 dB

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2) 2) Distortion Signal changes form or shape Each component has its own propagation speed,

therefore its own delay in arriving

3) Noise Thermal noise – random motion of electrons, creating

an extra signal Induced noise – outside sources such as motors and

appliances Crosstalk – effect of one wire on another Impulse noise – a spike for a short period from power

lines, lightning

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Digital Transmission Digital Transmission Ch4Ch4

Methods to transmit data digitally1) Line coding

Process of converting binary data to a digital signal2) Block coding

Coding method to ensure synchronization and detection of errors Three steps

Division Substitution Line coding

3) Sampling is process of obtaining amplitudes of a signal at regular intervals

Transmission modes Parallel Serial

Synchronous Asynchronous

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Signal Level versus Data Level Signal level

number of values allowed in a particular signal Data level

number of values used to represent data Note: figure b should say three signal levels, two data levels

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Pulse Rate versus Bit Rate Pulse

minimum amount of time required to transmit a symbol Pulse rate

defines number of pulses per second Bit rate

defines number of bits per second BitRate = PulseRate x log2L where LL is the number of data levelsdata levels

A signal has four data levels with a pulse duration of 1 ms. We calculate the pulse rate and bit rate as follows:

Pulse Rate = 1000 pulses/sPulse Rate = 1000 pulses/s Bit Rate = PulseRate x logBit Rate = PulseRate x log22 L L

= 1000 x log= 1000 x log22 4 = 2000 bps 4 = 2000 bps

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Line Coding

Process of converting binary data to a digital signal

Line Coding schemes Unipolar Polar Bipolar

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UnipolarUnipolar Uses only one voltage level Polarity is usually assigned to binary 1; a 0 is

represented by zero voltage

Potential problems: DC component Lack of synchronization

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PolarPolar

Uses two voltage levels, one positive and one negative

Alleviates DC component Variations

Nonreturn to zero (NRZ) Return to zero (RZ) Manchester Differential Manchester

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PolarPolar NRZ

Value of signal is always positive or negative NRZ-L

Signal level depends on bit represented positive usually means 0 negative usually means 1

Problem: synchronization of long streams of 0s or 1s NRZ-I (NRZ-Invert)

Inversion of voltage represents a 1 bit 0 bit represented by no change Allows for synchronization Long strings of 0s may still be a problem

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NRZ-L and NRZ-I EncodingNRZ-L and NRZ-I Encoding

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Return to Zero (RZ) May include synchronization as part of the signal for

both 1s and 0s How?

Must include a signal change during each bit Uses three values: positive, negative, and zero 1 bit represented by pos-to-zero 0 bit represented by neg-to-zero

Disadvantage Requires two signal changes to encode each bit; more bandwidth

necessary

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ManchesterManchester Uses an inversion at the middle of each bit interval for

both synchronization and bit representation Negative-to-positive represents binary 1 Positive-to-negative represents binary 0 Achieves same level of synchronization with only 2

levels of amplitude

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DifferentialDifferential ManchesterManchester Inversion at middle of bit interval is used for synch Presence or absence of additional transition at beginning of

interval identifies the bit Transition 0; no transition 1 Requires two signal changes to represent binary 0; only one to

represent 1

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Bipolar EncodingBipolar Encoding

Uses 3 voltage levels: pos, neg, and zero Zero level 0 1s are represented with alternating positive and negative

voltages, even when not consecutive Two schemes

Alternate mark inversion (AMI)

Bipolar n-zero substitution (BnZS)

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Block Coding Coding method to ensure synchronization and detection of errors Three steps

Division Substitution Line coding

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Sampling Analog data must often be converted to

digital format (ex: long-distance services, audio)

Sampling is process of obtaining amplitudes of a signal at regular intervals

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Pulse Amplitude Modulation

Analog signal’s amplitude is sampled at regular intervals; result is a series of pulses based on the sampled data

Pulse Coded Modulation (PCM) is then used to make the signal digital

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Pulse Coded Modulation

First quantizes PAM pulses; an integral value in a specific range to sampled instances is assigned

Each value is then translated to its 7-bit binary equivalent

Binary digits are transformed into a digital signal using line coding

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Digitization of an Analog SignalDigitization of an Analog Signal

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Sampling Rate: Nyquist TheoremSampling Rate: Nyquist Theorem Accuracy of digital reproduction of a signal depends on number

of samples Nyquist theorem

number of samples needed to adequately represent an analog signal is equal to twice the highest frequency of the original signal

ExampleExample

What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Each sample is 8 bits

SolutionSolutionThe sampling rate must be twice the highest frequency

in the signal Sampling rate = 2 x (11,000) Sampling rate = 2 x (11,000) = 22,000 samples/sec= 22,000 samples/sec Bit rateBit rate = = sampling rate x number of bits /samplesampling rate x number of bits /sample = 22000 x 8 = 22000 x 8 = 172 Kbps= 172 Kbps

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4.4 Transmission Mode Parallel

Bits in a group are sent simultaneously, each using a separate link

n wires are used to send n bits at one time Advantage: speedspeed Disadvantage: costcost; limited to short distanceslimited to short distances

Serial Transmission of data one bit at a time using only one single link Advantage: reduced cost Disadvantage: requires conversion devices Methods:

Asynchronous Synchronous

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Asynchronous Transmission Slower, ideal for low-speed communication when gaps may occur

during transmission (ex: keyboard) Transfer of data with start and stop bits and a variable time interval

between data units Timing is unimportant Start bit alerts receiver that new group of data is arriving Stop bit alerts receiver that byte is finished Synchronization achieved through start/stop bits with each byte

received Requires additional overhead (start/stop bits) Cheap and effective

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Synchronous Transmission

Bit stream is combined into longer frames, possibly containing multiple bytes

Requires constant timing relationship Any gaps between bursts are filled in with a special sequence of

0s and 1s indicating idle Advantage: speed, no gaps or extra bits Byte synchronization accomplished by data link layer

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Chapter 6

Multiplexing

Multiplexing A set of techniques that allows the simultaneous transmission of

multiple signals across a single data link Can utilize higher capacity links without adding additional lines for

each device – better utilization of bandwidth Multiplexer (MUX)

Combines multiple streams into a single stream (many to one). Demultiplexer (DEMUX)

Separates the stream back into its component transmission (one to many) and directs them to their correct lines.

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CATEGORIES OF MULTIPLEXING المجمعاتأصناف

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TIME DIVISION MULTIPLEXING

Digital process that allows several connections to share the high bandwidth of a link

Time Slots and Frames Each host given a slice of timeslice of time (time slot) A frame consists of one complete cycle of time slots, with one

slot dedicated to each sending device.

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TDM Frames

Mux-to-mux speed = aggregate terminal speeds data rate of the link that carries data from n

connections must be n times the data rate of a connection to guarantee the flow of data

i.e., the duration of a frame in a connection is n times the duration of a time slot in a frame

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ExampleExample Four 1-Kbps connections are multiplexed together.

A unit is 1 bit. Find the duration of 1 bit before multiplexing the transmission rate of the link the duration of a time slot, and the duration of a frame?

Solution The duration of 1 bit = 1/1 Kbps = (1 ms). The rate of the link = 4 * 1 Kbps =4 Kbps. Time slot duration = 1/4 ms = .25 ms Frame duration = 4 * .25 ms = 1 ms.

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INTERLEAVING

Process of taking a specific amount of data from each device in a regular order

May be done by bit, byte, or any other data unit Character (byte) Interleaving

Multiplexing perform one/more character(s) or byte(s) at a time Bit Interleaving

Multiplexing perform on one bit at a time

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example

• Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte/ channel

• show the size of the frame

• Frame rate

• Duration of a frame

• Bit rate for the link.

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Example A multiplexer combines four 100-Kbps channels using

a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? 400 Kbps/8 = 50K

frame/sec What is the frame duration? (1/50K) = .02 ms = 20

µs What is the bit rate? 4 * 100kbps = 400 Kbps What is the bit duration? ( 1/400 K) = 2.5 µs

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SYNCHRONIZINGSYNCHRONIZING Framing bit (s) is (are) added to each frame for synchronization between

the MUX and DEMUX synchronization bits allows the DEMUX to synchronize with the incoming

stream so it can separate time slots accurately If 1 framing bit per frame, framing bits are alternating between 0 and 1

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ExampleExample We have four sources, each creating 250 char/ sec. If the

interleaved unit is a character and 1 synchronizing bit is added to each frame, find

(1) Data rate of each source 2000 bps = 2 Kbps

(2) Duration of each character in each source 1/250 s = 4 ms

(3) Frame rate link needs to send 250 frames/sec

(4) Duration of each frame 1/250 s = 4 ms

(5) Number of bits in each frame 4 x 8 + 1 = 33 bits

(6) Data rate of the link. 250 x 33 = 8250 bps

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Example Example • 2 channels, one with a bit rate of 100 Kbps and another

with a bit rate of 200 Kbps, are to be multiplexed.

1. How this can be achieved?

2. What is the frame rate?

3. What is the frame duration?

4. What is the bit rate of the link?

Solution1. Allocate 1 slot to the 1st channel and 2 slots to the 2nd

channel. • Each frame carries 3 bits.

2. The frame rate is 100k frames/sec because it carries 1 bit from the first channel.

3. The frame duration is 1/100,000s= 10 us. 4. The bit rate is 100,000 frames/s x 3 bits/frame= 300

Kbps

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STDMSTDM

Mux-to-Mux speed < aggregate terminal/host speeds

Time slots allocated based on traffic patterns uses statistics to determine allocation among users must send port address with data (takes additional time slots)

May Potential loss of data during peak periods may use data buffering and/or flow control to reduce loss

Not always transparent to user terminals and host/FEP delays and data loss possible

So why use a stat mux? more economical - need fewer muxes, cheaper lines more efficient - allows more terminals to share same line OK to use in many situations (e.g., terminal users

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FREQUENCY DIVISION MULTIPLEXING

Assigns different analog frequencies to each connected device Like Pure TDM,

mux-to-mux speed = aggregate terminal speeds No loss of data so transparent to users and host/FEP

Channels must be separated by strips of unused B.W - guard B.W

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FDM FDM PORCESSPORCESS

Signals of each channel are modulated onto different carrier signal

The resulting modulated signals are then combined into a single composite signal that is sent out over a media link

The link should have enough bandwidth to accommodate it

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FDMFDM DEMULTIPLEXING

Demultiplexer uses a series of filters to decompose the multiplexed signal into its constituent component signals

The individual signals are then passed to a demodulator that separates them from their carriers and passes them to the waiting receivers

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ExampleExample• Assume that a voice channel occupies a B.W of 4 KHz. We

need to combine 3 voice channels into a link with a B.W of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain without the use of guard bands

Solution

Shift (modulate) each of the 3 voice channels to a different B.W

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ExampleExample 5 channels, each with a 100-KHz B.W, are to be multiplexed

together. What is the minimum B.W of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference?

Solution For 5 channels, we need at least 4 guard bands. the required B.W is at least 5 x 100 + 4 x 10 = 540 KHz

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Wave Division MultiplexingWave Division Multiplexing

An analog multiplexing technique to combine optical signals Multiple beams of light at different frequency Carried by optical fibber A form of FDM Each color of light (wavelength) carries separate data channel Commercial systems of 160 channels of 10 Gbps now available

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Digital Signal Service Hierarchy of digital signals

DS-0 single channel of 64 Kbps DS-1 single service or 24 DS-0 channels multiplexed 1.544Mbps

DS-2 single service or 4 DS-1 channels = 96 DS-0 channels

= 6.312 Mbps

DS-3 single service, 7 DS-2 channels = 28 DS-1 channels

= 672 DS-0 channels = 44.376 Mbps

DS-4 6 DS-3 channels = 42 DS-2 channel = 168 DS-1 channels = 4032 DS-0 = 274.176 Mbps

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DS Hierarchy

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TT LinesLines Digital lines designed for digital data, voice, or audio May be used for regular analog (telephone lines) if sampled then

multiplexed using TDM

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T-1T-1 frame structure