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1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

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Page 1: 1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

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Ch 6 Long-Distance Communication

Carriers, Modulation, and Modems

Page 2: 1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

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Sending Signals across Long Distances

Sending Signals across Long Distances

Important fact: Current becomes weaker as it travels (signal loss) A continuous, oscillating signal travels farther than

direct currentFor long-distance communication

Send a sine wave (called a carrier wave) Modifies (modulate) the carrier to encode date

modulated carrier technique used for telephone, radio, and television

Page 3: 1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

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Illustration of a CarrierIllustration of a CarrierCarrier

Usually a sine wave Oscillates continuously Frequency of carrier fixed (f = 1/T)

T

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Types of ModulationTypes of Modulation

Amplitude modulation (used in AM radio)Frequency modulation (used in FM radio)Phase shift modulation (used for data)

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Illustration of AMIllustration of AMStrength of signal encodes 0 (1/3 full strength) or 1 (2/3 strength)Receiver monitors incoming carrier, detects modulation,

reconstruct the original data, and discard the carrierAt least a half cycle of wave needed for each bitData rate limited by carrier bandwidth

Page 6: 1 Ch 6 Long-Distance Communication Carriers, Modulation, and Modems

6The answer is 01101001

Ex. The wave form diagram below shows the transmission of 8 bits using amplitude modulation where a high amplitude represents a 1 bit and a low amplitude represents a 0 bit. Each bit is transmitted in half of a wavelength. Give the value of the 8 bits being transmitted.

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Illustration ofPhase-Shift Modulation

Illustration ofPhase-Shift Modulation

A change in phase (phase shift) encodes K>1 bitsData rate higher than carrier bandwidth

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Phase-Shift ExamplePhase-Shift Example

Section of wave is omitted at phase shiftData bits determine size of omitted section

8 Patterns vs. 8 possibilities of 3-bit strings

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Ex. A communication system transmits information by sending one of 8 different voltage levels (patterns) down a wire every millisecond. (1) What is the baud rate of the system? (2) How many bits per second are being transmitted?

Answer:(1) The baud rate is the signals per second or 1000. (2) Each signal can be any one of 8 values. It takes 3 bits to represent the 8 different values since log2(8) = 3. There are 3000 bits per second being transmitted.

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ModemModem

Name abbreviates modulator / demodulator transforms data into a modulated signal extracts data from modulated signal

Hardware device that contains separate circuitry for

Modulation of outgoing signal Demodulation of incoming signal

Used for long-distance communication One modem at each end

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Modulator on one modem connects to demodulator on other (4-wire circuit)

Telephone companies allow companies to lease a circuit between any two locations

ModemModem

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Types of ModemsTypes of ModemsConventional

Use four wires Transmit modulated electrical wave

Optical Use glass fibers Transmit modulated light

Wireless Use air / space Transmit modulated RF wave

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Dialup ModemsDialup ModemsThe modem can simulate lifting the handset,

dialing, or hanging up the phone Transmit modulated audio tone

Carrier is the tone heard if one lifts the handset

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Modem TerminologyModem Terminology

Full-duplex modemProvides 2-way communicationAllows simultaneous transmissionUses four wires

Half-duplex modemDoes provide 2-way communicationTransmits in one direction at any timeUses only two wires

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RecallRecall

Bandwidth Maximum times per second signal can change

Throughput The number of bits per second that can be

transmitted Related to underlying hardware bandwidth

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Relationship Between Digital

Throughput and Bandwidth

Relationship Between Digital

Throughput and BandwidthGiven by Nyquist’s theorem:

D = 2 B log2 K

whereD is maximum data rateB is hardware bandwidthK is number of values used to encode data

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Applications of Nyquist’s Theorem

Applications of Nyquist’s Theorem

For RS-232K is 2 because RS-232 only uses two values,

+15 or –15 volts, to encode data bitsD is 2 B log2 2 = 2 B

For phase-shift encodingSuppose K is 8 (possible shifts)D is 2 B log2 8 = 2 B x 3 = 6 B

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More Bad NewsMore Bad News

Physics tells us that real systems emit and absorb energy (e.g., thermal)

Engineers call unwanted energy noiseNyquist’s theorem

Assumes a noise-free systemOnly works in theory

Shannon’s theorem corrects for noise

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Shannon’s TheoremShannon’s Theorem

Gives capacity in presence of noise:

C = B log2 (1 + S/N) ( B*(dB/3.01) ……Thompson

formula )where

C is the effective channel capacity in bits per secondB is hardware bandwidthS is the average power (signal)N is the noise

S/N is signal-to-noise ratio

B=log10(S/N) belsdecibel (dB): 1dB=0.1 beleg, 30dB3=log10(S/N)S/N=1000

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Application of Shannon’s Theorem

Application of Shannon’s Theorem

Conventional telephone systemEngineered for voiceBandwidth is 3000 HzSignal-to-noise ratio is approximately 30DB

(1000)Effective capacity is

3000 log2 (1 + 1000) = ~30000 bps Conclusion: dialup modems have little hope

of exceeding 28.8 Kbps

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Ex. Consider a network connection with 200K Hz of bandwidth. (1) If the channel is perfectly noiseless and you use a transmission method that uses 4 different states, what is the maximum transmission speed you can achieve? (2) If the channel has 28dB of noise, what it the maximum transmission speed you can achieve? (3) How can it be that the noisy channel can send more data than the ideal channel?

Answer:(1) Speed = 2 * 200K * log24 = 800K bits/sec(2) Speed = B * dB / 3.01 = 200K * 28 / 3.01 = 1.86 M bits/sec(3) The ideal channel is slower because it is not using enough different states. If it used more states, it would run faster.

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The Bottom LineThe Bottom Line

Nyquist’s theorem means finding a way to encode more bits per cycle improves the data rate

Shannon’s theorem means that no amount of clever engineering can overcome the fundamental limits of a real transmission system

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General Concept of Multiplexing

General Concept of Multiplexing

Separate pairs of communications travel across shared channel

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Multiplexing TerminologyMultiplexing Terminology

Multiplexor Device that accepts data from multiple sources Sends data across shared channel

Demultiplexor Device that extracts data from shared channel Sends to correct destination

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Time Division Multiplexing (TDM)

Time Division Multiplexing (TDM)

Slotted TDM Statistical TDM: if a given source does not have data to

send, the multiplexor skips the source (suitable for burst traffic)

M

U

X

M

U

X

A3B3C3D3 A2B2C2D2 A1B1C1D1

framesA4 A3 A2 A1

B4 B3 B2 B1

D4 D3 D2 D1

C4 C3 C2 C1

A4 A3 A2 A1

B4 B3 B2 B1

C4 C3 C2 C1

D4 D3 D2 D1

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Frequency Division Multiplexing (FDM)

Frequency Division Multiplexing (FDM)

Multiple items transmitted simultaneouslyUses different carrier frequencies, or

“channels” E.g., Radio, cable TV

M

U

X

M

U

X

guard band

guard band

channel 3

channel 2

channel 1

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Wave Division MultiplexingWave Division Multiplexing

When applied to light, FDM is called wavelength division multiplexing (WDM)

Informally called color division multiplexing

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Scientific Principle BehindFrequency Division

Multiplexing

Scientific Principle BehindFrequency Division

Multiplexing

Note: this is the same principle that allows a cable TV company to send multiple television signals across a single cable

Two or more signals that use different carrier frequencies can be transmitted over a single medium simultaneously without interference