# CS代考计算机代写 Note: We will start at 12:53 pm ET

Note: We will start at 12:53 pm ET

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18-441/741: Computer Networks Lectures 4: Physical Layer II

Swarun Kumar

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Physical Layer: Outline

• Digitalnetworks

• ModulationFundamentals

• CharacterizationofCommunicationChannels

• FundamentalLimitsinDigitalTransmission

• DigitalModulation

• LineCoding

• PropertiesofMediaandDigitalTransmission Systems

• ErrorDetectionandCorrection

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Transferring Information

• Information transfer is a physical process

• In this class, we generally care about

– Electrical signals (on a wire or wireless) – Optical signals (in a fiber)

– More broadly, EM waves

• Information carriers can be very diverse:

– Sound waves, quantum states, proteins, ink & paper, etc.

• Quote (usually attributed to Einstein):

– You see, wire telegraph is a kind of a very, very long

cat. You pull his tail in New York and his head is meowing in Los Angeles.

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Modulation

• Changing a signal to convey information

• Ways to modulate a sinusoidal wave

– Amplitude Modulation (AM) – Frequency Modulation (FM) – Phase Modulation (PM)

In music:

Volume Pitch Timing

• In our case, modulate signal to encode a 0 or a 1. (multi-valued signals sometimes)

– Analog is the same – value just changes continuously

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Modulation Examples

Amplitude

001100 1100011100011000 1110

Frequency

Phase

0110110001

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Why Different Modulation Methods?

• Offerschoiceswithdifferenttradeoffs: – Transmitter/Receiver complexity

– Power requirements

– Bandwidth

– Medium (air, copper, fiber, …) – Noise immunity

– Range

– Multiplexing

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Physical Layer: Outline

• Digitalnetworks

• ModulationFundamentals

• CharacterizationofCommunicationChannels

• FundamentalLimitsinDigitalTransmission

• DigitalModulation

• LineCoding

• PropertiesofMediaandDigitalTransmission Systems

• ErrorDetectionandCorrection

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Questions of Interest

• How long will it take to transmit a message?

– How many bits are in the message (text, image)?

– How fast does the network/system transfer information?

• Can a network/system handle a voice (video) call?

– How many bits/second does voice/video require? At what quality?

• How long will it take to transmit a message without errors?

– How are errors introduced?

– How are errors detected and corrected?

• What transmission speed is possible over radio, copper cables, fiber, infrared, …?

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Transmitter

Receiver

A Communications System

Communication channel

Transmitter

• Converts information into a signal suitable for transmission

• Injects energy into communications medium or channel

– Telephone converts voice into electric current

– Wireless LAN card converts bits into electromagnetic waves

Receiver

• Receives energy from medium

• Converts received signal into a form suitable for delivery to user

– Telephone converts current into voice

– Wireless LAN card converts electromagnetic waves into bits

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Digital Binary Signal

101101 +A

0 T 3T 4T 6T -A

Here, Bit Rate = 1 bit / T seconds

For a given communications medium:

• How do we increase the bit rate (speed) ?

• How do we achieve reliable communications?

• Are there limits to speed and reliability?

2T

5T

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Bandwidth

• Bandwidth is width of the frequency range in which the Fourier transform of the signal is non-zero.

• Sometimes referred to as the channel width

• Or, where it is above some threshold value (Usually, the half power threshold, e.g., -3dB)

• dB – short for decibel

– Defined as 10 * log10(P1/P2)

– When used for signal to noise: 10 * log10(S/N)

• Also: dBm – power relative to 1 milliwatt – Defined as 10 * log10(P/1 mW)

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Signal = Sum of Waves

≈

+ 1.3 X + 0.56 X + 1.15 X

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Closer look at waves

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The Frequency Domain

• A (periodic) signal can be viewed as a sum of sine waves of different strengths.

– Corresponds to energy at a certain frequency

• Every signal has an equivalent representation in the frequency domain.

– What frequencies are present

and what is their strength (energy)

• E.g., radio and TV signals, …

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• Spectrum of a signal: measures power of signal as function of frequency

• x1(t) varies faster in time & has more high frequency content than x2(t)

• Bandwidth Ws is defined as range of frequencies where a signal has non-negligible power, e.g. range of band that contains 99% of total signal power

Mini Quiz: Between [A] x1 and

[B] x2, which has more bandwidth?

Spectrum of x1(t)

Spectra & Bandwidth

1.2 1 0.8 0.6 0.4 0.2 0

Spectrum of x2(t)

frequency (kHz)

1.2 1 0.8 0.6 0.4 0.2 0

frequency (kHz)

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0

3

36

39

6

9

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0

3

36

39

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9

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Transmission Channel Considerations

• Every medium supports transmission in a certain frequency range.

– Outsidethisrange,effectssuchas attenuation, .. degrade the signal too much

• Transmission and receive hardware will try to maximize the useful bandwidth in this frequency band.

– Tradeoffsbetweencost,distance,bit rate

• As technology improves, these parameters change, even for the same wire.

Good Bad

Frequency

Signal

Attenuation & Dispersion

• Notnicelowpassfilters • Whydowecare?

Good Bad

+

= ???

Frequency

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Limits to Speed and Distance

• Noise: “random” energy is added to the signal.

• Attenuation: some of the energy in the signal leaks away.

• Dispersion: attenuation and propagation speed are frequency dependent.

(Changes the shape of the signal)

● Effects limit the data rate that a channel can sustain. » But affects different technologies in different ways

● Effects become worse with distance. » Tradeoff between data rate and distance

Pulse Transmission Rate

• Objective: Maximize pulse rate through a channel, that is, make T as small as possible

Channel

t

l If input is a narrow pulse, then typical output is a spread-out pulse with ringing

l Question: How frequently can these pulses be transmitted without interfering with each other?

l 2Wc pulses/sec with binary amplitude encoding where Wc is the bandwidth of the channel

T

t

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Bandwidth of a Channel

X(t) = a cos(2pft) Channel Y(t) = A(f) a cos(2pft)

• If input is sinusoid of frequency f, then

– output is a sinusoid of same frequency f

A(f)

– Output is attenuated by an amount A(f) that depends on f

– A(f)≈1, then input signal passes readily

– A(f)≈0, then input signal is blocked

• Bandwidth Wc is range of frequencies passed by channel

1

0

W f c

Ideal lowpass channel

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Multi-level Pulse Transmission

• Assume channel of bandwidth Wc, and transmit 2Wc pulses/sec (without interference)

• If pulses’ amplitudes are either -A or +A, then each pulse conveys 1 bit, so

Bit Rate = 1 bit/pulse x 2Wc pulses/sec = 2Wc bps

• If amplitudes are from {-A, – A/3, +A/3, +A}, then bit rate is 2x2Wc bps

• By going to M=2m amplitude levels, we achieve

Bit Rate = m bits/pulse x 2Wc pulses/sec = 2mWc bps

In the absence of noise,

the bit rate can be increased without limit by increasing m

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Noise & Reliable Communications

• All physical systems have noise

– Electrons always vibrate at non-zero temperature

– Motion of electrons induces noise

• Presence of noise limits accuracy of measurement of received signal amplitude

• Errors occur if digital signal separation is comparable to noise level

• Thus, noise places a limit on how many amplitude levels can be used in pulse transmission

• Bit Error Rate (BER) increases with decreasing signal-to- noise ratio

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Signal-to-Noise Ratio (SNR)

Signal

Noise

Signal + noise

t

No errors Signal + noise

t

error

High SNR

t

t

Low SNR

Signal

t

SNR =

Noise

t

Average signal power Average noise power

SNR (dB) = 10 log10 SNR

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Physical Layer: Outline

• Digitalnetworks

• ModulationFundamentals

• CharacterizationofCommunicationChannels

• FundamentalLimitsinDigitalTransmission

• DigitalModulation

• LineCoding

• PropertiesofMediaandDigitalTransmission Systems

• ErrorDetectionandCorrection

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The Nyquist Limit

• AnoiselesschannelofwidthHcanatmost transmit a binary signal at a rate 2 x H.

– Assumes binary amplitude encoding

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The Nyquist Limit

• A noiseless channel of width H can at most transmit a binary signal at a rate 2 x H.

– Assumes binary amplitude encoding

– E.g. a 3000 Hz channel can transmit data at a rate of at most 6000 bits/second

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Sample Quiz Question

• [True / False] The bandwidth of Wi-Fi (802.11ac, first-gen) is 80 MHz. So by

Nyquist theorem, it’s max speed is 160 Mbps

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Past the Nyquist Limit

• More aggressive encoding can increase the bandwidth

• Example: modulate multi-valued symbols

– Modulate blocks of “digital signal” bits, e.g, 3 bits = 8 values – Often combine multiple modulation techniques

PSK

PSK+AM

• Problem? Noise!

– The signals representing two symbols are less distinct

– Noise can prevent receiver from decoding them correctly

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Example: Modem Rates

100000 10000 1000 100

1975 1980 1985 1990 1995 2000

15-441 © 2008-10

Year

Lecture 30

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Modem rate

Capacity of a Noisy Channel

• Places upper bound on channel capacity, while considering noise

• Shannon’s theorem:

C = B x log2(1 + S/N)

– C: maximum capacity (bps)

– B: channel bandwidth (Hz)

– S/N: signal to noise ratio of the channel

Often expressed in decibels (db) ::= 10 log(S/N)

• Example:

– Local loop bandwidth: 3200 Hz (old school dialup) – Typical S/N: 1000 (30db)

– What is the upper limit on capacity?

C = 3200 x log2(1 + 1000) = 31.9 Kbps

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Shannon’s Channel Capacity Theorem

C=W log (1+SNR) bps c2

• Arbitrarily-reliable communications is possible if the transmission rate R < C
• If R > C, then arbitrarily-reliable communications is not possible

• “Arbitrarily-reliable” means the BER can be made arbitrarily small through sufficiently complex “coding”

• C can be used as a measure of how close a system design is to the best achievable performance

• Bandwidth Wc & SNR determine C

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Sample Quiz Question

• FindtheShannonchannelcapacityforaWiFi channel with Wc = 80 MHz and SNR = 40 dB

SNR (dB) = 40 dB corresponds to SNR = 10^(40/10) = 10000

C = 80 log2 (1 + 10000) Mbps

= 80 log10 (10001)/log102 = 1063 Mbps

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Physical Layer: Outline

• Digitalnetworks

• ModulationFundamentals

• CharacterizationofCommunicationChannels

• FundamentalLimitsinDigitalTransmission

• DigitalModulation

• LineCoding

• PropertiesofMediaandDigitalTransmission Systems

• ErrorDetectionandCorrection

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From Signals to Packets

Analog Signal

“Digital” Signal

BitStream 00101110001

Packets

Packet Transmission

0100010101011100101010101011101110000001111010101110101010101101011010111001

Header/Body

Sender

Header/Body

Header/Body

Receiver

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Baseband versus Carrier Modulation

• Basebandmodulation:sendthe“bare”digital signal

– Channel must be able to transmit low frequencies – For example, copper media

• Carriermodulation:usethesignalto modulate a higher frequency signal, called a carrier

– Can send the signal in a particular part of the spectrum

– Can modulate the amplitude, frequency or phase

– For example, wireless and optical

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Bandpass Channels

fc–Wc/2 fc fc+Wc/2

• Bandpass channels pass a range of frequencies around

some center frequency fc

– Radio channels, telephone & DSL modems

• Digital modulators embed information into waveform with frequencies passed by bandpass channel

• Sinusoid of frequency fc is centered in middle of bandpass channel

• Modulators embed information into a sinusoid

0

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Amplitude Carrier Modulation

Signal

Carrier Modulated Frequency Carrier

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Amplitude

Amplitude

Signaling rate and Transmission Bandwidth

• Frommodulationtheory:

If

Baseband signal x(t) with bandwidth B Hz

then

Modulated signal x(t)cos(2pfct) has bandwidth 2B Hz

B

f

f

•

• •

• •

If bandpass channel has bandwidth Wc Hz,

Then baseband channel has Wc/2 Hz available, so modulation system supports Wc/2 x 2 = Wc pulses/second That is, Wc pulses/second per Wc Hz = 1 pulse/Hz

Recall baseband transmission system supports 2 pulses/Hz

f

fc-B c fc+B

Frequency Division Multiplexing: Multiple Channels

Determines Bandwidth of Link

Determines Bandwidth of Channel

Different Carrier Frequencies

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Amplitude

Frequency Modulation

Information 1 0 1 1 0 1

+1

0

-1

• Use two frequencies to represent bits – “1”sendfrequencyfc+d

– “0”sendfrequencyfc–d

• Demodulator looks for power around fc + d or fc – d

Frequency Shift Keying

T 2T 3T 4T 5T 6T

t

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Phase Modulation

Information 1 0 1 1 0 1

Phase Shift Keying

+1

-1

0 T 2T 3T 4T 5T 6T

t

• Map bits into phase of sinusoid:

– “1” send A cos(2pft) , i.e. phase is 0 – “0” send A cos(2pft+p) , i.e. phase is p

• Equivalent to multiplying cos(2pft) by +A or -A

– “1” send A cos(2pft) – multiply by 1 – “0” send A cos(2pft+p) = – A cos(2pft) – multiply by -1

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Modulator & Demodulator

Modulate cos(2pfct) by multiplying by Ak for T seconds:

x cos(2pfct)

during kth interval

Demodulate (recover Ak) by multiplying by 2cos(2pfct)

for T seconds and lowpass filtering (smoothing):

Ak

Yi(t) = Ak cos(2pfct) Transmitted signal

Yi(t) = Akcos(2pfct)

Received signal during kth interval

x 2cos(2pfct)

Xi(t)

Lowpass Filter (Smoother)

2Ak cos2(2pfct) = Ak {1 + cos(2p2fct)}

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Example of Phase Modulation

Information

+A

-A

+A

-A

A cos(2pft)

101101

Baseband Signal

0 T

3T 4T 6T

2T

5T

Modulated Signal

x(t)

0 T

3T 4T 6T

-A cos(2pft)

2T

5T

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Example of Phase Demodulation

A {1 + cos(4pft)} +A

0 T

-A {1 + cos(4pft)}

After multiplication at receiver

x(t) cos(2pfct)

Baseband signal discernable after smoothing

Recovered Information

-A +A

-A

3T 4T 6T

2T

5T

0 T

3T 4T 6T

2T

101101

5T

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•

Quadrature Amplitude Modulation (QAM)

QAM uses two-dimensional signaling

– Ak modulates in-phase cos(2pfct)

– Bk modulates quadrature phase sin(2pfct)

– Transmit sum of inphase & quadrature phase components

x

cos(2pfct) + Y(t)

Ak

Bk

Yi(t) = Ak cos(2pfct)

x Yq(t) = Bk sin(2pfct) sin(2pfct)

Transmitted Signal

l l

Yi(t) and Yq(t) both occupy the bandpass channel QAM sends 2 pulses/Hz

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QAM Demodulation

Lowpass filter (smoother)

Y(t)

x 2cos(2pfct)

x 2sin(2pfct)

Ak

2cos2(2pfct)+2Bk cos(2pfct)sin(2pfct)

= Ak {1 + cos(4pfct)}+Bk {0 + sin(4pfct)}

smoothed to zero

Bk

2Bk sin2(2pfct)+2Ak cos(2pfct)sin(2pfct)

= Bk {1 – cos(4pfct)}+Ak {0 + sin(4pfct)} smoothed to zero

Lowpass filter (smoother)

Signal Constellations

• Eachpair(Ak,Bk)definesapointintheplane • Signalconstellationsetofsignalingpoints

Bk

(A, A)

(A,-A)

4 possible points per T sec. 2 bits / pulse

Bk

(-A,A)

(-A,-A)

Ak Ak

16 possible points per T sec. 4 bits / pulse

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Physical Layer: Outline

• Digitalnetworks

• CharacterizationofCommunicationChannels

• FundamentalLimitsinDigitalTransmission

• ModemsandDigitalModulation

• LineCoding(nextlecture)

• PropertiesofMediaandDigitalTransmission Systems

• ErrorDetectionandCorrection

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