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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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)
16
0
3
36
39
6
9
42
12
15
18
21
24
27
30
33
0
3
36
39
6
9
42
12
15
18
21
24
27
30
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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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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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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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