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Transcript of Computer Networks Group Universität Paderborn Computer Networks Chapter 2: Physical layer Holger...
Computer Networks GroupUniversität Paderborn
Computer Networks Chapter 2: Physical layer
Holger Karl
WS 05/06, v 1.1 Computer Networks - Physical layer 2
Goals of this chapter
Answer the basic question: how can data be transported over a physical medium?
Understand the basic service provided by a physical layer Different ways to put “bits on the wire” Reasons why performance of any physical layer is limited Reasons for errors A few examples of important physical layers
Note: This is vastly simplified material
WS 05/06, v 1.1 Computer Networks - Physical layer 3
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 4
Basic service of physical layer: transport bits
Physical layer should enable the transport of bits between two locations A and B
Abstraction: Bit sequence – correct, in order delivery
Layer 0 ”Physical connection”
Pair of copper wires
Layer 1 “Bit ! voltage conversion”
Layer 1 “voltage ! Bit conversion”
SAP of Layer 1
Bits Bits
SAP of Layer 1
WS 05/06, v 1.1 Computer Networks - Physical layer 5
A bit ! signal conversion rule
A simple conversion rule For a “1” bit, apply voltage to the pair of wires For a “0” bit, no voltage
Layer 0 ”Physical connection”
Layer 1 “Bit ! voltage conversion”
Layer 1 “voltage ! Bit conversion”
Bit=1: Close switchBit=0: Open switch
If voltage: Indicate a “1” bitIf no voltage: Indicate a “0” bit
This is called “Non return to
zero” NRZ
WS 05/06, v 1.1 Computer Networks - Physical layer 6
Example: Transmit bit pattern for character “b”
Character “b” needs a representation as a sequence of bits One option: Use the ASCII code of “b”, 98, as a binary
number 01100010 Resulting
voltage put on the wire:
Note: Abstract data is represented by physical signals – changes of a physical quantity in time or space!
WS 05/06, v 1.1 Computer Networks - Physical layer 7
What arrives at the receiver?
Typical pattern at the receiver:
What is going on here?
Note: this and the following examples are exaggerated!
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WS 05/06, v 1.1 Computer Networks - Physical layer 8
Some background: Fourier analysis
To understand signal propagation on a physical medium, some background is required how such signals can be analyzed/treated mathematically
First: Fourier’s theoremAny periodic function g(t) (with period T) can be written as a (possibly infinite) sum of sine and cosine functions; the frequencies of these functions are integer multiples of the fundamental frequency f = 1/T.
Constants c, an, bn are to be determined.
WS 05/06, v 1.1 Computer Networks - Physical layer 9
Fourier analysis – computing coefficients
Coefficients c, an, bn in the Fourier series can be computed:
Because of orthogonality of sines and cosines as basis functions
The nth summary terms are called harmonics The sum of the squares of the nth coefficients – an
2 + bn2 –
is proportional to the energy contained in this harmonic Usually, root of it is shown - (an
2 + bn2)1/2
WS 05/06, v 1.1 Computer Networks - Physical layer 10
Applying Fourier analysis to example
The transmitted waveform of ‘b’ is not a periodic signal – Fourier not applicable directly
Use a trick: Suppose waveform is repeatedinfinitely often, resulting in a periodic waveform with period 8 bit times
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
Repeated waveform for bit pattern 'b'
Time
Cur
rent
WS 05/06, v 1.1 Computer Networks - Physical layer 11
Applying Fourier analysis to example
Result of computing an, bn, c and using 512 terms to represent the signal:
Almost no discernible difference between original signal and Fourier series
Curves overlap
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rent
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Fact 1: Signals are attenuated in a physical medium
Attenuation : Ratio of transmitted to received power High attenuation ! little power arrives at receiver
Attenuation depends on Actual medium Distance between
sender and receiver … other factors
Normalized, typically given in dB
0 1 2 3 4 5 6 7 80
0.2
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Cur
rent
Time
Received attenuated signal
WS 05/06, v 1.1 Computer Networks - Physical layer 13
Fact 2: Not all frequencies pass through a medium
Previous picture assumed that all frequencies travel unhindered through a physical medium
This is not the case for real media! Simplified behavior: frequencies up to given upper bound fc
can pass; higher frequencies are suppressed Mathematically: the Fourier series is cut off at a certain harmonic High frequencies are attenuated to zero
This frequency fc is called the bandwidth of a physical medium (or channel)
Smaller fc means fewer harmonics can pass through
WS 05/06, v 1.1 Computer Networks - Physical layer 14
Bandwidth-limited medium – example
Result when fewer and fewer harmonics are transported
0 1 2 3 4 5 6 7 8-0.2
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Cur
rent
Fourier series with 128 harmonics
0 1 2 3 4 5 6 7 8-0.2
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1.2Fourier series with 32 harmonics
Cur
rent
Time0 1 2 3 4 5 6 7 8
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1.2
Cur
rent
Time
Fourier series with 8 harmonics
0 1 2 3 4 5 6 7 8-0.4
-0.2
0
0.2
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1
1.2
Cur
rent
Time
Fourier series with 4 harmonics
0 1 2 3 4 5 6 7 8-0.4
-0.2
0
0.2
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1
1.2
Cur
rent
Time
Fourier series with 2 harmonics
0 1 2 3 4 5 6 7 80
0.2
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0.8
1
Cur
rent
Time
Fourier series with 1 harmonic
WS 05/06, v 1.1 Computer Networks - Physical layer 15
Fact 3: Attenuation depends on frequency
Model just used: Cutoff Attenuation is 1 below
bandwidth, infinite above
More realistic: attenuation depends on frequency
Attenuation close to 1 below bandwidth limit and increases for higher frequencies
Both are examples of a bandwidth-limited medium / channel
1
fc
1
fc
WS 05/06, v 1.1 Computer Networks - Physical layer 16
Example with frequency-dependent attenuation
Suppose attenuation is 2, 2.5, 3.333… , 5, 10, 1 for the 1st, 2nd, … harmonic
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rent
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We have to explainthis behavior:
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rent
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Received signal with frequency-dependent attenuation
WS 05/06, v 1.1 Computer Networks - Physical layer 17
Fact 4: Media not only attenuates, but also distorts
In a physical medium other than vacuum, different frequencies have different propagation speed
Some wave lengths travel faster than others
Apparent result: Waves arrive at receiver out of phase Recall: a sine wave is determined by amplitude a, frequency f, and
phase
Amount of phase shift in the medium depends on frequency
This is called distortion
WS 05/06, v 1.1 Computer Networks - Physical layer 18
Example with frequency-dependent attenuation and jitter
Behavior of “real” medium already well matched! What about the “wriggling”?
0 1 2 3 4 5 6 7 8-0.2
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Received signal with frequency-dependent attenuation
and phase change
Time
Cur
rent
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1.2
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rent
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We have to explainthis behavior:
WS 05/06, v 1.1 Computer Networks - Physical layer 19
Fact 5: Real media are noisy
A physical medium, in combination with the receiver, exhibits random (thermal) noise
Fluctuations in the receiver circuitry, interference from nearby transmissions, etc.
Materializes as random fluctuations around the (noise-free) received signal
Typical model: noise as a Gaussian random variable of zero mean, uncorrelated in time
More sophisticated models exist
WS 05/06, v 1.1 Computer Networks - Physical layer 20
Example with frequency-dependent attenuation and distortion, random noise
When taking all five facts into account, the received wave form can be satisfyingly explained:
0 1 2 3 4 5 6 7 8-0.2
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1
1.2C
urre
nt
Time
WS 05/06, v 1.1 Computer Networks - Physical layer 21
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 22
Is it so simple? WHEN is the
middle of a bit? HOW does
receiver know? 0 1 2 3 4 5 6 7 8-0.2
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rent
Converting signals to data: Sampling
Suppose we have a channel with “sufficient” bandwidth available, free of noise
How does a receiver convert the signal back to data? Simple: Look at the signal
If high, bit is a 1 If low, bit is a 0
0 1 1 1 0000
WS 05/06, v 1.1 Computer Networks - Physical layer 23
0 1 2 3 4 5 6 7 8-0.2
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Vo
ltage
Time
Fourier series with 8 harmonics
Sampling over a noisy or bandwidth-limited channel
0 1 1 1 0000
In presence of noise or limited bandwidth (or both), signal will not likely be exactly 0 or 1
Or whatever 0 and 1 amounts to after attenuation
Instead of comparing to these precise values, receiver has to use some thresholds within which a signal is declared as a 0 or a 1
WS 05/06, v 1.1 Computer Networks - Physical layer 24
0 1 2 3 4 5 6 7 8-0.4
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ltage
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Fourier series with 2 harmonics
Sampling & low bandwidth
What happens when little bandwidth is available?
Assuming same thresholds as before
At some sampling points, the signal will be outside the thresholds!
No justifiable decision possible
What are possible ways out?
0 ? 1 ? 0??0
WS 05/06, v 1.1 Computer Networks - Physical layer 25
Possible way out: Make thresholds wider?
Wide thresholds would (apparently) reduce opportunity for confusion
E.g., +/- 0.4
But: what happens in presence of noise?
Wider thresholds lead to higher probability of incorrect decisions!
! Not good!
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ltage
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Fourier series with 2 harmonics
WS 05/06, v 1.1 Computer Networks - Physical layer 26
Way out 2: Increase time for a single bit
If bandwidth is limited, received signal cannot track very steep raises and falls in the signal
Hence: give the signal more time to reach the required level for a 0 or a 1 detection.
This means: Time for a single bit has to be extended! Useable data rate is reduced!
This is a fundamental limitation and cannot be circumvented
Formally: maximum data rate · 2H bits/s
where H is the channel bandwidth Basic reason: need to sample sufficiently often
WS 05/06, v 1.1 Computer Networks - Physical layer 27
Way out 3: Use more than just 0 and 1 in the channel
Who says we can only use 0 and 1 as possible levels for the transmitted signal?
Suppose the transmitter can generate signals (current, voltage, …) at four different levels, instead of just two
Then: to determine one of four levels, two bits are required Distinction:
“Bits” are 0 or 1, used in “higher” layers “Symbols” can have multiple values, are transmitted over the
channel Symbol rate: Rate at which symbols are transmitted
Measured in baud Data rate: Rate at which physical layer processes incoming data
bits Measured in bit/s
WS 05/06, v 1.1 Computer Networks - Physical layer 28
0 1 2 3 40
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1
1.5
2
2.5
3
Time
Sym
bol v
alu
e
Way out 3: Use four-level symbols to encode two bits
Example: Map 00 ! 0, 01 ! 1, 10 ! 2, 11 ! 3 Symbol rate is then only half the data rate as each symbol
encodes two bits
0 1 1 0 0 0 1 0
WS 05/06, v 1.1 Computer Networks - Physical layer 29
Data rate with multi-valued symbols – Nyquist
Using symbols with multiple values, the data rate can be increased
Nyquist formula summarizes:
where V is the number of discrete symbol values
maximum data rate · 2H log2 V bits/s
WS 05/06, v 1.1 Computer Networks - Physical layer 30
Unlimited data rate with many symbol levels?
Nyquist’s theorem appears to indicate that unlimited data rate can be achieved when only enough symbol levels are used
Is this plausible? More and more symbol levels have to be spaced closer
and closer together What then about noise?
Even small random noise would then result in one symbol being misinterpreted for another
So, not unlimited?
WS 05/06, v 1.1 Computer Networks - Physical layer 31
Shannon limit on achievable data rate
Achievable data rate is fundamentally limited by noise More precisely: by the relationship of signal strength S compared
to noise N The relatively fewer noise there is at the receiver, the easier it is
for the receiver to distinguish between different symbol levels Relationship characterized by Shannon, 1948
where S is signal strength, N is noise level
This theorem formed the basis for information theory
maximum data rate · H log2 (1 + S/N) bits/s
WS 05/06, v 1.1 Computer Networks - Physical layer 32
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 33
When to sample the received signal?
How does the receiver know WHEN to check the received signal for its value?
One typical convention: in the middle of each symbol But when does a symbol start?
The length of a symbol is usually known by convention via the symbol rate
The receiver has to be synchronized with the sender at the bit level
The link layer will have to deal with frame synchronization There is also “character” synchronization – omitted here
WS 05/06, v 1.1 Computer Networks - Physical layer 34
Overly simplistic bit synchronization
One simple option: Assume that sender and receiver at some point in time are
synchronized That both have an internal clock that tics at every symbol step
Usually, this does not work Clock drift is major problem – two different clocks never stay in
perfect synchrony
Errors if synchronization is lost:
01 1 1 1 10 0 0
Sender:
Channel
Receiver with a slightly fast clock:
0 1 01 1 1 0 1 0
WS 05/06, v 1.1 Computer Networks - Physical layer 35
Options to tell the receiver when to sample
Relying on permanently synchronized clocks does not work Provide an explicit clock signal
Needs parallel transmission over some additional channel Must be in synch with the actual data, otherwise pointless
! Useful only for short-range communication
Synchronize the receiver at crucial points (e.g., start of a character or of a block)
Otherwise, let the receiver clock run freely Relies on short-term stability of clock generators (do not diverge
too quickly)
Extract clock information from the received signal itself Treated next in more detail
WS 05/06, v 1.1 Computer Networks - Physical layer 36
Extract clock information from signal itself
Put enough information into the data signal itself so that the receiver can know immediately when a bit starts/stops
Would the simple 0 ! low, 1 ! high mapping of bit! symbol work?
It should – after all, receiver can use 0-1-0 transitions to detect the length of a bit
But it fails depending on bit sequences: think of long runs of 1s or 0s – receiver can loose synchronization
Not nice not to be able to transmit arbitrary data
1 0 1 1 0 0 0 1 1 0 1Daten:
NRZ-L
WS 05/06, v 1.1 Computer Networks - Physical layer 37
Extract clock information from signal itself – Manchester
Idea: At each bit, provide indication to receiver that this is where a bit {starts/stops/has its middle}
Example: Manchester encoding For a 0 bit, have the symbol change in the bit middle from low to
high For a 1 bit, have the symbol change in the bit middle from high to
low
Ensures sufficient number of signal transitions Independent of what data is transmitted!
1 0 1 1 0 0 0 1 1 0 1Daten:
Manchester
WS 05/06, v 1.1 Computer Networks - Physical layer 38
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 39
Baseband versus broadband transmission
The transmission schemes described so far: Baseband transmission
Baseband transmission directly puts the digital symbol sequences onto the wire
At different levels of current, voltage, … essentially, direct current (DC) is used for signaling
Baseband transmission suffers from the problems discussed above
Limited bandwidth reshapes the signal at receiver Attenuation and distortion depend on frequency and baseband
transmissions have many different frequencies because of their wide Fourier spectrum
Possible alternative: broadband transmission More correct name: bandpass transmission
WS 05/06, v 1.1 Computer Networks - Physical layer 40
Broadband transmission
Idea: get rid of the wide spectrum needed for DC transmission
Use a sine wave as a carrier for the symbols to be transmitted
Typically, the sine wave has high frequency But only a single frequency!
Pure sine waves has no information, so its shape has to be influenced according to the symbols to be transmitted
The carrier has to be modulated by the symbols (widening the spectrum)
Three parameters that can be influenced Amplitude a Frequency f Phase
WS 05/06, v 1.1 Computer Networks - Physical layer 41
Amplitude modulation
Given a sine wave f(t) and a time-varying signal s(t) Signal can be analog (i.e., a continuous function of time) or digital
(i.e., a discrete function of time) Signal can be e.g. the symbol levels discussed above
The amplitude modulated sine wave fA(t) is given as:
I.e., the amplitude is given by the signal to be transmitted
Receiver can extract s(t) from fA(t)
Special cases: s(t) is an analog signal – amplitude modulation s(t) is a digital signal – also called amplitude keying s(t) only takes 0 and 1 (or 0 and a) as values – on/off keying
WS 05/06, v 1.1 Computer Networks - Physical layer 42
Amplitude modulation – example
Question: How to solve bit synchronization here? Is Manchester applicable?
Binarydata
On/off modulated
carrier
WS 05/06, v 1.1 Computer Networks - Physical layer 43
Frequency modulation
The frequency-modulated sine wave fF(t) is given by
Modulation/keying terminology like for AM Example
Binarydata
Frequency-modulated
carrier
Note: s(t) has an additive constant in this example to avoid having frequency zero
WS 05/06, v 1.1 Computer Networks - Physical layer 44
Phase modulation
Similarly, a phase modulated carrier is given by
Modulation/keying terminology again similar Example:
s(t) is chosen such that there are phase changes when the binary data changes
Typical example for differential coding
Binarydata
Phase-modulated
carrier
WS 05/06, v 1.1 Computer Networks - Physical layer 45
Phase modulation with high multiple values per symbol
A receiver can usually quite well distinguish phase shifts Hence: Use phase shifts of 0, /2, , 3/2 to encode two
bits per symbol Even better: Use /4, 3/4, 5/4, 7/4 phase shifts for each bit Why better? Clock extraction! Result: Data rate is twice the symbol rate
Technique is called Quadrature Phase Shift Keying (QPSK)
Visualization as constellation diagram:
WS 05/06, v 1.1 Computer Networks - Physical layer 46
Combinations of different modulations
Amplitude, frequency, and phase modulations can be fruitfully combined
Example: 16-QAM (Quadrature Amplitude Modulation) Use 16 different combinations of phase change and amplitude for
each symbol Per symbol, 24 = 16 bits are encoded and transmitted in one step Constellation diagram:
WS 05/06, v 1.1 Computer Networks - Physical layer 47
Bit error rate as function of SNR
The higher the SNR, the better the reception The more reliably can be signals converted to bits at receiver Actually: Energy per bit – takes into account data rate,
#bits/symbol
Concrete bit error probability/rate (BER) depends on SNR and used modulation
Example: DPSK, data rate corresponds to bandwidth
WS 05/06, v 1.1 Computer Networks - Physical layer 48
Examples for SINR ! BER mappings
1e-07
1e-06
1e-05
0.0001
0.001
0.01
0.1
1
-10 -5 0 5 10 15
Coherently Detected BPSKCoherently Detected BFSK
BER
SNR
WS 05/06, v 1.1 Computer Networks - Physical layer 49
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 50
Digital vs. analogs signals
A sender has two principal options what types of signals to generate1. It can choose from a finite set of different signals – digital transmission
2. There is an infinite set of possible signals – analog transmission
Simplest example: Signal corresponds to current/voltage level on the wire In the digital case, there are finitely many voltage levels to choose from In the analog case, any voltage is legal
More complicated example: finite/infinitely many sinus functions In both cases, the resulting wave forms in the medium can well be
continuous functions of time!
Advantage of digital signals: There is a principal chance that the receiver can precisely reconstruct the transmitted signal
WS 05/06, v 1.1 Computer Networks - Physical layer 51
Structure of digital communication systems
How to put these functions together into a working digital communication system?
How to structure transmitter and receiver? How to bridge from a data source to a data sink? Essential functions for baseband transmission
Data source
FormatSourceencode
Channelencode
Physical
transmit
Physical medium
Data sink
FormatSourcedecode
Channeldecode
Physical
receive
Channelsymbols
Sourcebits
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Functions
Format: Bring source information in digital form E.g., sample and quantize an analog voice signal, represent text
as ASCII
Source encode: Remove redundant or irrelevant data E.g., lossy compression (MP3, MPEG 4); lossless compression
(Huffmann coding, runlength coding)
Channel encode: Map source bits to channel symbols Potentially several bits per symbol May add redundancy bits to protect against errors Tailored to channel characteristics
Physical transmit: Turn the channel symbols into physical signals
At receiver: Reverse all these steps
WS 05/06, v 1.1 Computer Networks - Physical layer 53
Structure of a (digital) broadband system
Previous example assumed a simple physical transmission in baseband
Using broadband transmission adds complexity to signal generation
Data source
FormatSourceencode
Channelencode
Physical
transmit
Physical medium
Data sink
FormatSourcedecode
Channeldecode
Physical
receive
Channelsymbols
Sourcebits
Modu-late
Demo-dulate
Discrete set of analog waveforms
WS 05/06, v 1.1 Computer Networks - Physical layer 54
Tricky part: Receiver!
Difficult: How to decide, given an incoming, noisy version of a channel symbol (=a waveform) what the originally sent symbol/waveform was?
Receiver (channel decoder) knows, for each channel symbol
All legal waveforms s1(t), …, sm(t)
The actual, incoming, distorted waveform r(t) = si(t) + n(t) Where n(t) is noise, i is unknown index of transmitted channel symbol
How to determine i?
WS 05/06, v 1.1 Computer Networks - Physical layer 55
Coherent receiver
Coherent receiver: Receiver has perfect time synchronization with transmitter, perfect phase
Not true in practice, a simplification
Conceptually: Receiver compares r(t) with all si(t), computes distance measure
T is length of a channel symbol
Result is i that minimizes this distance measure This waveform is assumed to be the one that the transmitter has
sent
WS 05/06, v 1.1 Computer Networks - Physical layer 56
Receiver as correlator/matched filter
Alternative: Compute the correlation of received signal with legal waveforms
Result is i that maximizes the correlation with the actually received channel symbol
Practical implementations have many additional constraints
WS 05/06, v 1.1 Computer Networks - Physical layer 57
Overview
Baseband transmission over physical channels Limitations on data rate: Nyquist and Shannon Clock extraction Broadband versus baseband transmission Structure of digital communication systems Examples
WS 05/06, v 1.1 Computer Networks - Physical layer 58
Example physical layers
Guided transmission media Copper wire – twisted pair Copper wire – coaxial cable Fiber optics
Wireless transmission Radio transmission Microwave transmission Infrared Lightwave
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Electromagnetic spectrum
Hz
103 105 107 109 1011 1013 1015
leitungsgebundene Übertragungstechniken
verdrillte DrähteKoaxialkabel Hohlleiter optischeFasern
sichtbaresLicht
InfrarotMikrowellen
Fernsehen
KurzwelleMittelwellen
-Radio
Langwellen-Radio
nicht-leitungsgebundene Übertragungstechniken
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Conclusion
The physical layer is responsible for turning a logical sequence of bits into a physical signal that can propagate through space
Many different forms of physical signals are possible Signals are limited by their propagation in a physical
medium (limited bandwidth, attenuation, dispersion) and by noise
Bits can be combined into multi-valued symbols for transmission
Gives rise to the difference in data rate and baud rate
Baseband transmission is fraught with problems, partially overcome by modulating a signal onto a carrier (broadband transmission)