encoding and framingweb.eecs.umich.edu/~zmao/eecs489/lectureslides/coding... · 2004-11-04 · -...
TRANSCRIPT
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1Mao F04
Encoding and Framing
EECS 489 Computer Networkshttp://www.eecs.umich.edu/~zmao/eecs489
Z. Morley MaoTuesday Nov 2, 2004
Acknowledgement: Some slides taken from Kurose&Ross and Katz&Stoica
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2Mao F04
Questions
§ Why are some links faster than others?§ What limits the amount of information we can send on a
link?§ How can we increase the capacity of a link?
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3Mao F04
Signals: Analog vs. Digital
§ Signal: a function s(t) that varies with time (t stands for time)§ Analog: varies continuously
- Example: voltage representing audio (analog phone call)
§ Digital: discrete values; varies abruptly- Example: voltage representing 0s an 1s
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4Mao F04
Signals: Periodic vs. Aperiodic
§ Period: repeat over and over again, once per period- Period (T) is the time it takes to make one complete cycle- Frequency (f) is the inverse of period, f = 1/T;measured in hz
§ Aperiodic: don’t repeat according to any particular pattern
T = 1/f
Sig
nal s
tren
gth
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5Mao F04
Data vs. Signal
signaldata data
communication medium
data signal
TelephoneAnalog Analog
ModemDigital Analog
CODECAnalog Digital
Digital Transmitter
Digital Digital
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6Mao F04
Attenuation
Links become slower with distance because of signal attenuationAmplifiers and repeaters can help
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7Mao F04
Noise
§ A signal s(t) sent over a link is generally- Distorted by the physical nature of the medium
• This distortion may be known and reversible at the receiver
- Affected by random physical effects• Fading• Multipath effects
- Also interference from other links• Wireless• Crosstalk
§ Dealing with noise is what communications engineers do
s(t) r(t)
n(t)
S
link
- noise
transmitted signal received signal
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8Mao F04
Noise Limits the Link Rate
§ Suppose there were no noise- Then, if send s(t) always receive s(t+?)- Take a message of N bits say b1b2….bN, and send a
pulse of amplitude of size 0.b1b2….bN
- Can send at an arbitrarily high rate- This is true even if the link distorts the signal but in a
known way
§ In practice the signal always gets distorted in an unpredictable (random) way- Receiver tries to estimate the effects but this lowers
the effective rate
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9Mao F04
Physical Layer Functions
§ Functions1. Encode bit sequence into analog signal2. Transmit bit sequence on a physical medium (Modulation)3. Receive analog signal4. Convert Analog Signal to Bit Sequence
AdaptorAdaptor AdaptorAdaptorSignal
Adaptor: convert bits into physical signal and physical signal back into bits
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10Mao F04
Block Diagram
NRZI
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11Mao F04
Modulation
§ The function of transmitting the encoded signal over a link, often by combining it with another (carrier signal)- E.g., Frequency Modulation (FM)
• Combine the signal with a carrier signal in such a way that the i frequency of the received signal contains the information of the carrier
- E.g. Frequency Hopping (OFDM)• Signal transmitted over multiple frequencies • Sequence of frequencies is pseudo random
1 1 0 0 1 1 1 0 0 11 1 0 0 1 1 1 0 0 1
Bit sequence Modulated signal Received signal Received bit sequence
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12Mao F04
Outline
Ø Relation between bandwidth and link rateØFourier transform- Nyquist’s Theorem- Shannon’s Theorem
§ Encoding§ Framing
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13Mao F04
§ Any periodic signal g(t) with period T (=1/f) can be constructed by summing a (possibly infinite) number of sines and cosines
§ To construct signal g(t) we need to compute the values a0, a1, …, b0, b1, …, and c !
- Compute coefficients using Euler’s formulae§ But it’s an infinite series...§ Often the magnitude of the an’s and bn’s get smaller as the
frequency (n times 2πf ) gets higher.§ Key point: a “reasonable reconstruction” can be often be
made from just the first few terms (harmonics)- Though the more harmonics the better the reconstruction…
Fourier Transform
( ) ( )∑∑∞
=
∞
=
++=11
2cos2sin21
)(n
nn
n nftbnftactg ππ
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14Mao F04
Fourier Transform: Example
sin(2p f t) +
1/3 sin(6p f t) =
g3(t)
Note: f = 1/T
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15Mao F04
Bandwidth & Data Rate
§ Physical media attenuate (reduce) different harmonics at different amounts
§ After a certain point, no harmonics get through.§ Bandwidth: the range of frequencies that can get through
the link§ Example:
- Voice grade telephone line 300Hz – 3300Hz- The bandwidth is 3000Hz
§ Data rate: highest rate at which hardware change signal
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16Mao F04
Outline
Ø Signal study- Fourier transformØNyquist’s Theorem- Shannon’s Theorem
§ Encoding§ Framing
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17Mao F04
§ Establish the connection between data rate and bandwidth (actually the highest frequency) in the absence of noise
- Developed in the context of analog to digital conversion (ACDs)
§ Say how often one needs to sample an analog signal to reproduce it faithfully
§ Suppose signal s(t) has highest frequency fmax
- Assume B = fmax, i.e., lowest frequency is 0
§ Then, if T= 1/(2B) then it is possible to reconstruct s(t) correctly§ Niquist’s Theorem: Data rate (bits/sec) <= 2*B (hz)
Nyquist’s Theorem(aka Nyquist’s Limit)
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18Mao F04
Why Double the Frequency?
§ Assume a sine signal, then- We need two samples in each period to identify sine function- More samples won’t help
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19Mao F04
Nyguist’s Theorem Revisited
§ If signal has V distinct levels, then Data rate <= 2*B*log2V- V distinct values can be used to encode log2(V) bits- Bi-level encoding V = 2 à Data rate <= 2*B- Example of achieving 2*B with bi-level encoding
§ Can you do better than Nyquist’s limit?- Yes, if clocks are synchronized sender and receiver, we only need
one sample per period- This is because the synchronized starting sample counts as one
of the two points
0 V
5 V
1/B
1/(2B)
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20Mao F04
Outline
Ø Signal study- Fourier transform- Nyquist’s TheoremØShannon’s Theorem
§ Encoding§ Framing
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21Mao F04
Shannon Theorem
§ Establish the connection between bandwidth and data rate in the presence of noise
§ Noisy channel- Consider ratio of signal power to noise power.- Consider noise to be super-imposed signal- Decibel (dB) = 10 log10 (S/N)- S/N of 10 = 10 dB- S/N of 100 = 20 dB- S/N of 1000 = 30 dB
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22Mao F04
Shannon Theorem (cont’d)
§ Data rate in the presence of S/N is bounded as follows
Data rate <= B log 2 (1 + S/N)
§ Example: - Voice grade line: S/N = 1000, B=3000, C=30Kbps- Technology has improved S/N and B to yield higher
speeds such as 56Kb/s
§ Higher bandwidth à higher rate; Intuition:- Signal has more space to “hide” from noise- Noise gets “diluted” across frequency space
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23Mao F04
Outline
§ Signal study- Fourier transform- Nyquist’s Theorem- Shannon’s Theorem
Ø Encoding§ Framing
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24Mao F04
Encoding
§ Specify how bits are represented in the analog signal
- This service is provided by the physical layer
§ Challenges: achieve: - Efficiency – ideally, bit rate = clock rate- Robust – avoid de-synchronization between sender
and receiver when there is a large sequence of 1’s or 0’s
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25Mao F04
Assumptions
§ We use two discrete signals, high and low, to encode 0 and 1
§ The transmission is synchronous, i.e., there is a clock used to sample the signal
- In general, the duration of one bit is equal to one or two clock ticks
§ If the amplitude and duration of the signals is large enough, the receiver can do a reasonable job of looking at the distorted signal and estimating what was sent.
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26Mao F04
Non-Return to Zero (NRZ)
§ 1 à high signal; 0 à low signal
0 0 1 0 1 0 1 1 0
NRZ(non-return to zero)
Clock
Disadvantages: when there is a long sequence of 1’s or 0’sSensitive to clock skew, i.e., difficult to do clock recovery Difficult to interpret 0’s and 1’s (baseline wander)
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27Mao F04
Non-Return to Zero Inverted (NRZI)
§ 1 à make transition; 0 à stay at the same level§ Solve previous problems for long sequences of 1’s,
but not for 0’s
0 0 1 0 1 0 1 1 0
Clock
NRZI(non-return to zero
intverted)
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28Mao F04
Manchester
§ 1 à high-to-low transition; 0 à low-to-high transition§ Addresses clock recovery and baseline wander problems§ Disadvantage: needs a clock that is twice as fast as the transmission
rate- Efficiency of 50%
0 0 1 0 1 0 1 1 0
Clock
Manchester
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29Mao F04
4-bit/5-bit (100Mb/s Ethernet)
§ Goal: address inefficiency of Manchester encoding, while avoiding long periods of low signals
§ Solution:- Use 5 bits to encode every sequence of four bits such that no 5 bit code has
more than one leading 0 and two trailing 0’s- Use NRZI to encode the 5 bit codes - Efficiency is 80%
0000 111100001 010010010 101000011 101010100 010100101 010110110 011100111 01111
1000 100101001 100111010 101101011 101111100 110101101 110111110 111001111 11101
4-bit 5-bit 4-bit 5-bit
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30Mao F04
Outline
§ Signal study- Fourier transform- Nyquist’s Theorem- Shannon’s Theorem
§ EncodingØ Framing
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31Mao F04
Framing
§ Specify how blocks of data are transmitted between two nodes connected on the same physical media
- This service is provided by the data link layer
§ Challenges- Decide when a frame starts/ends- If use special delimiters, differentiate between the true frame
delimiters and delimiters appearing in the payload data
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32Mao F04
Byte-Oriented Protocols: Sentinel Approach
§ STX – start of text§ ETX – end of text § Problem: what if ETX appears in the data portion of the
frame?§ Solution
- If ETX appears in the data, introduce a special character DLE (Data Link Escape) before it
- If DLE appears in the text, introduce another DLE character before it
§ Protocol examples- BISYNC, PPP, DDCMP
Text (Data)STX ETX
8 8
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33Mao F04
Byte-Oriented Protocols: Byte Counting Approach
§ Sender: insert the length of the data (in bytes) at the beginning of the frame, i.e., in the frame header
§ Receiver: extract this length and decrement it every time a byte is read. When this counter becomes zero, we are done
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34Mao F04
Bit-Oriented Protocols
§ Both start and end sequence can be the same- E.g., 01111110 in HDLC (High-level Data Link Protocol)
§ Sender: in data portion inserts a 0 after five consecutive 1s§ Receiver: when it sees five 1s makes decision on the next two bits
- If next bit 0 (this is a stuffed bit), remove it- If next bit 1, look at the next bit
• If 0 this is end-of-frame (receiver has seen 01111110) • If 1 this is an error, discard the frame (receiver has seen 01111111)
Text (Data)Start
sequence
8 8End
sequence
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35Mao F04
Clock-Based Framing (SONET)
§ SONET (Synchronous Optical NETwork)§ Developed to transmit data over optical links
- Example: SONET ST-1: 51.84 Mbps- Many streams on one link
§ SONET maintains clock synchronization across several adjacent links to form a path
- This makes the format and scheme very complicated
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36Mao F04
SONET Multiplexing
§ STS-3c has the payloads of three STS-1’s byte-wise interleaved.
§ STS-3 is a SONET link w/o multiplexing§ For STS-N, frame size is always 125 microseconds
- STS-1 frame is 810 bytes- STS-3 frame is 810x3 =2430 bytes
STS-1FH
STS-1FH
STS-1FH
STS-3cFH
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37Mao F04
STS-1 Frame
§ First two bytes of each frame contain a special bit pattern thatallows to determine where the frame starts
§ No bit-stuffing is used§ Receiver looks for the special bit pattern every 810 bytes
- Size of frame = 9x90 = 810 bytes
90 columns
9 ro
ws
Data (payload)overhead
SONET STS-1 Frame
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38Mao F04
Clock-Based Framing (SONET)
§ Details:- Overhead bytes are encoded using NRZ- To avoid long sequences of 0’s or 1’s the payload is
XOR-ed with a special 127-bit pattern with many transitions from 1 to 0
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39Mao F04
What do you need to know?
§ Concept of bandwidth and data rate§ Nyquist’s Theorem § Shannon’s Theorem§ Encoding
- Understand (not memorize) NRZ, NRZI, Manchester, 4/5 bit
§ Framing- Understand framing for bit/byte oriented protocols and
clock based framing