ee122: encoding & framing
DESCRIPTION
EE122: Encoding & Framing. November 17, 2003. EECS 122: Introduction to Computer Networks Encoding and Framing. Computer Science Division Department of Electrical Engineering and Computer Sciences University of California, Berkeley Berkeley, CA 94720-1776. Today’s Lecture: 23. 2. 17,18. - PowerPoint PPT PresentationTRANSCRIPT
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Katz, Stoica F04
EE122: Encoding & FramingNovember 17, 2003
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Katz, Stoica F04
EECS 122: Introduction to Computer Networks
Encoding and Framing
Computer Science DivisionDepartment of Electrical Engineering and Computer Sciences
University of California, BerkeleyBerkeley, CA 94720-1776
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Today’s Lecture: 23
Network (IP)
Application
Transport
Link
Physical
2
7, 8, 9
10,11
14, 15, 16
21, 22
23
6
17,18
19, 20
24
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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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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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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
treng
th
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Data vs. Signal
signaldata datacommunication medium
data signal
TelephoneAnalog Analog
ModemDigital Analog
CODECAnalog Digital
Digital Transmitter
Digital Digital
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Attenuation
Links become slower with distance because of signal attenuation
Amplifiers and repeaters can help
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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)
Σ
link
- noise
transmitted signal received signal
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Noise Limits the Link Rate
Suppose there were no noise- Then, if send s(t) always receive s(t+Δ)- 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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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
Adaptor AdaptorSignal
Adaptor: convert bits into physical signal and physical signal back into bits
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Block Diagram
NRZI
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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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Outline
Relation between bandwidth and link rateFourier transform- Nyquist’s Theorem- Shannon’s Theorem
Encoding Framing
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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 2f ) gets higher. Key point: a “reasonable reconstruction” can be often be
made from just the first few terms (harmonics)- Tough the more harmonics the better the reconstruction…
Fourier Transform
11
2cos2sin21)(
nn
nn nftbnftactg
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Fourier Transform: Example
sin(2πf t) +
1/3 sin(6πf t) =
g3(t)
Note: f = 1/T
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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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Outline
Signal study- Fourier transformNyquist’s Theorem- Shannon’s Theorem
Encoding Framing
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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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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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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 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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Outline
Signal study- Fourier transform- Nyquist’s TheoremShannon’s Theorem
Encoding Framing
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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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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 higher rate; Intuition:
- Signal has more space to “hide” from noise- Noise gets “diluted” across frequency space
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Outline
Signal study- Fourier transform- Nyquist’s Theorem- Shannon’s Theorem
Encoding Framing
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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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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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Non-Return to Zero (NRZ) 1 high signal; 0 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’s- Sensitive to clock skew, i.e., difficult to do clock recovery - Difficult to interpret 0’s and 1’s (baseline wander)
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Non-Return to Zero Inverted (NRZI)
1 make transition; 0 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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Manchester
1 high-to-low transition; 0 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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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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Outline
Signal study- Fourier transform- Nyquist’s Theorem- Shannon’s Theorem
Encoding Framing
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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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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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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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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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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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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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STS-1 Frame
First two bytes of each frame contain a special bit pattern that allows 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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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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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