intro lecture notes
TRANSCRIPT
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KEET 3107
Information Theory & CodingSemester 1
Session 2013/2014
Dr Norf izah Md A li
[email protected]@gmail.com
mailto:[email protected]:[email protected] -
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Introduction
Information theory- a discipline centered around common mathematicalapproach, to study the collection and manipulation of information.
-theoretical basis for observation, measurement, data compression, datastorage, communication, estimation, decision making and pattern
recognition.
-guide to the development of information- transmission systems based on astudy of possibilities and limitations inherent in natural law.
-a study of how the laws of probability, and of mathematics in general,
describe limits on the designs of information-transmission systems, but alsooffer opportunities.
-one may design strategies into a communication system to overcome noiseand errors that may occur in the communication channel.
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Introduction(Contd)
Information theory originated by Claude Shannon in his 1948 paper. Basic
theory underlying the task of communication through noisy channel. Heshowed that each channel is characterized by channel capacity such that
an arbitrarily small probability of error is achievable at any transmission
rate below the channel capacity.
Probability of error and rate of data transmission can be specifiedindependently
To achieve small error of probability is the method of coding the
transmitted information in blocks.
Information theory gives the insight into design of information-
transmission systems. By developing a clear concept of information and its
transmission, a much deeper understanding of the purposes and
limitations of a technique is obtained.
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Channel Capacity
C= W log2(1+P/WNo)bits/s
Where
P is the average transmitted power
W is the channel bandwidth
No/2 is the power spectral density of additive noise
OR
C= W log2(1+S/N) bits/s
S/N is the signal-to-noise power ratio
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Information and Sources
Let E be some event which occurs with probability P(E). I(E) = log 1/P(E) units of information
loga x = 1/ (logba) logb x choice of base for log determine the unitfor information
I(E) = log2 1/P(E) bits
I(E) = ln 1/P(E) nats (natural unit)
I(E) = log10 1/P(E) Hartleys
In general if we use a logarithm to the base r,
I(E) = logr 1/P(E) r-ary units
1 Hartley = 3.32 bits
1 nat = 1.44 bits
Take note that, If P(E) = 1/2 , then I(E) = 1 bit
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A source emits a sequence of symbols from a fixed finite source with
alphabet S ={s1, s2., sq} are statistically independent.Such an information source is termed as zero-memory source
The probabilities with which the symbols occur:
P(s1), P(s2),P(sq)
If symbol s1 occurs, the amount of information is:
I(s1) = log 1/P(s1) bits
Source Si,.
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The average amount of information obtained per symbol from
the source is:
(s)I(si) bitsS
This quantity, the average amount of information per source
symbol, is called the entropy H(S) of the zero memory source.
H(S) = (s) log 1/P(si) bit
H(S) is interpreted either as the average amount of information per
symbol provided by the source OR as the average amount of
uncertainty which the observer has before the inspection of the
output of the source.
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Example 1:
Consider the source S = { s1, s2 ,s3} with P(s1)
= , P(s2) = P(s3) = , then
H(S) = log 2 + log 4 + log 4
= 3/2 bits
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For zero memory information source with q-symbol source
alphabet, the maximum value of the entropy is exactly log q, and this
maximum value of the entropy is achieved if, and only if, all thesource symbols are equiprobable.
Binary entropy function is at
maximum value for p=0.5, ie whenboth 1 and 0 are equally likely.
In general, the entropy of a
discrete source is maximum
when the letters from the
source are equally probable.
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Example 2:Symbols of Sn 1 2 3 4 5 6 7 8 9
Sequence of
symbols
s1s1 s1s2 s1s3 s2s1 s2s2 s2s3 s3s1 s3s2 s3s3
Probability P(i) 1/4 1/8 1/8 1/8 1/16 1/16 1/8 1/16 1/16
H(S2) = (i) log 1/P (i) bits
=1/4 log4+ 4 x 1/8 log 8 + 4 x 1/16 log 16= 3 bits per symbol
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The Markov Information
Source A more general type of information source with q symbols in
which the occurrence of a source symbol si may depend on a
finite number m of preceding symbols.
M th-order Markov source
P(si/ sji, sj2,.. sjm) for i=1,2, q;jp= 1,2.,q
For an m th order markov source, the probability of emitting a
given symbol is known if we know the m preceding symbols
the state of the m th order Markov source, ie, qm possible
state.
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Example 3
Consider a second-order Markov source with the binary source
alphabet S = {0,1}. Assume the conditional symbol probabilities:
P(0/00) = P(1/11) =0.8
P(1/00) = P(0/11) = 0.2
P(0/01) = P(0/10) = P(1/01) = P(1/10) = 0.5
Since q is equal to 2, we assumed a second order Markov source,
hence there are four states of the source 00, 01, 10, 11
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State Diagram of a Second Order Markov Source. The possible states are indicated by four
possible dots. The possible state transitions are indicated by arrows from state to state, with
probability of a transition shown .
For example: if we are in state 00 we can go to either state 01 or 00 but not to state 10 or 11.
The probability of remaining in state 00 is 0.8 and the probability of going to state 01 is 0.2.