# 5.2 Representing symbols by bits  (Page 2/2)

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Let us assume that we represent a symbol ${x}_{n}$ , with probability ${p}_{n}$ , by ${l}_{n}$ bits. Then, the average number of bits spent per symbol will be

$\langle L\rangle =\sum_{n=1}^{N} {p}_{n}{l}_{n}$
We see that this equation is equal to the entropy if the code words are selected to have the lengths ${l}_{n}=-\lg {p}_{n}$ . Thus, if the source produces stochastically independent outcomes with probabilities ${p}_{n}$ , such that $\lg {p}_{n}$ is an integer, then we can easily find an optimal code as we show in the next example.

## Finding a minimal representation

A four-symbol alphabet produces stochastically independent outcomes with the following probabilities. $({x}_{1})=\frac{1}{2}$ $({x}_{2})=\frac{1}{4}$ $({x}_{3})=\frac{1}{8}$ $({x}_{4})=\frac{1}{8}$ and an entropy of 1.75 bits/symbol. Let's see if we can find a codebook for this four-letter alphabet that satisfies the Source CodingTheorem. The simplest code to try is known as the simple binary code : convert the symbol's index into a binary number and use the same number of bits for each symbol byincluding leading zeros where necessary.

$↔({x}_{1}, \mathrm{00})\text{}↔({x}_{2}, \mathrm{01})\text{}↔({x}_{3}, \mathrm{10})\text{}↔({x}_{4}, \mathrm{11})$
As all symbols are represented by 2 bits, obviously the average number of bits per symbol is 2.Because the entropy equals $1.75$ bits, the simple binary code is not a minimal representation according to the source coding theorem.If we chose a codebook with differing number of bits for the symbols, a smaller average number of bits can indeed be obtained. The idea is to use shorter bit sequences for the symbols that occur more often , i.e., symbols that have a higher probability. One codebook like this is
$↔({x}_{1}, 0)\text{}↔({x}_{2}, \mathrm{10})\text{}↔({x}_{3}, \mathrm{110})\text{}↔({x}_{4}, \mathrm{111})$
Now $\langle L()\rangle =1\frac{1}{2}+2\frac{1}{4}+3\frac{1}{8}+3\frac{1}{8}=1.75$ . We can reach the entropy limit! This should come as no surprise, as promised above, when $\lg {p}_{n}$ is an integer for all $n$ , the optimal code is easily found.

The simple binary code is, in this case, less efficient than theunequal-length code. Using the efficient code, we can transmit the symbolic-valued signal having this alphabet 12.5%faster. Furthermore, we know that no more efficient codebook can be found because of Shannon's source coding theorem.

## Optimality of the ascii code

Let us return to the ASCII codes presented in . Is the 7-bit ASCII code optimal, i.e., is it a minimal representation? The 7-bit ASCII code assign an equal length (7-bit) to all characters it represents. Thus, it would be optimal if all of the 128 characters wereequiprobable, that is each character should have a probability of $\frac{1}{128}$ . To find out whether the characters really are equiprobable an analysis of all English texts would be needed. Such an analysis is difficult to do. However, the letter "E" is more probable than the letter "Z", so the equiprobable assumption does not hold, and the ASCII codeis not optimal.

(A technical note: We should take into account that in English text subsequent outcomes are not stochastically independent. To see this, assume the first letter to be "b", then it is more probable that the next letter is "e", than "z". In the case where the outcomesare not stochastically independent, the formulation we have given of Shannon's source coding theorem is no longer valid, to fix this, we should replace the entropy with the entropy rate, but we will not pursue this here).

## Generating efficient codes

From Shannon's source coding theorem we know what the minimum average rate needed to represent a source is. But other than in the case when the logarithm of the probabilities gives an integer, we do not get any indications on how to obtain that rate. It is a large area of research to getclose to the Shannon entropy bound. One clever way to do encoding is the Huffman coding scheme.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
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Asali
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Samantha
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Asali
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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