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Only Size needs to be Huffman coded in the above scheme, since, within a given Size, all the input values havesufficiently similar probabilities for there to be little gain from entropy coding the Additional Bits (hence they are codedin simple binary as listed). Each coded Size is followed by the appropriate number of Additional Bits (equal to Size) todefine the sign and magnitude of the coefficient difference exactly.

There are only 12 Sizes to be Huffman coded, so specifying the code table can be very simple and require relatively few bitsin the header.

In JPEG all Huffman code tables are defined in the image header. Each table requires 16 n bytes, where n is the number of codewords in the table.

The first 16 bytes list the number of codewords of each length from 1 to 16 bits (codewords longer than 16 bits areforbidden). The remaining n bytes list the decoded output values of the n codewords in ascending codeword order ( n 256 ).

Hence 16 12 28 bytes are needed to specify the code table for DC coefficients.

The jpeg run-amplitude code

The remaining 63 coefs (the AC coefs) of each 64-element vector usually contain many zeros and so are coded with acombined run-amplitude Huffman code.

The codeword represents the run-length of zeros before a non-zero coef and the Size of that coef. This is then followed by the Additional Bits whichdefine the coef amplitude and sign precisely. Size and Additional Bits are defined just as for DC coefs.

This 2-dimensional Huffman code (Run, Size) is efficient because there is a strong correlation between the Size of acoef and the expected Run of zeros which precedes it - small coefs usually follow long runs; larger coefs tend to followshorter runs. No single 2-D event is so probable that the Huffman code becomes inefficient.

In order to keep the code table size n below 256, only the following Run and Size values are coded: Run 0 15 Size 1 10 These require 160 codes. Two extra codes, corresponding to (Run,Size) = (0,0) and (15,0) are used for EOB (End-of-block)and ZRL (Zero run length).

EOB is transmitted after the last non-zero coef in a 64-vector. It is the most efficient way of coding the finalrun of zeros. It is omitted in the rare case that the final element of the vector is non-zero.

ZRL is transmitted whenever Run 15 , and represents a run of 16 zeros (15 zeros and a zero amplitude coef) which can be part of a longer run of anylength. Hence a run of 20 zeros followed by -5 would be coded as

(ZRL) (4,3) 010

When the code tables are defined in the image header, each codeword is assigned to a given (Run,Size) pair by making thedecoded output byte Code Byte equal to ( 16 Run Size ).

The default JPEG code for (Run,Size) of AC luminance DCT coefficients is summarised below in order of decreasing codeprobability:

(Run,Size) Code Byte (hex) Code Word (binary) (Run,Size) Code Byte (hex) Code Word (binary)
(0,1) 01 00 (0,6) 06 1111000
(0,2) 02 01 (1,3) 13 1111001
(0,3) 03 100 (5,1) 51 1111010
(EOB) 00 1010 (6,1) 61 1111011
(0,4) 04 1011 (0,7) 07 11111000
(1,1) 11 1100 (2,2) 22 11111001
(0,5) 05 11010 (7,1) 71 11111010
(1,2) 12 11011 (1,4) 14 111110110
(2,1) 21 11100
(3,1) 31 111010 (ZRL) F0 11111111001
(4,1) 41 111011

As an example, let us code the following 8 8 block: -13 -3 2 0 0 0 1 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Concerting this to (DC Size) or (Run,Size) and values for the Additional Bits gives:

(4) -13 (0,2) -3 (0,3) 6 (2,2) 2 (3,1) -1 (ZRL) (1,1) 1 (EOB) 101 0010 01 00 100 110 11111001 10 111010 0 11111111001 1100 1 1010

The compressed bitstream for this block is listed on the lower line, assuming that the default Huffman code tables, givenabove, are used.

shows the histogram of probabilities for the (Run,Size) codewords used to code Lennausing the Q lum quantisation matrix. The bin number represents the decoded byte value.

shows the equivalent histogram when the quantisation matrix is 2 Q lum .

Histogram of the (Run,Size) codewords for the DCT of Lenna, quantised using Q lum .
Histogram of the (Run,Size) codewords for the DCT of Lenna, quantised using 2 Q lum .

Note the strong similarity between these histograms, despite the fact that represents only 2 3 as many events. Only the EOB probability changes significantly, because its probability goes up as the number of events (non-zero coefs) per block goes down.

It turns out that the (Run,Size) histogram remains relatively constant over a wide range of image material and acrossdifferent regions of each image. This is because of the strong correlation between the run lengths and expected coefsizes. The number of events per block varies considerably depending on the local activity in the image, but theprobability distribution of those events (except for EOB) changes much less.

and also give the mean bit rates to code Lenna for the two quantisation matrices. Comparingthese with the theoretical entropies from this figure (lower row) we get:

Q matrix Mean Entropy (b/pel) JPEG Bit Rate (b/pel) JPEG efficiency
Q lum 0.8595 0.8709 98.7%
2 Q lum 0.5551 0.5595 99.21%

Hence we see the high efficiency of the (Run,Size) code at two quite different compression factors. This tends to apply overa wide range of images and compression factors and is an impressive achievement.

There is even very Little efficiency lost if a single code table is used for many images, which can avoid the need totransmit the 16 n (168 bytes) of code definition in the header of each image. Using the recommended JPEG default luminance tables(Annex K.3.3) the above efficiencies drop to 97.35% and 95.74% respectively.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Image coding. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10206/1.3
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