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.TXT 3 1 122 0 0
Cg a73.000000,73.000000,160
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard {\b DCT (}{\b\i 
Discrete Cosine Transform}{\b )}\par \par Assume that we have an 8 x 8 
block of integers which represent the \par the gray scale values 
(0-255) of an 8 X 8 block of pixels.\par }
.EQN 16 0 124 0 0
{0:N2}NAME:8
.TXT 0 10 140 0 0
Cg a51.000000,51.000000,76
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard where {\b N1,} {
\b N}{\b 2} represent the size of the array and {\b A} is the\par 
array of integers.}
.EQN 3 -10 125 0 0
{0:N1}NAME:8
.EQN 20 0 126 0 0
{0:A}NAME:({8,8}ö100ö100ö9ö9ö9ö9ö9ö9ö100ö100ö9ö9ö9ö9ö9ö9ö100ö100ö9ö9ö9ö9ö9ö9ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100ö100)
.TXT 16 0 127 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard Then the DCT 
transform is given by:}
.EQN 4 0 128 0 0
{0:k1}NAME:0,1;7
.EQN 3 0 129 0 0
{0:k2}NAME:0,1;7
.EQN 9 0 130 0 0
({0:B}NAME)[({0:k1}NAME,{0:k2}NAME):((0,{0:N1}NAME-1,{0:i}NAME,((0,{0:N2}NAME-1,{0:j}NAME,4*({0:A}NAME)[({0:i}NAME,{0:j}NAME)*{0:cos}NAME({0:\p}NAME*({0:k1}NAME)/(2*{0:N1}NAME)*(2*{0:i}NAME+1))*{0:cos}NAME({0:\p}NAME*({0:k2}NAME)/(2*{0:N2}NAME)*(2*
{0:j}NAME+1))){64})){64})
.EQN 27 0 131 0 0
{0:B}NAME={19043}?_n_u_l_l_
.TXT 21 0 132 0 0
Cg a71.000000,71.000000,74
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard We can now 
eliminate any element which is less than an arbitrary value, {\b Q:}}
.EQN 7 0 133 0 0
{0:Q}NAME:100
.EQN 6 0 134 1 0
({0:C}NAME)[({0:k1}NAME,{0:k2}NAME):(({0:floor}NAME(({0:B}NAME)[({0:k1}NAME,{0:k2}NAME))){73}(({0:B}NAME)[(({0:k1}NAME,{0:k2}NAME))>{0:Q}NAME)){71}((0){73}(({0:B}NAME)[({0:k1}NAME,{0:k2}NAME)ó{0:Q}NAME))
.EQN 25 0 135 0 0
{0:C}NAME={19043}?_n_u_l_l_
.TXT 25 0 136 0 0
Cg a71.000000,71.000000,109
{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard Alternatively, 
we can divide each element by an arbitrary quantization factor {\b Q}, 
and take the integer result:}
.EQN 9 0 137 0 0
({0:D}NAME)[({0:k1}NAME,{0:k2}NAME):{0:floor}NAME((({0:B}NAME)[({0:k1}NAME,{0:k2}NAME))/({0:Q}NAME))
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{0:D}NAME={0}?_n_u_l_l_
.TXT 18 1 141 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0
\fcharset0\fnil MS Sans Serif;}}\plain\cf1\fs24 \pard Alternatively we 
can define a {\i quantization table} and divide our matrix by it. The 
following matrix has been defined by the JPEG group:}
.EQN 20 5 148 0 0
{0:Q}NAME:({8,8}ö64ö32ö32ö32ö16ö16ö16ö16ö32ö32ö16ö16ö16ö8ö8ö8ö32ö16ö16ö16ö16ö4ö4ö4ö16ö16ö16ö8ö8ö4ö4ö1ö16ö8ö8ö8ö8ö2ö1ö1ö8ö8ö8ö8ö8ö1ö1ö1ö8ö8ö8ö8ö8ö1ö1ö1ö8ö4ö4ö4ö2ö1ö1ö1)*100
.EQN 23 -1 149 0 0
({0:E}NAME)[({0:k1}NAME,{0:k2}NAME):{0:floor}NAME((({0:B}NAME)[({0:k1}NAME,{0:k2}NAME))/(({0:Q}NAME)[({0:k1}NAME,{0:k2}NAME)))
.EQN 19 1 151 0 0
{0:E}NAME={19043}?_n_u_l_l_