ELE 488 F06 ELE 488 Fall 2006 Image Processing and Transmission ( ) Image Compression Review of Basics Huffman coding run length coding Quantization independent samples uniform and optimum correlated samples vector quantization 10/19/06

ELE 488 F06 Why Compression? Storage and transmission –Large size of uncompressed multimedia data –High bandwidth to send uncompressed video in real time. Slow access to storage devices does not allow real time play back of uncompressed content –1 x CD-ROM transfer rate ~ 150 kB/s –320 x 240 x 24 fps video ~ 5.5MB/s => 36 seconds to transfer 1-sec uncompressed video from CD UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Example: Storing An Encyclopedia –500,000 pages of text (2kB/page) ~ 1GB => 2:1 3,000 color pictures (640  480  24bits) ~ 3GB => 15:1 –500 maps (640  480  16bits=0.6MB/map) ~ 0.3GB => 10:1 –60 minutes of stereo sound (176kB/s) ~ 0.6GB => 6:1 –30 animations with average 2 minutes long (640  320  16bits  16frames/s=6.5MB/s) ~ 23.4GB => 50:1 –50 digitized movies with average 1 minute long (640  480  24bits  30frames/s = 27.6MB/s) ~ 82.8GB => 50:1 Require a total of 111.1GB storage capacity without compression Reduce to 2.96GB with compression From Ken Lam’s DCT talk 2001 (HK Polytech) UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Sampling and Quantization Signal is sampled Each sample quantized to B bits (uniform steps)  B bit per sample How many samples? –Nyquist –In practice? B = ? –Signal to noise ratio: 6 dB per bit –Meaning in practice? CD music: 44100x2x16 = 1.41 Mb/sec – raw data rate Observations: –Not all quantization levels occur equally likely Should B bits be used for the frequently occurred samples as well as less likely occurred samples? –Why uniform quantization? sampler + quantizer encoder image capture

ELE 488 F06 Variable Length Codes Quantized PCM values may not occur with equal probability –Is it best to use same # of bits for each sample? Example: 4 levels with probabilities: P(level 1)=0.125, P(level 2)=0.5, P(level 3)=0.25, P(level 4)=0.125 –Need 2 bits to code each symbol IF using same # bits –If use less bits for levels with high probability and more bits for low probability (Variable Length Codes – VLC) level 2 => [0], level 3 => [10], level 1 => [110], and level 4 => [111 ] Average code length is  i p i l i =1.75 bits < 2 bits Compression ration = 2/1.75 UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Entropy Coding Use fewer bits for commonly occurred values Minimum # of bits for each level determined by “entropy of source” –Entropy is a measure of information of a source Definition: H =  i p i log 2 (1 / p i ) bits (entropy of previous example is 1.75) Can represent a source perfectly with avg. H+  bits per sample (Shannon Lossless Coding Theorem) –“Compressability” depends on the sources Need to design a codebook to decode the encoded stream efficiently and without ambiguity Huffman Coding – assigns ~ log 2 (1/p i ) bits for the i th symbol UMCP ENEE631 Slides (created by M.Wu © 2001/2004)

ELE 488 F06 Huffman Coding S0 S1 S2 S3 S4 S5 S6 S PCM Huffman (trace from root) UMCP ENEE631 Slides (created by M.Wu © 2001) Step 1 – Arrange pi in decreasing order Step 2 – Merge two nodes with smallest prob to a new node and sum up prob. Arbitrarily assign 1 and 0 to each pair of merging branch Step-3 – Repeat until only one node left Read out codeword sequentially from root to leaf

ELE 488 F06 Huffman Coding: Pros & Cons Pro –Simple to implement (table lookup) –Gives best coding efficiency Con –Need source statistics –Integer code length for each symbol  average code length usually larger than min (entropy) Improvement –Code groups of symbols –Arithmetic coding –Lempel-Ziv coding or LZW algorithm “universal”, no need to estimate source statistics fix-length codeword for variable-length source symbols UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Run-Length Coding How to efficiently encode a string of binary symbols (2 levels)? “ w w w w w w w b b w w w b w b w w w w w w b b b....” Run-length coding (RLC) –Code lengths of runs of w and of b good if often getting frequent large runs of “0” and sparse “1” –For the example (7w) (1b) (3w) (1b) (1w) (1b) (6w) (3b)... –Assign fixed-length codeword to run-lengths in a range (e.g. <7) –Or use variable-length code (Huffman) to obtain further improvement RLC also applicable to general a data sequence with many consecutive “w” or “b” UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Facsimile

ELE 488 F06 Probability Model of Run-Length Coding Assume w occurs independently with probability p Calculate prob of runs of length L. Restrict L to L=0,1, …, M P( L = l ) = p L (1 – p) for 0  l  M –1 (geometric distribution) P( L = M ) = p M (when having M or more “0”) Avg. # binary symbols per zero-run –S avg =  L (L+1) p L (1-p) + M p M = (1 – p M ) / ( 1 – p ) Compression ratio C = S avg / log 2 (M+1) = (1 – p M ) / [( 1–p ) log 2 (M+1)] Example –p = 0.9, M=15  S avg = 7.94, C = 1.985, H = bpp –Avg. run-length coding rate B avg = 4 bit / 7.94 ~ bpp –Coding efficiency = H / B avg ~ 91% UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Sampling and Quantization Signal is sampled Each sample quantized to B bits (uniform steps)  B bit per sample How many samples? –Nyquist –In practice? B = ? –Signal to noise ratio: 6 dB per bit –Meaning in practice? CD music: 44100x2x16 = 1.41 Mb/sec – raw data rate Observations: –Not all quantization levels occur equally likely Should B bits be used for the frequently occurred samples as well as less likely occurred samples? –Why uniform quantization? sampler + quantizer encoder image capture

ELE 488 F06 Quantization Quantizer x ↔ u correction

ELE 488 F06 Quantization Quantizer …

ELE 488 F06 Mean Square Error Mean square error: Optimization problem: Solution gives optimum thresholds (t k ) and reconstruction values (r k )

ELE 488 F06 Solution r k conditional mean (centroid)

ELE 488 F06 The Lloyd-Max Quantizer Optimal mean square error quantizer: Nonlinear equations. Closed form solution only for simple p(x). Resort to numerical solution in general.

ELE 488 F06 Numerical Solution of the LM Equations Iterative algorithm: a. Pick initial values for t (e.g. uniform grid) b. Find r values using (1). ( numerical integration) c. Find new t values using (2) d. Go to line b until the t and r values converge OR: Start with a=t1 and r1. Determine t2 using (1). Then determine r2 using (2) …. If tL>b, repeat with smaller r1. If tL < b, repeat with larger r1.

ELE 488 F06 Observations on the optimal quanitizer

ELE 488 F06 Example: Uniform Density Uniform quantizer is the optimal mean square quantizer

ELE 488 F06 Uniform Density, cont. Uniform density of interval:

ELE 488 F06 Example: Triangle Start with uniform quantizer Use iterative algorithm. optimum thresholds (red) reconstruction values (blue)

ELE 488 F06 Example: Gaussian Start with uniform quantizer Use iterative algorithm. optimum thresholds (red) and reconstruction values (blue)

ELE 488 F06 Example: Triangle 2 Start with uniform quantizer Use iterative algorithm. optimum thresholds (red) and reconstruction values (blue)

ELE 488 F06 Example: Gaussian Mixture Start with uniform quantizer Use iterative algorithm. optimum thresholds (red) and reconstruction values (blue)

ELE 488 F06 (Jain’s Fig.4.17) Piecewise Constant Approximation Approximation of Optimum Quantizer for Large L

ELE 488 F06 Implement Nonuniform Quantizer Using Companding Compressor-Expander –Uniform quantizer proceeded and followed by nonlinear operations Why use compandor? –Smaller values suffer more distortion under uniform quantizer –Nonlinearity in perceived luminance small difference in low luminance is more visible –Overall companding system may approximate Lloyd-Max quantizer f(.) compressor g(.) expander Uniform quantizer u w yu’ UMCP ENEE631 Slides (created by M.Wu © 2001)

ELE 488 F06 Compandor Nonlinear transformation –To make overall system approximate Lloyd-Max quantizer –Without iterative process to determine parameters (Jain’s Fig.4.19) UMCP ENEE631 Slides (created by M.Wu © 2001)