Chuong 4_Nen Anh

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    Xl nh s

    M ha v nnnh s

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    32 Ni dungTrong phn ny chng ta stm hiu vcc

    kthutc p dng nn nh:Mcch ca nnnh

    Cc kiu dtha dliu

    Cc bc trong m hnh nn v gii nnCc thut ton m ha v gii m

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    Mcch ca nnnh

    nh skhngc nn scn mt lng ln khngg an u r v ng ng ruy n

    V d nh mu kch thc 640 x 480 cn khng.

    Mc ch ca vic nn nh l gim slng dliu cnbiu dinnh. Do sgim khng giancnlu trv tngc tc ng truyn

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    Cc cch tip cn

    Phng php nn bo ton thng tin

    Thng tin c bo ton Tlnn thp

    Phng php nn lm mt thng tin Thn tin khn c bo ton

    Tlnn cao

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    Dliu vs. thng tin

    Dliu (data)l phng tin c dng

    truy n t ithng tin

    Nn d liu l phng php gim sl n d li u cn biu din 1 l nthng tin cho trc vi tiu ch gi li c cn nhiu thn tin cn tt

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    Tlnn

    compression

    Tlnn:

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    Dtha dliu

    a u ng :

    V d :

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    Cc kiu dtha dliu

    (1) Dtha m ha

    (2) Dtha khng gian & thi gian

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    M hnh thng tin

    Qu trnh thng tinc to ra c coi l mt qu

    Mt s kin ngu nhin E vi xc sut P(E) s cl n tin co bn :

    Entropy ca ngun tin: l lng tin trung bnh trn 1 ktca ngun tin (source symbol). Gisngun tin c

    1, 2,, J

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    Lng tin trn mi pixel

    Githit cc gi trmc xmc to ra bi mt bin, k

    bng:

    Lng tin trung bnh:

    n vthng tin

    units/pixelEntro

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    Dtha m ha

    Code: danh sch cc k hiu (symbol) dngxy, , ,

    Code word: Mt chui cc k hiuc sdngbiu din 1 thn tin ho c s ki n v d : thn tin vmc xm).

    Code word length: Scc k hiu trong mi tm

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    Dtha m ha (Cont)

    V d: cho bcnh c cc thng snhsaun c c x

    rk: mc xm thk-thp(rk): Xc sut xut hin mc xm rk

    k k

    Chiu di trung bnh ca cc tm:

    Sbit cnm ha bcnh:

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    Dtha m ha (Cont)

    Trng hp l(rk) = constant

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    Dtha m ha

    Trng hp l(rk) = variable length

    dtha m ha:

    Tlnn:

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    Dtha m ha

    dtha m ha:

    Trong:

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    Dtha khng gian

    a s ng n su xu n c ckhi pixel:

    image

    Tnh lng tin trung bnh

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    M hnh m hanh

    Mapper:Bininhu vo theo cch c thlmgim dtha thng tin gia cc pixel

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    M hnh m hanh

    Quantizer: Lng t ha l bc tng hoc gimcc gi tr u ra ca bc mapper theo cc mc

    mong mun

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    M hnh m hanh

    Symbol encoder:M ha k tl bc gn m cdi ngn nht cho k tc tn sut xy ra nhiu

    .

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    M hnh gii mnh

    ,

    c thc hin theo thtngc li.

    Ring bc gii lng tkhng thkhi

    lng t

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    Tiu chnh gi cht lng

    Trong qu trnh nnnh, ngoi vic quan tm ti tl nn cht l n nh sau khi ii nn cn hic xem xt.

    nh gi cht lng nh u ra bng cch so

    Tiu chnh gi: Ch uan:da vo kin n i uan st Khch quan:da vo cng cton hctnh

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    nh gi chquan

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    nh gi khch quan

    Mean-square signal-to-noise ratio(SNR)

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    Nn khng mt dliu

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    Huffman coding

    Backward Pass

    n c c u v c o c c n n

    theo chiu ngc li

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    Huffman coding

    avg us ng u man co ng:

    Lavg assuming binary codes:

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    Huffman decoding

    m

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    V d

    Cho cc k hiu c xc sut nhsau:

    K hiu Xc sut

    B 0,4

    C 0 1

    D 0,02

    E 0,08

    Lp bng m Huffman cho cc k hiu trnF 0,2

    Thc hin gii m trong trng hp nhnc dy k hiu: 00110001010101000110

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    cim ca m Huffman

    hin ch1 ln

    M Huffman c gi lm khi

    v mi ku c u n ng m m c n .

    Vic gii m tng k hiu c thc hintc thi v khng phthuc k hiu trc vsau

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    M ha shc

    Chui cc k hiu ngunc m hacng nhau

    Khng c stngng 1 -1 gia k hiu

    n un v t m

    nhng tlnn cao hn

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    M ha shc

    Chui k tngun c gn 1 tm shc tng ng vi 1 kho ng con trongon [0,1]

    Khi sl n k t tron bn tin tn lnkhong con biu din chui sthub l i

    m tchui cng nhiu

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    M ha shc

    1 2 3 3 4

    ,

    2) Subdivide [0, 1) based on the probability of i

    3) Update interval by processing source symbols

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    V d

    Encode

    a1a2a3a3a4

    [0.06752, 0.0688)

    ,

    0.068

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    32Gii m

    .

    0.8

    .

    0.72

    .

    0.688

    .

    0.5856

    .

    0.57152

    a4

    Decode 0.572

    a3

    0.4 0.56 0.624 0.5728 056896

    a2

    0.2 0.48 0.592 0.5664

    0.56768a1

    ..0.56 0.56 0.5664

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    32M LZW Lempel-Ziv-Welch

    Thut ton m ha:

    Re eat:

    Tm dy k tdi nht wc trong t in

    Cp nht dy k twa vo trong t in,

    trong wal dy k tkhng c trongt in

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    32M LZW Lempel-Ziv-Welch

    1. BA is not in the Dictionary; insertBA, output the code for its prefix:code(B)2. AB is not in the Dictionary; insertAB, output the code for its prefix:code(A)

    . .

    BAA is not in Dictionary; insertBAA, output the code for its prefix:code(BA)

    4. AB is in the Dictionary.

    ABA is not in the Dictionary; insertABA, output the code for its prefix:code(AB). s no n e c onary; nser , ou pu e co e or s pre x:co e

    6. AA is in the Dictionary and it is the last pattern; output its code:code(AA)

    The compressed message is:

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    32M LZW Lempel-Ziv-Welch

    V d2: M ha dy k tsau:

    1. BA is not in the Dictionary; insert BA, output the code for its prefix: code(B)

    2. AB is not in the Dictionary; insert AB, output the code for its prefix: code(A)

    3. BA is in the Dictionary.

    s no n c onary; nser , ou pu e co e or s pre x: co e

    4. AB is in the Dictionary.

    ABRis not in the Dictionary; insert ABR, output the code for its prefix: code(AB)

    5. RRis not in the Dictionary; insert RR, output the code for its prefix: code(R). .

    RRA is not in the Dictionary and it is the last pattern; insert RRA, output code for its prefix:

    code(RR), then output code for last character: code(A)

    The compressed message is:

    39

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    32M LZW Lempel-Ziv-Welch

    Example: Chui cha nn: aaabbbbbbaabaaba

    S ng ts = T ng s t 8

    = 16 * 8

    = s

    Tm s lng bit sau khi nn:

    S lng bits = Tng s t m* 12

    = 7 * 12

    = 84 bits

    Note: Mi nhca t in c kch thc 12 bit

    40

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    32M LZW Lempel-Ziv-Welch

    Use LZW to decompress the output sequence

    1. 66 is in Dictionary; output string(66) i.e. B2. 65 is in Dictionary; output string(65) i.e. A, insert BA

    3. 256 is in Dictionary; output string(256) i.e. BA, insert AB

    . . . ,

    5. 65 is in Dictionary; output string(65) i.e. A, insert ABA

    6. 260 is not in Dictionary; outputprevious output + previous output first character: AA, insert AA

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    32M hnh m hanh

    Mapper:Bininhu vo theo cch c thlmgim dtha thng tin gia cc pixel

    u am o: p: www. a er ar.org scre e-

    Cosine-Transform-Tutorial.htm

    44

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    32Bini cosine ri rc (DCT)

    DCT biu din tn hiu (tp cc s) di dngng c c m cos v c c n s c n au.

    V dtn hiu:

    Cc hsca php bini DCTc tnh bic ng c:

    45

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    32Bini cosine ri rc (DCT)

    Vi k = 0: ta c hsw0c gi l hs1 chi u (DC coefficient):

    46

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    32Bini cosine ri rc (DCT)

    Vi k khc 0: cc hswkc gi l hs xoay chi u (AC coefficient):

    47

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    32Bini cosine ri rc (DCT)

    Bini DCT 2 chiu:

    forward

    inverse

    if u=0 if v=0

    if u>0 v

    48

    Bi i i i (DCT)

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    32Bini cosine ri rc (DCT)

    Bini DCT vi n = 8:

    D = TT x A x T A l ma tr n cn bini

    49

    Bi i i i (DCT)

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    32Bini cosine ri rc (DCT)

    Mi quan hgia hm cos vnh

    Hs1 chiu Chiu tng ca tn s

    xoaytng

    tn s

    50

    Bi i i i (DCT)

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    32Bini cosine ri rc (DCT)

    V d:

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    Bi i i i (DCT)

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    32Bini cosine ri rc (DCT)

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    M h h h h

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    32M hnh m hanh

    Quantizer: Lng t ha l bc tng hoc gimcc gi tr u ra ca bc mapper theo cc mc

    mong mun

    53

    f Lng t ha

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    32Lng tha

    Quantization

    Q(u,v): quantization table

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    f Chun nn JPEG

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    32Chun nn JPEG

    Cu trc ca chun nn JPEG

    55

    of Zig zag scan

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    32Zig-zag scan