Ch03-1 - Xử lý nâng cao chất lượng ảnh

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  • 7/30/2019 Ch03-1 - X l nng cao cht lng nh

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    X l nng cao chtlngnh

    11/01/2013Phm Vit Hng Nhp mn x l nh 1

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    Cc kthutx l nhnhm nng cao chtlng

    Sdng ton tim

    Sdng ton t khng gian

    Khi phcnh do nhiu

    11/01/2013 Phm Vit Hng Nhp mn x l nh 2

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    L nhng php ton khng phthucvtrim

    nh:

    Tnggim sng

    Thng k tnsut

    Binitnsut

    11/01/2013 Phm Vit Hng Nhp mn x l nh 3

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    Vinh I vi kch thc m x n v s nguyn c th

    tnggim sng l:

    I [i, j] = I [i, j] + c; (i,j)

    Nu c > 0 th tng sng

    Nu c < 0 th gim sng

    11/01/2013 Phm Vit Hng Nhp mn x l nh 4

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    Vinh I vi kch thc m x n v hai s Min, Max

    v ngng th tch ngng l:

    I [i, j] = I [i, j] > =?Max: Min; (I,j)

    Vi Min = 0, Max = 1 th kthut ny l chuyn

    nhentrng

    11/01/2013 Phm Vit Hng Nhp mn x l nh 5

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    Nhmgimbtmc xm bng cch nhm cc

    mc xm gn nhau thnh mt nhm.

    Nu c 2 nhm th l tch ngng.

    11/01/2013 Phm Vit Hng Nhp mn x l nh 6

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    Biutnsutmc xm g canh I l sim

    nh c gi tr g canh I k hiu l h(g).

    11/01/2013 Phm Vit Hng Nhp mn x l nh 7

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    Da vo biutnsut c th xc nh cc cmmc xm gn nhau.

    Thng thng c nhiu nhm v kch thc khc

    nhau

    11/01/2013 Phm Vit Hng Nhp mn x l nh 8

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    khi qut th kch thc nhm bng nhau ta c

    I [i, j] = I [i, j] / bunch_size * bunch_size; (i,j)

    B cmvi kch thc nhm l 3

    11/01/2013 Phm Vit Hng Nhp mn x l nh 9

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    nh I cgi l cn bng l tng l nu

    h(g) = h(g); (g,g)

    Vinh I kch thc m n v c smc xm cn

    bng l no_level S im nh trung bnh ca mi mc xm ca nh

    cn bng l

    TB = (m n) / no_level S im nh c mc xm g l

    11/01/2013 Phm Vit Hng Nhp mn x l nh 10

    g

    0i

    h(i)t(g)

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    Cn bngtnsut l xc nh hm f: g f(g)sao

    cho

    Ch : nh cn bng xong chachc l cn bngl tng

    11/01/2013 Phm Vit Hng Nhp mn x l nh 11

    1

    TB

    t(g)round0,maxf(g)

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    V d cn bngnh I vino_level= 4

    11/01/2013 Phm Vit Hng Nhp mn x l nh 12

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    Ngng trong k thut tch ngng thng

    c ci bi ngi s dng.

    K thuttm tch ngng t ng nhm tm ra

    ngng mt cch t ng

    Da vo biutnsut

    Vt th tch lm 2 phn nu tng lnh trong tng

    phn ltithiu

    11/01/2013 Phm Vit Hng Nhp mn x l nh 13

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    nh I kch thc m n

    G l smc xm canhkckhuytthiu

    t(g) l s im nh c mc xm g

    Mmen qun tnh trung bnh c mc xm g l

    Hm f: g f(g)l

    11/01/2013 Phm Vit Hng Nhp mn x l nh 14

    g

    0i

    i.h(i)t(g)

    1m(g)

    21)m(Gm(g)t(g)m.n

    t(g)f(g)

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    Tm sao cho:

    11/01/2013 Phm Vit Hng Nhp mn x l nh 15

    f(g)max)f(1Gg0

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    Nubitnhgc v hm bini th ta s c nh

    bini v biutnsutcanhbini

    Nubitbiutnsutcanhgc v hm

    bini th c tnh cbiutnsutcanh

    bini khng?

    11/01/2013 Phm Vit Hng Nhp mn x l nh 16

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    V d: cn tnh h(q) vi q = f(g)

    11/01/2013 Phm Vit Hng Nhp mn x l nh 17

    )(

    1

    )()(

    qfi

    ihqh