17411 - Bai giang Xu ly anh

8/3/2019 17411 - Bai giang Xu ly anh

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TRNG I HC HNG HI VIT NAM

KHOA CNG NGH THNG TIN

B MN H THNG THNG TIN

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BI GING

X L NH

TN HC PHN : X L NH

M HC PHN : 17411

TRNH O TO : I HC CHNH QUY

DNG CHO SV NGNH : CNG NGH THNG TIN

HI PHNG - 2011

.


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MC LC

Chng I: Tng quan v x l nh s 7

1.1.X l nh s l g? 71.2.Ngun gc ca x l nh s 71.3.Cc ng dng ca x l nh s 71.4.Cc bc c bn trong x l nh s 81.5.Cc thnh phn ca mt h thng x l nh 11Chng II: Cc kin thc c bn v nh s 14

2.1. Cm nhn th gic 14

2.2. Ly mu v lng t ha nh 14

2.3. Mi quan h gia cc im nh 16Chng III: Nng cao cht lng nh trong min khng gian 19

3.1. Cc php bin i mc xm c bn 19

3.2. X l histogram 20

3.3. Nng cao cht lng nh s dng cc ton t s hc/logic 21

3.4. B lc trong min khng gian 25

3.5. Cc b lc lm mt nh trong min khng gian 29

3.6. Cc b lc lm sc nt nh trong min khng gian 31Chng IV: Nng cao cht lng nh trong min tn s 34

4.1. Php bin i Fourier v min tn s 34

4.2. Cc b lc lm mt nh trong min tn s 37

4.3. Cc b lc lm sc nt nh trong min tn s 38

Chng V: Nn nh 40

5.1. Cc kin thc c bn 40

5.2. Nn nh khng mt thng tin 425.3. Nn nh c mt thng tin 45

Chng VI: X l hnh thi nh 50

6.1. Php gin nh v php co nh nh phn 50

6.2. Php m nh v php ng nh nh phn 51

6.3. Mt s thut ton hnh thi c bn trn nh nh phn 51

6.4. X l hnh thi nh xm 53

Chng VII: Phn on nh 597.1. Pht hin tnh khng lin tc 59


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7.2. Phn on nh da vo cc vng nh con 60

Mt s thi mu 64


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4Tn hc phn: Nhn dng v x l nh Loi hc phn: 2

B mn ph trch ging dy: H thng Thng tin Khoa ph trch: CNTT.

M hc phn: 17411 Tng s TC: 4

Tng s tit L thuyt Thc hnh/Seminar T hc Bi tp ln n mn hc

75 45 30 0 khng khng

Hc phn hc trc: Khng yu cu.

Hc phn tin quyt: Khng yu cu.

Hc phn song song: Khng yu cu.

Mc tiu ca hc phn: Cung cp cc kin thc v lnh vc x l nh s; Gip cho sinh vin nm

c cc k thut x l nh c bn.

Ni dung ch yu:

Cc kin thc c bn v nh s; Cc k thut nng cao cht lng nh; Cc thut ton x

l hnh thi; Cc k thut phn on nh; Cc thut ton nn nh v chun nh nn.

Ni dung chi tit:

TN CHNG MCPHN PHI S TIT

TS LT TH BT KT

Chng I: Tng quan v x l nh s 5 3 2

1.1.X l nh s l g?1.2.Ngun gc ca x l nh s1.3.Cc ng dng ca x l nh s1.4.Cc bc c bn trong x l nh s1.5.Cc thnh phn ca mt h thng x l nhChng II: Cc kin thc c bn v nh s 5 3 2

2.1. Cm nhn th gic

2.2. Ly mu v lng t ha nh

2.3. Mi quan h gia cc im nh

Chng III: Nng cao cht lng nh trong min khng gian 15 9 6

3.1. Cc php bin i mc xm c bn

3.2. X l histogram

3.3. Nng cao cht lng nh s dng cc ton t s hc/logic

3.4. B lc trong min khng gian

3.5. Cc b lc lm mt nh trong min khng gian

3.6. Cc b lc lm sc nt nh trong min khng gian

Chng IV: Nng cao cht lng nh trong min tn s 15 9 6


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TN CHNG MCPHN PHI S TIT

TS LT TH BT KT

4.1. Php bin i Fourier v min tn s

4.2. Cc b lc lm mt nh trong min tn s

4.3. Cc b lc lm sc nt nh trong min tn sChng V: Nn nh 15 7 6 2

5.1. Cc kin thc c bn

5.2. Nn nh khng mt thng tin

5.3. Nn nh c mt thng tin

Chng VI: X l hnh thi nh 10 6 4

6.1. Php gin nh v php co nh nh phn

6.2. Php m nh v php ng nh nh phn6.3. Mt s thut ton hnh thi c bn trn nh nh phn

6.4. X l hnh thi nh xm

Chng VII: Phn on nh 10 4 4 2

7.1. Pht hin tnh khng lin tc

7.2. Phn on nh da vo cc vng nh con

Nhim v ca sinh vin:

Tham d cc bui hc l thuyt v thc hnh, lm cc bi tp c giao, lm cc bi thi giahc phn v bi thi kt thc hc phn theo ng quy nh.

Ti liu hc tp:

1. Lng Mnh B, Nguyn Thanh Thy,h h , Nh xut bn Khoa hc vK thut H Ni, 2003.

2. V c Khnh, i h h, Nh xut bn Thng k, 2003.3. Rafael C. Conzalez & Richard E. Woods, Digital Image Processing, 2nd edition, Pearson

Education, 2004.

Hnh thc v tiu chun nh gi sinh vin:

- Hnh thc thi: thi vit.

- Tiu chun nh gi sinh vin: cn c vo s tham gia hc tp ca sinh vin trong cc bui

hc l thuyt v thc hnh, kt qu lm cc bi tp c giao, kt qu ca cc bi thi gia hc

phn v bi thi kt thc hc phn.

Thang im: Thang im ch A, B, C, D, F.

im nh gi hc phn: Z = 0,3X + 0,7Y.


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6Bi ging ny l ti liu chnh thc v thng nhtca B mn H thng Thng tin, Khoa

Cng ngh Thng tin v c dng ging dy cho sinh vin.

Ngy ph duyt: / /

Trng B mn


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CHNG I: TNG QUAN V X L NH S

1.1. X l nh s l g?

X l nh l mt lnh vc mang tnh khoa hc v cng ngh. N l mt ngnh khoa hc mi

m so vi nhiu ngnh khoa hc khc nhng tc pht trin ca n rt nhanh, kch thch cc trung

tm nghin cu, ng dng, c bit l my tnh chuyn dng ring cho n. X l nh c a vo ging dy bc i hc nc ta khong chc nm nay. N l

mn hc lin quan n nhiu lnh vcv cn nhiu kin thc c s khc. u tin phi k n X

l tn hiu s l mt mn hc ht sc c bn cho x l tn hiu chung, cc khi nim v tch chp,

cc bin i Fourier, bin i Laplace, cc b lc hu hn Th hai, cc cng c ton nh i s

tuyn tnh, Sc xut, thng k. Mt s kin th cn thit nh Tr tu nhn tao, Mng n ron nhn

to cng c cp trong qu trnh phn tch v nhn dng nh.

1.2. Ngun gc ca x l nh sng dng u tin c bit n l nng cao cht lng nh bo c truyn quacp t Lun

n n New York t nhng nm 1920. Vn nng cao cht lng nh c lin quan ti phn b

mc sng v phn gii ca nh. Vic nng cao cht lng nh c pht trin vo khong nhng

nm 1955. iu ny c th gii thchc v sau th chin th hai, my tnh phttrin nhanh to

iu kin cho qu trnh x l nh s thun li. Nm 1964, my tnh c kh nng x l v nng

cao cht lng nh t mt trng v v tinh Ranger 7 ca M bao gm: lm ning bin, lu nh.

T nm 1964 n nay, cc phng tin x l, nng cao cht lng, nhn dng nh pht trin khng

ngng. Cc phng php tri thc nhn to nh mng n ron nhn to,cc thut ton x l hin i

v ci tin, cc cng c nn nh ngy cng c p dng rng ri v thu nhiu kt qu kh quan.

1.3. Cc ng dng ca x l nh s

Bin i nh (Image Transform)

Trong x l nh do s im nh ln cc tnh ton nhiu ( phc tp tnh ton cao) i hi dung

lng b nh ln, thi gian tnh ton lu. Cc phng php khoa hc kinh in p dng chox l

nh hu ht kh kh thi. Ngi ta s dng cc php ton tng ng hoc bin i sang min x

l khc d tnh ton. Sau khi x l d dng hn c thc hin, dng bin i ngc a v

min xc nh ban u, cc bin i thng gp trong x l nh gm:

- Bin i Fourier, Cosin, Sin

- Bin i (m t) nh bng tch chp, tch Kronecker (theo x l s tn hiu [3])

- Cc bin i khc nh KL (Karhumen Loeve), Hadamard

Mt s cc cng c sc xut thng k cng c s dng trong x l nh. Do khun kh ti liu

hng dn c hn, sinh vin c thm cc ti liu nm c cc phngphp bin i v mt s

phng php khc c nu y.


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8Nn nh

nh d dng no vn chim khng gian nh rt ln. Khi m t nh ngi ta a k thut

nn nh vo. Cc giai on nn nh c th chia ra th h 1, th h 2. Hin nay, cc chun MPEG

c dng vi nh ang pht huy hiu qu

1.4. Cc bc c bn trong x l nh s

Cc phng php x l nh bt u t cc ng dng chnh: nng cao cht lng nh v phn

tch nh.

d tng tng, xt cc bc cn thit trong x l nh. u tin, nh t nhin t thgii

ngoi c thu nhn qua cc thit b thu (nh Camera, my chp nh). Trc y, nh thu qua

Camera l cc nh tng t (loi Camera ng kiu CCIR). Gn y, vi s pht trin ca cng

ngh, nh mu hoc en trng c ly ra t Camera, sau n c chuyn trc tip thnhnh s

to thun li cho x l tip theo. (My nh s hin nay l mt th d gn gi). Mt khc,nh cng

c th tip nhn t v tinh; c th qut t nh chp bng my qut nh. Hnh 1.1 di y m t cc

bc c bn trong x l nh.

S ny bao gm cc thnh phn sau:

a) Phn thu nhn nh (Image Acquisition)

nh c th nhn qua camera mu hoc en trng. Thng nh nhn qua camera l nh

tng t (loi camera ng chun CCIR vi tn s 1/25, mi nh 25 dng), cng c loi camera

s ho (nh loi CCDChange Coupled Device) l loi photodiot to cng sng ti mi im

nh.

Camera thng dng l loi qut dng ; nh to ra c dng hai chiu. Cht lng mt nh thu

nhn c ph thuc vo thit b thu, vo mi trng (nh sng, phong cnh)

b) Tin x l (Image Processing)

Sau b thu nhn, nh c th nhiu tng phn thp nn cn a vo b tin x l nng

cao cht lng. Chc nng chnh ca b tin x l l lc nhiu, nng tng phn lm nh r

hn, nt hn.

c) Phn on (Segmentation) hay phn vng nh

Phn vng nh l tch mt nh u vo thnh cc vng thnh phn biu din phn tch,

nhn dng nh. V d: nhn dng ch (hoc m vch) trn phong b th cho mc ch phn loi


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9bu phm, cn chia cc cu, ch v a ch hoc tn ngi thnh cc t, cc ch, cc s (hoc cc

vch) ring bit nhn dng. y l phn phctp kh khn nht trong x l nh v cng d gy

li, lm mt chnh xc ca nh. Kt qu nhn dng nh ph thuc rt nhiu vo cng on ny.

d) Biu din nh (Image Representation)

u ra nh sau phn on cha cc im nh ca vng nh (nh phn on) cng vi m

lin kt vi cc vng ln cn. Vic bin i cc s liu ny thnh dng thch hp l cn thit cho x

l tip theo bng my tnh. Vic chn cc tnh cht th hin nh gi l trch chn c trng

(Feature election) gn vi vic tch cc c tnh ca nh di dng cc thng tin nh lng hoc

lm c s phn bit lp i tng ny vi i tng khc trong phm vi nh nhnc. V d:

trong nhn dng k t trn phong b th, chng ta miu t cc c trng ca tng kt gip phn

bitk t ny vi k t khc.

e) Nhn dng v ni suy nh (Image Recognition and Interpretation)

Nhn dng nh l qu trnh xc nh nh. Qu trnh ny thng thu c bng cch so snh

vi mu chun c hc (hoc lu) t trc. Ni suy l phn on theo ngha trn c snhn

dng. V d: mt lot ch s v nt gch ngang trn phong b th c th c ni suy thnh m in

thoi. C nhiu cch phn loai nh khc nhau v nh. Theo l thuyt v nhn dng, cc m hnh

ton hc v nh c phn theo hai loi nhn dng nh c bn:

-Nhn dng theo tham s.

-Nhn dng theo cu trc.

Mt s i tng nhn dng kh ph bin hin nay ang c p dng trong khoa hc v

cng ngh l: nhn dng k t (ch in, ch vit tay, ch k in t), nhn dng vn bn (T ext),

nhn dng vn tay, nhn dng m vch, nhn dng mt ngi

f) C s tri thc (Knowledge Base)

Nh ni trn, nh l mt i tng kh phc tp v ng nt, sng ti, dung lng

im nh, mi trng thu nh phong ph ko theo nhiu. Trongnhiu khu x l v phn tch

nh ngoi vic n gin ha cc phng php ton hc m bo tin li cho x l, ngi ta mong

mun bt chc quy trnh tip nhn v x l nh theo cch ca con ngi. Trong cc bc x l ,nhiu khu hin nay x l theo cc phng php tr tu con ngi. V vy, y cc c s tri

thc c pht huy. Trong ti liu, chng 6 v nhn dng nh c nu mt viv d v cch s

dng cc c s tri thc .

g) M t (biu din nh)

T Hnh 1.1, nh sau khi s ho s c lu vo b nh, hoc chuyn sang cc khu tip

theo phn tch. Nu lu tr nh trc tip t cc nh th, i hi dung lng b nh cc ln v

khng hiu qu theo quan im ng dng v cng ngh. Thng thng, cc nh th c ct(biu din) li (hay n gin l m ho) theo cc c im ca nh c gi l cc c trng nh


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10(Image Features) nh: bin nh (Boundary), vng nh (Region). Mt s phng php biudin

thng dng:

Biu din bng m chy (Run-Length Code)

Biu din bng m xch (Chaine -Code)

Biu din bng m t phn (Quad-Tree Code)

Biu din bng m chy

Phng php ny thng biu din cho vng nh v p dng cho nh nh phn. Mt vngnh R

c th m ho n gin nh mt ma trn nh phn:

- U(, ) = 1 u (m, n) thuc R

- U( m, ) = 0 u (m, n) khng thuc R

Trong : U(m, n) l hm m t mc xm nh ti ta (m, n). Vi cch biu din trn,mt

vng nh c m t bng mt tp cc chui s 0 hoc 1. Gi s chng ta m t nh nh phnca

mt vng nh c th hin theo to (x, y) theo cc chiu v c t ch i vi gi tr 1khi

dng m t c th l: (x, y)r; trong (x, y) l to , rl s lng cc bit c gi tr 1lin tc theo

chiu ngang hoc dc.

Biu din bng m xch

Phng php ny thng dng biu din ng bin nh. Mt ng bt k c chia thnh

cc on nh. Ni cc im chia, ta c cc on thng k tip c gn hng cho on thng

to thnh mt dy xch gm cc on. Cc hng c th chn 4, 8, 12, 24, mi hng c m

ho theo s thp phn hoc s nh phn thnh m ca hng.

Biu din bng m t phn

Phng php m t phn c dng m ho cho vng nh. Vng nh u tin c chia lm

bn phn thng l bng nhau. Nu mi vng ng nht (cha ton im en (1) haytrng (0)),

th gn cho vng mt m v khng chia tip. Cc vng khng ng nht c chia tip lm bn

phn theo th tc trn cho n khi tt c cc vng u ng nht. Cc m phn chia thnh cc vng

con to thnh mt cy phn chia cc vng ng nht.

Trn y l cc thnh phn c bn trong cc khu x l nh. Trong thc t, cc qu trnh s

dng nh s khng nht thit phi qua ht cc khu ty theo c im ng dng. Hnh 1.2 cho s

phn tch v x l nh v lu thng tin gia cc khi mt cch kh y . nh sau khi c

s ha c nn, lu li truyn cho cc h thng khc s dng hoc x l tip theo. Mt khc,

nh sau khi s ha c th b qua cng on nng cao cht lng (khi nh cht lng theo mt

yu cu no ) chuyn ti khu phn n hoc b tip khu h n chuyn trc tip ti

khu ch ch c trng. Hnh 1.2 cng chia cc nhnh song song nh: nngcao cht lng nh

c hai nhnh phn bit: nng cao cht lng nh (tng sng, tng phn,lc nhiu) hoc khiphc nh (hi phc li nh tht khi nh nhn c b mo) v.v


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1.5. Cc thnh phn ca mt h thng x l nh

Theo quan im ca quy trnh x l, chng ta th hin cc khi c bn trn Hnh 1.1, cc

khi chi tit v lung thng tin trn Hnh 1.2. Theo quan im ca h thng x l trn mytnh s,

h thng gm cc u o (thu nhn nh); b s ha ; my tnh s; B hin th; B nh. Cc thnh

phn ny khng nhc li y (c thm gio trnh cu trc my tnh).

Mt h thng x l nh c bn c th gm: my tnh c nhn km theo v mch chuyni

ho VGA hoc SVGA, a cha cc nh dng kim tra cc thut ton v mt mn hnhc h tr

VGA hoc SVGA. Nu iu kin cho php, nn c mt h thng nh Hnh 1.4. bao gm mt my

tnh PC km theo thit b x l nh. Ni vi cng vo ca thit b thu nhn nh l mtvideo camera,

v cng ra ni vi mt mn hnh. Thc t, phn ln cc nghin cu ca chng tac a ra trn

nh mc xm (nh en trng). Bi vy, h thng s bao gm mt thit b x lnh en trng v mt

mn hnh en trng.

nh mc xm c p dng trong nhiu lnh vc nh sinh vt hc hoc trong cng nghip.Thc t ch ra rng bt k ng dng no trn nh, mc xm cng ng dng c trnnhmu. Vi

l do , h thng ban u nn ch bao gm cc thit b thu nhn v hin th nh en trng. Vi nh


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12mu, nn s dng mt h thng mi nh Hnh 1.3, tr trng hp bn cn mtcamera TV mu

v mt mn hnh a tn s (v d nh NEC MultiSync, Sony Multiscan, hoc Mitsubishi Diamond

Scan) hin th nh mu. Nu kh nng hn ch, c th dng PC km theo v mch VGA v mn

hnh VGA, dng nh c.

CU HI N TP

1. Trnh by cc thnh phn v lu thng tin gia cc khi trong qutrnh x l nh.

2. Nu khi nim v nh ngha im nh.

3. Th no l phn gii nh, cho v d?

4. Trnh by nh ngha mc xm, cho v d.

5. Nu quan h gia cc im nh.

6. Trnh by v khong cch o v phn loi khong cch gia cc im nh.

7. Nu ngha ca cc php bin i nh, lit k mt s php bin i v cho v d.


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CHNG II: CC KIN THC C BN V NH S

2.1. Cm nhn th gic

Th gic my tnh l nh my tnh trong vic ng dng khng dnh cho th gic ngi. Mt

trong nhng ch chnh trong lnh vc th gic my tnh l phn tch nh.Phn tch nh bao hm vic xem xt d liu nh gii quyt d dng mt bi ton th gic. Qu

trnh phn tch nh bao hm hai vn khc: trch chn c trng v phn lp mu.

Trch chn c trng l qu trnh x l thng tin nh thu c mc cao nh l thng tin v

nhn hay mu, Phn lp mu l hot ng ly thng tin mc cao ny v t xc nh cc i

tng nm trong nh.

C nhiu ng dng ca th gic my tnh nh:

Trong cc h thng sn xut, th gic my tnh thng c s dng trong vic iu khin chtlng.

Trong nhiu lnh vc khc nhau trong cng ng y t, m chc chn cc kiu ng dng s

tip tc c pht trin. Cc v d hin nay v cc h thng y t ang c pht trin bao gm: cc

h thng chn on cc khi u da t ng, cc h thng tr gip gii phu thn kinh khi phu thut

no b, v cc h thng test bnh n t ng.

Lnh vc an ninh v php lut cng l mt lnh vc ha hn cho vic pht trin cc h thng th

gic my tnh, vi cc ng dng t nhn dng t ng vn tay cho n phn tch DNA. Cc hthng an ninh nhn dng ngi thng qua vic scan vng mc mt, scan khun mt, v cc ng

tnh mch tay c pht trin.

Chng trnh khng gian U.S. v BQP, vi vic pht trin cc kh nng th gic cho r bt ang

c nghin cu v pht trin. Cc ng dng t xe c t ch cho n bt bm mc tiu v nhn

dng. Cc v tinh theo qu o tri t thu thp nhng dung lng ln d liunh hng ngy, v cc

nh ny s t ng c scan h tr vic lp bn , d bo thi tit, v gip chng ta hiu c

nhng thay i ang xy trn hnh tinh chng ta.

2.2. Ly mu v lng t ha nh

Gii thiu

Mt nh g(x, y) ghi c t Camera l nh lin tc to nn mt phng hai chiu. nh cn

chuyn sang dng thch hp x l bng my tnh. Phng php bin i mt nh (hay mt

hm)lin tc trong khng gian cng nh theo gi tr thnh dng s ri rc c gi l s ho nh.

Vicbin i ny c th gm hai bc:

Bc 1: o gi tr trn cc khong khng gian gi l ly mu

Bc 2: nh x cng (hoc gi tr) o c thnh mt s hu hn cc mc ri rc gil

lng t ho.


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14Ly mu

Ly mu l mt qu trnh, qua nh c to nn trn mt vng c tnh lin tc c chuyn

thnh cc gi tr ri rc theo ta nguyn. Qu trnh ny gm 2 la chn:

- Mt l: khong ly mu.

- Hai l: cch th hin dng mu.

La chn th nht c m bo nh l thuyt ly mu ca Shannon. La chn th hai lin quan

n o (Metric) c dng trong min ri rc.

Khong ly mu (Sampling Interval)

nh ly mu c th c m t nh vic la chn mt tp cc v tr ly mu trong khng gian

hai chiu lin tc. u tin m t qua qu trnh ly mu mt chiu vi vic s dng hm delta:

Tip theo chng ta nh ngha hm rng lc vi cc khong x nh sau:

vi rl s nguyn, x : khong ly mu

Nh vy, hm rng lc l chui cc xung rng lc t (- n +). Gi s hm mtchiu g(x)

c m t (gn ng) bng g(rx ) tc l:

Khi tn hiu ly mu c m hnh ho

hoc tng ng

Trong thc t, rkhng th tnh c trong khong v hn (t n + ) m l mt slng

Nx mu ln c th. Nh vy, n gin c th ni hm lin tc g(x) c th biu dintrn mt

min vi diNx mu thnh chui nh sau:

Ch 1: Khong ly mu (Sampling Interval) x l mt tham s cn phi c chn nh,

thch hp, nu khng tn hiu tht khng th khi phc li c t tn hiu ly mu.


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15Ch 2: T l thuyt v x l tn hiu s [5], (2 -6) l tch chp trong min khng gianx.

Mt khc (2-6) tng ng vi tch chp trong min tn s tc l bin i Fourier ca gs(x) l

Gs(s).

trong xl gi tr tn s ng vi gii trx trong min khng gian.

iu kin khi phc nh ly mu v nh tht c pht biu t nh l ly mu ca Shannon.

Lng t ha

Lng t ho l mt qu trnh lng ho tn hiu tht dng chung cho cc loi x l tn hiu

trn c s my tnh. Vn ny c nghin cu k lng v c nhiu li gii l thuyt di

nhiu gi nh ca cc nh nghin cu nh Panter v Dite (1951), Max (1960), Panter (1965). Cc

gi tr ly muZl mt tp cc s thc t gi trZmin n ln nhtZmax. Mi mt s trong cc gitr muZcn phi bin i thnh mt tp hu hn s bit my tnh lu tr hoc x l.

nh ngha: Lng t ho l nh x t cc s thc m t gi tr ly mu thnh mt gii hu hn

cc s thc. Ni cch khc, l qu trnh s ho bin .

Gi sZl mt gi tr ly mu (s thc) ti v tr no ca mt phng nh, v Zi


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16trong : s 1 l gi tr logic;N4(p) tp 4 im ln cn cap.

- Cc ln cn cho: Cc im ln cn cho NP(p) (C th coi ln cn cho la 4 hng:

ng-Nam, ng-Bc, Ty-Nam, Ty-Bc)

Np(p) = { (x+1, y+1); (x+1, y-1); (x-1, y+1); (x-1, y-1)}

- Tp kt hp:N8(p) = N4(p) + NP(p) l tp hp 8 ln cn ca im nhp.

- Ch : Nu (x, y) nm bin (mp) nh; mt s im s nm ngoi nh.

b) Cc mi lin kt im nh.

Cc mi lin kt c s dng xc nh gii hn (Boundaries) ca i tng vt thhoc

xc nh vng trong mt nh. Mt lin kt c c trng bi tnh lin k gia cc im v mc

xm ca chng.

Gi s Vl tp cc gi tr mc xm. Mt nh c cc gi tr cng sng t thang mc xm t

32 n 64 c m t nh sau :

V={32, 33, , 63, 64}.

C 3 loi lin kt.

- Lin kt 4: Hai im nhp v q c ni l lin kt 4 vi cc gi tr cng sng V

nu q nm trong mt cc ln cn cap, tc q thucN4(p)

- Lin kt 8:Hai im nh p v q nm trong mt cc ln cn 8 ca p, tc q thuc N8(p)

- Lin ktm (lin kt hn hp): Hai im nh p v q vi cc gi tr cng sng V

c ni l lin kt m nu.

+. q huc N4() hc+. q huc NP(p)

c) o khong cch gia cc im nh.

nh ngha: Khong cch D(p, q) gia hai im nh p to (x, y), q to (s, t) l hm

khong cch (Distance) hoc Metric nu:

1. D(,q) 0 (Vi D(p,q)=0 nu v chnu p=q)

2. D(p,q) = D(q,p)

3. D(,z) D(,q) + D(q,z); z im nh khc.Khong cch Euclide: Khong cch Euclide gia hai im nh p(x, y) v q(s, t) c nh

ngha nh sau:


17/70

17De(p, q) = [(xs)2 + (yt)2]1/2

Khong cch khi: Khong cch D4(p, q) c gi l khong cch khi th (City- Block

Distance) v c xc nh nh sau:

D4(p,q) = | x - s | + | y - t |

Gi tr khong cch gia cc im nh r: gi tr bn knh r gia im nh t tm im nhn

tm im nh q khc. V d: Mn hnh CGA 12 (12*2,54cm = 30,48cm=304,8mm) phngii

320*200; t l 4/3 (Chiu di/Chiu rng). Theo nh l Pitago v tam gic vung, ngcho s

ly t l 5 phn (5/4/3: ng cho/chiu di/chiu rng mn hnh); khi di tht l

(305/244/183) chiu rng mn hnh 183mm ng vi mn hnh CGA 200 im nh theo chiu dc.

Nh vy, khong cch im nh ln cn ca CGA 12 l 1mm.

Khong cchD8(p, q) cn gi l khong cch bn c (Chess-Board Distance) gia imnhp,

q c xc nh nh sau:

D8(p,q) = max (| x-s | , | y-t |)

CU HI N TP

1. Ti sao phi s ha nh? Trnh by cch biu din nh s trn my tnh?

2. Lymu l g? Lng t l g? Khong ly mu l g? Cho v d minh ha?

3. Trnh by cc mi quan h gia cc im nh?


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18

CHNG III: NNG CAO CHT LNG NH TRONG MIN

KHNG GIAN3.1. Cc php bin i mc xm c bn

Nng cao cht lng l bc cn thit trong x l nh nhm hon thin mt s c tnhca nh.

Nng cao cht lng nh gm hai cng on khc nhau: tng cng nh v khi phcnh. Tng

cng nh nhm hon thin cc c tnh ca nh nh :

- Lc nhiu, hay lm trn nh,

- Tng tng phn, iu chnh mc xm ca nh,

- Lm ni bin nh.

Cc thut ton trin khai vic nng cao cht lng nh hu ht da trn cc k thut trong min

im, khng gian v tn s. Ton t im l php bin i i vi tng im nh ang xt, khng

lin quan n ccim ln cn khc, trong khi , ton t khng gian s dng cc im ln cn

quy chiu ti im nh ang xt. Mt s php bin i c tnh ton phc tp c chuyn sang

min tn s thc hin, kt qu cui cng c chuyn tr li min khng gian nh cc bin i

ngc.

Khi nim v ton t im:

X l im nh thc cht l bin i gi tr mt im nh da vo gi tr ca chnh n m khng

h da vo cc im nh khc. C hai cch tim cn vi phng php ny. Cch th nht dng mt

hm bin i thch hp vi mc ch hoc yu cu t ra bin i gi tr mc xm ca im nhsang mt gi tr mc xm khc. Cch th hai l dng lc mc xm (Gray Histogram). V mt

ton hc, ton t im l mt nh x t gi tr cng nh sng u(m, n) ti to (m, n) sang gi

tri cng nh sng khc v(m, n) thng qua hmf(.), tc l:

Ni mt cch khc, ton t im l ton t khng b nh, mt mc xc u [ 0 , N] c

nh x sang mt mc xm v [ 0 ,N] : v =f( u ) . ng dng chnh ca ccton t im lbin i

tng phn ca nh. nh x fkhc nhau ty theo cc ng dng. Cc dng ton tim c

gii thiu c th nh sau:

Tng tng phn

Cc cp , , xc nh tng phn tng i.L

l s mc xm cc iTch nhiu v phn ngng


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19

Trong a = b =t gi l phn ngng.

Bin i m bn.

Ct theo mc

Trch chn bit.

3.2. X l histogram

a) L thuyt

Lc mc xm ca mt nh, t nay v sau ta qui c gi l lc xm, l mt hm

cung cp tn sut xut hin ca mi mc xm (grey level).

Lc xm c biu din trong mt h to vung gc x,y. Trong h to ny, trc

honh biu din s mc xm t 0 n N, N l s mc xm (256 mc trong trng hp chng taxt). Trc tung biu din s im nh cho mt mc xm (s im nh c cng mc xm). Cng c

th biu din khc mt cht: trc tung l t l s im nh c cng mc xm trn tng s im nh.

S im nh S im nh

Mc xm Mc xm

a) nh m b) nh nht

Lc xm cung cp rt nhiu thng tin v phn b mc xm ca nh. Theo thut ng ca

x l nh gi l tnh ng ca nh. Tnh ng ca nh cho php phn tch trong khong no phn

b phn ln cc mc xm ca nh: nh rt sng hay nh rt m. Nu nh sng, lc xm nm

bn phi (mc xm cao), cn nh m luc xm nm bn tri(mc xm thp).Theo nh ngha ca lc xm, vic xy dng n l kh n gin. Thut ton xy dng

lc xm c th m t nh sau:


20/70

20Bt u

H l bng cha lc xm (l vec t c N phn t)

a. Khi to bng

t tt c cc phn t ca bng l 0

b. To bng

Vi mi im nh I(x,y) tnh H[I(x,y)] = H[I(x,y)] + 1

c. Tnh gi tr Max ca bng H. Sau hin bng trong khong t 0 n Max.

Kt thc

Lc xm l mt cng c hu hiu dng trong nhiu cng on ca x l nh

nh tng cng nh ( xem chng Bn). Di y ta xem xt mt s bin i lc xm hay

dng.

b) Bin i lc mc xm

Trong tng cng nh, cc thao tc ch yu da vo phn tch lc xm. Trc tin ta

xt bng tra LUT(Look Up Table). Bng tra LUT l mt bng cha bin i mt mc xm i sang

mc xm j nh ni trong phn a. Mt cch ton hc, LUT c nh ngha nh sau:

- Cho GI l tp cc mc xm ban u GI = {0, 1, ..., NI}

- Cho GF l tp cc mc xm kt qu GF = {0, 1, ..., NF}

cho tin ta cho NI = NF = 255.

- f l nh x t GI vo GF: giGi s gfGF m gf = f(gi)

Vi mi gi tr ca mc xm ban u ng vi mt gi tr kt qu. Vic chuyn i mt mc xm

ban u v mt mc xm kt qu tng ng c th d dng thc hin c nh mt bng tra.

Khi xy dng c bng, vic s dng bng l kh n gin. Ngi ta xem xt mc xm ca

mi im nh, nh bng tra tnh c mc xm kt qu. Gi l bng tra.

3.3. Nng cao cht lng nh s dng cc ton t s hc/logic

C hai nhm thao tc i s p dng ln nh l: s hc v logic. Cc thao tc s hc c: cng,

tr, chia, v nhn cn cc thao tc logic gm: AND, OR, v NOT. Cc thao tc ny c thc hin

trn hai nh ngoi tr thao tc NOT ch cn mt nh, v c thc hin trn c s pixel-pixel.

p dng cc thao tc s hc ln 2 nh, ta thao tc theo cc pixel tng ng. Chng hn,

cng hai nh I1 v I2 tao ra nh I3 ta c

),(),(),( 321 crIcrIcrI

1195

1167

13109

565432

652443

676463

;

553

624

666

;

642

543

743

321 III

Php cng c s dng kt hp thng tin trong hai nh. Cc ng dng n gm pht trin

cc thut ton khi phc nh m hnh ho nhiu cng, v to cc hiu ng c bit nh l


21/70

21morphing nh trong cc nh chuyn ng (Hnh 2.2-5). Ch rng php morphing nh ng c

th i hi s dng n cc php bin i hnh hc (xem phn 3.5), sp thng hai nh. Morphing

nh cng c s dng trong thao tc da trn thi gian, do mt dung lng cn xng tng ln

trong nh th hai thng c s dng cng vo trong nh th nht khi qu thi gian.

Hnh 2.2-5. Cng nh

Tr hainh thng c s dng pht hin chuyn ng. Xt trng hp khi khng c g

thay i trong mt cnh c; nh kt qu t vic tr hai nh thu nhn c lin tip s l mt nh

mu en ton 0. Nu c ci g chuyn ng trong cnh, th php tr nh s to ra cc gi tr khc

0 ti cc v tr c chuyn ng. Hnh 2.2-6 minh ho vic s dng php tr nh pht hin chuyn

ng.


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22

Hnh 2.2-6. Tr nh

Vic nhn v chia nh c s dng iu chnh v cng sng ca nh. Mt nh thng

thng cha mt s khng i ln hn 1. Vic nhn v chia nh thng c s dng iu chnh

sng ca nh. Mt nh thng thng cha cc s khng i ln hn 1. Vic nhn cc gi tr

pixel vi s ln hn 1 th lm cho nh sng ln, cn chia cho s ln hn 1 th lm ti nh i. iu

chnh sng thng c s dng nh l mt bc tin x l trong nng cao nh v c th hin

nh trong hnh 2.2-7.


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23

Hnh 2.2-7. Nhn v chia nh

Cc thao tc logic AND, OR, v NOT to nn mt tp hon chnh, ngha l cc thao tc logic

khc (XOR, NOR, NAND) u c th sinh ra c bng cch t hp cc thao tc c s trn. Chng

thao tc di dng bit-wise trn d liu pixel.

Cc thao tc logic AND v OR c s dng t hp thng tin trong hai nh. Vic ny c ththc hin cho cc hiu ng c th, nhng mt ng dng c ngha hn cho phn tch nh l thc

hin mt ton t mt n. S dng AND v OR lm phng php n gin trch chn ROI t mt

nh, nu cc phng php ho phc tp hn khng sn c. Chng hn, mt mt n hnh vung

mu trng (ton 1) dng cho php AND vi mt nh s ch cho phn nh m trng vi hnh vung

ny xut hin nh u ra vi nn phn cn li en, mt mt n hnh vung mu en (ton 0) dng

cho php OR vi mt nh s ch cho phn nh m trng vi hnh vung ny xut hin nh u ra

vi nn phn cn li trng. Qu trnh ny c gi l masking nh, v hnh2.2-8 minh ho cc kt

qu ca cc thao tc ny.


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24

Ton t NOT to ra i ca nh gc, bng cch nghch o mi bit trong gi tr pixel, v c

th hin trn hnh 2.2-9.

Hnh 2.2-9. nh b

3.4. B lc trong min khng gian

Lc khng gian thng thng c thc hin kh nhiu hoc thc hin mt s kiu nng cao

nh. Cc thao tc ny c gi l lc khng gian phn bit chng vi lc tn s, s c trnh

by trong phn 2.5.

C ba kiu lc c trnh by y l:

-

Lc trung bnh.- Lc trung v.

- Lc nng cao.


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25Hai kiu lc u tin ch yu c s dng che y hoc kh nhiu, mc d chng cng

c th c s dng cho cc ng dng c th khc. Chng hn, mt b lc trung bnh lm cho nh

"mn hn" nh hnh 2.2-10. Ccb lc nng cao thng lm sng cc cnh v cc chi tit nm

trong nh.

Hnh 2.2-10. Lc trung bnh

Nhiu b lc khng gian c thc hin thng qua php cun. Bi v thao tc mt n cun cung

cp kt qu l tng trng s ca cc gi tr ca mt pixel v cc lng ging ca n, nn c gi l

l mt b lc tuyn tnh.

Mt kha cnh th v ca cc mt n cun l hiu ng tng th c th d on c da trn

mu ca chng. Chng hn, nu cc h s ca mt n c tng bng 1, th sng trung bnh ca nh

s c gi nguyn. Nu tng cc h s bng 0, th sng trung bnh s mt i v tr v mt nhti. Hn na, nu cc h s c c m v dng, th mt n l mt b lc tr v ch thng tin cnh,

cn nu cc h s u dng, th n l mt b lc lm m (blur) nh the image.

Lc trung bnh:

Cc b lc trung bnh thao tc trn cc nhm pixel a phng c gi l vng lng ging

v thay th pixel trung tm bi trung bnh ca cc pixel trong cng lng ging . Vic thay th ny

c thc hin bng mt mt n cun chng hn nh mt n 3x3 sau y:

9/19/19/1

9/19/19/19/19/19/1

Ch rng cc h s trong mt n ny c tng bng 1, nn sng nh gi nguyn, v cc

h s u dng nn n c khuynh hng lm nho nh. Cng c cc b lc trung bnh khc phc

tp hn c thit k dng cho cc kiu nhiu c th. Chng s c trnh by trong chng 3.

Lc trung v:

Lc trung v l lc phi tuyn. Mt php lc phi tuyn l mt kt qu khng th thu c t

mt tng trng s ca cc pixel lng ging, nh thc hin vi mt n cun. Tuy nhin lc trung

v cng thc hin trn cng mt vng lng ging a phng. Sau khi nh ngha kch thc


26/70

26vng lng ging, pixel trung tm c thay bng trung v tc l gi tr chnh gia ca tt c cc

gi tr lng ging.

V D

Cho trc mt cng lng ging 3x3:

5 3 3

5 4 4

6 5 255

Trc ht ta sp xp cc gi tr ny t b n ln (3,3,4,4,5,5,5,6, 7) v chn gi tr chnh

gia, trng hp ny l 5. Gi tr 5 ny tip c t vo v tr trung tm.

Mt b lc trung v c th s dng mt vng lng ging c kch thc bt k, nhng ph

bin l cc kch thc 3x3, 5x5 v 7x7. Ch rng nh u ra phi c ghi vo mt nh ring (b

m). Hnh 2.2-11 minh ho vic s dng b lc trung v kh nhiu.

Hnh 2.2-11. Lc trung v

B lc nng cao:

Cc b lc nng cao c xt y c cc b lc kiu laplacian-type v lc sai phn

(difference filter). Cc kiu b lc ny c khuynh hng a ra, hoc nng cao cc chi tit trong

nh. Hai mt n cun s dng cho cc b lc kiu laplacian l

010

1-51

01-0

121

2-52

12-1

Cc b lc kiu laplacian s nng cao c cc chi tit u theo mi hng. Cn cc b lc

sai phn s nng cao cc chi tit theo hng xc nh theo mt n chn. C 4 mt n cun lc sa i

phn, tng ng vi theo cc hng dc, ngang v hng theo hai ng cho:

010

010010

000

1-11000

1-00

010001

001-

010100


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27Cc kt qu ca vic p dng cc b lc ny c th hin trn hnh 2.2-12.

Hnh 2.2-12. Cc b lc nng cao

3.5. Cc b lc lm mt nh trong min khng gian

Do c nhiu loi nhiu can thip vo qu trnh x l nh nn cn c nhiu b lc thch hp. Vinhiu cng v nhiu nhn ta dng cc b lc thng thp, trung bnh v lc ng

hnh(Homomorphie); vi nhiu xung ta dng lc trung b, gi trung v, lc ngoi (Outlier).


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28a. Lc trung bnh khng gian

Vi lc trung bnh, mi im nh c thay th bng trung bnh trng s ca cc im ln cn

v c nh ngha nh sau:

Nu trong k thut lc trn, ta dng cc trng s nh nhau, phng trnh trn s tr thnh:

vi :y(m, n): nh u vo,

v(m, n): nh u ra,

a(k, l) : l ca s lc.

Lc trung bnh c trng s chnh l thc hin chp nh u vo vi nhn chp H. Nhn chpH

trong trng hp ny c dng:

Trong lc trung bnh, thng ngi ta u tin cho cc hng bo v bin ca nh khi b m khilm trn nh. Cc kiu mt n c s dng ty theo cc trng hp khc nhau. Cc b lc trn l

b lc tuyn tnh theo ngha l im nh tm ca s s c thay bi t hp cc imln cn

chp vi mt n.

Gi s u vo biu din bi ma trnI:

nh s thu c bi lc trung bnh Y=HIc dng:


29/70

29Mt b lc trung bnh khng gian khc cng hay c s dng. Phng trnh ca b lc

c dng:

y, nhn chpHc kch thuc 2x2 v mi im nh kt qu c gi tr bng trung bnh cng can vi trung bnh cng ca 4 ln cn gn nht. Lc trung bnh trng s l mt trng hp ring ca

lc thng thp.

b. Lc thng thp

Lc thng thp thng c s dng lm trn nhiu.V nguyn l ca b lc thng thp

ging nh trnh by trn. Trong k thut ny ngi ta hay dng mt s nhn chp c dng sau:

Ta d dng nhn thy khi b =1, Hb chnh l nhn chp Ht1 (lc trung bnh). hiu r hnbn

cht kh nhiu cng ca cc b lc ny, ta vit li phng trnh thu nhn nh di dng:

Trong [, ] l nhiu cng c phng sai 2. Nh vy, theo cch tnh ca lc trung bnh ta

c:

Nh vy, nhiu cng trong nh gimiNw ln.c. Lc ng hnh hnh (Homomorphie Filter)

K thut lc ny hiu qu vi nh c nhiu nhn. Thc t, nh quan st c gm nh gc nhn

vi mt h s nhiu. GiX(m, n) l nh thu c,X(m, n) l nh gc v (, ) l nhiu,nh vy:

Lc ng hnh thc hin ly logarit ca nh quan st. Do vy ta c kt qu sau:

R rng, nhiu nhn c trong nh s b gim. Sau qu trnh lc tuyn tnh, ta chuyn v nh c bng

php bin i hm e m.


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30

3.6. Cc b lc lm sc nt nh trong min khng gian

Mc ch chnh ca lc nt l lm sc nt cc chi tit ni bt trong nh hoc lm ni chi tit

b nhe. Lm nt nh c dng trong nhng ng dng nh in n in t, y hc, my kim tra sn

phm trong cng nghip, pht hin mc tiu qun s.

C s ca lc thng cao khng gianHnh dng ca p ng xung c s dng trong lc thng cao (lm nt nh) khng gian ch

ra rng lc ny cn c cc h s gn tm dng v cc h s ngoi vi m. Vi mt n 3 x 3 ta c th

s dng h s dng tm, cn cc h s khc m.

Chng hnxt mt n Laplace lm nt nh:

Ch rng tng cc h s bng 0. Do khi mt n di chuyn trn vng c mc xm hng hay thay

i chm, th gi tr xut ra s bng khng hoc rt nh. Kt qu ny ph hp vi lc tng ng

min tn s. Hn na lc ny loi b cc thnh phn c tn s thp, do a n gi tr trung bnh

ca cc mc xm tin v khng v v vy gim tng phn tng th trong nh.

Vic gi tr trung bnh gim v khng khin cho nh c mt vi gi tr xm m. V chng ta ch

xt cc mc xm dng, nn kt qu ca lc thng cao cn c co gin hoc ct b kt qu

cui cng thuc khong [0,L-1]. Ch rng vic ly gi tr tuyt di ca nh c lc thng caokhng phi l gii php tt v cc gi tr c mc xm m ln s xut hin sng ln trong nh.

3.6.1. Lc c khuch i tn s cao.

Mt nh lc thng cao khng gian c th xem nh l hiu gia nh gc v nh qua lc thng

thp, tc l:

g(x,y) := f(x,y)fsm(x,y),

trong fsm(x,y) l nh c lm trn ca f(x,y) qua lc thng thp.

D dng kim tra nh ra g(x,y) nhn c bng cch tnh p ng ca nh f(x,y) vi mt nLaplace trn. Bng cch nhn nh gc vi h s khuch i A, ta c ci bin l lc co khuch i

tn s cao:

G(x,y) := Af(x,y)lc thng thp

= (A-1)f(x,y) + lc thng cao.

Ni cch khc, nh g(x,y) nhn c t f(x,y) bng cch tnh p ng ti mi im vi mt n


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31Trong w=9A-1. Vi A=1 ta c kt qu lc thng cao tiu chun. Vi A>1 ta c phn ca nh

gc c cng thm kt qu ca lc thng cao m phc hi cc thnh phn lc thng thp b mt

trong php ton lc thng cao. Kt qu cui cng ta c mt nh gn vi nh gc, vi cp lm ni

ng bin tng i ty theo h s khuch i A. Ni chung vic tr mt nh bnhe t nh gc

gi l mt n khng nt. y l mt trong nhng phng php c bn c s dng trong cng

ngh in n v xut bn.

Tng t nh lc thng thp, trong lc thng cao ta c th s dng cc mt n vi kch thc

ln hn. Chng hn, mt n 7x7c gi tr ti tm bng 48, cn cc gi tr khc bng -1 v cc h s

c chun ha vi h s bng 1/49. Tuy nhin, trong thc th cc mt n kch thc ln hn 3x3

him khi s dng.

CU HI N TP1. Trnh by v b lc trong min khng gian (spatial filtering), lc tuyn tnh (linear

Filtering) v cch x l b lc trong min khng gian (Spatial Filtering Process).

2. Khi nim v mt n? Cch s dng mt n trong x l nh?

3. Lm sc nt mt vng nh.

4. Lm mt mt vng nh cho trc.

5. Trnh by v b lc Median Filters.


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32

CHNG IV: NNG CAO CHT LNG NH TRONG

MIN TN S

4.1. Php bin i Fourier v min tn s

Bin i Fourrier cho mt tn hiu c th hnh dung nh sau:

Min thi gian Min tn s

Bin i Fourrier cho mt tn hiu mt chiu gm mt cp bin i:

- Bin i thun: chuyn s biu din t khng gian thc sang khng gian tn s

(ph v pha).

Cc thnh phn tn s ny c gi l cc biu din trong khng gian Fourrier ca tn hiu.

- Bin i ngc: chuyn i s biu din ca i tng t khng gian Fourrier

sang khng gian thc.

a) Khng gian mt chiu

Cho mt hm f(x) lin tc. Bin i Fourrier ca f(x), k hiu F(u), u biu din tn s khng

gian, c nh ngha:

2( ) ixuF u f x e dx

()

trong :

f(x): biu din bin tn hiu

e-2ixu : biu din pha.

Bin i ngc ca F(u) cho f(x) c nh ngha:

2( ) ixu f x F u e du

()

b) Khng gian hai chiu

Cho f(x,y) hm biu din nh lin tc trong khng gian 2 chiu, cp bin i Fourier cho

f(x,y) c nh ngha:

- Bin i thun 2 ( ), ( , ) i xu yvF u v f x y e dxdy

()

u,v biu din tn s khng gian.

- Bin i ngc

2 ( ), ( , ) i xu yv f x y F u v e dudv

()

4.1.1. Bin i Fourrier ri rc - DFT

x(t) TF X(f)


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33Bin i DFT c pht trin da trn bin i Fourrier cho nh s. y, ta dng tng

thay cho tch phn. Bin i DFT tnh cc gi tr ca bin i Fourrier cho mt tp cc gi tr

trong khng gian tn s c cch u.

a) DFT cho tn hiu mt chiu

Khai trin Fourrier ri rc DFT cho mt dy {u(n), n = 0, 1, ..., N-1} nh ngha bi:

1

.

0

( )N

k n

N

n

v k u n W

vi k =0, 1, ..., N-1 ()

vi WN = e-j2/N

v bin i ngc

1.

0

1( ).

Nk n

N

k

u n v k W N

WN-kn , k=0, 1, ..., N-1 ()

Thc t trong x l nhngi ta hay dng DFT n v:

1

.

0

1 ( ).

N

k nN

n

v k u n W N

, k=0, 1, ..., N-1 ()

1

.

0

1( ).

Nk n

N

k

u n v k W N

, k=0, 1, ..., N-1 ()

b) DFT cho tn hiu hai chiu (nh s)

DFT hai chiu ca mt nh M x N : {u(m,n) } l mt bin i tch c v c nh ngha:

1 1

.

0 0

, ( , ). .N N

km l n

N N

m n

v k l u m n W W

0 l, k N-1 ()

v bin i ngc:

1 1

. .

20 0

1, ( , ) . .

N Nk m l n

N N

k l

u m n v k l W W N

0 m, n N-1 ()

Cp DFT n v hai chiu c nh ngha:

1 1

. .

0 0

1, ( , ). .N N

k m l nN N

m n

v k l u m n W W N

0 l, k N-1 ()

1 1

. .

0 0

1, ( , ). .

N Nk m l n

N N

k l

u m n v k l W W N

0 m, n N-1 ()

Vit li cng thc 3.27 v 3.28, ta c:

1 1

. .

0 0

1, ( , ).

N Nk m l n

N

m n

v k l u m n W N

0 l, k N-1 ()

1 1 . .

0 0

1, ( , ).N N k m l n

N

k l

u m n v k l W N

0 m, n N-1 ()

y, WN(km+ln) l ma trn nh c s.


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34

Vi ej = cos() +jsin() (cng thc le). Do vy:

2 lnln 2 ln 2 ln

cos sin

j km

km NN

km kmW e j

N N

.

Nh vy, cc hm c s trong ma trn nh c s ca bin i Fourier l cc hm cosine v

hm sine. Theo tnh ton trn, ta thy bin i Fourrier biu din nh trong khng gian mi theo cchm sine v cosine.

4.1.2. Mt s tnh cht v p dng

a) Tnh cht

- i xng v n v

FT = F, F-1 = F*

- Chu k

v(k + N, l + N) = v(k,l) k, l ()u(k + N, l + N) = u(k,l) k, l ()

- Ph Fourier mu ho

, 0 , 1

,0

U m n m n N U m n

if not

khi :

2 2, , ,

k lU DFT u m n v k l

N N

vi 1 2,U l bin i Fourier ca u(m,n).- Bin i nhanh: V DFT hai chiu l tch c, do bin i V = FUF tng

ng vi DFT n v 1 chiu 2N.

- Lin hip i xng:

DFT v DFT n v ca mt nh thc c tnh i xng lin hp:

*, ,

2 2 2 2

N N N N v k l v k l

vi 0 l N/2-1 ()

hay *, ,v k l v N k N l vi 0k,lN/2-1 ()

b)Thut ton bin i nhanh -FFT(Fast Fourrier Transform)

- Trng hp 1 chiu

T cng thc

1.

0

1.

Nk n

N

n

v k u n W N

vi k=0, 1,...,N-1, ta nhn thy:

Nhn xt:

+ Vi mi gi tr k ta cn N php nhn v N php cng.

+ tnh N gi tr ca v(k) ta cn N2 php nhn.


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35

tnh ton mt cch hiu qu , ngi ta dng thut ton tnh nhanh gi l FFT vi

phc tp tnh ton l O(Nlog2N).

Thut ton tnh nhanh c th tm tt nh sau:

- gi s N = 2n

- gi s WN l nghim th N ca n v: WN = e-2j/N v M =

N

2 ta c:

v(k) =

1

2 0

2 1

M n

M

u(n)W2Mnk

- Khai trin cng thc trn ta c:

v(k) =(

1

0

1

M n

M

u(2n)W2M2nk +

1

0

1

M n

M

u(2n+1)W2M(2n+1)k )/2 ()

v W2M2nk = W 2Mnk, do :

v(k) =

1

2 [uchn(n) + ul(n)]

Ch rng v(k) vi k = [0, M-1] l mt DFT trn M = N/2. Thc cht thut ton FFT l

dng nguyn tc chia i v tnh chu k tnh DFT. Vi k = [0, M-1] ta dng cng thc 3.37; vi

k = [M, 2M-1] ta dng php tr trongcng thc 3.37. C th dng thut ton ny c sa i mt

cht tnh DFT ngc. Bn c coi nh mt bi tp.

- Trng hp 2 chiu

Do DFT 2 chiu l tch c nn t cng thc (3.29), ta c:

v(k,l) =

1

0

1

0

1

NW u m nN

km

m

N

n

N

( , )WNln ()

T cng thc 3.38, ta c cch tnh DFT hai chiu nh sau:

- Tnh DFT 1 chiu vi mi gi tr ca x (theo ct)

- Tnh DFT 1 chiu theo hng ngc li (theo hng) vi gi tr thu c trn .

4.2. Cc b lc lm mtnh trong min tn s - Lc thng thp

Cc ng bin v nhiu trong nh tp trung nhiu vo phn tn s cao ca php bin i

Fourier ca n. Do , lm trn nh bng phng php min tn s ta c th loi b cc thnh

phn tn s cao trong bin i Fourier ca nh

Nhc li l:

Trong F(u,v) l bin i Fourier ca nh c lm trn. Vn l la chn mt hm lc H(u,v)sao cho t c G(u,v) bng cch lm suy gim cc thnh phn c tn s cao ca F(u,v). Bin i

Fourier ngc G(u,v) ta c nh c lm trn g(x,y). V cc thnh phn tn s cao b loi b, v


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36thng tin trong vng tn s thp c cho qua, nn phng php ny gi l lc thng thp

(lowpass filtering). Di y l mt vi hm lc thng dng.

4.2.1. Lc l tng

Lc thng thp 2D l tng, vit tt ILHF, c hm lc

Trong D0> 0 l hng s cho trc, gi l ngng hay tn s ct, v

L khong cch t gc ta (0,0) n im (u,v). Thut ng l tng biu th tt c cc thnh

phn tn s nm trong hnh trn bn knh D0c gi nguyn, trong khi tt c cc tn s ngoi

ng trn hon ton b suy gim.

Ch rng, trong chng ny cc hm lc i xng qua gc. iu ny da trn gi thit gc

ca php bin i Fourier t ti tm ca hnh vung N x N trong min tn s.

Tn s ct D0c chn ty theo chng ta mun gi li bao nhiu phn trm ca ph cng sut

ton phn.

Trong P(u,v) l ph cng sut. S phn trm gi li v gi tr D0lin h vi nhau bi:

4.2.2 Lc Butterworth

Lc thng thp Butterworth bc n c hm lc

Hay ci bin

4.3. Cc b lc lm sc nt nh trong min tn s - Lc thng cao.

Ta bit rng, nh c th b nhe do lm suy gim cc thnh phn tn s cao trong bin iFourier ca n. V cc phn t bin v nhng cho thay i t ngt khc trong mc xm tng ng

cc thnh phn tn s cao, vic lm nt nh c th thc hin trong min tn s bng phng php


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37lc thng cao (highpass filtering): lm suy gim cc thnh phn tn s thp nhng khng ph

hy thng tin tns cao trong bin i Fourier.

4.3.1. Lc l tng.

Lc thng cao 2D l tng, vit tt ILHF, c hm lc

Trong D0 > 0

4.3.2 Lc Butterworth

Lc thng cao Butterworth bc n c hm lc

Hay ci bin

CU HI N TP

1. So snh s ging v khc nhau ca nng cao cht lng nh trong min khng gian v min tn

s.

2. Trnh by cc b lc lm mt nh trong min tn s?

3. Trnh by cc b lc lm sc nt nh trong min tn s?


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38

CHNG V: NN NH

5.1. Cc kin thc c bn

a)Nn d liu (Data Compression)

Nn d liu nhm lm gim lng thng tin dtha trong d liu gc v do vy, lng thng

tin thu c sau khi nn thng nh hn d liu gc rt nhiu. Vi d liu nh, kt qu thng l10:1. Mt s phng php cn cho kt qu cao hn. Theo kt qu nghin cu c cng b gn y

ti Vin K thut Georfie, k thut nn fratal cho t s nn l 30 trn 1Ngoi thut ng d

liu, do bn cht ca k thut ny n cn c mt s tn gi khc nh : gim d tha, m ha

nh gc.

T hn hai thp k nay, c rt nhiu k thut nn c cng b trn cc ti liu v nn v

cc phn mm nn d liu xut hin ngy cng nhiu trn thng trng. Tuy nhin, cha c

phng php nn no c coi l phng php vn nng (Universal) v n ph thuc vo nhiuyut v bn cht ca d liu gc. Trong chng ny, chng ta khng th hy vng xem xt tt c cc

phng php nn. Hn th na, cc k thut nn d liu chung c trnh by trong nhiu ti

liu chuyn ngnh. y, chng ta ch cp cc phng php nn c c th ring cho dliu

nh.

T l nn (Compression Rate)

T l nn l mt trong cc c trng quan trng nht ca mi phng php nn. Tuy nhin, v cch

nh gi v cc kt qu cng b trong cc ti liu cng cn quan tm xem x t.Nhn chung, ngi tanh ngha t l c bn ca phng php nn. Nhiu khi t l nn cao cngcha th ni phng

php hiu qu hn cc phng php khc, v cn cc chi ph nh thigian, khng gian v thm

ch c phc tp tnh ton na. Th d nh nn phc v trong truyn d liu: vn t ra l hiu

qu nn c tng hp vi ng truyn khng.Cng cn phn bit d liu vi nn bng truyn.

Mc ch chnh ca nn l gim lngthng tin d tha v dn ti gim kch thc d liu. Tuy

vy, i khi qu trnh nn cng lmgim bng truyn tn hiu s ha thp hn so vi truyn tn hiu

tng t.

b) Cc loi dtha dliu

Nh trn ni, nn nhm mc ch gim kch thc d liu bng cch loi b d tha d liu.

Vic xc nh bn cht cc kiu d tha d liu rt c ch cho vic xy dng cc phng php nn

d liu khc nhau. Ni mt cch khc, cc phng php nn d liu khc nhau l do sdng cc

kiu d tha khc nhau. Ngi ta coi c 4 kiu d tha chnh :

- S phn b k t :

Trong mt dy k t,c mt s k t c tn sut xut hin nhiu hn so vi cc dy khc. Do

vy, ta c th m ha d liu mt cch c ng hn. Cc dy k t c tn sut cao c thay bi


39/70

39mt t m nh phn vi s bt nh; ngc li cc dy c tn sut xut hin thp s c m ha

bi t m c nhiu bt hn. y chnh l bn cht ca phng php m ha t ha Huffman.

- S lp li ca cc k t :

K thut nn dng trong trng hp ny l thay dy lp bi dy mi gm hai thnh phn: s

ln lp v k hiu dng m. Phng php m ha kiu ny c tn l m ha lot di RLC (Run

Length Coding).

-Nhng mu s dng tn sut:

C th c dy k hiu no xut hin vi tn sut tng i cao. Do vy, c th m ha bi t

bt hn. y l c s ca phng php m ha kiu t in do Lempel-Ziv a ra v c citin vo

nm 1977, 1978 v do c tn gi l phng php nn LZ77,LZ78. Nm 1984, TeryWelch ci

tin hiu qu hn v t tn l LZW (Lempel-Ziv-Welch).

- d tha v tr:

Do s ph thuc ln nhau ca d liu, i khi bit c k hiu (gi tr) xut hin ti mtv tr,

ng thi c th on trc s xut hin ca cc gi tr cc v tr khc nhau mt cch ph hp.

Chng hn, nh biu din trong mt li hai chiu, mt s im hng dc trong mt khi d liu

li xut hin trong cng v tr cc hng khc nhau. Do vy, thay v lu tr d liu, ta ch cn lu

tr v tr hng v ct. Phng php nn da trn s d tha ny gi l phng php mha d on.

c) Phn loi phng php nn

C nhiu cch phn loi cc phng php nn khc nhau. Cch th nht da vo nguynl nn.

Cch ny phn cc phng php nn thnh hai h ln:

-Nn chnh xc hay nn khng mt thng tin: h ny bao gm cc phng php nn msau

khi gii nn ta thu c chnh xc d liu gc.

-Nn c mt thng tin: h ny bao gm cc phng php m sau khi gii nn ta khngthu

c d liu nh bn gc. Phng php ny li dng tnh cht ca mt ngi, chp nhn mts vn

xon trong nh khi khi phc li. Tt nhin, cc phng php ny ch c hiu qu khi m vn

xon chp nhn c bng mt thng hay vi dung sai no y.

Cch phn loi th hai da vo cch thc thc hin nn. Theo cch ny, ngi ta cngphn thnh hai h:

- Phng php khng gian (Spatial Data Compression): Cc phng php thuc h nythc

hin nn bng cc tc ng trc tip ln vic ly mu ca nh trong min khng gian.

- Phng php s dng bin i (Transform Coding): gm cc phng php tc ng ln s

bin i ca nh gc m khng tc ng trc tip nh h trn.

C mt cch phn loi khc na, cch phn loi th ba, da vo trit l ca s m ha.

Cch ny cng phn cc phng php nn thnh hai h:- Cc phng php nn th h th nht: Gm cc phng php m mc tnh ton l n

gin, th d vic ly mu, gn t m,.v.v.


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40- Cc phng php nn th h th hai: da vo bo ha ca t l nn.

5.2. Nn nh khng mt thng tin

a) Phng php m ha lot diPhng php m ha lot di lc u c pht trin dnh cho nh s 2 mc: mc en (1), v

mc trng (0) nh cc vn bn trn nn trng, trang in, cc bn v k thut.Nguyn tc ca phng

php l pht hin mt lot cc bt lp li, th d nh mt lot ccbt 0 nm gia hai bt 1, hay ngc

li, mt lot bt 1 nm gia hai bt 0. Phng php ny ch chiu qu khi chiu di dy lp ln hn

mt ngng no . Dy cc bt lp gi l lot hay mch (run). Tip theo, thay th chui bi mt

chui mi gm 2 thng tin: chiu di chui v bt lp (k t lp). Nh vy, chui thay th s c

chiu di ngn hn chui cn thay.Cn lu rng, i vi nh, chiu di ca chui lp c th ln

hn 255. Nu ta dng 1byte m ha th s khng . Gii php c dng l tch cc chui

thnh hai chui: mtchui c chiu di 255, chui kia l s bt cn li.

Phng php RLC c s dng trong vic m ha lu tr cc nh Bitmap theo dng PCX,

BMP.

Phng php RLC c th chia thnh 2 phng php nh: phng php dng chiu di tm c

nh v phng php thch nghi nh kiu m Huffman. Gi s cc mch gm M bits. tin trnh

by, tM = 2m1. Nh vy mch c c thay bi mch mi gm m bits. Vi cch thc ny, mi

mch u c m ha bi t m c cng di. Ngi ta cngtnh c, viM= 15,p = 0,9, ta

s c m = 4 v t s nn l 1,95.Vi chiu di c nh, vic ci t thut ton l n gin. Tuy

nhin, t l nn s khng ttbng chiu di bin i hay gi l m RLC thch nghi.

b) Phng php m ha Huffman

Nguyn tc

Phng php m ha Huffman l phng php da vo m hnh thng k. Da vo d liu gc,

ngi ta tnh tn sut xut hin ca cc k t. Vic tnh tn sut c thc hin bi cch duyt tun

t tp gc t u n cui. Vic x l y tnh theo bit. Trong phng php ny ngi ta gn cho

cc k t c tn sut cao mt t m ngn, cck t c tn sut thp t m di.Ni mt cch khc,

cc k t c tn sut cng cao c gn m cng ngn v ngc li. R rang vi cch thc ny, ta

lm gim chiu di trung bnh ca t m ha bng cch dng chiu di bin i. Tuy nhin,

trong mt stnh hung khi tn sut l rt thp, ta c th khng c li mt cht no, thm ch cn

b thit mt t bit.

Thut ton

Thut ton bao gm 2 bc chnh:


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41- Giai on th nht: tnh tn sut ca cc k t trong d liu gc: duyt tp gc mt

cch tun t t u n cui xy dng bng m. Tip sau l sp xp li bng m theo th t

tnsut gim dn.

- Giai on th hai: m ha: duyt bng tn sut t cui ln u thc hin ghp 2 phnt

c tn sut xut hin thp nht thnh mt phn t duy nht. Phn t ny c tn sut bng tng 2tn

sut thnh phn. Tin hnh cp nht li bng v ng nhin loi b 2 phn t xt. Qu trnh

c lp li cho n khi bng ch c mt phn t. Qu trnh ny gi l qu trnh to cy m

Huffman v vic tp hp ctin hnh nh mt cy nh phn 2 nhnh. Phn t c tn sut thp

bn phi, phn t kia bn tri. Vi cch to cy ny, tt c cc bit d liu/k t l nt l; cc nt

trong l cc nt tng hp. Sau khi cy to xong, ngi ta tin hnh gn m cho cc nt l. Vic

m ha rt n gin: mi ln xung bn phi ta thm 1 bit 1 vo t m; mi ln xung bn tri ta

thm mt bit 0. Tt nhin c th lm ngc li, ch c gi trn m thay i cn tng chiu di l

khng i. Cng chnh do l do ny m cy c tn gi l cy m Huffman nh trn gi. Qu

trnh gii nn tin hnh theo chiu ngc li kh n gin. Ngi ta cng phi da vo bng m to

ra trong giai on nn (bng ny c gi li trong cu trc ca tp nn cng vi d liu nn). Th

d, vimt tp d liu m tn sut cc k t cho bi.

c) Phng php LZW

M u

Khi nim nn t in c Jacob Lempel v Abraham Ziv a ra ln u tin vo nm 1997,

sau pht trin thnh mt h gii thut nn t in LZ. Nm 1984, Terry Welch ci tin gii

thut LZ thnh mt gii thut mi hiu qu hn v t tn l LZW. Phng php nn tin da

trn vic xy dng t in lu cc chui k t c tn sut lp li cao v thay th bng t m tng

ng mi khi gp li chng. Gii thut LZW hay hn cc gii thut trc n k thut t chc t

in cho php nng cao t l nn.

Gii thut nn LZW c s dng cho tt c cc loi file nh phn. N thng c dung

nn cc loi vn bn, nh en trng, nh mu, nh a mc xm v l chun nn cho cc dng nh

GIF v TIFF. Mc hiu qu ca LZW khng ph thuc vo s bt mu ca nh.Phng php

Gii thut nn LZW xy dng mt t in lu cc mu c tn sut xut hin cao trong nh. T

in l tp hp nhng cp vng v gha ca n. Trong , vng s l cc t mc sp xp

theo th t nht nh. gha l mt chui con trong d liu nh. T in c xydng ng thi

vi qu trnh c d liu. S c mt ca mt chui con trong t in khng nh rng chui

tng xut hin trong phn d liu c. Thut ton lin tc tra cu v cp nht t in sau mi

ln c mt k t d liu u vo.


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42Do kch thc b nh khng phi v hn v m bo tc tm kim, t in ch gii

hn 4096 phn t dng lu ln nht l 4096 gi tr ca cc t m. Nh vy di ln nht ca

t m l 12 bits (4096 = 212). Cu trc t in nh sau:

+ 256 t m u tin theo th t t 0255 cha cc s nguyn t 0255. y l m ca

256 k t c bn trong bng m ASCII.

+ T m th 256 cha mt m c bit l m xa (CC Clear Code). Mc ch vic dng

m xa nhm khc phc tnh trng s mu lp trong nh ln hn 4096. Khi mt nh c quan

nim l nhiu mnh nh, v t in l mt b t in gm nhiu t in con. C ht mt mnh nh

ngi ta ligi mt m xa bo hiu kt thc mnh nh c, bt u mnh nh ming thi

khi to li t in cho mnh nh mi. M xa c gi tr l 256.

+ T m th 257 cha m kt thc thng tin (EOI End Of Information). M ny c gi tr

l 257. Nh chng ta bit, mt file nh GIF c th cha nhiu nh. Mi mt nh s c m ha

ring. Chng trnh gii m s lp i lp li thao tc gii m tng nh cho n khi gp m kt thc

thng tin th dng li.

+ Cc t m cn li (t 258 n 4095) cha cc mu thng lp li trong nh. 512 phn t

u tin ca t in biu din bng 9 bit. Cc t m t 512 n 1023 biu din bi 10 bit, t 1024

n 2047 biu din bi 11 bit v t 2048 n 4095 biu din bi 12 bit.

V d minh ha c ch nn ca LZW

Cho chui u vo l ABCBCABCABCD (M ASCII ca A l 65, B l 66, C l 67) T in ban u gm 256 k t c bn.


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43

5.3. Nn nh c mt thng tin

Tuy bn cht ca cc phng php nn da vo bin i rt khc vi cc phng php trnh

by trn, song theo phn loi nn, n vn c xp vo vo h th nht. V c cc c th ringnn chng ta xp trong phn ny.

a) Nguyn tc chung

Cc phng php m ha da vo bin i lm gim lng thng tin d tha khng tcng

ln min khng gian ca nh s m tc ng ln min bin i. Cc bin i c dng y l cc

bin i tuyn tnh nh: bin i KL, bin i Fourrier, bin i Hadamard, Sin, Cosinvv

V nh s thng c kch thc rt ln, nn trong ci t ngi ta thng chia nh thnh cc

khi ch nht nh.Thc t, ngi ta dng khi vung kch thc c 16x16. sau bin itng

khi mt cch c lp.

Chng ta bit, dng chung ca bin i tuyn tnh 2 chiu l:

-x(k,1) l tn hiu vo

- a(m,n,k,1) l cc h s ca bin i l phn t ca ma trn bin iA.

Ma trn ny gi l nhn ca bin i. Cch xc nh cc h s ny l ph thuc vo tngloi

bin i s dng. i vi phn ln cc bin i 2 chiu, nhn c tnh i xng v tch c :

A[m,n,k,1] = A[,k] A[,1]


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44Nu bin i l KL th cc h s chnh l cc phn t ca vct ring.

b) Thut ton m ha dng bin i 2 chiu

Cc phng php m ha dng bin i 2 chiu thng c 4 bc sau:

B1. Chia nh thnh khi

- nh c chia thnh cc khi nh kch thc kx 1 v bin i cc khi mt cch

c

lp thu c cc khi Vi, i=0,1,,B viB = MxN/(k x1).

B2. c nh phn phi bit cho tng khi

- Thng cc h s hip bin ca cc bin i l khc nhau. Mi h s yu cu lng

ha

vi mt s lng bit khc nhau.

B3. Thit kb lng ha

- Vi phn ln cc bit i, cc h s v(0, 0) l khng m. Cc h s cn li c trung

bnh 0.

tnh cc h s, ta c th dng phn b Gauss hay Laplace. Cc h s c m ha bi s bit

khc nhau, thng t 1 n 8 bit. Do vy cn thit k 8 b lng ha. d ci t, tn hiu vo v1

(k, l) c chun ha c dng:

Trc khi thit k b lng ha, ngi ta tm cch loi b mt s h s khng cn thit. B4. M ha

- Tn hiu u vo ca b lng ha s c m ha trn cc t bit truyn i hay lu

tr li. Qu trnh m ha da vo bin i c th c tm tt trn hinh 7.4

-Nu ta chn php bin i KL, cho phng php s c mt s nhc im: khi lng

tnh ton s rt ln v phi tnh ma trn hip bin, tip sau l phi gii phng trnh tm tr ring v

vct ring xc nh cc h s. V l do ny, trn thc t ngi ta thch dng cc bin i khc

nh Hadamard, Haar, Sin v Cosin. Trong s bin i ny, bin i Cosin thng hay c dng

nhiu hn.

c) M ha dng bin i Cosin v chun JPEG


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45* Php bin i Cosin mt chiu

- Php bin i Cosin ri rc (DCT) c Ahmed a ra vo nm 1974. K t n

nay n c ng dng rt rng ri trong nhiu phng php m ha nh khc nhau nh hiu sut

gnnh ti u ca n i vi cc nh c tng quan cao gia cc im nh ln cn. Bin i

Cosin ri rc c s dng trong chun nh nn JPEG v nh dng phim MPEG.

Ph bi i Ci chiu

Php bin i Cosin ri rc mt chiu c nh ngha bi:

Khi dy u vox(n) l thc th dy cc h sX(k) cng l s thc. Tnh ton trn trngs thc

gim i mt na thi gian so vi bin i Fourier. t c tc bin i tha mn yu cu

ca cc ng dng thc t, ngi ta ci tin k thut tnh ton v a ra nhiu thut ton bin i

nhanh Cosine. Mt trong nhng thut ton c gii thiu di y.

* Php bin i Cosin nhanh

Php bin i Cosin nhanh vit tt l FCT (Fast Cosine Transform), da vo tng a biton ban u v t hp cc bi ton bin i FCT trn cc dy con. Vic tin hnh bin i trn cc

dy con s n gin hn rt nhiu so vi dy gc. V th, ngi ta tip tc phn nh dy tn hiu

cho n khi ch cn mt phn t.

Gii thut bin i Cosin nhanh khng thc hin trc tip trn dy tn hiu u vo x(n) m thc

hin trn dy() l mt hon v cax(n). Gi thit s im cn tnh FCT l ly thaca 2:N=2M.

D liu u vo s c sp xp li nh sau:

Nh vy, na u dy() l cc phn t ch s chn cax(n) xp theo chiu tng dn ca ch s.

Na sau ca() l cc phn t ch s l cax(n) xp theo chiu gim dn ca ch s.

Thay vo cng thc Cosin ri rcta c:

Rt gn biu thc:


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46

ChiaX(k) ra lm hai dy, mt dy bao hm cc ch s chn, cn dy kia gm cc ch s l.

Phn ch s chn

C th chuyn v dng:

Thut ton bin i nhanh Cosin c th m t bng cc bc sau:

Bc 1: Th dy h Cij.

Xc nh s tngM = log2 N

Tng hin thi m=1

Bc 2: u M hc hi bc 5. u khg k hc.

(Cha ht cc khi trong mt tng)

Bc 3: Khi hi hi k = 0.

Bc 4: u k


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47

CHNG VI: X L HNH THI NH

6.1. Php gin nh v php co nh nh phn

Vi nh nh phn, mc xm ch c 2 gi tr l 0 hay 1. Do vy, ta coi mt phn t nh nh

mt phn t l gc v c th p dng cc ton t hnh hc (morphology operators) da trn khi

nim bin i hnh hc ca mt nh bi mt phn t cu trc (structural element).

Phn t cu trc l mt mt n dng bt k m cc phn t ca n to nn mt m -tp.

Ngi ta tin hnh r mt n i khp nh v tnh gi tr im nh bi cc im ln cn vi m-tp

ca mt n theo cch ly hi hay ly tuyn. Hnh di y , ch ra mt phn t cu trc v cch ly

hi hay tuyn:

0 1 0 0 1

0 1 1 0 0 0

0 1 0 0 1

a) Phn t cu trc b) mt vng nh

0 1 0 0 1

0 1 1 0 0 0

0 1 0 0 1

c) Tuyn d) Hi

Hnh 4.16. Ci thin nh nh phn

Da vo nguyn tc trn, ngi ta s dng 2 k thut: dn nh (dilatation) v co nh

(erosion).

Dn nh nhm loi b im en b vy bi cc im trng. Trong k thut ny, mt ca s

N+1 x N+1 c r i khp nh v thc hin i snh mt pixel ca nh vi (N+1)2 -1 im ln

cn (khng tnh im tm). Php i snh y thc hin bi php tuyn lgc. Thut ton bin

i c tm tt nh sau:

For all pixels I(x,y) do

Begin

. Tnh FOR(x,y) {tnh or l gc }

- if FOR(x,y) then ImaOut(x,y)


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48

6.2. Php m nh v php ng nh nh phn

Co nh l thao tc i ngu ca gin nh nhm loi b im trng b vy bi cc im en.

Trong k thut ny, mt ca s (N+1) x (N+1) c r i khp nh v thc hin snh mt pixel ca

nh vi (N+1)2 -1 im ln cn. Snh y thc hin bi php hi lgc. Thut ton bin i c

tm tt nh sau:

For all pixels I(x,y) do

Begin

. Tnh FAND(x,y) {Tnh v l gc}

- if FAND(x,y) then ImaOut(x,y)


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49

Hnh 4-15 Khuch i bi lp 2 x 2

Phng php ni suy tuyn tnh

Trc tin, hng c t vo gia cc im nh theo hng. Tip sau, mi im nh

dc theo ct c ni suy theo ng thng. Th d vi khuch i 2x2, ni suy tuyn tnh theo

hng s tnh theo cng thc:

v1(m,n) = u(m,n)

v1(m,2n+1) = u(m,n) + u(m,n+1) (4-14)

vi 0 m M-1, 0 n N-1

v ni suy tuyn tnh ca kt qu trn theo ct:

v1(2m,n) = v1(m,n)

v1(2m+1,n) = v1(m,n) + v1(m+1,n) (4-15)

vi 0 m M-1, 0 n N-1.

Nu dng mt n:

4

1

2

1

4

12

11

2

14

1

2

1

4

1

H

ta cng thu c kt qu trn.

Ni suy vi bc cao hn cng c th p dng cch trn. Th d, ni suy vi bc p (p nguyn), ta

chn p hng cc s 0 , ri p ctcc s 0. Cui cng, tin hnh nhn chp p ln nh vi mt n H

trn [1].

6.4. X l hnh thi nh xm

ChpChn hng

0,

1 1 3 3 2 2

1 1 3 3 2 2

4 4 5 5 6 6

4 4 5 5 6 6

1 3 2

4 5 6

1 0 3 0 2 0

0 0 0 0 0 0

4 0 5 0 6 0

0 0 0 0 0 0


50/70

50Thng thng, trong phn tch nh, ta mun nghin cu t m mt vng c th trong nh,

ta gi l vng quan tm (Region of Interest-ROI).

lm iu , ta cn cc thao tc chnh sa cc to khng gian ca nh, v chng c

xp vo loi cc thao tc hnh hc nh.

Cc thao tc hnh hc nh trnh by y bao gm: ct xn, zoom, phng to (enlarge), thunh (shrink), tnh tin, v quay.

Qu trnh ct xn nh l qu trnh chn ra mt phn nh ca nh, mt nh con, v ct n ra

khi phn cn li ca nh.

Sau khi ta ct xn ra mt nh con t nh gc, ta c th zoom to n bng cch phng to

n. Qu trnh zoom ny c th c thc hin theo mt s cch thc khc nhau, nhng thng

thng l c thc hin x l bc 0 hoc bc nht (zero- or first-order hold ).

Mt x l bc 0 c thc hin bng cch lp li cc gi tr pixel trc , do to ra mt

hiu ng khi (block).

m rng kch thc nh bng x l bc nht, ta cn thc hin mt php ni suy tuyn

tnh gia cc pixel k nhau. Trn 2.2-1 l kt qu so snh gia hai phng php ny ln nh.

Hnh 2.2-1. Cc phng php Zoom

Bc 0: Vic thc thi x l bc 0 l r rng trongkhi x l bc nht phc tp hn. Cch d


51/70

51dng nht l tm gi tr trung bnh gia hai pixel v s dng n lm gi tr cho pixel xen vo

gia; ta c th lm iu ny cho cc hng trc (sau n cc ct), nh sau:

828

484

848

85258

46864

86468

Hai pixel u tin trong hng th nht c ly trung bnh (8 + 4)/2 = 6, v s ny c

chn vo gia hai pixel ny. Cng vic ny c lp li cho tt c cc cp pixel trong mi hng.

Tip theo ta thc hin cho cc ct vi cng cch thc, ta c:

8525846864

86468

8525865.555.56

46864

66666

86468

Phng php ny cho php m rng c mt nh kch thc N x N thnh nh c kch

thc (2N - 1) x (2N - 1).

Bc 1: Mt phng php khc (x l bc nht) cho php thu c cng kt qu trn l mt

qu trnh ton hc c gi l php cun.

Vi phng php cun dng m rng nh, c hai bc cn tin hnhm rng nh bng

cch chn thm cc hng v cc ct ton 0 vo gia cc hng v ct hin c v thc hin php

cun.

Thc hin bc 1 m rng nh bng chn thm 0 nh sau:

943

672

753

0000000

9004030

0000000

6007020

0000000

7005030

0000000

Tip theo thc hin bc 2, ta s dng mt mt n cun, n s c trt ln ton b nh

m rng, v thc hin mt php ton s hc n gin ti mi v tr ca pixel. Mt n cun cho

php x l bc nht c chn l:


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52

4

1

2

1

4

1

2

11

2

1

4

1

2

1

4

1

Qu trnh cun thc hin bng cch chng mt n ln nh, nhn cc cp gi tr trng khp

ln nhau, v cng tt c li. iu ny tng ng vi vic tm tch v hng ca vc t mt n vi

nh con ngay pha di. Chng hn, nu ta t mt n ln gc tri trn ca nh, ta thu c tnh

ton (th hin y l theo th t t tri qua phi, v t trn xung di):

000

030

000

nhn v hng vc t

4

1

2

1

4

1

2

1

12

1

4

1

2

1

4

1

=

=30

4

10

2

10

4

10

2

1310

2

10

4

10

2

10

4

1

Ch rng, gi tr hin c ca nh vn cha thay i. Bc tip theo l ta trt mt n sang

phi mt pixel v lp li qu trnh tnh ton nh trn, ta c:

000

503

000

*

4

1

2

1

4

1

2

11

2

1

4

1

2

1

4

1

=

=40

410

210

415

21013

210

410

210

41

Ch rng, y ton t cun thc hin mt php ly trung bnh 2 pixel lng ging. Qu

trnh ny tip tc cho n cui hng, mi ln t kt qu tnh ton c vo v tr tng ng vi tm

ca mt n. Khi kt thc mt hng, mt n c trt xung v v u hng tip theo v qu trnh

lp li theo tng hng cho n th tc thc hin xong cho ton b nh; qu trnh trt, nhn, ly

tng trn c gi l php cun (xem hnh 2.2-2). Ch rng nh u ra phi c t trong mt

mng nh ring khc, gi l b m, do cho cc gi tr hin c khng b ghi trong qu trnhcun. Nu ta gi mt n cunl M(r, c) v nh l I(r, c), th phng trnh cun c cho bi:


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53

x y

yxMycxrI ),(),(

Tnh tin v quay nh:

Hai thao tc ng ch khc trong hnh hc nh ROI l tnh tin v quay.

Qu trnh tnh tin c th c thc hin vi cc phng trnh sau:

0

0

'

'

ccc

rrr

trong r v c l cc to mi, r v c l cc to ban u, v r0 v c0 l cc khong

cch cn dch chuyn (tnh tin) nh.

Qu trnh quay cn s dng cc phng trnh sau:

cossin

sincos

crc

crr

trong cr , l cc to mi, r v c l cc to ban u, l gc quay. c nh ngha

theo chiu quay kim ng h tnh t trc honh trong nh m gc to nm gc tri trn.

C hai qu trnh quay v tnh tin c th c t hp vo thnh cng mt phng trnh nh

sau:

cos)(sin)(

sin)(cos)(

00

00

ccrrc

ccrrr

vi r' v c' l cc to mi, r, c, r0, c0 v nh nh ngha trn.

C mt s kh khn thc t khi p dng trc tip cc phng trnh trn. Khi tnh tin, s lm

g vi khng gian tha. Nu ta dch chuyn mi th trn hng i xung, th ta s t ci g vo

hng trn cng? C hai tu chn c bn: t y hng trn cng vi mt gi tr khng i, thng l

en (0) hoc trng (255), hoc qun li bng cch dch hng di y ln trn cng, nh hnh Hnh

2.2-3. Php quay cng tng t, nh hnh 2.2-4a minh ho nh c th c quay rotated off"screen" (mt phng nh). Mc d iu ny c th c nh li bng cch tnh tin tr li tm (Hnh

2.2-4b, c), nhng ta vn c nhng khng gian tha cc gc. Ta c th t y khng gian ny bng

hng s, hoc ct ra phn trung tm l phn hnh ch nht ca nh ri m rng ra kch thc nh

ban u.


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54

Hnh 2.2-3. Tnh tin

Hnh 2.2-4. Php quay

CU HI N TP

1. Trnh by php gin nh. Cho v d? Nhn xt kt qu?

2. Trnh by php co nh. Cho v d? Nhn xt kt qu?

3. Trnh by php m nh? Cho v d? Nhn xt kt qu?

4. Trnh by php ng nh? Cho v d? Nhn xt kt qu?


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55

CHNG VII: PHN ON NH

7.1. Pht hin tnh khng lin tc

Phn vng nh l bc then cht trong x l nh. Giai on ny nhm phn tch nh thnh

nhng thnh phn c cng tnh cht no da theo bin hay cc vng lin thng. Tiu chun

xc nh cc vng lin thng c th l cng mc xm, cng mu hay cng nhm... Trc ht cn

lm r khi nim "vng nh" (Segment) v c im vt l ca vng. Vng nh l mt chi tit, mt

thc th trng ton cnh. N l mt tp hp cc im ccng hoc gn cng mt tnh cht no :

mc xm, mc mu, nhm Vng nh l mttrong hai thuc tnh ca nh. Ni n vng nh l

ni n tnh cht b mt. ng bao quanh mtvng nh (Boundary) l bin nh. Cc im trong

mt vng nh c bin thin gi tr mc xmtng i ng u hay tnh kt cu tng ng.

Da vo c tnh vt l ca nh, ngi ta c nhiu k thut phn vng : phn vng da theo min

lin thng gi lphn vng da theo min ng nht hay min k ; phn vng da vobin gi l

phn vng bin. Ngoi ra cn c cc k thut phn vng khc da vo bin , phnvng da theo

kt cu.

a) Phn vng nh theo ngng bin

Cc c tnh n gin, cn thit nht ca nh l bin v cc tnh cht vt l nh : tng

phn, truyn sng, mu sc hoc p ng ph.Nh vy, c th dng ngng bin phn

vng khi bin ln c trng cho nh.Th d, bin trong b cm bin nh hng ngoi cth phn nh vng c nhit thp hay vngc nhit cao. K thut phn ngng theo bin

rt c li i vi nh nh phn nh vn bnin, ha, nh mu hay nh X-quang.

Vic chn ngng rt quan trng. N bao gm cc bc :

- Xem xt lc xm ca nh xc nh cc nh v cc khe. Nu nh c dng rn

ln (nhiu nh v khe), cc khe c th dng chn ngng.

- Chn ngng tsao cho mt phn xc nh trc ca ton b s mu l thp hnt.

- iu chnh ngng da trn lc xm ca cc im ln cn.- Chn ngng theo lc xm ca nhng im tha mn tiu chun chn. Th d, vi

nh c tng phn thp, lc ca nhng im c bin Laplace g(m,n) ln hn

gi trtnh trc (sao cho t5% n 10% sim nh vi Gradient ln nht scoi nh

bin) scho php xc nh cc c tnh nh lng cc tt hn nh gc.

- Khi c m hnh phn lp xc sut, vic xc nh ngng da vo tiu chun xc sut

nhm cc tiu xc sut sai s hoc da vo mt s tnh cht khc ca lut Bayes.

7.2. Phn on nh da vo cc vng nh con


56/70

56K thut phn vng nh thnh cc min ng nht da vo cc tnh cht quan trng no

ca min nh. Vic la chn cc tnh cht ca min s xc nh tiu chun phn vng. Tnh ng

nht ca mt min nh l im ch yu xc nh tnh hiu qu ca vic phn vng. Cc tiuchun

hay c dng l s thun nht v mc xm, mu sc i vi nh mu, kt cu si vchuyn ng.

Cc phng php phn vng nh theo min ng nht thng p dng l :

- Phng php tch cy t phn

- Phng php cc b

- Phng php tng hp

a) Phng php tch cy t phn

V nguyn tc, phng php ny kim tra tnh ng n ca tiu chun ra mt cch tng

th trn min ln ca nh. Nu tiu chun c tha mn, vic phn on coi nh kt thc. Trong

trng hp ngc li, chia min ang xt thnh 4 min nh hn. Vi mi min nh, p dng mt

cch quy phng php trn cho n khi tt c cc min u tha mn iu kin.

Phng php ny c th m t bng thut ton sau :

Procedure PhanDoan(Mien)

Begin

If i ag khg ha The

Begin

Chia i ag hh 4 i : Z1, Z2, Z3, Z4

For i=1 to 4 do PhanDoan (Zi)

End

Else exit

End

Tiu chun xt min ng nht y c th da vo mc xm. Ngoi ra, c th da vo lch

chun hay chnh gia gi tr mc xm ln nht v gi tr mc xm nh nht. Gi s Max vMin

l gi tr mc xm ln nht v nh nht trong min ang xt. Nu :

|MaxMin| < T(ngng)

ta coi min ang xt l ng nht. Trng hp ngc li, min ang xt khng l min ng nht

v s c chia lm 4 phn.

Thut ton kim tra tiu chun da vo chnh lch max, min c vit :

Function Examin_Criteria(I, N1, M1, N2, M2, T)

/* i hi h c i a 255 c .

(1, M1), (2, M2) a i u v i cui ca i; T gg. */Begin

1. Max=0 ; Min=255


57/70

572. For i = N1 to N2 do

If I[i,j] < Min

Then Min=I[i,j] ;

If I[i,j]


58/70

58

Da theo nguyn l ca phng php ni, ta c 2 thut ton :

- Thut ton t mu (Blob Coloring) : s dng khi nim 4 lin thng, dng mt ca s di

chuyn trn nh so snh vi tiu chun ni.

- Thut ton quy cc b: s dng phng php tm kim trong mt cy lm tng kch

thc vng.c) Phng php tng hp

Hai phng php ni (hp) v tch u c nhc im. Phng php tch s to nn mt

cu trc phn cp v thit lp mi quan h gia cc vng. Tuy nhin, n thc hin vic chia qu chi

tit. Phng php hp cho php lm gim s min lin thng xung ti thiu, nhng cu trc hng

ngang dn tri, khng cho ta thy r mi lin h gia cc min.V nhc im ny, ngi ta ngh

nphi hp c 2 phng php. Trc tin, dungphng php tch to nn cy t phn, phn

on theo hng t gc n l. Tip theo, tinhnh duyt cy theo chiu ngc li v hp cc vngc cng tiu chun. Vi phng php ny tathu c mt cu trc nh vi cc min lin thng c

kch thc ti a.

Gii thut tch hp gm mt s bc chnh sau:

1. Ki a iu chu g h.

a) Nu khng tha mn tiu chun ng nht v s im trong mt vng nhiu hn 1, tch

vng nh lm 4 min (trn, di, phi, tri) bng cch quy. Nu kt qu tch xong v khng tch

c na chuyn sang bc 2.

b) Nu tiu chun ng nht tha mn th tin hnh hp vng v cp nht li gi tr trung

bnh ca vng cho vng ny.

2. H vg

Kim tra 4 ln cn nh nu trn. C th c nhiu vng tha mn. Khi , chn vng ti u nht

ri tin hnh hp.

CU HI N TP CHNG 7

1. Th no l vng nh ? Mc ch ca phn vng nh l g ? 2. Th no l phn vng nh theo ngng bin ? Cho v d ?

3. Th no l phn vng nh theo min ng nht ? Cho v d ?


59/70

594. Trnh by phng php tch cy t phn phn vng nh ?

5. Trnh by phng php hp phn vng nh ?


60/70

60

MT S THI MU


61/70

61Trng i Hc Hng Hi Vit Nam

Khoa Cng ngh Thng tinB MN HTHNG THNG TIN

-----***-----

THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: 2010- 2011 thi s: K duyt :

1Thi gian: 75 pht

Cu 1: (2 im) (Chn 4 cu bt k)

- Ti sao cn phi x l nh s. ng dng ca x l nh. Cho v d?

- Nu cch biu din nh s trn my tnh?

- S ha nh l g? Ti sao cn phi s ha nh?

- Khi nim v mt n? Cch s dng mt n trong x l nh?

- Nu khi nim v mc xm (Gray level). Cho v d?

- Nu khi nim v di xm ca mt nh s? Cho v d?


- Trnh by hiu bit ca bn v b lc lm sc nt trong min khng gian?

- Trnh by hiu bit ca bn v php x l hnh thi: ng nh nh phn. ng dng trongthc t?

- Trnh by v b lc Median Filters?

- Nhn xt s ging v khc nhau gia 2 cch s ly Histogram: Histogram Equalization v

Histogram Matching?- Trnh by phng php m ha theo thut ton Shanno-Fano?

- Trnh by phng php m ha theo thut ton Huffman?

- Trnh by php gin nh. Cho v d? Nhn xt kt qu?

- Trnh by php co nh. Cho v d? Nhn xt kt qu?Cu 3: (2 im)

X l lm bng histogram (Histogram Equalization) cho vng nh sau:Cho nh s:

2 3 3 24 2 4 33 2 3 52 4 2 4

Gray Scale [0..7]V biu minh ha histogram trc vsau x l ca vng nh cho.

Cu 4: (2 im)

Cho vng nh sau:2 3 5 35 9 3 49 1 2 93 3 12 8

Mt n kch thc 3x3

1 1 1

1 8 11 1 1

Thc hin lm mt vng nh cho vi b lc trung bnh.


62/70

62

Trng i Hc Hng Hi Vit NamKhoa Cng ngh Thng tin

B MN H THNG THNG TIN-----***-----

THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: 2010- 2011

thi s: K duyt :

2Thi gian: 75 pht





- Khi nim v mt n? Cch s dng mt n trong x l nh?- Nu khi nim v mc xm (Gray level). Cho v d?






-

Nhn xt s ging v khc nhau gia 2 cch s ly Histogram: Histogram Equalization vHistogram Matching?

- Trnh by phng php m ha theo thut ton Shanno-Fano?





5 3 3 2

4 2 4 35 3 5 82 4 5 4

Gray Scale [0..9]V biu minh ha histogram trc v sau x l ca vng nh cho.

Cu 4: (2 im)

Cho vng nh sau:2 3 5 55 5 3 8

9 3 2 53 6 3 9

Mt n kch thc 3x3

1 1 1

1 -8 11 1 1

Thc hin lm sc nt vng nh vi b lc cho (Laplacian Filter)


63/70

63



THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: 2010- 2011

thi s: K duyt :

3Thi gian: 75 pht





- Khi nim v mt n? Cch s dng mt n trong x l nh?- Nu khi nim v mc xm (Graylevel). Cho v d?






- Nhn xt s ging v khc nhau gia 2 cch s ly Histogram: Histogram Equalization vHistogram Matching?






5 5 3 54 5 4 33 5 3 55 4 5 4


Cu 4: (2 im)

Cho vng nh sau:2 3 5 55 9 3 89 1 2 43 3 12 9

Mt n kch thc 3x3

1 2 1

2 4 21 2 1



64/70

64



THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: 2010 - 2011

thi s: K duyt :

4Thi gian: 75 pht





- Khi nim v mt n? Cch s dng mt n trong x l nh?- Nu khi nim v mc xm (Gray level). Cho v d?








- Trnhby phng php m ha theo thut ton Huffman?




3 3 3 44 3 4 33 5 3 53 4 5 5


Cu 4: (2 im)

Cho vng nh sau:2 3 5 55 7 3 89 9 9 43 3 6 9

Mt n kch thc 3x3

1 1 1

1 -8 11 1 1

Thc hin lm sc nt vng nh vi b lc cho (Laplacian Filter)


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65



THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: 2010 - 2011

thi s: K duyt :

5Thi gian: 75 pht





- Khi nim v mt n? Cch s dng mt n trong x l nh?

- Nu khi nim v mc xm (Gray level). Cho v d?







- Trnh by phng php m hatheo thut ton Shanno-Fano?




X l lm bng histogram (Histogram Equalization)cho vng nh sau::Cho nh s:

5 3 3 54 5 4 33 3 3 54 4 5 4


Cu 4: (2 im)

Cho vng nh sau:2 3 5 55 9 3 8

9 1 2 43 3 12 9

Mt n kch thc 3x3

2 1 2

1 4 1

2 1 2



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Khoa Cng ngh Thng tinB MN H THNG THNG TIN

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THI KT THC HC PHNTn hc phn: NHN DNG & X L NHNm hc: x

this: K duyt :

xx

Thi gian: 75 pht

Cu 1: (2 im)

Trnh by hiu bit ca bn v cc php bin i mc xm ca nh: theo hm logarit, theohm m?

Cu 2: (2 im)

- Trnh by hiu bit ca bn v b lc trung bnh (b lc lm mt tuyn tnh trong minkhng gian)?

- Trnh by hiu bit ca bn v b cc i (b lc lm mt phi tuyn trong min khnggian)?

Cu 3: (2 im)

a) Php tr nh thng c s dng trong cc ng dng cng nghip pht hin cc thnhphn cn thiu trong sn xut sn phm. Cch lm l lu tr mt nh chun ca mt snphm c sn xut ng n; nh ny sau c thc hin php tr t cc nh ca ccsn phm khc c sn xut. Trong trng hp l tng, kt qu ca php tr s bng 0nu cc sn phm c sn xut ng n. Kt qu i vi cc sn phm b thiu thnh phns khc khng ti nhng vng tng ng vi thnh phn cn thiu. The b, iu ki c c cch y c h hc hi c g hc h? ii hch?

b) Trong mt ng dng, ngi ta s dng mt n ca b lc trung bnh gim nhiu trn nhban u. Sau ngi ta s dng mt n Laplacian nng cao cht lng ca cc chi titnh trong nh.u a gc h ca cc ha c y h k qu c c gi guyhay khg? ii hch?

Cu 4: (2 im)Cho nh s:

0 2 6 43 5 1 71 7 3 62 6 5 4

Mt n kch thc 3x3:

1/161 2 12 4 21 2 1

Thc hin lm mt nh s dng b lc trung bnh c trng s vi mt n trn.Cu 5: (2 im)

Cho nh nh phn:

0 0 0 0 0 0

0 1 1 1 1 0

0 0 1 0 0 0

0 1 1 1 1 0

0 0 1 1 0 0

0 0 0 0 0 0

Phn t cu trc:

0 1 0

1 1 1

0 1 0

Thc hinphp ngnh nh phn vi phn t cu trc trn.


67/70



-----***-----

THI KT THC HC PHN

Tn hc phn: NHN DNG & X L NHNm hc: x

thi s: K duyt :

xx

Thi gian: 75 pht

Cu 1: (2 im)

Trnh by hiu bit ca bn v cc php bin i: to m bn nh, tng tng phn nh?Cu 2: (2 im)

a. Trnh by hiu bit ca bn v b lc trung v (b lc lm mt phi tuyn trong minkhng gian)?

b. Trnh by hiu bit cabn v b lc lm sc nt trong min khng gian?Cu 3: (2 im)

a. Php tr nh thng c s dng trong cc ng dng cng nghip pht hin ccthnh phn cn thiu trong sn xut sn phm. Cch lm l lu tr mt nh chun camt sn phm c sn xut ng n; nh ny sau c thc hin php tr t ccnh ca cc sn phm khc c sn xut. Trong trng hp l tng, kt qu ca phptr s bng 0 nu cc sn phm c sn xut ng n. Kt qu i vi cc sn phm

b thiu thnh phn s khc khng ti nhng vng tng ng vi thnh phn cn thiu.The b, iu ki c c cch y c h hc hi c g hc h?ii hch?

b. Trong mt ng dng, ngi ta s dng mt n ca b lc trung bnh gim nhiu trnnh ban u. Sau ngi ta s dng mt n Laplacian nng cao cht lng ca c cchi tit nh trong nh.u a gc h ca cc ha c y h k qu c c

gi guy hay khg? ii hch?Cu 4: (2 im)

Cho nh s:

0 2 6 43 5 1 71 7 3 62 6 5 4

Mt n kch thc 3x3:

1/91 1 11 1 11 1 1

Thc hin lm mt nh s dng b lc trung bnhkhng c trng svi mt n trn.Cu 5: (2 im)

Cho nh nh phn:

0 0 0 0 0 0

1 1 1 1 1 1

0 1 1 1 1 0

1 1 1 1 1 1

0 1 1 1 1 0

0 0 0 0 0 0

Phn t cu trc:

0 1 0

1 1 1

0 1 0

Thc hin php m nh nh phn vi phn t cu trc trn.


68/70



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THI KT THC HC PHN


thi s: K duyt :

xx

Thi gian: 75 pht

Cu 1: (2 im)

a) Trnh by hiu bit ca bn v qu trnh s ho nh?b) Trnh by hiu bit ca bn v cc im ln cn ca im nh?

Cu 2: (2 im)

a. Trnh by hiu bit ca bn v php x l hnh thi: gin nh nh phn?b. Trnh by hiu bitca bn v k thut phn on nh: pht hin ng thng?

Cu 3: (2 im)

a. Php tr nh thng c s dng trong cc ng dng cng nghip pht hin ccthnh phn cn thiu trong sn xut sn phm. Cch lm l lu tr mt nh chun camt sn phm c sn xut ng n; nh ny sau c thc hin php tr t ccnh ca cc sn phm khc c sn xut. Trong trng hp l tng, kt qu ca phptr s bng 0 nu cc sn phm c sn xut ng n. Kt qu i vi cc sn phm

b thiu thnh phn s khc khng ti nhng vng tng ng vi thnh phn cn thiu.The b, iu ki c c cch y c h hc hi c g hc h?ii hch?

b. Trong mt ng dng, ngi ta s dng mt n ca b lc trung bnh gim nhiu trnnh ban u. Sau ngi ta s dng mt n Laplacian nng cao cht lng ca c cchi tit nh trong nh.u a gc h ca cc ha c y h k qu c c

gi guy hay khg? ii hch?Cu 4: (2 im)

Cho nh s:

0 2 6 43 5 1 71 7 3 62 6 5 4

Mt n kch thc 3x3:

1/60 1 01 2 10 1 0

Thc hin lm mt nh s dng b lc trung bnh c trng s vi mt n trn.Cu 5: (2 im)

Cho nh nh phn:

0 0 0 0 0 0

0 1 1 1 1 0

0 0 1 0 0 0

0 1 1 1 1 0

0 0 1 1 0 0

0 0 0 0 0 0

Phn t cu trc:

1 1 1

1 1 1

1 1 1

Thc hinphp ngnh nh phn vi phn t cu trc trn.


69/70



-----***-----

THI KT THC HC PHN


thi s: K duyt :

xx

Thi gian: 75 pht

Cu 1: (2 im)

Trnh by hiu bit ca bn v php tr nh v php trung bnh nh?Cu 2: (2 im)

a. Trnh by hiu bit ca bn v php x l hnh thi: ng nh nh phn?b. Trnh by hiu bit ca bn v k thut phn on nh: pht hin im phn bit

Cu 3: (2 im)a. Php tr nh thng c s dng trong cc ng dng cng nghip pht hin cc

thnh phn cn thiu trong sn xut sn phm. Cch lm l lu tr mt nh chun camt sn phm c sn xut ng n; nh ny sau c thc hin php tr t ccnh ca cc sn phm khc c sn xut. Trong trng hp l tng, kt qu ca phptr s bng 0 nu cc sn phm c sn xut ng n. Kt qu i vi cc sn phm

b thiu thnh phn s khc khng ti nhng vng tng ng vi thnh phn cn thiu.The b, iu ki c c cch y c h hc hi c g hc h?ii hc

17411 - Bai giang Xu ly anh

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