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
-----***-----
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:
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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:
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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
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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
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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:
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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
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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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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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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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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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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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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
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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
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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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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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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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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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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
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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
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572. For i = N1 to N2 do
If I[i,j] < Min
Then Min=I[i,j] ;
If I[i,j]
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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 ?
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594. Trnh by phng php tch cy t phn phn vng nh ?
5. Trnh by phng php hp phn vng nh ?
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MT S THI MU
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61Trng i Hc Hng Hi Vit Nam
Khoa Cng ngh Thng tinB MN HTHNG THNG TIN
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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?
Cu 2: (4 im) (Chn 4 cu bt k)
- 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.
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THI KT THC HC PHN
Tn hc phn: NHN DNG & X L NHNm hc: 2010- 2011
thi s: K duyt :
2Thi 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?
Cu 2: (4 im) (Chn 4 cu bt k)
- 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 vHistogram 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:
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)
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THI KT THC HC PHN
Tn hc phn: NHN DNG & X L NHNm hc: 2010- 2011
thi s: K duyt :
3Thi 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 (Graylevel). Cho v d?
- Nu khi nim v di xm ca mt nh s? Cho v d?
Cu 2: (4 im) (Chn 4 cu bt k)
- 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 vHistogram 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:
5 5 3 54 5 4 33 5 3 55 4 5 4
Gray Scale [0..8]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 9 3 89 1 2 43 3 12 9
Mt n kch thc 3x3
1 2 1
2 4 21 2 1
Thc hin lm mt vng nh cho vi b lc trung bnh.
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THI KT THC HC PHN
Tn hc phn: NHN DNG & X L NHNm hc: 2010 - 2011
thi s: K duyt :
4Thi 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?
Cu 2: (4 im) (Chn 4 cu bt k)
- 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 vHistogram Matching?
- Trnh by phng php m ha theo thut ton Shanno-Fano?
- Trnhby 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:
3 3 3 44 3 4 33 5 3 53 4 5 5
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 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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THI KT THC HC PHN
Tn hc phn: NHN DNG & X L NHNm hc: 2010 - 2011
thi s: K duyt :
5Thi 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?
Cu 2: (4 im) (Chn 4 cu bt k)
- 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 vHistogram Matching?
- Trnh by phng php m hatheo 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:
5 3 3 54 5 4 33 3 3 54 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 9 3 8
9 1 2 43 3 12 9
Mt n kch thc 3x3
2 1 2
1 4 1
2 1 2
Thc hin lm mt vng nh cho vi b lc trung bnh.
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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.
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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.
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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)
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.
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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 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