3. Tang Cuong Anh

7/31/2019 3. Tang Cuong Anh

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1

CHNG 3

TNG CNG NHTRONG MIN KHNG GIAN

Ths. Phm Vn TipKhoa Cng ngh thng tini hc i Nam

2

NI DUNG BI GING1. Kin thc cs2. Mt s php bin i cp xm cbn

a. Phnh nh - nh m bnb. Php bin i logc. Php bin i ly thad. Cc php bin i tuyn tnh tng phn

3. X l lc xm v cn bng lc xm4. Tng cng nh s dng ton t s hc v logic

a. Php trnhb. Trung bnh nh

5. Cc khi nim cbn v lc khng gian6. Cc b lc khng gian lm trn

Lc tuyn tnh Lc phi tuyn

7. Cc b lc khng gian lm nt Kin thc cs

S dng o hm bc hai ton t Laplacian S dng o hm bc nht ton t Gradient

3

MIN KHNG GIAN VMIN TN S

Min khng gian:L tp hp cc im nh trong nh.Phng php min khng gian l mt th tc tc ng ln ccim nh trong min khng gian .

Min tn s:c biu din theo tn s thng qua cc php bin i.Phng php min tn s da trn php bin i Fourier canh.Tng ng vi cc min ny cc phng php tng cngnh

4

MIN KHNG GIAN

Cc php x l trong min khng gian tc ng trctip ln im nh c k hiu l:

g(x, y) = T[f(x, y)]

f(x, y) l nh u vo.

g(x, y) l nh u ra.

T l ton t tc ng ln f trong ln cn ca im (x,y).


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MT N/B LC

Ln cn ca im (x, y) l hnh vung hoc hnh ch nht ctm l (x, y).

Gc ta

nh f(x, y)

x

y

(x, y)

Ln cn ca im (x, y)trong hnh vung 33

Tm ca ln cn (x, y) di chuyn theo tng im nh bt u tgc tri pha trn (gc ta ). 6

XL IM

Ton tT hot ng ti mi vng ln cn ca v trim nh (x,y) trong nh f cho nh u ra g tng ng.

T tc ng ln vng ln cn c kch thc 11 (tc ng lnim n) g ch ph thuc vo gi tr ca f ti im (x, y), vT trthnh hm bin i cp xm c dng:

s = T(r)

r = f(x, y)

s = g(x, y)

K thut ny c gi l k thut x l im

7

BIN I NG NHT

V d: Hm bin i ng nht cc im nh

r

s=T(r)

m

T(r)

Ti SngTi

Sng

Hm bin i ng nht T(r).nh kt qu c tng phnging vi nh gc.

m

8

TNG TNG PHN

V d: Hm tng cng tng phn ca nh

r

s=T(r)

m

T(r)

Ti SngTi

Sng Hm tng tng phn T(r). nh

kt qu c tng phn cao hnnh gc nhlm ti nhng mcxm nh hn m v tng sngnhng cp xm ln hn m

m


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TNG TNG PHN

V d: Hm tng cng tng phn ca nh

r

s=T(r)

m

T(r)

Ti SngTi

Sng Hm tng tng phn T(r). nh

kt qu c tng phn cao hnnh gc nhlm ti nhng mcxm nh hn m v tng sngnhng cp xm ln hn m

m

10

PHN NGNG

V d: Hm phn ngng

r

s=T(r)

m

T(r)

Ti SngTi

Sng Hm phn ngng T(r) cho kt

qu l nh c hai mc xm (nhnh phn). Nhng im nh c cpxm nh hn m sc quy vmu en, nhng im nh c gitr ln hn hoc bng m c quyv mu trng.

11

XL MT N/B LC

i vi nhng ln cn ln hn 11 vic x l im nh phctp hn nhiu.

Mt ln cn c kch thc ln hn 11 c gi l mt mt n,hoc b lc, hoc mu, hoc ca s.

Cc gi tr trong mt nc gi l cc h s ca mt n.

K thut ny c gi l k thut x l mt nhay k thut lc

12

NI DUNG BI GING

1. Kin thc cs2. Mt s php bin i cp xm cbn





Lc tuyn tnh

Lc phi tuyn7. Cc b lc khng gian lm nt Kin thc cs S dng o hm bc hai ton t Laplacian S dng o hm bc nht ton t Gradient


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MT S PHP BIN I CP XMC BN

Quy c:Cc gi trim nh trc khi x l k hiu l r.

Cc gi trim nh sau khi x l k hiu l s.

r v s quan h vi nhau qua biu thc s = T(r).

r

s

00

T(r)

L-1

L-1 14

MT S PHP BIN I CP XMC BN

Ba loi hm cbn thngs dng tng cng nh.

Php bin i m bn vng nht.

Php bin i logarit (log vlog ngc)

Php bin i ly tha(ly tha bc n v cn bc n)

Cp xm u vo, r

m bn

Log

Cn bc n

ng nht

Ly tha bc n

Log ngc

15

PHNH NH -PHP BIN I M BN

Phnh ca mt nh vi cc cp xm nm trong phm vi [0, L-1] c c bng cch s dng php bin i m bn:

s = L - 1 r

Hnh bn m t php bin i

m bn. m bn

0 L-1L/4 L/2 3L/4

L/4

L/2

3L/4

L-1

Cpxm

ura

Cp xm u vo

16


Hnh di m tnh gc v nh phnh bng cch s dngphp bin i m bn


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Cho nh a cp xm I, vi cc cp xm nm trong on [0, 7].Tm nh m bn ca I. s = 7 - r

0 2 3 4 0 3 5 6 7

5 2 5 6 7 7 0 3 4

2 3 4 1 6 2 1 0 4

7 4 6 2 3 7 1 3 3

2 3 1 0 4 5 6 2 5

7 0 1 2 7 0 0 0 3

4 5 2 4 5 6 7 0 3

2 1 6 3 4 5 6 2 7

3 6 2 5 3 7 0 3 1 18


Cho nh a cp xm I, vi cc cp xm nm trong on [0, 7].Tm nh m bn ca I.

0 2 3 4 0 3 5 6 7

5 2 5 6 7 7 0 3 4

2 3 4 1 6 2 1 0 4

7 4 6 2 3 7 1 3 3

2 3 1 0 4 5 6 2 5

7 0 1 2 7 0 0 0 3

4 5 2 4 5 6 7 0 3

2 1 6 3 4 5 6 2 7

3 6 2 5 3 7 0 3 1

7 5 4 3 7 4 2 1 0

2 5 2 1 0 0 7 4 3

5 4 3 6 1 5 6 7 3

0 3 1 5 4 0 6 4 4

5 4 6 7 3 2 1 5 2

0 7 6 5 0 7 7 7 4

3 2 5 3 2 1 0 7 4

5 6 1 4 3 2 1 5 0

4 1 5 2 4 0 7 4 6

19

CI T THUT TON BIN I M BN

u vo: Ma trn nh u vo X(m,n) c kch thc MxN vmc sng ti a l L.

u ra: Ma trn nh u ra Y(m,n) qua php ly m bn.

Qu trnh thc hin:

for(i=0;i


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PHP BIN I LOG

Php bin i Log nh x mtkhong hp cc gi tr cp xmthp trong nh u vo thnh mt

khong rng hn cc gi tr cpxm ca nh u ra.

Ngc li n nh x mt khongrng cc gi tr cp xm cao trongnh u vo thnh mt khong hphn cc gi tr cp xm ca nhu ra.

22

PHP BIN I LOG NGC

i ngu vi php bin i Log

Php bin i Log ngc nh xmt khong rng cc gi tr cpxm thp trong nh u vo thnhmt khong rng hn cc gi trcp xm ca nh u ra.

Ngc li n nh x mt

khong hp cc gi tr cp xmcao trong nh u vo thnh mtkhong hp hn cc gi tr cpxm ca nh u ra.

23

PHP BIN I LOG

Hnh di l ph Fourier v php bin i log ca vi c = 1s = log (1 + r)

24

PHP BIN I LOG

Cho nh a cp xm I, vi cc cp xm nm trong on [0, 255].Dng bin i s = Log(1+r) tm nh u ra.

10 10 10 10 10 10 10 10

10 20 20 20 20 20 20 10

10 20 130 13 0 1 30 13 0 20 10

10 20 130 25 0 2 50 13 0 20 10

10 20 130 25 0 2 50 13 0 20 10

10 20 130 13 0 1 30 13 0 20 10

10 20 20 20 20 20 20 10

10 10 10 10 10 10 10 10


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PHP BIN I LOG

Cho nh a cp xm I, vi cc cp xm nm trong on [0,255]. Dng bin i s = Log(1+r) tm nh u ra.

10 10 10 10 10 10 10 10

10 20 20 20 20 20 20 10

10 20 1 30 13 0 1 30 130 20 10

10 20 1 30 25 0 2 50 130 20 10

10 20 1 30 25 0 2 50 130 20 10

10 20 1 30 13 0 1 30 130 20 10

10 20 20 20 20 20 20 10

10 10 10 10 10 10 10 10

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 2 2 2 2 1 1

1 1 2 2 2 2 1 1

1 1 2 2 2 2 1 1

1 1 2 2 2 2 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

26

PHP BIN I LY THA

Dng chung ca php bin i ly tha l:s = cr

c, l nhng hng s dng.

Hnh bn ch ra cc cung tng ng

ca php bin i ly tha vi t nh

n ln v c = 1.

Khi c = = 1 Php ng nht.

Cp xm u vo, r

27

(a) nh cng hng tchp xng sng ngi.

(b-d) Kt qu sau khi pdng php bin i theophng trnh s = crvi c= 1 v- = 0.6,

- = 0.4,- = 0.3.

a b

c d

MRI

28

(a) nh chp t trn caomt vng t.

(b-d) Kt qu sau khi pdng php bin i theophng trnh s = crvi c= 1 v- = 3.0,

- = 4.0,- = 5.0.

a b

c d

PHP BIN I LY THA


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25 26 45 18 90 45 54 42

15 2 25 214 97 54 54 120

18 154 14 201 98 65 54 201

19 254 13 201 48 32 24 12

200 2 10 2 54 2 31 47 201 8 120

2 1 2 18 2 17 120 10 2 1 56 5 8 2 1

0 236 208 1 0 12 95 4 36

154 243 201 12 12 65 5 54

Cho nh a cp xm I, vi cc cp xm nm trong on [0,255]. Dng bin i s = r0.3 tm nh u ra.

PHP BIN I LY THA

30

25 26 45 18 90 45 54

15 2 25 214 97 54 54

18 154 14 201 98 65 54

19 254 13 201 48 32 24

200 2 10 2 54 2 31 47 201 8

2 1 2 18 2 17 1 20 1 02 1 56 5 8

0 236 208 1 0 12 95 4

3 3 3 2 4 3 3

2 1 3 5 4 3 3

2 5 2 5 4 3 3

2 5 2 5 3 3 3

5 5 5 5 3 5 2

2 5 5 4 4 5 3

0 5 5 2 2 4 2

Cho nh a cp xm I, vi cc cp xm nm trong on [0,255]. Dng bin i s = r0.3 tm nh u ra.

PHP BIN I LY THA

31

CC PHP BIN I TUYN TNH TNG PHN

Thay i tng phn

Lm mng cp xm

Lm mng mt phng bit

32

THAY I TNG PHN

nh s l tp hp cc im, m mi im c gi tr sngkhc nhau. y, sng mt ngi d cm nhn nh songkhng phi l quyt nh.

Thc t ch ra rng hai i tng c cng sng nhng ttrn hai nn khc nhau s cho cm nhn khc nhau.

Nh vy, tng phn biu din s thay i sng ca nhso vi nn. Ni mt cch khc, tng phn l ni caim nh hay vng nh so vi nn.


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THAY I TNG PHN

Cc nh c tng phn thp c th do cng nh sngkm, hoc do b cm ng khng tt.

tng thay i tng phn l gia tng cc khong cp xmtrong nh.

34

THAY I TNG PHN

V tr ca im (r1, s1),(r2, s2) s quy nh hnh dngca php bin i.

Kt qunh u ra ca phpbin i bn:

xm trong khong (0, 0)n (r1, s1) gim.

xm trong khong t(r1, s1) n (r2, s2) tng.

xm trong khong(r2, s2) n (L-1, L-1) ciu ha.

35

THAY I TNG PHN

V tr ca im (r1, s1),(r2, s2) s quy nh hnhdng ca php bin i.

Kt qunh u ra ca phpbin i bn:

xm trong khong (0, 0)n (r1, s1) tng.

xm trong khong t(r1, s1) n (r2, s2) gim.

xm trong khong(r2, s2) n (L-1, L-1) ciu ha.

36

THAY I TNG PHN

Nu r1 = s1, r2 = s2, lc php bin i trthnh phpng nht. Kt qu ca nhu ra khng thay i so vinh u vo.

(r1,s1)

0 L-

L/ L/ 3L

L/4

L/2

3L/4

L-1

Cp xmu vo

(r2,

s2)


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THAY I TNG PHN

Nu s1 = 0, s2 = L-1, lc php bin i trthnh hmphn ngng. Kt qunhu ra s l nh nh phn.

38

THAY I TNG PHN

Thng thng chng ta sdng hm vi r1 r2 vs1 s2 nh hnh bn gin tng phn. Lc hm ln iu tng.

39

THAY I TNG PHN

nh gcc tng phn thp

nh sau khi gin tng phn.

(r1, s1) = (rmin, 0)(r2, s2) = (rmax, L-1)rmin cp xm nh nhtrmax cp xm ln nht

nh sau khiphn ngng.

(r1, s1) = (m, 0)(r2, s2) = (m, L-1)m l gi tr cp xm trungbnh

40

LM MNG MC XM

Chiu sng cao trong mt di c trng cc cp xm i vimt nh l vic thng xy ra.

Ngi ta thng s dng phng php ny lm ni lnnhng c trng cn quan tm ca nh.

C hai cch tip cn lm mng mc xm:

Hin th mt gi tr cao cho tt c cc mc xm trong di cn quan tmv hin th mt gi tr thp cho nhng mc xm cn li

Hin th mt gi tr cao cho tt c cc mc xm trong di cn quan tm

v gi nguyn cc mc xm cn li.


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LM MNG MC XM

Hin th mt gi tr cao chott c cc mc xm trong dicn quan tm v hin th mtgi tr thp cho nhng mcxm cn li. p dng cho nh

khng nn.

42

LM MNG MC XM

Hin th mt gi tr caocho tt c cc mc xmtrong di cn quan tmv gi nguyn cc mcxm cn li. p dngcho nh c nn. T(r)

0 L-1L/4 L/2 3L/4

L/4

L/2

3L/4

L-1

Cpxm

ura

Cp xm u vo

43

THUT TON LT CT MC XM

u vo: Ma trn nh u vo X(m,n) c kch thc MxN, mc sngti a l L.

u ra: Ma trn nh u ra Y(m,n) qua php ct lt mc xm.

Qu trnh thc hin:

Trng hp c nnfor(i=0;i


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LM MNG MT PHNG BIT

Gi s mi im nh c biu din bi 8 bit. Ta tng rngnh cu to t 8 mt phng 1 bit.

Mt phng bit 0 c ngha t nht

Mt phng bit 7 c ngha nhiu nht

bit thp nht

bit th 0

bit cao nhtbit th 7

46

LM MNG MT PHNG BIT

7

nh nh phn i vi mt phng bit th 7c th ly c bng cch x l nh uvo vi php phn ngng:

nh x tt c cc cp xm t 0 n 127 thnh0.

nh x tt c cc cp xm t 128 n 255thnh 255.

47

LM MNG MT PHNG BIT

7 6

5 4 3 48

LM MNG MT PHNG BIT

2 1 0


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3. Xl lc xm v cn bng lc xm4. Tng cng nh s dng ton t s hc v logic





S dng o hm bc hai ton t Laplacian S dng o hm bc nht ton t Gradient 50

XL LC XM

Lc xm ca mt nh vi cc cp xm nm trong khong[0, L-1] l mt hm ri rc:

h(k) = nk.Trong :

k l mc xm, k = 0, 1, 2, ..., L-1.

nk l s lng im nh trong nh c mc xm bng k.

h(k) l lc xm ca nh vi cp xm bng k.

51

V D LC XM

Cho nh sau I, hy tnh lc xm ca nh I.

0 1 1 3 5 4

5 1 4 7 4 6

3 2 1 3 2 5

1 3 4 5 6 7

0 2 3 4 3 6

k 0 1 2 3 4 5 6 7h(k) 2 5 3 6 5 4 3 2

0 1 2 3 4 5 6 7012

34567

Cpxm

Simnh

52

CHUN HA LC XM

Trong thc t, ta chun ha lc xm bng cch chia gi tr

nk cho tng sim nh trong nh (k hiu l n). V vy, lc

xm chun ha c cho bi cng thc:

p(k) = nk/n.

p(k) l c lng xc sut xy ra cp xm th k.

Tng cc thnh phn ca lc xm chun ha bng 1.

=

=

1

0 1)(

L

k kp


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XL LC XM

Trc honh tng ng vi cc gi tr cp xm k.

Trc tung tng ng vi cc gi tr ca h(k) hoc p(k).

h(k)/p(k)

k

54

XL LC XM

i vi nh ti, lc xm ca nh s phn b ch yu v phagi tr cp xm thp, tc l pha bn tri ca th.

h(k)/p(k)

rk

nh ti

55

h(k)/p(k)

rk

XL LC XM

i vi nh sng, lc xm ca nh s phn b ch yu v phagi tr cp xm cao, tc l pha bn phi ca th.

nh sng

56

XL LC XM

i vi nh c tng phn thp, lc xm ca nh s phnb ch yu cc gi tr cp xm trung bnh, tc l pha gia caca th.

h(k)/p(k)

rk

nh c tng phn thp


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XL LC XM

i vi nh c tng phn cao, lc xm ca nh s phnbu trn tt c cc gi tr cp xm, tc l trn ton th.

h(k)/p(k)

rk

nh c tng phn cao

58

CN BNG LC XM

nh u vo c th:

Ti khng nhn r nt,

Sng m,

tng phn thp kh nhn thy cc i tng.

Chng ta phi x l nh u ra r hn, c nhiu thng tin hn.

Qu trnh x l l nh x mi im nh vi cp xm k trong nhu vo thnh im nh tng ng vi cp xm sk trong nh ura.

59

PHNG PHP CN BNG LC XM Vo: nh I c L cp xm, k=0, 1, ..., L-1. Sim nh trong nh

I l n.

Ra: nh J vi cc cp xm sk [0, L-1], k=0, 1, ..., L-1.

(Lu : nh u ra J c tng phn tt hn nh u vo I)

Phng php:

1. Tnh sim nh c cp xm k l nk (k=0, 1, ..., L-1) trong nh I.

2. Tnh p(k) xc sut xy ra cp xm k trong nh I:

( ) kn

p kn

= k = 0, 1, 2, ..., L-1.

60

PHNG PHP CN BNG LC XM

[ ]( 1) ( )ks round L T k = k = 0, 1, 2, ..., L-1.

4. Cp xm sk nh u ra J tng ng vi cp xm k ca nh u vo I ctnh theo cng thc:

5. Xy dng nh u ra J bng cch nh x tng ng cp xm k ca nh

u vo thnh cp xm sk ca nh u ra.

3. Tnh T(k) l xc sut xy ra cp xm nh hn hoc bng k:

0 0

1( ) ( )

k k

j

j j

T k p j nn= =

= = k = 0, 1, 2, ..., L-1.


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2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4

V D CN BNG LC XM

Lc xm

0 1

1

2

2

3

3

4

4

5

5

6

6

7 8 9

S lng im nh

Cp xm

k 0 1 2 3 4 5 6 7 8 9

nk 0 0 6 5 4 1 0 0 0 0

Cho nh u vo I c 10 cp xm.Cn bng lc xm nh I

62

k 0 1 2 3 4 5 6 7 8 9

nk 0 0 6 5 4 1 0 0 0 0

0 0 6 11 15 16 16 16 16 16

0 06

/16

11

/16

15

/16

16

/16

16

/16

16

/16

16

/16

16

/16

0 03.3

3

6.1

6

8.4

89 9 9 9 9

0

k

j

j

n=

( )T k

V D CN BNG LC XM

9 ( )k

s T k=

63

3 6 6 3

8 3 8 66 3 6 9

3 8 3 8

V D CN BNG LC XM

Lc xm ca nh sau khi cn bng

0 1

1

2

2

3

3

4

4

5

5

6

6

7 8 9

S lng im nh

Cp xm

2 3 3 2

4 2 4 3

3 2 3 5

2 4 2 4

k sk

0 0

1 0

2 3

3 6

4 8

5 9

6 9

7 9

8 9

9 9 64

THUT TON CN BNG HISTOGRAMu vo: Ma trn nh u vo X(m,n) c kch thc MxN, c L mc xm, ccmc xm nm trong khong [a,b].u ra: Ma trn nh u ra Y(m,n) qua php san bng Histogram t cmc sng nm trong khong [a1,b1]Qu trnh thc hin://Tnh Histogramfor(k=0;k


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CN BNG LC XM

nh trc khi cn bng nh sau khi cn bng Lc xm cn bng

66

CN BNG LC XM


67

CN BNG LC XM


68

CN BNG LC XM



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CN BNG LC XM

0 1 2 3 2

2 2 2 3 4

3 0 2 3 3

0 2 0 2 0

1 3 2 2 2

Cn bng lc xm cho nh I c cho di yvi 8 cp xm.

70

V D CN BNG LC XM

Lc xm

k 0 1 2 3 4 5 6 7

nk 5 3 8 8 1 0 0 0

Cho nh u vo I c 8 cp xm.Cn bng lc xm nh I

0 1 2 3 2

3 2 2 3 4

3 0 2 3 3

0 3 0 2 0

1 3 1 2 2

0 1

1

2

2

3

3

4

4

5

5

6

6

7

S lng im nh

Cp xm

78

71

k 0 1 2 3 4 5 6 7

nk 5 3 8 8 1 0 0 0

5 8 16 24 25 25 25 25

5/25 8/25 16/25 24/25 25/25 25/25 25/25 25/25

1.4

12.242

4.484

6.727

7 7 7 7

0

k

j

j

n=

( )T k

V D CN BNG LC XM

7 ( )ks T k=

72

V D CN BNG LC XM

Lc xm

0 1 2 3 2

3 2 2 3 4

3 0 2 3 3

0 3 0 2 0

1 3 1 2 2

0 1

1

2

2

3

3

4

4

5

5

6

6

7

S lng im nh

Cp xm

7

8

1 2 4 7 4

7 4 4 7 7

7 1 4 7 7

1 7 1 4 1

2 7 2 4 4

9k sk

0 1

1 2

2 4

3 7

4 7

5 7

6 77 7


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3. X l lc xm v cn bng lc xm4. Tng cng nh sdng ton ts hc v logic






TON TS HC V LOGIC

Ton t s hc, c 4 ton t cbn Ton t cng + Ton t tr - Ton t nhn * Ton t chia /

Ton t logic, c 4 ton t cbn Ton t phnh NOT Ton t v AND Ton t hoc OR Ton t loi tr XOR

75

TNG CNG NH SDNGTON TS HC V LOGIC

Ton t s hc v logic i hi phi thc hin trn tng imgia hai hay nhiu nh.

Ngoi tr php ton NOT thc hin trn mt nh.

Ton t logic thc hin trn nh cp xm, cc im nh c xl nh l cc s nh phn.

Mu trng c biu din bi s 1 v mu en c biu dinbi s 0.

Ton tNOT ng ngha vi php bin i m bn.

76

V D V TON TAND

nh gc Mt n AND Kt qu khi p dngton t AND


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V D V TON TOR

nh gc Mt n OR Kt qu khi p dngton t OR78

PHP TRNH

S khc bit gia hai nh f(x, y) v h(x, y) c tnh bi cngthc:

g(x, y) = f(x, y) h(x, y)

Nh vy s khc bit ca hai nh chnh l s khc bit tngng ca tng cp im nh.

ng dng ca php trnh c s dng rng ri trong lnhvc y hc tm ra s sai khc ca cc nh X-quang.

79

V D V PHP TRNH

a) nh gc Fractal

b) nh kt qu sau khi thit lp 4mt phng bit thp nht ca nhgc v gi tr 0.

Dng k thut lm mngmt phng bit

nh kt qu b) gnging vi nh gc.

c) S khc bit gia hai nh a) vb) gn nh l mu en.

d) nh sau khi cn bng lc xm ca nh c).

a b

c d

80

TRUNG BNH NH

Xt nh nhiu g(x, y) vi cc thnh phn nhiu (x, y) nm ln

trong nh gc f(x, y) nh sau:

g(x, y) = f(x, y) + (x, y)

Trong : Trung bnh theo cc ta ca nhiu c gi tr bng 0.

gim nhiu trong nh, ta thm vo cc nh nhiu c tp

cc nh nhiu {gi(x, y)}, sau tnh gi tr trung bnh ca tp

nh nhiu ny.


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81

TRUNG BNH NH

Vi K nh nhiu khc nhau gi(x, y), i =1, 2, ..., K, ta c:

1

1( , ) ( , )

K

i

i

g x y g x yK =

= Lc :

{ }( , ) ( , )E g x y f x y=

Theo lut s ln, khi K ln th tim cn n( , )g x y ( , )f x y

Trung bnh nh c ng dng rt nhiu trong lnh vc thinvn hc. Chng hn nh vic chp cc i tng trong di ngnh.

82

V D V TRUNG BNH NH

Hnh nh ca mt thin h, thin

h ny c tn l NGC 3314. y

l nh do knh vin vng khng

gian Hubble ca NASA chp.

Thin h NGC nm cch tri t

khong 140 triu nm nh sng.

83

V D V TRUNG BNH NH

Hnh nh ca thin h NGC 3314

c sa i thng qua php

lm nhiu Gaussian vi trung

bnh cc gi tr nhiu bng 0.

84

V D V TRUNG BNH NH

Hnh (a) n (d) a ra kt quca vic ly trung bnh 8, 16, 64,128 nh nhiu.

Kt qunh trung bnh tc vi K = 128, nh (d) gnging vi nh gc (e).

a b

c d e


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85









TON TTUYN TNH:

Gi X(m,n) l nh vo v Y(m,n) l nh ra ca php x l cdng:

Khi , c gi l ton t tuyn tnh nu tho mn:

(a(X1(m,n))+b(X2(m,n))) = aY1(m,n)+bY2(m,n)

vi Y1(m,n) = (X1(m,n)) v Y2(m,n) = (X2(m,n)).

H tuyn tnh c ngha quan trng l: khi c nhiu tn hiu voh thng c th x l c lp tng tn hiu sau t hp kt qu

li. 86

TON TTUYN TNH V PHP NHNCHP KHNG GIAN

Y(m,n)X(m,n) Php

x l

NHN CHP KHNG GIAN:

a.Nhn chp i vi hm lin tc

- Vi hm mt bin:

Cho f(x) v g(x) l cc hm lin tc, php nhn chp i vihm mt bin c nh ngha v k hiu nh sau:

- Vi hm hai bin

Cho f(x,y) v g(x,y) l cc hm lin tc, php nhn chp i

vi hm hai bin c nh ngha v k hiu nh sau:

87


+

= dttxgtf )()(g(x)*f(x)

=

d)dy,g(x*),f(y)g(x,*y)f(x,

b. Nhn chp i vi tn hiu s

- Tn hiu smt chiu

Cho f(k) v g(n) vi k,n Z l cc tn hiu s mt chiu. Phpnhn chp c nh ngha nh sau:

- Tn hiu shai chiu

Cho f(k,l) v g(m,n) vi m,n,k,l Z l cc tn hiu s haichiu. Php nhn chp c nh ngha nh sau:

88


+

=

=k

kngkfgf )(*)(*

+

=

+

=

=k l

lnkmglkfgf ),(*),(*


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c. Nhn chp i vi nh s

Cho X(m,n) l nh u vo vi m [0,M-1], n [0,N-1] vH(k,l) l mt n ca php nhn chp k[0,K-1], l[0,L-1](thng K=L


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Nhiu trng:

- L nhiu c p h nng lng khng i;

- Cng nhiu trng khng i kh i tn s tng;

-Thng thng nhiu trng c s dng xp x th tpnhiu trong nhiu trng hp;

- H m t tng quan ca nhiu trng l hm del-ta. Nh vynhiu trng khng tng quan ti ha i mu bt k;

- S dng nhiu trng lm hnh nhiu n gin nht v cli v mt tnh ton.

93

KHI NIM TP NHIU

Nhiu Gauss: Ltrng hp c bit; Nhiu Gauss l dng xp x nhiu tt trong nhiu trng

hp thc t;Mt phn b xc sut ca nhiu c c trng bng

hm Gauss; Trong trng hp mt chiu, nhiu Gauss c c trng

bnggi tr trung bnh v lch tiu chun ca binngu nhin( phng sai 2)

94

KHI NIM TP NHIU

( )2

2

2

1( )

2

x

p x e

=

nhchu nh hng ca nhiu Gauss vi tr trungbnh khng v phng sai bng 13:

95

KHI NIM TP NHIU

Mt s dng nh hng nhiu: Nhiu cng:

f( m, n ) = g( m, n ) + ( m, n )Trong nhiu ( m, n ) c lp thng k vi tn hiu; Nhiu nhn: nhiu l h m ca bin tn hiu

f( m, n ) = g( m, n ) + ( m, n )g( m, n ) == g( m, n )( 1+ ( m, n ) ) == g( m, n )n( m, n )

Nhiu xung: khi trn nh xut hin cc im nhiu ring

bit c sng khc bit ln so vi cc im l ncn; Nhiu dng mui tiu: xut hin khi nh b bo ha binhiu xung. Khi nh s b nh hng ca cc imnhiu en trng. 96

KHI NIM TP NHIU


25/44

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V d nhiu dng mui tiu: vi t l nhiu l 1% v5% tng ng. Gi tr ca cc im nh trong khong[0, 255].

97

KHI NIM TP NHIU

98

LC KHNG GIAN

B lc l mt hnh vung hoc hnh ch nht cha cc gi trv hng. B lc cn c gi bng cc thut ng khc nh:mt n, nhn, mu, ca s.

Cc gi tr trong b lc c gi l cc h s ca b lc.

Thng thng ngi ta s dng cc b lc c kch thc l sl nh: 33, 55, ...

w1 w2 w3

w4 w5 w6

w7 w8 w9

V d minh ha b lc c kch

thc 33. w5 l tm ca b lc

99

C CH CA LC KHNG GIAN

Qu trnh x l l qu trnh di chuyn mt n lc theo tng im

mt trong nh.

Ti mi im (x, y), p ng ca b lc ti im c tnh

bng cch s dng cc mi quan h xc nh trc.

Vi b lc khng gian tuyn tnh, p ng chnh l tng ca cc

tch gia h s ca b lc vi cp xm ca im nh tng ng

trong vng p dng lc.

( 1, 1) ( 1, 1) ( 1, 0) ( 1, ) ...

(0, 0) ( , ) ... (1, 0) ( 1, ) (1,1) ( 1, 1)

R w f x y w f x y

w f x y w f x y w f x y

= + +

+ + + + + + +

100

C CH CA LC KHNG GIANGc ca nh

nhf(x, y)

w(-1,-1) w(-1,0) w(-1,1)

w(0,-1) w(0,0) w(0,1)

w(1,-1) w(1,0) w(1,1)

f(x-1,y-1) f(x-1,y) f(x-1,y+1)

f(x,y-1) f(x,y) f(x,y+1)

f(x+1,y-1) f(x+1,y) f(x+1,y+1)

Mt n

x

y


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101

C CH CA LC KHNG GIAN

Thng thng h s w(0, 0) trng khp vi gi tr f(x, y) ca

nh, ngha l mt n c tm ti im (x, y) trong qu trnh tnh

ton.

8 5 6 5f z z z z +

102









103

LC KHNG GIAN TUYN TNH

i vi mt n c kch thc mn, vi m = 2a+1 vn = 2b+1, vi a, b l nhng s nguyn khng m.

Vi iu kin trn, chng ta s c nhng mt n vi kch thcl nhng s l.

Khi , b lc tuyn tnh ca nh f c kch thc MN vi mtn lc mn p dng ti im (x, y) c cho bi cng thc sau:

( , ) ( , ) ( , )a b

s a t b

g x y w s t f x s y t = =

= + + Trong : a=(m-1)/2, b=(n-1)/2.

104


c c nh lc, chng ta phi p dng cng thc trn lnton b cc im nh, tc l x = 0, 1, ..., M vy = 0, 1, ..., N.

Phng trnh trn cn c gi l tch chp ca mt n vi nh.

Mt n b lc nhiu lc cng cn c gi l mt n chp hayl nhn chp.

( , ) ( , ) ( , )a b

s a t b

g x y w s t f x s y t = =

= + +


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105


Chng ta cng c th k hiu p ngR ca nh ti im (x, y)

vi mt n chp c kch thuc mn theo cng thc:

1 1 2 2

1

...mn

mn mn i i

i

R w z w z w z w z=

= + + + =

Trong , cc gi trw l cc h s mt n, cc gi trz l cc

gi tr cp xm ca nh tng ng vi cc h s, mn l s lng

cc h s trong mt n.

106

V D LC KHNG GIAN

Mt n 33 trong hnh trn p ng ti im (x, y) ca nh

c tnh bi cng thc9

1 1 2 2 9 9

1

...i i

i

R w z w z w z w z=

= + + + =

w1 w2 w3

w4 w5 w6

w7 w8 w9

107

XL TRNG HP C BIT

Trng hp thc thi cc php ton ln cn khi tm ca b lc

nm trn bin ca nh?

w1 w2 w3

w4 w5 w6

w7 w8 w9

f(0,0) f(0,1) f(0,2) f(0,3) f(0,4)

f(1,0) f(1,1) f(1,2) f(1,3) f(1,4)

f(2,0) f(2,1) f(2,2) f(2,3) f(2,4)

f(3,0) f(3,1) f(3,2) f(3,3) f(3,4)

f(4,0) f(4,1) f(4,2) f(4,3) f(4,4)

Mt n

108

XL TRNG HP C BIT

Khi tm mt n di chuyn gn n bin ca nh th mt hoc

mt s dng/ct ca mt n s nm ngoi nh.

v d di cc h sw1, w4, w7nm ngoi nh.

w1 w2 w3

w4 w5 w6

w7 w8 w9


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109

XL TRNG HP C BIT

Cch 1: Gi s mt n c kch thc nn.Cho v tr tm ca mt n khng c nh hn (n-1)/2 im nh

k t bin nh sau khi lc c kch thc nh hn nh gc,

nhng tt c cc im nh u c x l.

w1 w2 w3

w4 w5 w6

w7 w8 w9

110

XL TRNG HP C BIT

Cch 2: Yu cu nh kt qu c kch thc bng nh gca thm cc dng m v ct m mang gi tr 0 vo quanh binca nh.

w1 w2 w3

w4 w5 w6

w7 w8 w9

0 0 0 0 0 0 0

0 0

0 0

0 0

0 0

0 0

0 0 0 0 0 0 0

111

LC KHNG GIAN LM TRN

Cc b lc lm trn c s dng khi vt mv khnhiu.

Kh mthng c s dng trong cc bc tin x l nh.

Chng hn xa bi nhng chi tit nh khi nh trc khi trch chn cci tng.

Lp li cc l hng nh ca cc ng thng hay cc cung

Kh nhiu dng khi nhng vt mca nh.

112

B LC TUYN TNH LM TRN

u ra (p ng) ca b lc tuyn tnh lm trn n gin ltrung bnh cc im nh cha trong ln cn ca mt n lc.

Nhng b lc ny c gi l lc trung bnh v lc thngthp.


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113


u ra (p ng) ca b lc tuyn tnh lm trn n gin ltrung bnh cc im nh cha trong ln cn ca mt n lc.

Nhng b lc ny c gi l lc trung bnh v lc thngthp.

114

LC THNG THP

Dng b lc thng thp kh nhiu cng.

Gi s, nh u vo c nhiu cng k hiu l X(m,n). lmtrn nhiu, ta nhn chp nh vi mt n H(k,l), lc ny nh ura c tnh nh sau:

vi r=(L-1)/2

H(k,l) c gi l b lc thng thp.

c im ca b lc thng thp l:

=

=

++=

1L

0k

1L

0l

r)lnr,kl)X(mH(k,n)Y(m,

= =

=

K

i

K

j

jiH1 1

1],[

LC TRUNG BNH Vi b lc ny, kh nhiu ta tnh trung bnh cc im nh trong

mt ln cn nh. Phng php n gin nht l cng tt c cc imnh trong mt mt n lc (ca s lc), sau chia cho tng simnh trong mt n v gi l b lc trung bnh. Vi mt n c kchthc 3x3, ta nhn chp nh vi ma trn sau:

Ngoi ra, khi nhn chp ti tm, nhn mnh n vai tr ca imnh ang xt, ngi ta a ra b lc trung bnh c trng s nh sau:

Lc trung bnh trng s l mt trng hp ring ca lc thng thp.115

=

111

111

111

9

13]][[ lkH

=

111

121

111

1013 ]][[ lkH

116


Bi v nhiu ngu nhin c c trng l cha nhng im nhtrong cc cp xm ng dng r rng nht ca lm trn l khnhiu.

Tuy nhin, cc cnh ca i tng cng c c trng bi ccim nh trong cc cp xm, v vy lc trung bnh s c tcng ph ln cc cnh ca cc i tng trong nh n s lmnhe (m) i cc cnh.

ng dng chnh ca cc lc trung bnh l khi cc chi titkhng thch hp trong nh.


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117


Vi b lc khng gian kch thc 33, th cch sp xp ngin nht l cho cc h s bng 1/9.

B lclm trn33

111

111

111

9

1

f(x-1,y-1) f(x-1,y) f(x-1,y+1)

f(x,y-1) f(x,y) f(x,y+1)

f(x+1,y-1) f(x+1,y) f(x+1,y+1)

1( , ) [ ( 1, 1) ( 1, ) ( 1, 1)

9

( , 1) ( , ) ( , 1)

( 1, 1) ( 1, ) ( 1, 1)]

g x y f x y f x y f x y

f x y f x y f x yf x y f x y f x y

= + + + +

+ + + +

+ + + + + +118


= =

++=

1

1

1

1

),(9

1),(

i j

jyixfyxg

Nh vy, vi b lc trn, p ng ca qu trnh lc l gi trtrung bnh ca cc im nh trong ln cn b lc.

1( , ) [ ( 1, 1) ( 1, ) ( 1, 1)

9

( , 1) ( , ) ( , 1)

( 1, 1) ( 1, ) ( 1, 1)]


f x y f x y f x y

f x y f x y f x y

= + + + +

+ + + +

+ + + + + +

119


Hnh di y l mt b lc c cc h s khc nhau c gi lb lc trung bnh c trng s.

Ta thy h s trung tm ca b lc c gi tr ln hn so vi cch s khc im nh ng vi tm b lc c tm quan trnghn cc im nh khc.

121

242

121

16

1

1( , ) [ ( 1, 1) 2 ( 1, )

16

( 1, 1) 2 ( , 1) 4 ( , )2 ( , 1) ( 1, 1) 2 ( 1, )

( 1, 1)]

g x y f x y f x y

f x y f x y f x yf x y f x y f x y

f x y

= + +

+ + + +

+ + + + + +

+ +B lc lm trn 33

120


Theo phng trnh trn, vic thc hin lc tng qut trn nh ckch thc MN vi b lc trung bnh c trng s, kch thcb lc l mn (m v n l s l), cc h s l ws,t c thc hintheo cng thc sau:

( , ) ( , )

( , )

( , )

a b

s a t b

a b

s a t b

w s t f x s y t

g x y

w s t

= =

= =

+ +

=

Trong :

a = (m-1)/2b = (n-1)/2


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121


nh gc v nh c lc vi kch thc mt n l 33

122



123



124




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125



126

Example

a) nh gc kch thc 500500

b) - f) Kt qu ca vic lm trn vi blc trung bnh hnh vung kch thc n= 3, 5, 9, 15 v 35.

Ch :

Mt n ln c s dng c lngcc i tng nh trong nh.

Kch thc mt nc chn linquan n kch thc ca cc i tng

cn ng nht vi nn.

a bc d

e f

127

V D LC TUYN TNH LM TRN

Cho nh I kch thc 77 v b lc tuyn tnh lm trn c kchthc 33 nh sau. Tm nh c c sau khi lc.

10 15 15 45 25 14 23

12 14 255 12 15 12 45

25 26 25 12 45 255 12

14 48 98 51 12 15 20

12 32 36 34 25 26 2412 14 5 7 54 12 51

14 56 25 14 20 47 12

111

111

111

9

1

128

V D LC TUYN TNH LM TRN

Cho nh I kch thc 77 v b lc tuyn tnh lm trn c kchthc 33 nh sau. Tm nh c c sau khi lc.


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129


Lc trung bnh khng gian s lm mnh.

xa i cc i tng nh c cng sng ln trong nhta cho n c sng ln vi mu nn bng cch lm mnthng qua lc trung bnh khng gian.

Kch thc mt n c quan h mt thit vi kch thc ca cci tng m chng ta cn xa i lm nn.

Sau khi lc xong, ta c th tin hnh phn ngng mt cch hpl lm ni bt cc i tng cn quan st.

130


a) nh c chp t knh vin vng Hubble v cc hnh tinh quay quanh tri t(Ngun NASA)

b) nh kt qu sau khi p dng lc trung bnh vi mt n lc c kch thc1515.

c) nh sau khi phn ngng trn nh b) vi gi tr ngng bng 25% gi trcng im nh sng nht.

a b c

131

LC PHI TUYN

L b lc khng gian phi tuyn vi p ng (u ra) da trnth t (xp loi) cc im nh trong vng c lc, v sau thay th gi tr ca im nh trung tm bng gi tr c xcnh thng qua kt qu xp loi.

Cc b lc theo thng k th t gm:

B lc trung v (Median Filter)

B lc gi trung v (Pseude-Median Filter)

132

LC TRUNG V

Thc cht y l b lc phi tuyn gim nhiu xung. Trong phplc ny, mi im nh u vo sc thay th bi trung v ccim nh trong mt ca s no , tc l:

Y(m,n) = TrungVi {X(m-k,n-l) | (k,l)W}

Trong : W l ca sc chn trc.

Cch thc hin:

tnh gi tr mi ca im nh ti v tr (m,n), ta t mt ca shnh vung c kch thc l (3x3, 5x5, ) ln nh ban u saocho tm ca ca s ny trng vi im (m,n). Gi tr mi ti im(m,n) chnh l phn t trung v ca cc phn t nm trong ca s. Tc l ta phi sp xp cc phn t trong ca s ny theo th ttng dn (hoc gim dn), lc ny phn t trung v chnh l phntgia ca dy sau khi sp xp.


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133

LC TRUNG V

V d: Vi nh u vo c cho nh sau:

nh u ra qua b lc trung v vi ca s 3x3 c tnh nhsau:

- tnh Y[1,1], ly cc phn t trong ca s 3x3 c tm ti vtr (1,1) ta c dy: 1 5 7 6 20 7 4 3 5. Sau khi s p xptng dn ta c dy: 1 3 4 5 5 6 7 7 20. Trung v ca dy lphn t th nm, tc l Y[1,1]=5.

- Cui cng ta c nh u ra nh sau:

=

56475

350649

78534

587206

109751

],[ NMX

=

56475

36659

77764

57756

109751

],[ NMY

134

Ci t minh ho thut ton lc trung v u vo: Ma trn nh u vo X(m,n) c kch thc MxN.

u ra: Ma trn nh u ra Y(m,n) qua b lc trung v vi ca s l KxK.For (i=0;i


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137

Ci t minh ho thut ton lc gi trung v u vo: Ma trn nh u vo X(m,n) c kch thc MxN.

u ra: Ma trn nh u ra Y(m,n) qua b lc gi trung v vi ca s lKxK.

di dy con d.

Qu trnh thc hin:

138

Ci t minh ho thut ton lc gi trung v

//Gi li bin ca nhr=K/2;//Bin ca nhFor (i=r;i


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141

LC TRUNG V

K thut lc trung v dng lc nhiu bng cch trt trn mt

phng nh, mi ln di chuyn qua trn mt im nh.

Nhng phn t trong ca sc xem nh l mt chui {xn} v

im quan tm c thay th bi gi tr trung v ca chui.

V d: Chui {xn} = {1, 2, 9, 4, 5} c trung v = 4. im trung

tm sc thay th bi gi tr = 4. Kt qu ta c: {1, 2, 4, 4,

5}

142

LC TRUNG V

K thut lc trung v thng dng mt n c kch thc l 33,

55.

Vic lc s dng li khi qu trnh lc khng lm thay i kt

qu ca nh cn lc.

Lc trung v vi mt n c kch thc nn c tnh nh chui

mt chiu. Ta tin hnh sp xp dy ri thay th phn t tm

bng trung v ca dy va tm c.

143

LC TRUNG V

V d: Lc trung v trn nh vi ca s lc l 33.

15 17 18

16 78 17

17 15 20

15 15 16 17 17 17 18 20 78

1 2 3 4 5 6 7 8 9

15 17 18

16 17 17

17 15 20

144

V D LC TRUNG V

a) nh gc X-quang chp mt bo mch ca mt thit b (c nhiu).b) nh sau khi lc trung bnh tuyn tnh lm trn vi mt n 33.

c) nh sau khi lc trung v vi mt n 33.

a b c


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145






Lc tuyn tnh Lc thng k th t



LC KHNG GIAN LM NTSHARPING 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.

Lc khng gian lm nt s dng trong cc lnh vc khc nhau

nh to bn in in t, nh y hc, hoc trong cc h thng qun

s.

Cc b lc lm nt ch yu da trn o hm bc nht v o

hm bc hai.

147

O HM

Xem xt o hm trong ng cnh s ha.

Tp trung vo o hm mt chiu nh sau:

Vng cp xm khng thay i (cc phn on phng).

Ti nhng v tr bt u v kt thc khng lin tc (ti bc ln,bc xung khng lin tc).

Dc cc on dc xung cp xm.

148

O HM

C nhiu cch nh ngha o hm.

Chng ta s dng tnh cht sai phn nh ngha o hm.Ngha l o hm bc nht i vi hm mt bin f(x) l saiphn:

( 1) ( )f

f x f xx

= +

Chng ta s dng o hm tng phn tin cho vic xem xti vi hm hai bin f(x, y).


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149

O HM

Tng t, o hm bc hai i vi hm mt bin f(x) l saiphn:

2

2( 1) ( 1) 2 ( )

ff x f x f x

x

= + +

( 1) ( )f

f x f xx

= +

o hm bc nht i vi hm mt bin f(x) l sai phn:

150

151

NHN XT VO HM BC NHTV O HM BC HAI

Tm li, so snh gia o hm bc nht v o hm bc haichng ta c mt s kt lun sau y:

o hm bc nht cho kt qu bin dy hn trong nh.

o hm bc hai p ng mnh hn i vi chi tit mn, chnghn nhon thng mng, im ring l.

o hm bc nht p ng mnh i vi dc ln.

o hm bc hai cho kt qup ng kp ti bc nhy thay i

cp xm. (Chng ta cng ch rng, o hm bc hai p ngmnh i vi on thng hn bc nhy, im nh hn on

thng).

152

NHN XT VO HM BC NHTV O HM BC HAI

o hm bc hai thng ph hp hn so vi o hm bc nhti vi vic tng cng nh bi v kh nng ca o hm bchai tng cng chi tit mn tt hn.


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SDNG O HM BC HAITON TLAPLACIAN

S dng o hm bc hai trn hm hai bin tng cng nh.

nh ngha cng thc ri rc ca o hm bc hai.

Xy dng mt n lc

154


Ton to hm ng hng l ton t Laplacian i vi nhf(x, y) c nh ngha nh sau:

2 22

2 2

f ff

x y

= +

Ton t Laplacian i vi nh f(x, y) l ton t tuyn tnh.

155


Trong : o hm bc hai tng phn theo hng x:

2

2( 1, ) ( 1, ) 2 ( , )

ff x y f x y f x y

x

= + +

2

2( , 1) ( , 1) 2 ( , )

ff x y f x y f x y

y

= + +

o hm bc hai tng phn theo hng y:

156


2 22

2 2

f ff

x y

= +

Thay vo cng thc trn ta c:

2 [ ( 1, ) ( 1, )

( , 1) ( , 1)] 4 ( , )

f f x y f x y

f x y f x y f x y

= + + +

+ +

2

2( 1, ) ( 1, ) 2 ( , )

ff x y f x y f x y

x

= + +

2

2( , 1) ( , 1) 2 ( , )

ff x y f x y f x y

y

= + +


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Theo cng thc trn ta c th xy dng mt n lc L nh sau:

0 1 0

1 -4 1

0 1 0

2 [ ( 1, ) ( 1, )

( , 1) ( , 1)] 4 ( , )

f f x y f x y

f x y f x y f x y

= + + +

+ +

Mt n ny bt bin vi php quay 90o.

158


Theo cng thc trn ta c th xy dng mt n lc L nh sau:

0 -1 0

-1 4 -1

0 -1 0

2 4 ( , ) [ ( 1, )

( 1, ) ( , 1) ( , 1)]

f f x y f x y

f x y f x y f x y

= + +

+ + +


159


Chng ta cng c th xem xt o hm bc hai theo ngcho, khi mt n lc L thu c l:

1 1 1

1 -8 1

1 1 1


2 [ ( 1, 1) ( 1, )

( 1, 1) ( 1, 1)

( 1, ) ( 1, 1)

( , 1) ( , 1)] 8 ( , )

f f x y f x y

f x y f x y

f x y f x y

f x y f x y f x y

= + + + + +

+ + + +

+ +

+ +

160


Chng ta cng c th xem xt o hm bc hai theo ngcho, khi mt n lc L thu c l:

-1 -1 -1

-1 8 -1

-1 -1 -1


2 8 ( , ) [ ( 1, 1)

( 1, ) ( 1, 1)

( 1, 1) ( 1, )

( 1, 1) ( , 1) ( , 1)]

f f x y f x y

f x y f x y

f x y f x y

f x y f x y f x y

= + + +

+ + + +

+ + +

+ + +


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TNG CNG NH VITON TLAPLACIAN

Nu ton t Laplacian c h s tm mt n m th chng ta trnh gc cho i lng nh thu c thng qua vic p dngton t Laplacian (f*L), ngc li ta cng vi i lng .

2

2

( , ) ( , )( , )

( , ) ( , )

f x y f x yg x y

f x y f x y

=

+

Nu h stm mt n m

Nu h stm mt n dng

( , ) ( , )( , )

( , ) ( , )

f x y f x y Lg x y

f x y f x y L

=

+

Nu h stm mt n m

Nu h stm mt n dng

162


Hoc:

( , ) ( , ) [ ( 1, ) ( 1, )

( , 1) ( , 1) 4 ( , )]


f x y f x y f x y

= + +

+ + +

( , ) ( , ) [4 ( , ) ( 1, )

( 1, ) ( , 1) ( , 1)]


f x y f x y f x y

= + +

+

Nu h stm mt n m

Nu h stm mt n dng

163


n gian ngi ta thng ly gi tr tuyt i ca php nhnchp cng vi nh gc.

g f f L= +

164

a) nh chp b mt camt trng

b) nh vi b lcLaplacian

c) nh c thay it l

d) nh sau khi cng nhgc vi nh to ra do

ton t Laplacian

a b

c d


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Ta c th bin i cng thc theo cch sau:

( , ) ( , ) [ ( 1, ) ( 1, )

( , 1) ( , 1) 4 ( , )]

5 ( , ) [ ( 1, ) ( 1, )

( , 1) ( , 1)]


f x y f x y f x y

f x y f x y f x y

f x y f x y

= + +

+ + +

= + +

+ + +

0 -1 0

-1 5 -1

0 -1 0 166

167

CH

0 - 1 0

-1 5 -1

0 - 1 0

0 0 0

0 1 0

0 0 0

2

2

( , ) ( , )( , )

( , ) ( , )

f x y f x yg x y

f x y f x y

=

+

= +

0 - 1 0

-1 4 -1

0 - 1 0

-1 -1 -1

-1 9 -1

-1 -1 -1

0 0 0

0 1 0

0 0 0= +

-1 -1 -1

-1 8 -1

-1 -1 -1

168

SDNG O HM BC NHTTON TGRADIENT

o hm bc nht trong x l nh l c thc hin bng cchs dng ln ca gradient.

Vi hm f(x, y), gradient ca f ti ta (x, y) c nh nghal mt vectct hai chiu:

x

y

f

G xf

fG

y

= =


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ln ca vect cchobi:

x

y

f

G xf

fG

y

= =

12 2 2

12 22

( )x y

f mag f G G

f f

x y

= = +

= +

x yf G G +

Xp x

170


( 1, ) ( , ) ( , 1) ( , )

x y

f ff G G

x y

f x y f x y f x y f x y

+ = +

= + + +

( 1, ) ( , )xG f x y f x y= +

Nh vy:

( , 1) ( , )y

G f x y f x y= +

171


Trong vng 33, s dng k hiu z5 ca mt ntng ng vi im nh f(x, y), im z1 ca mtn tng ng vi im nh f(x-1, y-1), ...

Xp xn gin nht i vi o hm bc nhttha mn iu kin:

z1 z2 z3

z4 z5 z6

z7 z8 z9

( 1, ) ( , )xG f x y f x y= +

( , 1) ( , )yG f x y f x y= +

Gx = (z6 z5) v Gy = (z8 z5).172


Khi :

z1 z2 z3

z4 z5 z6

z7 z8 z9

8 5 6 5( ) v ( )

x yG z z G z z= =

1 12 2 2 22 2

8 5 6 5[ ] [( ) ( ) ]

x yf G G z z z z = + = +

8 5 6 5f z z z z +


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Hai ton t khc do Roberts ngh nh sau:

z1 z2 z3

z4 z5 z6

z7 z8 z9

9 5 8 6( ) v ( )x yG z z G z z= =

1 12 2 2 22 2

9 5 8 6[ ] [( ) ( ) ]x yf G G z z z z = + = +

9 5 8 6f z z z z +

174


Hai ton t khc do Sobel ngh nh sau:

z1 z2 z3

z4 z5 z6

z7 z8 z9

7 8 9 1 2 3

3 6 9 1 4 7

( 2 ) ( 2 )

( 2 ) ( 2 )

x

y

G z z z z z z

G z z z z z z

= + + + +

= + + + +

x yf G G +

175

CH

Tng tt c cc h s trong mt n bng 0. iu ny nhm lmcho p ng ti nhng vng cp xm khng thay i c gi trbng 0.

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