Phat Hien Muc Tieu Di Dong Su Dung Bo Loc Kalman Tran Nhat Quang

7/31/2019 Phat Hien Muc Tieu Di Dong Su Dung Bo Loc Kalman Tran Nhat Quang

1/87

B Gio Dc V o To Trng i Hc S Phm K Thu t TP H Ch Minh

Khoa Cng Ngh Thng Tin

KHA LUN TT NGHIP

PHT HIN MC TIU DI NGS DNG B LC KALMAN

Sinh vin thc hin:TRN NHT QUANGMSSV:06102070

Gio vin hng dn:THS.TRN TIN C

TP H Ch Minh, 2011


2/87

I HC S PHM K THUT TP.H CH MINH CNG HA X HI CH NGHA VIT NAM

KHOA CNG NGH THNG TIN c lp T do Hnh phc

NHIM V THC HIN KHA LUN TT NGHIP

H tn sinh vin: Trn Nht Quang MSSV: 06102070 Chuyn ngnh: Cng Ngh Thng Tin Tn ti: Pht hin mc tiu di ng sdng b lc Kalman

Ni dung thc hin:

L thuyt:

Tm hiu phng php pht hin vt th da trn mu sc. Tm hiu v b lc Kalman ng dng trong pht hin mc tiu di ng.

Thc hnh:

Xy dng chng trnh pht hin mc tiu di ng da trn phn tch mu sc vs dng b lc Kalman ti u c on.

Kt hp vic theo di i tng vi cc thut ton ha my tnh to ra chcnng vv iu khin nh t camera.

Thi gian thc hin:t2.9.2010 n 1.1.2011

Ch k ca sinh vin: _______________________________________________________

TP H Ch Minh, ngy thng nm 2011

TRNG KHOA CNTT GING VIN HNG DN

(K{ v ghi r h tn) (K{ v ghi r h tn)

ng Trng Sn Trn Tin c


3/87

ti Pht Hin Mc Tiu Di ng S Dng Mt N Kalman

GVHD: ThS. Trn Tin c

SVTH: Trn Nht [email protected] ii

LI CM N

Trong suy ngh ca em, bn nm i hc ca mt sinh vin c tng kt bi n ttnghip. Tu em xc nh s quan trng ca n ny nh th.

Nhn c ti mnh thch l mt may mn. V em c nhn ti ny.

n ny c th l khng qu phc tp so vi nhng thnh tu v tin hc hin nay, nhngvi mt sinh vin nh em th khng th trnh khi nhng kh khn trong qu trnh thchin.

May mn thay em nhn c s gip nhit tnh ca thy Trn Tin c, H S

Phm K Thut TP HCh Minh, gio vin hng dn v cng l c vn hc tp ca emtrong sut nhng nm i hc. Thy khng ch dy em v kin thc m cn cho em nhiuhiu bit qu bu khc. Em rt cm n thy! Xin c gi ti thy li cm n chn thnh!Chc thy lun khe mnh tip tc hng dn thm nhiu lp sinh vin na.

Bn cnh , cng nh nhiu bn sinh vin khc, hon thnh c khng ch n nym cn c kha hc i hc, em cn phi nh vo s quan tm, ng h ht sc ln lao ca

cha m em v mi ngi trong gia nh. Nn ni y, ty lng mnh, em xin c gili cm n chn thnh nht ti cha m em v mi ngi! H lun l nhng ngi m emyu qu. Cu mong nhng iu tt p nht n vi h!

Ngoi ra, khng th khng nh ti cc thy c trong khoa Cng Ngh Thng Tin, khoa oTo Cht Lng Cao, H S Phm K Thut TP H Ch Minh, v cc thy c khc tn tmdy v quan tm gip em cng cc bn trong thi gian qua. Nh h em mi c cnhng kin thc nh hm nay hon thnh n ny. Em xin c cm n cc thy c!

Mt ln na, em cm n mi ngi!

TP HCh Minh, thng 12 nm 2010

Trn Nht Quang


4/87



SVTH: Trn Nht [email protected] iii

NHN XT CA GIO VIN HNG DN

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

TP H Ch Minh, ngy thng nm 2011

Gio vin hng dn


5/87



SVTH: Trn Nht [email protected] iv

NHN XT CA GIO VIN PHN BIN

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

_________________________________________________________________________

TP HCh Minh, ngy thng nm 2011

Gio vin phn bin


6/87



SVTH: Trn Nht [email protected] v

MC LC

NHIM V THC HIN KHA LUN TT NGHIP ................................................................ i

LI CM N ....................................................................................................................... ii

NHN XT CA GIO VIN HNG DN .......................................................................... iii

NHN XT CA GIO VIN PHN BIN .............................................................................. iv

MC LC............................................................................................................................. v

DANH MC HNH............................................................................................................. viii

DANH MC BIU .......................................................................................................... ix

Mu ............................................................................................................................... 1

Vn thc t................................................................................................................. 1

Mc ch, i tng v phm vi nghin cu .................................................................... 1

B cc ca bo co .......................................................................................................... 5

1 Chng 1 Nn Tng .................................................................................................... 6

1.1 Cc khi nim c bn v xc sut .......................................................................... 7

1.1.1 Cng thc tnh xc sut .................................................................................. 7

1.1.2 Xc sut c iu kin ...................................................................................... 8

1.2 Bin ngu nhin .................................................................................................... 9

1.3 Hm xc sut ......................................................................................................... 9

1.4 Hm tch ly ........................................................................................................ 10

1.5 Cc c trng s ca bin ngu nhin ................................................................. 10

1.5.1 Kz vng ........................................................................................................ 10

1.5.2 Phng sai ................................................................................................... 12

1.6 Moment ca bin ngu nhin ............................................................................. 13

1.7 Hip phng sai .................................................................................................. 17

1.7.1 nh ngha.................................................................................................... 17

1.7.2 Tnh cht ...................................................................................................... 18

1.8 Ma trn hip phng sai ..................................................................................... 19

1.9 Phn phi chun (phn phi Gauss) .................................................................... 19


7/87



SVTH: Trn Nht [email protected] vi

1.10 c lng ....................................................................................................... 21

1.11 Nguyn tc trc giao (Orthogonality principle) ................................................ 24

2 Chng 2 B Lc Kalman .......................................................................................... 262.1 Gii thiu v lc .................................................................................................. 27

2.2 Gii thiu b lc Kalman ..................................................................................... 27

2.2.1 Tng quan .................................................................................................... 27

2.2.2 i nt v Rudolf Emil Kalman ..................................................................... 28

2.2.3 ng dng ca b lc Kalman ........................................................................ 29

2.2.4 V d dn nhp ............................................................................................. 29

2.2.5 M hnh tng qut ca c lng dng b lc Kalman ................................ 31

2.3 B lc Kalman ..................................................................................................... 32

2.3.1 Cc k hiu s dng ..................................................................................... 32

2.3.2 Xy dng b lc Kalman ............................................................................... 33

2.3.3 Tm tt cc phng trnh ca b lc Kalman ............................................... 38

2.4 B lc Kalman trong OpenCV .............................................................................. 39

3 Chng 3 Pht Hin Vt Th Da Trn Mu Sc ...................................................... 42

3.1 H mu HSV ........................................................................................................ 43

3.1.1 nh ngha.................................................................................................... 43

3.1.2 Chuyn t mu RGB sang HSV ...................................................................... 44

3.2 Moment ca nh (image moment) ...................................................................... 46

3.3 Thut ton pht hin i tng theo mu sc .................................................... 46

3.4 Ci t thut ton pht hin vt th da theo mu sc vi OpenCV.................... 47

4 Chng 4 Thit KV Ci t Chng Trnh Pht Hin Di ng ............................... 49

4.1 Thit k............................................................................................................... 50

4.1.1 M hnh h thng ........................................................................................ 50

4.1.2 Bo ........................................................................................................... 52

4.2 Ci t................................................................................................................. 53

4.2.1 Hot ng ca chng trnh ........................................................................ 53

4.2.2 Cc on m chnh ....................................................................................... 56

Kt Lun ............................................................................................................................ 70

V l thuyt ................................................................................................................... 70


8/87



SVTH: Trn Nht [email protected] vii

Vng dng .................................................................................................................. 70

Hng pht trin........................................................................................................... 70

Ph Lc.............................................................................................................................. 71Hng dn ci t OpenCV 2.1 cho project Winform CLR, Visual Studio 2008 C++ ........ 71

TI LIU THAM KHO ....................................................................................................... 77


9/87



SVTH: Trn Nht [email protected] viii

DANH MC HNH

Hnh 0-1 Phn mu vng v hng ca cy bt sc theo di .......................................... 2

Hnh 0-2 Theo di i tng ............................................................................................... 2

Hnh 0-3 Dng theo di chuyn ng v hnh t camera. ............................................... 3

Hnh 0-4 nh kt qu ca thao tc v qua camera ............................................................... 3

Hnh 0-5 Zoom nh bng cch iu khin cc vt th .......................................................... 4

Hnh 0-6 Kt qu ca vic zoom nh .................................................................................... 4

Hnh 1.6-1 Cc phn phi xc sut v " nhn". ng gia: phn phi chun. ........... 17Hnh 2.3-1 Tm tt qu trnh lm vic ca lc Kalman ....................................................... 39

Hnh 3.1-1 Hnh nn ngc biu din h mu HSV ............................................................ 43

Hnh 3.1-2 Hnh trn biu din cc sc mu (H: 0-360) v bo ha (S: 0-1). Mu : H=0,

mu xanh l: H=120, mu xanh dng: H=240 .................................................................. 43

Hnh 4.2-1 Ca siu khin ............................................................................................. 53

Hnh 4.2-2 Ca s camera hin kt qu ca chc nng theo di ........................................ 53

Hnh 4.2-3 Ca s camera hin kt qu ca chc nng v ................................................. 54

Hnh 4.2-4 nh kt qu ca thao tc v qua camera trn Khung nhn ................................ 54

Hnh 4.2-5 Zoom nh bng cch iu khin cc vt th ..................................................... 55

Hnh 4.2-6 Kt qu ca vic zoom nh trong Khung nhn ................................................... 55

Hnh PL-0-1 Cu hnh CMake 2.8.3..................................................................................... 71

Hnh PL-0-2 To project CLR Windows Forms Application .................................................. 72

Hnh PL-0-3 Thm item ...................................................................................................... 73

Hnh PL-4 Thm menu ....................................................................................................... 73

Hnh PL-5 Kt ca vic ci t v chy thnh cng mt chng trnh Winform dngOpenCV ............................................................................................................................. 76


10/87



SVTH: Trn Nht [email protected] ix

DANH MC BIU

Biu 1.6-1 Phn phi xc sut ca X hp 1, phng sai ln hn: cc gi tr phn tn xakz vng hn ...................................................................................................................... 15

Biu 1.6-2 Phn phi xc sut ca X hp 2, phng sai nhhn: cc gi tr tp trung

hn quanh kz vng ............................................................................................................ 15

Biu 1.6-3 Moment trung tm bc 3 ln hn 0: lch dng (ui bn phi di hn) .... 16

Biu 1.9-1 th phn phi xc sut ca X ................................................................... 20

Biu 1.9-2 th ca mt s phn phi chun ............................................................. 21

Biu 4.1-1 Mt s phn phi Gauss, vi phng sai 2 cng ln th cc gi tr x c gi trln (xa kz vng) s c xc sut cao hn ............................................................................. 51


11/87


SVTH: Trn Nht [email protected] 1

Mu

Vn thc t

Ngy nay, mi ngi nghe ni ngy cng nhiu v cc t t li c khnng thay thcon

ngi trong vic vn hnh xe, mt chng trnh my tnh siu khin xe. Nhng chic xeny thm ch cn c gii thiu l an ton hn xe li bi con ngi trong mt strnghp.

Trong cc ng dng bo vnh camera an ninh, nhng thit b ny hot ng 24/24. Tuynhin, ch mt vi on phim c quay l c { ngha quan trng, chng hn nh onphim ghi li cnh c ngi t nhp vo hnh lang (ni t camera) vo bui ti. Vy lmsao chng ta c thtm ra on video ? S rt mt thi gian (v km hiu qu) nu ta

ngi xem ton b 24 giphim do camera quay. Cc chng trnh s gip chng ta.

Mt ng dng khc, c th chng ta khng thch th lm, l cc tn la khng ikhng (air-to-air missile: AAM). cc l tn la dn hng c bn t mt my bay tiu dit my bay khc. Tn la dn hng hot ng theo nguyn l pht hin mc tiu(thng thng bng ra a hoc hng ngoi, i khi cng s dng Lazer hoc quang hc)sau tng dn n mc tiu [1].

V cn nhiu ng dng khc na.

Vy lm sao my tnh (cc chng trnh) lm c nhng vic trn? C nhiu th linquan cn phi gii quyt, nhng mt phn quan trng trong l vn pht hin ra ccvt thchuyn ng.

V vic pht hin ra cc i tng di ng c s dng rt rng ri nh thnn c thni ti ny rt c { ngha trong khoa hc ln thc tin.

Mc ch, i tng v phm vi nghin cu

n ny s tm hiu v ci t thut ton pht hin mc tiu di ng da trn phn tchmu sc v s dng b lc Kalman. ng dng c vit s dng th vin OpenCV 2.1 vngn ng lp trnh C++, IDE Visual Studio 2008.

C nhiu thut ton pht hin chuyn ng. Mt s thut ton ph bin c kn sauy:

Lucas-Kanade: phn tch cc b (mt vng nh) tm ra v tr mi ca nhng imc trng (thng l cc im gc (corner)).


12/87

MU



Horn-Schunck: mt thut ton da trn phn tch ton cc, kh chm hn Lucas-Kanade v ngy nay t dng [2].

Mean-Shift: mt k thut phn tch d liu tng qut. c dng trong nhiu lnh

vc khng ch ring trong x l nh [2]. Mt pht trin ca Mean-shift dng cho xl nh l Cam-shift (continuously adaptive mean-shit).

Nhng thut ton ny kh phc tp i vi mt sinh vin i hc.

V vy, mt phng php n gin hn c tm hiu v s dng y. l phngphp pht hin vt th da trn mu sc (Chng 3) kt hp vi b lc Kalman (Chng 2) theo di chuyn ng.

C thhn, chng ta s thc hin mt gii php pht hin v theo di cc i tng cmu sc chnh trc (Hnh 0-2). y i tng l mt phn cy bt c mu hng v

vng (Hnh 0-1).

Hnh 0-1 Phn mu vng v hng ca cy bt sc theo di

Hnh 0-2 Theo di i tng


13/87


14/87

MU



Sau khi theo di c chuyn ng ca vt th, ta s s dng cc thut ton trong ha my tnh to ra mt ng dng v (Hnh 0-3 v 0-4) v iu khin nh t camera (dichuyn v zoom nh) (Hnh 0-5 v 0-6).

Hnh 0-5 Zoom nh bng cch iu khin cc vt th

Hnh 0-6 Kt qu ca vic zoom nh


15/87

MU



ng dng c vit y dng li vi cc chc nng va trnh by. Tuy nhin, vi kinthc v pht hin vt th v b lc Kalman, ni dung l thuyt c tm hiu hon ton cthdng pht trin nhiu ng dng thch hp khc.

B cc ca bo co

Bo co ca n ny gm 2 phn chnh: Phn l thuyt v Phn vit ng dng.

Phn l thuyt gm 3 chng:

Chng 1: Nn Tng. Trnh by v cc ni dung ton v xc sut thng k c linquan ti xy dng b lc Kalman.

Chng 2: B Lc Kalman. Gii thiu v lc, gii thiu b lc Kalman, xy dng blc Kalman, b lc Kalman trong OpenCV 2.1.

Chng 3: Pht Hin Vt ThDa Trn Mu Sc. Trnh by vphng php phthin i tng bng phn tch mu sc v ci t vi OpenCV 2.1.

Phn vit ng dng gm 1 chng:

Chng 4: Thit KV Ci t Chng Trnh. Trnh by v thit kca chng trnhcng nh trch dn v gii thch mt son code quan trng.

Bo co cn c phn Ph lc trnh by Cch ci t v cu hnh OpenCV 2.1 dng phttrin chng trnh ca n ny.


16/87

6

1Chng1Nn Tng

Cc ni dung chnh:

Bin ngu nhin v cc c trng s ca n Hm xc sut v hm tch ly Hip phng sai v ma trn hip phng sai c lng im Nguyn tc trc giao


17/87

CHNG 1.NN TNG



Cc ni dung v xc sut v ton c trnh byy nhm mc ch phc v vic xy dng b lcKalman (Chng 2). Nu bn quan tm n cc vn v kha cnh su hn ca l thuyt xc sut

v ton, bn hy tm c cc ti liu chuyn v xc sut thng k v ton.

1.1 Cc khi nim c bn v xc sut

Trong cuc sng hng ngy, khi thc hin mt vic lm m c nhiu kt qu c th xy ra,ngi ta sngh ti khnng xy ra ca mi kt qu. Xc sut ra i gip chng tanh gi khnng .

Mt v d l vic tung xc xc. Kt qu sl 1 trong 6 trng hp: ta c mt 1 chm hay

2, 3, 4, 5 v 6 chm.

Trong xc sut thng k, vic tung xc xc c gi l mtphp th. Php thn gin lmt hnh ng m ta mun tnh ton khnng xy ra gia cc kt qu ca n.

Kt qu ca php thc gi l mt bin c(hay s kin (event)). Trong v d trn, ta cmt s bin c:

c mt 1 chm

c mt 3 chm

c mt 6 chm

Mi mt bin c c khnng xy ra khc nhau, v ta c mt i lng lxc sutdng do t khnng ny.

V d: nu ta dng mt b bi gm c 52 l, v thc hin php th rt ra mt l t b bi,nh vy c tt c 52 khnng u ra. Nh vy, ta c mt vi bin c v xc sut ca n:

"Rt ra l bi va va en cng mt lc" (0 phn t), xc xut = 0/52 = 0

"L bi rt ra l con 5 c" (1 phn t), xc xut = 1/52

"L bi rt ra l con Gi" (4 phn t), xc xut = 4/52

"L bi rt ra l mt l bi" (52 phn t), xc xut = 52/52 = 1

1.1.1 Cng thc tnh xc sutThng th (nu khng i su vo cc vn phc tp), ta c thtnh c xc sut bngtrc gic, n gin xc sut trong cc trng hp trn l ly t l.

Cng thc cin

Tng qut, ngi ta nh ngha xc sut nh sau:

P(A) = mA / n


18/87

CHNG 1.NN TNG



vi mA l s bin cs cp thun li cho A, tc l, mt cch tng i1, cc bin c m n

xy ra th A xy ra. Cn n l s bin cs cp ng khnng, tc l tt ccc trng hpc th xy ta khi thc hin php th.

Cng thc tnh xc sut theo thng k

nh ngha xc sut c in c nhng hn ch (nhng trng hp khng tnh c xcsut) l:

N ch xt trong trng hp s bin cs cp (n) l hu hn.

V chxt trong trng hp h bin cl ng khnng2.

Do , ngi ta a ra thm nh ngha xc sut theo thng k:

P(A) = lim (m/n), n

trong , n l s ln thc hin php th cn m l s ln bin c A xut hin.

Trong thc t, n khng cn ti v cng, ty theo trng hp m ngi ta chn mt s nln l c. Khi , ta c

P(A) m/n1.1.2 Xc sut c iu kinQuan st lc bn cnh.

Ly ngu nhin mt im M trong hnh E. Gi A lbin c M A v B l bin c M B.

Gi s bin cB xy ra, bng trc gic ta c ththy cc bin c s cp thun li cho bin c Achnh l phn giao gia A v B: A B.

V vy xc sut A xy ra khi B xy ra l:

P(A|B) = S(AB) / S(B), vi S l din tch.

P(A|B) c gi l xc xut c iu kin, c l Xc xut A xy ra khi B xy ra.

C l khng phi ni, P(A|B) v P(A) = S(A) / S(E) khng bng nhau.

Tip theo, chia t v mu cho S(E) ta c:

P(A|B) = [S(AB) / S(E)] / [S(B) / [S(E)] = P(AB) / P(B), vi P(B) > 0

trong AB l bin c xy ra khi A v B ng thi xy ra.

1 gii thch y thno l mt bin cs cp thun li s phi i su vo khi nim bin c. Ta skhng lm iu y. Nu quan tm bn c thc cc ti liu v xc sut.2 H bin c ng kh nng l h bin c m cc bin c xy ra hon ton khng c th t u tin no.

Chng xy ra mt cch hon ton ngu nhin, cng bng.


19/87

CHNG 1.NN TNG



Trn y ch l mt v d, nhng ta s thy cng thc tnh xc sut c iu kin cng gingnh vy.

Mt cch tng qut, ngi ta nh ngha cng thc xc sut c iu kin nh sau:

P(A|B) = P(AB) / P(B), vi P(B) > 0

Ta ni n xc sut c iu kin y v di y ta s cp n khi nim kzvng ciu kin.

Nh vy n y ta c cc khi nim c bn vphp th, bin cv xc sut.

Tuy nhin, cc bin cnh vy vn mi chc miu t bng li, khng th dng trongtnh ton nh lng. V bin ngu nhin ra i gii quyt vn ny.

1.2 Bin ngu nhinBin ngu nhin (random variable) c th hiu n gin l mt nh x. N nh x mt binc vi gi tr s.

V d: Trong php th gieo mt ng xu, ta c th c mt nh xnh sau:

X(bin c) = 0 nu bin c="c mt hnh"1 nu bin c="c mt s"

Mt v d khc l php th giao xc xc, ta c th c nh xnh sau:

X(s chm) = s chm

trong s chm l s chm gieo c ca xc xc.

Mi X trong cc v dtrn c gi l mt bin ngu nhin.

Nh vy, bin ngu nhin khng c ngha l mt bin nh cc bin ton hc khc. Thc

cht n l mt hm s (hay nh x).

Cc kt qu u ra c nh x bi bin ngu nhin c gi l gi tr ca bin ngunhin. Trong v dgieo ng xu, bin ngu nhin X c 2 gi tr 0 v 1. Trong v d tung xcxc, bin ngu nhin X c 6 gi tr 1, 2, 3, 4, 5 v 6.

1.3 Hm xc sut

i vi bin ngu nhin ri rc, hm xc sut, cn gi l hm khi xc sut (ting Anh:

Probability mass function), k hiu l p(x) c nh ngha nh sau:

p(x) =

1 , = 12 , = 23 , = 3 , = 0 ,


20/87

CHNG 1.NN TNG



trong , x1, x2, xn l cc gi tr ca bin ngu nhin, v p1, p2, pn l xc sut ti cc binc c gi trtng ng. p(x) = 0 ti cc x khc v l cc bin c khng th xy ra (gingnh bin cbc c l bi va va en trong b bi).

Hm ny l hm c trng cho mt bin ngu nhin ri rc.

Ni thm: i vi bin ngu nhin lin tc, c mt hm tng ng lhm mt . Vtrc gic, hm mt chnh l hm xc sut c lm mn, tc l cc gi trx gn nhlin tip nhau (xk+1 xk 0). Hm mt c k hiu l f(x) (khc vi hm tch ly F(x)).1.4 Hm tch ly

Hm tch ly cn gi l hm phn phi xc sut, c nh ngha nh sau (chung cho cbin ngu nhin ri rc v lin tc):

F(x) = P[X


21/87

CHNG 1.NN TNG



E(X) = 70 * 1/100 + -1 * 99/100 = -0.29

Nh vy c ngha l trung bnh ct 1 ng, ngi chi sthu v -0.29 ng (mt i0.29 ng).

Ni thm: mt tr chi c kzvng khc 0 c coil tr chi khng cng bng. Trong tr

chi strn, ngi chi l.

1.5.1.1 Cc tnh cht ca kvngKz vng c tnh tuyn tnh, ngha l n c tnh cht sau:

E(aX + bY) = aE(X) + bE(Y), vi X, Y l 2 bin ngu nhin; a, b l 2 hng s

Ni cch khc, n bao gm cc tnh cht sau:

E(c) = c, vi c = hng s

E(c.X) = c.E(X)

E(X + Y) = E(X) + E(Y), vi X, Y l 2 bin ngu nhin

Kz vng khng c tnh nhn, ngha l E(X.Y) khng bng E(X).E(Y). Lng sai khc giaE(X.Y) v E(X).E(Y) l hip phng saicov(X,Y) (sni bn di):

cov(X.Y) = E(X.Y) - E(X).E(Y)

E(X.Y) ch bng E(X).E(Y) khi 2 bin ngu nhin l c lp:

E(X.Y) = E(X).E(Y), vi X v Y l 2 bin ngu nhin c lp

Ta s ni li vn ny trong phn Hip phng sai.

Ngoi ra d thy mt tnh cht na l v mt n vth E(X) c cng n v vi X, v d nu

X c n vl gram th E(X) cng c n v l gram.

1.5.1.2 Kvng c iu kinCho 2 bin ngu nhin X v Y, ta c khi nim kz vng c iu kin c nh ngha nhsau:

E[X|Y](y) = E[X|Y = y] = .P( X=x|Y=y )trong y l mt bin c m bin ngu nhin Y nh x, k hiu Y=y ngha l xy ra bin cy. P( X=x|Y=y ) l xc sut c iu kin (xem mc 2.1.2).

Ni thm: Kzvng lp l mt khi nim da trn kzvng c iu kin:

E( E[X|Y] ) = E(X)

E( E[X|Y] ) gi l kzvng lp.

Chng minh:


22/87

CHNG 1.NN TNG



1.5.1.3 Kvng ca ma trnNu bin ngu nhin X l mt ma trn th kz vng ca n l ma trn cc kz vng ca ccphn t ca X:

E(X) = E( 1,1 1,2 1,2,1 2,2 2, ,1 ,2 , ) = (1,1) (1,2) (1, )(2,1) (2,2) (2, ) ( ,1) ( ,2) ( , )

Tnh cht ny c dng trong ma trn hip phng sai (s cp bn di).

1.5.2 Phng saiPhng sai (variance) ca mt bin s ngu nhin ri rc X c nh ngha nh sau:

var(X) = E[(X - )2]

trong l kz vng ca X, = E(X).

Mt cch trc gic, nu coi kz vng l trung bnh, th phng sai chnh l khong cch

bnh phng trung bnh t cc gi tr ca X ti kz vng. Hay ni cch khc, nu cc gi tr(ca bin ngu nhin X) cng phn tn (nm xa) gi tr trung bnh (kz vng) th var(X) cngln v ngc li.

Mt cng thc phng sai tng ng l:

var(X) = E(X2) - 2

Chng minh: Sdng cc tnh cht ca kzvng ta c thchng minh cng thc ny nhsau:

var(X) = E[(X - )2]

= E[ X2 2X + 2 ]

= E(X2) E(2X ) + E(2)

= E(X2) 2 E(X) + 2

= E(X2) 2 2 + 2


23/87

CHNG 1.NN TNG



= E(X2) - 2

Ni thm: Theo kin c nhn ca ti, c thngi ta mun o phn tn quanh gi trtrung bnh ca cc gi trca bin ngu nhin nn mun tnh E(|X-|). Tuy nhin vic dnggi trtuyt i dn ti kh khn no trong vic pht trin cc cng thc nn ngi

ta thay trtuyt i bng bnhphng. D sao y chl suy din c nhn v ngun gcphng sai.

V mt n v, phng sai c n vl bnh phng n v ca X. V dX c n v l cmth var(X) sc n v cm2.

Ngi ta a ra thm khi nim lch chun, k hiu (X), c tnh bng cn bc 2 caphng sai:

(X) =

var(x)

Vi nh ngha ca phng sai, ta d thy n lun 0, v vy khng cn iu kin g y.

Nh vy, lch chun, do ktha tnh cht ca phng sai, cng c thdng so snhs phn tn ca cc gi tr quanh gi tr kz vng: cng ln th cc gi tr cng nm xa

(phn tn) gi tr kz vng.

Mt thun li ca lch chun l n c cng n v vi lch chun.

Ni thm: C v nh lch chun l s quay tr v mong mun ban u: tnh tonkhong cch phn tn trung bnh (ch khng phi khong cch bnh phng trung

bnhphng sai). Khong cch phn tn trung bnh m ti ni l E(|X-|). Mc d lch chun khng bng khong cch phn tn trung bnhny (E(|X-|) ) nhng cngchng c vn g khi dng n gii quyt vn m ngi ta mong mun khong

cchphn tn trung bnh gii quyt: tnh ton sphn tn ca cc gi trquanh gi trkzvng ca mt bin ngu nhin.

1.6 Moment ca bin ngu nhin

Moment c cp mc ny l moment trong ton v xc sut thng k, n khc vikhi nim moment ca nh (sni bn di) cng nh moment trong vt l.

Cho F l mt bin ngu nhin, ta c i lng E(Fk) c gi l moment bc k ca F, v ilng E[ ( F-E(F) )k] c gi l moment trung tm bc k ca F.

R rng, kz vng m chng ta cp trong cc mc trc chnh l moment bc 1 ca binngu nhin. Cn phng sai chnh l moment trung tm bc 2.

Ni thm: Thng thng tip cn mt vn mi l, chng ta thng (v c l l nn)bt u ttrc gic ri n mt stng qut n gin (gn trc gic) v sau l stngqut ha v mrng. C thkhi nim moment ra i sau cc khi nim phng sai v lch chun, v bn c ththy n l stng qut ha v mrng ca cc khi nim ny.


24/87

CHNG 1.NN TNG



Chng ta c tht cu hi cc moment khc kz vng v phng sai c { ngha g khng.Hu nh mi s tng qut ha u rt c { ngha i vi s pht trin. Cc moment cabin ngu nhin cho ta cc thng tin vdng iu ca phn b xc sut (hnh dng ca

th hm xc sut (hay hm mt i vi bin ngu nhin lin tc)) ca bin ngu nhin.

V d, nu moment trung tm bc 2 (phng sai) ca bin ngu nhin F nh, th c ngha lcc gi tr ca F ni chung t b sai lch so vi gi tr kz vng ca n, hay ni cch khc phnln xc sut ca phn b xc sut ca F tp trung trong mt khong nhxung quanh imgi tr kz vng. Ngc li, nu moment trung tm bc 2 ln, th phn b xc sut ca F nitrung s phn tn ra xa im gi tr kz vng hn. y l { ngha ta ni trn.

V d: C 2 hp bi, mi hp ng 12 vin bi. Gi bin l bin ngu nhin nh x khi lng

(n v gram) cc vin bi, ni cch khc X l khi lng cc vin bi. Ta c:

Hp 1:

X (gram) 30 40 50 60 70

Slng 3 4 1 1 3

Hp 2:

X (gram) 30 40 50 60 70

Slng 2 3 4 2 1

C hai hp u c E(X) = 47.5 (gram) (kz vng: khi lng trung bnh)

Nhng phng sai th khc nhau: (nhc li: var(X) = E(X2) (E(X))2)

Hp 1: var(X) = 2491.67 47.52 = 235.42 Hp 2: var(X) = 2391.67 47.52 = 135.42

Suy ra: cc gi tr (khi lng) ca hp 2 tp trung gn gi tr kz vng (47.5) hn.


25/87

CHNG 1.NN TNG



Biu 1.6-1 Phn phi xc sut ca X hp 1, phng sai ln hn: cc gi tr phn tn xa k vng hn

Biu 1.6-2 Phn phi xc sut ca X hp 2, phng sai nhhn: cc gi tr tp trung hn quanh k vng

Moment trung tm bc 3 ca F thhin lch (skewness) ca phn b xc sut ca F:

Nu F c phn b xc sut i xng quanh im gi tr k vng(c ngha l F v 2E(F) - Fc cng phn b xc sut), th moment trung tm bc 3 ca n bng 0. Nu nh momenttrung tm bc 3 ln hn khng th phn b xc sut ca F c gi l lch dng hay lchv bn phi (phn ui bn phi di hn), cn nu moment trung tm bc 3 nh hnkhng th phn b xc sut ca F c gi l lch m hay lch v bn tri (phn ui bntri di hn).

0.25

0.33

0.08 0.08

0.25

30 40 50 60 70

XcsutP(X)

X (khi lng)

K vng

47.5|

0.17

0.25

0.33

0.17

0.08

30 40 50 60 70

XcsutP(X)

X (khi lng)

K vng

47.5

|


26/87

CHNG 1.NN TNG



V d: Gi s c mt bin ngu nhin F vi phn b xc sut ri rc sau:

F -2 1 3

p(F) 1/2 1/4 1/4

Khi gi tr kz vng ca F: E(F) = 0, moment trung tm bc 3 ca F bng:

E[ (F-E(F))3 ] = -23.1/2 + 13.1/4 + 33.1/4 = 3 > 0

th phn b xc sut ca F c phn ui lch v bn phi so nu ly im gi tr kz vng(F=0).

Biu 1.6-3 Moment trung tm bc 3 ln hn 0: lch dng (ui bn phi di hn)

Cn moment trung tm bc 4 c lin quan n cng thc tnh kurtosis m ta c th coi l nhn ca phn b xc xut so vi phn b chun3, nu nhn ny ln hn 0 thbin ngu nhin c th ca phn b xc sut nhn hn phn b chun, ngc li thn bt hn. Xem thm ti ti liu [3].

3 Phn b chun (normal distribution) l phn b xc sut thng dng trong thc t. Ta sni n phn b

ny trong cc mc sau.

0.50

0.25 0.25

-2 1 3

Xcsut

P(F)

F

K vng|0

(+) Phn b lch dng(ui phi di hn)

(-) Phn b lch m(ui tri di hn)


27/87

CHNG 1.NN TNG



Hnh 1.6-1 Cc phn phi xc sut v " nhn". ng gia: phn phi chun.

1.7 Hip phng sai

1.7.1 nh nghaCho 2 bin ngu nhin X v Y, ta c nh ngha hip phng sai (covariance) ca X v Y , khiu Cov(X,Y):

Cov(X,Y) = E[ (X-x)(Y- y) ]

trong x, y ln lt l kz vng ca X, Y.

Mt cng thc tng ng ca hip phng sai:

Cov(X,Y) = E (X Y) - xy

Chng minh: Sdng cc tnh cht ca kzvng.

Cov(X,Y) = E[ (X-x)(Y- y) ]

= E( XY - xY - X y + xy )

= E(XY) - E(xY) - E(X y) + E(xy)

= E(XY) - xE(Y) - y E(X) + xy

= E(XY) - 2xy + xy

= E (X Y) - xy

Theo cng thc th nht, mt cch trc gic, ta c th thy { ngha ca hip phng sai,

l s bin thin cng nhau ca 2 bin ngu nhin: Nu 2 bin c xu hng thay icng nhau (ngha l, khi mt bin c gi trcao hn gi tr kz vng th bin kia c xu hngcng cao hn gi tr kz vng), th hip phng sai gia hai bin ny c gi trdng. Mtkhc, nu mt bin nm trn gi tr k vng cn bin kia c xu hng nm di gi tr kvng, th hip phng sai ca hai bin ny c gi tr m.


28/87

CHNG 1.NN TNG



Nu 2 bin ngu nhin l c lp (s bin thin ca chng khng lin quan nhau) th hipphng sai ca chng bng 0.

X,Y c lp Cov(X,Y) = 0

Tuy nhin iu ngc li khng ng: nu hip phng sai ca X, Y bng 0 th khng nht

thit 2 bin ny c lp. Cc bin ngu nhin m c hip phng sai bng 0 c gi lkhng tng quan (uncorrelated), chng c thc lp nhau hoc khng.

Chng ta tng cp trong phn kz vng: Kz vng khng c tnh nhn, v lng khcbit gia kz vng ca tch v tch cc kz vng l hip phng sai:

Cov(X,Y) = E(X Y) - xy

y cng l cng thc m chng ta va trnh by.

Nh vy, nu X, Y c lp, ta c Cov(X,Y) = 0 E(X,Y) = xy

Cng t cng thc nh ngha hip phng sai, ta thy n v ca hip phng sai l tchn v ca X v Y, v dX c n vl m, Y c n v l kg th cov(X,Y) c n v m.kg.

1.7.2 Tnh chtVi X, Y l 2 bin ngu nhin, a, b l cc hng s(theo ngha a, b khng l bin ngu nhin)ta c cc tnh cht sau c suy ra tnh ngha hip phng sai v tnh cht kz vng:

Cov(X,X) = Var(X)

Cov(X,Y) = Cov(Y,X) Cov(aX, bY) = abCov(X,Y)

Cov(X1+X2, Y1+Y2) = Cov(X1, Y1) + Cov(X2, Y1) + Cov(X1, Y2) + Cov(X2, Y2)

Tng qut, vi cc bin ngu nhin X1, X2, Xn v Y1, Y2, Ym ta c:

Cov(x Xi, Y Yj) = XY Cov(Xi, Yj)

Var(X1 + X2) = Var(X1) + Var(X1) + 2Cov(X1, X2)

Tng qut, vi cc bin ngu nhin X1, X2, Xn ta c:

Var(x X) = x Var(X) + 2

Cov(Xi, Xj)i, j:i


29/87

CHNG 1.NN TNG



1.8 Ma trn hip phng sai4

Nh chng ta va trnh by, hip phng sai l i lng tnh ton stng quan gia 2bin ngu nhin.

Vy gi s chng ta c mt vector bin ngu nhin5 c 3 phn t X1, X2, X3. Nu ta muntnh s tng quan gia tt c cc cp bin ngu nhin th ta phi tnh tt c l 3 hipphng sai: Cov(X1, X2), Cov(X1, X3), Cov(X2, X3).

Mt cch tng qut, ma trn hip phng sai ra i cho php ta tnh tt c cc Cov

gia 2 bin ngu nhin trong mt vector bin ngu nhin.

Cho mt vector bin ngu nhin X cha n bin ngu nhin, ma trn hip phng sai ca X,k hiu l , c nh ngha l:

= 1 , 1 1, 2 1 , 2, 1 2 , 2 2, , 1 , 2 , vi X = 1 Ni d hiu, ma trn hip phng sai l ni cha cc hip phng sai. Mi phn t can l mt hip phng sai ca 2 bin ngu nhin Xi v Xj vi i l ch s hng v j l ch sct (i, j = 1..n).

Quan st trn ng cho ca ma trn hip phng sai (i=j) ta thy ti l cc phngsai, v: Cov(Xi, Xi) = Var(Xi).

1.9 Phn phi chun (phn phi Gauss)

Trong cc mc va ri chng ta tm hiu kh cc kha cnh c bn ca mt bin ngunhin.

Chng ta cng cp n hm xc sut (hm mt ) l hm c trng cho mtphnphi xc sut. Ni mt cch n gin phn phi xc sut (hay phn b xc sut) l mtthut ngdng ch s phn b cc xc sut ca mt bin ngu nhin.

Trong v d v hp bi mc 1.6, chng ta c mt hp ng 12 vin bi. Gi bin l bin

ngu nhin nh x khi lng (n v gram) cc vin bi, ni cch khc X l khi lng ccvin bi. Ta c:

4 Ti liu [4] Lindsay I Smith, A tutorial on Principal Components Analysis, 2002, c ni v ma trn hipphng sai v ng dng ca n trong phn tch thnh phn chnh (mc 2.1.4 The covariance Matrix). Bncng c th tham kho cch ni d hiu v cc khi nim xc sut ng dng trong tin hc trong quyn ny.5

Vector bin ngu nhin c th hiu n gin l mt mng (tp) cc bin ngu nhin.


30/87

CHNG 1.NN TNG



X (gram) 30 40 50 60 70

Slng 3 4 1 1 3

Bng trn cng l mt cch biu din phn phi xc sut ca X, bi v nu ta ly s

lng chia cho 12 (tng s bi) ta s xc sut ca bi c khi lng tng ng. Hoc ta c th:

Biu 1.9-1 th phn phi xc sut ca X

thtrn cng l mt cch biu din mt phn phi xc sut.Ni tm li, khi ta ni n mt phn phi xc sut l ta mun ni v phn phi cc xc sutca cc gi tr ca mt bin ngu nhin, tc l phn phi xc sut cho bit mi gi tr cabin ngu nhin s nhn xc sut l bao nhiu.

C rt nhiu phn phi xc sut, v d ca chng ta va ni ti cng l mt phn phi xcsut. Tuy nhin, trong thc t, ngi ta thng s dng mt phn phi xc sut c tn lphn phi chun (normal distribution), hay phn phi Gauss.

Mt bin ngu nhin X c ni l c phn phi Gauss khi n c hm mt l hm

Gauss, k hiu l XN(, ), c l X c phn phi chun (hay phn phi Gauss) vi thams, (2 tham sc t mt hm Gauss). Khi hm mt ca X l:f(x; , ) = , 2 (x)= 12 e-(x-)222

Vi phn phi xc sut nh trn, ngi ta tnh c , ln lt chnh l kz vng v lch chun ca X.

Di y l th ca mt s phn phi chun.

0.25

0.33

0.08 0.08

0.25

30 40 50 60 70

Xcs

utP(X)

X (khi lng)

Kvng

47.5|


31/87

CHNG 1.NN TNG



Biu 1.9-2 th ca mt sphn phi chun

Vi = 0 v = 1, phn phi c gi lphn phi chun chun ha.

Quan st th ta thy phn phi chun c dng i xng. Gi tr kz vng ca X l , chnh

l trc i xng. V lch chun (hay phng sai 2) cng ln th th ca n cngbt, iu ny ph hp vi hiu bit ca chng ta: khi lch chun ln th cc gi trcng phn tn ra xa kz vng.

1.10 c lng

Trong mc ny ta s cp mt s khi nim lin quan n vn c lng (estimation).

Khi nim

c lng l phng on gi trcha bit da vo cc quan st. C 2 loi: c lng im: Gi trc lng l mt s. V d, mt chic xe ang chy vi vn

tc khong 40km/h.

c lng khong: Gi trc lng l mt on [a, b]. V d, mt chic xe angchy vi vn tc khong 35 n 40 km/h.

Trong bo co ny ta ch cp n c lng im.

c lng im


32/87

CHNG 1.NN TNG



Gi a l mt gi trcha bit, khi ngi ta k hiu c lng im ca a l (c thcl a m), v ta vit a .Thng k

Trong c lng, ngi ta c khi nim thng k: Thng k l mt i lng m gi tr ca

n ch ph thuc vo cc quan st, khng ph thuc vo cc tham scha bit.

V d: Cc biu thc dng tnh cc iu sau l cc thng k:

T l hc sinh gii.

im trung bnh ca sinh vin.

Nh vy, chng ta da vo cc thng k tm ra gi trc lng, v cy ngi ta cn githng k l hm c lng.

Mt cch tng qut nu ta c mu6 {x1, x2, xn}, th hm : = (x1, x2, xn)

dng tnh ra gi trc lng cho tham s cgi l hm c lng.c lng khng chch (unbiased estimators)

Hm c lng gi l c lng khng chch ca tham s nu nh:E() = .

Ngi ta chng minh c cc c lng sau y l cc c lng khng chch ca ctham stng ng:

Tham s K hiu c lng khng chch

Trung bnh X

Phng sai Var (hay 2) S2

lch chun 2 S

6 Mu: c th hiu l b d liu quan st. c mt nh ngha y , xin tham kho cc ti liu v thng

k.


33/87

CHNG 1.NN TNG



T l cc phn tc tnh cht no

p f

Bng 1.10-1 Bng cc c lng khng chch

vi X, f, S2, S, ln lt l trung bnh, t l cc phn t c tnh cht quan tm, phng sai v lchchun ca mu quan st.

Hoc ni cch khc:

Trung bnh mu Xl c lng khng chch cho trung bnh ca m ng7.

Phng sai mu S28( lch chun S) l c lng khng chch cho phng sai (lch chun) ca m ng.

T l cc phn t c tnh cht quan tm f

l c lng khng chch cho t l cc phnt ca m ng.

V d: tm t lngi bit ni 2 ngoi ng ca mt nc (m ng), ngi ta kho stmt b phn dn c (mu). T lngi bit ni 2 ngoi ng trong b phn dn c ny lmt c lng khng chch ca t lngi bit ni 2 ngoi ng trong cnc.

Ni thm: Thng th m ng rt ln nn kho st ton b (cho kt qu chnh xc) gnnh khng th, do ngi ta dng mu, l mt tp con ca m ng m ngi hyvng kho st n scho kt qu mun bitm ng.

V d: kho st chiu cao trung bnh ca ngi t tui 15-18, ngi ta khng thdo

ht tt c nhng ngi trong tui ny (m ng), do ngi ta chn ra mt sngi o chiu cao (mu). Bng nhiu phng php ngi ta c th m bo hockhng nh rng kt qu tmu l ng vi m ng. c ng khng chch lm c

phn no mc ch .

Ngoi ra ngi ta cn c khi nim c lng hiu qu, c lng vng. Xin xem thm ccti liu v thng k nu mun bit chi tit.

chch (bias)

Gi l mt c lng ca , ta c nh ngha chch l:Bias() = E()

Vi c lng khng chch ta c Bias bng 0.

7m ng l tp hp cc phn t cn kho st.8 Chnh xc lphng sai mu hiu chnh. Vi slng cc phn t ca mu ln th phng sai v phnghiu chnh gn bng nhau. Tuy nhin khi chng minh th ngi ta chng minh c phng sai mu hiuchnh l c lng khng chch (ch khng phi phng sai). Xin xem thm cc ti liu v thng k nu

mun bit chi tit.


34/87

CHNG 1.NN TNG



Ngoi ra cn c cc khi nim c lng hiu qu tt nht, c lng nht qun vng,c th tham kho ti liu [5] Cao Ho Thi, c lng cc tham sthng k.

Sai sbnh phng trung bnh (Mean Squared Error- MSE)

Sai sbnh phng trung bnh ca c lng c nh ngha l:MSE() = E[ ( )2 ]

Ngi ta chng minh c rng:

MSE() = Var() + [ - E() ]2Hay

MSE() = Var() + [ Bias() ]2Nu l c lng khng chch th:

MSE() = Var(),v Bias() = 01.11 Nguyn tc trc giao (Orthogonality principle)

Nguyn tc trc giao c dng nhiu nht trong thit lp c lng tuyn tnh.[6]

Nguyn tc trc giao c dng ti u c lng. N c thc dng tm clng c sai sbnh phng trung bnh nh nht (Minimum MSE - MMSE).[6]

Gi x l mt vector ngu nhin9cha bit sc c lng da trn vector quan st10 y.

Nu c lng x ca x v y c quan h tuyn tnh:x = Hy + c, vi H l mt ma trn v c l mt vector no .

Khi ,nguyn tc trc giao pht biu rng c lng

s c sai sbnh phng trung

bnh nh nht (MMSE) khi v ch khi:

E[ ( - x)yT ] = 0, vE( - x) = 0

9 Nhc li: vector ngu nhin n gin l mt b cc bin ngu nhin. V d ta c th gi ta mt imtrong khng gian 2 chiu l mt vector ngu nhin, v 2 ta x v y ca n c th xem l 2 bin ngu nhin.10 Vector quan st y l ma trn cng kch thc vi vector ngu nhin x cn tm. V d: vector x cn tm lmt vector ct 2x1 th vector quan st y cng l vector ct 2x1. N chnh l b d liu m ta dng clng x. V d: ta mun c lng vn tc ca mt vt khi bit ta ca n qua thi gian, khi vn tc l

x, ta l vector quan st y.


35/87

CHNG 1.NN TNG



Nu x v y c trung bnh (mean, hay kz vng) bng 0 th ch cn iu kin th nht l c(v iu kin th2 tng tha).

Mt cht ghi ch: yTl ma trn chuyn vca y.

Vy nguyn tc trc giao thc cht l hphng trnh dng tm cc tham scho mthm c lng (y l H v c)n t ti u(theo ngha c lng ny c sai s bnhphng trung bnh nh nht MMSE).

Bn c th xem v d v pht biu cho trng hp tng qut ca nguyn tc trc giaotrong ti liu [6] Orthogonality principle Wikipedia.

Nguyn tc ny cng nh hu ht cc ni dung trnh by trong chng ny s cdng trong xy dng b lc Kalman (Chng 2).


36/87

26

2Chng2B Lc Kalman

Cc ni dung chnh:

Gii thiu v lc Gii thiu b lc Kalman Xy dng b lc Kalman B lc Kalman trong OpenCV


37/87

CHNG 2.BLC KALMAN



2.1 Gii thiu v lc

Mt cch tng qut, lc l loi b nhng phn khng quan tm hoc khng c { ngha hocnhng phn sai st thu c mt u ra tt hn (theo ngha hp hn vi mong mun,quan tm ca ta).

V d: Trong x l nh ta c lc lm trn lm mn nh, lc trung v kh nhiu muitiu. Trong microphone hoc cc my thu m cng c mt b phn lc loi b tp m.

Hnh thc hn mt cht, gi s ta c trong mt tn hiu X bao gm tn hiu S (signal) ta

quan tm v nhiu N (noise):

X(k) = S(k) + N(k)

Nu ta bit nhiu N dao ng quanh 0 v c gi tr trung bnh bng 0:

N(k) / m = 0, m: s ln quan st N(1), N(2), N(m)

cng tc l N(k) = 0

Vy khi ta c:

X(k) = (S(k) + N(k)) = S(k) + N(k) = S(k)

Nhn theo mt kha cnh no ta loi c nhiu N.

Tuy nhin, trn y ch l mt v d. Trong thc tcc nhiu rt a dng v phc tp nnrt kh (i khi l khng th) loi b hon ton nhiu. B lc chlm sch tn hiu n

mt mc no v trong a strng hp ta cng ch cn c th.

2.2 Gii thiu b lc Kalman

2.2.1 Tng quanNm 1960, R.E. Kalman cng b bi bo ni ting v mt gii php quy gii quyt biton lc thng tin ri rc tuyn tnh (discrete data linear filtering).[13] Bi bo c ta ANew Approach to Linear Filtering and Prediction Problems. Khong 50 nm tri qua, blc Kalman tr nn ph bin. N xut hin trong rt nhiu ng dng (xem mc 2.2.2) v

bi v vn lc m n gii quyt l mt vn c bn trong rt nhiu lnh vc, nn n cth vn lun cn c dng nhiu ng dng mi na tng lai (tr khi c mt b lchay mt gii php no tt hn ra i).

Mt cch tng qut, b lc Kalman l mt tp hp cc phng trnh ton hc gip ti uc lng trng thi ca mt h(theo ngha gi trc on c sai s bnh phng trungbnh nh nht MMSE) da trn m hnh h thng (system model), gi tr o(measurement value) v cc hiu bit v nhiu (ca h thng ln php o).

Biu sau y l mt m t hnh thc ca mt b lc ni chung.


38/87

CHNG 2.BLC KALMAN



Trong biu trn, b lc Kalman cng nh cc b lc khc s nm vo v tr ca hnh chnht B lc. im khc bit ca b lc Kalman l s hiu qu ca n.

Nh phn sau s trnh by, ta sphng trnh ha cc tn hiu: tnh hiu u vo (cn o),tn hiu nhiu (bao gm sai s cm bin), tn hiu o c v tn hiu u ra.

Ngoi ra b lc Kalman cn cho php thm mt tn hiu iu khin v cng kn nhiuh thng (do m hnh h thng, khc nhiu ca php o ni trn).

2.2.2 i nt v Rudolf Emil KalmanRudolf Emil Kalman sinh nm 1930 ti th Budapest, Hungary.Nm nay (2011) ng 81 tui. ng tt nghip bng c nhn v thc s v Electrical engineering ti hc vin MITnm 1953 v 1954. ng nhn bng tin s ti i hc Columbia nm 1957.

Nhng v tr chnh ng ph trch:

Nghin cu ton hc ti RIAS (Research Institute for Advanced Study) ti Baltimoret 1958 1964.

Gio s ti i hc Standford t 1964-1971.

Gio s nghin cu sau i hc (Graduate Research Professor) v Gim c Centerfor Mathematical System Theory, i hc Florida, Gainesville t 1971 1992.

Tnm 1973, ng tr thnh thnh vin ca Mathematical System Theory ti ETH(Swiss Federal Institute of Technology), Zurich.

ng cng nhn nhiu gii thng nh:

Huy chng danh d ca IEEE (1974) Huy chng Centennial ca IEEE (1984)

Gii thng Kyoto trong cng ngh cao ca t chc Inamori, Nht Bn (1985)

Gii thng Steele ca Hi Ton hc M (American Mathematical Society) (1987) Gii thng Bellman (1997)

ng cng l vin sca Vin Hm lm Khoa hc Quc gia (M), Vin Hn lm K thut Qucgia (National Academy of Engineering) (M) v Vin Hn lm Khoa hc v Ngh thut

(American Academy of Arts and Sciences) (M).

Cm bin(Bo) B lc

Tn hiucn o

Tn hiuo

Tn hiunhiu

Sai scacm bin

Tn hiu lc


39/87

CHNG 2.BLC KALMAN



ng cn l vin s ca cc vin hm lm cc nc khc nh Vin Hn lm Khoa hcHungary, Php v Nga. ng cng nhn nhiu bng tin s danh d.[7]

2.2.3

ng dng ca b lc KalmanBi v b lc Kalman gii quyt mt vn c bn l lc nhiu v ti u cho cc c lngnn n c ng dng rt rng ri.

Mt sng dng c lit k t bi vit Kalman Filter trn Wikipedia [8]:

Li tng ca my bay (Autopilot)

c lng trng thi sc ca pin (Battery state of charge (SoC) estimation)

Giao din tng tc vi my tnh bng no (Braincomputer interface) Chaotic signals

nh v chuyn ng (Dynamic positioning) Cc ng dng trong kinh t, c bit l kinh tv m, time series, v econometrics

H thng dn ng qun tnh (Inertial guidance system)

Theo di bng radar (Radar tracker) H thng nh v v tinh (Satellite navigation systems)

Speech enhancement

D bo thi tit (Weather forecasting)

H thng nh v (Navigation Systems)

M hnh ha 3 chiu (3D-Modelling)

Vit Nam c mt sng dng nh:

ng dng lc Kalman trong phn tch bin dng nh cao tng do bc x nhit mttri.[9]

Ci thin cht lng truyn ng khng ng b bng cu trc tch knh trc tips dng kalman filter quan st t thng.[10]

ng dng Kalman Filter cho d bo nhit 2m t sn phm m hnh HRM.[11] H thng dn ng qun tnh INS/GPS.[12]

2.2.4 V d dn nhpChng ta s xem xt mt v d vc lng sau.

Gi s ta c mt vt th chuyn ng thng u (vn tc khng i) theo phng x no. Ta cn xc nh v tr ca vt sau thi gian t.

Cc thng tin ta bit l:

Vn tc ca vt khng i: v = 70 m/s.

Vtr ban u ca vt x(0) = 0


40/87

CHNG 2.BLC KALMAN



Chng ta cng c mt b phn nh v gn trn vt th cho chng bit v tr x(t) ca vt thlc t.

Vy l chng ta c 2 cch xc nh v tr vt:

Cch 1: Dng phng trnhng hc tnh vtr vt

Ta c v tr ca vt ti thi im t l x(t) = x(0) + v.t

Hay vit di dng truy hi l:1(0) = 01(1) = 1(0) + v = v

1

(2) =

1

(1) + v = 2v

(k hiu 1c ngha l c lng v tr cch 1)Cch 2: Sdng b phn nh vgn trn vt

Khi cc vtr n gin l gi tr do b phn ny cung cp:2(0) = z(0)2(1) = z(1)

2

(2) = z(2)

Trong (k) l gi tr do b phn nh v cung cp ti thi im k.

n y ta thy, nu nh 1(k) = 2(k), vi mi k, th c vnh kt ca tnh ton v gi tro c ca b phn nh v l c thtin tng. Tuy nhin, bn s tht vng khi bit rng ch l cm gic ca bn, cn thc tkhng phi vy!

Bi l php tnh ca bn (trong nhng trng hp phc tp) ln gi tr cung cp bi bphn nh vkhng c c sno khng nh l chnh xc vi thc t. Cho d 2 kt quny hon ton trng khp th cng c th ch l ngu nhin.

Mt trng hp khc thng xy ra hn l khi 2 kt qu ny khng ging nhau. Vy bn stin tng kt qu no?

Cu hi ny a n mt gii php th3, l kt hp c 2 kt qu.

Cch 3: Tnh ton kt qu bng cch tng hp 2 phng php3(k) = . 1(k) + .2(k)


41/87

CHNG 2.BLC KALMAN



Cch ny nhn c vn. V c thc trng hp 1chnh xc hn, c trng hp 2 chnhxc hn, nn s l tt nht nu ty vo trng hp m ta chn (hoc kt hp) kt qu mt

cch thch ng.

Tuy nhin, n y li ny sinh 2 vn mi:

1. v phi chn ra sao c kt qu ti u?2. V nh ni, v c th s thay i ( ty trng hp m kt hp 2 c

lng cho hp l nht), vy th u sl cn c, tiu chun ta cp nht 2 h sny?

n y, b lc Kalman xut hin. 2 vn chng ta va nu chnh l i tng gii quytca b lc Kalman.

2.2.5

M hnh tng qut ca c lng dng b lc KalmanS khi sau s m t m hnh tng qut ca mt c lng dng b lc Kalman.

Chng ta c th nh x cc phn trong v d mc 2.2.4 vo s trn nh sau:

M hnh h thng: phng trnh ng hc x(k) = x(k-1) + v.

Cm bin: bnh v gn trn vt.

Gi trc lng ti u:

3.

B lc Kalman: hp en tnh cc h sv .

Mt c lng thc s s dng b lc Kalman cng bao gm cc phn nh trn.

Nh vy n y chng ta c y cc khi nim cn thit bc vo xy dng blc Kalman (thc cht l tm v gii cc phng trnh tnh ton cc tham scho c

lng ti u).

M hnh h thng

Cm bin(bo)

B lc Kalman

Gi tr tnh ton

Gi tro

Gi trc lngti u


42/87

CHNG 2.BLC KALMAN



2.3 B lc Kalman

2.3.1 Cc k hiu s dngMc ny lit k cc k hiu dng trong biu din cc phng trnh qu trnh xy dng cngnh kt qu ca b lc Kalman. Cc k hiu ny c s dng thng nht trong bo cony nhng cc sch v ti liu khc c th c h thng k hiu khc. Bn c th tham

kho vn k hiu ny trong mc 1.3 ON THE NOTATION USED IN THIS BOOK ca ti liu[14].

K hiu ngha

H Ma trn o nhy c dng xc nh quan h tuyn tnh ca vector trngthi v vector o.

K Ma trn li KalmanP Ma trn hip phng sai ca sai sc lng

Q Ma trn hip phng sai ca sai s h thng

R Ma trn hip phng sai ca sai so t

x Vector trng thi

z Vector o (measurement vector)

Ma trn chuyn trng thi

Bng 2.3-1 K hiu chun ca b lc Kalman

K hiu ngha

x Vector trng thi

xk Phn t th k ca chui cc trng thi x1, x2, x3,

x

c lng ca x

xk() Tin nghim ca c lng xkxk(+) Hu nghim ca c lng x ti thi im tk

x o hm ca x theo thi gian tBng 2.3-2 Cc k hiu lin quan n vector trng thi

K hiu Tn Kch thc (hng x ct)

x Vector trng thi h thng n x 1


43/87

CHNG 2.BLC KALMAN



w Vector nhiu h thng r x 1

u Vector iu khin r x 1

z Vector gi tro t l x 1

v Vector nhiu o t l x 1

Vector chuyn trng thi n x n

Q Ma trn hip phng sainhiu h thng

r x r

H Ma trn h s quan h tuyntnh

l x n

R Ma trn hip phng sainhiu o t

l x l

Bng 2.3-3 K hiu v kch thc cc ma trn

Tn Biu thc

Trng thi h thng xk= xk-1 + wk-1Gi tro zk= H xk+ vk

Bng 2.3-4 Biu thc ca trng thi v gi tro

2.3.2 Xy dng b lc Kalman11Theo nh m hnh ca c lng dng b lc Kalman cp trn, trc tin, mhnh h thng s cho ta mt gi tr gi l c lng tin nghim (c lng cha qua blc Kalman). N c tnh bng phng trnh:

xk() =

xk-1

(+) + wk-1

trong :

xk() : gi tr tin nghim ti bc k x

k-1(+): gi tr hu nghim ca bc k-1 (gi tr ti u ca bc trc)

wk-1: nhiu h thng ti bc k-1

11Phn ny l phn din dch li v gii thch chi tit hn ni dung t mc 4.2 KALMAN FILTERtrong ti liu [14] Mohinder S. Grewal and Angus P. Andrews, KALMAN FILTERING - Theory and

Practice Using MATLAB, third edition, John Wiley & Sons, 2008.


44/87

CHNG 2.BLC KALMAN



Mt ch quan trng: Tt c cc nhiu trong b lc Kalman u c ginh l tun theophn phi chun (Gauss) (xem 1.9, chng 1).

Sau , gi tr tin nghim xk() si qua b lc Kalman kt hp vi gi tro (zk) tora mt c lng ti u.

C th tm tt qu trnh lm vic ca lc Kalman nh sau:

Tnh tin nghim (thu nghim bc trc) Lc Hu nghim (c lng ti u)

Ta c phng trnh quan h ca c lng tin nghim v hu nghim:

trong :

xk() : gi tr tin nghim xk(+) : gi tr hu nghim K1 v Kk : cc ma trn cha bit, Kk l li ca b lc Kalman

Theo quy tc trc giao, sai s ca c lng bnh phng trung bnh l nh nht(MMSE) ta phi c:

Thay xk =

xk-1 + wk-1 v biu thc ca xk(+)vo phng trnh th nht ta c:

Tip tc thay zk = H xk + vk vo ta thu c:

S dng cc iu kin sau:

v

(do vk v z l 2 bin ngu nhin c lp v vk c phn phi chun vi kz vng bng 0 nn tac E(v.z) = E(v) x E(z) = 0 x E(z) = 0)

Ta bin i phng trnh trn:


45/87

CHNG 2.BLC KALMAN



Nh vy ta c:

t cc sai s:

Sai sk(+) v k(-) l cc sai s ca c lng ng vi tin nghim v hu nghim. Sai sk l sai s ca o t.Tr2 phng trnh sau: +() = 0(hai phng trnh ny c l v ta mun tha quy tc trc giao)

ta thu c: + = 0Thay xk, k v kvo phng trnh trn ta c:Lu { l v cc bin ngu nhin w, v c kz vng bng 0 v chng c lp vi x, z nn ta c:

v thay K1 , zk v k(-) vo ta c:


46/87

CHNG 2.BLC KALMAN



Nhc li:ta c theo nh ngha

Cov(XY) = E[( X-x )( Y-y )]

= E(XY) - xy

vi x, y ln lt l kz vng ca 2 bin ngu nhin X v Y.

Vy nu X hoc/v Y c kz vng bng 0 th:

Cov(XY) = E(XY)

T tnh cht trn, ta c ma trn hip phng sai ca sai s ca c lng tin nghim l:

(thc cht P chnh l sai sbnh phng trung bnh MSE)

bi v k(-) l bin ngu nhin c kz vng bng 0.12Khai trin phng trnh pha trn v thPk vo ta thu c:

trong : Rk l ma trn hip phng sai ca sai s nhiu ot (cng thu c do kz vngca nhiu o t v bng 0) = Nh vy ta thu c li ca b lc Kalman:

Tng t, i vi hu nghim ca c lng ta cng c ma trn hip phng sai ca sai s

c lng hu nghim:

Thay biu thc ca K1 tm c trn

12Chng minh: do l c lng khng chch nn E[k(-)] = E[xk]Ta c: E[

k(-)] = E[

k(-)-xk] = E[

k(-)] E[xk] = 0


47/87

CHNG 2.BLC KALMAN



vo phng trnh quan h ca tin nghim v hu nghim:

ta thu c + = + ,+ = + .Tr 2 vcho xk v thay zk = H xk + vkvo phng trnh trnta c :

Thbiu thc va thu c vo Pk(+) v s dng:

ta c:

Phng trnh ta va thu c gi l dng Joseph (Joseph form) ca phng trnh cp nhtma trn hip phng sai.

Tip tc khai trin phng trnh ny ta c:

Phng trnh cui cng thng c dng cp nht Pk(+) nht.

Ma trn hip phng sai ca sai s:

Ta c:

Tr 2 vca phng trnh sau cho xk v thxk = xk-1 + wk-1 vo ta thu c:


48/87

CHNG 2.BLC KALMAN



Thay biu thc ca sai sc lng tin nghim k(-) va thu c vo Pk(-) v s dng:ta c:

y chnh l biu thc ca ma trn hip phng sai ca sai s ca c lng tin nghim.

2.3.3 Tm tt cc phng trnh ca b lc KalmanTrong phn trn chng ta xy dng xong cc phng trnh ca b lc Kalman. Phn nys tm tt li cc phng trnh cho vic tra cu khi s dng.

M t Phng trnh

M hnh h thng xk=

xk-1 + wk-1

wk N(0, Qk)M hnh o zk= H xk+ vk

vk N(0, Rk)Gi tr khi to E(x0)=0

E(x0x0T)=P0Gi thit c lp E(wkvj

T)=0 vi mi k v j

c lng trng thi () = 11(+)Ma trn hip phng sai ca sai s () = 11(+)1 + 1Cp nht trng thi (+) = () + [ ()]Cp nht ma trn hip phng sai ca sai s (+) = [ ]()Ma trn li Kalman = ()() + 1

Bng 2.3-5 Tm tt cc phng trnh ca b lc Kalman

V qu trnh lm vic ca b lc Kalman c th tm tt trong hnh sau:


49/87

CHNG 2.BLC KALMAN



Hnh 2.3-1 Tm tt qu trnh lm vic ca lc Kalman

2.4 B lc Kalman trong OpenCV

thc hin lc Kalman, OpenCV cung cp cho chng ta cc hm sau:

Hm to v hy cu trc CvKalman:

cvCreateKalman(int nDynamParams, //S chiu vector trng thiint nMeasureParams, //S chiu vector oint nControlParams //S chiu vector iu khin

);

cvReleaseKalman(CvKalman** kalman

);

Cu trc CvKalman cha cc thng tin dng trong b lc Kalman: vector trngthi, ma trn li (Kalman gain) v cc ma trn lin quan.

typedef struct CvKalman {int MP; // measurement vector dimensionsint DP; // state vector dimensionsint CP; // control vector dimensionsCvMat* state_pre; // predicted state:

// x_k = F x_k-1 + B u_kCvMat* state_post; // corrected state:

// x_k = x_k + K_k (z_k- Hx_k)

() = 11(+)1 + 1

Tnh trng thi t m hnh h thng

(1) Cp nht trng thi:() = 11(+) + Buk-1 (iu khin)(2) Tnh ma trn hip phng sai ca sai s:

= ()() + 1 (+) = () + [ ()]

(+) = [

]

()

Kt hp gi tro to c lng ti u

(3) Tnh li Kalman:

(4) Kt hp gi tro to c lng ti u:

(5) Cp nht ma trn hip phng sai sai s:

Khi to 1(+) v 1(+)


50/87

CHNG 2.BLC KALMAN



CvMat* transition_matrix; // state transition matrix// F

CvMat* control_matrix; // control matrix// B

// (not used if there is nocontrol)

CvMat* measurement_matrix; // measurement matrix// H

CvMat* process_noise_cov; // process noise covariance// Q

CvMat* measurement_noise_cov; // measurement noise covariance// R

CvMat* error_cov_pre; // prior error covariance:// (P_k=F P_k-1 Ft) + Q

CvMat* gain; // Kalman gain matrix:// K_k = P_k H^T (H P_k H^T

+ R)^-1CvMat* error_cov_post; // posteriori error covariance

// P_k = (I - K_k H) P_kCvMat* temp1; // temporary matricesCvMat* temp2;CvMat* temp3;CvMat* temp4;CvMat* temp5;

} CvKalman;

Hm tin on trng thi, kt qu ca hm ny l tin nghim state_pre.

cvKalmanPredict(CvKalman* kalman,const CvMat* control = NULL

);

Hm tnh ton ti u cho c lng ca trng thi (kt hp gi tr ca phpo), kt qu ca hm ny l hu nghim stat_post.

cvKalmanCorrect(CvKalman* kalman,CvMat* measured

);

Sau khi thc hin 2 hm trn, chng ta thc hin xong mt chu kz ca bc lng

dng b lc Kalman.

Tm tt lm vic ca b lc Kalman trong OpenCV:

1. Khi to cc tham s v ma trn cn thit (trong cu trc CvKalman):

MP: kch cvector o DP: kch cvector o CD: kch cvector iu khin transition_matrix (F): ma trn chuyn trng thi control_matrix (B): ma trn iu khin (nu c)

measurement_matrix (H): ma trn o


51/87

CHNG 2.BLC KALMAN



process_noise_cov (Q): ma trn hip phng sai ca nhiuh thng

measurement_noise_cov (R): ma trn hip phng sai canhiu o

l cc tham s bt bin. Ngoi ra ta cn phi khi to trng thi ban u cho h thng:

state_post (x_k+): hu nghim (coi nh trng thi u tin),c th khi to ngu nhin.

error_cov_post (P_k+): ma trn hip phng sai ca sai s hunghim (coi nh sai s ca trng thi u tin)

2. Sau khi cc thng strn c khi to ta gi hm

x_k- = cvKalmanPredict( cautrucCvKalman, vectordieukhien )

ly v trng thi tin nghim. Tham svectordieukhien bng 0 nu khng c iu

khin.Khi gi cvKalmanPredict, trng thi tin nghim sc tnh bng cch ly hu nghimca bc trc nhn vi ma trn chuyn trng thi cng vi ma trn iu khin (nu c):

x_k- = F* x_k-1- + B*vectordieukhien

3. Sau ta ly gi tro _k v gi tip hm cvKalmanCorrect ly hu nghim:

x_k+ = cvKalmanCorrect(cautrucCvKalman, z_k)

Khi gi hm cvKalmanCorrect, cc thng s: li K_k, trng thi hu nghim x_k+ vma trn hip phng sai ca sai s hu nghim P_k+ v sc tnh.

Nh vy, s dng b lc Kalman trong OpenCV c th tm tt trong lc sau:

Khi to Lp

[ GicvKalmanPredict Cung cp gi tro GicvKalmanCorrect]


52/87

42

3Chng3

Pht Hin Vt ThDa Trn Mu Sc

Cc ni dung chnh:

H mu HSV Cc moment ca nh Thut ton pht hin vt th da trn

mu sc Ci t thut ton vi OpenCV


53/87

CHNG 3.PHT HIN VT THDA TRN MU SC



3.1 H mu HSV

3.1.1 nh nghaC nhiu hmu khc nhau dng biu din mu sc trong my tnh. Trong s thngdng l h mu RGB (trong mt s mn hnh LCD th dng BGR, y ch l sthay i v tr3 mu c bn).

Trong m thut ngi ta thng dng mu HSV (hay cn c tn HSB) hoc HSL v cc hmu ny trc quan, d cm nhn. Trong mc ny ta s trnh by v h mu HSV.

Hmu HSV c Alvey Ray Smith a ra nm 1978.

Hnh 3.1-1 Hnh nn ngc biu din h mu HSV

Hnh trn bn di y l mt mt ct ngang biu din H v S.

Hnh 3.1-2 Hnh trn biu din cc sc mu (H: 0-360) v bo ha (S: 0-1).Mu : H=0, mu xanh l: H=120, mu xanh dng: H=240

H mu HSV c 3 tham s:

H (hue) (0-360): sc mu. V dmu c H = 0, mu vng c H = 60.


54/87




S (saturation) (0-1): bo ha. Ch ra mc thun ca mu, mu c S cng lnth cng r, ngc li th cng m. Ti S=0 sc mu khng cn gi tr, mi H uch cho mu xm hoc trng.

V (value, c ni ghi B: Brightness) (0-1): pht sng. l cng nh sng mti tng pht ra ch khng phi do phn x t cc ngun sng khc.

Hmu HSV ny c biu din bi mt hnh nn ngc (Hnh 3.1-1).

Trong H (mu sc) c biu din bng gc quay quanh trc ng ca hnh nn, v vyH c gi tr 0-360o, vi:

Mu c H = 0o

Mu xanh l c H = 120o

Mu xanh dng c H = 240o

S l khong cch t trc hnh nn ti im xt mu. S c gi tr thc nm trong on 0-1.Ti trc: S = 0, ti vnh ngoi cng: S = 1. Ta c th quan st thy mu ca S cng r khicng nm gn vnh (S>0). Vi S = 0 th H khng cn { ngha, nh trong hnh , ti trc cahnh nn ch c mu xm v trng.

pht sng V (hay B) l khong cch tnh di ca hnh nn (v hnh nn b lt ngc)

ti mt ct ngang hnh nn cha im xt mu. Ti nh (bn di) ca hnh nn, V c gitr l 0, ti l mu en (khng c nh sng). V = 1 ti tm ca mt y pha trn, ti lmu trng (H: khng xc nh, S=0, V= 1).

Mt tnh cht quan trng ca h mu HSV l cc mu c c trng bng sc (H). Tcl mt vt th c mt mu nht nh th H ca n thay i rt t trong cc sng khcnhau ca mi trng. Ngha l, v d, mt vt th c mu vng vi H = 32. 3 th khi mitrng thay i sng, ch c gi tr S v V ca n (mu ca vt th) l thay i, cn H chdao ng trong khong 32.3 6. y l tnh cht khin ta chn HSV lm h mu cho

nh mun phn ngng (mc 3.3).

3.1.2 Chuyn t mu RGB sang HSVNh mc trn va trnh by, trong l thuyt ca h mu HSV th:

H: Sc mu [0-360] vi mu ti im 0 S: bo ha [0-1]

V: Gi trcng sng [0-1]

Nhng trong OpenCV, HSV c lu ging nh RGB, tc 1 byte cho mt gi tr H, S v V,nn khong c gi tr 0-255.

Bi v H c gi trln hn 255 nn nc lu l H/2, v d: H = 240 (mu xang dng)th OpenCV lu H = 240/2 = 120. Cn S v V th c gin thnh 0-255 (snguyn) tng


55/87




ng cho khong 0-1 (s thc). y l mt lu { quan trng biu din mu HSV trongOpenCV.

Thut ton Chuyn tmu RGB sang HSV:

(Thut ton ny tnh ton trn l thuyt. Sau khi c c gi trtrn l thuyt (Hl, Sl, Vl)

(vi Hl: 0-360, Sl: 0-1, Vl: 0-1) ta chuyn v gi trc lu bi OpenCV: H=Hl/2, S = Sl*255,V = Vl*255)

t:

M = max (R, G, B)

m = min (R, G, B)

C = M m

Gi tr H:

H = 60 *

, = 0 6, = + 2, = + 4, =

Gi tr S =

0, = 0

,

Gi tr V = M/255

Thut ton hon thnh.

Ch : mod l php chia ly d trn S THC (nh hm MOD trong MS Excel hoc modtrnhttp://www.wolframalpha.com).

Ni thm: vy trong ti ca chng ta, ly c khong ngng ti uta lm nh

sau:

Chp mt hnh cha vt th.

Pick mt mu ra (dng MS Paint hoc phn mm rgbhsv)

Tnh H, S, V theo thut ton trn (nhi sang dng ca OpenCV sau khi tnh (H: 0-

180))

Ch :

Nht thit phi tnh H theo thut ton trn, chH tcc chng trnh hoc code

khc lkhng ng.

H th thng bin thin rt t i vi mt mu, ta chcn +, - 3 vo H tnh l c

khong ti u.
http://www.wolframalpha.com/http://www.wolframalpha.com/http://www.wolframalpha.com/http://www.wolframalpha.com/


56/87




Cn S, V th chng trnh rgbhsv tnh ng, c thdng tra khong bin thin

ca 2 i lng ny cho thun tin.

V d: ta c mu vi gi tr RGB = (195, 125, 43) (mu vng hi ti nu).Ta tnh:

M = 195 (R)

m = 43

C = M-m = 152

Gi tr H:

(G-B)/C = 0.5395 0.5395 mod 6 = 0.5395

0.5395*60 = 32.37o (gi tr l thuyt) Gi trlu trong OpenCV l: H = 32.37 / 2 = 16

Gi tr S:

C/M = 0.7795 (gi tr l thuyt) Gi trlu trong OpenCV l: S = 0.7795*255 = 198

Gi tr V:

M/255 = 0.7647

Gi trlu trong OpenCV l: V = 0.7647*255 = 195

3.2 Moment ca nh (image moment)

Moment 2 chiu ca bc (p+q) ca mt nh f(x, y) c kch thc MxN c nh ngha l:

Mpq = (, )1 =01 =0 trong p=0, 1, 2, v q=0, 1, 2, l s nguyn.

Mt s tnh cht nh n gin c c trng bi moment nh l:

Din tch (cho nh nh phn13) hoc tng mc xm(i vi nh xm14): M00.

Tm nh: {xc, yc} = {M10/ M00, M01/ M00}

2 tnh cht trn ca moment sc dng trong phn pht hin i tng theo mu sc.

3.3 Thut ton pht hin i tng theo mu sc

T nhng kin thc trnh by trong cc phn trn, ta c th xy dng mt thut ton phthin i tng theo mu sc n gin nh sau:

13nh nh phn: l nh c mu trng (f = 255) v en (f = 0).14

nh xm: nh c mc xm f=0..255.


57/87




(1)Chuyn tnh RGB (hoc cc h mu khc) v HSV.(2)Thc hin phn ngng nh: to nh nh phn tnh HSV vi ngng trn v di

thch hp (tham kho cch ly ngng trong mc 3.1.2). Cc im nh nm trong

khong ngng trn v ngng di s c gi tr 1 (trng), cn cc im cn li sc gi tr0 (en).

(3)Lm mn nh phn ngng loi bcc im nhiu (c th dng lc trung v).(4)Tnh ton cc moment M00, M10 v M01 ly tm ca vt th (xc, yc) = (M10/ M00,

M01/ M00).

Thut ton hon thnh.

C mt sim cn lu { trong thut ton trn:

H mu c chn bc 1 phi l HSV. C mt h mu tng t c dng

trong nhiu h thng l HSL. Mc d 2 h mu ny da trn cng mt { tng vSc mu, bo ha v sng, nhng chng c nhng khc bit khin cho hmu HSL khng phi l ng vin tt nht cho phn ngng nh theo mu (sc ).

Khi ly cc gi trngng, nu tm c mt ngng (mt mu) c trong h RGB taphi chuyn n v h HSV theo theo thut ton trnh by mc 3.1.2. C nhiu

chng trnh (v c code) chuyn tmu RGB sang HSV nhng chng khng chnhxc v iu ny dn ti tht bi trong vic phn ngng nh. Ngoi ra ta cng philu cch biu din mu HSV trong OpenCV l HSV = (0-180, 0-255, 0-255) agi trvo chng trnh cho ng.

im lu { sau cng l vbc (3). Trong thc tci t, nu ta tin hnh lm mnnh (kh nhiu) snh hng rt nhiu ti hiu nng ca chng trnh (chngtrnh s chm i r rt). Bn cnh vic ny cng khng cn thit lm chngtrnh ca chng ta bi v mu ca vt th m ta chn kh khc bit vi mi trng,nn nu ta chn ngng tt th hu nh nh sau phn ngng s khng c nhiu15.Cho nn gii php y (cho phn ci t) l bqua bc 3. Thay vo , n

gin ta ch dng mt cu lnh iu kin:

nu (M00 < din_tch_min) th quay li bc (1)

Cu lnh trn c ngha l nu din tch ca vt th (M00) nh hn mt gi tr no

(din_tch_min) th ta b qua vic tnh ta tm, xem nh l mt nhiu ch khngphi vt th (v vt th ca ta lun c kch thc ln hn din_tch_min ny).

3.4 Ci t thut ton pht hin vt th da theo mu sc viOpenCV

Trong OpenCV c h tr chng ta cc hm dng cho thut ton pht hin vt th theo musc va trnh by.

15

Nhiu y l cc im nh khng thuc v vt th ta quan tm m vn lt c vo vng phn ngng.


58/87




Hm cvCvtColor

void cvCvtColor(const CvArr* src,

CvArr* dst,int code);

Hm ny dng chuyn nh t mt h mu ny sang hmu khc. Trong src l nh

ngun, dst l nh u ra, v code l m chuyn. Mt scode thng dng:

CV_BGR2RGB: chuyn t BGR sang RGB CV_RGB2BGR: chuyn t RGB sang BGR CV_RGB2RGBA: thm tham s Alpha cho h mu RGB CV_BGR2BGRA: thm tham s Alpha cho h mu BGR CV_BGR2HSV: chuyn t BGR sang HSV (ta dng m ny trong chng trnh)

Hm cvInRangeS

void cvInRangeS(const CvArr* src,CvScalar lower,CvScalar upper,CvArr* dst

);

Hm ny to ra nh phn ngng. Trong , src l nh ngun v dst l nh u ra. lower

v upper l 2 ngng ca ta, n c dng l mt 3 gi tr (H, S, V).

Hm cvMomentsvoid cvMoments(

const CvArr* image,CvMoments* moments,int isBinary = 0

);

Hm ny thc hin tnh ton cc moment ca nh, trong image l nh cn tnh momentv moments l mt bin cu trc cha cc moment ca nh.

typedef struct CvMoments {// spatial moments

double m00, m10, m01, m20, m11, m02, m30, m21, m12, m03;// central momentsdouble mu20, mu11, mu02, mu30, mu21, mu12, mu03;// m00 != 0 ? 1/sqrt(m00) : 0double inv_sqrt_m00;

} CvMoments;

Bng vic s dng cc hm trn chng ta c th ddng ci t thut ton pht hin vt

th da trn mu. Code c th cho phn ny sc trnh by trong chng 4.


59/87

49

4Chng4

Thit KV Ci tChng Trnh Pht Hin Di ng

Cc ni dung chnh:

Thit k chng trnh: M hnh hthng v Bo.

Ci t: Hot ng ca chng trnhv Cc on m quan trng.


60/87

CHNG 4.THIT KV CI T CHNG TRNH PHT HIN DING



4.1 Thit k

4.1.1 M hnh h thngNh trnh by trong chng 2, cc c lng s dng b lc Kalman phi gm 2 phn:M hnh h thng (tnh ton gi trc lng cho trng thi h thng) v Cm bin (ly gitro thc tca trng thi). Trong mc ny chng s trnh by v m hnh ca h thng.

Mc ch ca chng ta l xy dng mt m hnh theo di vt th chuyn ng, nh vychng ta c th m hnh h thng nh sau:

4.1.1.1 Vector trng thiVector trng thi l mt vector ct cha 4 thnh phn:

xk = Trong x, y l ta ca vt th, vx, vy l vn tc theo phng x v y. y chng ta xemvt th c mt chuyn ng vi vn tc khng i nn ta c ma trn chuyn trng thi nhsau:

=

1 0 1 0

0 1 0 1

0 0 1 0

0 0 0 1

Nu ly ma trn chuyn trng thi nhn vi vector trng thi ta s c trng thi bcsau:

xk+1 = xk =1 0 1 00 1 0 10 0 1 00 0 0 1

= + +

Tc l ta c xk+1 = xk + vx v yk+1 = yk + vy.

4.1.1.2 Nhiu h thngBi v vector trng thi c kch thc 4x1 nn vector nhiu h thng wk cng phi cng kchthc 4x1. Do ma trn hip phng sai ca nhiu h thng Q l mt ma trn 4x4:

Q = 103 0 0 00 103 0 00 0 103 00 0 0 103

N l mt ma trn ch c cc phn ttrn ng cho chnh (ch s hng = ch s ct) khckhng, bi v ta gi thit rng nhiu ca cc gi trkhc nhau l c lp thng k. V d

nhiu ca x v y l c lp thng k nn Cov(x, y) = 0. i vi cc gi trtrn ng cho


61/87




chnh, l phng sai ca bin ngu nhin , v d: phn t 0, 016 chnh l Var(X). Ccgi tr ny th hin ln ca khong gi tr ca nhiu (hay ni cch khc chnh l lnca nhiu). Nh ta bit, b lc Kalman ginh cc nhiu c phn phi chun vi kz

vng bng 0 (hm mt i xng qua trc Oy), v vy, nu phng sai (hay lchchun) cng ln th cc gi tr x ln s c xc sut cao hn. Ngc li, nu phng sai cngnh th cc gi tr cng tp trung quanh gi tr kz vng (y l 0), cc im x c gi tr lns c xc sut gn bng 0.

Nh vy, tm li, nu ta mun (hay ngh rng) nhiu c gi trln th chophng sai (cc

gi trtrn ng cho) ln v ngc li.

Biu 4.1-1 Mt sphn phi Gauss,vi phng sai 2 cng ln th cc gi tr x c gi tr ln (xa k vng) s c xc sut cao hn

4.1.1.3 Vector oMc d vector trng thi c 4 thnh phn, nhng cc gi trta c on chc 2, l ta x v y. V vy vector o k ca chng ta cng ch c 2 thnh phn x, y:

zk = Bi v vector trng thi c kch thc 4x1, vector o c kch thc 2x1 nn ma trn o H(th hin quan h tuyn tnh gia x v z) sc kch thc 2x4: (2x1) = (2x4)(4x1)

16

Ch s hng v ct trong OpenCV bt u t0 (nh mng trong C++).


62/87




H = 1 0 0 00 1 0 0

Ma trn H trn th hin ng quan h ca x v z, bi v x v z u l cc ta :

zk = Hxk = 1 0 0 00 1 0 0

= Ni thm v ma trn o: Ma trn o l ma trn thhin quan h tuyn tnh gia trng thiv gi tro. Xem li V d dn nhp mc 2.2.4, chng 2, ta c b phn nh vtr v ta

nn gi tro c quan h vi trng thi quan tm x - ta ca vt th, l:

z = a.x, a l h s tuyn tnh (m y l ma trn H)

Trong trng hp khc, nu ta khng c b phn nh vtr v ta m c ng h vntc tr v vn tc, th khi vector o l vn tc v d nhin n khng bng x. Gia z v xkhi sc mt h stuyn tnh a khc. Chng hn trong chuyn ng u th:

z = (1/t) * x, a = (1/t), t l thi gian tnh t lc x = 0.

Nh vy, ta li c thm mt lu { na, h sa (hayy l ma trn H) c ththay i

theo thi gian.Tuy nhin trong trng hp ca chng ta th n phi l khng i.

4.1.1.4 Nhiu oBi v vector o c kch thc 2x1 nn vector nhiu o wkcng c kch thc l 2x1, do

ma trn hip phng sai ca nhiu o R s l ma trn c kch thc 2x2.

R = 104 00 104

Nh trong mc 4.1.2 ni, cc gi tr khng nm trn trn ng cho s bng 0 bi vnhiu ca cc thnh phn khc nhau l c lp nhau.

Cc gi trtrn ng cho chnh cng ln th nhiu c th cng ln. y ta cho nhiucng nh chng tta cng tin tng php o (nhiu y c th coi l sai s ca php

o).

4.1.2 BoBo cho ra gi tr ca vector o z y chnh l hm pht hin vt th da trn mu scvit theo thut ton trnh by trong chng 3. u ra ca hm ny l ta tm ca vtth quan tm.

Ngoi ra cn c 2 gi tr cn khi to l:

Hu nghim ca bc 0: ta khi to vi gi tr ngu nhin. Ma trn hip phng sai sai s hu nghim bc 0: khi to l ma trn n v.


63/87




4.2 Ci t

Chng trnh vit cho n ny s dng th vin OpenCV 2.1 (xem cch ci t v thitlp trong phn Ph lc) vi ngn ng lp trnh C++, CLR Windows Forms Applicationproject, IDE Visual Studio 2008.

4.2.1 Hot ng ca chng trnhChng trnh c 3 form chnh: Ca siu khin, Camera v Khung nhn.

Ca siu khin: form ny th hin cc chc nng ca chng trnh: Theo di (mc nh): theo di v tr ca vt th c mu quan tm (vng v

hng). V: vcc ng theo di chuyn ca vt th.

iu khin: di chuyn, zoom mt nh theo chuyn ng v s xut hin cacc vt th. Thot: dng v thot chng trnh. Gii thiu: hin ca s thng tin vchng trnh.

Hnh 4.2-1 Ca siu khin

Camera: l ca s ghi hnh, n hin hnh thu tcamera cng nh kt qu ca chcnng v v theo di.

Hnh 4.2-2 Ca s camera hin kt qu ca chc nng theo di


64/87




Hnh 4.2-3 Ca s camera hin kt qu ca chc nng v

Khung nhn: ni hin th kt qu ca chc nng vv iu khin nh.

Hnh 4.2-4 nh kt qu ca thao tc v qua camera trn Khung nhn


65/87




Hnh 4.2-5 Zoom nh bng cch iu khin cc vt th

Hnh 4.2-6 Kt qu ca vic zoom nh trong Khung nhn


66/87




4.2.2 Cc on m chnhPhn ny s ghi li code v gii thch km theo ca mt s hm quan trng.

4.2.2.1Hm StartCamera_MotionTrackingHm ny s ly tng frame nh v gi cc hm khc xl{, sau n th hin kt qu.

public: void StartCamera_MotionTracking(){

/*Ta s tin hnh detect vt th da trn mu (color

based tracking)Sau dng b lc Kalman to ra kt qu ti u

theo Gi tro, m hnh h thng v nhiu*/

static CvScalar mauVang = CV_RGB(255,255,0);static CvScalar mauHong = CV_RGB(255,0,225);

//Khi to capture cho cameraCvCapture* capture;capture = cvCreateCameraCapture(0);

//Nu khng c camera th thot chng trnhassert( capture != NULL );

//Ly tng frame v x l

//To nh cha ng v quo ca vt th (tvic bt mu, cha lc Kalman)

IplImage* imgScribbleYellow = NULL;IplImage* imgScribblePink = NULL;

//nh cha 2 quo vIplImage* imgYellowAndPink = NULL;

//to nh cha frame ly t cameraIplImage* frame = NULL;

//ly ra mt frame t cameraframe = cvQueryFrame( capture );if( !frame ) return;

//v khi to cc nh cng kch thc vi frameimgScribbleYellow =

cvCreateImage(cvGetSize(frame), IPL_DEPTH_8U, 3);imgScribblePink = cvCreateImage(cvGetSize(frame),

IPL_DEPTH_8U, 3);imgKalmanYellow = cvCreateImage(cvGetSize(frame),

IPL_DEPTH_8U, 3);imgKalmanPink = cvCreateImage(cvGetSize(frame),

IPL_DEPTH_8U, 3);imgYellowAndPink = cvCreateImage(cvGetSize(frame),

IPL_DEPTH_8U, 3);

//Khi to tt c phn t v 0


67/87




cvZero(imgScribbleYellow);cvZero(imgScribblePink);cvZero(imgKalmanYellow);cvZero(imgKalmanPink);

cvZero(imgYellowAndPink);//Khai bo nh phn ngng mu vng v mu hngIplImage* imgYellowThresh = NULL;IplImage* imgPinkThresh = NULL;

//Khai bo cc bin cha v tr ca vt thstatic int posXYellow, XYellow_Kalman;static int posYYellow, YYellow_Kalman;static int posXPink, XPink_Kalman;static int posYPink, YPink_Kalman;

//v vtr trc (trong frame trc)int lastXYellow, lastYYellow, lastXPink,lastYPink;int XYellow_Kalman_last, YYellow_Kalman_last;int XPink_Kalman_last, YPink_Kalman_last;

//Cc ngng mu cn thit (h mu HSV)//TT NHT MU VNG://MU VNG: H = 17.4//MU HNG: H = 168.5//New: nh sng nhiu (1/1/2011)//MU VNG: H = 12.1//MU HNG: H = 166.6

//Ti u c//CvScalar yellowLower =cvScalar(14, 150, 130),

yellowUpper=cvScalar(20, 250, 240);//CvScalar pinkLower =cvScalar(166, 150, 150),

pinkUpper=cvScalar(171, 255, 235);

//Ti u mi://CH : KHNG GIM H THP QU (Vng: ko di 14).//v KHNG M KHONG S,V RNG QU, v s trng mu

da v cc vt khc

//TT NHT MU VNG:CvScalar yellowLower =cvScalar(11, 130, 150),

yellowUpper=cvScalar(20, 240, 255);//TT NHT MU HNG:CvScalar pinkLower =cvScalar(165, 150, 150),

pinkUpper=cvScalar(171, 255, 250);

//ngng S v V nh sau (rng qu) ko tt bng.(xem phin bn 0.4)

//Th vi gi tr scale to 255 (ch khng chia 2): KHNG chnh xc. (xem phin bn 0.4)

//=>(QUAN TRNG) Kt lun: H: tnh bng mod vchia 2 l chnh xc (nh wiki)

// S, V: lynh bn ngoi dng chng trnh RGB to HSV (chnh xc nht)

//Mu XANH DUONG (test)


68/87




//CvScalar pinkLower =cvScalar(100, 100,50),pinkUpper=cvScalar(140, 240, 150);

//tm vt th

CvPoint yellowCenter , pinkCenter;//bin bool chnh c gi tr mi hay ko (c

pht hin vt th trong frame mi ko)bool

coGiaTriMoiYellow=false,coGiaTriMoiPink=false;

//Khi to cu trc CvKalmanCvKalman* kalmanYellow;KhoiTaoKalman(kalmanYellow);CvKalman* kalmanPink;KhoiTaoKalman(kalmanPink);

//Vector o : vector ct 2x1: cha ta x v yCvMat* vectorDo_z_k =cvCreateMat( 2, 1, CV_32FC1

);cvZero( vectorDo_z_k );

const CvMat* trangThaiPre_UocDoan;const CvMat* trangThaiPost_ToiUu;

/*Bt u vng lp ly tng frame t camera v xl*/

while(tiepTuc){

//ly mt frame x lframe = cvQueryFrame( capture );if( !frame ) break;

//Lt ngc frame li (nh gng)cvFlip(frame,0 ,1);

/*X l bt i tng chuyn ng da trnmu sc*/

//To nh phn ngng mu vng v hngimgYellowThresh =

TaoAnhPhanNguong(frame,yellowLower,yellowUpper);imgPinkThresh=

TaoAnhPhanNguong(frame,pinkLower, pinkUpper);

//lm trn lc nhiu//HM NY LM

Phat Hien Muc Tieu Di Dong Su Dung Bo Loc Kalman Tran Nhat Quang

Documents

Transcript of Phat Hien Muc Tieu Di Dong Su Dung Bo Loc Kalman Tran Nhat Quang