Tom Tat LV ThS Luu Van Quyen

HC VIN CNG NGH BU CHNH VIN THNG

------------------------------------

LU VN QUYN

S DNG B LC KALMAN

TRONG BI TON BM MC TIU

CHUYN NGNH: K THUT VIN THNG

Ms: 60.52.02.08

TM TT LUN VN THC S

H NI - NM 2013

Lun vn c hon thnh ti: HC VIN CNG NGH BU CHNH VIN THNG

Ngi hng dn khoa hc: GS.TSKH. NGUYN NGC SAN

Phn bin 1:.

Phn bin 2:.

Lun vn s c bo v trc hi ng chm lun vn thc s ti

Hc vin Cng ngh Bu chnh Vin thng

Vo lc:.gi..ngy.thng.nm..

C th tm hiu lun vn ti:

- Th vin ca Hc vin Cng ngh Bu chnh Vin thng

1

M U Vi s pht trin ca khoa hc k thut. Nhu cu s dng h thng nh v v

dn ng tr thnh mt nhu cu khng th thiu trong cuc sng ngy nay. T

ngun gc vic theo di bm st mc tiu hin i ngy nay l s kt hp ca cc

khoa hc k thut nhn dng mc tiu v phng php theo di i tng, bao gm

v tr, kch thc, hnh dng vn tc ca i tng. Bm mc tiu s dng cho nhiu

mc ch khc nhau nh v tinh gim st khng gian c s dng theo di cc

chuyn ng ca mc tiu nht nh.

Thi gian gn y vic ng dng b lc Kalman c lng qu o ca i

tng qua cc khung hnh c s dng nhiu trong cc thit b in t dn dng nh

Camera gim st, iu hng Robot, d tm mn, thit b kim tra hnh l. Cho n

nay c nhiu cng trnh nghin cu v lnh vc bm bt mc tiu trn c s x l

nh v cc thut ton bm theo i tng chuyn ng nh: So khp mu, Mean-

shift, Camshift, Particle, Kalman Mi phng php c cc u im v nhc im

khc nhau v cho hiu qu nht nh vi tng loi i tng v mc tiu theo di

khc nhau.

Ni dung ca lun vn c cu trc thnh cc phn nh sau:

Chng I. Tng qut v l thuyt bm mc tiu

Chng II. S dng b lc Kalman trong bi ton bm mc tiu

Chng III. V d minh ho

2

CHNG I: TNG QUT V L THUYT

BM MC TIU 1.1. Nhng khi nim c bn.

1.1.1 nh ngha v bi ton bm mc tiu Cho mt i tng (S) c u ra l y(t) trc tc ng ca u vo u(t)

Hnh 1.1. S ca i tng iu khin vi u vo v u ra

Bi ton bm mc tiu cn phi tm tn hiu iu khin u vo sao cho tn

hiu u ra ( )y t bm theo mc tiu.

1.1.2 nh ngha v sai s Hu nh cc phng php nh gi, c lng tham s m hnh c xy

dng trn c s p dng nguyn l v k thut tham chiu, trong xc nh mt

hm sai s phn nh s khc lch gia m hnh v h ng hc thc.

1.1.2.1 Phng php sai s u ra

Hm sai s c nh ngha:

0 ( ) ( ) ( )e t y t y t= (1.1) 1.1.2.2 Phng php sai s u vo

Trong phng php sai s u vo khng s dng trc tip d liu o lng v

o hm cc bc theo thi gian ca tn hiu u vo h ng hc nn khng cn phi

quan tm n c tnh kch thch lin tc nn bi ton c lng tham s m hnh ni

ring, nhn dng h ng hc ni chung bao gi cng c nghim.

1.1.2.3 Phng php sai s phng trnh.

Sai s phng trnh c nh ngha trc tip t phng trnh ng hc ca

m hnh nh sau:

( ) ( ) ( ) ( ) ( )ce t H s y t K s u t= (1.2)

31.1.2.4 Phng php sai s d bo

Sai s d bo c nh ngha nh sau:

( ) ( )( ) ( ) ( ) ( )( )pe

C s K se t y t u tH sD s

= (1.3)

1.1.3 S dng tiu ch ti u

1.1.3.1 Khi nim

Ch tiu cht lng J ca mt h thng c th c nh gi theo sai lch ca

i lng iu khin, thi gian qu hay theo mt ch tiu hn hp trong iu kin

lm vic nh hn ch v cng sut, tc , gia tc

1.1.3.2 Tiu ch ti u tc ng nhanh (thi gian ti thiu)

i vi bi ton ti u tc ng nhanh th ch tiu cht lng J c dng.

0

1T

J dt T= = (1.6) 1.1.3.3 Tiu ch nng sut ti u.

Nng sut y c xc nh bi cht lng ca h thng bm theo mc tiu

trong thi gian T nht nh. Khi ch tiu cht lng J c dng.

00 0

[ ( ), ), ]T T

TJ L x t ut t dt tdt = = = (1.7) 1.1.3.4 Tiu ch nng lng ti u.

Ch tiu cht lng J i vi tiu chi nng lng ti thiu c dng.

2

0

( )T

J u t dt= (1.10) 1.1.4 Xy dng khu phn hi. Xt mt h thng c m t bi cc phng trnh u ra v trng thi:

x Ax Buy Cx= + =

& (1.11)

Chn mt lut iu khin c dng:

( )u r Kx= (1.12)

4

x&

Hnh 1.3: iu khin s dng phn hi bin trng thi

1.1.5 Xy dng iu khin bm bng phn hi trng thi. Bi ton t ra l iu khin i tng c m t:

A

T

dx x budty c x

= +=

(1.17)

0 1 1

1 2

0 0 ... 0 00 0 ... 0

01

( )n

n

dx z udt

a a a

y c c c z

= + =

MM M M

% % %L

(1.18)

Taz

Hnh 1.4 iu khin bm vi i tng (1.18)

i tng(1.17)

STaz x

u y

Hnh 1.5 iu khin bm vi i tng (1.17)

51.1.6 Bi ton tng hp h thng

Bi ton tng hp h thng l ton b qu trnh tnh ton, la chn b sung

thm cc khu ph hp vo h thng. h thng khi hot ng t c nhng

yu cu cht lng ra v sai lch, thi gian p ng qu .

1.2 Phng php bm mc tiu truyn thng.

1.2.1 Tho lun phng php iu khin truyn thng s dng thng tin, d liu ca tn hiu u ra ca i

tng iu khin lm tn hiu u vo a ra tn hiu iu khin i tng.

H thng ny c biu din bi cc phng trnh sau:

( ) ( ) ( ) ( ) ( )x t A t x t B t u t= +& (1.25) '( ) ( ) ( )y t C t x t= (1.26)

Hnh 1.8 H thng iu khin c in

1.2.2 B iu khin PID: (Proportional-Integral-Derivative) Tn gi PID l ch vit tt ca ba thnh phn c bn c trong b iu khin

(hnh 1.10a) gm khu khuch i (P), khu tch phn (I), khu vi phn (D).

sDT

1

IT s

Hnh 1.10 iu khin vi b iu khin PID

B iu khin PID c m t c dng tng qut sau.

0

1 ( )( ) [ ( ) ( ) ]t

P DI

de tu t k e t e d TT dt

= + + (1.28) T m hnh vo ra tng qut ta c c hm truyn ca b iu khin PID

61( ) [1 s]P DI

R s k TT s

= + + (1.29)

1.2.3 Chn tham s PID ti u theo sai lch bm Bi ton c nhim v xc nh cc tham s ca b iu khin PI, gm kp, TI

trong cng thc (1.29) hoc kp, TI, TD trong cng thc (1.31) sao cho tn hiu ra y(t)

bm c vo hiu lnh (t) mt cch tt nht theo ngha.

2 2( ) ( ) ( ) minQ t y t e t= = (1.32)

V bi ton thit k b iu khin PID ti u tr thnh

arg min ( )p f p = (1.34) 1.3. Phng php bm mc tiu hin i.

1.3.1. Tho lun phng php. L thuyt iu khin hin i s dng m t khng gian trng thi trong min

thi gian, mt m hnh ton hc ca mt h thng vt l nh l mt cm u vo, u

ra v cc bin trng thi quan h vi phng trnh trng thi bc mt.

Xut pht t quan im d trin khai l trng hp lut iu khin tuyn

tnh, c cho bi:

u(t) = K(x(t),t) (1.36)

Hnh 1.12: M hnh phng php iu khin hin i.

71.3.2. B quan st trng thi.

B C

A

B C

A

L

System

u ++

x& xy

x&++

Observer

y

x

x

Hnh 1.14 B quan st trng thi Luenberger bc y

H thng trong hnh 1.14 c nh ngha bi:

x Ax Bu= +& (1.38) y Cx= (1.39)

1.3.3 iu chnh trng thi (LQR) (Linear Quadratic Regulator) Kho st vn duy tr trng thi ca h thng gi tr l 0, chng tc ng

nhiu, ng thi vi cc tiu tiu hao nng lng 0, (0)x Ax Bu x x= + =& (1.43)

y Cx=

0

1min ,2

T TJ x Qx u Ru dt = + (1.44)

Chn lut iu khin hi tip trng thi u = - Kx, K l hng s, thay vo biu

thc ca J

0

1 ( )2

T TJ x Q K RK xdt

= + (1.45) 1.3.4 Gii thut thit k LQG (Linear Quadratic Gausian)

Gi s phng trnh o lng ng ra c cho bi.

8wx Ax Bu = + +& (1.57)

y Cx v= + Gi s phng trnh hi tip c trng thi y .

u = -Kx + r (1.58)

Nu K c chn s dng phng trnh Riccati LQR v L c chn bi s

dng phng trnh Riccati ca b lc Kalman. iu ny c gi l thit k LQG.

iu quan trng ca cc kt qu ny l trng thi hi tip ca K v li ca b quan

st L c th c thit k ring r.[4]

1.3.5 M t b c lng trng thi gim bc Cho mt h ng hc S tuyn tnh bc n m t bi

(S) n n n n nx A x B u= +& (1.62) n n ny C x= Vi , bc ca phn ng thi iu khin v quan st c ca S, cc

vector v ma trn c kch thc ph hp.

Hy xc nh b nh gi trng thi (SE) bc e,

(SE) e e e e ex A x B u= +& (1.63) e e ey C x=

1.3.6 Gim bc phn t iu khin Cc bi ton lin quan n phn t iu khin da vo tn hiu phn hi lm

c s ra chin lc iu khin v cn phi x l trong khu khp kn. T v tr

xut pht ca tn hiu phn hi m trong l thuyt h thng chia ra thnh iu khin

truyn thng v iu khin hin i

1.4. Vai tr ca b lc Kalman

1.4.1 t vn . Phng trnh trng thi ca i tng

wx Ax Bu = + +& (1.72) y Cx v= + (1.73)

9

x&

x& x

x

y

Hnh1.15 B quan st trng thi ca Kalman

Phng trnh trng thi ca khu lc Kalman:

( ) x Ax Bu L y yy Cx= + + =

& (1.74)

Mc tiu ca thit k b lc Kalman l tm li c lng L c s c lng ti

u trong s hin din ca nhiu w(t) v v(t)

Sai s c lng:

( ) ( ) ( )x t x t x t= (1.75) 1.4.2 M hnh ton hc.

1.4.3 Qu trnh c lng trng thi. Qu trnh c lng s dng phng php mch lc Kalman trong gim st bm mc

tiu dc chia thnh hai giai on.

10

1k x F x

=

1T

k kP FP F Q

= +

1T Tk k kK P H ( HP H R )

= +k k k k k x x K ( z Hx )

= +

1k k kP ( K H )P=

Thc cht ca gii thut Kalman tuyn tnh l mt phng php c tnh

quy tuyn tnh cho php c lng trng thi ca mt h thng c nhiu sao cho

lch gia gi tr c lng v gi tr thc t l b nht.

1.4.4 Vai tr ca b lc Kalman Lc Kalman nhm c lng gi tr ch thc ca mt ci g , bng cch d

on gi tr ca n v tnh tin cy (hay bt nh) ca d on , ng thi o

c gi tr (nhng b sai s v c cc nhiu), sau ly mt trung bnh c trng gia

gi tr d on v gi tr o c c, lm gi tr c lng. C th coi n l mt

trng hp ca suy din c iu kin kiu bayes Cc thuc tnh c bn ca b

lc Kalman c bt ngun t cc yu cu ca c lng trng thi.

1.5 Kt lun chng Trong chng ny lun vn nu ra c cc khi nim c bn v bi ton bm

mc tiu, trn nhng khi nim c bn nu ra c phng php bm mc tiu

truyn thng, phng php bm mc tiu hin i. T tm ra vai tr ca b lc

Kalman trong bi ton bm mc tiu.

11

CHNG II: S DNG B LC KALMAN

TRONG BI TON BM MC TIU 2.1. Cc bin th ca b lc Kalman

2.1.1. Nguyn tc c bn Trong ng dng gim st, bm mc tiu di ng, mch lc Kalman l qu trnh lp i

lp li bc d on v hiu chnh trng thi ca h thng [13]. Xt mt h thng i

din bi mt khng gian trng thi nh phng trnh (2.1) v (2.2).

xk = Fxk-1 + vk (2.1) zk = Hxk + ek (2.2)

2.1.2. Mch lc Kalman tuyn tnh Mch lc Kalman tuyn tnh a ra mt c lng ti u cho trng thi k -tip s

dng cng thc tuyn tnh, gi s cc bin c phn b xc sut Gaussian.

- Gi tr trung bnh cho trng thi k tip:

1 k kx Fx = (2.3) - Hip phng sai ca c lng k tip:

1 Tk kP FP F Q = + (2.4) - Tnh ton li mch lc Kalman:

1( )T Tk k kK P H HP H R = + (2.5) - Gi tr hiu chnh trung bnh:

( )k k k k kx x K z Hx = + (2.6) - Hiu chnh hip phng sai:

( )k k kP I K H P= (2.7) 2.1.3. Mch lc Kalman m rng B lc Kalman m rng thc hin theo cc bc c lng

- Gi tr trung bnh cho trng thi k tip:

1 k kx x = (2.8) - Hip phng sai ca c lng k tip:

1k kP P Q = + (2.9)

12- Tnh ton li mch lc Kalman:

1( )k k kK P P R = + (2.10) - Gi tr hiu chnh trung bnh:

( )k k k k kx x K z x = + (2.11) - Hiu chnh hip phng sai:

( )k k kP I K P= (2.12) 2.1.4. Mch lc Unscented Kalman

Nguyn tc c bn ca Unscented Kalman l bin i Unscent. V c bn, y l mt

phng php tnh ton thng k mt bin ngu nhin sau khi bin i khng tuyn

tnh. Cho bin ngu nhin n chiu: xk-1 vi gi tr trung bnh $ 1kx v ma trn hip

phng sai Pk-1

Mch lc Unscented Kalman m t trng thi vi mt tp hp ti thiu cc im

(sigma) mu c chn lc cn thn. 2n+1 im sigma c chn xung quanh c

lng trc , vi n l kch thc ca khng gian trng thi. Sau mt trng s

xc sut c gn cho nhng im sigma. Tip theo, cc im sigma ny bin i

bng cch s dng bin i Unscent a ra mt c lng mi cho bin trng thi.

Bin trng thi sau c hiu chnh bng cch bin i cc im sigma thng qua

cc m hnh o lng tnh ton li Kalman. Cui cng, c lng c hiu

chnh s dng li Kalman

2.2. Lc Kalman trong bi ton bm mc tiu theo phng php

phn on.

2.2.1. Tho lun bi ton.

2.2.2 M hnh bi ton. u vo l mt chui cc khung hnh, gi nh rng khng c s thay i v

cng nh sng v khng c hin tng che khut. Ta c th vit nh sau:

yk(x) = yk-1(x dk(x)) ( 2.35) M hnh quan st cho khung hnh th k tr thnh.

gk(x) = yk(x) + nk(x) ( 2.36)

13 Cn phi c lng phn phi xc xut c iu kin kt hp ca trng vecter

chuyn ng dk, trng phn on cng sk, v trng phn on i tng (hay

video) zk. Dng lut Bayes ta c:

( ) ( )( )1 11 1 1 1, , , ,

, , , ,, ,

k k k k kk k k k k k

k k k

p d s g g gp d s z g g g

p g g g +

+ +

= (2.37)

M hnh mng Bayes th hin s tng tc gia 1 1, , , , ,k k k k k kd s z g g g +

gk

sk

dk zk

gk-1, gk+1

Hnh 2.1 M hnh mng Bayes cho bi ton phn on video

2.3 Bm mc tiu theo quy trnh ng thi. Trong phn ny cp h thng gim st mc tiu 3D nh hnh 2.2 H thng gim

st mc tiu, mc tiu theo di l ngi di chuyn trc ng knh camera, thu nh,

lu thnh file .avi v a vo h thng nhn dng v theo vt s dng tng mch lc

Kalman bm theo i tng cn theo di.

Hnh 2.2. H thng bm mc tiu

M t h thng:

S h thng gim st mc tiu hnh 2.2. H thng gm tn hiu vo v b phn

pht hin, bm mc tiu v a ra kt qu hin th.

14

Hnh 2.3. S nhn dng nh

B lc Kalman c coi nh b c lng trng thi h thng, c cu trc lc n

gin v hi t tt cng vi kh nng lc nhiu cao [9]. M hnh cn c c

lng d bo c m t bi h phng trnh trng thi :

xk = Fxk-1 + vk

zk = Hxk + ek Vector trng thi xk=[x, y, vx , vy], vector o lng zk = [x, y]T, ng vi ta v vn

tc ca nh i tng trn mt phng nh thi im k. vk, ek l vector nhiu trong

qu trnh chuyn ng v sai s php o.

2.4. Kt lun chng Trong chng ny lun vn nu tng v cc bin th ca b lc Kalman ng dng

thut ton mch lc Kalman trong bi ton bm mc tiu theo phng php phn

on, bm mc tiu theo quy trnh ng thi. a ra gii php ng dng thut ton

lc Kalman theo vt i tng, t file video thc hin tng b lc Kalman bm

theo ngi di chuyn.

15

CHNG 3: V D MINH HO 3.1. Bi ton bm mc tiu

3.1.1 t vn . Mt h thng bm mc tiu bng hnh nh l mt tp hp cc bi ton nh.

u vo ca h thng s l hnh nh thu c ti cc im quan st.

u ra ca h thng s l thng tin v chuyn ng ca cc i tng c

gim st

M hnh khi qut chung cho h thng bm mc tiu.

Hnh 3.1 H thng bm mc tiu tng qut

3.1.2 Bi ton pht hin i tng chuyn ng u vo ca bi ton pht hin i tng chuyn ng l cc khung hnh video

thu c t cc im quan st, theo di. Nh vy c th gii quyt bi ton ny ta

cn nghin cu mt s c im ca video.

3.1.2.1 Cc khi nim c bn v video.

3.1.2.2 Mt s thuc tnh c trng ca video

3.1.3 Bi ton phn loi i tng

3.1.3.1 Phn loi da trn hnh dng.

3.1.3.2 Phn loi da trn chuyn ng.

3.1.4 Bi ton theo vt i tng

3.1.4.1 t vn

u vo ca bi ton theo vt i tng l cc vt i tng, cc c trng ca

i tng c pht hin thng qua khi x l pht hin i tng, phn loi i

tng. Nh vy nhim v ca vn theo vt i tng l chnh xc ha s tng

16ng ca cc vt i tng trong cc khung hnh lin tip t d on hng chuyn

ng ca i tng.

3.1.4.2 Cc vn gii quyt

- Theo vt mc tiu da trn m hnh

- Theo vt mc tiu da trn min.

- Theo vt mc tiu da tn ng vin

- Theo vt mc tiu da vo c trng

* Chnh xc ho i tng tng ng (Object matching):

* D on chuyn ng

Nu gii quyt bi ton bm theo mc tiu t hiu qu v tin cy cao, c th ng

dng trong rt nhiu lnh vc.

3.2 Chng trnh m phng bm mc tiu

3.2.1 Qa trnh thu nhn v nhn dng nh M hnh h thng Camera gim st mc tiu:

Qu trnh ghi hnh c thc hin bng Webcam ca my Laptop thng qua chc nng h tr Image Acquistion ca phn mm matlab v lu li vi dng .avi hoc

.mat. Sau s dng file ny input cho module nhn dng nh v bm theo vt mc

tiu c thc hin bng b lc Kalman

Chng trnh m phng qu trnh nhn dng v bm mc tiu thc hin theo lu

hnh 3.6.

17

3.2.2. Bm mc tiu s dng thut ton Kalman Sau khi mc tiu c nhn dng, pht hin chuyn ng t rt trch c trng s

c thut ton Kalman bm theo vt i tng thc hin theo lu hnh 3.7.

18

Rt trch c trng

Kt thc

Hin th

N

Y

Hnh 3.7 Lu thut ton lc Kalman

D on

Tnh li Kalman

Hiu chnh

3.2.2.1 Thut ton mch lc Kalman tuyn tnh

Bc d on:

$ $ 1k kx F x

= , (3.1) 1 Tk kP FP F Q = + .

li Kalman: 1T T

k k kK P H ( HP H R ) = + . (3.2)

Bc hiu chnh:

$ $ $k k kk kx x K ( z H x ) = + , (3.3)

k k kP ( I K H )P= . 3.2.2.2 Thut ton mch lc Kalman m rng

19 Bc d on:

$ $ 1k kx x

= , (3.4) 1k kP P Q = + .

li Kalman: 1

k k kK P ( P R ) = + . (3.5)

Bc hiu chnh:

$ $ $k k kk kx x K ( z x ) = + , (3.6)

k k kP ( I K )P= . 3.2.2.3 Thut ton mch lc Unscented Kalman

Bc d on:

$ 20

L ( m )k i k ii

x W ( X )

== , (3.7) $ $2

0

TL ( c )

k kk i k i k iiP W ( X ) x ( X ) x

=

= . (3.8) Bin i Unscented:

k i k i( Z ) h(( X ) )= , i = 0,...,2L (3.9) 2

0k k

TL ( c )

k ki k i k iz z iP W ( Z ) z ( Z ) z R

=

= + $ $ $ $ , (3.10) $

$20kk

TL ( c )

kki k i k ix z iP W ( X ) x ( X ) z

=

= $ $ , (3.11) li Kalman:

$1

k k kkk x z z zK P P= $ $ $ . (3.12)

Bc hiu chnh:

$ $ kk k k kx x K ( z z ) = + $ , (3.13)

k k

Tk k k kz z

P P K P K= $ $ . (3.14)

3.3 Kt qu thc nghim M phng mch lc Kalman bm mc tiu. Cc khung nh c ly ngu

nhin l frame 72, 81, 93, 125 ca mt on video .avi khc nhau.

20

Frame 72

Frame 81

Frame 93

Frame 125

Hnh 3.8: Kt qu bm mc tiu

21Gi tr o lng thc t ti v tr i tng c biu din bi mt hnh ch

nht mu en trong khi mu xanh th hin cho d on v tr ca mc tiu. Ngoi ra

kch thc ng bao c th hin qua nh xm nn en, bng nh m mu xanh

bm theo ng bao.

T hnh 3.5 cho thy mch lc Kalman c th d on v nh hng chnh

xc cao. hnh ch nht mu xanh (d on) ph hp tng i chng kht vi hnh

mu en (o lng). Mch lc Kalman c p ng nhanh. iu ny c ngha l khi

i tng di chuyn th cho php o thay i t ngt.

Kt qu m phng ca phng php bm mc tiu s dng mch lc Kalman,

gi tr sai s c lng RMSE c tnh theo cng thc.

( )RMSE MSE GT ES= (3.15) 2

1

( )n

t

GT EMSEn== (3.16)

3.4 Kt lun chng

Bi ton nghin cu mt s k thut pht hin v bm mc tiu, ng thi

tin hnh x l cho ra kt qu l i tng ang cn theo vt ang v tr no

nh du. Sau khi xc nh v tr i tng, s tip tc iu khin thit b ti v tr

mong mun (v tr ca i tng ang theo vt), ng thi quyt nh ra s kin

22

KT LUN V HNG PHT TRIN 1. Kt lun.

* V mt l thuyt.

Lun vn nu ln tng quan v bm mc tiu, cc khi nim lin quan n

x l hnh nh trong bm mc tiu, phn tch cc loi nhiu trng thi v m hnh nh

hng n qu trnh theo di mc tiu di chuyn. Cc phng php bm mc tiu

nh So khp mu, dng quang, Meanshift, Camshift, tr nh nn, lc Particle, c

lng Kalman. Mi phng php c nhng im mnh v hn ch khc nhau. Tuy

nhin mch lc Kalman vn l la chn ti u cho qu trnh bm mc tiu xut pht

t cc u nhc im ca n.

* V mt thc tin.

Lun vn a ra hng tip cn ng dng mch lc Kalman trong bi ton

bm mc tiu c th nh s dng phng php nhn dng hnh nh ca phng php

tr nh nn, trch chn c trng, s dng thut ton mch lc Kalman bm mc

tiu chuyn ng. a ra kt qu m phng, nh gi kt qu sai s c lng v o

c.

2. Hng pht trin Trong qu trnh thc hin ti, do nhng hn ch v trnh v thi gian

thc hin ti, chng trnh c xy dng ch l cc thut ton pht hin chuyn

ng v theo vt mc tiu da vo video. trin khai trong thc t n i hi cn

phi ci tin hn na. Hy vng trong tng lai, nhng pht trin di y s gip

ti hon thin hn.

- Kt hp vic pht hin khun mt vi vic pht hin mt, pht hin hnh dng

ca con ngi.

- Xy dng c thut ton ci thin cht lng ca video nh loi tr nhiu,

loi tr bong v ti u ha cc thut ton tng tc ca chng trnh.

- Nghin cu mch lc Unscented Kalman phi tuyn cn chnh h thng v

tinh.