1

Bo co tun 3

Tm hiu v phng php

bnh phng ti thiu v ng

dng trong php o GPS

Sinh vin thc hin: Nguyn Nam Tin

SHSV: 20092705

GV Hng dn: TS.T Hi Tng

anh Trng Minh c

anh Trn Trung Hiu

H Ni, thng 03/2014

2

Mc lc

Mc lc 2

Phn 1: Gii thiu v phng php bnh phng ti thiu 3 1.1. Khi nim 3 1.2. ng dng 4 1.2.1. Lp cng thc hi quy dng y=ax+b 4 1.2.2. Lp cng thc hi quy tng qut dng a thc bc m 5 1.2.3. Bnh phng ti thiu tuyn tnh 7

Phn 2: ng dng phng php bnh phng ti thiu x l tn hiu trong php o GPS 11 2.1. Quy trnh x l tn hiu GPS 11 2.2. X l tn hiu bng phng php bnh phng ti thiu 12 2.2.1. Thi gian 12 2.2.2. Gi khong cch 12

2.2.3. p dng bnh phng ti thiu tuyn tnh 12

Ph lc 15

Ti liu tham kho 16

3

PHN 1: Gii thiu v phng php bnh phng ti thiu

1.1. Khi nim:

Phng php bnh phng ti thiu l k thut c lng thng k c s dng ph bin nht trong cc m hnh hi quy tuyn tnh. Mc ch ca phng php l t cc mu ri rc quan st c trn thc nghim, xc nh mt hm biu din gn ng s phn phi ca cc mu , t c th c lng c cc gi tr cha th o c trn thc t (ni suy).

Gi s o c cc mu (xi , yi), vi i=1,2,3,,n. Mc tiu l xc nh hm f(x) tha mn:

f(xi) yi

Gi s hm f c th thay i hnh dng, ph thuc vo mt s tham s, pj vi j = 1, 2,..., m :

f(x) = f(pj, x)

Sai s gia gi tr thc v gi tr c lng theo hm f(pj, x) ti x=xi :

yi f(xi)

Xc nh cc gi tr pj sao cho biu thc sau t cc tiu:

4

iu ny gii thch ti sao tn ca phng php l bnh phng ti thiu.

i khi thay v tm gi tr nh nht ca tng bnh phng, ngi ta c th tm gi tr nh nht ca bnh phng trung bnh:

iu ny dn n tn gi bnh phng trung bnh ti thiu.

1.2. ng dng:

1.2.1. Lp cng thc hi quy dng y=ax+b:

Gi s bit c n gi tr thc nghim yi (i=1,2,3,,n) ca hm f(x) ti cc im xi tng ng. Tm hm xp x f(x) l mt a thc cp m c dng:

Pm(x)=ax+b

Theo nh ngha, ta c:

Coi S l hm c 2 bin a v b, nh vy S t cc tiu ti im m o hm ca S theo a v theo b ng thi bng 0:

Rt gn v chuyn v ta c:

5

Gii ra ta c:

1.2.2. Lp cng thc hi quy tng qut dng a thc bc m:

Gi s bit c n gi tr thc nghim yi (i=1,2,3,,n) ca hm f(x) ti cc im xi tng ng. Tm hm xp x f(x) l mt a thc cp m c dng:

Khi cc h s ai (i=0,1,2,m) s l nghim ca h phng trnh chun c dng:

Sai s trung bnh:

6

Trong thc hnh, cc tham s ai c xc nh bng bng sau:

V d: S dng phng php bnh phng ti thiu tm a thc bc hai:

xp x vi hm cho bi bng sau:

y m=2, n=5 v ta thu c bng sau:

7

T ta c h phng trnh:

Gii ra ta c:

Vy a thc cn tm c dng:

so snh cc yi vi P2(xi) v chun b tnh sai s trung bnh, ta thc hin tnh ton v c bng sau:

Ta c:

8

Sai s trung bnh:

1.2.3. Bnh phng ti thiu tuyn tnh:

Bnh phng ti thiu tuyn tnh l k thut s dng phng php bnh phng ti thiu tm nghim gn ng cho h phng trnh tuyn tnh khng c nghim chnh xc, khi s phng trnh (m) nhiu hn s bin (n).

Gi s cn tm nghim ca phng trnh:

(tng ng vi: f(xi) yi) trong : A l ma trn c m.n (m.n), theo nh ngha v phng php bnh phng ti thiu ta cn tm gi tr nh nht ca:

Bnh phng chun ca v l vTv, trong vT l ma trn chuyn ca v, ta c:

Ta c: (Ax)T(Ax) = ATAx2, cn 2 s hng gia th bng nhau, do ta c th coi - bTAx (Ax)Tb = -2ATbx. Cn bTb l hng s, do vy ta cn tm min ca:

(AT)(A)x2 2(AT)(b)x i vi hm bc hai y=ax2+bx+c, th ca n l mt ng parabol ngha l hm s s t cc tiu ti im m o hm bng 0. Do ta cn tm im c:

((AT)(A)x2 2(AT)(b)x)=0

9

hay:

Cui cng ta c:

V d: Cho cc im (0, 3), (2, 3), (4, 4), (1, 2). Cn tm mt phng trnh c dng x + = y, hay vit theo dng ma trn:

Ta c:

v vect b:

v sau :

Ta c phng trnh:

10

Sau :

Cui cng ta tm c:

Vy phng trnh ng thng cn tm l: (20/59)x + 152/59 = y

11

PHN 2: ng dng phng php bnh phng ti thiu aaaaaaaa x l tn hiu trong php o GPS

2.1. Quy trnh x l tn hiu GPS:

Quy trnh x l tn hiu GPS c nu r trong s sau:

C th cc bc thc hin nh sau:

1. Tnh gi khong cch (caculate pseudoranges): 2. Tm v tr v tinh pht tn hiu v hiu chnh sai s thi gian (find satellite position and clock offsets): qu trnh Actiquision.

3. S dng phng php bnh phng ti thiu tnh hm hi quy (least squares filter (az, ele, DOP, X, Y, Z):

4. Lu tr cc kt qu thu c (Store results):

12

2.2. X l tn hiu bng phng php bnh phng ti thiu:

2.2.1. Thi gian:

Gi s thi gian truyn tn hiu gia v tinh k v b thu i l ki. Khi vi c l vn tc nh sng trong chn khng, ta c:

2.2.2. Gi khong cch:

Khong cch a l gia v tinh v b thu nu khng tnh n sai s:

Trong Xk, Yk, Zk, Xi, Yi, Zi ln lt l ta ca v tinh v b thu trong khng gian theo 3 trc x,y,z.

Tuy nhin trn thc t, vic tnh khong cch b nh hng bi rt nhiu yu t dn n cc sai lch trong php o. Do cng thc tnh tnh gi khong cch s l:

Trong :

dti v dtk l ln lt l lch v thi gian (clock offset) ca ng h b thu v ng h v tinh. c(dti - dtk) l sai s khong cch gy ra bi s bt ng b ny.

Tki l sai s gy ra do tng i lu

Iki l sai s gy ra do tng in li

eki l sai s gy ra do quan st (observational error of the pseudorange) (?)

2.2.3. p dng bnh phng ti thiu tuyn tnh:

Xt v tr tng i ca b thu:

Chn im mc c ta : (Xi,0, Yi,0, Zi,0). Thng thng l chn tm Tri t c ta (0,0,0). Khi cc gia s X, Y, Z c xc nh bi:

13

p dng khai trin Taylor ta c:

Tnh o hm ca f(Xi,1, Yi,1, Zi,1) theo Xi,1, Yi,1, Zi,1 ta c:

Do ta c:

Cc sai s Tki, Iki, eki c th c xc nh v x l ring, nn ta coi chng nh hng s. Nh vy c cc bin s: Xi, Yi, ZI, dti.

Bin i di dng ma trn:

14

p dng phng php bnh phng ti thiu tuyn tnh, ta thu c:

Gii ra ta tm c hm tuyn tnh gn ng cho tn hiu.

15

Ph lc

1. Khai trin Taylor:

ngha:

nh l cho ta thy mt a thc xp x mt hm kh vi ti mt im cho trc (gi l a thc Taylor ca hm ) c h s ch ph thuc vo cc gi tr ca o hm ti im

nh gi chnh xc sai s xp x:

2. Ma trn chuyn v:

Ma trn chuyn v l mt ma trn cc hng c thay th bng cc ct, v ngc li. Cc tnh cht:

(A + B)t = AT + BT v (cA)T = c(AT)

(AB)T = (BT)(AT)

(AT)B = (BT)A

Nu ma trn A nghch o c th AT cng nghch o c, v (A-1)T = (AT)-1

16

Ti liu tham kho

1. Thit k b thu mm cho cc h thng dn ng s dng v tinh tin tin TS.T Hi Tng

2. n tt nghip: Nghin cu pht trin b thu mm GPS thi gian thc Trng Minh c, Trn Trung Hiu

3. Trang web: https://vi.wikipedia.org

4. Ti liu: http://doc.edu.vn/tai-lieu/do-an-phuong-phap-binh-phuong-toi-thieu-de-xap- xi-ham-trong-thuc-nghiem-6863/

5. A Software-Defined GPS and Galileo Receiver - Kali Borre, Dennis M.Akos

6. Ton hc cao cp Tp 1,2 Nguyn nh Tr, T Vn nh, Nguyn H Qunh