Bài tập dài đo lường

7/30/2019 Bi tp di o lng

1/20

Bi tp ln o Lng TTCN Khoa in t

Cho mt trm bin p c cc tham s nh bng 1, bng 2.Bng 1: Tham s ca trm bin p

Ph ti Loi s Ch non ti cos28000KVA-35/10KV III U=Um; I=0.72Im; cos=0.75 0.92

Bng 2: Cp chnh xc ca cc thit b o v yu cu truyn dnCp chnh xc cc thit b S

knh chnh

xcPhng php

truyn dnTI TU V A KWh KVARh2 2 2 2 2 1.5 15 2 Thi gian

Phn I:Cu 1: V s o dng, p, cos, tn s, nng lng tc dng, nng lng

phn khng cho trm pha cao pS III: cng t tc dng 3 pha 3 phn t, cng t phn khng 3pha 2 phn t to gc lch pha 60o.

+ Cng t tc dng:.

AI,

.

AU.

BI ,.

BU.

CI ,.

CU+ Cng t phn khng:

.

AI,

.

ABU.

BI ,.

CAU

Chng minh:

+Cng t o nng lng tc dng

P3f = PA + PB + PC= UAIAcos(.

AU.

AI ) + UBIB cos(.

BU.

BI ) + UCIC cos(.

CU.

CI )

+Cng t o nng lng phn khngMq=K(ABA sin1 + BCA sin2 )

Trong :A=CAIA 2=(BCA)

B=CBIB 1=(ABA)BA=CBAUCA

GVHD: Phm Ph Thim - 1 - SVTH: Mai Vn Tuyn


2/20


CA=CCAUCA1=210o-=>sin1=sin(30o- )2=150o-=>sin2=sin(30o+)

=>Mq=KIdUd[sin(30o+) + sin(30o-

)] =KIdUd2cos30osin= KIdUdsin=KQ3f (pcm)

Cu 2: Chn thang o cho cc thit btrn s 1, Chn thang o bin dng in.

Dng in nh mc pha s cp l:

( )28000

461,883. 3.35

dmdm

dm

SI A

U= = =

Mun o c dng in ny ta phi s dng my bin dngI

B vi yu

cu t ra: Iscm 461,88 (A); Itcm= 5 (A)

Vy chn BI c : KI =500

52, Chn thang o cho bin in p .

in p nh mc pha s cp ca my bin p l: UmBA=35 (KV)

Cho nn chn BU c h s bin in p: KU=35000

100=

35

0,1

3, Chn thang o cho ampemet.V dng in pha th cp ca BI nh mc l 5(A) nn chn ampemet c

thang o l: DA=5(A)

Dng ampemet in t:

Ampemet in t c ch to da trn c cu ch th in t.Ampemetny o c c dng mt chiu v dng xoay chiu

B phn chnh l mt cun dy dn, c th quay quanh mt trc, nm trongt trng ca mt nam chm vnh cu. Cun dy c gn vi mt kim ch gcquay trn mt thc hnh cung. Mt l xo xon ko cun v kim v v tr skhng khi khng c dng in

Nguyn l hot ng:Khi dng in mt chiu chy qua cun dy, dng in chu lc tc ng

ca t trng (do cc in tch chuyn ng bn trong dy dn chu lcLorentz) v b ko quay v mt pha, xon l xo, v quay kim. V tr ca u

kim trn thc o tng ng vi cng dng in qua cun dyy l mt thit b thc t o dng in c bn ngoi th trng:



3/20


=> DA=5(A)

4, Chn thang o cho volmet.V in p pha th cp ca BU nh mc l 100(V) nn chn volmet c

thang o l: DV=100(V)

Cu to: Mch t gm li t bng tn silic c c tnh t tt v cun dycch in cao. C cu chnh Kim v 0 bng c kh. Cn du n nh kim bngnam chm vnh cu, bng nha baklit c cch in cao. V bng nhaABS. Ca s bng nha PC trong sut (hn ch chy). 2 ci v d cho viclp t Vn mt vo bng in.



4/20


=>DV=100(V)

5, Chn thang o cho cosmet .V cosmet c cun dng mc vo th cp BI , cun p mc vo th cpBU nn ta chn cosmet c:

Imcosmet = 5(A)Umcosmet = 100(V)

Do gc lch pha gia in p v dng in khng ph thuc vo h sbin i ca BU v BI nn ta chn thang o ca cosmet l (0.5 1)



5/20


Nguyn l hot ng:Cun tnh c mc ni tip vo pha A, hai cun dy ng c mc vi

hai in tr R v c t vo cc in p UAB v UAC

Gc quay ca c cu l:

Ch :



6/20


+ Trong s ny cc cun p u ni tip vi in tr R nn khng

ph thuc tn s, hay 21

I

I = const.

+ Cun dng c th mc vo cc pha B, C ty .

6, Cng t o nng lng tc dng .V cng t o nng lng tc dngc cun dng mc vo th cp BI ,cun p mc vo th cp BU nn tachn:

Idm = 5(A)Udm = 100(V)=0,1(kV)

.3

W P ttddm pha dm

=

Vi Cos dm = 1, tdm = 30 ngy =720h

3.0,1.5.1.720 623,54( )W kWhtddm

=

Chn thang o cng t nng lng tc dng DWtd=625 KWh

7, Cng t o nng lng phn khng.V cng t o nng lng phn khng c cun dng mc vo B I , cun p

mc vo BU nn ta chn:Imctpk = 5(A)Umctpk = 100(V)=0,1(kV)

3 . 3. . .720pha dm dm dmW Q t U I Sinpkdm= = (ly Sin=1)

3.0,1.5.1.720 623,54Wpkdm

= AR(kV h)

Chn thang o ca cng t phn khng DWpk= 625 KVARh

8, Chn thang o cho tn s met.V thit b o tn s mc vo th

cp ca BU nn ta chn : Um = 100(V).Vi tn s ca mng in l 50HZ v t dao ng nn ta chn thang o trongkhong 45 55 HZ


3. . . .W U I Cos t tddm dm dm dm=


7/20


PhotoHnh nh

Cat.NoM s

TypePhn loi

RangeThang o

NG H O TN S - HIU MUNHEAN SINGAPORE Mt 96 x96 mm

CP-96FQP

Kim 220V ; 45 55Hz

ng h o tn s met cng hng

Nguyn l hot ng:Tn s mt cng hng gm mt nam chm in,to ra bi cun dy qun

trn li st t hnh ch U,mt ming thp nm trong t trng ca nam chmin,gn cht vo thanh l co l thp rung c tn s dao ng ring khcnhau. Tn s dao ng ring ca hai l thp k nhau hn km nhau l 0,25hoc 0,5Hz.in p ca tn hiu cn do tn s s c a vo cun dy canam chm in s to ra s dao ng ca tt c cc l thp.Tuy nhin l thpno c tn s dao ng ring bng tn s f th s dao ng cc i do cng

hng ring,cn cc thanh khc khng cng hng th khng dao ng cci.Nh vy ta s c kt qu ti tr s tng ng vi thanh rung cc i.



8/20


Cu 3: Tnh s ch ca mi cng t trong thi gianmt thng, bit rng c80% thi gian my bin p lm vic ch nh mc , 20% thi gian my

bin p lm vic ch non ti. Xc nh cos TB ca ph ti.

1, Tnh s ch ca cng t o nng lng tc dng.a, S ch ca cng t tc dng trong ch nh mc:Thi gian nh mc:

Tm = 0,8.24.30 = 576(h)Im = IAm = IBm = ICm = 461,88(A)Um= 35(KV)cos m = 0,92 m = 23,074o

Ta c:

. .3 . .cos .,

dmIU AAW U I T AActtddm dmdmK K

U I

=

=35.0,1 461,88.5

3. . .0,92.57650035. 3

( )423,94 KWhb, S ch ca cng t tc dng trong ch non ti:

Thi gian non ti:Tnt = 0,2.24.30 = 144(h)Int = IAnt = IBnt = ICnt = 0,72.Idm =0,72.461,88 = 332,554(A)Unt = Udm = 35(KV)cos nt = 0,75 nt = 41,41o

Ta c:

. .

,cttdntW 3 . .cos .nt

nt

AAAA nt

U I

IUU I T

K K

=

=35.0,1 332,554.5

3. . .0, 75.14450035. 3

( )62,21 KWhc, S ch ca cng t tc dng trong mt thng.

( )cttd cttddm cttdntW W W 423, 94 62, 21 486,15 WhK= + = + =

2, Tnh s ch ca cng t o nng lng phn khng .a, S ch ca cng t phn khng trong ch nh mc.



9/20


Thi gian nh mc:Tdm = 0,8.24.30 = 576(h)Idm = IAdm = IBdm = ICdm = 461,88(A)Udm= 35(KV)

cos dm = 0,92 sindm= sin(arccos dm) = 0,392

Ta c:

ctpkdm

35.0,1 461,88.5W 3. . . . 3. . .0,392.576

35 500

dm dmdm dm

U I

U ISin T h

K K= =

( )180,63 KVARh

b,S ch ca cng t phn khng trong ch non ti.Thi gian non ti:

Tnt = 0,2.24.30 = 144hInt = 0,61Idm = 0,72.461,88= 332,554(A)Unt = Udm = 35(KV)Cosnt = 0,75 sindm= sin(arccos dm) = 0,66

Ta c:

ctpknt35.0,1 332,554.5

W 3. . . . 3. . .0,66.14435 500

nt nt nt nt

U I

U ISin T

K K= =

( )54,743 KVARh=

c,S ch ca cng t pkn khng trong mt thng.

( )ctpkW 180, 63 54, 743 235, 373ctpkdm ctpknt W W KVARh= + = + =3, Xc nh costb ca ph ti.

Ta c :

( )2 2 2 2

486,15cos 0,9

486,15 (235,373)

cttdtb

cttd ctpk

W

W W = =

+ +

Cu 4: Tnh sai s tuyt i v sai s tng i ca cc php o trong haitrng hp ph ti.**C s l thuyt tnh sai s tng i , tuyt i

chnh xc l l tiu chun quan trong nht ca thit b o

xx dii =Trong :xi l kt qu o ca ln o th i

xd l gi tr ng ca i lng o i l sai lch ca ln o th i



10/20


Sai s php o l sai lch gia gi tr thc v gi tr o c , sai sphp o ch c th xc nh 1 cch tng i v ta khng th bit c gi trthc ca i lng cn o

Sai s ca php c th biu din lm 2 dng

+Sai s tuyt i+Sai s tng iSai s tuyt i ca mt thit b o c nh ngha l gi tr ln nht ca

sai lch gy nn bi thit b trong khi o.]max[ i=

Tuy nhin sai s tuyt i cha nh gi c chnh xc v yu cu cacng ngh thit b o.Thng thng chnh xc ca mt php o hoc mtthit b o c nh gi bng sai s tng i

Sai s tng i c tnh vi mt php o

= vi x l gi tr i lng o

Sai s tng i c tnh vi mt thit b o

D=

vi D l khong gi tr nh nht ti gi tr ln nhtGi tr % gi l sai s tng i quy i dng xp xp cc thit b o

thnh cp chnh xcTheo quy nh hin hnh ca nh nc, cc dng c o c in c cp

chnh xc :0,05;0,1;0,2;0,5;1;1,5;2;2,5 v 4Thit b o s c cp chnh xc: 0,005;0,01;0,02;0,05;0,1;0,2;0,5;1Vy khi bit cp chnh xc ca mt thit b o ta c th xc nh c sai

s tng i quy i v suy ra sai s tng i ca thit b trong php o cth

X

D =

*Tnh ton sai s ca php o gin tip:

Gi s c 1 php o gin tip i lng y thng cc php o trc tip 1x , 2x

,.. nx . y=f( 1x , 2x ,.. nx ).

Ta c:

dy= 1 2. . ... . ny y y

dx dx dxx x x

+ + +

- Sai s tuyt i ca php o gin tip c nh gi:

y= 2 2 2 21 211 2

( . ) ( . ) ... ( . ) ( . )n

n k

kn k

y y y yx x x x

x x x x=

+ + + =



11/20


1x , 2x ,.. nx : Sai s tuyt i ca php o cc i lng trc tip 1x , 2x ,..

nx .

- Sai s tng i ca php o gin tip c nh gi:

y =y

y

=

2 2 2 2 2 2 2 2 21 21 2

1 2

( ) .( ) ( ) .( ) ... ( ) .( ) ...nn

n

xx xy y yx x x

y x y x y x

+ + + = + + +

1x , 2x ,.. nx : Sai s tng i ca php o trc tip cc i lng 1x ,

2x ,..

n

x .

-Bng tnh sai s ca hm y= 1x . 2x

Hm y Sai s tuyt i y Sai s tng i y = y/y

1x . 2x2 2 2 2

1 2 2 1( ) ( )x x x x +

2 21 21 2

( ) ( )x x

x x

+

1, Sai s ca cc php o ch nh mc.a, Sai s ca php o dng in.

Dng in qua cc pha A,B,C ( ba pha nh nhau ) c xc nh.Idm = KI.IAdm

Vi IAm l dng m ampemet o c ch nh mc.

( )461,88

.5 4, 62500

dmAdm

I

II A

K= = =

Ta c:Sai s tuyt i ca ampemet l:

( )ADI AAAdm 1,05.100

2. ===

Sai s tuyt i ca bin dng in l:

2 500

. . 2100 5

I KI IK K = = =

Vy sai s tuyt i ca php o dng in ch nh mc l :



12/20


( ) ( ) 22 .. AdmIAdmIdm IKIKI +=

( ) ( )

( )

2 22.4,62 100.0,1

13,62 A

= +

Sai s tng i ca php o dng in ch nh mc l:

%13,62

.100% .100% 2, 95%461,88dm

dmI

dm

I

I

= = =

b, Sai s ca php o in p.Sai s tuyt i ca bin in p:

2 35. . 7

100 0,1U KU U

K K = = =

Sai s tuyt i ca volmet :

( )VDU VVV 2100.100

2. ===

Kt qu o in p dy :

( )KVUKU VUd 35100.1,0

35. ===

Vy sai s tuyt i ca php o in p dy l:

( ) ( ) 22 .. VUVUd UKUKU +=

( ) ( )

( )

2 27.100 350.2

989,95 V

= +

Sai s tng i ca php o in p dy:

%989,95

.100% .100% 2,83%35000d

dU

d

U

U

= =

Sai s tuyt i ca php o in p pha:



13/20


( ) 22

.3

. VUV

Uf UKU

KU +

=

( )

( )

2

21007. 350.23

808,29 V

= + =

Sai s tng i ca php o in p pha :

%

808,29.100% . 3.100% 4%

35000ff

U

f

U

U

= =

c, Sai s ca php o nng lng tc dng.Nng lng tc dng ca ph ti ch nh mc l:

( )tddm cttddm35 500

W . .W . .423,94 14837900 Wh0,1 5

U IK K K= = =

Sai s tuyt i ca cng t tc dng ch nh mc:

cttddm WW .cttd td D =

Ta c : ( )W 625tdD KWh=

( )cttddm2

W .625 12,5100

KWh = =

Vy sai s tuyt i ca php o nng lng tc dng ch nh mcl :

( ) ( ) ( )

222

tddm ......W cttddmIUcttddmIUcttddmIU WKKWKKWKK ++=

( ) ( ) ( )2 2 2

7.100.423,94 350.2.423,94 350.100.12,5= + +

( )606248,1927 KWh=

Sai s tng i ca php o nng lng tc dng l :

tddmW

606248,1927.100% .100% 4%

14837900

tddm

tddm

W

W

= = =

d, Sai s ca php o nng lng phn khng .



14/20


Nng lng phn khng ca ph ti ch nh mc l :

( )pkdm ctpkdmW . .W 350.100.180, 63 6322050 AR U IK K KV h= = = Sai s tuyt i ca cng t phn khng :

ctpkdm WW .

ctpk pk

D

=Ta c:( )W 625 AR pkD KV h=

( )1, 5

.625 9, 375 AR 100

ctpkdmW KV h = =

Vy sai s tuyt i ca php o nng lng phn khng l :

( ) ( ) ( )222pkdm ......W ctpkdmIUctpkdmIUctpkdmIU WKKWKKWKK ++=

( ) ( ) ( )2 2 2

7.100.180, 63 350.2.180, 63 350.100.9,375= + +

( )373685, 2534 ARKV h=

Sai s tng i ca php o nng lng phn khng l:

pkdmW

373685,2534.100% .100% 5,9%

6322050

pkdm

pkdm

W

W

= =

2, Sai s ca cc php o ch non ti.a, Sai s ca php o dng in

Dng in ch non ti c xc nh :Int = KI.IAnt

C : IAnt = 0,72.IAdm=0,72.4,62=3,3264

Sai s tuyt i ca ampek :

( )AII AdmAnt 1,0==

Vy sai s tuyt i ca php o dng in ch non ti l :

( ) ( )22 .. AntIAntInt IKIKI +=

( ) ( )

( )

2 22.3,3264 100.0,1

12,01 A

= +

=

Sai s tng i ca php o dng in ch non ti l :



15/20


%12,01

.100% .100% 3, 61%332,554nt

ntI

nt

I

I

= = =

b, Sai s ca php o nng lng tc dng .Nng lng tc dng ca ph ti ch non ti l:

( )tdnt cttdnt35 500

W . .W . .62, 21 2177350 Wh0,1 5

U IK K K= = =

Sai s tuyt i ca cng t tc dng ch non ti :

( )cttdnt cttdntW W 12,5 KWh = =Vy sai s tuyt i ca php o nng lng tc dng ch non ti l

( ) ( ) ( ) 222tdnt ......W cttdntIUcttdntIUcttdntIU WKKWKKWKK ++=

( ) ( ) ( )2 2 2

7.100.62, 21 350.2.62, 21 350.100.12,5= + +

( )441813, 2325KWh=

Sai s tng i ca php o nng lng tc dng ch non ti:

tdntW

441813, 2325.100% .100% 20,3%

2177350

tdnt

tdnt

W

W

= = =

c, Sai s ca php o nng lng phn khng.Nng lng phn khng ca ph ti ch non ti l :

( )pknt ctpkntW . .W 350.100.54, 743 1916005 AR U IK K KV h= = =

Sai s tuyt i ca cng t phn khng ch non ti :

ctpknt WW .ctpk pk D =Ta c :

( )W 625pkD KVarh= ( )9,375 AR ctpknt ctpkdmW W KV h = =

Vy sai s tuyt i ca php o nng lng phn khng ch non ti l :

( ) ( ) ( )222pknt ......W ctpkntIUctpkntIUctpkntIU WKKWKKWKK ++=

( ) ( ) ( )2 2 2

7.100.54, 743 350.2.54, 743 350.100.9,375= + +

( )332570,1065 ARKV h=



16/20


Sai s tng i ca php o nng lng phn khng ch non ti l:

pkntW

332570,1065.100% .100% 17,36%

1916005

pknt

pknt

W

W

= = =

Phn II: o lng v bo v cho trm ngi ta cn thu thpcc tn hiu in v khng in. Hy v s truyn dn chotrm.

II.1. Thu thp tn hiu :II.1.1.M hnh

Ta s dng m hnh ni tip songII.1.2. S lng knh: 15

- Knh 1 (S1) : Tn hiu dng pha A : IA

- Knh 2 (S2) : Tn hiu dng pha B : IB

- Knh 3 (S3) : Tn hiu dng pha C : IC

- Knh 4 (S4) : Tn hiu p pha A : UA

- Knh 5 (S5) : Tn hiu p pha B : UB

- Knh 6 (S6) : Tn hiu p pha C : UC- Knh 7 (S7) : Tn hiu p dy AB : UAB

- Knh 8 (S8) : Tn hiu p dy BC : UBC

- Knh 9 (S9) : Tn hiu p dy CA : UCA

- Knh 10(S10) : Tn hiu hin th h s cos

- Knh 11(S11) : Tn hiu hin th tn s pha A

- Knh 12(S12) : Tn hiu hin th tn s pha B

- Knh 13(S13) : Tn hiu hin th tn s pha C- Knh 14(S14) : Tn hiu cng sut tc dng

- Knh 15(S15) : Tn hiu cng sut phn khng

II.1.3. Chn cc Mux

Trong thc t nhiu i lng o l cc i lng khng in, v d nh:

p sut, nhit V vy c th truyn ti cc thng tin, o c v kim

tra c ta cn phi c cc thit b chuyn i chun ha trong o lng. u

vo l cc tn hiu khng in, u ra l cc tn hiu in vi chun p 0-10V,

chun dng 4-20mA. Do s lng knh lin lc ln, v th s dng cc thit



17/20


b i ni (MUX) thc hin phn knh lin lc, tin cho vic truyn tn

hiu v kim tra sa cha khi c s c. gim sai s ca cc b i ni,

trong s ta b tr cc b i ni theo nhm v phn theo tng, nh m

tc x l thng tin nhanh hn.- Tn hiu dng : Mux1 (2 bit)

- Tn hiu p : Mux2 (3 bit)

- Cc tn hiu cn li : Mux3(3 bit)

- Mux (2bit)

II.1.4.Lp bng trng thi

II.1.4.1.Bng trng thi ca Mux1 (2bit)

- Dng in pha A : Chn D0

- Dng in pha B : Chn D1

- Dng in pha C : Chn D2

- Chn a ch : A1, A2

Chn a chChnclock

u ra

A1 A2 C1 Q1

0 0 1 D0

0 1 1 D1

1 0 1 D2

x x 0 0

=>Q1=C1(D0. 1 2A .A +D1. 1 2A .A +D2 1 2A .A )

II.1.4.2.Bng trng thi ca Mux2 (3bit)- in p pha A : chn D3

- in p pha B : chn D4

- in p pha C : chn D5

- in p dy UAB : Chn D6

- in p dy UBC : Chn D7

- in p dy UCA : Chn D8

- Chn a ch : A3, A4, A5



18/20


Chn a chChnclock u ra

A3 A4 A5 C2 Q2

0 0 0 1 D3

0 0 1 1 D4

0 1 0 1 D5

0 1 1 1 D6

1 0 0 1 D7

1 0 1 1 D8

x x x 0 0

x x x 0 0

=>Q2=C2(D3. 53 4A .A .A +D4. 53 4A .A .A +D5. 53 4A .A .A +D6. 53 4A .A .A +D7.

53 4A .A .A

+D8. 53 4A .A .A

)

II.1.4.3.Bng trng thi ca Mux3 (3bit)- Tn hiu hin th h s cos : chn D9

- Tn hiu hin th tn s pha A : chn D10

- Tn hiu hin th tn s pha B : chn D11

- Tn hiu hin th tn s pha C : chn D12

- Tn hiu cng sut tc dng : chn D13- Tn hiu cng sut phn khng : chn D14

- Chn a ch : A6, A7, A8

Chn a ch Chnclock u ra



19/20


A6 A7 A8 C3 Q3

0 0 0 1 D9

0 0 1 1 D10

0 1 0 1 D11

0 1 1 1 D12

1 0 0 1 D13

1 0 1 1 D14

x x x 0 0

x x x 0 0

=>Q3=C3(D9. 76 8A .A .A +D10. 76 8A .A .A +D11. 76 8A .A .A +D12. 76 8A .A .A +D13.

76 8A .A .A +D14. 76 8A .A .A )

II.1.4.4. Bng trng thi ca Mux ( 2bit )

Chn a chChnclock u ra

A9 A10 C Q

0 0 1 Q1

0 1 1 Q2

1 0 1 Q3

x x 0 0



20/20


9 10 1 9 10 2 9 10 3Q C (A .A .Q +A .A .Q +A .A .Q =

II.1.4.5. Bng trng thi tng hp cc Mux

Cc chn a ch Mux1 Mux2 Mux3 Mux

A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 C1 Q1 C2 Q2 C3 Q3 C Q

0 0 x x x x x x 0 0 1 0D 0 0 0 0 1 1Q

0 1 x x x x x x 0 0 1 1D 0 0 0 0 1 1Q

1 0 x x x x x x 0 0 1 2D 0 0 0 0 1 1Q

x x 0 0 0 x x x 0 1 0 0 1 3D 0 0 1 2Q

x x 0 0 1 x x x 0 1 0 0 1 4D 0 0 1 2Q

x x 0 1 0 x x x 0 1 0 0 1 5D 0 0 1 2Q

x x 0 1 1 x x x 0 1 0 0 1 6D 0 0 1 2Q

x x 1 0 0 x x x 0 1 0 0 1 7D 0 0 1 2Q

x x 1 0 1 x x x 0 1 0 0 1 8D 0 0 1 2Q

x x x x x 0 0 0 1 0 0 0 0 0 1 9D 1 3Q

x x x x x 0 0 1 1 0 0 0 0 0 1 10D 1 3Q

x x x x x 0 1 0 1 0 0 0 0 0 1 11D 1 3Q

x x x x x 0 1 1 1 0 0 0 0 0 1 12D 1 3Q

x x x x x 1 0 0 1 0 0 0 0 0 1 13D 1 3Q

x x x x x 1 0 1 1 0 0 0 0 0 1 14D 1 3Q

II.2. S truyn dn cho trm:

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