Giai Tich Mang Dien Doc

8/4/2019 Giai Tich Mang Dien Doc

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GII TCH MNG

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GII TCH MNG

LI NI U

H thng in bao gm cc khu sn xut, truyn ti v phn phi in nng. Kt cumt h thng in c th rt phc tp, mun nghin cu n i hi phi c mt kin thc tnghp v c nhng phng php tinh ton ph hp.Gii tch mng l mt mn hc cn c tn gi Cc phng php tin hc ng dng trong tnhton h thng in. Trong , cp n nhng bi ton m tt c sinh vin ngnh h thngno cng cn phi nm vng. V vy, c mt cch nhn c th v cc bi ton ny, gio trnhi t kin thc cs hc nghin cu l thuyt cc bi ton cng nh vic ng dng chngthng qua cng c my vi tnh. Phn cui, bng ngn ng lp trnh Pascal, cng vic m phngcc phn mc ca bi ton c minh ho.

Ni dung gm c 8 chng.1. i s ma trn ng dng trong gii tch mng.2. Phng php s dng gii cc phng trnh vi phn trong gii tch mng.3. M hnh ha h thng in.4. Graph v cc ma trn mng in.5. Thut ton dng tnh ma trn mng.6. Tnh ton tro lu cng sut.7. Tnh ton ngn mch.8. Xt qu trnh qu ca my pht khi c s c trong mng.

II. Phn lp trnh: gm c bn phn mc:1. Xy dng cc ma trn ca 1 mng c th

2. Tnh ton ngn mch.3. Tnh ton tro lu cng sut lc bnh thng v khi s c.4. Xt qu trnh qu ca cc my pht khi c s c trong mng in.

GV: L Kim Hng


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GII TCH MNG

CHNG 1

I S MA TRN NG DNG TRONG GIITCH MNG

Trong chng ny ta nhc li mt s kin thc vi s ma trn thng thng c ng dngtrong gii tch mng.

1.1. NH NGHA V CC KHI NIM CBN:

1.1.1. K hiu ma trn:

Ma trn ch nht A kch thc m x n l 1 bng gm m hng v n ct c dng sau:

[ ]ji

mnmm

n

n

a

aaa

aaa

aaa

A ==

...

............

...

...

21

22221

11211

Nu m = 1 v n >1 th A gi l ma trn hng hoc vecthng.Ngc li n = 1 v m > 1 th A gi l ma trn ct hoc vectct.

3

1

2

=A 132=A vV d:

1.1.2. Cc dng ma trn:

Ma trn vung: L ma trn c s hng bng s ct (m = n).V d:

333231

232221

131211

aaa

aaa

aaa

A =

Ma trn tam gic trn: L ma trn vung m cc phn t di ng cho chnh a j ca matrn bng 0 vi i > j.

33

2322

131211

00

0

a

aa

aaa

A =

Ma trn tam gic di: L ma trn vung m cc phn t trn ng cho chnh aj ca ma trnbng 0 vi i < j.

333231

2221

11

0

00

aaa

aa

a

A =

Trang 2


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GII TCH MNGMa trn ng cho: L ma trn vung nu tt c cc phn t trn ng cho chnh khc 0,cn cc phn t khc ngoi ng cho chnh ca ma trn bng 0 (a = 0 vi ).ji j

33

22

11

00

00

00

a

a

a

A =

Ma trn n v: L ma trn vung m tt c cc phn t trn ng cho chnh ca ma trnbng 1 cn tt c cc phn t khc bng 0 (a

Trang 3

ij = 1 vi i = j v a = 0 vi ).ji j

100

010

001

=U

Ma trn khng: L ma trn m tt c cc phn t ca ma trn bng 0.Ma trn chuyn v: L ma trn m cc phn t a = a j ji(i hng thnh ct v ngc li).

3231

2221

1211

aa

aa

aa

A =

322212

312111

aaa

aaaAT =v

, AT hoc ACho ma trn A th ma trn chuyn v k hiu l AtMa trn i xng: L ma trn vung c cc cp phn ti xng qua ng cho chnh bngnhau aj = aji.V d:

463

625

351

=A

Chuyn v ma trn i xng th AT = A, ngha l ma trn khng thay i.

Ma trn xin - phn i xng: L ma trn vung c A = - AT. Cc phn t ngoi ng chochnh tng ng bng gi tri ca n (aj = - aji) v cc phn t trn ng cho chnh bng0.V d:

063

605

350

=A

Ma trn trc giao: L ma trn c ma trn chuyn v chnh l nghch o ca n. (AT .A = U =A .AT vi A l ma trn vung v cc phn t l s thc).Ma trn phc lin hp: L ma trn nu th phn t a + jb bi a - jb th ma trn mi A* l matrn phc lin hp.Cho ma trn A th ma trn phc lin hp l A*

1124

53

jj

jA

++

=

1124

53

jj

jA

=

v

-Nu tt c cc phn t ca A l thc, th A = A*

-Nu tt c cc phn t ca A l o, th A = - A*.

Ma trn Hermitian (ma trn phc i): L ma trn vung vi cc phn t trn ng chochnh l s thc cn cc cp phn ti xng qua ng cho chnh l nhng s phc linhp, ngha l A = (A*)t.

532

324

j

j

A +

=


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GII TCH MNGMa trn xin - Hermitian (ma trn xin - phc i): L ma trn vung vi cc phn t trnng cho chnh bng 0 hoc ton o cn cc cp phn ti xng qua ng cho chnh lnhng s phc, tc A = - (A* t) .

032

320

j

jA

=

*

Trang 4

Nu ma trn vung phc lin hp c (A )

t

. A = U = A. (A

* t

) th ma trn A c gi l ma trnn v. Nu ma trn n v A vi cc phn t l s thc c gi l ma trn trc giao.

Bng 1.1: Cc dng ma trn.K hiu Dng ma trn K hiu Dng ma trnA = -AA = At

A = - At

A = A*

A = - A*

Khngi xngXin-i xngThcHon ton o

A = (A* t) HermitianA = - (A*)t Xin- HermitianAt A = U Trc giao(A*)t A = U n v

1.2. CC NH THC:

1.2.1. nh ngha v cc tnh cht ca nh thc:

Cho h 2 phng trnh tuyn tnha11x1 + a12x = k2 1 (1) (1.1)a21x1 + a22x = k2 2 (2)

t phng trnh (2) th vo phng trnh (1), gii c:Rt x2

21122211

2121221

aaaa

kakax

=

Suy ra:

21122211

1212112

aaaa

kakax

=

Biu thc (a11a22 - a12a21) l gi trnh thc ca ma trn h s A. Trong |A| l nh thc.

2221

1211||aa

aaA =

Gii phng trnh (1.1) bng phng php nh thc ta c:

21122211

212122222

121

1 ..

..

aaaa

kaka

A

ak

ak

x

==

21122211

121211221

111

2 ..

..

aaaa

kaka

A

ka

ka

x

==v

Tnh cht ca nh thc:

a. Gi tr ca nh thc bng 0 nu:- Tt c cc phn t ca hng hoc ct bng 0.

- Cc phn t ca 2 hng (ct) tng ng bng nhau.- Mt hng (ct) l tng ng t l ca 1 hoc nhiu hng (ct).

b. Nu ta i ch 2 hng ca ma trn vung A cho nhau ta c ma trn vung B v c det(B)= - det(A).

c. Gi tr ca nh thc khng thay i nu:

- Tt c cc hng v ct tng ng i ch cho nhau.- Cng thm k vo 1 hng (ct) th t tng ng vi cc phn t ca hng (ct) .


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GII TCH MNGd. Nu tt c cc phn t ca hng (ct) nhn vi tha s k, th gi tr ca nh thc l cnhn bi k.

e. Tch ca cc nh thc bng tch ca tng nh thc. | A.B.C| = |A| .|B| .|C|.f. nh thc tng khc tng cc nh thc. |A + B - C| = |A| + |B| -|C|.

1.2.2. nh thc con v cc phn phi s.Xt nh thc:

333231

232221

131211

aaa

aaa

aaa

A =

Chn trong nh thc ny k hng, k ct bt k vi 1 [ k[ n. Cc phn t nm pha trn k tgiao ca hng v ct chn to thnh mt nh thc cp k, gi l nh thc con cp k ca A.B k hng v k ct chn, cc phn t cn li to thnh 1 nh thc con b ca nh thc A.

Phn phi sng vi phn t aij ca nh thc A l nh thc con b c km theo

du (-1)i+j

.

3332

1312

3332

13121221 )1(

aa

aa

aa

aaA == +

Mi lin h gia cc nh thc v phn ph:- Tng cc tch ca cc phn t theo hng (ct) vi phn ph tng ng bng nh thc |A|.- Tng cc tch ca cc phn t theo hng (ct) vi phn ph tng ng trong hng (ct) khc

bng 0.

1.3. CC PHP TNH MA TRN.

1.3.1. Cc ma trn bng nhau:

Hai ma trn A v B c gi l bng nhau nu tt c cc phn t ca ma trn A bng tt c ccphn t ca ma trn B (a i, j; i, j = 1, 2, .. n).ij =bj

1.3.2. Php cng (tr) ma trn.

Cng (tr) cc ma trn phi c cng kch thc m x n. V d: C hai ma trn A[a

Trang 5

ij ] v B[bmn ij] th tng v hiu ca hai ma trn ny l ma trn C[cmn ij ] vi cmn ij = aij6 bijMrng: R = A + B + C +..... + N vi rij = aij 6 bij6 cij 6 ...6 nij .Php cng (tr) ma trn c tnh cht giao hon: A + B = B + A.Php cng (tr) ma trn c tnh cht kt hp: A + (B + C) = (A + B) + C.

1.3.3. Tch v hng ca ma trn:

k.A = B. Trong : bij = k .aij i & j .Tnh giao hon: k.A = A.k..Tnh phn phi: k (A + B) = k.A + k..B = (A + B) k.

(vi A v B l cc ma trn c cng kch thc, k l 1 hng s ).

1.3.4. Nhn cc ma trn:

Php nhn hai ma trn A.B = C. Nu ma trn A c kch thc m x q v ma trn B c kchthc q x n th ma trn tch C c kch thc m x n. Cc phn t cij ca ma trn C l tng cctch ca cc phn t tng ng vi i hng ca ma trn A v j ct ca ma trn B l:


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GII TCH MNGc

Trang 6

ij = ai1 .b1j + a .bi2 2j + ... + aiq .bqjV d:

2212121121321131

2212121121221121

2212121121121111

2221

1211

....

....

....

babababa

babababa

babababa

bb

bb

++

++

++

=

3231

2221

1211

.

aa

aa

aa

BA = x

B.APhp nhn ma trn khng c tnh cht hon v: A.B Php nhn ma trn c tnh cht phn phi i vi php cng:A (B + C) = A.B + A.C.Php nhn ma trn c tnh cht kt hp: A (B.C) = (A.B) C = A.B.C.Tch 2 ma trn A.B = 0 khi A = 0 hoc B = 0.Tch C.A = C.B khi A = B.

Nu C = A.B th CT T= B .AT

1.3.5. Nghch o ma trn:

Cho h phng trnh:

a11x1 + a12x2 + a13x3 = y1a21x1 + a22x2 + a23x3 = y2 (1.2)a31x1 + a32x2 + a33x3 = y3

Vit di dng ma trn A.X = YNu nghim ca h trn l duy nht th tn ti mt ma trn B l nghch o ca ma trn A.Do : X = B.Y (1.3)

Nu nh thc ca ma trn A 0 th c th xc nh x nh sau:i

331

221

111

1 yA

Ay

A

Ay

A

Ax ++=

332

222

112

2 y

A

Ay

A

Ay

A

Ax ++=

333

223

113

3 yA

Ay

A

Ay

A

Ax ++=

Trong : A11, A12, .... A33 l nh thc con ph ca a11, a12, a13 v |A| l nh thc ca ma trnA. Ta c:

A

AB

ji

ji = i, j = 1, 2, 3.

Nhn ma trn A vi nghch o ca n ta c A.A-1 = A-1.A = URt X t phng trnh (1.3) sau khi nhn c hai v cho A-1.

A.X = YA-1 -1.A.X = A .YU.X = A-1.YSuy ra: X = A-1 .Y

Nu nh thc ca ma trn bng 0, th ma trn nghch o khng xc nh (ma trn suy bin).Nu nh thc khc 0 gi l ma trn khng suy bin v l ma trn nghch o duy nht.Gi s 2 ma trn A v B cng cp v l kho lc :

-1(A.B) = B-1.A-1

Nu AT kho th (AT -1) cng kho:(At -1) = (A-1 t)


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GII TCH MNG

1.3.6. Ma trn phn chia:

AA1

A3

A2

A4=

Tng cc ma trn phn chia c biu din bi ma trn nh bng tng cc ma trn nhtng ng.

A1

A3

A2

A4

B1

B3

B2

B4

A16B1

A36B3

A26B3

A46B36 =

Php nhn c biu din nh sau:

A1

A3

A2

A4

B1

B3

B2

B4

C1

C3

C2

C4=

Trong := A .B + A .BC1 1 1 2 3

C = A .B + A .B2 1 2 2 4C = A .B + A .B3 3 1 4 3C = A .B + A .B4 3 2 4 4

Tch ma trn chuyn v nh sau:

AA1

A3

A2

A4= A

TA 1

AT3

A 2

AT4=

T T

Tch ma trn nghch o nh sau:

AA1

A3

A2

A4= A

-1B1

B3

B2

B4=

Trong :-1 -1= (A - A .A .A )BB1 1 2 4 3

-1B = -B

Trang 7

2 1.A .A2 4-1B = -A .A .B3 4 3 1

-1 -1B = A - A .A .B4 4 4 3 2(vi A v A phi l cc ma trn vung).1 4

1.4. SPH THUC TUYN TNH V HNG CA MATRN:

1.4.1. Sph thuc tuyn tnh:

S ct ca ma trn A(m x n) c th vit theo n vectct hoc m vecthng.

{c1}{c } ..... {c1 1}{r1}{r } ...... {r1 1}Phng trnh vectct thun nht.


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GII TCH MNGp {c } + p {c

Trang 8

1 1 2 2} + .... + p {c } = 0 (1.4)n nKhi tt c Pk= 0 (k = 1, 2, ...., n).Tng t vecthng l khng ph thuc tuyn tnh nu.qr= 0 (r = 1, 2, ..., n).

{r } + qq1 1 2{r2} + ...... + q {r } = 0 (1.5)n n 0 tha mn phng trnh (1.4), th vectct l tuyn tnh.Nu pk

Nu qr 0 tha mn phng trnh (1.5), th vecthng l tuyn tnh.Nu vectct (hng) ca ma trn A l tuyn tnh, th nh thc ca A = 0.

1.4.2. Hng ca ma trn:

Hng ca ma trn l cp cao nht m tt c cc nh thc con khc 0.0 [ r(A) [ min(m, n) vi A l ma trn kch thc m x n.

1.5. H PHNG TRNH TUYN TNH:

H phng trnh tuyn tnh ca m phng trnh trong n h sc vit:a11x1 + a12x + .... + a2 1nx = yn 1a21x1 + a22x2 + .... + a2nx = yn 2........................................ (1.6)a xm1 1 + a xm2 2 + .... + a xmn n = ym

Trong :ai j: L h s thc hoc phc ; x : L bin s ; y : L hng s ca h.j j

H phng trnh c biu din dng ma trn nh sau:A. X = Y (1.7)Ma trn mrng:

mmnmm

n

n

yaaa

yaaa

yaaa

A

....

....................

....

....

21

222221

111211

=

Nu y = 0 th h phng trnh gi l h thun nht, ngha l: A.X = 0.i0 th h gi l h khng thun nht.Nu mt hoc nhiu phn t ca vecty i

nh l:iu kin cn v h phng trnh tuyn tnh c nghim l hng ca ma trn h s bnghng ca ma trn mrng.H phng trnh tuyn tnh v nghim khi v ch khi hng ca ma trn h s nh hn hng ca

ma trn mrng.Nu hng ca ma trn r(A) = r() = r = n (sn) ca h phng trnh tuyn tnh (1.6) th h cnghim duy nht (h xc nh).

Nu r(A) = r() = r < n th h phng trnh tuyn tnh c v s nghim v cc thnh phn canghim ph thuc (n - r) tham s ty .


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GII TCH MNG

Trang 12

CHNG 2

GII PHNG TRNH VI PHN BNG

PHNG PHP S2.1. GII THIU.

Nhiu h thng vt l phc tp c biu din bi phng trnh vi phn n khng c th giichnh xc bng gii tch. Trong k thut, ngi ta thng s dng cc gi tr thu c bngvic gii gn ng ca cc h phng trnh vi phn bi phng php s ha. Theo cch , ligii ca phng trnh vi phn ng l mt giai on quan trng trong gii tch s.Trong trng hp tng qut, th t ca vic lm tch phn s l qu trnh tng bc chnh xcchui gi tr cho mi bin ph thuc tng ng vi mt gi tr ca bin c lp. Thng th

tc l chn gi tr ca bin c lp trong mt khong cnh. chnh xc cho li gii bi tchphn s ph thuc c hai phng php chn v kch thc ca khong gi tr. Mt s phngphp thng xuyn dng c trnh by trong cc mc sau y.

2.2. GII PHNG TRNH VI PHN BNG PHNGPHP S.

2.2.1 Phng php Euler:

Cho phng trnh vi phn bc nht.

),( yxfdx

dy

= (2.1)

y = g(x,c)y

y

x

y0

x00

Hnh 2.1: thca hm s tbi gii phng trnh vi phn

x

Khi x l bin c lp v y l bin ph thuc, nghim phng trnh (2.1) s c dng:y = g(x,c) (2.2)

Vi c l hng s c xc nh t l thuyt trong iu kin ban u. ng cong miut phng trnh (2.2) c trnh by trong hnh (2.1). T ch tip xc vi ng cong, onngn c th gi s l mt on thng. Theo cch , ti mi im ring bit (x0,y0) trn ngcong, ta c:

xdx

dyy

0

Vi0dx

dyl dc ca ng cong ti im (x0,y0). V th, ng vi gi tr ban u x0 v y0, gi

tr mi ca y c th thu c t l thuyt l x:


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GII TCH MNG

Trang 13

yyy += 01 hay hdx

dyyy

001 += (t h = x)

Khi y l s gia ca y tng ng vi mt s gia ca x. Tng t, gi tr th hai ca y c thxc nh nh sau.

h

dx

dyyy

1

12 +=

Khi ),( 111

yxfdx

dy= x

y

0

Hnh 2.2 : thca li gii xp xcho phng trnh vi phn bng

phng php Euler

y= g(x,c)

hhh

y3

y0y1

y2

x3x2x1x0

Qu trnh c th tnh tip tc, ta c:

hdx

dyyy

223 +=

hdx

dyyy

334 +=

...........................Bng gi tr x v y cung cp cho ton b bi gii phng trnh (2.1). Minh ha phng phpnh hnh 2.2.

2.2.2. Phng php bin i Euler.

Trong khi ng dng phng php Euler, gi tr dy/dx ca khong gi thit tnh ton bt uvt ra ngoi khong cho php. S thay th c th thu c bng cch tnh ton gi tr mica y cho x1 nh trc.x1 = x0 + h

hdxdyyy

00

)0(1 +=

Dng gi tr mi x1 v y1(0) thay vo phng trnh (2.1) tnh ton gn ng gi tr ca

1dx

dyti

cui khong.

),( )0(11

)0(

1

yxfdx

dy=

Sau tn dng gi tr y1(1) c th tm thy bi dng trung bnh ca

0dx

dyv

)0(

1dx

dynh sau:


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GII TCH MNG

Trang 14

hdx

dy

dx

dy

yy

+

+=2

)0(

100

)1(1

Dng x1 v y1(1), gi tr xp x th ba y1

(2) c th thu c bi qu trnh tng t nh sau:

hdx

dy

dx

dy

yy

+

+=2

)1(

100

)2(1

Ta c:

hdx

dy

dx

dy

yy

+

+=2

)2(

100

)3(1

Qu trnh c th tnh tip tc cho n khi hai s lin nhau c lng cho y l ngang bng nmtrong phm vi mong mun. Qu trnh hon ton lp li thu c gi tr y2. Kt qu thu c cs chnh xc cao hn t s bin i ca phng php Eulerc minh ha trong hnh 2.3.

+

2

)0(

10 dxdy

dxdy

y = g(x,c)

y1

y

x0 x1

h

y0

0dx

dy

0

dy(0)

dx1y2 Hnh 2.3 : thca ligii xp xcho phng

trnh vi phn bng phng

php bin i Euler.

x

Phng php Euler c thng dng gii h phng trnh vi phn cng lc. Cho hai phngtrnh:

)zy,,(

)zy,,(

2

1

xfdx

dz

xfdx

dy

=

=

Vi gi tr ban u x0, y0 v z0 gi tr mi y1 s l:

hdx

dzyy

001 +=

Vi: )z,y,( 00010

xfdx

dy=

Tng t.


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GII TCH MNG

Trang 15

hdx

dzzz

001 +=

Vi: ),,( 00020

zyxfdx

dz=

Cho s gia tip theo, gi tr x1 = x0 + h, y1 v z1 dng xc nh y2 v z2. Trong phng php

bin i Euler y1 v z1 dng xc nh gi tro hm ti x1 cho nh gi gn ng cp haiy1

(1) v z1(1).

2.2.3. Phng php Picard vi sxp x lin tc.

Csca phng php Picard l gii chnh xc, bi s thay th gi tr y nh hm ca xtrong phm vi gi tr x cho.y g(x)y l biu thc c lng bi s thay th trc tip gi tr ca x thu c gi trtng ng ca y. Cho phng trnh vi phn (2.1).dy = f(x,y)dxV tch phn gia khong gii hn cho x v y.

=1

0

1

0

),(y

y

x

xdxyxfdy

Th =1

0

),(01x

xdxyxfyy

Hay (2.3)+=1

0

),(01x

xdxyxfyy

S hng tch phn trnh by s thay i trong kt qu ca y vi s thay i ca x t x0n x1. Li gii c th thu c bi snh gi tch phn bng phng php xp x lintc.

Ta c th xem gi tr ca y nh hm ca x c th thu c bi s thay th y didng tch phn vi y0, cho gi tr ban u nh sau:

+=1

0

),( 00)1(

1

x

xdxyxfyy

Thc hin biu thc tch phn vi gi tr mi ca y by gic thay th vo phngtrnh (2.3) thu c ln xp x th hai cho y nh sau:

+=1

0

),( )1(10)2(

1

x

xdxyxfyy

Qu trnh ny c th lp li trong thi gian cn thit thu c chnh xc mongmun..

Tht vy, c lng tch phn lun lun phc tp th nhng phi gi thit cho bin cnh. Kh khn v cn thc hin nhiu ln tch phn, nn y l mt hn ch s p dngca phng php ny.Phng php Picard c th p dng gii ng thi nhiu phng trnh nh sau:

),,(1 zyxfdx

dy=

),,(2 zyxfdx

dz=

Theo cng thc, ta c:

+=

1

0

),,( 00101x

x

dxzyxfyy

+=1

0

),,( 00201x

xdxzyxfzz


13/126


14/126

GII TCH MNG

Trang 17

k2 = f(x0 + b1h, y0 + b2k1)hk3 = f(x0 + b3h, y0 + b4k2)hk4 = f(x0 + b5h, y0 + b6k3)h

Tip theo th tc ging nh dng cho ln xp x bc hai, h s trong phng trnh (2.8)thu c l:

a1 = 1/6; a2 = 2/6; a3 = 2/6; a4 = 1/6.V b1 = 1/2; b2 = 1/2; b3 = 1/2; b4 = 1/2; b5 = 1; b6 = 1.

Thay th cc gi tr vo trong phng trnh (2.8), phng trnh xp x bc bnRunge-Kutta trthnh.

)22(61

432101 kkkkyy ++++=

Vi k1 = f(x0,y0)h

hk

yh

xfk )2

,2

( 1002 ++=

hk

yh

xfk )2

,2

( 2003 ++=

hkyhxfk ),( 3004 ++=

Nh vy, s tnh ton ca y theo cng thc i hi s tnh ton cc gi tr ca k1, k2,k3 v k4 :y = 1/6(k1+2k2+2k3+k4)Sai s trong s xp x l bc h5.Cng thc xp x bc bn Runge-Kutta cho php gii ng thi nhiu phng trnh vi

phn.

),,( zyxfdx

dy=

),,( zyxgdxdz =

Ta co:y1 = y0+1/6 (k1+2k2+2k3+k4)z1 = z0+1/6 (l1+2l2+2l3+l4)

Vi: k1= f(x0,y0,z0)h

hl

zk

yh

xfk )22

,2

( 101

002 +++=

hl

zk

yh

xfk )22

,2

( 202

003 +++=

k4 = f(x0 + h, y0 + k3,z0 + l3)hl1 = g(x0,y0,z0)h

hl

zk

yh

xgl )22

,2

( 101

002 +++=

hl

zk

yh

xgl )22

,2

( 202

003 +++=

l4 = g(x0 + h, y0 + k3,z0 + l3)h


15/126

GII TCH MNG

Trang 18

2.2.5. Phng php don sa i.

Phng php da trn csngoi suy, hay tch phn vt trc, v lp li nhiu lnvic gii phng trnh vi phn.

),( yxfdx

dy= (2.9)

c gi l phng php don sa i. Th tc cbn trong phng php don sa i l xut pht tim (xn,yn) n im (xn+1, yn+1). Th thu c

1+ndx

dyt

phng trnh vi phn v sa i gi tr yn+1 xp x cng thc chnh xc.Loi n gin ca cng thc don phng php ca Euler l:

yn+1 = yn + ynh (2.10)

Vi:n

ndx

dyy ='

Cng thc chnh xc khng dng trong phng php Euler. Mc d, trong phng php

bin i Euler gi tr gn ng ca yn+1 thu c t cng thc don (2.10) v gi trthay th trong phng trnh vi phn (2.9) chnh l yn+1. Th gi tr chnh xc cho yn+1thu c t cng thc bin i ca phng php l:

2)''( 11

hyyyy nnnn ++= ++ (2.11)

Gi tr thay th trong phng trnh vi phn (2.9) thu c c snh gi chnh xc hncho yn+1, n lun lun thay th trong phng trnh (2.11) lm cho yn+1 chnh xc hn.Qu trnh tip tc lp li cho n khi hai gi tr tnh ton lin tip ca yn+1 t phngtrnh (2.11) trng vi gi tr mong mun chp nhn c.Phng php don bin i kinh in ca Milne. Don ca Milne v cng thc

bin i, theo ng l:)'2''2(

3

4123

)0(1 nnnnn yyy

hyy ++= +

V )''4'(3 1111 ++

+++= nnnnn yyyh

yy

Vi: ),(' )0( 111 +++ = nnn yxfyBt u ca s tnh ton i hi bit bn gi tr ca y. C th tnh ton bi Runge-Kutta hay mt s phng php s trc khi s dng cng thc don sa i caMilne. Sai s trong phng php l bc h5.Trong trng hp tng qut, phng php mong mun chn h nh nn ch vi lnlp l i hi thu c yn+1 hon ton chnh xc nh mong mun.

Phng php c th mrng cho php gii mt s phng trnh vi phn ngthi. Phng php don sa i l p dng c lp i vi mi phng trnh vi phnnh mt phng trnh vi phn n gin. V vy, thay th gi tr cho tt c cc bin phthuc vo trong mi phng trnh vi phn l i hi snh gi o hm ti (xn+1, yn+1).


16/126

GII TCH MNG

Trang 19

2.3. GII PHNG TRNH VI PHN BC CAO.

Trong k thut trc y m t cho vic gii phng trnh vi phn bc nht cng c thp dng cho vic gii phng trnh vi phn bc cao bng sa vo ca bin ph. Vd, cho phng trnh vi phn bc hai.

02

2

=++ cydxdyb

dxyda

Vi iu kin ban u x0, y0, v0dx

dyth phng trnh c thc vit li nh hai

phng trnh vi phn bc nht.

'ydx

dy=

a

cyby

dx

dy

dx

yd +==

''2

2

Mt trong nhng phng php m t trc y c th l vic lm i tm li giicho hai phng trnh vi phn bc nht ng thi.Theo cch tng t, mt vi phng trnh hay h phng trnh bc cao c th quy v h

phng trnh vi phn bc nht.

2.4. V D V GII PHNG TRNH VI PHN BNGPHNG PHP S.

Gii phng trnh vi phn s minh ha bng s tnh ton dng in cho mch RL nitip.

t = 0 R

e(t)i(t)

L

Hnh 2.4: Sbiu din ca mchin RL

Cho mch in RL trong hnh 2.4 sc in ng hiu dng khi ng kha l:e(t) = 5t 0 [ t [ 0,2

e(t) = 1 t > 0,2in trcho theo n v ohms l.R = 1+3i2

V in cm theo n v henrys l.L = 1

Tm dng in trong mch in theo cc phng php sau:EulersBin i Euler.Xp x bc bn Runge-KuttaMilnes

Picards


17/126

GII TCH MNG

Trang 20

Bi gii:Phng trnh vi phn ca mch in l.

)(teRidt

diL =+

Thay th cho R v L ta c:

)()31( 2 teiidt

di

=++ iu kin ban u ti t = 0 th e0 = 0 v i0 = 0. Khong chn cho bin c lp l:t = 0,025.

a. Phng trnh theo phng php Euler l.

tdt

dii

n

n =

in+1 = in +in

Vi nnnn

iiedt

di

)31( 2+=

Thay th gi tr ban u vo trong phng trnh vi phn, 00

=dt

dyv i0. V th, dng

in i1 = 0. Ti t1 = 0,025; e1 = 0,125 v 125,00})0(31{125,02

1

=+=dt

di

i1 = (0,125)0,025 = 0,00313Thi2 = 0 + 0,00313 = 0,00313

Lp bng k kt qu li gii a vo trong bng 2.1Bng 2.1:Gii bng phng php Euler

nThi giantn

Sc in ngen

Dng

012

3456789101112

0,0000,0250,050

0,0750,1000,1250,1500,1750,2000,2250,2500,2750,300

0,0000,1250,250

0,2500,3750,5000.6250,7500,8751,0001,0001,0001,000

0,000000,000000,00313

0,009300,018440,030480,45340,062950,083230,106110,128370,150000,17100

0,000000,125000,24687

0,365700,481540,594440,704380,811300,915040,890310,865280,83988

nnn

n

iiedt

di)31(

2+= t

dt

diii

n

nn+=

1

1


18/126

GII TCH MNG

Trang 21

b. Phng trnh ca phng php bin i Euler l.

tdt

dii

n

n =)0(

)0()0(1 nnn iii +=+

tdtdi

dtdi

i nnn

+= +

2

)0(

1)1(

)1()1(1 nnn iii +=+

Vi )0( 12)0(

11

)0(

1

})(31{ ++++

+= nnnn

iiedt

di

Thay th gi tr ban u e0 = 0 v i0 = 0 vo trong phng trnh vi phn 00

=

dx

di

Do : ; .0)0(0 =i 0)0(

1 =i

Thay th vo trong phng trnh vi phn v e0)0(1 =i 1 = 0,125

125,00})0(31{125,0 2)0(

1

=+=dt

di

V 00156,0025,0)2

0125,0()1(0 =

+=i

Nn00156,000156,00)1(1 =+=i

Trong li gii v d cho phng php, khng thc hin lp li . Bi gii thuc bng phng php bin i Eulerc a vo trong bng 2.2.

1)1(1 ++ = nn ii

Bng 2.2: Bi gii bng phng php bin i Euler.

nThi Sc DngGian in in intn ng en

0123456789101112

0,000 0,000 0,00000 0,00000 0,00000 0,125 0,00000 0,12500 0,001560,025 0,125 0,00156 0,12344 0,00309 0,250 0,00465 0,24535 0,004610,050 0,250 0,00617 0,34383 0,00610 0,375 0,01227 0,36272 0,007580,075 0,375 0,01375 0,36124 0,00903 0,500 0,02278 0,47718 0,010480,500 0,02423 0,47573 0,01189 0,625 0,03612 0,58874 0,013310,625 0,03754 0,58730 0,01468 0,750 0,05222 0,69735 0,016060,750 0,05360 0,69594 0,01740 0,875 0,07100 0,80293 0,018740,175 0,875 0,07234 0,80152 0,02004 1,000 0,09238 0,90525 0,021330,200 1,000 0,09367 0,90386 0,02260 1,000 0,11627 0,87901 0,022290,225 1,000 0,11596 0,87936 0,02198 1,000 0,13794 0,85419 0,021670,250 1,000 0,13763 0,85455 0,02136 1,000 0,15899 0,82895 0,021040,275 1,000 0,15867 0,82935 0,02073 1,000 0,17940 0,80328 0,020410,300 1,000 0,17908

)0(

1+ndt

di

ndt

di

1+ne )0(

ni )0(1+ni )1(ni


19/126

GII TCH MNG

Trang 22

c. Phng trnh dng phng php Runge-Kutta gii.

iitedt

di)31()( 2+=

Ta c:tiitek nnn += })31()({

21

tk

ik

it

tek nnn

+

++

+=

2.

231)

2( 1

21

2

tk

ik

it

tek nnn

+

++

+=

2.

231)

2( 2

2

23

tkikittek nnn ++++= )}(.)(31)({ 32

34

)22(61

4321 kkkkin +++=

in+1 = in + inVi:e(tn) = en

2

)2

( 1++

=

+ nnneet

te

e(tn + t) = en+1Thay th gi tr ban u tm c k1:k1 = 0.

Tm c k2:

[ ] 00156,0025,00)0(312

125,00 22 =

++

=k

Tm c k3:

00154,0025,02

00156,0

2

00156,031

2

125,002

3 =

++

=k

Tm c k4:00309,0025,000154,0)00154,0(31125,00 24 =++=k

Th00155,0)00309,000308,000312,00(6

10 =+++=i

V i1 = i0 +

i0 = 0+ 0,00155 = 0,00155Bi gii thu c bng phng php Runge-Kutta c a vo trong bng 2.3.d. Cng thc don sa i ca phng php Milne l.

)'2''2(3

4123

)0(1 nnnnn iii

tii +

+= +

)''4'(3 1111 ++

++

+= nnnnn iiit

ii

Vi

n

ndt

dii ='

V


20/126

GII TCH MNG

Trang 23

nnn

n

iiedt

di)31( 2+=

Cc gi tr ban u i hi phi thu c t li gii ca phng php Runge-Kutta.Vi i0 = 0; i1 = 0,00155; i2 = 0,00615; i3 = 0,01372.Thay th vo phng trnh vi phn, ta c:

i0 = 0; i1 = 0,12345; i2 = 0,23485; i3 = 0,36127.Bt u ti t4 = 0,100 v thay th vo trong cng thc don, c lng u tin choi4 l:

[ ] 02418,0)36127,0(224385,0)12345,0(2)025,0(340)0(4 =++=i

Thay th e4 = 0,500 v i4 = 0,02418 vo trong phng trnh vi phn, ta c:i4 = 0,500 [ 1 + 3(0,02418)

2]0,02418 = 0,47578Don v gi tr chnh xc, ch khc nhau mt s hng thp phn v vy khng i hilp li nhiu ln. Kt qu sau tng bc c ghi vo bng 2.4. Ti t9 gi tr donca dng in l 0,11742 nhng trong khi gi tr chnh xc l 0,11639. Vic thc hinlp li bi s thay th gi tr chnh xc trong phng trnh vi phn thu c i9 =0,87888. C ln lt dng trong cng thc sa i thu c c lng th hai cho i9= 0,11640, trc khi kim tra gi tr chnh xc. Thc hin lp li trong tt c cc bcm bo yu cu chnh xc.


21/126

GII TCH MNG

Trang 24

Thi

Sc

Dng

en

+en+1

k1

k2

gian

in

in

k1

--------in+

---

k2

in+

---

k3

en+1

in+k3

k4

in

tn

ng

in

2

2

2

en

0,0

00

0,0

00

0,0

0000

0,0

0000

0,0

625

0,00000

0,0

0156

0,0

0078

0,0

0154

0,1

25

0,0

0154

0,0

03

09

0,0

0155

0,0

25

0,1

25

0,0

0155

0,0

0309

0,1

875

0,00310

0,0

0461

0,0

0386

0,0

0459

0,2

50

0,0

0614

0,0

06

10

0,0

0460

0,0

50

0,2

50

0,0

0615

0,0

0610

0,3

125

0,00920

0,0

0758

0,0

0994

0,0

0756

0,3

75

0,0

1371

0,0

09

03

0,0

0757

0,0

75

0,3

75

0,0

1372

0,0

0903

0,4

375

0,01824

0,0

1048

0,0

1896

0,0

1046

0,5

00

0,0

2418

0,0

11

89

0,0

1047

0,1

00

0,5

00

0,0

2419

0,0

1189

0,5

625

0,03014

0,0

1331

0,0

3084

0,0

1329

0,6

25

0,0

3748

0,0

14

68

0,0

1330

0,1

25

0,6

25

0,0

3749

0,0

1468

0,6

875

0,04483

0,0

1606

0,0

4552

0,0

1604

0,7

50

0,0

5353

0,0

17

40

0,0

1605

0,7

50

0,0

5354

0,0

1740

0,8

125

0,0

6224

0,0

1874

0,0

6291

0,0

1872

0,8

75

0,0

7226

0,0

2004

0,01873

0,1

75

0,8

75

0,0

7227

0,0

2004

0,9

375

0,08229

0,0

2134

0,0

8294

0,0

2132

1,0

00

0,0

9359

0,0

22

60

0,0

2133

0,2

00

1,0

00

0,0

9360

0,0

2260

1,0

000

0,10490

0,0

2229

0,1

0475

0,0

2230

1,0

00

0,1

1590

0,0

21

99

0,0

2230

0,2

25

1,0

00

0,1

1590

0,0

2199

1,0

000

0,12690

0,0

2167

0,1

2674

0,0

2168

1,0

00

0,1

3758

0,0

21

37

0,0

2168

0,2

50

1,0

00

0,1

3758

0,0

2137

1,0

000

0,14827

0,0

2105

0,1

4811

0,0

2105

1,0

00

0,1

5863

0,0

20

73

0,0

2105

0,2

75

1,0

00

0,1

5863

0,0

2073

1,0

000

0,16900

0,0

2041

0,1

6884

0,0

2042

1,0

00

0,1

7905

0,0

20

09

0,0

2041

Bng2.3:

GiibngphngphpRunge-Kutta

n 0 1 2 3 4 5 6 7 8 9 10

11

12

Bng 2.4:Bi gii bng phng php ca Milne.


22/126

GII TCH MNG

Trang 25

NThi gian Sc in Dng in Dng in

tn ng en (don) in in (sa i)in

456789

10

11

12

0,100 0,500 0,02418 0,47578 0,024190,125 0,625 0,03748 0,58736 0,037480,150 0,750 0,05353 0,69601 0,053530,175 0,875 0,07226 0,80161 0,072260,200 1,000 0,09359 0,90395 0,093580,225 1,000 0,11742 0,87772 0,11639

0,87888 0,11640+0,250 1,000 0,13543 0,85712 0,13755

0,85464 0,13753+0,275 1,000 0,16021 0,82745 0,15911

0,82881 0,15912+

0,300 1,000 0,17894 0,80387 0,178980,80382 0,17898+

+ : gi tr sa i th hai thu c bi vng lpd. Phng trnh dng phng php Picard hm tng ng khi u cho i, cn i0 = 0l:

[ ]dtiiteii t += 03

0 3)(

Thay th e(t) = 5t v gi tr ban u i0 = 0

==t t

dtti0

2)1(

2

55

Thay i(1) cho i trong phng trnh tch phn, thu c:

56

375

6

5

2

5

8

375

2

55

732

0

62)2( tttdt

ttti

t

=

=

Qu trnh tip tc, ta c:

dtttttt

tit

+++=

0

87632)3( ....

8

125

7

375

8

375

6

5

2

55

....56

375

24

5

6

5

2

5 7432++=

tttt

dtttttt

tit

+++=

0

76432)4( ....7

3758

375245

65

255

....56

375

2424

5

6

5

2

5 75432++=

ttttt

Gii hn chui sau s hn bc bn l:

24

5

6

5

2

5 432 ttti +=

Nu hm dng xp x i chnh xc bn s thp phn vi s hn xp xu tin khng ch n sai s ln th .

5log t [ log0,00120log t [ 9,415836 - 10t [ 0,2605


23/126

GII TCH MNG

Trang 26

Gi tr gii hn l hm xp x hp l. V vy, trong v d ny hm c th dng chthu c y cho trong khong 0 [ t [ 0,2; Bi v cho t > 0,2 th e(t) = 1. Cho nn, hmxp x khc phi chnh xc cho trong khong 0,2 [ t[ 0,3 nh sau:

( ) dtiiit

+= 2,033109367,0

( ){ } 0,2)-0,90386(t0,09367+=+= dtit

2,0

3)1(

09367,0309367,0109367,0

( ) [ ]{ }dttti t ++= 2,03)2( )2,0(90386,009367,032,090386,009367,0109367,0

( ){ }dttttt += 2,032 )2,0(45089,22,076189,0)2,0(07897,1190386,009367,0

dtttt

tx

x

+=

4

)2,0(45089,2

3

)2,0(76189,0

2

)2,0(07897,1)2,0(

90386,009367,0432

Cui cng, ta c:

i(3)

= 0,09367 + 0,90386(t - 0,2) - 0,48762(t - 0,2)2

-- 0,05420(t - 0,2)3 - 0,30611(t - 0,2)4 + 0,86646(t - 0,2)5 ....Chui gii hn, hm xp x l:i = 0,09367 + 0,90386(t - 0,2) -

- 0,48762(t - 0,2)2 - 0,05420(t - 0,2)3 - 0,30611(t - 0,2)4Cho i hiu chnh trong bn s thp phn, ta c:0,86646(t - 0,2)5[ 0,00005(t - 0,2) [ 0,14198Hm hp l cho trong khong 0,2 [ t [0,342Gi tr thu c bng phng php Picard c a vo trong bng 2.5.

2.5. SO SNH CC PHNG PHP.

Trong bi gii ca phng trnh vi phn hm quan h gia bin ph thuc y v bin clp x cn tm tha mn phng trnh vi phn. Bi gii trong gii tch l rt kh v cmt s vn khng th tm c. Phng php s dng tm li gii bng cch biudin y nh mt s hm ca bin c lp x t mi gi tr xp x ca y c th thu c

bng s thay th hon ton hay biu din tng ng quan h gia cc gi tr lin tipca y xc nh cho vic chn gi tr ca x. Phng php Picard l phng php s kiuu tin. Phng php Euler, Runge-Kutta, v Milne l v d cho kiu th hai.

Kh khn ch yu pht sinh t phng php xp x y bng hm s, nh phng phpPicard, tm thy trong ln lp li s tch phn hin ti phi thc hin thu c hmtha mn. V vy phng php ny l khng thc t trong hu ht cc trng hp v tc dng.


24/126

GII TCH MNG

Trang 27

Bng 2.5: Gii bng phng php Picard.n Thi gian tn Sc in ng en Dng in in0123456789101112

00,0250,0500,0750,1000,1250,1500,1750,2000,2250,2500,2750,300

00,1250,2500,3750,5000,6250,7500,8751,0001,0001,0001,0001,000

00,001550,006150,013720,024190,037490,053540,072290,093670,115960,137640,158680,17910

Cc phng php theo kiu th hai i hi php tnh s hc n gin o thch hpcho vic gii bng my tnh s ca cc phng trnh vi phn. Trong trng hp tngqut, n gin quan hi hi dng trong mt khong nh cho cc bin c lp nhngngc li nhiu phng php phc tp c th dng trong khong tng i ln tnnhiu cng sc trong vic chnh xc ha li gii. Phng php Euler l n gin nht,nhng tr khi khong tnh rt nh th dng n cng khng ng vi thc t. Phng

php bin i Euler cng s dng n gin v c thm thun li kim tra h thng vnc trong qu trnh thu c ci thin sc lng cho y. Phng php c s chnhxc gii hn, v vy i hi dng khong gi tr nh cho bin c lp. Phng phpRunge-Kutta i hi s rt ln ca php tnh s hc, nhng kt qu cng khng chnhxc.Phng php don sa i ca Milne l t kh khn hn phng php Runge-Kutta v sosnh c chnh xc ca bc h5. V vy, phng php ca Milne i hi c bn gi tr banu cho bin ph thuc phi thu c bng mt s phng php khc, hu nh phng php

bin i Euler hay phng php Runge-Kutta, l nh nhau. Trong sng dng my tnh chophng php s. Chng trnh i hi bt u li gii nh phng php ca Milne. Li giitip tc dng cng thc khc cho d on v sau sa cha gi tr ca y cung cp qu trnh

h thng cho kim tra tt bng sa cha c lng ban u. Nu s khc nhau gia don vgi tr chnh xc l ng k, khong tnh c thc rt gn li. Kh nng trong phng phpca Milne khng c hiu lc trong phng php Runge-Kutta.


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Bi tp:

2.1. Gii phng trnh vi phn.

yx

dx

dy= 2

Cho 0 [ t [ 0,3; vi khong phng trnh 0,05 v gi tr ban u x0 = 0 v y0 = 1, bngcc phng php s sau y.EulerBin i Euler.PicardXp x bc bn Runge-KuttaMilne dng gi tr bt u thu c phng php Runge-Kutta2.2. Gii bng phng php bin i Euler h phng trnh vi phn.

y

dt

dx2=

2

x

dt

dy=

Cho 0 [ t [ 1,0; Vi khong phng trnh 0,2 v gi tr ban u i0 = 0,x0 = 0 vy0 = 12.3. Gii bng xp x bc bn Runge-Kutta phng trnh vi phn bc hai.y = y + xyCho 0 [ x [ 0,4; Vi khong phng trnh 0,1 v gi tr ban ux0 = 0,y0 = 1, v y0 = 0


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CHNG 3

M HNH HA CC PHN TTRONG H

THNG IN3.1. GII THIU:

Trong h thng in gm c cc thnh phn cbn sau:a. Mng li truyn ti gm:

- ng dy truyn ti.- Bin p.- Cc b tin tnh, khng in.

b. Ph ti.

c. My pht ng b v cc b phn lin hp: H thng kch t, iu khin....Cc vn cn xem xt y l: Ngn mch, tro lu cng sut, n nh qu . Mng litruyn ti c gi thit l trng thi n nh v thi hng ca n nh hn nhiu so vi my

pht ng b.

3.2. M HNH NG DY TRUYN TI.

3.2.1. ng dy di ng nht.

ng dy di ng nht l ng dy c in tr, in khng, dung khng, in dnr phn bu dc theo chiu di ng dy, c th tnh theo tng pha v theo n v di.

Trong thc tin dn r rt nh c th b qua. Chng ta ch quan tm n quan h gia inp v dng in gia hai u ng dy, mt u cp v mt u nhn. Khong cch tnh tu cp n u nhn.

tnh ton v xem xt mi quan h gia in p v dng in trn tng im cang dy ta c m hnh ton hc nh sau: (xem hnh 3.1). Ti ta x ly vi phn dx trnmi pha so vi trung tnh v kho st phn t dx.

I + dI IRIS

Hnh 3.1 : Quan hin pv dngin phn t dica ng dy truyn ti

Vi phn t dx ny ta c th vit:

x =1u cp

+VR-

+VS-

VV + dV

dx x = 0u nhn

dV = I .z .dx

Hay zIdx

dV.= (3.1)

V dI = V. y . dxVi z: Tng trni tip ca mi pha trn mi n v diy: Tng dn r nhnh ca mi pha trn mi n v di

Hay yVdx

dI.= (3.2)


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Ly vi phn bc 2 ca (3.1) v (3.2) theo x ta c:

dx

dIz

dx

Vd.

2

2

= (3.3)

dx

dVy

dx

Id.

2

2

= (3.4)

Th (3.1) v (3.2) vo (3.3) v (3.4) ta c:Vyz

dx

Vd..

2

2

= (3.5)

Iyzdx

Id..

2

2

= (3.6)

Gii (3.5) ta c dng nghim nh sau:).exp().exp( 21 xzyAxzyAV += (3.7)

Thay (3.7) vo o hm bc nht (3.1) ta c dng in

).exp(1

).exp(1

21 xzyA

yz

xzyA

yz

I = (3.8)

A1 v A2c xc nh tiu kin bin:V = VR v I = IR x = 0;Thay vo (3.7) v (3.8) cn bng ta c:

2

.

1

RR Iy

zV

A

+

= (3.9)

2

.

2

RR Iy

zV

A

= (3.10)

ty

zZc = : Gi l tng trng dy

yz.= : Gi l hng s truyn sng

Vy (3.9) v (3.10) c vit gn nh sau:

).exp(2

.).exp(

2

.)( x

ZIVx

ZIVxV cRRcRR

+

+= (3.11)

).exp(

2

).exp(

2

)( xI

ZV

x

IZ

V

xIR

c

RR

c

R

+

= (3.12)

Cng thc (3.11) v (3.12) dng xc nh in p v dng in ti bt cim no cang dy theo ta x.Ta vit (3.11) li nh sau:

[ ] [ ]

).(..).(.

).(exp).(exp21..).(exp).(exp.2

1.)(

xshZIxchV

xxZIxxVxV

CRR

CRR

+=

++=

(3.13)

Tng t (3.12)

).(.).()( xshZ

VxchIxI

C

RR += (3.14)

Khi x = 1 ta c in p v dng in u cp:


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).(..).(. xshZIxchVV CRRS += (3.15)

).(.).(. xchIxshZ

VI R

C

RS += (3.16)

3.2.2. S tng ng ng dy di (l > 240):

S dng cng thc (3.15) v (3.16) lp s tng ng ca ng dy di nhhnh 3.2 (gi l s hnh ).

ZIS IR

+VS- Y1 Y2

+VR-

Hnh 3.2 : Sca ng dytruyn ti

T s hnh 3.2 ta c:RRRRRS IZVZYZYVIZVV .).1(... 22 ++=++= (3.17)

12 ).( YVYVII SRRS ++= (3.18)

Thay VS(3.17) vo (3.18) v n gin ha ta c:[ ] RRS IYZYYYZYYI ).1(...)( 12121 ++++= (3.19)

ng nht (3.17) v (3.19) tng ng vi (3.15) v (3.16) ta c:Z = ZC sh ( .l) (3.20)Y1 = Y2 = Y (3.21)(1+Z.Y) = ch ( .l) (3.22)

Vy: == 2..1).(. 1).(lth

ZlshZlchY

CC

(3.23)

Vit gn (3.20) v (3.23) li ta c:

l

lshlz

l

lshlyZZ C .

).(..

.

).(..

== (3.24)

2.

)2.(.2

.

2.

)2.(.2.

l

lthly

l

lth

Z

lyY

C

== (3.25)

S dng s hnh (3.3) v khai trin sh v ch ta c th tnh Y v Zn chnh xc cnthit. Thng thng trong s ni tip ch cn ly 2 hay 3 phn t l t yu cu chnh xc:

.............!5!3

)(53

++++=xx

xxSh

.............!4!2

1)(42

++++=xx

xCh (3.26)

.........315

17

15

2

3)( 75

3

++= xxx

xxTh


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l

lshlz

.

).(..

+

V-

R

IRIs

2.)2.(.)2(.

l

lth

Z

ly

c

)2(.)2.(.

2.

l

lthly

+

V-

S

Hnh 3.3 : Sca mng tuyn ti

Nu ch ly hai s hng u.

+ 6

).(1..

2llzZ

=

22

2

.1

2

.

2

.

3

11

2

. llllY

(3.27)

3.2.3. S tng ng ca ng dy trung bnh:

Gm cc ng dy c .l


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GII TCH MNG

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+VR-

IRIS+

-VS

Z

VR-

+VS-

+

IRIS Z/2

Y

Z/2

Hnh 3.7 : S tngng ca ngdy tuyn ti ngnHnh 3.6 : Si xng T

3.2.4. Thng s A, B, C, D:

Cc thng s A, B, C, D c s dng thit lp cc phng trnh quan h gia in p vdng in u cung cp v u nhn ca ng dy truyn ti.

Bng 3.1 : Tham s A, B, C, D cho tng loi s

Loi ng dy A B C D-ng dy ding nht

-ng dy trungbnh.Si xng T.Si xng p-ng dy ngn

...24

.2

.1).(

22

++

+=

ZY

ZYlch

2

.1

ZY+

2

.1

ZY+

...240

.

6

.

1().(.22

++

+=

ZYZY

ZlshZC

...120

.

6

.

1().(

22

++

+=

ZYZY

YZ

lsh

C

Y

)4.1( ZYY +

0

Alch =).(

A

A)

4

.1(

ZYZ +

Z

Z

AV d: ng thc 3.15 v 3.16 c vit li nh sau:

1

VS = A.VR+ B.IRIS = C.VR+ D.IRBng 3.1 cho gi tr A, B, C, D ca tng loi ng dy truyn ti. ng dy di, ng dytrung bnh v ng dy ngn, cc thng s ny c c tnh quan trng l:

A.D - B.C = 1 (3.28)iu ny c chng minh.

3.2.5. Cc dng tng trv tng dn:

Xt cc ng dy truyn ti theo cc tham s A, B, C, D cc phng trnh c vit didng ma trn:

=

R

R

S

S

I

V

DC

BA

I

V(3.29)

Phng trnh 3.29 c vit li theo bin IS v IRs dng kt qu:

A.D - B.C = 1Nh sau:


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GII TCH MNG

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=

R

S

RRRS

SRSS

R

S

I

I

ZZ

ZZ

V

V(3.30)

Vi ZSS = A/C; ZSR= -1/C; ZRS = 1/C; ZRR= -D/CCng thc (3.30) c vit di dng k hiu:V = Z.I (3.31)

Thm mt cch biu din IS, IRtheo bin VS, VR nh sau:

=

R

S

RRRS

SRSS

R

S

V

V

YY

YY

I

I(3.32)

Hay I = Y. VVi: YSS = D/B; YSR= -1/B; YRS = 1/B; YRR= -A/By ma trn Z l ma trn tng trmch h, ma trn Y l ma trn tng dn ngn mch v m

bo Z = Y-1 ca mng hai ca. chng sau s tnh mrng cho mng n ca.

3.2.6. Cc thng s Z v Y dng cho cc gii thiu khc:

T bng 3.1 cc ng thc 3.30 v 3.31 thng s Z v Y c tnh nh sau (dng cho s p)

)221(/)

2

.1(

21;2

11

221/)

2.1(

YZZY

BAY

YB

Y

YZZYB

DY

RR

RSSR

SS

+=+==

===

+=+==

(3.33) Cc

tham s ny c th tnh trc tip t s hnh 3.4 vit ra cc phng trnh nt v loi dngnhnh gia.

3.3. MY BIN P:

3.3.1. My bin p 2 cun dy:

S tng ng ca my bin p (MBA) nh hnh 3.8. Cc tham sc quy v pha scp (pha 1).

I1

I2 +

-

2

2

2

1 XN

N

2

2

2

1 RN

N

X1R1

XmRm

+

V-

1 V2

Hnh 3.8 :S tngng ca my bin p


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GII TCH MNG

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Trong MBA lc, nhnh t ha c dng kh nh c th lt i v s tng ng c rtgn nh hnh 3.9

I12

2

2

11 R

N

NR

+ 2

2

2

11 X

N

NX

+

I2

+V1-

+V2-

+

V2-

I2I1XR

+V1-

Hnh 3.9 : S tngngn gin ha ca MBA

3.3.2. My bin p tngu:

My bin p t ngu (MBATN) gm c mt cun dy chung c s vng N1 v mt cun dy

ni tip c s vng N2, s 1 pha v 3 pha di.u cc a-n i din cho pha in p thp v u cc a-n i din cho pha in p cao. T lvng ton b l:

NaN

N

Va

Va=+=+= 11

'

1

2 Ia(a)

IN2

(a)

N1

N2

(n)

(a)VaN1

N2

(b)

(c)

(a)

(c)

(b)

IN1

Va

(n)

S tng ng ca MBATN c m phng nh hnh 3.12, trong Zex l tng troc pha h khi pha cap p ngn mch.

Hnh 3.11 : S 1 pha ca MBATNHnh 3.10 : MBA tngu 3 pha

Hai tng trngn mch na c tnh l:- ZeH: Tng tro c pha cao p khi s vng N1 b ngn mch ni tt cc a-n. V d

dng chng minh t hnh 3.12 (php quy i)ZeH = Zex N

2 (3.34)- ZeL: Tng tro c pha h p khi s vng N2 b ngn mch ni tt cc a-a


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hnh 3.13.

+

-Va

1:NIa

a+

Va-

Ia Zex

Zex+Va

-n

aI1 Ia

1:N a

n

+Va-

Iaa

n n Hnh 3.13 : S tngng khi

ni a-a ca MBATNHnh 3.12 : S tngng ca MBATNT s hnh 3.13 ta c:Va = Va

exaexa

a ZN

NVZ

N

VVI /

)1(/)( '1

== (3.35)

i vi my bin p l tng s ampe vng bng zero cho nn chng ta c:I1 = Ia NHay Ia = I1/NVi: Ia + Ia = I1V vy:

N

NIIa

1.1

=

Tng tr:

exa

a

aeL Z

N

N

N

N

I

V

I

VZ

2

1 1)1(

=

==

Do :

eLex ZN

NZ

21

= (3.36)

S dng (3.34) ta c:ZeH = (N-1)

2 Z eL = a2ZeL

* Nhc im ca MBATN:- Hai pha cao v h p khng tch nhau vin nn km an ton- Tng trni tip thp hn MBA 2 cun dy gy ra dng ngn mch ln

* u im ca MBATN:- Cng sut n v ln hn MBA 2 cun dy nn ti c nhiu hn

- li cng ln khi t s vng l 2:1 hoc thp hnV d minh ha: Cho mt MBA 2 cun dy c thng snh mc l 22KVA, 220/110V, f =50Hz. Cun A l 220V c Z = 0,22 + j0,4 () cun B l 110V c tng trl Z = 0,05 + j0,09().MBA u theo dng t ngu cung cp cho ti 110V vi ngun 330V. Tnh Zex, ZeL, ZeH dng

ph ti l 30A. Tm mc iu tit in p.Gii:Cun B l cun chung c N1 vng, cun A l cun ni tip c N2 vng.Vy N2 /N1 = 2 = a v N = a+1 = 3, do ZA = 0,24 + j0,4 (), ZB = 0,05 + j0,09 ()

Nn:ZeH = ZA + a

2ZB = 0,44+ j0,76 ()

ZeL = ZB + ZA/a2 = 0,11+j0,19 ()


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GII TCH MNG

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)(08,0049,01

2

2+=

== jN

NZ

N

ZZ eL

eHex

Mc iu chnh in p = %100.sin..cos..

V

XIRI +

%21,2%100.330

437,0.76,09,0.44,0

.3

30

=

+

=

3.3.3. My bin p c biu p:

Do ph ti lun thay i theo thi gian dn n in p ca h thng in cng thay i theo. gi cho in p trn cc dy dn nm trong gii hn cho php ngi ta iu chnh in pmt hoc hai pha ca MBA bng cch t b phn p vo MBA ni chung l t pha cao piu chnh mm hn. Khi t s vng N bng t sin p nh mc ta ni l t lngnht. Khi chng khng bng ta ni t l l khng ng nht. Biu p c hai loi:-Biu p di ti-Biu p khng ti

Biu p di ti c thiu chnh tng hoc bng tay, khi iu chnh bng tay phi davo kinh nghim v tnh ton tro lu cng sut trc . T su phn p c th l s thchay s phc trong trng hp l s phc in p hai pha khc nhau v ln v gc pha.MBA ny gi l MBA chuyn pha.

3.3.4. My bin p c t s vng khng ng nht:

Chng ta xt trng hp t s vng khng ng nht l s thc cn xt hai vn sau:- Gi tr tng i ca tng trni tip ca MBA t ni tip trong my bin p l tng cho

php c s khc nhau trong in p, t l khng ng nht c m t trn s bng ch a vgi thit rng a nm xung quanh 1 (a 1)

- Gi thit tng trni tip ca MBA khng i khi u phn p thay i v tr.MBA khng ng nht c m t theo hai cch nh hnh 3.14, tng dn ni tip trong haicch c quan h l Y1 = Y1/a

2.

Vi t l bin p bnh thng l a:1 pha a gi l pha iu p. V vy trong s 1 tng dnni tip c ni n pha 1 cn s 2 th c ni n pha a.

Xt hnh 3.15 ca MBA khng ng nht y tng trni tip c ni n pha n v cabiu p.

Y1

(2)

q

q

Y1a:1

Hnh 3.14 : Hai cch gii thiumy bin p khngng nht

(1)

qpa

Hnh 3.15 : S tngng ca MBA khngng nht

Y1a:1

a:1

p

p

Mng hai ca tng ng ca n l:


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GII TCH MNG

Trang 38

nt p:

a

YV

a

YV

aYaVVI

qp

qppq

1

2

1

21 /)(

=

=(3.37)

nt q:

a

YVYV

Ya

VVI

p

q

pqpq

11

1'

..

)(

=

=(3.38)

+

-Vp

q

0

+

-Vq

q

0

+

-Vq

Y1

Y2 Y3

p

0

+

-Vp

(b)

21

)1(

a

aY

21

)1(

a

aY

(c)

(1-a)Y1

aY1 Ipqq

0

a(a-1)Y1+

-Vq

Ipqp

0

+

-Vp

(a)

Y1/aIpq Ipq Ipq Ipqp

0

Hnh 3.16 : S tngng ca MBA khngng nht

s hnh 3.16a ta c:Ipq = VpY2 + (Vp-Vq)Y1 (3.39)Ipq = VqY3 + (Vq-Vp)Y1 (3.40)ng nht (3.39) v (3.40) vi (3.37) v (3.38) ta c:Y1 + Y2 = Y1/a

2

Y1 =Y1/aY1 + Y3 = Y1

Gii ra ta c:a

YYY

a

Y

a

YY

a

YY 113

121

21

1 ;; ===

S l hnh 3.16b. Ch tt c tng dn trong s tng ng l hm ca t s vng a. Vdu lin hp gia Y2 v Y 3 lun ngc. V d: Nu Y1 l in khng a > 1; Y2 l in khng;Y3 l in dung; nu a < 1; Y2 l dung khng v Y3 l in khng.S hnh 3.16c l s tng ng theo Y1 khi a 1 th tng trmch r v tngdn ni tip tin n Y1.

3.3.5. My bin p chuyn pha:

Trong h thng in lin kt c mch vng hay ng dy song song, cng sut tht truyntrn ng dy c iu khin bng my bin p chuyn pha, MBA c t s vng l s phcth ln v gc pha in p ph thuc vo v tr ca biu p.Khi cun scp v cun th cp c qun trn cng mt li th chng c cng pha v t l

phn p l thc. Tuy nhin trong my bin p t ngu chuyn pha cun scp v cun th cp

c b tr ty theo lch pha khi thay i u phn p th gc pha cng thay i theo. S minh ha hnh 3.17a, sn gin ha ch c mt pha ca MBATN chuyn pha l y cho gn gng, d thy cun dy th 2 ca pha a b lm lch in p i 900 so vi pha a.


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GII TCH MNG

Trang 39

s vecthnh 3.17b khi u phn p chy t R A th in p thay i t zero n aakt qu l in p th cp thay i t oa n oa.

ARc

A R

AR

b

a a

cb

b

c

a a

(b)(a)

Hnh 3.17 : My bin p tngu chuyn pha gm c ba phaa. Su dy

b. S vectNh hnh 3.17 ta thy rng in p cun ni tip cao hn bnh thng cho php cng sut lnhn chy trn ng dy ngha l: Thay v lp my bin p thng ta lp my bin p chuyn

pha s cho php nng cao in p cp v ng dy mang ti nhiu hn.

3.3.6. My bin p ba cun dy.

My bin p ba cun dy s dng trong nhng trng hp cn cung cp cho ph ti hai cp in p t mt cun dy cung cp. Hai cun dy ny gi l cun th hai v cun th ba(hnh 3.18). Cun th 3 ngoi mc ch trn cn c mc ch khc, chng hn c ni vo t chn sng bc 3. Trn s ta k hiu 11 l cun scp (P), 22 l cun th 2 (S), 33 lcun th 3 (T).

P S

T

Hnh 3.18 : My bin p bacun dy

Cc tham so c t th nghim l:ZPS: L tng trcun scp khi ngn mch cun 2 v hmch cun 3ZPT: L tng trcun scp khi ngn mch cun 3 v hmch cun 2ZST: L tng trcun th cp khi cun scp hmch v cun 3 ngn mch

ZST quy i v pha scp l: STS

PST Z

N

NZ '.

2

=

S tng ng ca MBA ba cun dy hnh 3.19 ZPS, ZPT, ZST, quy i v pha scp.Theo cch o ngn mch ta c:ZPS = ZP + ZS (3.41)ZPT = ZP + ZT (3.42)ZST = ZS + ZT (3.43)Tr (3.42) i (3.43) ta c:


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GII TCH MNG

Trang 40

ZPT - ZST = ZP - ZS (3.44)T (3.41) v (3.44) ta c:ZP =1/2 (ZPS + ZPT -ZST) (3.45)ZS =1/2 (ZPS + ZST -ZPT) (3.46)ZT =1/2 (ZST + ZPT - ZPS) (3.47)

Zp ZS

ZT

Hnh 3.19 : S tngng ca MBA ba cun dyB qua tng trmch r nn nt t q tch ri u cc 1 ni vi ngun cung cp, u cc 2 v3 ni n ti, nu cun 3 dng chn sng hi th th ni.

3.3.7. Ph ti:

Chng ta nghin cu v ph ti lin quan n tro lu cng sut v n nh. iu quan trng lphi bit s thay i ca cng sut tc dng v cng sut phn khng theo in p. cc ntin hnh cc loi ti gm c:

- ng ckhng ng b 5070 %- Nhit v nh sng 2030 %

- ng cng b 510 % tnh chnh xc ngi ta dng c tnh P-V v Q-V ca tng loi ti nhng x l phn tchrt phc tp. V vy ngi ta a ra ba cch gii thiu chnh v ti dng cho mc ch phn

tch.- Gii thiu theo cng sut khng i: C lng MVA v MVARu bng hng s

thng dng nghin cu tro lu cng sut.- Gii thiu theo dng in khng i: Dng in ti I trong trng hp ny c tnh

)(||

= VV

jQPI

V = |V|q v = tan-1 (Q/P) l gc h s cng sut, ln ca I c gi khng i.- Gii thiu theo tng trkhng i: y l cch gii thiu thng xuyn khi nghin

cu n nh nu lng MVA v MVAR bit v khng i th tng trti tnh nh sau:

jQP

V

I

VZ

==

2||

V tng dn:

2||

1

V

jQP

ZY

==

3.4. KT LUN:

Trong chng ny ta xem xt cc phn t ca h thng in nhng dy truyn ti, bin p,ph ti. M hnh ha chng trong h thng in vi trng thi n nh nghin cu cctrng thi cbn ca h thng: Ngn mch, phn b dng chy cng sut, v n nh qu .


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GII TCH MNG

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CHNG 4

CC MA TRN MNG V PHM VI NG

DNG

4.1. GII THIU:

S trnh by r rng chnh xc ph hp vi m hnh ton hc l bc u tintrong gii tch mng in. M hnh phi din tc c im ca cc thnh phnmng in ring bit nh mi lin h chi phi gia cc thnh phn trong mng. Phngtrnh ma trn mng cung cp cho m hnh ton hc nhng thun li trong vic gii bng

my tnh s.Cc thnh phn ca ma trn mng ph thuc vo vic chn cc bin mt cchc lp, c th l dng hoc p. V l, cc thnh phn ca ma trn mng s l tngtrhay tng dn.

c im ring ca cc thnh phn mng in c thc trnh by thun litrong hnh thc h thng ma trn gc. Ma trn din tc c im tng ng cami thnh phn, khng cung cp nhiu thng tin lin quan n kt ni mng in. N lcn thit, v vy bin i h thng ma trn gc thnh ma trn mng l din tc ccc tnh quan h trong li in.

Hnh thc ca ma trn mng c dng trong phng trnh c tnh ph thuc

vo cu trc lm chun l nt hay vng. Trong cu trc nt lm chun bin c chnl nt p v nt dng. Trong cu trc vng lm chun bin c chn l vng in pv vng dng in.S to nn ma trn mng thch hp l phn vic tnh ton ca chng trnh my tnh scho vic gii bi ton h thng in.

4.2. GRAPHS.

din t cu trc hnh hc ca mng in ta c th thay th cc thnh phn camng in bng cc on ng thng n khng kc im ca cc thnh phn.

ng thng phn on c gi l nhnh v phn cui ca chng c gi l nt. Ntv nhnh ni lin vi nhau nu nt l phn cui ca mi nhnh. Nt c th c nivi mt hay nhiu nhnh.

Graph cho thy quan h hnh hc ni lin gia cc nhnh ca mng in. Tphp con ca cc graph l cc nhnh. Graph c gi l lin thng nu v ch nu cng ni gia mi cp im vi nhau. Mi nhnh ca graph lin thng c n nhhng th n snh theo mt hng nht nh. S biu din ca h thng in vhng tng ng ca graph trnh by trong hnh 4.1.Cy l mt graph lin thng cha tt c cc nt ca graph nhng khng to thnh mtvng kn. Cc thnh phn ca cy c gi l nhnh cy n l tp h p con cc nhnh

ca graph lin thng chn trc. S nhnh cy b qui nh cho mi cy l: b = n - 1 (4

Vi: n l s nt ca graph


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GII TCH MNG

Trang 43

0

2

1

Hnh 4.1 :Mth thngin.

(a) S

mt pha.(b) S tht thun.(c) Graph nh hng.

(c)

7

5

12

3

43

4

6

(b)0

2

3

4

(a)

G

G

G

1

Nhnh ca graph lin thng khng cha trong cy c gi l nhnh b cy, tp

hp cc nhnh ny khng nht thit phi lin thng vi nhau c gi l b cy. B cyl phn b ca cy. S nhnh b cy l ca graph lin thng c e nhnh l:l = e - b

T phng trnh (4.1) ta cl = e - n + 1 (4.2)

Cy v b cy tng ng ca graph cho trong hnh 4.1c c trnh by trong hnh 4.27

2

6 44

e = 7n = 5

b = 4l = 3

32

1

5

0

3

Nhnh b cy

Nhnh cy

1

Hnh 4.2 :Cy v b cy ca graph lin thngnh hng

Nu nhnh b cy c cng thm vo cy th kt qu graph bao gm mt

ng kn c gi l vng. Mi nhnh b cy c cng thm vo s to thnh mthay nhiu vng. Vng ch gm c mt nhnh b cy c lp th gi l vng cbn. Bivy, s vng cbn ng bng s nhnh b cy cho trong phng trnh (4.2). Snh


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GII TCH MNG

Trang 44

hng ca vng cbn c chn ging nh chiu ca nhnh b cy. Vng cbn cagraph cho trong hnh 4.2 c trnh by trong hnh 4.3.

7

6 4

32

1

5

2 3

E GF

1

4

0

Hnh 4.3 :Vng cbn nh hng theo graph lin thng

Vt ct l tp hp ca cc nhnh, nu bi hoc chia graph lin thng thnh hai graphcon lin thng. Nhm vt ct c th chn c lp duy nht nu mi vt ct ch bao gmmt nhnh cy. Vt ct c lp nh vy gi l vt ct cbn. S vt ct cbn ng

bng s nhnh cy. Snh hng ca vt ct cbn c chn ging nh hng canhnh cy. Vt ct cbn ca graph cho trong hnh 4.2 c trnh by trong hnh 4.4

7

6 4

32

1

5B

D

C

3

2

A

41

0

Hnh 4.4 :Vt ct cbn nh hng theo graph lin thng

4.3. MA TRN THM VO.

4.3.1. Ma trn thm vo nhnh - nt .

S lin h gia nhnh v nt trong graph lin thng trnh by bi ma trn thmvo nhnh nt. Cc thnh phn ca ma trn c trnh by nh sau:

aj = 1 : Nu nhnh th i v nt th j c chiu hng t nhnh i vo nt jaj = -1: Nu nhnh th i v nt th j c chiu hng t nhnh i ra khi nt jaj = 0 : Nu nhnh th i v nt th j khng c mi lin h vi nhau.

Kch thc ca ma trn l e x n, vi e l s nhnh v n l s nt ca graph. Ma trn

thm vo nhnh nt cho trong graph hnh 4.2 trnh by nh trn. Vi:


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GII TCH MNG

Trang 45

eiaj

ji ...,2,104

0

===

n

40 1 2 3e

=

1

7

6

5

4

32

1

1 -1

1 -1

-1 1

-1

-1

-1

1

-11

1

Cc ct ca ma trn l ph thuc tuyn tnh. V vy hng ca < n.

4.3.2. Ma trn thm vo nt A.

Cc nt ca graph lin thng c th chn lm nt qui chiu. Nt qui chiu c ththay i, n c xem nh mt nt trong graph c th cn nhc khi n nh c th mtnt no lm nt qui chiu. Ma trn thu c t ma trn bi ct tng ng vint chn lm nt qui chiu l ma trn nhnh - nt A, n sc gi l ma trn nt. Kchthc ca ma trn l e x (n-1) v hng l n-1 = b.

Vi: b l s nhnh cy ca graph. Chn nt 0 lm nt qui chiu th hin trn graphtrong hnh 4.2. nt

e 41 2

1

2

3

4

5

6

7

A

=

-11

1 -1

1 -1

-1 1

-1

-1

-1

3

Ma trn A l hnh ch nht v l duy nht. Nu hng ca A sp xp theo mt cy ringbit th ma trn trn c th phn chia thnh cc ma trn con Ab c kch thc b x (n-1)v At c kch thc l l x (n-1). S hng ca ma trn Ab tng ng vi s nhnh cy vs hng ca ma trn At tng ng vi s nhnh b cy. Ma trn phn chia ca graphtrn hnh 4.2 c trnh by nh sau:


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GII TCH MNG

Trang 46

nt nt2 3 4

1

23

4

5

6

7

e

A =

1

-1-1

-1 1

-11

-1

1 -1

Cc nt

Nhnhbcy

Nhnh

cy

Ab

At

e

-1

1

=

Ab l ma trn vung khng duy nht vi hng (n -1).

4.3.3. Ma trn hng ng - nhnh cy K:

Hng ca cc nhnh cy n cc ng trong 1 cy c trnh by bng ma trnhng ng - nhnh cy. Vi 1 ng c nh hng t 1 nt qui chiu. Cc phnt ca ma trn ny l:kij = 1: Nu nhnh cy i nm trong ng t nt j n nt qui chiu v c nh hngcng hng.

kij = -1: Nu nhnh cy i nm trong ng t nt j n nt qui chiu nhng cnh hng ngc hng.kij = 0: Nu nhnh cy i khng nm trong ng t nt j n nt qui chiu.

Vi nt 0 l nt qui chiu ma trn hng ng - nhnh cy lin kt vi cy c trnhby hnh 4.2 c dng di y.

ng

1

2

3

4

Nhnh cy

K =

-1

-1

-1-1

-1

41 2 3

y l ma trn vung khng duy nht vi cp l (n-1). Ma trn hng - ng nhnhcy lin h nhnh cy vi cc ng nhnh cy ni n nt qui chiu v ma trn Ab linkt cc nhnh cy vi cc nt. V vy c t l tng ng 1:1 gia cc ng v cc nt.

Ab.Kt = 1 (4.3)

Do : Kt

= Ab-1

(4.4)


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GII TCH MNG

Trang 47

4.3.4. Ma trn vt ct cbn B.

Lin h gia nhnh vi vt ct cbn ca graph lin thng c th hin trongma trn vt ct cbn B. Cc thnh phn ca ma trn l.

bj = 1 : Nu nhnh th i v hng cng chiu vi vt ct cbn th jbj = -1 : Nu nhnh th i v hng ngc chiu vi vt ct cbn th j

bj = 0 : Nu nhnh th i khng lin quan vi vt ct th jMa trn vt ct cbn c kch thc l e x b ca graph cho trn hnh 4.4 l:

DA BVt ct c bn

Ce b

1

2

3

4

5

6

7

B =

1 1

1 1

-1 1

1

1

1

1

1

Ma trn B c th phn chia thnh cc ma trn con Ub v Bt. S hng ca ma trnUb tng ng vi s nhnh cy v s hng ca ma trn Bt tng ng vi s nhnh bcy. Ma trn phn chia c biu din nh sau:

b

1

2

3

4

5

6

7

eA

B CVt ct c bn

Db

Vt ct c bn

=

Nhnhbcy

Nh

nhcy

Ub

Bt

e

B =

11

-1 1

-1 1

1

1

1

1

1


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GII TCH MNG

Trang 48

Ma trn n v Ub cho ta thy quan h tng ng ca mt nhnh cy vi mt vt ct cbn..

Ma trn con Bt c th thu c t ma trn nt A. Lin h gia nhnh b cy vint cho thy bi ma trn con At v gia nhnh cy vi nt l ma trn con Ab. Tytng ng quan h ca mt nhnh cy vi mt vt ct cbn, Bt.Ab cho thy quan hgia cc nhnh b cy vi cc nt nh sau:

Bt.Ab = AtV vy

Bt = At .Ab-1

Theo phng trnh (4.4) ta cAb

-1 = Kt

V vy ta cBt = At .K

t (4.5)

4.3.5. Ma trn vt ct tng thmB .

Vt ct gi thit c gi l vt ct rng buc c tha vo sau tng bc s vt ct ng bng s nhnh. Mi vt ct rng buc ch gm mt nhnh b cy cagraph lin thng. Vt ct rng buc ca graph cho trn hnh 4.4 c trnh by tronghnh 4.5. 7

G

Vt ct rng buc

1 6 4

32

1

5B

D

C

0

2

34

F E

Vt ct c bnA

Hnh 4.5 :Vt ct cbn v rng buc nh hng theo graph lin thng

Ma trn vt ct tng thm c hnh thc biu din nh ma trn vt ct cbn cng thms ct ca vt ct rng buc. Vt ct rng buc c nh hng ph thuc vo hng

ca nhnh b cy. Ma trn vt ct tng thm ca graph trnh by trn hnh 4.5 l matrnB nh sau:


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GII TCH MNG

Trang 49

1

E FC DA Be

e

7

6

5

4

3

2

1

1-1

1

-1

1

1

-1 1

1

11

1 1

Vt ct c bn Vt ct gi to

G

=B

B : L ma trn vung c kch thc e x e v khng duy nht. Ma trnB c thphn chia nh sau:

Vt ct c bn Vt ct gi to

1

=B

DCBe Ae

7

6

5

43

2

1

1-1

1

-1

1

1

-1 1

1

11

1 1

G eVt ct gi

toVt ct c

bne

Nh

nhcy

Nhnhbcy

=

Bt

Ub 0

Ut

E F

4.3.6. Ma trn thm vo vng cbn C.

Tc ng ca nhnh cy vi vng c bn ca graph lin thng th hin bi ma trnvng cbn. Thnh phn ca ma trn l:

cj = 1 : Nu nhnh cy th i v hng cng chiu vi vng cbn th jcj = -1: Nu nhnh cy th i v hng ngc chiu vi vng cbn th jcj = 0 : Nu nhnh cy th i khng lin quan vi vng cbn th j

Ma trn vng cbn c kch thc e x l theo graph cho trn hnh 4.3 nh sau:


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GII TCH MNG

Trang 50

le E F G

1

2

3

4

5

6

7

C =

1

-1

-1

-1

1

1

1

1

Vng c bn

Ma trn C c th phn chia thnh cc ma trn con Cb v Ut. S hng ca ma trn Cbtng ng vi s nhnh cy v s hng ca ma trn Ut tng ng vi s nhnh b cy.Ma trn phn chia nh sau:

Vng c bnl

1

2

3

4

5

6

7

e E F Gl Vng c bn

e

Nhnhcy

Nh

nhbcy

=

Cb

Ut

C =

1

-1

-1

-1

1

1

1

1

Ma trn n v Ut cho thy mt nhnh b cy tng ng vi mt vng cbn.

4.3.7. Ma trn s vng tng thm .C

S vng c bn trong graph lin thng bng s nhnh b cy. c tng svng bng s nhnh, thm vo (e-l) vng, tng ng vi b nhnh cy, gi l vng h.Vng hc v bn cc nt ni bi nhnh cy. Vng hca graph cho trn hnh 4.3

c trnh by trong hnh 4.6. Hng ca vng hc xc nh theo nh hng canhnh cy.


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GII TCH MNG

Trang 51

7

6 4

32

1

5F

GE

3

0

24

Vng h

Vng c bnB

D

A C

1

Hnh 4.6 :Vng cbn v vng hnh hng theo graph lin thng

Ma trn vng tng thm c hnh thc nm bn cnh ma trn vng cbn, cc ctca n biu din mi quan h gia cc nhnh vi vng h. Ma trn ca graph trnh bytrong hnh 4.6 c biu din di y.

C : L ma trn vung, kch thc e x e v khng duy nht.

1

E F

=C

C DBAe

e

7

6

5

4

3

2

1

1

-1

1

1 -1

1

1

1

1-1

-1

1

1

Vng h Vng c bn

G


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GII TCH MNG

Trang 52

Ma trn C c th phn chia nh sau:

1

=C

e

7

6

5

4

3

2

1

1

1

1

-1

1

1

1

-1

-1

-1

11

1

A B C D E F G

=

ee

Nh

nhbcy

Nhnhcy

Cb

Ut

Ub

0

Vng h Vng c bn

Vng h Vng c bne

4.4. MNG IN GC.Thnh phn ca mng in l tng trv tng dn c trnh by trong hnh 4.7.

c tnh ca cc thnh phn c th biu din trong mi cng thc. Bin v tham s l:vpq: L hiu in th ca nhnh p-qepq: L ngun p mc ni tip vi nhnh p-qipq: L dng in chy trong nhnh p-q

jpq: L ngun dng mc song song vi nhnh p-qzpq: L tng trring ca nhnh p-qypq: L tng dn ring ca nhnh p-qMi mt nhnh c hai bin vpq v ipq. Trong trng thi n nh cc bin v tham

s ca nhnh zpq v ypq l mt s thc i vi dng in mt chiu v l mt s phci vi dng in xoay chiu.


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GII TCH MNG

Trang 67

CHNG 5

CC THUT TON DNG CHO VIC

THNH LP NHNG MA TRN MNG5.1. GII THIU.

Nhng phng php trnh by trong cc mc trn i hi mt s chuyn i vo ngc nhng ma trn c c nhng ma trn mng. Mt phng php thay thda trn mt thut ton c thc dng thnh lp trc tip ma trn tng trnt tnhng thng s h thng v s nt c m ho. Nguyn tc ca thut ton l thnhlp ma trn tng trnt theo tng bc, m phng cu trc ca mng bng cch thmvo tng nhnh mt. Mt ma trn c thnh lp cho mng ring c biu th sau khi

mi phn tc ni vi mng.Ngoi ra, mt thut ton c biu th chuyn ha ma trn tng dn vng t

ma trn tng trnt nh.Cc phng trnh mng:

INt = YNt .ENtENt = ZNt .INtYNt = A

t.y. A

ZNt = (YNt)-1

5.2. XC NH MA TRN YNT BNG PHNG PHPTRC TIP.

Gi Ei, Ej, Ek l in p ti cc nt khi bm mt dng vo nt i.

yjjij

EjIiyij

k

yik

ykki

yiik

Ei

yiiji

Yii

yii

Ek

Hnh 5.1 :S m t mngin ti 1 nt

Ij = 0; j i

ijij

jiij

iiiji yEEEyI += )().(


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GII TCH MNG

Trang 68

+=ij ij

jijiij

ij

iiij EyEyEy ).(

)()( ijij

j

ij ij

ijiiji yEyyE ++=

)().( ijij

j

ij

ijiii yEyyE +=

Ta c: +=+= ijiiijiijii yyyyY

ijij yY =

Do :

=+=ij

jijjijiiii EYEYEYI .

Vy : YNt l ma trn c cc thnh phn trn ng cho chnh l Yii thnh phn ngoing cho l Yij.Ch : Nu c tng h th chng ta phi tnh thm cc thnh phn tng h.

++=++=

rsijijiirsijijiijii yyyyyyY ,, += )( ,, rsijijijij yyY

5.3. THUT TON THNH LP MA TRN TNGTRNT:

5.3.1. Phng trnh biu din ca mt mng ring.

Gi thit rng ma trn tng trnt ZNtc bit t mt mng ring m ntv mt nt qui chiu 0. Phng trnh biu din ca mng ny cho trong hnh (5.2)l:

NutNutNut IZErr

.=

Mngring

1

2

m

0H qui chiu

I1

I2

Im

E1

E

E2

Hnh 5.2 : Sbiu din ca mtmng ring

Trong : NutEr

= m x 1 vectca cc in p nt c o i vi nt qui chiu.

NutIr

= m x 1 vectca cc dng in c bm vo nt khi mt nhnh p - q c

thm vo mng ring, n c th l mt nhnh cy hoc mt nhnh b cy nh cho hnh (5.3)


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GII TCH MNG

Trang 69

(a)S thm vo ca mt nhnh cy(b)S thm vo ca mt nhnh b cy- Nu p - q l mt nhnh cy, mt nt mi q c thm vo mng ring v to

thnh ma trn tng trnt kch thc l (m + 1) x (m + 1). Cc vectin p mi vdng in mi c kch thc l (m + 1) x 1. xc nh ma trn tng trnt mi yucu ch tnh cc phn t trong hng v ct mi.- Nu p - q l mt nhnh b cy, khng c nt mi c thm vo mng ring.Trong trng hp ny, kch thc ca cc ma trn trong phng trnh biu dinc gi nguyn, nhng tt c cc phn t ca ma trn tng trnt phi ctnh li bao hm nh hng ca nhnh b cy c thm vo.

1(a) (b)

Hnh 5.3 : Sbiu din ca mt mng ring vi mt nhnh c thm vo

Nhnh p-q

M

M

M

Mngin

p

2

1

m

q

Nhnh p-q

0

H quichiu

H quichiu

M

M

0

m

p q

Mngin

2

5.3.2. Sthm vo ca mt nhnh cy.

Gi s ma trn ZNt ban u c kch thc m x m, sau khi thm 1 nhnh cy kchthc m m +1. Gi s ta thm vo 1 nt q ta c phng trnh biu din ca mngring vi mt nhnh cy p - q c thm vo l nh (5.1). iu c ngha l mng tnti cc nhnh bng c hai pha.

1

H qui chiuM

M

M

M

0

Ii = 1

i

p

Mngin

2

q

Nhnh p-

qvpq

EpEq

Hnh 5.4 : Dngin c bmvo v stnh ton cc in p nt

ca Zqi


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GII TCH MNG

Trang 70

Do : Zqi = Ziq, vi i = 1, 2, ..., m v c lin quan n cc nt ca mng ring,nhng khng kn nt mi q.

Nhnh cy p - q thm vo c xem l c h cm vi mt hoc nhiu nhnh camng in.

=

q

m

p

qqqmq

mqmmm

pqpmp

qm

qm

q

m

p

I

I

I

I

I

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

E

E

E

E

E

*

**

**

**

***** **

**

*2

1

1

1

1

2221

1111

2

1

(5.1)

Cc phn t Zqi c thc xc nh bng cch bm vo mt dng in ti nt i v tnhin p ti nt q vi im qui chiu nh trnh by hnh (5.4).Gi s ta bm

dng I = 1A vo nt i (Ij = 0 j i) v tt c cc dng in ti cc nt khc bng

0, t phng trnh (5.1) suy ra:Eq = Zqi .Ii = Zqi

Tng t nh trn ta bm vo cc nt cn liE1 = Z1i .IiE2 = Z2i .Ii...............Ep = Zpi .Ii (5.2)................Em = Zmi .IiEq = Zqi .IiCho Ii = 1 trong phng trnh (5.2), Zqi c th thu c trc tip bng cch tnh

EqCc in p nt lin kt vi nhnh thm vo v in p qua nhnh c th hin

bi:Eq = Ep - vpq (5.3)

Cc dng in trong cc nhnh ca mng trong hnh (5.4) c din t trong ccs hng ca cc tng dn ban u v cc in p qua cc nhnh l:

=

yrs,pq

ypq,pq

yrs,rs

ypq,rs

Vrs

vpq

irs

ipq(5.4)

Trong phng trnh (5.4), pq l mt ch s cnh v lin quan vi nhnh thm vo, vrs l ch s bin i, lin quan n cc nhnh khc. Trong :

- ipq v vpq: L dng in v in p chy qua tng ng vi nhnh thm vo.- irs v vrs: L cc vectdng in v in p trong cc nhnh ca mng ring.- ypq,pq: L tng dn ring ca nhnh thm vo.- ypq,rs : L vectca cc tng dn tng h gia nhnh thm vo p - q v cc

nhnh r - s ca mng ring.

- yrs,pq : L vectchuyn v ca ypq,rs- [yrs,rs]: L ma trn tng dn ban u ca mng ring.Dng in chy trong nhnh cy thm vo cho trong hnh 5.4 l:


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ipq = 0 (5.5)Tuy nhin, vpq khng bng 0 v nhnh cy thm vo h cm vi mt hoc nhiu nhnhca mng ring. Ngoi ra:

srrs EEvrrr

= (5.6)Trong : Er v Es l cc sut in ng ti cc nt trong mng ring. T phng trnh

(5.5) ta c: =+= 0.. ,, rsrspqpqpqpqpq vyvyirr

Do :

= rsrspqpqpq

pq vyy

vrr

.1

,,

Th t phng trnh (5.6) ta c:rsvr

= )(1

,,

srrspq

pqpq

pq EEyy

vrrr

(5.7)

Th vpq vo trong phng trnh (5.3) t (5.7) ta c:

+= )(1 ,,

srrspq

pqpq

pq EEyy

EE rrr

Cui cng, th Ep, Eq, rEr

v sEr

t phng trnh (5.2) vi Ii = 1, ta c:

+= )(1

,,

rsrirspq

pqpq

piqi ZZyy

ZZrrr i = 1, 2, ....m i j (5.8)

Phn t Zqq c thc tnh bng cch bm mt dng in ti nt q v tnh in p tint . Gi s ta bm dng I = 1A vo nt q (Ij = 0 j q) v tt c cc dng in ticc nt khc bng 0, t phng trnh (5.1) ta suy ra.

Eq = Zqq .Iq = ZqqTng t nh trn ta bm vo cc nt cn li

E1 = Z1q.IqM Ep = Zpq.Iq (5.9)M Em = Zmq.Iq

Trong phng trnh (5.9), Zqq c th thu c trc tip bng cch tnh Eq.Tng t ta c in p gia 2 nt p v q l:

Eq = Ep - vpqin p ti cc nt p v q c lin kt vi nhau bi phng trnh (5.3) v dng inchy qua nhnh thm vo l:

ipq = -Iq = -1 (5.10)Cc in p qua cc nhnh ca mng ring c cho bi phng trnh (5.6) v cc dngin chy qua cc nhnh cho bi phng trnh (5.4) v (5.10) ta c:

=+= 1.. ,, rsrspqpqpqpqpq vyvyirr

Do :

pqpq

rsrspq

pqy

vyv

,

, .1 =

rr

Th t phng trnh (5.6) ta c:rsvr

pqpq

srrspq

pqy

EEyv

,

, ).(1 =

rrr

(5.11)


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GII TCH MNG

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Th vpq vo trong phng trnh (5.11) t (5.3) ta c:

pqpq

srrspq

pqy

EEyEE

,

, ).(1 ++=

rrr

Cui cng, th Ep, Eq, vrEr

sEr

t phng trnh (5.9) vi Iq = 1, ta c:

pqpq

sqrqrspqpqqq

yZZyZZ

,

, )(1 +

+=

rrr

(5.12)

Nu khng c h cm gia nhnh cy thm vo v cc nhnh khc ca mng ring, thcc phn t ca ypq,rs bng 0.V ta c:

pqpq

pqpqy

Z,

,

1=

T phng trnh (5.8), ta suy ra rng:Zqi = Zpi , i = 1, 2, ....m i j

V t phng trnh (5.12), ta c:Zqq = Zpq + Zpq,pqHn na, nu nh khng c h cm v p l nt qui chiu

Zpi = 0, i = 1, 2,......m i q Nn: Zqi = 0, i = 1, 2,......m i q Tng t: Zpq = 0V v vy: Zqq = Zpq,pq

5.3.3. Sthm vo ca mt nhnh b cy.

Nu nhnh p - q thm vo l mt nhnh b cy, phng php tnh cc phn t ca

ma trn tng trnt l mc ni tip vi nhnh thm vo mt sut in ng el nh chotrong hnh 5.5.Vic ny to thnh mt nt gi l m nt sc loi tr ra sau . Sut in ng elc chn nh th no m dng in chy qua nhnh b cy thm vo bng 0.

l

ipq =0 Eqel

Ypq,pqEp

pq

Gi s ma trn ZNt ban u c kch thc m x m, khi ta thm nhnh b cy v to ntgi l th ma trn ZNt c kch thc l (m+1) x (m+1).


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GII TCH MNG

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1

M

M

M

Mng in

2ip

Ii = 1

q

0

lvpqipq

el

H quichiu

Eq

El

EpHnh 5.5 :Dngin bm vo,sutin ng trong mch nitip vi nhnh b cy thm vov cc in p nt cho vic tnh

ton ca Zli

Phng trnh t trng cho mng ring vi nhnh p-l thm vo v mch ni tip scin ng el l .

=

l

m

lllml

mlmmm

l

lm

l

m

I

I

I

I

ZZZ

ZZZ

ZZ

ZZZ

e

E

E

E

*

**

**

*****

***

**

*2

1

1

1

212

1111

2

1

(5.13)

V: el = El - Eq

Phn t Zli c thc xc nh bng cch bm vo mt dng in ti nt i v tnh inp ti nt l thuc v nt q. V tt c cc dng in ti cc nt khc bng 0, t phngtrnh (5.13) ta suy ra:

Ek= Zki .Ii = ZkiTng t nh trn ta bm vo cc nt cn li

E1 = Z1i .IiM Ep = Zpi .IiM el = Zli.Ii , i =1, 2, ....m (5.14)

Cho Ii = 1 trong phng trnh (5.14), Zli c th thu c trc tip bng cch tnh el.Sut in ng trong mch ni tip l:el = Ep - Eq - vpl (5.15)

V dng in chy qua nhnh b cy thm vo l:ipq= 0

Nhnh p - l c thc l gii nh mt nhnh cy. Dng in trong nhnh ny, ng vicc s hn ca tng dn ban u v in p qua cc nhnh l:

=+== 0.. ,, rsrspqplplpqplpq vyvyiirr

Vi: ypq,pq: L tng dn ring ca nhnh p - qypq,rs: L tng dn tng h ca nhnh p - q vi nhnh r - sipl = ipq = 0

V vy:


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GII TCH MNG

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= rsrsplplpl

pl vyy

vrr

.1

,,

Do : vrspqrspl yy ,,rr

= pqpqplpl yy ,, =

Nn ta c:

=

rsrspqpqpq

pl vyyv

rr

.

1,, (5.16)

Th ln lt phng trnh (5.16), (5.6) v (5.14) vi Ii = 1 vo phng trnh (5.15) tac:

+= )(1

,,

sirirspl

plpl

qipili ZZyy

ZZZrrr i = 1, 2, .....m,i (5.17)l

Phn t Zll c thc tnh bng cch bm vo mt dng in ti nt l vi nt q lim nt qui chiu v tnh in p ti nt th l thuc v nt q. Gi s ta bm dng I =1A vo nt l (Ij = 0 i l), v tt c cc dng in ti cc nt khc bng 0. T phngtrnh 5.13) ta suy ra:

Ek= ZklIl = Zkl k = 1, 2, .....mTng t nh trn ta bm vo cc nt cn li.E1 = Z1l.IlM Ep = Zpl.Il (5.18)M el = Zll.Il = Zll

Tng t ta c in p gia 2 nt p v l l:el = Ep - Eq - vpl

Cho Il = 1 phng trnh (5.18), Zll c th thu c trc tip bng cch tnh el.

Dng in trong nhnh p - l l:ipl = -Il = -1

Dng in ny trong cc s hng ca cc tng dn ban u v cc in p qua ccnhnh l:

=+== 1.. ,, rsrspqplplpqplpq vyvyiirr

Vi: ypq,pq: L tng dn ring ca nhnh p - qypq,rs: L tng dn tng h ca nhnh p - q vi nhnh r - s

Tng t, v:vrspqrspl yy ,,

rr= pqpqplpl yy ,, =

Nn:plpl

rsrsplpl

yvyv

,

, .1 +=rr

(5.19)

Th ln lt phng trnh (5.19), (5.6) v (5.18) vo phng trnh (5.15) vi Il = 1 tac:

pqpq

slrlrspq

qlpllly

ZZyZZZ

,

, )(1 ++=

rrr

(5.20)

Nu nhnh thm vo khng h cm vi cc nhnh khc ca mng ring, th cc phn typq,rs = 0

V:pqpq

pqpq

yZ

,,

1=

T phng trnh (5.17) ta suy ra:


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GII TCH MNG

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Zli = Zpi - Zqi, i = 1, 2, ....m i l V t phng trnh (5.20):

Zll = Zpl - Zql + Zpq,pqHn na, nu s thm vo m khng h cm v p l nt qui chiu th:

Zpi = 0, i = 1, 2, .....m li V: Z

li= -Z

qi, i = 1, 2, .....m li

V tng t:: Zpl = 0V vy: Zll = - Zql + Zpq,pqCc phn t trong hng v ct th l ca ma trn tng trnt vi mng ring thm voc tm thy t cc phng trnh (5.17) v (5.20). Vic cn li ca tnh ton i hima trn tng trnt bao hm nh hng ca nhnh b cy thm vo. iu ny c thhon thnh bng cch bin i cc phn t Zij, trong i, j = 1, 2, .....m, v loi tr hngv ct l tng ng vi nt gi.

Nt gic loi tr bng cch ngn mch ngun sut in ng mch ni tip el. Tphng trnh (5.13) ta c:

lilNutNutNut IZIZE ..

rrr

+= (5.21)V: i, j = 1, 2, ....m (5.22)0.. =+= lllNutljl IZIZe

rr

Gii Il t phng trnh (5.22) v th vo (5.21):

Nut

ll

ljil

NutNut IZ

ZZZE

rrr

r).

.( =

y l phng trnh biu din ca mng ring bao hm nhnh b cy. T suy ra yucu ca ma trn tng trnt l:

ZNt (c bin i) = ZNt (trc lc loi tr) -ll

ljil

Z

ZZrr

.

Vi : Bt k phn t ca ZNt (c bin i) l:

Zij (c bin i) = Zij (trc lc loi tr) -ll

ljil

Z

ZZrr

.


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GII TCH MNG

Trang 76

END

ThmNhnh b cy

Da vo bng s liunhp li tng tr ban u Z

Tnh ZNt

Thmnhnh cy

k = e

Hnh thnh ma trn

S

S

Tnh ZNt

D

a vo b

ng s

li

unhp tng tr ban u Z

Thmnhnh cy

Nt qui chiuk := 1

Vo s liu

BEGIN

LU THNH LP MA TRN TNG TR NT


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GII TCH MNG

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CHNG 6

TRO LU CNG SUT

6.1. GII THIU:Nhim v ca gii tch mng l tnh ton cc thng s ch lm vic, ch yu

l dng v p ti mi nt ca mng in. Vic xc nh cc thng s ch mng inrt c ngha khi thit k, vn hnh v iu khin h thng in.

Mt s ln cc thut ton c xut trong 20 nm trli y. Trong chngny ta gii thiu cc phng php trn cc kha cnh nh: D chng trnh ha, tc gii, chnh xc....

Vic tnh ton dng cng sut phi c tin hnh tng bc v hiu chnh dn.Bn cnh mc ch xc nh trng thi tnh th vic tnh ton dng cng sut cn l mt

phn ca cc chng trnh v ti u v n nh. Trc khi c s xut hin ca my tnhs, vic tnh ton dng cng sut c tin hnh bng thit b phn tch mng. T nm1956, khi xut hin my tnh su tin th phng php tnh dng cng sut ng dngmy tnh sc xut v dn dn c thay th cc thit b phn tch mng. Ngynay cc thit b phn tch mng khng cn c dng na.

6.2. THIT LP CNG THC GII TCH.

Gi s mng truyn ti l mng 3 pha i xng v c biu din bng mng ni

tip dng nh trn hnh 6.1a. Cc phn t ca mng c lin kt vi nhau nn matrn tng dn nt YNt c th xc nh t s.Theo s 6.1a ta c:INt = YNt .VNt (6.1)

1p..0

+Vp-

IpP

Sp

(b)(a)

Hnh 6.1 : Sa cng ca ng dy truyn ti

YNt l mt ma trn tha v i xng. Ti cc cng ca mng c cc ngun cngsut hay in p. Chnh cc ngun ny ti cc cng lm cho p v dng lin h phituyn vi nhau theo (6.1) chng ta c th xc nh c cng sut tc dng v phnkhng bm vo mng (quy c cng sut dng khi c chiu bm vo mng) di

dng hm phi tuyn ca Vp v Ip. Ta c th hnh dung ngun cng sut bm vo mngni ngang qua cng ti u dng ca ngun bm nh hnh 6.1b.


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GII TCH MNG

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Phn loi cc nt:- Nt P -Q l nt m cng sut tc dng P v cng sut phn khng Q l cnh,

nh nt P 6.1 chng hn)()( SPLP

SPGP

SPLP

SPGP

SPp

SPppp QQjPPjQSIV +=+= (6.2)

Vi Vp = ep +jfp

Ch s GP v LP ng vi cng sut ngun pht v cng sut tiu thP. S cho bitcng sut cnh (hay p t).- Nt P -V tng t l nt c cng sut tc dng P cnh v ln in p

c gi khng i bng cch pht cng sut phn khng. Vi nt ny ta c:SPLP

SPGP

SPppp PPPIV ==]Re[

* (6.3)SP

pppp VfeV =+= )(22 (6.4)

- Nt V-q (nt h thng) r rng nt ny in p v gc pha l khng i. Vica ra khi nim nt h thng l cn thit v tn tht I2R trong h thng l khng xcnh trc c nn khng th cnh cng sut tc dng tt c cc nt. Nhn chung

nt h thng c ngun cng sut ln nht. Do ngi ta a ra nt iu khin in pni chung l n c cng sut pht ln nht. nt ny cng sut tc dng PS (s k hiunt h thng) l khng cnh v c tnh ton cui cng. V chng ta cng cn mt

pha lm chun trong h thng, gc pha ca nt h thng c chn lm chun thng mc zero radian. in p phc V cnh cn Ps v Qsc xc nh sau khi gii xongtro lu cng sut cc nt.

6.3. CC PHNG PHP GII QUYT TRO LUCNG SUT:

Theo l thuyt th c hai phng php tn ti l phng php s dng ma trnYNt v phng php s dng ma trn ZNt. V bn cht c hai phng php u sdng cc vng lp. Xt v lch s phng php th phng php YNta ra trc v matrn YNt d tnh v lp trnh, thm ch ngy nay n vn s dng vi h thng khng lnlm, phng php ny gi l phng php Gauss -Seidel. ng thi phng php

Newton cng c a ra phng php ny c u im hn v mt hi t. Sau khi cchloi tr trt t ti u v k thut lp trnh ma trn vevttha lm cho tc tnh tonv s lng lu tr t hn, th phng php Newton trnn rt ph bin. Ngy nay vih thng ln ti 200 nt hay hn na th phng php ny lun c dng. Phng

php dng ma trn ZNt vi cc vng lp Gauss - Seidel cng c tnh hi t nh phng php Newton nhng ma trn ZNt l ma trn y nn cn b nh hn ct gichng, l hn ch chnh ca phng php ny

Trong chng ny chng ta ch gii thiu nguyn l ca cc phng php, cncc phng php c bit nh: S l ma trn tha, sp xp ti u php kh, lc ,..... khng c cp n.


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GII TCH MNG

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6.4. LCH V TIU CHUN HI T.

Php gii tro lu cng sut c coi l chnh xc khi tha mn iu kin t(6.2) n (6.4) m ch yu l phi m bo chnh xc (6.4), hai tiu chun hi t ph

bin l:

- Mc cng sut tnh ton nt no theo Vp v Ipbn tri ng thc(6.2) n (6.4) ph hp tng ng vi gi tr cho sn bn phi. S sai khc ny gi l lch cng sut nt.

- lch in p nt gia 2 vng lp k tip nhau.Sau y ta xt tng tiu chun c th:+ Tiu chun lch cng sut nt:T (6.1) v (6.2) ta c

=

+==n

qqpqp

SPp

SPppp

SPpp VYVjQPIVSS

1

*** (6.5)

Tch phn thc v phn o ca (6.5) ta c lch cng sut tc dng v

lch cng sut phn khng thch hp cho c (6.2) v (6.3). Biu din trong ta vunggc nh sau: Ta s dng k hiu sau:

ppppp VjfeV =+=

qppq

pqpqpq jBGY

=

+=

Vi tng nt P -V hay P - QDng ta vung gc:

]))(()Re[(1

=

+=n

q

qqpqpqppSPPP jfejBGjfePP (6.6a)

Dng ta cc:

+=

=

n

q

qpqpqpqpqpSPpp VBGVPP

1

||)sincos(|| (6.6b)

Vi tng nt P - QDng ta vung gc:

]))(()Im[(1

=

+=n

qqqpqpqpp

SPpp jfejBGjfeQQ (6.7a)

Dng ta cc:

= =

n

qqpqpqpqpqp

SPpp VBGVQQ

1 ||)cossin(|| (6.7b)

Tiu chun hi t chung nht c dng trong thc t l:Pp Cp cho tt c nt P -V v P -QQp Cq cho tt c nt P -Q

Gi tr Cp v Cqc chn t 0,01 - 10 MVA hay MVAR ty theo trng hp.+ Tiu chun lch in p:Gi s bc lp l k, lch in p gia hai vng lp k v k +1 l:

( ) ( )kkp VVV =

+1 cho tt c cc nt P - Q

Tiu chun hi t l:Vp Cv cho tt c cc nt P - Q


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GII TCH MNG

Trang 80

Gi tr Cv t 0,01 n 0,0001

6.5. PHNG PHP GAUSS - SEIDEL S DNG MATRN YNT:

d hiu phng php ny ta gi thit tt c cc nt l nt P-Q tr nt h thngV - q. V in p ca nt h thng hon ton bit nn khng c vng lp no tnh chont ny. Ta chn nt h thng l nt cn bng. Do Vq (q s) coi l p ca nt q sovi nt s (k hiu nt s l nt h thng). Vi tt c cc nt, tr nt th s l nt h thngta rt ra c t (6.1) v (6.2):

=

===n

qqpq

P

PP npVYV

SI

1*

*

...2,1 ; p s (6.8)

Tch Ypq, Vp trong ra ri chuyn v ta c:

npVYV

S

YV

n

pqq

qpqP

P

ppp ...2,1

1

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