curcuit theory

L THUYT MCH IN

Bin son: Cung Thnh LongB mn K thut o v Tin hc cng nghipKhoa inTrng i hc Bch Khoa H Ni

Bi ging

H Ni - 2006

Kt cu chng trnh:A. Hc k 1

Mch in tuyn tnh

B. Hc k 2

+ Mch in phi tuyn

+ L thuyt ng dy di

C. Hc k 3

L thuyt trng in t

Ti liu tham kho

[1]. PGS. Nguyn Bnh Thnh & cc cng s, C s k thutin (quyn 1, 2, 3), Nh xut bn i hc v trung hc chuynnghip (1971)

[2]. Norman Balabanian, Electric Circuits, McGraw-Hill, Inc (1998)

MCH C THNG S TP TRUNG

MCH IN TUYN TNH

Chng 1. Khi nim v m hnh mch in

Chng 2. c im ca mch in tuyn tnh ch xc lp iu ho

Chng 3. Phng php gii mch in tuyn tnh ch xc lp iu ho

Chng 4. Quan h tuyn tnh v cc hm truyn t ca mch in tuyn tnh

Chng 5. Mng mt ca v mng hai ca tuyn tnh

Chng 6. Mch in tuyn tnh vi kch thch chu k khng iu ha

Chng 7. Mch in ba pha

Chng 8. Mch in tuyn tnh ch qu


Chng I

KHI NIM V M HNH MCH IN

I.1. Hin tng in t - M hnh m t h thng in t

I.2. nh ngha v cc yu t hnh hc ca mch in

I.3. Cc phn t c bn ca mch in Kirchhoff

I.4. Hai nh lut Kirchhoff m t mch in

I.5. Graph Kirchhoff

I.6. Phn loi cc bi ton mch


I.1. HIN TNG IN T - M HNH M T H THNG IN T

in t l hin tng t nhin, mt th hin ca vt cht di dngsng in t

M t cc h thng in t: m hnh mch v m hnh trng

u

i

TiNgunE(x,y,z,t)

H(x,y,z,t)


1. M hnh mch+ Ch c thng tin ti mt s hu hn im trong h thng

+ Cc phn t c bn: R, L, C, g

+ Da trn c s 2 nh lut thc nghim ca Kirchhoff

Vi m hnh mch, chng ta tp trung mi hin tng in t lin tc trongkhng gian vo mt phn t c th, do khng thy c hin tng truynsng trong h thng!

M hnh mch l m hnh gn ng ca qu trnh in t, b qua yu t khnggian



2. iu kin mch ho

Bc sng ca sng in t rt ln hn kch thc thit b in

dn in ca dy dn rt ln hn dn in ca mitrng ngoi


I.2. NH NGHA V CC YU T HNH HC CA MCH IN

1. nh nghaMch in:

+ mt tp hu hn cc phn t c bn l tng ghp vi nhau mtcch thch hp sao cho m t c truyn t nng lng in t

+ bin c trng: dng in v in p (trn cc phn t ca mch)


I.2. NH NGHA V CC YU T HNH HC CA MCH IN

2. Cc yu t hnh hc ca mch in

R1R3 R2

L1

L3

L2

C3e1

je2

i1 i2

i3

Cc phn t mch

Nhnh

Nt (nh)

Vng


I.3. CC PHN T C BN CA MCH IN KIRCHHOFF

Phn t c bn

+ i din cho mt hin tng in t trn vng xt

+ c biu din bng phn t mt ca

+ c 1 cp bin bin c trng dng in v in p trn ca

+ ni ti cc phn khc ca mch in qua ca.



1. in tr R, in dn g

R

i

u

+ in tr c trng cho qu trnh tiu tn trn vng xt

+ Quan h dng p:

+ n v: Ohm () v cc dn xut: k, M,Nu quan h u(i) l phi tuyn: in tr phi tuyn

Nu quan h u(i) l tuyn tnh: in tr tuyn tnh

u = Ri

+ Nghch o ca in tr R l in dn g. n v indn l Siemen (S)

( )ru u ir r=

i(A)

u(V)



2. in dung C

C

qi

u

u

qq(u)

+ in dung C c trng cho hin tng tch phng nnglng in trng trong vng xt

+ Quan h dng p:

+ tn s thp, in tch q ph thuc in p t vovng xt. a s quan h q(u) l tuyn tnh

+ Khi q(u) tuyn tnh: in dung C tuyn tnh

+ n v in dung: Farad (F) v cc dn xut ca F

, dqq q u iC dt = =

1u idtC

= ,q Cu= ,dui C dt=



3. in cm L

L

i

u

i

+ c trng cho hin tng tch phng nng lng ttrng trong vng xt

+ Quan h dng p: Khi (i) phi tuyn: in cm L l phi tuyn

Khi (i) tuyn tnh: in cm L l tuyn tnh

+ n v ca in cm: Henry (H) v cc dn xut

( ),i = du dt=

,Li = didt

u L=



4. H cm M

u2

u1i2i1 21

12

22

11

, ,1 1 1 2i i = ,2 2 1 2i i =

' ' ' '2 2 22 1 2 21 1 2 2

1 2

du i i M i L idt i i = = + = +

' ' ' '1 1 11 1 2 1 1 12 2

1 2

du i i L i M idt i i = = + = +

M12 = M21 = M gi l h s h cm gia 2 cun dy xc nh du ca in p h cm phi bit v tr khng gian ca cc cun dy

Khi nim cc tnh ca cc cun dy



4. H cm MNguyn tc: Khi chiu dng ging nhau vi mi cc tnh ca cc cun dyc lin h h cm th trong mi cun dy chiu t thng t cm v h cmtrng nhau.

*i1 L1

u1*

u2

L2 i2M

' '1 1 1 2u L i Mi= ' '2 2 2 1u L i Mi= +

Du ca in p t cm v h cm ph thuc vo chiu dngin p quy c tnh cho nhnh cha phn t h cm



5. Ngun p, ngun dng

e ik

j uk

5.1. Ngun p

-Ngun p c lp

-Ngun p ph thuc

5.2. Ngun dng

-Ngun dng c lp

-Ngun dng ph thuc

Thc t vnhnh khngc phpngn mch

ngun p, hmch ngun

dng!



6. M hnh phn t thc

+ tp hu hn cc phn t l tng ghp vi nhau 1 cch thch hp

R L

+ c nhiu m hnh tip cn mt phn t thc

+ sai s m hnh ho phn t thc:

MH TT = +



7. Vn trit tiu ngun trong mch

Ch trit tiu ngun trn s , phc v vic phn tch mch!

+ Ngun c lp:

- ngn mch ngun p

- h mch ngun dng

+ Ngun ph thuc:

- trit tiu nguyn nhn gy ra ngun ph thuc


I.4. HAI NH LUT KIRCHHOFF M T MCH IN

1. Lut Kirchhoff 1

i1

i2i3C

L

R

+ Pht biu:1

0n

kki

==

+ ngha: th hin tnh lin tc ca dngin qua mt mt kn (trng hp ring lqua mt nh ca mch)

2. Lut Kirchhoff 2

+ Pht biu: R1R4

L3C

L2e1 e2

10

m

kku

==

+ ngha: th hin tnh cht th ca qutrnh nng lng in t trong mt vngkn


I.4. HAI NH LUT KIRCHHOFF M T MCH IN

3. S phng trnh Kirchhoff c lp m t mch

R1R3 R2

L1

L3

L2

C3e1

je2

i1 i2

i3

Vi mch c n nhnh, d nh th:

-S phng trnh Kirchhoff 1 c lpl d -1 phng trnh

-S phng trnh Kirchhoff 2 c lpl n d + 1 phng trnh

Phn tch mch da trn h cc phng trnh Kirchhoff mt mch!


I.5. GRAPH KIRCHHOFF

**

e1 e5

R1

R2

R4 R5

R3

L2

L4

C3

j

i1 i2i3 i4 i5

12

3

4 5

+ nh ngha

+ Cy (ca Graph)

+ Cnh (s phng trnh K1 clp)

+ B cnh (s phng trnh K2 c lp)

+ Vit phng trnh K1 t Graph Kirchhoff

+ Vit phng trnh K2 t Graph Kirchhoff


I.6. PHN LOI CC BI TON MCH

+ Bi ton phn tch

+ Bi ton tng hp



Chng IIC IM CA MCH IN TUYN TNH CH

XC LP IU HO

II.1. Khi nim chung

II.2. Hm iu ho v cc i lng c trng

II.3. Phn ng ca nhnh R, L, C, R-L-C vi kch thch iu ho

II.4. Quan h dng, p dng phc trn cc nhnh c bn R, L, C, R-L-C

II.5. Hai nh lut Kirchhoff dng phc

II.6. Cng sut

C IM CA MCH IN TUYN TNH CH XC LP IU HO

II.1. KHI NIM CHUNG

+ Mch in tuyn tnh

+ Ch qu

+ Ch xc lp

+ Tn hiu dao ng iu ho

+ Mch in tuyn tnh ch xc lp iu ho

+ Tnh cht xp chng mch in tuyn tnh

II.2. HM IU HO V CC I LNG C TRNG

0 5 10 15 20 25 30 35 40 45-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Im

Ti

Xt dng iu ho i(t) = Imsin(t + i)

- Bin dao ng cc i Im- Tn s gc

- Gc pha ban u i

-Gi tr hiu dng:

12 ,f fT

= =

2

0

12

TmII i dt

T= =


II.3. C IM CA MCH IN TUYN TNH CH XC LP IU HO

1. ch xc lp iu ho, trong mch tuyn tnh dng v p bin thin iuho cng tn s

1.1. Vi in tr

1.2. Vi in cm

( ) ( )sinm ii t I t = +

( ) ( ) ( )( ) 2 sin 2 sinR i uu t Ri t RI t U t = = + = +

( ) ( )2 cos 2 sin2L i i

diu t L LI t LI tdt

= = + = + + ( ) ( )2 sinL L uu t U t = +



1.3. Vi t in

( ) ( )1 1 2 sinC iu t idt I tC C = = + ( ) ( )1 12 cos 2 sin

2C i iu t I t I t

C C

= + = + ( ) ( )2 sinC uu t U t = +1.4. Vi mch RLC ni tip

( ) ( )1 2 sin udiu t Ri L idt U tdt C = + + = +



2. ch xc lp iu ho, cc i lng dng v p ch c trng bi haithng s l tr hiu dng v gc pha u. Do , c th biu din bng s phchoc vector.

i

j

a

b

Aja jb Ae + = cos ; sina A b A

2.1. S phc

= =

-Cc i lng vt l (dng, p, sc in ng, ngun dng): dng ch in hoa c du chm phatrn

- Cc gi tr tng tr, tng dn, dng ch in hoa

2.2. Biu din phc cc i lng in



2.2. Biu din phc cc i lng in

V d:

( ) ( )0 05 30 5 2 sin 30I i t t= = +( ) ( )45 050 50 2 sin 45jU e u t t= =

Chuyn h phng trnh vi phn thnh h phngtrnh i s tuyn tnh!



2.3. Quan h dng, p dng phc trn cc phn t R, L, C, RLC

Ri

uRRI

U

( )2 sinR iu Ri RI t = = +2.3.1. Phn t R

Biu din: iI I=Ta c: R i uU RI U RI = = =




2.3.2. Phn t L

L

i

uL

Biu din iI I=Ta c 2 sin

2L idiu L LI tdt

= = + + Do :

2ij

L iU LI j LIe = + =

;L L L LU Z I Z j L jX= = = Vi ZL l tng tr phc ca in cm L, XL l cm khng

LU

I

LZ




2.3.3. Phn t C

Biu diniI I=

Ta c1 1 2 sin

2C iu idt I t

C C

= = + Do :

21 1i ij jCU Ie j IeC C

= = 1

C C CU j I Z I jX IC= = = ZC l tng tr phc ca in dung C, XC l dung khng

CU CZ

IiCuC




2.3.4. Phn t RLC

R L C

iu

R ZCZL

IU

1diu Ri L idtdt C

= + + ( )L C L CU RI jX I jX I R j X X I ZI= + = + =

Z l tng tr ca mch, X = XL XC l in khng

u

i

jj

jU UeZ Z eI Ie

= = =

2 2 ; XZ R X artgR

= + =


2.3. C IM CA MCH IN TUYN TNH CH XC LP IU HO


2.3.4. Phn t RLC

RU

LU

CU

XU Ui

j

cosR Z = sinX Z =cosRU U =sinXU U =

Tam gic tng tr

Tam gic in p

Ch cc miquan h nyv tam giccng sut phn sau!


II.4. LUT KIRCHHOFF DNG PHC

ch xc lp iu ho:

10

n

kk

I=

=

10

n

kkU

==


II.5. CNG SUT

1. Cng sut tc thi: p =uiV d nhnh gm 3 phn t RLC ni tip

2 sin . 2 sin12 sin . 2 cos 2 sin . 2 cos

R L Cp p p p I t IR t

I t LI t I t I tC

= + + =+

( ) ( )2 21 cos 2 sin 2L Cp RI t I X X t = + 2. Cng sut tc dng

2

0 0 0

1 1 1T T Tr XP pdt p dt p dt RIT T T

= = + = 2 . .cos cosP RI ZI I UI = = =

n v: Wat(W)

v dn xut


II.5. CNG SUT

3. Cng sut phn khng2 sin . . sinQ XI Z I I UI = = = VAr

4. Cng sut biu kin

S UI= VA2 2S P Q= +

4. Cng sut phc

S UI P jQ= = +

S

P

Q



Chng III

CC PHNG PHP PHN TCH MCH TUYN TNH CH XC LP IU HO

III.1. Khi nim chung

III.2. Phng php dng in nhnh

III.3. Phng php dng in vng

III.4. Phng php in th nh

III.5. Ba phng php c bn dng ma trn

III.1. KHI NIM CHUNG

- Da trn hai nh lut Kirchhoff

- Nguyn tc: i bin v bin i s mch

- Ba phng php c bn: dng nhnh, dng vng, th nh

Gii mchtrong minnh phc!

CC PHNG PHP PHN TCH MCH IN TUYN TNH CH XC LP IU HA

III.2. PHNG PHP DNG IN NHNH

1. Nguyn tc:

-Chn n l dng in cc nhnh

- Lp v gii h phng trnh i s trong minphc m t mch theo 2 nh lut Kirchhoff

2. Lu :

- V h cm (K2)

- V ngun dng (2 cch vit)



3. V d

**

e1 e5

R1

R2

R4 R5

R3

L2

L4

C3

j

i1 i2i3 i4 i5

*

*Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1EMZ



3. V d

1 2 3 0I I I J + = 3 4 5 0I I I J + = 1 1 2 2 4 1MZ I Z I Z I E+ =

2 2 3 3 4 4 4 2 0M MZ I Z I Z I Z I Z I + + + = 2 4 4 5 5 5MZ I Z I Z I E + + =


III.3. PHNG PHP DNG IN VNG

1. Nguyn tc:

- Chn n l dng in khp kn cc vng c lp camch

- Vit phng trnh theo lut Kirchhoff 2 cho cc dngvng

2. Lu :

- V ngun dng

- V dng in nhnh

- V h cm



3. V d

*

*Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1EMZ

- Xt mch nh hnh v trn

- Chiu vng chn nh cc mi tn m t trong hnh



3. V d

( ) ( )1 2 1 2 2 3 1v M v M vZ Z I Z Z I Z I E+ + = ( ) ( ) ( )2 1 2 3 4 2 4 3 32M v M v M vZ Z I Z Z Z Z I Z Z I JZ + + + + + + + =

( ) ( )1 4 2 4 5 3 5M v M v vZ I Z Z I Z Z I E + + + + =


III.4. PHNG PHP IN TH NH

1. Nguyn tc:

+ Chn n l th cc nh c lp. Vit (h) phng trnh K1 theoth cc nh chn

+ Gii (h) phng trnh thu c nghim l th cc nh c lp

+ Tnh dng in trong cc nhnh theo lut m tng qut

A BI EZ

ABU

Xt lut m:

ABZI E U = AB A BU = ( )A B A BEI Y EZ + = = +



2. Lu :

+ Khng tin s dng phng php in th nh chomch c h cm (khi gii tay)

3. V d

Xt mch in:

*

*Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1EMZ

ZM = 0!



3. V d

Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1E

A B

C

Chn 2 nh c lp: ,A B 0C = - Gc

1 1 1ACZ I U E+ = ( )11 1 1

1

AA

EI Y EZ = =

2 22

AAI YZ

= = 4 44

BBI YZ

= =

( )3 3 A BI Y = ( )5 5 5 BI Y E =



3. V d

Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1E

A B

C

Phng trnh K1 cho 2 nh A v B:

1 2 3

3 4 5

0

0

I I I JI I I J

+ = + =

T c h phng trnh th nh:

( )( )

1 2 3 3 1 1

3 3 4 5 5 5

A B

A B

Y Y Y Y Y E J

Y Y Y Y Y E J

+ + = + + + + =



4. Tng qut

Mch c 2 nh c lp:

AA A AB B kA kA lAk l

AB A BB B nB nB mBn m

Y Y Y E J

Y Y Y E J

= + + = +

Mch c 3 nh c lp:

AA A AB B AC C kA kA lAk l

AB A BB B CB C mB mB nBm n

AC A CB B CC C hC hC gCh g

Y Y Y Y E J

Y Y Y Y E J

Y Y Y Y E J

= + + = + + = +


III.5. BA PHNG PHP C BN DNG MA TRN

Xt mch in nh hnh v:

*

*Z1 Z2 Z4

Z5

Z3

J

1I 2I 3I

4I 5I

5E1EMZ

A B

C

N A B C

1 1 0 -12 -1 0 13 -1 1 04 0 -1 15 0 1 -1

VN V1 V2 V3

1 1 0 02 1 -1 03 0 1 04 0 1 15 0 0 1

+ Bng s nhnh nh v ma trn nhnh nh A

+ Bng s nhnh vng v ma trn nhnh vng C CZ

A



1. Vi ma trn nhnh nh A

+ Vector dng in nhnh nI+ Vector in p nhnh nU+ Vector sc in ng nhnh nE+ Vector th nh d+ Vector ngun dng nh dJ

( )( )

1

0 2n d

T n d

U A

A I J

= + =

1

2

...n

n

II

I

I

=

1

2

...n

n

UU

U

U

=

1

2

...n

n

EE

E

E

=

1

dd

...d

d

JJ

J

=

1

dd

...d

d

=

(1) Lut Ohm cho cc nhnh

(2) Lut Kirchhoff 1



1. Vi ma trn nhnh nh A

Xt v d minh ha cho u bi, ta c:

1 01 01 1

0 10 1

A

=

1

5

000

n

E

E

E

=

1

2

3

4

5

n

II

I III

=

Ad

B

= d

JJ

J

=

1

2

3

4

5

n

UU

U UUU

=

1 2 3

3 4 5

00

0T n dI I I J

A I JI I I J

+ + = = +

1

2

3

4

5

A

A

d n A B

B

B

UU

A U UUU

= =



2. Vi ma trn nhnh vng C

Cng vi cc vector lp vi ma trn nhnh nh A, ta lp thm:

+ Vector Jn:- Nhnh c ngun dng khp qua, cng chiu dng introng nhnh, ghi J; ngc chiu dng ghi - J

- Nhnh khng c ngun dng khp qua ghi 0

- Dng ma trn ct

+ Vector vI - Dng ma trn ct- Mi phn t l mt dng vng c lp chn+ Ta c: ( )

( )3

0 4n V n

T n

I CI J

C U

= +=

(3) chuyn i dng nhnh dng vng

(4) phng trnh Kirchhoff 2



2. Vi ma trn nhnh vng C

V d, trong mch in xt u bi

1 0 01 1 00 1 00 1 10 0 1

C

=

1

2

3

v

v v

v

II I

I

=

00

00

nJ J

=



3. Ma trn tng tr nhnh Z

+ Nguyn tc lp: Zkk tng tr trn cc nhnh

Zij tng tr h cm gia hai nhnh i v j

Chiu dng nhnh vo ccphn t h cm ngc nhau so vi cc cc cng tnh th zijmang du m!

+ Ta c:

( )5n n nU ZI E= (5) lut m cho nhnh c ngun

+ V d:

Z1 0 0 0 0

0 Z2 0 -ZM 0

0 0 Z3 0 0

0 -ZM 0 Z4 0

0 0 0 0 Z5



4. H phng trnh dng nhnh dng ma trn

+ Phng trnh K1: kt qu (2)

+ Phng trnh K2: thay (5) vo (4) ta c 0T n n T n T nC ZI E C ZI C E = =

+ V ta c h phng trnh dng nhnh:

0T n d

T n T n

A I JC ZI C E

+ = = (6)



5. H phng trnh dng vng dng ma trn

Thay (3) vo (6) ta c: ( )T v n T nC Z CI J C E+ = cng chnh l h phng trnh dng vng dng ma trn

T v T n T nC ZCI C E C ZJ=



6. H phng trnh th nh dng ma trn

T (5): ( )1 1 1n n n n n n n n nU ZI E Z U I Z E I Z U E = = = +

Theo (1) n dU A= Thay vo (2) ta c: 1 1 0T n T n dA Z U A Z E J

+ + = (*)

Thay (1) vo (*) ta c phng trnh th nh:

1 1T d T n dA Z A A Z E J = +




7. Lu

+ S dng Matlab gii mch in (chun b lm th nghim)

+ Cch vit dng ma trn cho php gii mch c h cm ddng theo c 3 phng php


Chng IVQUAN H TUYN TNH

V CC HM TUYN T CA MCH IN TUYN TNH

IV.1. Khi nim

IV.2. Phng php xc nh h s truyn t trong QHTT

IV.3. Mt s hm truyn t thng gp

IV.4. Truyn t tng h v truyn t khng tng h

IV.5. Bin i tng ng s mch in

QUAN H TUYN TNH V CC HM TRUYN T CA MCH IN TUYN TNH

IV.1. KHI NIM

Quan h tuyn tnh- Trong mch tuyn tnh, cc i lng dng, p nu coi mt nhm lkch thch, mt nhm l p ng th chng quan h tuyn tnh vinhau.

V d: ( )1 3 13 3 10I E Y E I= + 1I 2I

1U 2U1 11 1 12 2 10

2 21 1 22 2 20

U Z I Z I UU Z I Z I U

= + += + +

- H s trong quan h tuyn tnh: h s truyn t hay hm truyn t

- H s truyn t ph thuc kt cu mch, tn s ngun. Chng cth nguyn Ohm, Siemen hoc khng th nguyn


IV.2. PHNG PHP XC NH H S TRUYN T TRONG QHTT

1. Xc nh hm truyn t trong quan h tuyn tnh (QHTT)

Nguyn tc: da vo 2 nh lut K1, K2

Phng php:

+ Phng php th nht: Vit phng trnh phc chomch ri gii tm cc h s QHTT (cc hm truyn t HT)

+ Phng php th hai: Xt cc ch c bit trongmch tm HT (thng l cc ch cho php xtQHTT on gin hn)


2. V d xc nh hm truyn t trong quan h tuyn tnh (QHTT)

Tm quan h: ( ) ( )1 1 2 3 1, ,I E E I E Dng tng qut: 1 11 1 12 2 10I Y E Y E I= + + Cch 1: Gii trc tip mch xc nh cc h s

+ Vit phng trnh th nh:

1E 2E

1Z 2Z

3Z

1I

3IA

1 1 2 2

1 2 3A

Y E Y EY Y Y

+= + + ( ) ( )1 2 3 1 21 1 1 1 2

1 2 3 1 2 3A

Y Y Y YYI Y E E EY Y Y Y Y Y

+= = + + + +

Do : ( )1 2 3 1 2

11 12 101 2 3 1 2 3

, , 0Y Y Y YYY Y IY Y Y Y Y Y

+= = =+ + + +




1E 2E

1Z 2Z

3Z

1I

3IA Cch 2: Xt cc ch c bit

+ Cho trit tiu 2 ngun p, t mch suy ra:

10 0I =+ Cho: 1 20, 0E E=

Khi , t phng trnh, ta c: 11 12 2 122

II Y E YE

= = T mch: 2 2

1 2 3A

Y EY Y Y

= + + V: 1 21 1 2

1 2 3A

YYI Y EY Y Y

= = + +

Do : 1 2121 2 3

YYYY Y Y= + +




1E 2E

1Z 2Z

3Z

1I

3IA Cch 2: Xt cc ch c bit

+ Cho: 1 20, 0E E = T phng trnh: 11 11 1 11

1

II Y E YE

= = T mch:

( ) ( )1 2 31 1 1 1 1 11 2 3 1 2 3

,A AY Y YY E I Y E E

Y Y Y Y Y Y += = =+ + + +

Do : ( )1 2 31

111 1 2 3

Y Y YIYE Y Y Y

+= = + +

Nguyn l xp chng mch in tuyn tnh



IV.3. MT S HM TRUYN T THNG GP

1. Tng dn vo ca mt nhnh

1E

1Z 2Z

3Z

1I

3IA

( )1 2 3111

1 1 2 3

Y Y YIYE Y Y Y

+= = + +

Y11 gi l tng dn vo ca nhnh 1

Tng qut:

nnn

n

IYE

= Vi iu kin cc ngun khc ca mch trit tiu



2. Tng tr vo ca mt nhnh

1E

1Z 2Z

3Z

1I

3IA ( )1 1 2 311 11 1 2 3

Y Y YZ YY Y Y

+ += = +Tng qut:

0,

0,

n nnn k

n nl

E EZ E k nI I

J l

= = = =



1E

1Z 2Z

3Z

1I

3IA

3. Tng dn tng h

331

1 0, 0IYE E J

= = =

ngha: kh nng gy dng trn nhnh 3 cangun trn nhnh 1

Tng qut:0, 0

llk

k

IYE E J

= = =



4. Tng tr tng h

1E

1Z 2Z

3Z

1I

3IA

Ni chung: 1lk lkZ Y

1 113

3 3 0U UZI I E

= = =

Tng qut: ( ), 0; ,llk l kk

UZ J E l m kJ

= = = Zlk bng p truyn n cp ca l bi ngun dng Jk=1A ti ca k


IV.4. TRUYN T TNG H V TRUYN T

KHNG TNG H

1E2I

1I 2E

Gi thit phn mch gia hai nhnh 1 v 2 khng cha ngun, ta c:

2 21 1I Y E= v 1 12 2I Y E= Nu 1 2E E= v 1 2I I= th 12 21Y Y=Khi , ni mch truyn t tng h, ngcli c mch truyn t khng tng h

ngha: Nu thun tin, c th o ngun t nhnh ny sang nhnh khc tnh dng khi gia 2 nhnh c quan h truyn t tng h.


IV.4. TRUYN T TNG H V TRUYN T

KHNG TNG H

Mch cha cc phn t R, L, C, M tuyn tnh v cc ngunc lp th c tnh truyn t tng h.


IV.5. BIN I TNG NG S MCH IN

1. Mc ch: gip vic tnh ton phn tch mch in n gin hn

2. Nguyn tc: dng in v in p trn ca ca phn mch trcv sau bin i phi gi nguyn gi tr



3. Mt s php bin i c bn

3.1. Bin i nhnh cc phn t mc ni tip

1Z 2Z nZ

UI

IU

Z

( )1 2 ... nU Z Z Z I= + + +

U ZI= 1n

kk

Z Z=

= EIU

UI 1E 2E nE

1

n

kk

E E=

=




3.2. Bin i cc nhnh khng ngun mc song song

1Z 2Z nZ

I

U

Z

I

U

1 2 n

U U UIZ Z Z

= + + +

UIZ

= 11 21 1 1 1 n

kkn

Y YZ Z Z Z == + + + =

Tng t cho cc ngun dng cng u vo hai nh xc nh no :

1

n

kk

J J=

=




V d:Cho mch in nh hnh v, tnh dng in trong cc nhnh ca mch?

U1Z 1I

2I 3I3Z2Z

1Z

23Z

1I

U

Bin i s mch in, ta c:

2 323

23 2 3 2 3

.1 1 1 Z ZZZ Z Z Z Z

= + = +

11 23

UIZ Z

= + 3 1

22 3

Z IIZ Z

= +

2 13

2 3

Z IIZ Z

= +




3.3. Bin i sao tam gic

13Z

23Z

12Z

1Z

2Z 3Z

:Y12 13

112 13 23

.Z ZZZ Z Z

= + +12 23

212 13 23

.Z ZZZ Z Z

= + +13 23

312 13 23

.Z ZZZ Z Z

= + +

:Y 1 2

12 1 23

Z ZZ Z ZZ

= + + 1 313 1 32

Z ZZ Z ZZ

= + +

2 323 2 3

1

Z ZZ Z ZZ

= + +




V d:

1E 5E

1Z2Z

3Z4Z

5Z

6Z

1I2I

3I

4I5I

6I

AB

C

Chuyn Z2, Z4, Z6 ni tam gic thnh nisao, mch s d phn tch hn.

2 6

2 4 6A

Z ZZZ Z Z

= + +2 4

2 4 6B

Z ZZZ Z Z

= + +4 6

2 4 6C

Z ZZZ Z Z

= + +

Mch s c dng:




V d:

Trong mch mi tm cc dng: , 1,3,5kI k =T tm nt cc dng: 2 4 6, ,I I I/

1 3AB AO OB A BU U U Z I Z I= + = +

22

ABUIZ

= 6 2 1I I I= 4 3 2I I I= 1E

5E1Z

3Z

5ZBZ

CZAZ

1I

3I5I

A

B

C




3.4. Bin i tng ng cc nhnh song song cha ngun

J2Z

1Z

1EU

I

1I2I

UI Z

E

Ta c, trong s trc bin i:

11 2

1 2

E U UI I I J I JZ Z= + + = +

1

1 2 1

1 1 EI U JZ Z Z

= + + +

Trong s sau bin i:U EIZ Z

= +

Ta c: 1 2

1 1 1Z Z Z= + 1 1

1 1 2

Y E JEE Z JZ Y Y

+= + = +


Chng V.MNG MT CA V MNG HAI CA TUYN TNH

V.1. Khi nim v mng mt ca Kirchhoff

V.2. Phng trnh c trng ca mng mt ca

V.3. nh l Thevenin v Norton

V.4. iu kin a cng sut cc i ra khi mng 1 ca

V.5. Khi nim v mng hai ca Kirchhoff

V.6. Cc dng phng trnh mng hai ca

V.7. Ghp ni cc mng hai ca

V.8. Mng hai ca hnh T v

V.9. Cc hm truyn t p v hm truyn t dng

V.10. Phn tch mch c cha phn t phc hp

MNG MT CA V MNG HAI CA TUYN TNH

V.1. KHI NIM V MNG MT CA KIRCHHOFF

1. nh ngha

+ Mng mt ca l mt phn ca mch in tn cng bng mt ca

+ Bin trng thi ca mng mt ca: cp bin dng v p trn ca

2. Phn loi

- Mng mt ca khng ngun

- Mng mt ca c ngun


V.2. PHNG TRNH C TRNG CA MNG MT CA

IU

+ Phng trnh c trng:0U ZI U= +

Z v U0 xc nh t 2 ch sau:

+ H mch ca, tm U0+ Ngn mch ca, tm Z 0 hU U=

0 h

ng ng

U UZI I

= = + Mng 1 ca khng ngun: thay tngng bng mt tng tr duy nht. (Xemli php bin i tng ng mch in)

+ Mng mt ca c ngun: thay tngng theo nh l Thevenin hoc Norton

Z

0U

IU


V.3. NH L THEVENIN V NH L NORTON

Z

0U

IU

1. nh l Thevenin

C th thay tng ng mng mt ca phctp, c ngun bng s my pht in ngin c tng tr trong bng tng tr vo camng 1 ca khi trit tiu cc ngun v sc inng bng in p trn ca ca mng khi hmch ngoi.

2. nh l Norton

C th thay tng ng mng mt ca cngun bng s my pht in ghp bi mtngun dng (bng dng ngn mch mng mtca) ni song song vi tng dn bng tng dnvo ca mng khi trit tiu cc ngun.

U

I

Y J


V.3. NH L THEVENIN V NH L NORTON

Z

0U

IU

U

I

Y J( )2I UY J=

( )0 1UUIZ Z

= : T (1) v (2), ta c ththy tnh tng ng ca hainh l v cch chuyn ithng s gia hai s Thevenin v Norton!


V.4. IU KIN A CNG SUT CC I RA KHI MNG MT CA

Nguyn tc: dng nh l Thevenin chuyn mng mt ca v s my pht tng ng n gin ni vi ti.

I

tZngZ

E

Cng sut:

( ) ( )2

2 22 2

tt t t

ng t ng t

rEP r I r EZ r r x x

= = = + + + P ln nht khi: ng tx x=

V: ( )2 0tt ng trd

dr r r

= + Tm c iu kin cng sut pht ln ti ln nht nh sau:

t ngZ Z=


V d

5I

5Z

5EtdEtdZ

1Z2Z

3Z

4Z5Z

1E 5E

5IB

C

A

Cn tnh dng trong nhnh 5 ca mch in. p dng nh l Thevenina mch v dng hnh v pha bn phi.

( )4 3 1 2tdZ Z ss Z nt Z ssZ= 5 0

td BCE U I= =

( )2 3 4ACZ Z ss Z ntZ= 11

AC ACAC

EU ZZ Z

= +

43 4

ACBC td

UU E ZZ Z

= = +

tdE c th tnh nh sau:


V.5. KHI NIM V MNG HAI CA KIRCHHOFF

1. nh ngha

- L mt phn ca mch in tn cng bng hai ca Kirchhoff

- Bin trn ca: 1 1 2 2, , ,U I U I

1I 2I

1U 2U

2. Phn loi

- Mng hai ca c ngun v mng hai ca khng ngun

- Mng hai ca tuyn tnh v phi tuyn


V.6. CC DNG PHNG TRNH MNG HAI CA

1I 2I

1U 2UTrn c s quan h tuyn tnh, c th k ra6 dng phng trnh c bn

1. B s AKhng ngun: 10 10 0U I= = Dng ma trn:

101 211 12

21 221 2 10

UU UA AA AI I I

= +

1 11 2 12 2 10

1 21 2 22 2 10

U A U A I UI A U A I I

= + + = + +

Mng 2 ca tuyn tnh, tngh th det 1A =Xc nh b s c trng A qua cc ch : ngn mch ca 1, h mchca 2



2. B s B1I 2I

1U 2U2 11 1 12 1 202 21 1 22 1 20

U B U B I UI B U B I I

= + + = + +

202 111 12

21 222 1 20

UU UB BB BI I I

= +

Dng ma trn:1B A=

Xc nh b s B theo hai ch : ngn mch ca 1 v h mch ca 1



3. B s Z1I 2I

1U 2U1 11 1 12 2 102 21 1 22 2 20


= + + = + +

Dng ma trn:

101 111 12

21 222 2 20

UU IZ ZZ ZU I U

= +

Mng khng ngun:

10 20 0U U= =

Mng hai ca tuyn tnh, tng h th: 12 21Z Z= Xc nh b s Z qua vic xt hai ch : h mch ca 1 v h mch ca 2



4. B s Y

1I 2I

1U 2U1 11 1 12 2 10

2 21 1 22 2 20

I Y U Y U II Y U Y U I

= + + = + +

Dng ma trn:

101 111 12

21 222 2 20

II UY YY YI U I

= +

1Y Z =

Xc nh b s Y qua vic xt hai ch : ngn mch ca 1 v ngnmch ca 2



5. B s H

1I 2I

1U 2U1 11 1 12 2 10

2 21 1 22 2 20

U H I H U UI H I H U I

= + + = + +

Dng ma trn:

101 111 12

21 222 2 20

UU IH HH HI U I

= +

Xc nh b s H qua hai ch : h mch ca 1 v ngn mch ca 2



6. B s G

1I 2I

1U 2U 1 11 1 12 2 102 21 1 22 2 20

I G U G I IU G U G I U

= + + = + +

Dng ma trn

101 111 12

21 222 2 20

II UG GG GU I U

= +

Xc nh b s G qua hai ch : ngn mch ca 1 v h mch ca 2

1G H =



7. Nhn xt

- Mi mng 2 ca c mt b s c nh, ph thuc vo kt cu v thng sca mch.

- Mng 2 ca c ngun, khng tng h c 6 h s c lp

- Mng 2 ca khng ngun, khng tng h c 4 h s c lp

- Mng 2 ca khng ngun, tng h c 3 h s c lp ( Z12 = Z21, detA = 1)

- Mng 2 ca i xng c 2 h s c lp



8. V dXc nh b s Z ca mng 2 cahnh bn?

Phng trnh tng qut:

1I 2I

1U 2U1Z

2Z

E

1 11 1 12 2 10

2 21 1 22 2 20


= + + = + +

Xc nh qua 3 ch :

+ H mch 2 ca 1 2 0I I= = 10 1

20 2

0U UU U E

Ta c:

= = = =



8. V d

1I 2I

1U 2U1Z

2Z

E

1 0I+ H mch ca 1: =- T phng trnh, ta c:

112

1 12 2 2

2 22 2 222

2

UZU Z I IU Z I E U EZ

I

= = = + =

- T mch, tm 1 2,U I ( )1 222 1 1 2

1 2 1 2

,Z U EU EI U Z I

Z Z Z Z= = =+ +

Do : 1 212 1 22 1 22 2

,U U EZ Z Z Z ZI I

= = = = +



8. V d

1I 2I

1U 2U1Z

2Z

E

+ H mch ca 2: 2 0I =- T phng trnh, ta c:

( )1

1111 11 1

22 21 121

1

UZ IU Z IU EU Z I E Z I

= = = + =

- T mch, tm 2 1,U I

12 1 1 1

1

, UU Z I E IZ

= + =

Do : 1 2 1 111 1 21 11 1 1

,U U E Z I E EZ Z Z ZI I I

+ = = = = =



8. V d

1I 2I

1U 2U1Z

2Z

E

B s Z ca mng 2 ca hnh bntm c nh sau:

1 11 1

1 1 22 2

0U IZ ZZ Z Z EU I

= + +

Lu :T dng phng trnh ny c th suy ra dngphng trnh khc ca mng 2 ca


V.7. GHP NI CC MNG HAI CA

1. Ni xu chui

1I 2I1U 2U1A 2A

1I 2I

1U 2UA

1 2.A A A=



2. Ni ni tip

1I 2I

1I 2I1U

2U

2Z

1Z1I 2I

1U 2UZ

1 2Z Z Z= +



3. Ni song song

1I 2I1U 2U

1I 2I

1U 2U

1Y

2Y

1I 2I

1U 2UY

1 2Y Y Y= +


V.8. MNG HAI CA HNH T V

1U 2UnZ

1dZ 2dZ1I 2I Tnh b s A ca mng 2 ca hnh T. Phng trnh b s A:

1 11 2 12 2

1 21 2 22 2

U A U A II A U A I

= + = +

H mch ca 2:

T mch, ta c: 111 2 11 1

, nnd n d n

U ZUI U I ZZ Z Z Z

= = =+ +

Do : 11112

1 dn

ZUAU Z

= = + 1212

1

n

IAU Z

= =



1U 2UnZ

1dZ 2dZ1I 2I Ngn mch ca 2:T mch, ta c:

( )2 111

2 1 2 1 21

2

n d

n d d n d n d dd

n d

Z Z UUI Z Z Z Z Z Z Z ZZZ Z

+= = + ++ +

12

1 2 1 2

n

d n d n d d

Z UIZ Z Z Z Z Z

= + +

Do : 1 2112 1 22

d dd d

n

Z ZUA Z ZI Z

= = + +21

222

1 dn

ZIAI Z

= = +



1U 2UnZ

1dZ 2dZ1I 2I 1 1 21 2

2

1

1 1

d d dd d

n n

d

n n

Z Z ZZ ZZ Z

AZ

Z Z

+ + + = + C th thay mng hai ca Kirchhoff bt k cng b s A ca n bng 1 mng hnh T tng ng, vi cc thng s nh sau:

21

1nZ A= ( )1 11

21

1 1dZ AA= ( )2 22

21

1 1dZ AA=



1nZ 2nZ

dZ

1U 2U

Mng hnh

C th thay tng ng mng 2 ca Kirchoff khng ngun bt kvi b s A ca n bng mng hnh . Trong :

12dZ A= 12122 1

nAZ

A=

122

11 1n

AZA

=


V.9. CC HM TRUYN T P V HM TRUYN T DNG

Khi quan tm n vic truyn tn hiu i l mt trong hai trng thidng hay p trn ca v qu trnh truyn chng i qua mng, khi ch cn xt cc hm truyn t (khng cn xt c h 2 phng trnhvi 4 thng s c trng ca mng)

+ Hm truyn t p: 2

1u

UKU

=

+ Hm truyn t dng: 2

1i

IKI

=

+ Quan h cng sut: 2

1s

SKS

=



+ Khi ti bin thin th cc hm truyn t ph thuc vo c thng sca mng v ti

2 2

1 21 2 22 2 21 2 22

1i

I IKI A U A I A Z A

= = =+ +

2 2 2 2

1 11 2 2 12 2 11 2 12u

U I Z ZKU A I Z A I A Z A

= = =+ +

2 2 2

1 11s u i

S U IK K KU IS

= = =



Khi quan tm ti vic trao i nng lng tn hiu vi mch ngoi, khng xt s truyn t gia hai ca, ta cn dng khi nim tng trvo ca mng.

1U 2U2Z1I 2I1vZ1 11 2 121

1 21 2 22v

U A Z AZI A Z A

+= = +

+ Tng tr vo ngn mch v h mch

+ Ha hp ngun v ti bng mng hai ca


V.9. PHN TCH MCH CHA PHN T PHC HP

Vi mng mt ca:

S dng nh l Thevenin v Norton

Vi mng hai ca:

S dng b s Y, Z v da vo h phng trnh c trng camng gii (thay th bng ngun dng v ngun p ph thuc)

+ Dng phng php th nh cho ngun dng

+ Dng phng php dng vng cho ngun p



Khuych i thut ton:

+

- in tr vo

- in tr ra

- H s khuych i trong

S thay th tng ng:

+

vR

rR

( ) + +



S thay th ca khuych i mc vi sai:

+

rR

( ) + vR

vR


Chng VIMCH IN TUYN TNH VI KCH THCH CHU K

KHNG IU HA

VI.1. Nguyn tc chung

VI.2. Gii mch in c kch thch mt chiu

VI.3. Tr hiu dng v cng sut ca hm chu k

VI.4. V d p dng

VI.5. Ph tn ca hm chu k khng iu ha

MCH IN TUYN TNH VI KCH THCH CHU K KHNG IU HA

VI.1. NGUYN TC CHUNG

- Thc t cn tnh mch in c kch thch chu k khng sin (hthng in c cu chnh lu c ln, h quang in, bin tn,)

- Phng php gii:

+ Phn tch ngun chu k thnh tng cc thnh phniu ha (khc tn s)

( ) ( ) ( )2 sinT k k ke t e t E k t = = + + Cho tng thnh phn kch thch tc ng, tnh p ngca mch

+ Tng hp kt qu ( ) ( )ki t i t= ( ) ( )ku t u t=


VI.1. NGUYN TC CHUNG

- Ngun chu k c chuyn sang thnh tng cc tn hiu iuha da vo chui Fourier

( ) ( )0 k1

os k t+ck kmk

f t f F c =

= +k Z + - Tn s s bn ca cc thnh phn iu ha


VI.2. GII MCH IN C KCH THCH MT CHIU

1. c im ca mch mt chiu

+ Ngun mt chiu: gi tr khng i theo thi gian

+ ch xc lp:

0, 0di dudt dt= =

Do : R0I0 0U RI=

L0I 00 0L

dIU Ldt

= =

0I C0 0C

duI Cdt

= =


VI.2. GII MCH IN C KCH THCH MT CHIU

2. Cch gii mch in mt chiu ch xc lp

+ B qua nhnh cha t khi gii mch

+ B qua cun cm trong nhnh cha cun cm

+ Mch ch cn cc phn t in tr

+ H phng trnh lp theo phng php dng nhnh, dng vng, th nh DNG I S

+ Cc php bin i mch vn ng cho mch mt chiu


VI.3. TR HIU DNG V CNG SUT CA HM CHU K

1. Tr hiu dng

Vi dng in: ( ) ( ) ( )2 sink k kk k

i t i t I k t = = + ( ) 22

0 0

1 1 2 sinT T

k kI i dt I k t dtT T = = +

2 2

0 0

1 1T Tk k l k

kI i dt i i dt I

T T= + =

Tng t: 2k

kU U= 2k

kE E=


VI.3. TR HIU DNG V CNG SUT CA HM CHU K

2. Cng sut

Cng sut a vo phn t:

TuTi

0 0

1 1T TT T k kP u i dt u i dtT T

= =

0 0

1 1T Tk k k l kP u i dt u i dt PT T

= + =

+ H s mo:

2

1

1

kk

meo

IK

I=

+ H s nh: mdinhIKI

=


VI.4. V D P DNG

1. V d th nht

A

VR L

Cu

( )020 100 2 sin 314 20 2 sin 3.314 20u t t V= + + 610 ; 0,1 ; 10R L H C F= = =

Tnh s ch ca ampemet, vonmet v cng sutngun?

Gii

+ Cho thnh phn mt chiu tc ng:

Do C h mch nn: 0 0I A= 0 0CU V= 0 0 0P U I=+ Cho thnh phn xoay chiu th nht tc ng: ( )314 /rad s =


VI.4. V D P DNG

1. V d th nht

R

1U

j L1jC

11Z R j LC

= +

01 100 0U = (

11

1

UIZ

= 1 11CU j IC= { }1 1 1ReuP U I= + Cho thnh phn xoay chiu th hai tc ng ( )3.314 /rad s =

S tnh ton vn nh trn nhng tng tr ca cun cm v t C thay i

03 20 20U V= ( 3 3

3

1Z R j LC

= +

33

3

UIZ

=

3 33

1CU j IC= { }3 3 3ReuP U I=


VI.4. V D P DNG

1. V d th nht

+ Tng hp kt qu:

- S ch ca ampemet:0 1 3 1 3I I I I I I= + + = +

- S ch ca vonmet: 0 1 3c C C CU U U U= + +

- Cng sut tc dng ca ngun:

0 1 3 1 3u u u u u uP P P P P P= + + = +


VI.4. V D P DNG

2. V d th hai

V

A

1e3e

1R2R 3R

1L

2C

2i1i 3i

31 10 100 2 sin10e t= + ( ) ( )3 0 3 03 220 2 sin 10 20 150 2 sin 3.10 40e t t= + +

V

41 1 2 2 3100 ; 0, 2 ; 50 ; 10 ; 50R L H R C F R

V

= = = = = Tnh s ch ca vonmet, ampemet, ( )3 1 3, , ?e ei t P P

+ Cho thnh phn mt chiu tc ng:

Gii

1020 10 30

1 3

0 ; EI A I IR R

= = = + 0 30 3CU I R= 10 10 10 30; 0E EP E I P= =


VI.4. V D P DNG

2. V d th hai

+ Cho thnh phn xoay chiu th nht tc ng: ( )310 /rad s =

1R2R 3R

31E11E

21I31I11I 11LZ

21CZ

A11 11 31 31

111 21 31

AY E Y EY Y Y

+= + +

( )11 11 11 1AI Y E = 21 21 1AI Y = ( )31 31 31 1AI Y E = 11 21 21C CU Z I= { }11 11 11ReEP E I= { }31 31 31ReEP E I=


VI.4. V D P DNG

2. V d th hai

1R2R 3R

33E23CZ

33I13LZ13I23I

+ Cho thnh phn xoay chiu th hai tc ng ( )33.10 /rad s =33 33

313 23 33

AY E

Y Y Y = + +

13 13 3AI Y = 23 23 3AI Y = ( )33 33 33 3AI Y E = 23 23 33C CU Z I= 13 0EP = { }33 33 33ReEP E I=


VI.4. V D P DNG

2. V d th hai

+ Tng hp kt qu:

- S ch ca ampemet: 2 2 21 10 11 13I I I I= + +- S ch ca vonmet: 2 2 2

20 21 23C C C CU U U U= + +( )- Gi tr tc thi ca dng in i3: ( ) ( )3 30 31 33i t I i t i t= + +

1 10 11- Cng sut cc ngun: E E EP P P= +3 31 33E E EP P P= +

A

AV

WW


VI.5. PH TN CA HM CHU K KHNG IU HA

1. Ph bin v ph pha

- Tn hiu chu k c phn tch thnh: ( ) ( )k0

os k t+kmf t F c

=,km kF phn b theo tn s v ph thuc vo dng ca ( )f t

( ) ( ),km km k kF F = = c gi l ph bin v ph pha ca hm chu k- Vi cc hm chu k: Fkm(), km() c gi tr khc khng ti cc im rirc k trn trc tn s, ta gi l ph vch hay ph gin on.

- Tn hiu khng chu k (xung n hoc tn hiu hng), c th coi T, do 0. Cc vch ph xt nhau, phn b lin tc theo tn s, ta c ph chay ph lin tc.

- Vi cc tn hiu chu k dng i xng qua trc thi gian, chui Fourier khng c thnh phn iu ha chn, ph s trit tiu cc im 2k



2. Dng phc ca ph

+ Tn hiu biu din di dng ph tn qua cc cp ph: ( ) ( ),km kF k k + mi tn s k, ph tn xc nh bng mt cp: ,km kF

Biu din cc cp s module gc pha ny di dng phc. Cc gi tr nyphn b ri rc theo tn s, to thnh ph tn phc.

kjkm kmF F e

=( ) ( )0 k 0

1 1 1

1 1os k t+2 2

k kj jjk t jk tkm km kmf t f F c f F e e F e e

= + = + + ( ) 1 1

2 2kj jk t jk t

km kmf t F e e F e

= = ( )*

(*) l cng thc lin h gia hm thi gian v ph tn ca n



2. Dng phc ca ph

( ) 1 12 2

kj jk t jk tkm kmf t F e e F e

= = ( )*( )* c gi tr phc ri rc theo tn s . Tr tuyt i ca moule

hm s l ph bin , cn argumen l ph pha.

Quy c: 00

12

jomF e f

=

0 0 02 ; 0mF f = =



3. Tnh ph phc theo tn hiu cho

Nhn hai v ca (*) vijk te

ly tch phn trong mt chu k:

( ) ( )2 20 0

1 1 12 2

j l k tjk tkm km

lf t e d t F d t F e d t

=

= +

( )20

1 jk tkmF f t e d t

= ( )*


Chng VII.MCH IN BA PHA

VII.1. Khi nim v h thng ba pha

VII.2. Mch ba pha c ti tnh i xng

VII.3. Mch ba pha c ti tnh khng i xng

VII.4. o cng sut mch ba pha

VII.5. Phng php cc thnh phn i xng

MCH IN BA PHA

VII.1. KHI NIM V H THNG BA PHA

N

+

+

+

SA X

B

Y C

Z

- H thng in ba pha c s dng rng ri

- H thng ba pha gm ngun ba pha v ti ba pha

- Ngun ba pha gm 3 sc in ng mt pha, c to bi my pht in 3 pha.

1. Cu to ca my pht in ba pha:

+ Stato: hnh tr rng, ghp t cc l thp kthut in, t 3 cun dy AX, BY,CZ, lchnhau i mt mt gc 1200

+ Roto: hnh tr, l nam chm in nui bngngun mt chiu, quay t do trong lng stato

MCH IN BA PHA


N

+

+

+

SA X

B

Y C

Z

Khi roto quay, trong cc dy A, B, C xut hin cc sc in ngxoay chiu

( )AAX:e 2 sinAt E t=( ) ( )0BBY:e 2 sin 120Bt E t= ( ) ( )0: 2 sin 240C CCZ e t E t=

A B CE E E= =

MCH IN BA PHA


2. M hnh ni ngun ba pha (c ti)

Ae

Be

Ce CZ

BZ

AZA

B

C

X

Y

Z

+ y l m hnh ni ring bit bapha

+ Thc t ngi ta ni sao (Y) hoc tam gic () c pha ngun vti

MCH IN BA PHA


Ni dy h thng ba pha

- Ni sao:AZ

BZ

CZ

ABC

X Y Z

- Ni tam gic:A

B CX Y

Z

ABZ ACZ

BCZ

MCH IN BA PHA


3. Biu din phc cc s ngun ba pha

AE

BECE

o

0

Vi ngun i xng th:

jA AE E e=

0120jB AE E e

= 0240j

C AE E e=

V do : 0A B CE E E+ + = + Xt mch ba pha tuyn tnh ch XLH

+ Ba phng php c bn bit vn c th dng gii mch ba pha

MCH IN BA PHA


4. Khi nim ti tnh v ti ng

+ Ti tnh: gi tr hon ton xc nh, khng ph thuc votnh cht ca ngun

+ Ti ng: gi tr thay i ty theo tnh bt i xng cangun. Chng c gi tr xc nh khi t di cc ngun ixng. (C phng php gii ring cho mch ba pha ti ng)

MCH IN BA PHA

VII.2. MCH BA PHA I XNG, TI TNH

1. c im

+ Mch ba pha i xng l mch c c ngun v ti i xng

0,A B C A B CE E E E E E+ + = = = A B CZ Z Z= =

+ c im:

- Bit dng, p trn mt pha c th suy ra cc i lngtng ng trn cc pha cn li

- Mi lin h gia dng, p:

MCH IN BA PHA


1. c im

* Mch ba pha i xng u Y-Y: AZ

BZ

CZ

ABC

X Y Z d fI I= 0303 jd fU U e=

* Mch ba pha i xng u -:

d fU U= 0303 jd fI I e

= A

B CX Y

Z

ABZ ACZ

BCZ

fIdI

dU

MCH IN BA PHA


2. Phng php phn tch

- Tch ring tng pha tnh do th cc im trung tnh bng nhau

2.1. V d 1

AZ

BZ

CZO

AEBE

CE

AIBI

CI1O

NZ NI

Cho mch in nh hnh bn. Tnhdng, p trn cc pha ca ti vcng sut ngun?

Gii

( )1

03A B CA A B B C C

OA B C N N

Y E E EY E Y E Y EY Y Y Y Y Y

+ ++ += = =+ + + +

MCH IN BA PHA


2.1. V d 1

AZ

BZ

CZO

AEBE

CE

AIBI

CI1O

NZ NI

Nh vy, th O v O1 bng nhau. C th ni bng dy khng tr khnghai im v tch ring tng pha tnh.

1OOAE Z AI

AA

EIZ

= 0 0120 240;j jB A C AI I e I I e = = Cng sut ngun: { }3 3ReE A A AP P E= = I (v mch ba pha i xng)

MCH IN BA PHA


2.2. V d 2

O

AEBE

CE

AIBI

CI

dZ

dZ

dZ

2Z2Z 2Z

1CI2CI

2O

13Z13Z13Z

A

BC

Cho mch ba pha i xng nh hnh v trn. Tnh in p ri trn dy, cng sut tiu tn trn cc b ti mt v hai?

MCH IN BA PHA


2.2. V d 2

O

AEBE

CE

AIBI

CI

dZ

dZ

dZ

1Z

1Z

1Z

2Z2Z 2Z

1CI2CI

1O

2O

Dng bin i Y-, amch v dng nh hnh bn

Chp cc im trung tnh v tch ring tng pha tnh.

MCH IN BA PHA


2.2. V d 2

O AE AI dZ 1Z

2Z1O

2O

1AI2AI

A

Gi s tnh cho pha A:

1 2

1 2

AA

d

EI Z ZZZ Z

=+ +

dA d AU Z I=

T suy ra: ;B CI I

St p trn dy:

Dng in trn cc nhnh: 11

;A dAAE UI

Z= 2

2

;A dAAE UI

Z=

MCH IN BA PHA


2.2. V d 2

Cng sut tiu tn trn b ti th nht:

{ }1 1 1 1 13 3Ret A A A AP P Z I I= = Cng sut tiu tn trn b ti th nht:{ }2 2 2 2 23 3Ret A A A AP P Z I I= =

MCH IN BA PHA

VII.3. MCH BA PHA KHNG I XNG, TI TNH

1. Nguyn tc

- Phn tch mch ging nh mch in tuyn tnh c nhiungun kch thch ch XLH

- Thng c gng bin i v dng mc Y-Y, ri dngphng php th nh gii

2. V dCho mch in:

dZ

dZ

dZZZ Z

A

B

C

A

BC

040

220V 220V

MCH IN BA PHA


2. V d

Ngun ba pha khng i xng, cho di dng tam gic in pdy. Tnh cng sut tiu tn trn ti?

Gii

+ Chuyn ngun d cho v dng cc in p pha, trung tnh gi chntrng im A. Ta c:

0 00; 220 0 ; 220 40A B BA C CAE E U E U= = = = = ( (+ Chuyn ti ni thnh ni Y, ta c s nh sau:

MCH IN BA PHA


2. V d

dZ

dZ

dZ

tdZ

tdZ

tdZ1OO

BE

CE

+ T mch hnh bn, c th dngcc phng php phn tch mch bit gii. Nn dng th nh

+ Tnh dng trong cc nhnh

+ T tnh cng sut tiu tntrn ti

+ Ch : Tnh dng in qua cc ti ni tam gic?

MCH IN BA PHA

VII.4. O CNG SUT MCH BA PHA

1. Nguyn tco, tnh ri cng cng sut tng pha li

A B Cos os osA A B B C C A B CP U I c U I c U I c P P P = + + = + +A B Csin sin sinA A B B C C A B CQ U I U I U I Q Q Q = + + = + +

A A B B C CS U I U I U I P jQ= + + = +

2. Vi mch ba pha i xng: ch cn o trn mt pha ri suy ra c ba pha

A3 3 osA f fP P U I c = =A3 3 sinA f fQ Q U I = =

Thng tnh theo cc in p v dng in dy (cho c u Y v )

3 osd dP U I c =3 sind dQ U I =

MCH IN BA PHA

VII.4. O CNG SUT MCH BA PHA

3. Mch ba pha ba dy: dng phng php 2 wattmet

W

W**

*

*

A

B

C

AI

BI { } { } Re Retai AC A BC BP U I U I= + Vi mch ba pha 4 dy khng i xng: dng 3 wattmet

W

W**

*

*

A

C

AI

BI

W*

*

B

N

AZ

BZ

CZCI

MCH IN BA PHA

VII.4. PHNG PHP CC THNH PHN I XNG

+ Dng gii mch in ba pha c ti ng

+ Th nghim thc t: gi tr ca ti ng l tnh i vi mi thnhphn i xng ca ngun

+ Phng php phn tch mch ba pha c ti ng:

- Phn tch ngun ba pha KX thnh cc thnh phn i xng

- Gi mch ba pha X vi tng thnh phn ngun X

- Xp chng kt qu

MCH IN BA PHA


1. Phn tch ngun ba pha KX thnh cc TPX

+ Phn tch thnh cc thnh phn X thun, nghch v zero

o2AU

2CU2BUo

1BU

1AU

1CU 0CU0B

U0AU

1 2 0

1 2 0

1 2 0

A A A A

B B B B

C C C C

U U U UU U U UU U U U

= + + = + + = + +

( )1

MCH IN BA PHA



+ t ton t xoay 01 120a = (1

1 12

1 1

A

B A

C A

UU aUU a U

==

- H X thun: -H X nghch:2

22 2

2 2

A

B A

C A

UU a UU aU

==

- H X zero:0

0 0

0 0

A

B A

C A

UU UU U

==

- H thng 3 pha KX c biu din:1 2 0

21 2 0

21 2 0

A A A A

B A A A

C A A A

U U U UU aU a U UU a U aU U

= + += + += + +

( )2

MCH IN BA PHA



+ Tm c cc thnh phn i xng thun, nghch, zero ca pha A:

( )0 13A A B CU U U U= + + ( )22 13A A B CU U aU a U= + + ( )21 13A A B CU U a U aU= + +

+ Cc pha cn li:

21 2 0 1 2 0B B B B A A AU U U U aU a U U= + + = + +

21 2 0 1 2 0C C C C A A AU U U U a U aU U= + + = + +

MCH IN BA PHA


2. V d

+ Phn tch ngun ba pha khng i xng thnh cc thnh phn i xng

+ Gii bi ton mch ba pha KX, ti ng

(xt k hn trong gi bi tp)


Chng VIII

MCH IN TUYN TNH CH QU

VIII.1. Khi nim

VIII.2. Cc gi thit n gin ha m hnh qu trnh qu (QTQ)

VIII.3. Biu din hm theo thi gian v m rng tnh kh vi ca hm s

VIII.4. S kin v phng php tnh s kin

VIII.5. Phng php tch phn kinh in

VIII.6. Phng php tch phn Duyamen

VIII.7. Phng php hm Green

VIII.8. Phng php ton t Laplace


VIII.1. KHI NIM

U

K R L

i

UR

i

t

0

0L

L

WW

+

+ Thc t vn hnh thit b in: thay i t ngt kt cu v thngs mch, dn ti thay i v quylut phn b nng lng in t

+ Sau thi im thay i t ngtv kt cu v thng s: mch tinti trng thi xc lp no, qutrnh din ra nhanh hay chm

=


VIII.1. KHI NIM

1. nh ngha QTQ

L qu trnh xy ra trong mch k t sau khi c s thay it ngt v kt cu v thng s ca n

2. S tn ti ca QTQ

+ Do h thng cha cc phn t c qun tnh nng lng

+ Trong k thut in: cc phn t L, C l nguyn nhn gy ra qutrnh Q. Mch thun tr: ko c QTQ

+ Nghin cu QTQ: cn thit cho cng tc thit k, hiu chnh, vn hnh thit b in


VIII.1. KHI NIM

3. M hnh ton ca QTQ

0

0

kk

kk

i

u

= =

( )1

+ QTQ nghim ng (1), khi u t ln cn ca thi im c s thayi t ngt v kt cu v thng s ca mch

+ Nh vy, m hnh ton ca QTQ:

- H phng trnh vi phn m t mch theo 2 lut Kirchhoff

- Tha mn s kin ca bi ton quanh thi im xy ra s thayi v kt cu v thng s ca mch (t0)


VIII.1. KHI NIM

3. M hnh ton ca QTQ

+ Bi ton hay gp trong LTM: tnh cc p ng Q u(t), i(t),di kch thch ca ngun p hoc ngun dng

+ Hnh ng lm thay i kt cu v thng s ca mch: ng tcng m

+ Thng chn thi im ng m t0 = 0 (gc thi gian tnh QTQ)

4. Bi ton mch CQ

+ C hai dng: bi ton phn tch v bi ton tng hp

+ Mt s phng php phn tch: tch phn kinh in, tnh png xung ca hm qu v hm trng lng, ton t Laplace


VIII.2. CC GI THIT N GIN HA M HNH QTQ

Mc ch: n gin ha m hnh QTQ, c th dng m hnh mch xt v qu trnh tnh ton mch n gin hn

Cc gi thit n gin ha m hnh QTQ:

+ Cc phn t R, L, C l l tng

+ ng tc ng m l l tng: qu trnh ng ct coi l tc thi

+ Lut Kirchhoff lun ng

Ch

- Gi thit n gin ha th 2: khng phn nh ng hin tng vtl xy ra trong mt s trng hp

- Khc phc: cc lut ng m


VIII.2. CC GI THIT N GIN HA M HNH QTQ

V d K

u

1R 1L

2L2R 2i1i

(Trc khi m K:

) ( )1 20 0; 0 0i i =( ) ( ) ( )21 1 1 210 0 0; 0 02M MW L i W = =

Sau khi m K: ( ) ( )1 20 0i i i+ = + = (lut Kirchhoff)Vy chn i2 bng bao nhiu?

Nu ( )2 0 0i + cn mt cng sut v cng ln cp cho L2Nu ( )2 0 0i + = cng sut pht ra trn L1 v cng ln Cc gi thit vi phm lut qun tnh ca thit b


VIII.3. BIU DIN HM THEO THI GIAN

V M RNG TNH KH VI CA HM S

Do gi thit n gin ha qu trnh ng m nn c th khin iL, uC b nhy cp, trong khi thc t chng bin thin lin tc. Ta xt hai hm ton hc m ng dngca n c th kh vi ha cc hm gin on, tin xt bi ton QTQ trong nhiutrng hp

1. Hm bc nhy n v v ng dng

( ) 0 : 011: 0

tt

t = + Nhy cp: (-0;+0)

( ) 000

0 :1

1:t T

t Tt T = +

Nhy cp (-T0;+T0)

1

0

1

0

t

t0T

1.1. nh ngha





1.2. ng dngBiu din mt s qu trnh qua hm bc nhy n v

( ) ( ) ( ) ( ) ( )0 : 0

1: 0

tt t t f t

f t t = = +

( )f t( )t

1T 2T

( ) ( ) 1 21 2

:0 : ;f t T t T

tt T t T

= < >

+ Xt mt on tn hiu:

( ) ( ) ( ) ( ) ( )1 21 1t f t t T f t t T =




+ Biu din xung vung:


T

0U( ) ( )1 0 1 1u U t t T= + Biu din xung tam gic:

0U

T

( ) ( )02 1 1Uu t t t TT=





+ Mt s dng tn hiu khc:

0U

T

T

0U

( ) ( ) ( )03 01 1 1Uu t t t T U t TT= +

( ) ( )04 0 1 1Uu t U t t TT = +




2. Hm xung dirac v ng dng

2.1. nh ngha

t

0

( )t

0

t

( )0t T

0T

Hm xung dirac c nh ngha l o hmca hm bc nhy n v

( ) ( ) ( )( )0 : 0; 0

1: 0; 0

tdt tdt t

+= = +

( ) ( ) ( )( )0 0

0 00 0

0 : ;1

: ;

t T Tdt T t Tdt t T T

+ = = +Vi nh ngha ny, hm bc nhy n v c m rng tnh kh vi, n c ohm ti bc nhy




2. Hm xung dirac v ng dng

2.1. nh ngha

T

1Tt

( )+ C th coi:0

1limT

tT

= + Xung lng ca hm dirac bng 1: ( ) 1t dt+

=

+ V d:

0U

T

( ) ( )02 1 1Uu t t t TT=

( ) ( ) ( ) ( )0 02 0 00 0

1 1U Udu t t T t t t Tdt T T

= + Ta c:

( ) ( ) ( )0 0 0 00

1 1U t t T U t TT

=




2. Hm xung dirac v ng dng2.1. nh ngha

+ Tnh cht: ( ) ( ) ( ) ( ). .f t t T f T t T = 2.2. ng dng

0T

( )e t0T

+ Biu din cc xung hp. V d xung st e(t)

( ) ( )e t S t T= + Biu din cc tn hiu ri rc ( )'x t ( )x t

tT( ) ( ) ( )'

0

N

kx t x kT t kT

==

( )'x t( )x t ( )*x tLM MH MT


VIII.4. S KIN V PHNG PHP TNH S KIN

1. nh nghaSK l gi tr ca bin v o hm ti cp n-1 ca bin trong phngtrnh vi phn m t mch, ti thi im (+0)

2. ngha

+ V mt ton hc: QTQ m t bi h phng trnh vi phn thng, theo 2 lut Kirchhoff cho tha mn s kin. Nghim tng qut cha hs hay hng s tch phn. Dng s kin tnh gi tr cc HSTP, ngvi iu kin u ca bi ton.

+ V mt vt l: SK chnh l trng thi ca mch ngay sau ng tcng m. Trng thi ny nh hng ti QTQ



3. Hai lut ng m

+ Khi l tng ha qu trnh ng ct: c th vi phm tnh lin tc caqu trnh nng lng trong mch c nhy cp nng lng, xuthin cc gi tr VCL trong phng trnh vi phn m t mch.

+ Khc phc nhc im trn: 2 lut ng m

3.1. Lut ng m th nht

i1

i2i3C1

L

R

C31 2 3 0i i i+ =

' '1 1 2 3 3 0C CC u i C u + =

Theo lut K1:

Phi m bo: ( ) ( )1 1 3 30 0 0C CC u C u =( ) ( ) ( ) ( )1 1 1 3 3 30 0 0 0 0C C C CC u u C u u+ + =

( ) ( )0 0k Ck k CkC u C u+ = hay ( ) ( )0 0k kq q+ =



3. Hai lut ng m3.1. Lut ng m th nht

+ Pht biu: Tng in tch ti mt nh c bo ton trong qu trnhng m nhng ti cc phn t ring bit c th c nhy cp

+ Nu ti mt nh ch c duy nht mt phn t in dung C th:

( ) ( )0 0C Cu u+ = 3.2. Lut ng m th hai

R1R4

L3C

L2e1 e2

V bo ton t thng trn mt vng knbt k trong qu trnh ng m



3.2. Lut ng m th hai

R1R4

L3C

L2e1 e2

Theo lut K2:' '

1 1 3 3 4 4 2 2 1 2CR i u L i R i L i e e+ + = + Ti ti im ng m, nu ngun chaxung lng ( )S t( )3 3 2 2 3 3 2 2L i L i S L i L i S = =

+ Nu ngun khng c s hng VCL ti thi im ng m th:

( ) ( )0 0k k k kL i L i+ = hay ( ) ( )0 0k k + = + Pht biu: Tng t thng trn mt vng kn bo ton trong qu trnhng m

+ Trng hp vng kn ch cha mt cun cm: ( ) ( )0 0L Li i+ =



4. Cc bc tnh s kin

Bc 1: + Tm uC(-0), iL(-0) trc ng m

+ p dng lut ng m tm cc SK c lp uC(+0), iL(+0)

Bc 2: + Vit h phng trnh vi phn m t mch, sau ng m

+ Cho t = 0, tnh cc s kin theo yu cu

Bc 3: + Nu cha SK, o hm HPT bc 2

+ Thay t = 0, tm tip cc SK cn li

+ C th o hm nhiu cp nu cn

Ch : mch cp n nu c n phn t qun tnh c lp. Khi cntnh o hm ti cp n-1 ca 1 bin



4. Cc bc tnh s kin

V d

K

3R

4R

2R1R

1L

2Ce

1i2i

3iTnh cc s kin ik(+0), k=1,2,3 cng ohm cp 1 ca chng?

Gii

+ Trc khi ng K: mch c dng

Gii mch ch xc lp, tm iL(t), uC(t)

Thay t = 0, tnh cc gi tr iL(-0), uC(-0)

+ p dng lut ng m

3R

4R

2R1R

1L

2Ce

1i2i

3iA



4. Cc bc tnh s kin

3R

4R

2R1R

1L

2Ce

1i2i

3iA - Ti A: ( ) ( )2 20 0C Cu u+ = - Trong vng 1: ( ) ( ) ( )10 0 0L Li i i+ = = +Bc 2: Vit phng trnh sau khi ng kha K

3R2R1R

1L

2Ce

1i2i

3iA( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )

1 2 3

'1 1 1 1 3 3 1

2 2 2 3 3

0 0 0 0

0 0 0 0

0 0 0 0C

i i i

R i L i R i e

R i u R i

+ + + = + + + + + = + + + + + =+ o hm h trn mt ln na tm nt ccs kin theo yu cu bi.


CC PHNG PHP TNH QU TRNH QU TRONG MCH IN TUYN TNH

TB MTT PTVP+SK ( )x t( )X pPTSSH mch

TT

+ Phng php tch phn kinh in

+ Phng php tnh p ng xung hm qu v hmtrng lng

+ Phng php ton t Laplace


VIII.5. PHNG PHP TCH PHN KINH IN

1. Ni dungTm nghim qu dng: ( ) ( ) ( )cb tdx t x t x t= +

( )cbx t + V mt ton hc: nghim ring cho tha mn kch thch+ V mt vt l: l qu trnh c ngun nui duy tr

N l nghim ca qu trnh xc lp( )tdx t + V mt ton hc: nghim ring cho tha mn s kin

ca phng trnh vi phn thun nht

+ V mt vt l: p ng ca mch khng c ngunnui duy tr

Nu kch thch l chu k th xcb(t) chnh l nghim xc lp sau ngm bit cch tm.Vn ca phng php TPK l i tm nghim t do: xtd(t)



1. Ni dung

Xc nh nghim t do:Nghim tng qut ca phng trnh vi phn c dng: ( ) pttdx t Ae=Do : pttd td

dx pAe pxdt

= =pt

tdtd

xAex dtp p

= =(H) phng trnh vi phn thun nht vit thnh:

( )11 0... 0n nn n tdA p A p A x+ + + = khng c nghim tm thng th: ( ) 0.... 0nnp A p A = + + = ( )1Gii (1) c h s m pk ca cc nghim t do

Nghim qu : ( ) ( )1

k

np t

xl kk

x t x t A e=

= +Cc hng s tch phn Ak c xc nh nh s kin



2. Lp v gii phng trnh c trng

2.1. Lp phng trnh c trng

Cch 1: + Vit phng trnh vi tch phn m t mch sau ng m

+ Trit tiu cc ngun

+ Thay dxtd/dt bi pxtd; xtddt bi xtd/p v nhm cc tha schung xtd

+ Phng trnh theo p thu c l phng trnh c trng

V d: Lp phng trnh c trngcho mch sau

C

Le 2

R

1R K1i

2i3i



2. Lp v gii phng trnh c trng2.1. Lp phng trnh c trng

C

Le 2

R

1R K1i

2i3i

+ H phng trnh dng nhnh m t mch:

1 2 3

21 1 2 2

1 1 3

0

1

i i idiR i R i L edt

R i i dt eC

= + + = + = + Trit tiu ngun v dng ton t p: ( )

1 2 3

1 1 2 2 3

1 1 2 3

00 0

10 0

td td td

td td td

td td td

i i iR i R Lp i i

R i i iCp

= + + + = + + =




C

Le 2

R

1R K1i

2i3i

+ H khng c nghim tm thng khi:

1 2

1

1 1 1det 0 0

10

R R Lp

RCp

+ = ( )2 11 2 0R Lp RR R LpCp Cp

+ + + + =( ) ( )21 1 2 1 2 0R LCp R R C L p R R + + + + = ( )2

(2) Chnh l phng trnh c trng cn tm




Cch 2: + i s ha s sau ng m (L Lp; C 1/Cp), trittiu cc ngun

+ Tm tng tr vo ca mch nhn t mt nhnh bt k

+ Cho trit tiu tng tr vo, thu c phng trnh c trng

V d: Lp phng trnh c trng chomch sau

C

Le 2

R

1R K1i

2i3i




+ i s ha s ta c:

2R

1R

Lp

1Cp

( )( )

+ Tng tr vo nhn t nhnh 1:

2

1 1

2

1

1vR Lp

CpZ p RR Lp

Cp

+= +

+ +( ) ( )21 1 2 1 2

22 1

R LCp R R C L p R RLCp R Cp+ + + += + +

+ Phng trnh c trng:

( ) ( ) ( )21 1 1 2 1 20 0vZ p R LCp R R C L p R R= + + + + =




2R

1R

Lp

1Cp

+ C th tm tng tr vo nhn tnhnh 2

( ) ( )1

2 2

11v

RCpZ p R Lp

RCp

= + ++

( ) ( )21 1 2 1 21 1

RCLp R R C L p R RRCp

+ + + += ++ Phng trnh c trng thu c:

( ) ( ) ( )22 1 1 2 1 20 0vZ p R LCp R R C L p R R= + + + + =



2. Lp v gii phng trnh c trng2.2. Vit dng nghim t do+ Nu phng trnh c trng c nghim n p1, p2,, pn:

1 21 2

1

... n kn

p t p tp t p ttd n k

kx Ae A e A e A e

== + + + =

+ Nu phng trnh c trng c nghim bi n th:

11 2 ...

pt pt n pttd nx Ae A te A t e

= + + ++ Nu phng trnh c trng c nghim phc p j =

( )costtdx Ae t = ++ Nu phng trnh c trng c c nghim n, bi v phc thnghim t do l xp chng ca cc thnh phn



2. Lp v gii phng trnh c trng2.3. S m c trng v dng iu ca qu trnh t do

+ Khi pk l nghim n: A

A

t0; 0kp A< >

0; 0kp A> 0 th QTQ khng ti qutrnh xc lp

- Nu pk



2. Lp v gii phng trnh c trng2.3. S m c trng v dng iu ca qu trnh t do

+ Khi pk l nghim phc: k k kp j = ( )k

1osk

nt

td k kk

x A e c t =

= +- Nghim t do dao ng theo hm cos

- Nu k 0 bin dao ng ln dn, QTTD khng tt, QTQ khng tin c ti QTXL

+ Khi pk l nghim bi (thc hoc phc) th ch khi Re(pk) < 0 nghimqu mi dn ti xc lp



3. Xc nh cc hng s tch phn Ak

+ Vit nghim qu : ( ) ( )1

k

np t

xl kk

x t x t A e=

= ++ Tm s kin ti cp thch hp

+ Thay t = 0 nghim qu x(t)

+ Cn bng vi cc gi tr s kin tm cc hng s Ak



4. Trnh t gii bi ton QTQ bng phng php TPK

Bc 1: Tm nghim xc lp sau ng m

Bc 2: + Lp phng trnh c trng v gii tm s m c trng

+ Xp chng nghim

Bc 3: + Tnh s kin v tm cc h s Ak+ Vit nghim y ca QTQ



5. V d p dng

Xt QTQ khi ng in p U mt chiu co mch R-C, ban u tcha c np. Tm dng v p qu trn C

C

RK

U

Bc 1: Tnh nghim xc lp0

( )xl

xl

i Au U V==

Bc 2: Lp PTT v vit dng nghim Q

( ) 1 1v RCpZ p R Cp Cp+= + = Nghim ca PTT: 1p

RC=

Ta c cc nghim qu nh sau:

( ) 1 tRCCqd Cxl Ctd uu t u u U A e= + = + ( )1

0t

RCqd xl td ii t i i Ae

= + = +



5. V d p dng

C

RK

U( )0 0Cu V =( )

Bc 3: Tnh s kin v tm Ai, Au

+ C

nn ( )0 0 0C Cu u V+ = =( ) ( ) ( )+ Do , sau khi ng K th: 0 0 0R Uu Ri U i R+ = + = + =

+ Cho tha mn nghim qu :1 0

0RCu uU A e A U+ = =

1 0RC

i iU UAe AR R

= =+ Vy dng in Q trong mch v p Q trn t C c dng:

( ) 1 tRCCqdu t U Ue= ( )1 tRC

qdUi t eR

=


VIII.6. PHNG PHP TCH PHN DUYAMEN

1. Ni dung phng php

+ Hn ch ca phng php TPK: ch gii mch c kch thch xoaychiu iu ha hoc kch thch hng

+ Nguyn tc ca phng php tch phn Duyamen:

Kch thch c chia thnh chui cc bc nhy n v Tm p ng cho tng thnh phn bc nhy Xp chng cc p ng thu c nghim ca QTQ

+ Phng php tch phn Duyamen ch p dng cho bi ton cs kin zero



2. Khai trin bc nhy h-vi-xaid

( )f t( )df

+ Kch thch f(t) c khai trin thnhtng cc nguyn t n v, dng:

( ) ( )1 t df

+ Nu hm f(t) l tng ca nhiu hm, dng khi nim hm 1(t) ta c:

( ) ( ) ( )

(bt u t thi im , bin bnglng tng vi phn ti im

( ) ( ) ( ) ( ) ( ) ( )1 1 1 2 2 31 1 1 1 1 1t f t t t T f t t T t T t T f t= + + ( ) ( ) ( ) ( )1

01 1k k k k

kt T t T f t t

== =

( ) ( ) ( )' '1 kt f t t= ( ) ( ) ( )'0

1t

tDo : V: f t d

=



3. p ng H-vi-xaid

MHM( )h t( )1 t

0SK =+ p ng H-vi-xaid ca mch (k hiuh(t)) chnh l p ng Q ca i lngx(t) (dng hoc p) khi kch thch vo mchhm bc nhy 1(t) vi s kin zero

+ C th khai trin 1 hm gii tch bt k thnh tng cc bc nhy nv v tnh p ng qu ca mch i vi kch thch . (C s caphng php TP Duyamen)

Nu: ( ) ( )1 t h t > th ( ) ( ) ( ) ( ) ( )1 1t df t df h t > Hay vi phn p ng Q x(t) c dng:

( ) ( ) ( ) ( )'1dx t t f h t d =



4. Cng thc tch phn Duyamen

+ Ta c: ( ) ( ) ( ) ( )'1dx t t f h t d = l p ng qu ca kch thch: ( ) ( )'1 t f d + Di kch thch f(t) th p ng qu ca mch l:

( ) ( ) ( ) ( )'0

1t

t x t f h t d

= ( )*



5. V d

t

uU

T

C

RK

uTm in p qu trn t C khi ng mch RC vo ngun p xung ch nht nh hnh di?

Bc 1: Tm p ng h-vi-xaid

+ Kch thch hng 1(t)

+ Mch xc lp: i = 0, uC = u =1

+ Phng trnh c trng:1 10R pCp RC

+ = = + Nghim qu : ( ) 11 tRCC uu t A e= +

( )+ Do SK zero nn: ( )11 tRCC uu t e h t= =



5. V d

C

RK

u

t

uU

T

+ Ngun p: ( ) ( ) ( )1 1u t U t t T= ( ) ( ) ( )'u t U t U t T =

Bc 2: p dng cng thc tch phn Duyamen

( ) ( ) ( )'0

t

C uu t u h t d

= ( ) ( ) ( )1

0

1t T

RCU U T e d

= ( ) ( ) ( )1 11 1 1 1t t TRC RCU e t U e t T =

Ch : ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

1 ; 1 1t t

f d t f t f T d T f t T = =


VIII.7. PHNG PHP HM GREEN

1. Ni dung phng php

+ Khai trin kch thch 1(t)f(t) thnh cc xung dirac nguyn t

+ Tm p ng qu x(t) nh l tng cc p ng nguyn t y

( )f d + Cch phn tch 1(t)f(t) thnh cc xung nguyn t:

Mi phn lng ti t = l:( ) ( )f d t

Ly tng v hn cc phn lng :

( ) ( ) ( ) ( )0

1t

t f t f t d

=



2. Hm green v cng thc tnh QTQ+ Hm green l p ng Q ca mch khi c kch thch dirac tc ngvo mch vi s kin zero

+ K hiu: ( )xg t - hm trng lng ca p ng Q x(t) Kch thch: ( ) ( )f t d p ng: ( ) ( ) ( )xdx t f g t d = Do , p ng Q l:

( ) ( ) ( )0

t

xx t f g t d

= + Tm hm g(t) qua hm h(t): ( ) ( )dh tg t

dt =

( )**



3. Cc bc tnh QTQ bng phng php hm Green

+ Bc 1: Cho kch thch 1(t), tm p ng hx(t)

+ Bc 2: o hm hx(t) theo t, ta c gx(t)

+ Bc 3: Tnh p ng qu x(t) vi kch thch f(t) bng cng thc (**)



4. V d

C

RK

u

t

uU

T

( )

Tnh uC(t) qu khi ng in p xung chnht vo mch RC bng phng php hmGreen?

+ c:1

1 1t tRC

uh t e e = =

( ) tug t e =( ) ( ) ( ) ( ) ( ) ( )

0 0

1 1t t

tC uu t u g t d U T e d

+ Dng cng thc hm Green, ta c:

= = ( ) ( ) ( ) ( )

0 0

1 1t t

t tU e d U T e d

=



4. V d

( ) ( )1 11 10

t tt tt Ue e t T Ue eT

=

( ) ( ) ( ) ( )( ) ( )1 1 1 1t TtCu t U e t U e t T =


VIII.8. PHNG PHP TON T LAPLACE

1. Ni dung phng php

+ Khng tm nghim trc tip trong min thi gian

+ C s ca phng php l s dng ton t Laplace

chuyn bi ton trong min thi gian v min ton th phng trnh vi phn + SK vi gc f(t) chyn thnh HPT is vi nh F(p)

+ Gii PT (HPT) i s trong min ton t, bin i ngc cnghim Q trong min thi gian



2. Php bin i Laplace

+ Bin i Laplace thun: ( ) ( )0

ptX p x t e dt

= ( )1+ Tn hiu c bin i Laplace nu (1) hi t: x(t) tng khng nhanhhn hm m Met

+ Tnh cht ca php bin i Laplace:

Tnh tuyn tnh: ( ){ } ( )k k k kL C x t C X p= Bin i Laplace ca o hm: ( ){ } ( ) ( )' 0L x t pX p x=

( ) ( ){ } ( ) ( ) ( ) ( ) ( )11 2 '0 0 ... 0n nn n nL f t p X p p x p x x = ( ) ( ) ( )1 L pTt T x t T X p e R Tnh cht tr:



3. Tm gc t nh Laplace

+ Bin i Laplace ngc: ( ) ( )12

a jpt

a j

x t X p e dpj

+

=

+ Dng cng thc khai trin:

Vit nghim dng: ( ) ( )( )M p

X pN p

= (bc M(p)




Nu N(p) c nghim bi n: p1 = p2 = = pn = p th:

( ) ( ) ( )2 131 21 ...0! 1! 2! 1 ! n ptnA AA At x t t t t e

n = + + + +

Trong : ( )( ) ( ) ( )( )

1!

n nn

n nn

M pdA p pp pn n dp N p

= =

( )( ) ( ) ( )( )

1

11

1 !

n nn

n nn

M pdA p pp pn n dp N p

+

= = +

( )( ) ( ) ( )( )

1

11

1 !

nn

nn

M pdA p pp pn dp N p

= =




Nu N(p) c nghim phc p j =

( ) ( ) ( )1 2 os t+tt x t Be c =Trong : ( )

( )' kM p

a jbp pN p

= +=2 2B a b= +

bartana

= Nu N(p)=0 c nhiu loi nghim th tm gc cho tng loi v xp chng



4. ng dng bin i Laplace tnh QTQ trong mch in

4.1. S ton t

R( )i t( )Ru t

R

( )RU p( )I p ( ) ( )RU p RI p=

L( )Li t( )Lu t ( )LU p

( )I p Lp ( )0Li ( ) ( ) ( )0LU p LpI p Li=

C( )i t( )Cu t

( )I p1Cp ( )0Cu

p

( )CU p ( ) ( )( )01 C

C

uU p I p

Cp p= +




4.2. Algorithm gii

+ Tnh mch ch c, tm iL(-0), uC(-0)

+ Lp s ton t theo phng php gii thiu trong 4.1

+ Dng cc phng php c bn gii tm nh Laplace canghim Q

+ Suy ra nghim Q t nh tm c bc trn



4.3. V d


R3

R2

R1

C

e1E2

AK

B

Vi , E2 =20V (mtchiu), R1 = 40, R2 = 10, R3 = 10, C = 4.10-4 F. Tnh in p qu uAB(t) khi chuyn kho K ngt ngun e1 vng ngun E2 vo mch, bit trc khichuyn kho K mch ch xclp, chn t = 0 ti thi im chuyn khoK.

1 40 2 sin100 ( )e t V=




4.3. V d

+ Trc khi chuyn kho K:Gii mch xc lp vi kch thch iu ho, tm c:

Do :+ Sau khi chuyn kho K ng ngun e2, ta c:

05.2687 3.7935 6.4923 35.75CU j = = ( ) ( )00 6.4923 2 sin 35.75 5.3643Cu = =

( )( ) ( )

( ) ( )2

12

2

1 2 2 3

01

20 250 2,3841. 11 1 9 138,89 138,891

C

pTAB

E p uR R p pCU p eC p p p p

R R C p R

++ += = + ++ ++

+ Nghim: ( ) ( ) ( ) ( )( ) ( )138,89138,894 4,16 1 4 1,78 1t TtABu t e t e t T =

L THUYT MCH IN

curcuit theory

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