giải tích mạch chương 2

Chng 2 : Mch xc lp iu ha

Bi ging Gii tch Mch 2012

2.1 Qu trnh tun hon 2.2 Qu trnh iu ha 2.3 Phng php bin phc 2.4 Gii bi ton mch dng nh phc 2.5 Quan h dng p trn cc phn t mch 2.6 Cc nh lut mch dng phc 2.7 th vect 2.8 Cng sut 2.9 H s cng sut & cch hiu chnh 2.10 Phi hp tr khng

1

Tn hiu kho st : dng in i(t) , in p u(t)

Bi ging Gii tch Mch 2012 2

Tun hon : f(t) = f(t+T)

2.1 Qu trnh tun hon

Dao ng k quanst, o tr tc thi

Volt , Amper o tr hiu dng

o c


Tr hiu dng

2 2

0 0

1 1( ) ( )T T

RMS RMSI i t dt U u t dtT T= =

2.1 Qu trnh tun hon

Dng in (in p) tun hon s c tr hiudng IRMS (URMS) l bng vi tr s dng (p) DCkhi cng sut tiu tn trung bnh do 2 dng in(in p) gy ra trn cng in tr R l nh nhau

Biu thc tnh tr hiu dng( RMS Root Mean Square )


M t ( ) sin( )( ) sin( )

m

m

i t I tu t U t

= +

= +

2.2 Qu trnh iu ha

Dng in , in p

Im , Um : bin : tn s gc , : pha ban u

Tr hiu dng2

2

mRMS

mRMS

II

UU

=

=


: pha ban u, ta c th ni u2(t) sm pha so viu1(t), hoc u1(t) chm pha so vi u2(t).

0 ta ni u1(t) v u2(t) lch pha.

=0 ta ni u1(t) v u2(t) ng pha

2.2 Qu trnh iu ha

Bi ging Ton K Thut 2012 6

1 1 1( ) ( )mu t U sin t = +

2 2 2( ) ( )mu t U sin t = +

Cng tn s. Cng dng lng gic. Cng dng bin (cc i hay hiu dng)

Ta ni u1(t) nhanh pha hn u2(t) mt gc th =1-2 (hay ta c th ni 2 chm pha hn 1 mt gc ).

Nu ta ni u2(t) nhanh pha hn u1(t) mt gc th =2-1

So snh pha hai tn hiu iu ha


( ) ( )mu t U sin t = +

2.3 Phng php bin phc

Vct quay


( ) ( )mu t U sin t = +

Biu din di dng vct quay

1 1 1( ) ( )mu t U sin t = +

2 2 2( ) ( )mu t U sin t = +

Biu din di dng vct quay


1 1 1( ) ( )mu t U sin t = +

2 2 2( ) ( )mu t U sin t = +

1 2

1 1 2 2

( ) ( ) ( )( ) ( )m m

u t u t u tU sin t U sin t

= +

= + + +

)()( 21 tutu +Vct quay

nh phc nh phc cho tn hiu iu ha

Min t Min phc

Cc quan h


( ) sin( )mf t F t = + jm mF F e F

= =

{ }1( ) Im sin( )m mf t F F t = = +{ }2 ( ) Re cos( )m mf t F F t = = +

Hiu dng phc 2RMS

FF

=

Cc tnh cht ca vct bin phc


: ( ) ; ( )Cho f t F g t G

( )kf t k F

( ) ( )f t g t F G

: ( ) 3cos(2 30 ) 3 30o oVD f t t F= + =

3 ( ) 3 9 30of t F =

3 30 4 60 5 23,13o o oF G

+ = + =

( ) 4cos(2 60 ) 4 60o og t t G= =

Tnh t l

Tnh xp chng

0( ) ( ) 5cos(2 23,13 )f t g t t+ =


Cc tnh cht ca vct bin phc

( )df t j Fdt

1( ) jf t dt F

( ) 6sin(2 30 ) 6cos(2 120 ) 2 6 120df t o o odt t t j F= + = + =

3 3 312 2 2 2( ) sin(2 30 ) cos(2 60 ) 60

o o ojf t dt t t F= + = =

: ( ) ; ( )Cho f t F g t G

: ( ) 3cos(2 30 ) 3 30o oVD f t t F= + =

( ) 4cos(2 60 ) 4 60o og t t G= =

Tnh o hm

Tnh tch phn

2.4 Gii bi ton mch dng nh phc


( )R L Cu u u e t+ + =1 ( )didt CRi L idt e t+ + =

( ) jme t E E e

=

1j CR I j L I I E

+ + =

1( )m

C

EIR j L

=

+

1( )CR j L j I E

+ =

e(t) = 10 cos 2t (V)R = 4; L = 2H; C = 0,5F

01

2.0,5

10 0 10 2 36,874 (2.2 ) 4 3

o

Ij j

= = =

+ +

Vay : i(t) = 2 cos (2t - 36,87o) A

R L

Ce(t)

uR uLuC

i(t) Min t

Min phcGii pt vi phn tm i(t)

Pt i s

Phng php vct bin phc


Min thi gian Min phc

PP ny do Charles Proteur Steinmetz tm ra vo nm 1897 .

Mch xc lp iu ha

Mch phc

H phng trnh vi tch phn

H phng trnh i s phc

nh phcTn hiu iu ha

2.5 Quan h dng p trn cc phn t mch

in tr


R RU R I

=

Cung pha

R

u(t)

i(t)

cos( )R mi I t = +

cos( )R R mu Ri RI t = = +

Min phcRIR

UR

IR

UR

R mI I

=

R mU RI

=

IL

UL

jL


in cm


L LU j L I

=

Lch pha 900

cos( )L mi I t = +0cos( 90 )LL m

diu L LI tdt

= = + +

Min phc

L

u(t)

i(t)

IL

UL

L mI I

=

L mU j LI

=

IC

UC

-j/C


in dung


C CjU I

C

=

Lch pha 900

cos( )C mu U t = +

0cos( 90 )CC mdui C CU tdt

= = + +

Min phc

C

u(t)

i(t)

IC

UCC mU U

=

C mI j CU

=


2.6 Cc nh lut dng phc

in trR RU R I

=

RIR

UR

L L L LU j L I jX I

= =IL

UL

jL in cm

in dungIC

UC

-j/C

C C C CjU I jX I

C

= =

RU

-j/C

jLI

Bi ging Gii tch Mch 2012

2.6 Cc nh lut dng phc

Tr khng

U Z I

=

Dn np

R

U

-j/CjLI I

U

Z

Z R jX Z = + =

I

U

Y

I Y U

= Y G jB Y = + =

1YZ

=

Z: Tr khng (impedance) R: in tr (resistance) X: in khng (reactance) n v tnh []


= u i

Y: Dn np (admittance) G: in dn (conductance) B: in np (susceptance) n v tnh [S]

Z R jX Z = + =

| Z |: module cua Z: goc lech pha gia u va i

Y G jB Y = + =

| Y |: module cua Y: goc lech pha gia i va u

Tr khng & Dn np

= i u

nh lut Kirchhoff dng phc

nh lut Kirchhoff dng phc v dng: Tng cc dng in phc ti mt nt bng khng. Qui c dng i vo nt mang du dng, i ra nt mang du m

nh lut Kirchhoff dng phc v p: Tng cc p phc trong mt vng kn bng khng.


0Knt

I

=

0Kvngkn

U

=

V d Tm dng in trong cc nhnh

v in p trn cc phn t

S phc ha


1H1/9F

1

35cos3t [V]

i1(t)

i2(t) i3(t)

Gii

1

3

I1I2 I3

j3

-j35 0o VI II

a

b

Nt a Vng I Vng II

1 2 3 0I I I

=0

1 25 0 (3 3) 0I j I

+ + + =

2 3(3 3) ( 3) 0j I j I

+ + =

0 01 2

03

1 36,87 ; 1 53,13

2 81,87

I I

I

= =

=

V d


1

3

I1I2 I3

j3

-j35 0o VI II

a

b

01

02

03

1 36,87

1 53,13

2 81,87

I

I

I

=

=

=

01 1

02 2

02

03

1 1 36,87

3 3 53,13

3 3 36,87

3 3 2 8,13

R

R

L

C

U I

U I

U j I

U j I

= =

= =

= =

= =

i1(t) = cos (3t + 36,87o) [A]i2(t) = cos (3t - 53,13o) [A]i3(t) = 1,41 cos (3t + 81,87o)[A] uR1(t) = cos (3t + 36,87o) [V]uR2(t) = 3 cos (3t - 53,13o) [V]uL(t) = 3 cos (3t + 36,87o) [V]uC(t) = 4,24 cos (3t - 8,13o) [V]

V d ngun ph thuc


Tm i1 ? i2 ? 1 23 ( 1)I I K+ =

1 22 0,5 4 0 ( 2)Rj I U I K + =

24 ( )RU I Ohm=

S phc ha

1( ) 3 2 sin(4 45 )oi t t A= +

2 ( ) 3 2 sin(4 45 )oi t t A=

1 2 0jI I+ = 1 2 3I I+ =

1 3 2 45oI =

2 3 2 45oI =

4 +-3sin4t

[A]

18

F

12 R

uuR

i2

i1

Gii

I4+-

12UR

I2

-j2 I1

UR3 0o A

a

b

2.7 th vect (vector diagram) nh ngha :

biu din hnh hc ca cc nh lut mch dng phc Phn loi:

th vet p, vect dng th vet tr khng, dn np th vet cng sut

Cng dng : thng dung cho cac bai toan: Mo ta ro hn quan he gia cac ai lng ien trong mach. Tm hieu s anh hng cua mot thong so mach len cac ai

lng ien. Cho phep xac nh module va pha cac ai lng da tren

mot so so lieu o ( thng dung kem vect hieu dung phc).


Biu din hnh hc ca nh phc


03 4 5 53,13U j

= + = 5

3

4

53,130

j

+1Re

Im

th vect v nh phc


Cho Tm

1 212 5 ; 9 12U j U j

= + = +

1 2U U U

= + Dng nh phc

01 2 21 17 27 39U U U j

= + = + = 0

1

02

12 5 13 22,62

9 12 15 53,13

U j

U j

= + =

= + = Dng th vect

Re

Im

0 9

12

53,130

12

5

22,62

Re

Im

0

53,130

22,620

=390


V d th vect v nh phc Tm R v XL nu bit I=2A,

Uac=100V, Uab=173V, Ubc=100V (RMS)

R jXL

jXC

I

Uac

c

a b

I

Ubc

Uac Uab

2 2 2

0 0

cos 0,8652

30 60

ab bc ac

ab bc

U U UU U

+ = =

= =

0 02 0 173 60abI U

= =

43,25 75

43,2575

abL

L

UR jX jI

RX

+ = = +

= =

th vng ca tr khng v dn np

Khi nim: tr khng Z v dn np Y l cc s phc dng th kho st khi thng s nhnh thay i


Z1Z2 Z

Y1 Y2 Y

Z1

Y1

Z2Y2

Z

Y

R

jX

G

jB

th vng ca tr khng v dn np Nhnh R-L ni tip


Z

Y

R

jX

G

jB

R

jXL

Z,Y Biu thc tr khng & dn np

LZ R jX= +

B < 0 v G > 0 qu tch l ng trn

2 2

GRG B

=+

2 221 1

2 2G B

R R + =

12R

1R

2 2

1 1 G jBZY G jB G B

= = =

+ +

th vng ca tr khng v dn np Nhnh R-L ni tip


Z

Y

R

jX

G

jB

R

jXL

Z,Y Biu thc tr khng & dn np

LZ R jX= +

B < 0 v G > 0 qu tch l ng trn

2 2LBX

G B

=+

2 221 1

2 2L LG B

X X

+ + =

12 LX

1

LX

2 2

1 1 G jBZY G jB G B

= = =

+ +



Mch Qu tch tr khng Qu tch dn np

R

jXC

Z,Y

R

jXC

Z,Y

Z

R

jX

12R

Y G

jB

12R

1R

Z

R

jX

Y

G

jB

12 CX

1

CX




R jXLZ,Y

R jXLZ,Y

Y

G

jB1R

YG

jB

1

LXZ

R

jX

12 LX

1

LX

ZR

jX

12R

1R




R jXCZ,Y

R jXCZ,Y Y

G

jB1

CX

YG

jB

1R

Z

RjX 1

2R1R

Z

R

jX

12 CX

1

CX

Chng 2 : Mch xc lp iu ha Tn hiu kho st : dng in i(t) , in p u(t)Slide Number 3Slide Number 4Slide Number 5So snh pha hai tn hiu iu haSlide Number 7Vct quayVct quaynh phcCc tnh cht ca vct bin phcCc tnh cht ca vct bin phc2.4 Gii bi ton mch dng nh phcPhng php vct bin phc2.5 Quan h dng p trn cc phn t mch2.5 Quan h dng p trn cc phn t mch2.5 Quan h dng p trn cc phn t mchSlide Number 18Slide Number 19Slide Number 20nh lut Kirchhoff dng phcV d V d V d ngun ph thuc2.7 th vect (vector diagram)Biu din hnh hc ca nh phc th vect v nh phcV d th vect v nh phc th vng ca tr khng v dn np th vng ca tr khng v dn np th vng ca tr khng v dn np th vng ca tr khng v dn np th vng ca tr khng v dn np th vng ca tr khng v dn np

giải tích mạch chương 2

Documents