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1 2.2 Phöông phaùp aûnh phöùc Moãi ñaïi löôïng ñieàu hoøa seõ ñöôïc bieåu dieãn baèng 1 soá phöùc (aûnh phöùc) 2.2.1 Soá phöùc Moät soá phöùc C coù theå ñöôïc bieåu dieãn döôùi caùc daïng sau: j: ñôn vò aûo j 2 = -1 a: Laø phaàn thöïc cuûa soá phöùc a= ReC Re: Real b: Laø phaàn aûo cuûa soá phöùc b= ImC Im: Imaginary C a jb Ví duï: C = 6 + j8. 6: Laø phaàn thöïc cuûa soá phöùc C 8: Laø phaàn aûo cuûa soá phöùc C Daïng ñaïi soá

Chuyen Doi So Phuc Dung May Tinh CA Nhan121

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  • 12.2 Phng phap anh phc

    Moi ai lng ieu hoa se c bieu dien bang 1 so phc (anh phc)

    2.2.1 So phc

    Mot so phc C co the c bieu dien di cac dang sau:

    j: n v ao j 2 = -1

    a: La phan thc cua so phc a= ReC Re: Real

    b: La phan ao cua so phc b= ImC Im: Imaginary

    C a jb

    V du: C = 6 + j8.

    6: La phan thc cua so phc C

    8: La phan ao cua so phc C

    Dang ai so

  • C = a + jb (Descartes) th C co the bieu dien dang cc la:

    Dang cc

    C C

    2 2

    cos

    sin

    C m

    C a b

    a C

    b C

    barctg

    a

    Vi la oun

    =

    =

    =

    la argument hay goc

  • C = a + jb

    Dang lng giac

    Cong thc Euler

    eJ = cos + jsin

    Dang Mu:

    A = a + jb = C (cos + jsin) = C eJ

  • Cac phep tnh ve so phc.

    C1 = a1 + jb1.

    C2 = a2 + jb2.

    Phep cong: C1 + C2 = (a1 + a2) + (b1 + b2)j

    Phep tr: C1 - C2 = (a1 - a2) + (b1 - b2)j

    Phep nhan:C1.C2 = (a1a2 - b1b2) + (b1a2 + a1b2)j

    Phep chia :

    2

    C

    C1 1 2 1 2 1 2 1 2

    22 2 2

    (a a b b ) + (b a - a b )j=

    a + b

  • 5 Phep nhan, phep chia dung dang cc:

    1 1 1 1 1

    2 2 2 2 2

    1 2 1 2 1 2

    111 2

    2 2

    C a jb C

    C a jb C

    C C C C

    CC

    C C

  • V du

  • 72.2.2 Bieu dien ai lng ieu hoa bang so phc.

    ( ) sin( ) ( )

    ( ) sin( ) ( )

    ( ) sin( ) ( )

    ( ) sin( ) ( )

    m i m i

    m u m u

    m e m e

    m j m j

    ai lng ieu hoa ai lng phc

    i t I t I I A

    u t U t U U V

    e t E t E E V

    j t J t J j A

    V du

  • Cach s dung may tnh e chuyen oi So Phc (Casio FX 500MS, fx570ES)

    Pol ( 6 , 8 ) = 10

    Ket qua

    Alpha tan = 53,1

    6 + 8j = 10/53,10

    Chuyen oi dang Descartes sang dang cc

  • 10/53,10= 6 + 8j

    10 shitf - 53,1 = 6

    Ket qua

    shift [( = 8

    Chuyen oi dang cc sang dang Descartes

  • May tnh Casio fx 570MS

    03 4 5 53,13j

    Chuyen sang ON - CPLX

    Chuyen oi dang Descartes sang dang cc

  • May tnh Casio fx 570ES

    Chuyen may tnh sang tnh toan so phc: MODE - 2 - CMPLX

    Chuyen oi dang cc sang dang Descartes

  • Chuyen oi dang Descartes sang dang cc

  • The set of equations can be arranged in matrix form, as shownin equation A.6.

    HE 2 PHNG TRNH

    11 12 1 1

    21 22 2 2

    a a b

    a a b

    x

    x

    11 12

    11 22 12 21

    21 22

    det ( )a a

    A a a a aa a

  • AP DUNG NH LUAT CRAMER

  • HE 3 PHNG TRNH

    1 111 12 13

    21 22 23 2 2

    31 32 33 3 3

    x ba a a

    a a a x b

    a a a x b

  • P847 Fundermantal of Electric circuit

  • 1 12 13

    2 22 23

    3 32 331 det( )

    b a a

    b a a

    b a ax

    A

    11 1 13

    21 2 23

    31 3 332 det( )

    a b a

    a b a

    a b ax

    A

    11 12 1

    21 22 2

    31 32 33 det( )

    a a b

    a a b

    a a bx

    A