Tín hiệu hệ thống-Lecture 06

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  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    Lecture-6

    "0"00! T#n $i%u ' $% t$(ng

    Biu din tn hiu bng chui FourierBiu din tn hiu bng chui Fourier

    )i*u +in t#n $i%u -.ng t t#n $i%u tr1 gia3)i*u +in t#n $i%u -.ng t t#n $i%u tr1 gia3

    C$u4i F3urier l56ng gi7C$u4i F3urier l56ng gi7C$u4i F3urier $'m m8 $C$u4i F3urier $'m m8 $7 ng ;a $% t$(ng i t#n $i%u tu?n $3'n7 ng ;a $% t$(ng i t#n $i%u tu?n $3'n

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    )i*u +in t#n $i%u -.ng t t#n $i%u tr1 gia3

    Biu din tn hiu da vo khng gian tn hiu trc giao:

    Tm cntha iu kin nng lng sai s min:

    Sai s:1

    ( ) ( ) ( )N

    n n

    n

    e t f t c x t =

    =

    2

    1

    1 ( ) ( )t

    n nt

    n

    c f t x t dt E

    = Nng lng ca thnh phn sai s min: 2

    1

    N

    e f n n

    n

    E E c E=

    =

    Nng lng ca thnh phn sai s

    0 nu N

    t!p c"s#

    1 1 2 2

    1( ) ( ) ( ) ... ( ) ( )

    N

    N N n n

    nf t c x t c x t c x t c x t

    =+ + + = $

    %hi N & ta c': lu ( )*u +,- .ng / mt nng lng

    1 2

    1

    ( ) ( );n nn

    f t c x t t t t

    =

    = 1hu2i34u5i65

    2

    1

    *1 ( ) ( )t

    n nt

    n

    c f t x t dt E

    = Th7c: 8h9c:

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    Khng gian tn hiu lng gic trc giao:

    0 0 0 0 0 0{1, cos( ), cos(2 ),..., cos( ),....; sin( ), sin(2 ),..., sin( ),...}t t n t t t n t

    0n : gi l hi th9 n 00

    2,

    T

    =

    Biu din f(t) trong khng gian tn hiu lng gic trcgiao:

    0 0 0 1 1 0

    1 1

    ( ) cos( ) sin( );n nn n

    f t a a n t b n t t t t T

    = =

    = + + + 1 0

    1 0

    1

    1 01

    1

    0

    02

    00

    ( )cos( ) 2( )cos( )

    cos ( )

    t T

    t Tt

    n t T t

    t

    f t n t dt

    a f t n t dt Tn t dt

    +

    +

    += = 1 0

    10

    0

    2; ( )sin( )

    t T

    nt

    b f t n t dt T

    +

    = 1 0

    10

    0

    1( )

    t T

    ta f t dt

    T

    +=

    ;it # )

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    Biu din f(t) trong khng gian tn hiu lng gic trcgiao:

    0 0 0 1 1 0

    1 1

    ( ) cos( ) sin( );n nn n

    f t a a n t b n t t t t T

    = =

    = + + + ;it # )

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    Biu din f(t) trong khng gian tn hiu lng gic trcgiao:;> )?: khai t5i@n ABtC,6

    DtEF

    t54ng kh4Gng 0t0 0

    2 / 2T = =

    / 2

    00

    10,504ta e dt

    = =

    / 220

    2 2cos(2 ) 0,504

    1 16tna e nt dt

    n

    = =

    + / 2

    20

    2 8sin(2 ) 0,504

    1 16

    t

    n

    nb e nt dt

    n

    = = +

    ( )21

    2( ) 0.504 1 cos2 4 .s in2nt ; 01 16n

    f t nt n tn

    =

    = + + +

    0 0 0,504C a= =

    2

    20,504

    1 16nC

    n

    = +

    1tan 4n n =

    1

    21

    2( ) 0.504 1 cos(2 tan 4 ) ; 0

    1 16nf t nt n t

    n

    =

    = +

    +

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    nh tu!n hon c"a chu#i $ouri%r lng gic:

    1hu2i 34u5i65 ch4 ABtC chH .ng t54ng kh4Gng tIttIJT0Ng4i kh4Gng tIttIJT0K

    0 0

    1

    ( ) cos( );n nn

    t C C n t for all t

    =

    = + +0( ) ( );t T t for all t = BtC Li@u )iMn ch4 tin hiu tunh4n

    ;!= nu BtC$ tIttIJT0Li@u )iMn ch4 ABtC$ tIttIJT0BtC Li@u)iMn ch4 t>n hiu tun h4n )4 l!p l )?:

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    &hu#i $ouri%r lng gic c"a tn hiu tu!n hon:

    0 0 0

    1 1

    ( ) cos( ) sin( )n nn n

    f t a a n t b n t

    = =

    = + +

    0 0

    1

    ( ) cos( )n nn

    f t C C n t

    =

    = + +

    00

    0

    2( )cos( )n

    Ta f t n t dt

    T

    =

    00

    0

    2( )sin( )n

    Tb f t n t dt

    T=

    00

    0

    1( )

    Ta f t dt

    T=

    0 0C a=

    n n nC a b= +

    1tan nnn

    b

    a

    =

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    ' d :

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    ' d *:

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier l56ng gi7

    +i,u kin t-n t.i chu#i $ouri%r:

    0( )

    Tf t dt<

    TQn t

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier $'m m8 $

    /0 tn hiu h1 12 trc giao:

    { }0 ; 0, 1, 2,....jn te n =

    &hu#i $ouri%r h1 12:

    0

    ( ) jn t

    nn

    f t D e

    ==( )0

    0

    *1( )

    jn t

    nT

    n

    D f t e dtE

    =

    00

    0

    1( )

    jn t

    nT

    D f t e dtT

    =

  • 7/13/2019 Tn hiu h thng-Lecture 06

    12/16

    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier $'m m8 $

    34i li5n h gi6a chu#i $ouri%r h1 12 0h7c 8 chu#i$ouri%r

    lng gic:0cos( )n nC n t + ( )0 0( ) ( )

    2n nj n t j n tnC e e

    + += +

    nD nD

    0 01

    ( ) cos( )n nn

    f t C C n t

    == + +( )0 00

    1

    ( ) jn t jn t

    n n

    n

    f t D D e D e

    =

    = + + 0jn tnn

    D e

    =

    =

    9.ng h1 12 8lng gic l

    tng ;ngth

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier $'m m8 $

    0 0

    2

    2

    n

    n

    jnn

    jnn

    D C

    CD e

    CD e

    =

    =

    =

    1,2,3,...n=

    >h? $ouri%r:

    0 0;D C=

    1

    2n n n

    D D C= =8h\ Li[n ]: Bch@nC

    8h\ pha: BlAC; ;n n n nD D = =

    34i li5n h gi6a chu#i $ouri%r h1 12 0h7c 8 chu#i$ouri%r

    lng gic:

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    C$u4i F3urier $'m m8 $

    +nh lC >arD%val :

    0 0

    1

    ( ) cos( )n nn

    f t C C n t

    == + +

    2 2

    0

    1

    12

    f n

    n

    P C C

    =

    = +

    0( ) jn t

    n

    n

    f t D e

    =

    = 2 2201

    2f n nn n

    P D D D

    = =

    = = +

    1Wng su*t ca t>n hiu tun h4n LXng t\ng cWng su*t ca t*tcG c^c hi

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems- Tran Quang Viet FEEE, HCMUT Semester: 02/09!0

    7 ng ;a $% t$(ng i t#n $i%u tu?n $3'n

    00

    0

    2( ) ;

    jn t

    n

    nf t D e T

    == =

    ( )j t j te H j e

    input

    _utput

    0 0

    0( )jn t jn t

    n n

    n n

    D e D H jn e

    = =

    `nput ABtC _utput=BtC

    (EF)

    u!n honc=ng chukG vHi f(t)

    +0 7ng c"a h th4ng EF& vHi tn hiu tu!n hon :

  • 7/13/2019 Tn hiu h thng-Lecture 06

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    Signal & Systems - Tran Quang Viet FEEE HCMUT Semester: 02/09!0

    7 ng ;a $% t$(ng i t#n $i%u tu?n $3'n

    ' d :

    2

    2

    2

    ( ) (1 4 )

    j nt

    nf t en

    == (3 1) ( ) ( )D y t f t+ =( ) 1/(3 1)

    ( ) 1/(3 1)

    H s s

    H j j

    = += +

    2

    2

    2( )

    (1 4 )( 6 1)

    j nt

    n

    y t en j n

    =

    = +

    2

    r 2 2 2 21 1

    8 1P 2 | |

    (1 4 ) (36 1)ipple n

    n n

    Dn n

    = =

    = = +

    rP 0.025

    ipple= rP 0.05ipple =