tín hiệu hệ thốnglecture 04

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tài liệu khoa điện trường đh Bách khoa tphcm. hcmut

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  • 7/18/2019 Tn hiu h thngLecture 04

    1/18

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

    Lecture-4

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

    Phn tch h thng LTIC trongPhn tch h thng LTIC trongmin thi gianmin thi gian

    )i*i t$i%u)i*i t$i%u+- .ng *i ng ' 1ng 3$4ng+- .ng *i ng ' 1ng 3$4ng+- .ng 5ung 67n 8+- .ng 5ung 67n 8 tt+- .ng *i ng ' 1;t 3

  • 7/18/2019 Tn hiu h thngLecture 04

    2/18

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

    +- .ng *i ng ' 1;t 3$ >$@- >nlutin

    Tnh cht c'! tch chp

    Tnh cht k$t h5p ./t tch chp c'! ,(t)3 g(t) v h(t)( ) [ ( ) ( )] [ ( ) ( )] ( ) [ ( ) ( )] ( ) ...f t g t h t f t g t h t f t h t g t = = =

    4m 1 1 1 1( ) [ ( ) ( )] ( )[ ( ) ( )]f t g t h t f g t h t d

    =

    1 2 1 2 2 1( ) ( ) ( )f g h t d d

    = 0t 3 1 2 3 2d d = + =

    1 3 1 3 3 1( ) [ ( ) ( )] ( ) ( ) ( )f t g t h t f g h t d d

    =

    1 3 1 1 3 3( ) ( ) ( )f g d h t d

    = [ ( ) ( )] ( )f t g t h t=

  • 7/18/2019 Tn hiu h thngLecture 04

    11/18

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

    T#>$ >$@- >nlutin

    Tnh cht c'! h6 th2ng tch chp

    76 th2ng tch chp l tu+$n tnh

    1 1 2 2 1 1 2 2( ) [ ( ) ( )] ( ) ( ) ( ) ( )h t a f t a f t a h t f t a h t f t + = +

    76 th2ng tch chp l bt bi$n ( ) ( )f t T y t T

    8h/p n2i ti$p 9 h6 th2ng tch chp

    1( )h t 2 ( )h t

    ( )f t ( )y t1( )y t

    1 2( ) ( )h t h t ( )f t ( )y t

  • 7/18/2019 Tn hiu h thngLecture 04

    12/18

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

    T#>$ >$@- >nlutin

    Tnh cht c'! h6 th2ng tch chp

    : ;< v= o p ng .ung c'! h6 th2ng tch chp

    ( )h t

    ( )h t

  • 7/18/2019 Tn hiu h thngLecture 04

    13/18

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

    +- .ng *i t#n $i%u $'m mA -$.>

    Dch 1i vE .i+u i-n tn hiu + tnh p ng cF4 h thngGchi ti!t

    vE cch .i-u i-n tn hiu th7o Hng h1 1 phc (chuIiJouri7r,

    .i!n Ki Jouri7r) 8L 9c E cp nhNng ch9Ong 84u

    56t p ng vi h1 1 78t, 8: .i!n phc

    ( )( ) ( ) ( )st s ty t h t e h e d

    = = ( ) ( )

    st sy t e h e d

    =

    PQt: ( ) ( ) sH s h e d

    = ( ) ( ) sty t H s e =

    R1 tru@En: ( ) stinput signal e

    output signalH sinput signal =

    =

    ( )( )

    ( )

    P sH s

    Q s=( ) ( ) ( ) stQ D y t P D e=

    = (

  • 7/18/2019 Tn hiu h thngLecture 04

    14/18

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

    T#n$ =n 68n$ >?a $% t$(ng

    Snh Kn ;nh cF4 h thng phC thu

  • 7/18/2019 Tn hiu h thngLecture 04

    15/18

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

    T#n$ =n 68n$ >?a $% t$(ng

    Pp ng cF4 h thng vi ngW vo .; gii hHn:

    ( ) ( ) ( ) ( ) ( )y t h t f t h f t d

    = = ( ) ( ) ( )y t h f t d

    1( )f t K < < 1( ) . ( )y t K h d

    2( )h d K

    < <

    1 2( )y t K K <

    56t p ng cF4 RS >SXD vi tn hiu vo (t)

    !u (t) .; gii hHn:

    BIBO

    !u h thng Kn ;nh:

    !u h thng 2hng Kn ;nh hoQc Kn ;nh .iAn @(t) 2hng .;gii hHn vi 1

  • 7/18/2019 Tn hiu h thngLecture 04

    16/18

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

    T#n$ =n 68n$ >?a $% t$(ng

    Zng Cng [4 vo tnh Kn ;nh

    Sh[c t! h thng x\ l] tn hiu c^n ph$i Kn ;nhY R thng2hng Kn ;nh g"@ r4 ngW r4 2hng .; gii hHn (thc t: .$o h_4) vi 1

  • 7/18/2019 Tn hiu h thngLecture 04

    17/18

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

    T#n$ =n 68n$ >?a $% t$(ng

    '( )y t

    ''( )y t

    ( )y t

    2 2

    0 0''( ) ( ) ( )y t y t t + =

    Zng Cng [4 vo tnh Kn ;nh

    M

    ''( ) ( ) ( )K K

    y t y t x tM M

    + =

    0

    K

    M = 2 2

    0 0''( ) ( ) ( )y t y t x t + =

    0 0( ) ( ) ( ) ( ) .sin( ) ( )x t t y t h t t u t = = =

    = (

  • 7/18/2019 Tn hiu h thngLecture 04

    18/18

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

    T#n$ =n 68n$ >?a $% t$(ng

    Zng Cng [4 vo tnh Kn ;nh:

    1 2 1 2;C C C R R R= = = =

    Bi(t)

    2

    2 2 2

    ( ) ( )3 1( )o o o

    d v t dv t Kv t

    dt RC RC dt R C

    + +

    2

    2 2 2

    ( ) ( )3 1( )i i i

    d v t dv t K v t

    dt RC dt R C

    = + +

    3

    4

    1 R

    KR

    = + 3K=Dhan: PQt: 01

    RC =

    2 2 2 2

    0 0 0( ) ( ) 3( 3 ) ( )o iD v t D v t + = + +