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  • 1

  • LOGO

    iugaza2010.blogspot.commelasmer@gmail.com

    Discrete Fourier Transform

    (DFT)

    Part Two

  • 3

    Use the DFT in linear filter

    Suppose thar we have a finite-duration sequence x(n) of length L ,

    which excites an FIR filter ,with impulse response h(n) of length M

    The output sequence y(n) of the FIR filter can be expressed in time

    domain as the linear convolution of x(n) and h(n).

    If size of x(n)=

    If size of h(n)=

    ( ) ( ) ( )

    To avoid aliasing , we need to take DFT of N points

    where: N L+M 1

    L

    M

    Y k X k H k

  • 4

    2

    2

    ( ) ( ) |

    ( ) ( ) |

    k

    N

    k

    N

    Y k Y k = 0,1,.......,N 1

    X H k = 0,1,.......,N 1

  • 5

    Zero padding

  • 6

    Zero padding of analyzed sequence results in approximating

    its DTFT better

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    Determine the response of

    FIR filter with impulse response: h(n)={ ,2,3}

    to the input x(n)={ ,2,

    1

    1 2,1}

    Ex:

    (a) Using linear convolution(chapter 2)

    ] ;

    ] ;

    1221

    0321

    0000

    03663

    002442

    0001221

    038119411

  • 11

    h(n)={ ,2,3} , 1 1x(n)={ ,2,2,1}

    (b) Using DFT&IDFT (without zero padding )

    2 2

    44

    2 3

    4 4 4

    4 2 4 6

    4 4 4

    3 6 9

    4 4 4

    1 1 1 1 1 1 1 1

    1 1

    1

    1

    j jN

    NW e W e j

    W W WW

    W W W

    W W W

    2 3

    2 4 6

    3 6 9

    1 1 1 1

    ( ) ( ) 1 1

    1 11 ( ) ( ) ( )

    1 ( ) ( ) ( )

    j j j j j

    j j j

    j j j

    1 1

    1 1

    1 1

    1 1 1 1

    1 1

    1 1 1 1

    1 1

    N N N

    j j

    j jX W x

    j

    1

    1 6

    2 2 2( )

    3 2

    0 2 2

    jX k

    j j

  • 12

    2 2 2

    1 1 1 1 1 6

    1 1 2 1 ( )

    1 1 1 1 2 0

    1 1 1 1

    N N N

    j j jX W x X k

    j j j

    1 2

    1 2

    ( ) ( ) ( )

    ( 0) (0) (0) (6)(6) 36

    ( 1) ( 1 )( 2 2 ) 4

    ( 2) (0)(2) 0

    ( 1) ( 1 )( 2 2 ) 4

    ( ) {36,4 ,0, 4 }

    Y k X k X k

    Y k X X

    Y k j j j

    Y k

    Y k j j j

    Y k j j

  • 13

    1

    1

    4 4 ,4

    1 1 1 1 36

    1 1 4

    1 1 1 1 0

    1 1 4

    m

    j j jx W X

    j j j

    1/ 4 1/ 4 1/ 4 1/ 4

    1/ 4 / 4 1/ 4 / 4 =

    1/ 4 1/ 4 1/ 4 1/ 4

    1/ 4 / 4 1/ 4 / 4

    j j

    j j

    3 1 2

    36 9

    4 7

    0 9

    4 11

    ( ) ( ) ( ) {9,7,9,11}

    j

    j

    x m x n x n

    2

  • 14

    h(n)={ ,2,3} , 1 1x(n)={ ,2,2,1}

    (c) Using circular convolution formula

    1 * 1 2 2 1 1 2 2 1

    2 * 1 1 2 2 2 2 4 4

    3 * 2 1 1 2 6 3 3 6

    0 * 2 2 1 1 0 0 0 0

    9 7 9 11

    3

  • 15

    1* 1 2 2 1 0 0 1 2 2 1 0 0

    2* 0 1 2 2 1 0 0 2 4 4 2 0

    3* 0 0 1 2 2 1 0 0 3 6 6 3

    0* 1 0 0 1 2 2 0 0 0 0 0 0

    0* 2 1 0 0 1 2 0 0 0 0 0 0

    0* 2 2 1 0 0 1 0 0 0 0 0 0

    1 4 9 11 8 3

    (d) Using circular convolution (with zero padding)

    4

  • 16

    L=3,M=4 N=3+4 1 6 N at least must be 6 take N=8

    h(n)={ ,2,3,0,0,0} , x(n)={ ,2,2,11 1 ,0,0}

    (e) Using DFT&IDFT (with zero padding)

    27

    8

    0

    3

    4 2 4( ) 1 2 2 , k=0,1,2,...

    : ( ) ( ) , k=0,1,2,.......,7

    6 , 1 1.707 4.21

    2 1 , 3 0.29 0.1

    ...

    2132

    4

    .,7k k k

    j j

    nj

    n

    j

    k

    DFT X k x n e

    X(k = 0) = X(k = ) = j

    X(k = ) = j X

    X k e e

    (k = ) =

    X )

    e

    j

    (k =

    0 , 5 0.29 0.12132

    6 1 , 7 1.707 4.21

    = X(k = ) = j

    X(k = ) = j X(k = ) = j

  • 17

    27

    8

    4

    0

    2

    : ( ) ( ) , k=0,1,2,.......,7

    6 , 1 2.414 4.414

    2 2 2 , 3 0.4

    ( ) 1 2 3 , k=0,1

    14 1.585

    ,2,......

    8

    4

    .

    2

    ,7

    nkj

    n

    k kj j

    DFT H k h n e

    H(k = 0) = H(k = ) = j

    H(k = ) = j H(k = ) = j

    X k e e

    H(k = ) =

    , 5 0.414 1.5858

    6 2 2 , 7 2.414 4.414

    H(k = ) = j

    H(k = ) = j H(k = ) = j

  • 18

    ( ) ( ) ( )

    ( 0) (0) (0) 36 , (1) 14.07 17.48

    (2) 4 , (3) 0.07 0.515

    (4) 0 , (5) 0.07 0.515

    (6) 4

    Y k X k H k

    Y k X H Y j

    Y j Y j

    Y Y j

    Y j

    , (7) 14.07 17.48 Y j

    27

    8

    0

    1: y(n)= ( ) , n=0,1,2,.......,7

    8

    ( ) {1, 4,9,11,8,3,0,0}

    nkj

    k

    IDFT X k e

    y n

    5

  • 19

    Using linear convolution

    ( ) { ,4,9,11, }1 8,3y n

    Using DFT&IDFT or circular Con.(without zero padding)

    ( ) { ,7,9 11}9 ,y n

    Using DFT&IDFT or circular Con.(with zero padding)

    ( ) { ,4,9,11, }1 8,3y n

  • (0) : (0) (4) 1 8 9

    (1) : (1) (5) 4 3 7

    (2) : (2) (6) 9 0 9

    (3) : (3) (7) 11 0 11

    index y y

    index y y

    index y y

    index y y

    ( ) { ,7,9 11}9 ,y n

    ( ) { ,4,9,11,8,3,1 0,0}y n

    Aliasing occurs when the size of DFTs is smaller than L+M-1

  • 21

    DFT repeats itself every N points (Period = N) but we usually display it for n = 0 ,, N-1

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