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DIGITAL SIGNAL PROCESSING

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Let x(n) be a sequence of length N and X(k) is N ... X(z) = z-n , ROC include unit circle. by defining zk = ej2 k/N, k = 0, 1, 2, ..., N-1 ... – PowerPoint PPT presentation

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Title: DIGITAL SIGNAL PROCESSING


1
DIGITAL SIGNAL PROCESSING
  • CHAPTER 8
  • DISCRETE FOURIER TRANSFORM
  • (DFT)

2
CHAPTER 8 DFT DEFINATION
  • To perform frequency analysis on a discrete-time
    signal, x(n),
  • need to convert the time-domain to an
    equivalent frequency-domain representation. In
    order to this, need to use a powerful
    computational tool to perform frequency analysis
    called
  • Discrete Fourier Transform or DFT.
  • The continuous Fourier Transform is defined as
    below
  • However, this integral equation of Fourier
    Transform is not suitable to perform frequency
    analysis due to this 2 reasons
  • Continuous nature can be handled by Computer
  • The limits of integration cannot be from minus
    infinity to infinity. There should be a finite
    length sequences that can be handled by computer.

3
CHAPTER 8 DFT
DEFINATION
  • Let x(n) be a finite length sequences.
  • Thus, the N-point DFT of x(n) defined as X(k)
    is
  • X(k) e-j2?nk/N, k 0,1,
    , N-1
  • k represent the harmonic number of the transform
  • component.
  • n is the finite length sequence interval defined
    as
  • 0 n N-1, N is the sequence length
  • Thus X(k) being complex and has real imaginary
    component, so that the kth harmonic of X(k) is
  • X(k) R(k) jI(k)

4
CHAPTER 8 DFT PROPERTIES
  • The are 4 properties of DFT
  • 1. Periodicity
  • If X(k) is the N-point DFT of x(k),
  • x(nN) x(n), for all n
  • X(kN) X(k), for all k
  • It shows that DFT is periodic with period
    N, also known
  • as Cyclic property of the DFT
  • 2. Linearity
  • If X1(k) and X2(k) are the N-point DFT
  • of x1(n) and x2(n),
  • ax1(n) bx2(n) DFT aX1(k)
    bX2(k)

5
CHAPTER 8 DFT PROPERTIES
  • 3. Circular Shifting
  • Let x(n) be a sequence of length N and
    X(k) is N
  • -point DFT, thus the sequence, x,(n)
    obtained from
  • x(n) by shifting x(n) cyclically by m
    units. Then,
  • x,(n) DFT
    X(k)e-j2?km/N
  • 4. Parsevals Theorem
  • if x(n) DFT X(k) and
  • y(n) DFT Y(k)
  • thus, y(n) 1/N
    Y(k)

6
CHAPTER 8 DFT
RELATIONSHIP WITH z-TRANSFORM
  • The z-transform of the sequence, x(n) is given
    by
  • X(z) z-n , ROC include unit
    circle
  • by defining zk ej2?k/N, k 0, 1, 2, ,
    N-1
  • X(k) X(z) zk ej2?k/N , k 0,1,2, ,
    N-1
  • e-j2?nk/N
  • where ?k 2?k/N, k 0,1,2,,N-1

7
CHAPTER 8 DFT
CONVOLUTION
  • To perform convolution for DFT, need to do the
    following
  • 1. Find N-point DFT of the sequence h(n) and
  • x(n).
  • 2. Multiply DFT to form Y(k) H(k)X(k)
  • 3. Perform inverse DFT to obtain y(n).

8
CHAPTER 8 DFT EXAMPLES
  • EXAMPLE 1
  • Find the DFT for the following finite length
    sequence,
  • x(n) ¼, ¼, ¼
  • Solution
  • 1. Determine the sequence length, N
  • N 3, k 0,1,2
  • 2. Use DFT formula to determine X(k)
  • X(k) e-j2?nk/N , k 0,
    1, 2
  • X(0) ¼ ¼ ¼ ¾
  • X(1) ¼ ¼e-j2?/3 ¼e-j4?/3
  • ¼ ¼ cos (2?/3) jsin(2?/3)
    ¼ cos (4?/3)
  • jsin(4?/3)
  • ¼ ¼ -0.5 j0.866 ¼
    -0.5 j0.866
  • ¼ ¼ -1 0

9
CHAPTER 8 DFT EXAMPLES
  • Continued from Examples 1
  • X(2) ¼ ¼ e-j4?/3 ¼ e-j8?/3
  • ¼ ¼ cos (4?/3) jsin(4?/3)
  • ¼ cos (8?/3) jsin(8?/3)
  • ¼ ¼ -0.5 j0.866 ¼-0.5
  • j0.866
  • 0
  • Thus,
  • X(k) ¾, 0, 0

10
CHAPTER 8 DFT EXAMPLES
  • Examples 2
  • Given the following the finite length
    sequences,
  • x(n) 1,1,2,2,3,3
  • Perform DFT for this sequences.
  • Solution
  • 1. Determine the sequence length, N 6.
  • 2. Use DFT formula to determine X(k).
  • X(k) e-j2?nk/N , k
    0,1,2,3,4,5
  • X(0) 12, X(1) -1.5 j2.598
  • X(2) -1.5 j0.866, X(3) 0
  • X(4) -1.5 j0.866, X(5) -1.5 j2.598
  • Thus,
  • X(k) 12, -1.5 j2.598, -1.5 j0.866, 0,
    -1.5 j0.866,
  • -1.5 j2.598

11
CHAPTER 8 DFT EXAMPLES
  • Examples 3
  • Find the DFT for the convolution of 2
    sequences
  • x1(n) 2, 1, 2, 1 x2(n) 1, 2, 3, 4
  • Solution
  • 1. Determine the sequence length for each
  • sequence, N 4. Thus, k 0,1,2,3
  • 2. Perform DFT for each sequences,
  • (i) X1(0) 6, X1(1) 0, X1(2) 2, X2(3)
    0
  • X1(k) 6,0,2,0
  • (ii) X2(0) 10, X2(1) -2j2, X2(2)
    -2, X2(3) -2-j2
  • X2(k) 10,-2j2,-2,-2-j2
  • 3. Perform Convolution by
  • X3(k) X1(k) X2(k)
  • 60, 0, -4, 0

12
CHAPTER 8 IDFT DEFINATION
  • IDFT is the inverse Discrete Fourier Transform.
  • The finite length sequence can be obtained from
    the Discrete Fourier Transform by performing
    IDFT.
  • The IDFT is defined as
  • x(n) 1/N e-j2?nk/N,
  • where n 0,1, , N-1

13
CHAPTER 8 IDFT EXAMPLES
  • EXAMPLES 4
  • Determine the IDFT for the following DFT
    sequence,
  • X(k) 1, 2, 3, 4
  • SOLUTION
  • 1. Determine the length of the sequence, N
    4
  • 2. Calculate the IDFT by the IDFT formula
  • x(n) 1/4 e-j2?nk/4,
  • x(0) ¼(1 2 3 4) 5/2
  • x(1) -0.5 j0.5, x(2) -0.5
  • x(3) -0.5 j0.5
  • 3. Thus the finite length sequence, x(n) is
  • x(n) 2.5, -0.5-j0.5, -0.5, -0.5j0.5

14
CHAPTER 8 IDFT EXAMPLES
  • EXAMPLES 5
  • Obtain the finite length sequence, x(n) from
    the DFT sequence in Example 3.
  • Solution
  • 1. The sequence in Example 3 is
  • X3(k) 60, 0, -4, 0
  • 2. Use IDFT formula to obtain x(n)
  • x3(n) 1/4 e-j2?nk/4,
  • x3(0) 14, x3(1) 16, x3(2)
    14, x3(3) 16
  • Thus the finite length sequences are
  • x3(k) 14, 16, 14, 16

15
CHAPTER 8 DFT IDFT
COMPLEXITY OF DFT
  • A Large number of multiplications and additions
    are required to compute DFT.
  • To compute the 8-point of DFT of the sequence,
    x(n),
  • The X(k) will be the summation of
  • x(0)e-j2?(0)k/8 until x(7)e-j2?(7)k/8
  • For the eight terms, there will be 64
    multiplication (82) and 56 addition (8 x (8-1))
  • Hence, for N-point DFT, there will be N2
    multiplication and N(N-1) addition.

16
CHAPTER 8 DFT IDFT
COMPLEXITY OF DFT
  • Thus, need one algorithm to reduce the number of
    calculation and speeds up the computation of DFT.
    The algorithm is called Fast Fourier Transform
    (FFT). It utilizes special properties of the DFT
    to construct a computational procedure that
    requires Nlog2N complex multiplication and Nlog2N
    complex addition to carry out N-point DFT.
  • The MATLAB command use to perform DFT IDFT is
  • 1. fft2 () - for DFT
  • 2. ifft2 () - for IDFT

17
CHAPTER 8 FFT
DEFINATION
  • FFT is the algorithm that efficiently computes
    the DFT.
  • In applying FFT, the DFT expression can be
    written as
  • X(k) WNnk , k 0, 1,,N-1
  • where, WN e-j2?/N

18
CHAPTER 8 IFFT DEFINATION
  • The FFT for IDFT can be defined as
  • x(n) 1/N WNnk
  • WN2 (e-j2?/N)2 e-j2?2/N WN/2

19
CHAPTER 8 DFT IDFT
SUMMARY
  • The DFT IDFT can be summarized below
  • 1. It is a powerful method to perform frequency
  • analysis which are used widely in digital
  • image processing including blurring and
    enhancing.
  • 2. Since the DFT IDFT will become tedious
  • when the length of the sequence become
  • big, one algorithm is develop to overcome
  • this problem.
  • 3. The algorithm can be found in MATLAB. The
    function
  • are
  • 1. FFT2 to perform DFT
  • 2. IFFT2 to perform IDFT

20
DIGITAL SIGNAL PROCESSING
  • THATS ALL.
  • I HOPE YOU CAN LEARN SOMETHING FROM THIS COURSE
    ESPECIALLY FOR APPLICATION RELATED TO DSP.
  • HAVE A NICE LONG HOLIDAY.
  • GOOD LUCK FOR YOUR FINAL EXAM.
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