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Homework Help: DFT of a finite lengt sequence

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the finite length x[n]= 2δ[n]+δ[n-1]+δ[n-3]

    We perform the following operation on this sequence:

    (i) We compute the 5-point DFT X[k]
    (ii) We compute a 5-point inverse DFT of Y[k]=X[k]2

    a) Determine the sequence y[n] for n= 0, 1, 2, 3, 4
    b) If N-point DFTs are used in the two step pocedure, how should we choose N so that y[n]=x[n]*x[n] for 0 ≤ n ≤ N-1?

    2. Relevant equations

    DFT transform pair:

    [tex]X[k]=\sum_{n=0}^{N-1}x[n]e^{(-j\frac{2π}{N}kn)}[/tex]

    [tex]x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{(j\frac{2π}{N}kn)}[/tex]

    3. The attempt at a solution

    a)

    [tex]X[k]=\sum_{n=0}^{4}x[n]e^{(-j\frac{2π}{5}kn)}=2 + e^{-j\frac{2π}{5}k} + e^{-j\frac{2π}{5}3k}[/tex]

    Then for calculation for Y[k] we substitute (2*pi/5)=a

    [tex]Y[k]= X[k]^{2} = 4 + 4e^{-jak}+2e^{-j4ak}+4e^{-j3ak}+e^{-j2ak}+e^{-j6ak}[/tex]

    Now to get y[n] through:

    [tex]y[n]=\frac{1}{N}\sum_{k=0}^{4}X[k]^{2}e^{(j\frac{2π}{5}kn)}[/tex]

    gives us a very complicated expression contained with sums of complex exponentials. It seems extremely difficult from here to simplify it so i can calculate y[n] for n=0,1,2,3,4
     
  2. jcsd
  3. Aug 20, 2012 #2
    Please post the expression that you obtained but found difficult to simplify. From what I remember in early college, these problems are usually a ton of tedious algebra, so you may not be far off.
     
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