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DFT symmetry property?

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    As an example, if we find a DFT of x[n]={1,1,0,1}
    the result will be X(m)={3,1,-1,1}


    2. Relevant equations

    My Question is that as we know DFT holds symmetry property, why this answer does not void for that property?
     
  2. jcsd
  3. Feb 21, 2012 #2
    Let's take the definition from Wikipedia:
    [tex]X_k=\sum_{n=0}^{N-1}x_n\,e^{-i2\pi kn/N}[/tex]
    For real data you have then the symmetry:
    [tex]X_k=\bar{X}_{N-k},\, (k=1,...,N).[/tex]
    In your case [itex]N=4[/itex], and you have [itex]X_0=3,X_1,=1,X_2=-1,X_3=1,[/itex] all real.

    From the symmetry property you should have [itex]X_4=X_0,X_3=X_1,X_2=X_2[/itex].
    And you have it ([itex]X_4[/itex] can be thought of as defined by the symmetry property).
     
    Last edited: Feb 21, 2012
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