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Homework Help: DFT Properties

  1. Jan 17, 2014 #1

    The first six values of the 10-point DFT of a real-valued sequence x(n) are given by

    {10, −2 + j3, 3 + j4, 2 − j3, 4 + j5, 12}

    Determine the DFT of x[n] = x[n+5] (10 point sequence)

    Relevant Equations:

    DFT(x[n-m]) = exp(-j*(2pi/N)*k*m) * X(k)
    where N = 10 ; m = -5

    Using the relevant equation calculating 6 points of the DFT from the shifted input is straight forward as X(k) is given 0<=k<=5 from the six point DFT series given.
    I am failing to see how to calculate the 10 point series however. My intuition is that it has something to do with the periodicity of the DFT but I cannot see any patterns emerging from shifting the input. The magnitudes of the coefficients don't seem to change but their angles due, Still I can not spot any patterns...
  2. jcsd
  3. Jan 17, 2014 #2
    I figured it out! In case anyone gets held up by something similar,

    By paying attention to the symmetry property of DFT for a real valued input X(n) = X*(N-n) where X represents the DFT coefficients and N is the fundamental period, in this case 10, the remaining coefficients (n = 6:9) can be determined
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