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Problems with DFT

  1. Jan 13, 2017 #1
    1. The problem statement, all variables and given/known data
    The 2-point DFT of x[n] is given by the expression X[k] = 2δ[k], for k = 0, 1.

    2. Relevant equations
    What is x[n]?

    3. The attempt at a solution
    i know the answer but i don't know how they calculate it
    because a delta pulse is only one at k=zero.
    [itex] x\left[n\right]=\frac{1}{N}\sum _{k=0}^{N-1}X\left[k\right]\cdot e^{j\cdot \frac{2\pi }{N}kn}\: [/itex]
    because k=0
    [itex] e^{j\cdot \frac{2\pi }{N}kn}\: = 1 [/itex]
    So i come to an answer where x[n] = δ[n]
    But it is: x[n] = δ[n]+δ[n−1] ??? How to they calculate that, i don't understand it
     
  2. jcsd
  3. Jan 13, 2017 #2

    Henryk

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    Gold Member

    i'm not sure if I understand the problem correctly, but it looks to me that the FT of the signal has only two frequency component, at k = 0 and k=1.
    If that is correct, the sum in your attempted solution contains only two non-zero terms: k = 0, and k = 1, both of them are equal to 2.
    So, ##x(n) = 2 \cdot e^{j \frac {2 \pi}N 0 \cdot n} + 2 \cdot e^{j \frac {2 \pi}N1 \cdot n}##

    Hope that helps
     
  4. Jan 14, 2017 #3
    uhh no because the answer is x[n] = δ[n]+δ[n−1]
     
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