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Fourier transform Discreet time-shift

  1. May 21, 2006 #1
    suppose we have this discreet fucntion:

    x(k)=rect(k+N/2)= 1 ; when -N/2=<K<=N/2-1
    x(k)=0; otherwise

    This is discreet function(not continuous) of k shifted forward by N/2, we need to find fourier transform for it ..

    anyway let N=6 for simplicity, then:

    x(k)=rect(k+3)= 1 ; when -3=<K<=2
    x(k)=0; otherwise

    i think there's a law for finding FT of shifted signals but i can't remember ,and i need a guidance to get the solution

  2. jcsd
  3. May 21, 2006 #2
    First you must be careful and assert that N is even, because the function would be nothing if N was odd and it was shifted over by a fractional amount, which you did specify.

    To answer your question though, a positive phase shift of p corresponds to a shift of the transform by [tex]e^{\frac{2 \pi i k p}{N}}[/tex] where k is the counting variable, N is the period, so if x[n] [tex]\rightarrow X[k][/tex] then [tex]x[n+\frac{p}{2}] \rightarrow X[k] e^{\frac{\pi i k p}{N}}[/tex]. See http://en.wikipedia.org/wiki/Discrete_Fourier_transform

    Well I shouldn't have used the same counting variables for both x and its transform. Let me clean this up some.
    Last edited by a moderator: May 21, 2006
  4. May 22, 2006 #3
    so the solution would be like this:

    x(k)<<DFT...<< X(f)
    x(K+N/2) <<DFT....<< X(f).e^{(j*2*pi*k*N/2)/N}=X(f)*e^(j*pi*k)

    where k is counting variable,N number of samples, j: imaginary unit

    so the next step is to find X(f) of non-shifted disctreet function x(k)

    if you have more comments or corrections about the problem , please do post again here

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