DFT of some functions

Asked Apr 02 '13 at 01:05

Active Apr 02 '13 at 01:05

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Given the DFT pair $x[k]$ and $X[r]$, for a sequence of length N, express the DFT of the following sequences as a function of X[r]:

$$ y[k]=x[2k]$$

I guess this is a simple question, but I can't find the trick to solve it. What I can do is:

$$ Y[k']=\sum_{k'=0}^{N-1}y[k']e^{-2j\pi k'r/N}=\sum_{2k=0}^{2(N-1)}x[2k]e^{-2j\pi (2k)r/N}=\sum_{k=0}^{(N-1)}x[k]e^{-2j\pi (k)r/N}=X[r]$$ I don't know if this is correct? Anyone help? Thanks.

asked Apr 02 '13 at 01:05

Cheung

Is $N$ divisible by 2? – copper.hat Apr 02 '13 at 01:09
Let's say yes... – Cheung Apr 02 '13 at 01:12
If $2\not\mid n$, the following looks relevant: http://www.mirlab.org/conference_papers/International_Conference/ICASSP%202010/pdfs/0003658.pdf – copper.hat Apr 02 '13 at 02:58

DFT of some functions

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