Flip operator's equality to Double Fourier

  • Thread starter valjok
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  • #1
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Main Question or Discussion Point

Some authors say that applying Fourier transform twice flips the vector, F2[x(t)] = x(-t). Yet, the simple checks proves this wrong. For instance, take 2x2 DFT:

[tex]\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]^2 = \left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]
[/tex]

The Identity is different from counter-identity. It cannot therefore flip the vector elements. Where is the mistake?
 

Answers and Replies

  • #2
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For the size-n DFT, (F2x)(t) = x(-t), where x is a periodic function with period n.

In the case n = 2, there is no mystery, since x(-0) = x(0) and x(-1) = x(-1 + 2) = x(1), since x has period 2.
 
  • #3
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This is a great answer!
 

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