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- Thread starter seamie456
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jasonRF

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[tex]

X_k = \sum_{n=0}^{N-1} x_n e^{-i 2 \pi \frac{k}{N} n} = DFT\left(x\right)_k

[/tex]

and it inverse DFT by

[tex]

x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k e^{+i 2 \pi \frac{k}{N} n} = IDFT\left(X\right)_n,

[/tex]

then the way to get [itex]x_n[/itex] from [itex]X_k[/itex] using a DFT would be something like,

[tex]

x_n = \frac{1}{N} \left(\sum_{k=0}^{N-1} X_k^* e^{-i 2 \pi \frac{k}{N} n}\right)^* = \frac{1}{N} \left( DFT \left( X^* \right) \right)_n^*

[/tex].

where the asterix represents conjugation. Does that make sense?

jason

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