[Math] Are DFT(x) and IDFT(x) complex conjugate

fourier analysis

I was playing a little with an FFT program I downloaded from the web, taking its source code as a basis for some experimentation. After reading a few texts on DFT/FFT, I was a little confused as to the sign of the exponent – some make it negative and some positive, for the DFT.

So, I decided to go and exchange the signs of the exponents (i.e., switch $\omega$ and its conjugate) in the FFT and IFFT functions. Then, I noticed that the resultant (supposedly frequency domain) vectors are just the conjugates of each other, and applying the inverse transform gave my the original, time domain vectors.

So, is it true to say that the DFT of a vector is the conjugate of the IDFT of the same vector, multiplied by the scale-factor?

$DFT(\hat{x}) = \alpha * conj(IDFT(\hat{x}))$

where ${IDFT(DFT(\hat{x})) = \hat{x}}$

and the $conj(\hat{y})$ operator is an array operation, working on each element of vector $\hat{y}$, and $\alpha$ is the appropriate scale factor (like $1/n$).

Best Answer

Not in general. If the sequence you are transforming is real, then yes (hint: look carefully at the definitions of the forward and inverse transforms), but the relation you propose is not true when the sequence you are transforming is complex.

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