# Are DFT(x) and IDFT(x) complex conjugate?

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 $1/n$).

• What happens when you just work out an example? Choose a dimension $>2$! Commented Apr 8, 2012 at 20:41
• @draks - as I wrote, when working on an example where n=8, I got the result as described. I am trying to understand if this a coincidence or a rule. Unfortunately, I don't have Matlab available to do some massive checks.
– ysap
Commented Apr 8, 2012 at 21:29
• To clarify - is it a coincidence, like depending on the type of FFT algorithm [here, DIT DFTrn as far as I could identify the code] or a rule.
– ysap
Commented Apr 8, 2012 at 21:36