0
$\begingroup$

I was looking into proof and find something strange:

Proof

The last part we obtain from DFT definition. $$X[k] = \sum^{N-1}_{n=0}x[n]W^{kn}_N, \quad\text{Where}\quad W^{kn}_N = e^{-j\frac{2\pi}{N}nk}$$

So, looking at $W^{-k}_NX[N-k]$ we can see that when $k = 0$ we will have $W^{0}_NX[N]$ and we know that our DFT coefficients only can be from $[0,N-1]$.

So my question is: what does it mean when we have $X[N]$ DFT coefficient? Or did i understand something wrong?

|improve this question
$\endgroup$
  • $\begingroup$ There are two different visions of what time reversal means in the discrete domain. In one, $x_r[n] = x[N-1-n]$ which replaces $\mathbf x = \big[x[0], x[1],\ldots, x[N-1]\big]$ with $\mathbf x_r = \big[x[N-1], x[N-2],\ldots, x[1], x[0]\big]$ and the other which has $x_r[n] = x[-n]$ which replaces $\big[x[0], x[1],\ldots, x[N-1]\big]$ with $\big[x[0], x[N-1], x[N-2],\ldots, x[1]\big]$ $\endgroup$ – Dilip Sarwate Dec 11 '17 at 3:16
2
$\begingroup$

You can evaluate $X[k]$ (and $x[n]$) outside the domain $[0,N-1]$. The sequences are periodic with period $N$, so $X[N]=X[0]$, and, more generally, $X[k]=X[k+N]$.

|improve this answer
$\endgroup$
  • $\begingroup$ yeah, thank you, this came to my head after posting a question) $\endgroup$ – qqffx Dec 10 '17 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.