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Let $X=\{X_k\}^{N-1}_{k=0} $ be the Discrete Fourier Transform of $\{ x\} = \{ x_n \}^{N-1}_{n=0} $

what is the relationship between $X$ and Discrete Fourier Transform of $\overline{x}$


Definition Discrete Fourier

Let $a$ discrete function of $N$ samples $\forall n = 0,1, \dots , N-1, f[n].$

The DFT of $f$ denotes $\widehat{f} [x]$ is defined by $$ \widehat{f}[x] = \sum^{N-1}_{n=0}f[x] e^{-2i \pi \frac{nk}{N}}$$

Inversion Discrete Fourier

Given discrete fourier transfomr $\widehat{f}[x]$ of a discrete function $f[n]$ the inverse of $DFT$ is given by

$$f[n]=\frac{1}{N} \sum^{N-1}_{k=0} \widehat{f}[k] e^{2 i \pi \frac{nk}{N}} $$


Scratchwork

$$X = \sum^{N-1}_{n=0} x_n e^{e i \pi \frac{nk}{N}}$$

now finding the relationship

$$ \sum^{N-1}_{n=0} \overline{x_n} e^{-2i \pi \frac{nk}{N}} =\sum^{N-1}_{n=0} \overline{x_n e^{2i \pi \frac{nk}{N}}} = \overline {\sum^{N-1}_{n=0}x_n e^{2i \pi \frac{nk}{N}}} $$

and we have that

$$x= \frac{1}{N} \sum^{n-1}_{N=0} X_n e^{2i \pi \frac{nk}{N}} $$

Not sure where to go from there besides double summation

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    $\begingroup$ When you write $\overline{x}$ do you mean the average value of the $x_n$, $\frac{1}{N} \sum_{n=0}^{N-1} x_n$ or do you mean the pointwise complex conjugate, $\{\overline{x_n}\}_{n=0}^{N-1}$? $\endgroup$ – Eric Towers Apr 24 '18 at 1:58
  • $\begingroup$ it means conjugation in complex analysis @EricTowers. $\endgroup$ – Tiger Blood Apr 24 '18 at 3:24
  • $\begingroup$ looking at it back not sure X is the sum it might just be like the term $\endgroup$ – Tiger Blood Apr 24 '18 at 3:30
  • $\begingroup$ Your definitions are not stated correctly. The DFT seems to be just a complex number multiplied by $f(x)$. Instead of $f(x)$ on the right side there should be something depending on $n$. $\endgroup$ – Kavi Rama Murthy Apr 24 '18 at 6:14
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The DFT of an $f:{\mathbb Z}_N\to{\mathbb C}$ is defined as $$\widehat f(k):=\sum_{n=0}^{N-1}f(n)e^{-2\pi i n k/N}\ .$$ Consider now the function $\bar f$ defined by $\bar f(n):=\overline{f(n)}$. It is a simple calculation involving complex conjugates only to show that $$\widehat{\bar f}(k)=\overline{\widehat f(-k)}\qquad(k\in{\mathbb Z}_N)\ .$$

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