$X$ represents sample in frequency domain and $x$ represents samples in time domain.


$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $


$ X[k] = \frac{1}{N} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $


$ X[k] = \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1} x[n] \ e^{-j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} X[k] \ e^{j \frac{2\pi}{N} n k} $


$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{j \frac{2\pi}{N} n k} $

$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{-j \frac{2\pi}{N} n k} $

Please observe the scaling factor $\frac{1}{N}$ and change of a negative sign $-$ over the exponent term ${e^{\pm j\frac{{2\pi }}{N}nk} }$. Why is every notation of DFT valid?

  • $\begingroup$ Who told you they are all valid definitions? $\endgroup$ Aug 19 '14 at 14:32
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    $\begingroup$ I have seen the above notations in some research papers (IEEE letters & transactions on wireless, signal processing). Please could you explain why the above notations are not valid for DFT? $\endgroup$ Aug 20 '14 at 3:20

It simply does not matter and it is just a matter of convention which version is used. I always use the first definition, which is also the one you will find in (almost) all textbooks on DSP. If you take always the same time domain signal $x[n]$ then it is obvious that $X_2[k]$ (according to notation 2) and $X_3[k]$ (according to notation 3) are simply scaled versions of $X[k]$ (according to notation 1). The one with scaling $1/\sqrt{N}$ is the unitary version of the DFT (i.e. the DFT becomes a unitary transformation). It's just notation number 4 that is a bit weird. In this case, $X_4[k]=X[-k]$, but I have never seen the sign convention like this. In this case, $x[n]$ can be interpreted as the Fourier series coefficients of $X[k]$. But again, all four versions are OK and you're free to use any of them as long as everybody knows which version of the DFT you mean.


likewise to @MattL.'s answer, just observe that $x[n]$ in NOTATION 2 is scaled to be $N$ times larger than the $x[n]$ in NOTATION 1. and $x[n]$ in NOTATION 3 is scaled to be $\sqrt{N}$ times larger than the $x[n]$ in NOTATION 1. that's all that change of convention means.

in NOTATION 4, just note that there is no operational difference between $-j$ and $+j$. both numbers are purely imaginary and they both have equal claim to squaring to be $-1$ or have equal claim to being the $\sqrt{-1}$ (stating it loosely). these two numbers are not equal, but they are equivalent. they have identical properties (which are simply that they are not real numbers and they square to be $-1$).

so as long as all occurrences of $j$ are replaced by $-j$ (which means that all occurances of $-j$ are replaced by $+j$) in all of the theorems that come with the DFT, then NOTATION 4 will be fine as a convention and is just as valid as NOTATION 1.

in both cases, the scaling case and the case of swapping of $-j$ and $+j$, note that the theorems associated with the DFT must also be modified slightly for them to be "equally valid" conventions.

  • $\begingroup$ Although Notation 4 isn't quite the normal, you can relate it to the other definitions by letting $m=-n$ and then rewriting the equation. Essentially it swaps what are normally the time domain and the frequency domain. You can also show that the DFT(DFT(x(n)))=x(-n). That is, if you take the DFT of the DFT you end up with a time reversed sequence. $\endgroup$
    – David
    Aug 21 '14 at 12:13

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