I want to ask a simple maybe stupid question, why in DFT the frequency bin size is limited to $n/2+1$


What I see from Wikipedia and also in complex discrete fourier transform, the sum should be from $0 \rightarrow n-1$, here $n$ is the sample size.

  • $\begingroup$ You may use $N$ for the length and $n$ for indexing. $\endgroup$
    – jomegaA
    Feb 11, 2020 at 21:03

1 Answer 1


The sum in the definition of a length $N$ DFT always goes from $n=0$ to $n=N-1$:

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N},\qquad k=0,1,\dots,N-1\tag{1}$$

However, if the sequence $x[n]$ is real-valued, which is the case for most applications, then the $N$ DFT bins $X[k]$ are not independent of each other:

$$X[k]=X^*[N-k],\qquad k=0,1,\dots,N-1\tag{2}$$

So for even $N$, only the first $N/2+1$ bins carry information, the remaining $N/2-1$ bins can be computed from $(2)$.

Note that for real-valued $x[n]$, the values $X[0]$ and $X[N/2]$ are real-valued, and the remaining $N/2-1$ values for $k=1,2,\ldots,N/2-1$ are generally complex-valued. This means that $N$ real-valued samples $x[n]$ are represented by $N$ real-valued numbers in the frequency domain (2 real numbers plus $N/2-1$ complex numbers). This shouldn't come as a surprise.


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