0
$\begingroup$

I want to ask a simple maybe stupid question, why in DFT the frequency bin size is limited to n/2+1, http://support.ircam.fr/docs/AudioSculpt/3.0/co/FFT%20Size.html. What I see from Wikipedia and also in complex discrete fourier transform, the sum should be from 0 -> n-1, here n is the sample size.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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.