I've came across the author saying that

... for a real cosine input having k cycles in the N-point input time sequence, the amplitude response of an N-point DFT bin in terms of the bin index m is approximated by the sinc function $$X(m)=\frac{A_{0}\color{red}{N}}{\color{red}{2}}\cdot\frac{\sin[\pi(k-m)]}{\pi(k-m)}$$ where $A_{0}$ is the peak value of the DFT’s input sinusiod.

I have no idea where $\frac{N}{2}$ came from. I'm wondering if you can give me some hints.

  • 1
    $\begingroup$ Does this help at all? $\endgroup$
    – Peter K.
    Apr 24, 2016 at 17:28
  • $\begingroup$ @PeterK. That seems what I've been looking for. Thank you very much. $\endgroup$
    – bae
    Apr 24, 2016 at 17:45

1 Answer 1


The common formulation of the forward DFT preserves energy (Parseval's theorem). This means that a longer constant magnitude sinewave input to a DFT, which has proportionally more energy, must be represented by a proportionally larger value in the DFT result, preserving that greater energy. Thus a factor of N, which scales with the length of the input.

For strictly real input, this energy is divided evenly between the positive frequency bin (k) and the negative frequency bin (N-k) of a DFT result (for a DFT of length N). These two complex result bins are always complex conjugates of each other (thus equal in magnitude) for strictly real input to a DFT. Thus another factor of 1/2 is included to do this even energy split.

N and 1/2 results in a factor of N/2 which has to appear somewhere in the result of an energy preserving formulation of a forward DFT.

Note that the 1/2 does not appear for bin 0 or for bin N/2 of even length DFTs, as these have no complex conjugate image (or rather they include both of their complex conjugates summed together into a single result bin).

Also note that this Sinc function result is an approximation. The actual result is closer to the sum of two Dirichlet functions (also called periodic Sinc functions). See here for more details.


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