I have just started working on FFT and its related phase factors.

I have read for an N=2 point FFT, We have two phase factors, one on the positive axis and one on negative axis.

similarly for an 4 point FFT, we have 4 phase factors as theta=(pi/2), at

 0'  W 40=1
-90' W 41=-j
180' W 42=-1
 90' W 43= j

here we can see the values that are 180 degree apart are negative to each other.

Here I wanted to know how it works for an 32 point FFT.

As theta=(pi/16) do we have 16 phase factors on the positive axis and 16 phase factors on the negative axis with the interval of (pi/16) ?

so will it follow the pattern as I mentioned below?

 0'       W16 0 =1
-11.25'   W16 1 =0.98 -j0.195
-22.5'    W16 2 =0.92 -j-0.38
and so on
-168.75'  W16 15=
and the negative parts of the above values from **W16 16** to **W16 32**

Am I correct? please verify the above details and let me know how the phase factors works for an 32 point FFT

Thanks in advance

  • $\begingroup$ please let me know if the question is not clear or understandable $\endgroup$
    – aram
    Commented Nov 28, 2013 at 16:56
  • 2
    $\begingroup$ What do you mean by the term "phase factors"? The cosine and sine (or real and imaginary) terms in any FFT bin can be combined to represent any phase in any length FFT. $\endgroup$
    – hotpaw2
    Commented Nov 28, 2013 at 23:07
  • $\begingroup$ @hotpaw2 Phase factor here i mean the 'twiddle factor' W, which describes a "rotating vector", which rotates in increments according to the number of samples, N. $\endgroup$
    – aram
    Commented Nov 29, 2013 at 10:16
  • $\begingroup$ more specifically it refers root-of-unity complex multiplicative constants in the butterfly operations of the Cooley-Tukey FFT algorithm, used to recursively combine smaller discrete Fourier transforms $\endgroup$
    – aram
    Commented Nov 29, 2013 at 10:30

1 Answer 1


The base functions of the discrete time Fourier transform takes on values from the set $$\{z\in\mathbb{C}:z^N=1\}$$ where $N$ is the dimension of the vector space. The solutions can be parametrized to be in the form $$z_k = \exp(2 \pi i \frac{k}{N})$$ for $k=0..N-1$ because $$z_k^N = \exp(2 \pi i k \frac{N}{N}) = \exp(2\pi i k) = 1$$

In other words, the values the base functions components are chosen from are the $N$ non-identical $N$-th roots of 1, which are equidistantly distributed on the unit circle and include $1$ itself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.