FFT function and buffer size

[luke514]

Watching some videos online about FFT and trying to use (in STM32CubeIDE) the arm_rfft_fast_f32 function of the arm_math.h DSP library, it is not clear to me why the buffer output produced by that function is N/2 long and not N

Is it something related to Shannon's theorem? To the symmetry of the spectrum? (i.e. is it symmetrical around the origin).

You are welcome to share external links if there are detailed explanations that I did not see during my internet search.

[Sarra.S · Accepted solution]

Hello @luke514, 

The output buffer length is N/2 due to the properties of the FFT itself 

>>To the symmetry of the spectrum?

yes, the spectrum is symmetric around the Nyquist frequency, which is half the sampling rate. so, the second half of the FFT output is a mirror image of the first half.

Typically, The buffer's size must correspond to a power of 2