How could it be that by FFT is not symmetric?
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Answers (2)
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See earlier Answer; the FFT is two-sided but there's only 1 DC component; not two so doubling 1:L/2 doubles it twice as it is already full magnitude which is where the initial asymmetry came from. If there is a significant DC component in the input signal, then that doubling is enough to be apparent as in your case; if it were tiny or took the DC component out before applying FFT the effect would likely not be noticed.
From the signal in the doc example, look at P2=abs(S/L) --
>> [P2(1:10);P2(end:-1:end-9)] ans = 1.0e-15 * 0.0600 0.0372 0.3311 0.3308 0.1617 0.2704 0.0999 0.1121 0.0886 0.2677 0.0372 0.3311 0.3308 0.1617 0.2704 0.0999 0.1121 0.0886 0.2677 0.1087 >>
NB: the symmetry after the first element (DC) to the end; there's no DC component at the end of the returned FFT vector.
Similarly for Fmax the symmetry in the complex FFT is about the L/2+1 point, not L/2 and so:
>> P2((L/2+1)-5:(L/2+1)+5) ans = 1.0e-15 * 0.3544 0.3263 0.1500 0.6822 0.2950 0.4871 0.2950 0.6822 0.1500 0.3263 0.3544 >>
NB: the Fmax element at L/2+1 is unique; not two of 'em and note its magnitude compared to each side.
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