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I've got a 4096 samples long 44.1 kHz audio-clip. After applying the FFT to it I get 4096 frequency bands.

Each band would then span 10.77 Hz (44100 / 4096).

I've been told the 2nd half of the frequencies is conjugate symmetric to the first half.

Considering this is my calculation above correct or did I miss something important?

Paul R
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1 Answers1

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That's pretty much correct - for most common complex-to-complex FFTs with purely real inputs (i.e. all imaginary parts zero) the first N/2 output bins (0..2047 in your case) are typically the only bins that you will be interested in. The first bin is DC (0 Hz), and bin N/2 corresponds to Nyquist (Fs/2 = 22.05 kHz), which is not normally of interest. Bins above N/2 are just complex conjugate "mirror images" of the bottom N/2-1 bins.

See this answer for more details.

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Paul R
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  • do you mean real-to-complex fft? the conjugate symmetry property is from the fact that the signal is real-valued. – thang Feb 16 '13 at 00:07
  • No, I was talking about a complex-to-complex FFT where the input is purely real, i.e. all imaginary parts zero. For real-to-complex FFT implementations the symmetry is implict as there are only N/2+1 outputs. I'll edit the answer to clarify this. – Paul R Feb 16 '13 at 08:19