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643 B
643 B
Output of FFT:
{[DC,NY],C[1],C[2],...,C[n/2]}
On a sample basis
Let ps stand for "per sample"
C[1]
gives the 1/N
ps coefficient
C[2]
gives the 2/N
ps coefficient
C[N/2]
gives the 1/2
ps coefficient
In seconds
Let f
(=44100) be the samplerate. Then:
C[1]
gives the f/N
hz coefficient
C[2]
gives the 2f/N
hz coefficient
C[N/2]
gives the f/2
hz coefficient
From herz to position
Let h
be some frequency. Then from the above we see that for m := Nh/f
we get the mf/N = h
hz coefficient in C[m]
.