1. III Digital Audio
III.4 (Mo Oct 16) Fast Fourier Transform (FFT)

2. Recall these facts: The representation
wr = w(rΔ) = w(r/N) = ∑m = 0, 1, 2, 3, ... N-1 cm e i2.mr/N
identifies the sequence w = (w0,w1,w2,…,wN-1) as a vector in the N-dimensional complex space ¬N. So our samples of fundamental frequency f = 1 are identified with the vectors w ∈ ¬N. We have a scalar product — similar to the highschool formula (u,v) = |u|.|v|.cos(u,v):
Have N exponential functions e0, e1, e2,... eN-1 that are represented as vectors in ¬N
em = (em(r) = ei2.mr/N)r = 0,1,2,...N-1
〈em, em〉 = 1, 〈em, ek〉 = 0 m ≠ k = orthogonality relations mentioned above!
The e0, e1, e2,... eN-1 = orthonormal basis like for normal 3 space! (ortho ~ perpendicular, normal ~ length 1)
They replace the sinusoidal functions!
90o em el ek u v

3. cm = 〈w, em〉 = 1/N.∑r = 0, 1, 2, 3, ... N-1 wr e-i2.mr/N
And this:
Every sound sample vector w = (w0,w1,w2,…,wN-1) in ¬N can be written as a linear combination
w = ∑m = 0, 1, 2, 3, ... N-1 cm em
of the exponential functions, and the (uniquely determined) coefficients cm are calculated via
cm = 〈w, em〉 = 1/N.∑r = 0, 1, 2, 3, ... N-1 wr e-i2.mr/N
90o em el ek
So we have to calculate the coefficients cm! The big question is:How fast can we do this?
N can be very large!! Think of CD: samples/second

4. cm = 〈w, em〉 = 1/N.∑r = 0, 1, 2, 3, ... N-1 wr e-i2.mr/N
The Fast Fourier Transform is an algorithm for a faster calculation of the coefficients
cm = 〈w, em〉 = 1/N.∑r = 0, 1, 2, 3, ... N-1 wr e-i2.mr/N
It was published 1965 in a 5-page (!) paper in Math. Comput. 19: 297–301 entitled
An Algorithm for the Machine Calculation of Complex Fourier Series
by James W. Cooley (IMB T.J. Watson Research Center) and John W. Tuckey (Princeton University and AT&T Bell Labs). This is one of the most cited papers.
The algorithm was used already 1805 by the great mathematician Carl Friedrich Gauss for astronomical calculations, but not recognized since his text was written in Latin…
Let us now see how much faster the FFT algorithm is. And faster than what?
John W. Tuckey
Carl F. Gauss

5. For the calculation of the N values
cm = 〈w, em〉 = 1/N.∑r = 0, 1, 2, 3, ... N-1 wr e-i2.mr/N
we suppose that the values
wr for all r = 0,1,... N-1
E(N) = e-i2/N
are all known. We have to calculate the powers
e-i2.2/N = E(N)2,... e-i2.(N-1)/N = E(N)N-1
by N-2 multiplications.
And then for all m, r the N2 multiplications
wr × e-i2.mr/N
And we have to add N-1 times the products and make one division by N per m, adding up to N2 additional operations.
This all adds up to
2N2 + N operations, which grows in a quadratic manner with respect to N.
Example: N = 44100: 2N2 + N-2 = ≈ Billions

6. Compared to 3.889 × 109, we only get 2.048 × 105 here!
The FFT algorithm allows instead of
2N2 + N operations, which grows in a quadratic manner with respect to N.
a growth of only
N.log(N),
which means a much lesser growth ans shown here for the quotient
(2N2 + N-2)/N.log(N)
Compared to × 109, we only get × 105 here!

7. The FFT algorithm supposes that N is a power of 2, N = 2M, not only an even number as before.
We then have this fundamental fact when stepping to the double 2N of the sample size:
Fourier(2N) ≤ 2.Fourier(N) + 8N
where Fourier(N) means the growth function of N, we leave this as an intuitive concept for our needs, ok?
Then the FFT growth formula Fourier(N) ≤ N.log(N) can be proved without efforts inductively (supposing N is large!), i.e. for 2N if we suppose it true for N:
Fourier(2N) ≤ 2.Fourier(N) + 8N ≤ 2N.log(N) + 8N = 2N(log(N) +4) ~ 2N.(log(N)+log(2)) = 2N.log(2N)
So we are left with the proof of the above fundamental fact!

8. Fourier(2N) ≤ 3 × 2N + 2N + 2.Fourier(N) = 2.Fourier(N) + 8N, QED.
And this is the trick:
The sound sample vector w = (w0,w1,w2,…,w2N-1) in ¬2N is split into two parts of length N each:
w = (w0,w1,w2,…,w2N-1)
¬2N
w+ = (w0, w2, w4,…,w2N-2)
even
w− = (w1, w3,…,w2N-1)
odd ¬N
c+ = (c+0, c+1, c+2,…,c+N-1)
Fourier(N)
c− = (c−0, c−1,…, c−N-1)
Fourier(N)
Have this decisive formula: cr = (c+r + e(N)r . c−r)/2, c±N+r = c±r, i.e. per r we have 2 multiplications, one addition and a total of 2N powers of e(N). Together with the 2.Fourier(N) operations for the c+ and c− we have
Fourier(2N) ≤ 3 × 2N + 2N + 2.Fourier(N) = 2.Fourier(N) + 8N, QED.