III Digital Audio III.3 (Fr Oct 04) Complex Fourier representation (preliminaries to FFT)

1. III Digital Audio III.3 (Fr Oct 04) Complex Fourier representation (preliminaries to FFT)

2. Have made calculations of this type in finite Fourier theory for the Nyquist theorem w(t) = A0 + A1 sin(2.ft+Ph1) + A2 sin(2.2ft+Ph2) + A3 sin(2.3ft+Ph3) +... Amsin(2.mft+Phm) = amcos(2.mft) + bmsin(2.mft) w(t) = A0 + a1cos(2.ft) + b1sin(2.ft)+ a2cos(2.2ft) + b2sin(2.2ft)+... For complex calculations, the calculus with sinusoidal functions are usless, need more elegant approach!

3. Recall the circle representation of sinusoidal functions: ¬ = plane of complex numbers cos(x)+i.sin(x) sin(x) i = √-1 1 cos(x) Have the famous Euler formula: cos(x)+i.sin(x) = eix cos(x+y) + i.sin(x+y) = ei(x+y) = eix . eiy = [cos(x) + i.sin(x)] . [cos(y) + i.sin(y)]= [cos(x).cos(y) - sin(x).sin(y)] + i[sin(x).cos(y) +cos(x).sin(y)]

4. Translate Fourier’s formula into the complex number representation: cos(x)+i.sin(x) = eix cos(-x)+i.sin(-x) = e−ix= cos(x)−i.sin(x) cos(x)+i.sin(x) = eix+ cos(x)−i.sin(x) = e−ix = 2cos(x)= eix +e−ix cos(x)= (eix + e−ix)/2 sin(x)= (eix − e−ix)/2i w(t) = = a0 + a1cos(2.ft) + b1sin(2.ft)+ a2cos(2.2ft) + b2sin(2.2ft)+...= c0+ c1 e i2.ft + c−1 e−i2.ft + c2 e i2.2ft + c−2 e−i2.2ft + c3 e i2.3ft + c−3 e−i2.3ft +... a0 = c0 n > 0: an = cn + c−n bn = i(cn− c−n) w(t) = ∑n = 0, ±1, ±2, ±3, ... cn e i2.nft

5. Translate the finite Fourier’s formula into the complex number representation: w(rΔ) = a0 + ∑m = 1,2,3,...n-1 amcos(2.mf. rΔ) + bmsin(2.mf. rΔ) + bnsin(2.nft. rΔ) We only consider a special case, which is easy to write down, but it shows the general situation! Namely: a sound sample from t = 0 to t = 1, period P = 1 sec, i.e. fundamental frequency f = 1 Hz whence Δ = 1/N = 1/2n and rΔ =r/N, r = 0,1,2,... N-1 We may then write: wr = w(rΔ) = w(r/N) = ∑m = 0, 1, 2, 3, ... N-1 cm e i2.mr/N Why no negative indices? In fact, we have them, but they are somewhat hidden: e i2.mr/N . e i2.m(N-r)/N = e i2.mr/N +i2.m(N-r)/N = e0 = 1, so e i2.m(N-r)/N = e i2.-mr/N −m = negative Also, cN-m = complex conjugate to cm since the am, bm are all real numbers. Therefore we have a total of N/2 independent complex coefficients, i.e. N real coefficients as required from the original formula.

6. 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 in ¬N. On this space, we have a scalar product — similar to the highschool formula (u,v) = |u|.|v|.cos(u,v): v = complex conjugate u 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, eq〉 = 0 m ≠ q = orthogonality relationsmentioned above! The e0, e1, e2,... eN-1 = orthonormal basis like for normal 3 space! (ortho ~ perpendicular, normal ~ length 1) They replace the sinusoidal functions! eq 90o 90o 90o el em

7. 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 cmem 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 eq 90o 90o 90o el em