Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology IIIDigital Audio III.3 (Fr Oct 10) Complex Fourier representation (preliminaries to FFT)

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology w(t) = A 0 + A 1 sin(2 .ft+Ph 1 ) + A 2 sin(2 .2ft+Ph 2 ) + A 3 sin(2 .3ft+Ph 3 ) +... w(t) = A 0 + a 1 cos(2 .ft) + b 1 sin(2 .ft)+ a 2 cos(2 .2ft) + b 2 sin(2 .2ft)+... A m sin(2 .mft+Ph m ) = a m cos(2 .mft) + b m sin(2 .mft) For complex calculations, the calculus with sinusoidal functions are usless, need more elegant approach! Have made calculations of this type in finite Fourier theory for the Nyquist theorem

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology ¬ = plane of complex numbers Recall the circle representation of sinusoidal functions: cos(x) sin(x) cos(x)isin(x) cos(x) + i. sin(x) Have the famous Euler formula: cos(x)isin(x) = e ix Have the famous Euler formula: cos(x) + i. sin(x) = e ix cos(x+y) + i.sin(x+y) = e i(x+y) = e ix. e iy = [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)] Have the famous Euler formula: cos(x)isin(x) = e ix Have the famous Euler formula: cos(x) + i. sin(x) = e ix cos(x+y) + i.sin(x+y) = e i(x+y) = e ix. e iy = [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)] 1 i = √-1

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology Translate Fourier’s formula into the complex number representation: cos(x)isin(x) = e ix cos(x) + i. sin(x) = e ix cos(-x)isin(-x) = e −ix = cos(x)−isin(x) cos(-x) + i. sin(-x) = e −ix = cos(x) − i. sin(x) cos(x)isin(x) = e ix + cos(x)−isin(x) = e −ix cos(x) + i. sin(x) = e ix + cos(x) − i. sin(x) = e −ix = 2cos(x)= e ix +e −ix = 2cos(x) = e ix +e −ix cos(x)isin(x) = e ix cos(x) + i. sin(x) = e ix cos(-x)isin(-x) = e −ix = cos(x)−isin(x) cos(-x) + i. sin(-x) = e −ix = cos(x) − i. sin(x) cos(x)isin(x) = e ix + cos(x)−isin(x) = e −ix cos(x) + i. sin(x) = e ix + cos(x) − i. sin(x) = e −ix = 2cos(x)= e ix +e −ix = 2cos(x) = e ix +e −ix w(t) = = a 0 + a 1 cos(2 .ft) + b 1 sin(2 .ft)+ a 2 cos(2 .2ft) + b 2 sin(2 .2ft)+... = c 0 + c 1 e i2 .ft + c −1 e − i2 .ft + c 2 e i2 .2ft + c − 2 e − i2 .2ft + c 3 e i2 .3ft + c − 3 e − i2 .3ft +... cos(x)= (e ix + e −ix )/2 sin(x)= (e ix − e −ix )/2i w(t) = ∑ n = 0, ±1, ±2, ±3,... c n e i2 .nft a 0 = c 0 n > 0:a n = c n + c −n b n = i(c n − c −n )

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology Translate the finite Fourier’s formula into the complex number representation: w(rΔ) = a 0 + ∑ m = 1,2,3,...n-1 a m cos(2 .mf. rΔ) + b m sin(2 .mf. rΔ) + b n sin(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: w r = w(rΔ) = w(r/N) = ∑ m = 0, 1, 2, 3,... N-1 c m 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 = e 0 = 1, so e i2 .m(N-r)/N = 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 = e 0 = 1, so e i2 .m(N-r)/N = e i2 .-mr/N −m = negative Also, c N-m = complex conjugate to c m since the a m, b m 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.

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology The representation w r = w(rΔ) = w(r/N) = ∑ m = 0, 1, 2, 3,... N-1 c m e i2 .mr/N identifies the sequence w = (w 0,w 1,w 2,…,w N-1 ) as a vector in the N-dimensional complex space So our samples of fundamental frequency f = 1 are identified with the vectors w ∈ identifies the sequence w = (w 0,w 1,w 2,…,w N-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. On this space, we have a scalar product — similar to the highschool formula (u,v) = |u|.|v|.cos(u,v): Have N exponential functions e 0, e 1, e 2,... e N-1 that are represented as vectors in Have N exponential functions e 0, e 1, e 2,... e N-1 that are represented as vectors in ¬ N e m = (e m (r) = e i2 .mr/N ) r = 0,1,2,...N-1 〈 e m, e m 〉 = 1, 〈 e m, e q 〉 = 0 m ≠ q = orthogonality relations mentioned above! e 0, e 1, e 2,... e N-1 The e 0, e 1, e 2,... e N-1 = orthonormal basis like for normal 3 space! (ortho ~ perpendicular, normal ~ length 1) They replace the sinusoidal functions! 90 o emememem elelelel eqeqeqeq u v = complex conjugate

Guerino Mazzola (Fall 2014 © ): Introduction to Music Technology Every sound sample vector w = (w 0,w 1,w 2,…,w N-1 ) in ¬ N can be written as a linear combination w = ∑ m = 0, 1, 2, 3,... N-1 c m e m of the exponential functions, and the (uniquely determined) coefficients c m are calculated via c m = 〈 w, e m 〉 =(1/N). ∑ r = 0, 1, 2, 3,... N-1 w r e -i2 .mr/N Every sound sample vector w = (w 0,w 1,w 2,…,w N-1 ) in ¬ N can be written as a linear combination w = ∑ m = 0, 1, 2, 3,... N-1 c m e m of the exponential functions, and the (uniquely determined) coefficients c m are calculated via c m = 〈 w, e m 〉 =(1/N). ∑ r = 0, 1, 2, 3,... N-1 w r e -i2 .mr/N 90 o emememem elelelel eqeqeqeq