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

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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

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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

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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 )

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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.

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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

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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

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