Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology IIIDigital Audio III.2 (M Oct 12) Finite Fourier Theory

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Recall: Classical Joseph Fourier (partials/overtones) wave w 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 ) +... every periodic function w(t) !!! Three big problems: 1.The functions in real life are not periodic. 2.When dealing with real life, we do not know all values of a wave, but only a finite number of values that can be measured. 3.The formula is an infinite sum, also not controllable in real life.

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Solution of the first problem: The functions in real life are not periodic: Just make them periodic by periodic extension of the observed sound vibration! one period P Therefore the Fourier formula will start from the fundamental frequency being f = 1/P Hz Example: If we have a sample duration of P = 0.5 sec, then the fundamental frequency is f = 1/P = 2 Hz

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Solution of the second problem: When dealing with real life, we do not know all values of a wave, but only a finite number of values that can be measured. We are measuring a finite number of function values in a regular way: w(t)t Δ -5Δ-4Δ-3Δ-2Δ-Δ0Δ2Δ3Δ4Δ We are measuring N = 2n function values w(rΔ) for r = -n, -(n-1),...0,1,...n-1, the even number N of values is chosen in order to make the following calculations. example n = 5 N = 10 Δ = sample period 1/Δ = sample rate/frequency P = NΔ = 2nΔ fundamental frequency = f = 1/P = 1/(2nΔ) Δ = sample period 1/Δ = sample rate/frequency P = NΔ = 2nΔ fundamental frequency = f = 1/P = 1/(2nΔ)

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Solution of the third problem: The formula is an infinite sum, also not controllable in real life. We only use a finite number of overtones to approximate the wave. To this end we use the following well-known formula of goniometry: a.cos(x) + b.sin(x) = A.sin(x + arccos(b/A)), where A = √(a 2 + b 2 ) This allows us to rewrite the m-th overtone A m sin(2 .mft+Ph m ) by A m sin(2 .mft+Ph m ) = a m cos(2 .mft) + b m sin(2 .mft) We then look at the finite Fourier expression at time rΔ: These are N = 2n linear equations in the N unknowns a 0, b n, and a m, b m, m = 1,2,...n-1. It can be shown by use of the so-called orthogonality relations of sinusoidal functions that these equations have always exactly one solution. 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 .nf. rΔ)

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Summary of the above method (Nyquist theorem) Given a wave w(t) defined in a period P of time as above, if we measure ist values every Δ sample period, dividing the period P into N = 2n intervals, i.e. P = N. Δ = 2n. Δ, then there is exactly one set of coefficients a 0, b n, and a m, b m, m = 1,2,...n-1 solving the finite Fourier expression 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 .nf. rΔ) for all sample times rΔ, r = -n, -(n-1),... n-1. The maximal overtone of this Fourier expression has index n = N/2, half the sample number N. Its frequency is half the sample frequency: n.f = n/P = n/(N. Δ) = 1/(2Δ), it is called the Nyquist frequency. This fact is known as the Nyquist sample theorem. for all sample times rΔ, r = -n, -(n-1),... n-1. The maximal overtone of this Fourier expression has index n = N/2, half the sample number N. Its frequency is half the sample frequency: n.f = n/P = n/(N. Δ) = 1/(2Δ), it is called the Nyquist frequency. This fact is known as the Nyquist sample theorem. w(t ) t

Guerino Mazzola (Fall 2015 © ): Introduction to Music Technology Intuitively speaking, one must have the double frequency of samples to get the maximal frequency in a finite Fourier analysis. Most famous example: Since the human ear perceives roughly maximally Hz sinusodial waves, a Fourier analysis of sound should have at least the double, Hz sampling rate. This is the reason, why CDs have sampling rate Hz. Exercises: (1)Suppose we have a wave of period 20 seconds and a sample rate of 80 Hz. What is the fundamental frequency? How many overtones do we have in the finite Fourier analysis? (2) Look at the Spectrum tool in Audacity and discuss the „size“ button. Exercises: (1)Suppose we have a wave of period 20 seconds and a sample rate of 80 Hz. What is the fundamental frequency? How many overtones do we have in the finite Fourier analysis? (2) Look at the Spectrum tool in Audacity and discuss the „size“ button.