Fourier Analysis: Solving Real-Life Wave Problems

1. III Digital Audio III.2 (W Oct 02) Finite Fourier Theory

2. wavew Recall: Classical Joseph Fourier (partials/overtones) every periodic function w(t) !!! w(t) = A0 + A1 sin(2.ft+Ph1) + A2 sin(2.2ft+Ph2) + A3 sin(2.3ft+Ph3) +... Three big problems: The functions in real life are not periodic. 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. The formula is an infinite sum, also not controllable in real life.

3. Solution of the first problem: The functions in real life are not periodic: Just make them periodic by periodic extensionof 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

4. 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) Δ = sample period1/Δ = sample rate/frequency P = NΔ = 2nΔ fundamental frequency =f = 1/P = 1/(2nΔ) t example n = 5 N = 10 Δ -5Δ -4Δ -3Δ -2Δ -Δ 0 Δ 2Δ 3Δ 4Δ 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.

5. 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 = √(a2 + b2) This allows us to rewrite the m-th overtone Amsin(2.mft+Phm) by Amsin(2.mft+Phm) = amcos(2.mft) + bmsin(2.mft) We then look at the finite Fourier expression at time rΔ: w(rΔ) = a0 + ∑m = 1,2,3,...n-1 amcos(2.mf. rΔ) + bmsin(2.mf. rΔ) + bnsin(2.nf. rΔ) These are N = 2n linear equations in the N unknowns a0,bn, and am, bm, 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.

6. 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 coefficientsa0,bn, and am, bm, m = 1,2,...n-1 solving the finite Fourier expression w(t) t w(rΔ) = a0 + ∑m = 1,2,3,...n-1 amcos(2.mf. rΔ) + bmsin(2.mf. rΔ) + bnsin(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.

7. 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 perceivesroughly maximally 20 000 Hz sinusodial waves, a Fourier analysis of sound should have at least the double, 40 000 Hz sampling rate.This is the reason, why CDs have sampling rate44 100 Hz. • Exercises: • Supposewehave a waveofperiod 20 secondsand a sample rate of 80 Hz. Whatisthe fundamental frequency? Howmanyovertones do wehave in the finite Fourier analysis? • Look at theSpectrumtool in Audacityanddiscussthe „size“ button.