How Fourier series coefficients are calculated

A periodic signal x(t) with period T (ω₀ = 2π/T) can be written as a sum of harmonically related sinusoids. The Fourier series coefficients tell how much of each harmonic is present.

Exponential Fourier series

x(t) = Σ(k = −∞ to ∞) cₖ e^(jkω₀t)

Process to calculate cₖ:

1. Find the period T and ω₀ = 2π/T. Check that x(t) satisfies the Dirichlet conditions (finite number of maxima, minima and discontinuities per period, and absolutely integrable).

2. Multiply both sides by e^(−jmω₀t) and integrate over one period. Because ∫ e^(j(k−m)ω₀t) dt over T equals T for k = m and 0 otherwise (orthogonality), only one term survives.

3. This gives the analysis formula:

   cₖ = (1/T) ∫(over one period) x(t) e^(−jkω₀t) dt

4. c₀ = (1/T)∫x(t)dt is the average (DC) value; |cₖ| and ∠cₖ give the magnitude and phase spectra.

Trigonometric form

x(t) = a₀ + Σ(n=1 to ∞) [aₙ cos(nω₀t) + bₙ sin(nω₀t)]

• a₀ = (1/T)∫x(t)dt
• aₙ = (2/T)∫x(t) cos(nω₀t) dt
• bₙ = (2/T)∫x(t) sin(nω₀t) dt

with cₖ = (aₖ − jbₖ)/2 for k > 0 and c₋ₖ = cₖ* for real x(t). Symmetry simplifies the work: even x(t) has bₙ = 0; odd x(t) has a₀ = aₙ = 0.

Discrete-time (DTFS) for a sequence of period N

cₖ = (1/N) Σ(n = 0 to N−1) x[n] e^(−j2πkn/N), with x[n] = Σ(k=0 to N−1) cₖ e^(j2πkn/N). The cₖ repeat with period N.

Example

x(t) = +1 for 0 < t < T/2 and −1 for T/2 < t < T. Then

cₖ = (1/T)[∫₀^(T/2) e^(−jkω₀t)dt − ∫_(T/2)^T e^(−jkω₀t)dt] = (1 − (−1)ᵏ)/(jπk)

So cₖ = 2/(jπk) for odd k and 0 for even k, c₀ = 0. In trigonometric form, bₙ = 4/(πn) for odd n: x(t) = (4/π)[sin ω₀t + (1/3) sin 3ω₀t + (1/5) sin 5ω₀t + …].