Fourier series coefficients of a square wave

Assumption (the figure's marks are -2, 2, 4): a square wave of amplitude ±1 with period T = 4 s, odd about the origin: x(t) = +1 for 0 < t < 2 and x(t) = -1 for -2 < t < 0 (repeating every 4 s; it flips sign again at t = 4). So ω₀ = 2π/4 = π/2 rad/s.

Exponential coefficients

cₖ = (1/T) ∫{-2}^{2} x(t) e^(-jkω₀t) dt = (1/4) [ -∫{-2}^{0} e^(-jkπt/2) dt + ∫₀^2 e^(-jkπt/2) dt ]

∫₀^2 e^(-jkπt/2) dt = (1 - e^(-jkπ)) / (jkπ/2)

∫_{-2}^{0} e^(-jkπt/2) dt = (e^(jkπ) - 1) / (jkπ/2)

cₖ = (1/4) · (2/(jkπ)) · [ (1 - e^(-jkπ)) - (e^(jkπ) - 1) ] = (1/(2jkπ)) [ 2 - 2cos kπ ] = (1 - cos kπ)/(jkπ)

• For even k: cos kπ = 1, so cₖ = 0 (including c₀ = 0, since the mean is zero).
• For odd k: cos kπ = -1, so cₖ = 2/(jkπ) = -j 2/(kπ).

Trigonometric form

x(t) = (4/π) [ sin(πt/2) + (1/3) sin(3πt/2) + (1/5) sin(5πt/2) + ... ]

That is, bₖ = 4/(kπ) for odd k, and aₖ = 0.

General result: for amplitude ±A and period T the coefficients are cₖ = -j 2A/(kπ) for odd k, 0 for even k. If the wave is instead shifted so that it is an even function, cₖ = (2A/kπ) sin(kπ/2) is obtained, and the magnitude |cₖ| = 2A/(kπ) is the same.