Short notes

Fourier Series Short Notes

Feed a 1 kHz square wave into a spectrum analyser and you'll see spectral lines at 1 kHz, 3 kHz, 5 kHz, 7 kHz — odd harmonics only, with amplitudes falling as 1/n. That harmonic structure is the Fourier series of the square wave in action, and the reason there are no even harmonics is the half-wave symmetry of the waveform: x(t + T/2) = −x(t). Understanding this saves you from computing every coefficient individually.

EEE, ECE, EI

How it works

The trigonometric Fourier series is x(t) = a₀ + Σ[aₙcos(nω₀t) + bₙsin(nω₀t)] where ω₀ = 2π/T. The DC coefficient a₀ = (1/T)∫x(t)dt over one period. For a square wave with amplitude A and 50% duty cycle: aₙ = 0 for all n (even symmetry eliminates cosine terms when properly centred), and bₙ = 4A/(nπ) for odd n and 0 for even n. The complex exponential form uses coefficients cₙ = (1/T)∫x(t)e^(−jnω₀t)dt, where |cₙ| versus n is the amplitude spectrum and ∠cₙ versus n is the phase spectrum.

Key points to remember

Dirichlet conditions for Fourier series existence: the signal must be periodic, have a finite number of maxima, minima, and discontinuities in one period, and be absolutely integrable over one period. Even symmetry (x(−t) = x(t)) means bₙ = 0, so only cosine terms remain. Odd symmetry means aₙ = 0 and a₀ = 0. Half-wave symmetry eliminates all even harmonics. Parseval's theorem states that average power equals the sum of squared Fourier coefficients: P = Σ|cₙ|². For the square wave, power convergence is slow because coefficients fall only as 1/n, not 1/n².

Exam tip

Every Anna University paper has a Fourier series coefficient derivation — identify the symmetry first, because it halves your integration work and examiners specifically check whether you used the symmetry shortcut.

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