Fourier series of periodic signals

Why it matters

Mains-driven instruments, PWM drives, rotating machinery and clock lines all produce periodic signals, and their behaviour is described by harmonics. The Fourier series tells you how much of a periodic signal sits at each multiple of the fundamental frequency, which is exactly what a harmonic analyser, a THD meter or a filter design needs.

Key ideas

The idea. A periodic signal x(t) with period T0 can be written as a sum of sinusoids at the fundamental frequency f0 = 1/T0 and its integer multiples (harmonics). There is no energy at any other frequency: the spectrum is a set of lines at k·f0.

Two equivalent forms

• Trigonometric: DC term a0 plus ak cos(kω0t) + bk sin(kω0t).
• Complex exponential: Σ ck e^(jkω0t) for k from −∞ to ∞. For a real signal, c(−k) = ck*, so the two-sided magnitude spectrum is even and the phase spectrum is odd.
• Polar (amplitude–phase) form: a0 + Σ Ak cos(kω0t + φk), with Ak = 2|ck|.

Dirichlet conditions (sufficient for convergence): over one period x(t) is absolutely integrable and has a finite number of maxima, minima and finite discontinuities. At a point of continuity the series converges to x(t); at a jump it converges to the mid-point of the jump.

Gibbs phenomenon. A truncated series overshoots near a discontinuity by about 9 % of the jump height. Adding more terms makes the ripple narrower but does not reduce the peak overshoot.

Symmetry shortcuts (they save most of the integration in exams)

• Even x(−t) = x(t): bk = 0 — cosine terms only (plus DC).
• Odd x(−t) = −x(t): a0 = 0 and ak = 0 — sine terms only.
• Half-wave symmetry x(t ± T0/2) = −x(t): only odd harmonics are present, and a0 = 0.
• Removing a DC offset changes only a0; shifting in time changes only the phases.

Decay of coefficients. Coefficients of a signal with jumps (square wave) fall as 1/k; a continuous signal with slope jumps (triangle wave) falls as 1/k². Smoother signals have fewer significant harmonics.

Parseval's relation. The average power equals the sum of the powers in each harmonic — the basis of THD calculations.

Properties (complex form, x ↔ ck): linearity; time shift x(t − t0) ↔ ck e^(−jkω0t0) (magnitudes unchanged); differentiation dx/dt ↔ jkω0 ck; time reversal x(−t) ↔ c(−k).

LTI response. If a periodic x(t) drives an LTI system with frequency response H(jω), the output is periodic with coefficients ck·H(jkω0) — each harmonic is scaled and phase-shifted separately.

Formulas

• x(t) = a0 + Σ from k=1 to ∞ of [ak cos(kω0t) + bk sin(kω0t)] — trigonometric series.
• a0 = (1/T0) ∫ over T0 of x(t) dt — DC (average) value.
• ak = (2/T0) ∫ over T0 of x(t) cos(kω0t) dt, bk = (2/T0) ∫ over T0 of x(t) sin(kω0t) dt.
• x(t) = Σ ck e^(jkω0t), ck = (1/T0) ∫ over T0 of x(t) e^(−jkω0t) dt — exponential series.
• c0 = a0, ck = (ak − j·bk)/2 for k ≥ 1.
• P = (1/T0) ∫ over T0 of |x(t)|² dt = Σ |ck|² = a0² + Σ (ak² + bk²)/2 — Parseval.
• x(t) = (4A/π) Σ over odd k of sin(kω0t)/k — odd square wave between +A and −A.
• ck = (A·τ/T0) · sinc(kτ/T0), with sinc(u) = sin(πu)/(πu) — rectangular pulse train of height A and width τ, centred at t = 0.
• THD = √(P − P1)/√P1 (powers of the AC signal, DC excluded) — total harmonic distortion.

Symbols: T0 fundamental period (s); f0 = 1/T0 (Hz); ω0 = 2π/T0 (rad/s); k harmonic number; a0, ak, bk, ck, A in the unit of x (V, A, …); τ pulse width (s); P normalised power (unit of x)²; P1 power of the fundamental.

Worked examples

Example 1 (standard). A ±5 V square wave with period 2 ms is odd about t = 0. Find the amplitudes of the fundamental and third harmonic, and the fraction of the power in the fundamental.

1. Odd and half-wave symmetric ⇒ only bk, odd k: bk = 4A/(kπ).
2. f0 = 1/(2 ms) = 500 Hz. b1 = 4 × 5/π = 6.366 V at 500 Hz.
3. b3 = 4 × 5/(3π) = 2.122 V at 1.5 kHz.
4. Total power P = A² = 25 V² (the square of ±5 V is always 25).
5. P1 = b1²/2 = 6.366²/2 = 20.26 V², so P1/P = 8/π² = 0.811.
6. b1 ≈ 6.37 V, b3 ≈ 2.12 V; about 81.1 % of the power is in the fundamental (THD = √(1 − 0.811)/√0.811 ≈ 48.3 %).

Example 2 (GATE level). A pulse train has height 1, width τ = T0/4, centred at t = 0. Find c0, c1, c2, c4, and the fraction of the total power contained in the DC term plus the fundamental.

1. ck = (τ/T0) sinc(kτ/T0) = (1/4) · sin(kπ/4)/(kπ/4) = sin(kπ/4)/(kπ).
2. c0 = τ/T0 = 0.25; c1 = sin(π/4)/π = 0.2251; c2 = sin(π/2)/(2π) = 0.1592; c4 = sin(π)/(4π) = 0 (every 4th harmonic is missing).
3. Total power P = (1/T0) ∫ x² dt = τ/T0 = 0.25.
4. DC + fundamental (both k = ±1): c0² + 2|c1|² = 0.0625 + 2 × 0.05066 = 0.1638.
5. Fraction = 0.1638/0.25. About 65.5 % of the power is in DC plus the fundamental.

Common mistakes

• Using 2/T0 for a0. The DC term is the average, 1/T0. (Some books write the series with a0/2; then a0 = (2/T0)∫x dt. Check which convention the question uses.)
• Forgetting the negative-frequency coefficients in Parseval's sum: a harmonic contributes 2|ck|².
• Claiming that more terms remove the Gibbs overshoot.
• Expecting a square wave to contain even harmonics — with half-wave symmetry it has none.
• Writing the period of cos(3πt) + sin(6πt) as anything but 2/3 s; ω0 = 3π, so the 6π term is the second harmonic.
• Mixing ck (two-sided, half amplitude) with Ak (one-sided amplitude) when reading a spectrum.

For GATE IN

Typical questions: identify which harmonics or coefficients are zero from symmetry; compute a0, a specific ck or bk; find total or fractional power with Parseval; find the output of an LTI system (often an ideal filter) for a periodic input by keeping only the harmonics inside its passband. Practise the standard pairs (square, triangle, pulse train, half- and full-wave rectified sine) and the sinc envelope of a pulse train.

Quick check

1. Which coefficients vanish for an even signal?
2. What value does the series converge to at a jump from −1 to +3?
3. A periodic signal has c0 = 1, c±1 = 0.5, all others zero. What is its power?
4. A full-wave rectified sine contains which harmonics of the mains frequency?
5. What is a0 of x(t) = 4 + 3 cos(2πt)? Answers: 1. All bk. 2. The mid-point, 1. 3. 1 + 0.25 + 0.25 = 1.5. 4. DC and even harmonics only (its own fundamental is twice the mains frequency). 5. 4.