Fourier Series of Triangle Wave

The triangle wave is the smoothest common periodic waveform — it has no discontinuities in its value, only a kink in its slope. This smoothness makes its harmonic amplitudes fall off as 1/k², far faster than a square or sawtooth wave, which is why triangle oscillators are used when low harmonic distortion is needed in audio and test equipment.

Core Concept

A triangle wave rises and falls linearly, forming a zigzag. It is continuous everywhere, meaning there are no sudden jumps in value. The only abruptness is in the slope, which reverses at the peaks. This is called a first-order discontinuity, as opposed to the zeroth-order (value) discontinuity of a square or sawtooth wave.

The triangle wave with peaks on the positive axis and zero crossings is an even function: x(-t) = x(t). Even symmetry means all sine coefficients b_k are zero. Only cosine terms survive. Together with half-wave symmetry (the shape of the second half is a flipped first half), only odd cosine harmonics appear.

Because the waveform is continuous, integrating the coefficient formula by parts twice produces a 1/k² decay instead of 1/k. Each extra degree of smoothness — no jump in value, no jump in slope, and so on — multiplies the decay exponent by one. This is the mathematical reason smooth signals have compact spectra.

Key Formula

For the triangle wave defined as x(t) = (4A/T)|t| - A for -T/2 < t < T/2 (or equivalently, x(t) = A - (4A/T)|t - T/4| centred differently), the Fourier series is x(t) = (8A/π²) · Σ [(-1)^((k-1)/2) / k²] · cos(kω0t), summed over odd k = 1, 3, 5, ... Coefficients: a_0 = A/2 if the wave stays positive; a_k = 8A/(π²k²) · (alternating sign) for odd k; b_k = 0.

Given: Triangle wave symmetric about t = 0.
         x(t) = A(1 - 2|t|/T) ... peak A at t=0, zero at t=±T/2.
         Even function, odd harmonics only (half-wave symmetry).

Step 1: Verify symmetry.
         x(-t) = x(t) → even. a_0 = average = A/2? No.
         For this centred form: average = A/2.
         Wait — check: ∫₋T/2^(T/2) A(1-2|t|/T) dt
         = 2A ∫₀^(T/2) (1-2t/T) dt = 2A[t - t²/T]₀^(T/2)
         = 2A[T/2 - T/4] = 2A · T/4 = AT/2
         So a_0 = (1/T)(AT/2) = A/2.

Step 2: Check half-wave symmetry.
         x(t + T/2) = A(1 - 2|t + T/2|/T) ≠ -x(t) in general.
         For the odd-centred version x(t) = -(4A/T)|t| + A with mean zero:
         half-wave holds, giving only odd k.

Step 3: Compute a_k for even-centred version.
         a_k = (4/T) ∫₀^(T/2) x(t) cos(kω0t) dt
         Integrate by parts twice.
         a_k = 8A/(π²k²) for odd k, 0 for even k.

Final Answer:
  x(t) = A/2 + (8A/π²)[cos(ω0t) - cos(3ω0t)/9 + cos(5ω0t)/25 - ...]

Exam Tip: The 1/k² amplitude decay is the identifying mark of the triangle wave. If a GATE problem shows a spectrum with coefficients proportional to 1/k² and only odd harmonics, think triangle wave. Compare with square wave (1/k, odd harmonics) and sawtooth (1/k, all harmonics). The triangle wave can be obtained by integrating a square wave — integration multiplies coefficients by 1/(jkω0), adding one power of 1/k to the decay.

Properties Summary

• Even symmetry: all b_k = 0; only cosine terms and DC term appear.
• Half-wave symmetry: only odd harmonics survive; even k coefficients are zero.
• Amplitude decay: a_k proportional to 1/k² — faster than square or sawtooth due to continuity.
• Integration link: the triangle wave is the integral of the square wave; this adds one extra 1/k factor.
• Low harmonic distortion: 1/k² decay means third harmonic is 1/9 the fundamental, versus 1/3 for square wave.
• Average power: P = A²/3 for a triangle wave between 0 and A; verify using Parseval.

Quick Revision

• Triangle wave: even + half-wave symmetry → only odd cosine terms.
• Amplitude decay: 1/k² — the defining characteristic.
• Coefficients: a_k = 8A/(π²k²) with alternating sign for k = 1, 3, 5,...
• No Gibbs phenomenon: triangle is continuous so no overshoot appears near peaks.
• Integrating a square wave gives a triangle wave — coefficients gain one factor of 1/k.
• Differentiating a triangle wave gives a square wave — coefficients gain one factor of k.
• Third harmonic power is (1/9)² = 1/81 of fundamental power — very low distortion.
• Exam trap: writing a_k = 4A/(π²k²) instead of 8A/(π²k²) — the factor of 8 comes from twice the integration over half the period for even functions.