Fourier Series of Sawtooth Wave

The sawtooth wave appears in cathode-ray oscilloscope sweep generators, music synthesizers, and power electronics because it contains every harmonic of the fundamental frequency. Its Fourier series is the canonical example of a signal with all integer harmonics present.

Core Concept

A sawtooth wave rises linearly over one period, then drops instantly back to its starting value. That linear ramp is an odd function if centred on zero, meaning x(-t) = -x(t). Odd symmetry guarantees the DC term is zero and only sine terms appear in the Fourier series.

Unlike the square wave, the sawtooth does not have half-wave symmetry. Its second half is not a flipped copy of the first — it is a continuation of the same ramp. This means all harmonics are present: k = 1, 2, 3, 4, and so on. The amplitude still falls as 1/k, but now every harmonic contributes.

The 1/k amplitude decay results from the single discontinuity per period. A linear ramp is smooth everywhere except at the jump. Smooth segments suppress high harmonics, but the sharp jump keeps them alive at 1/k amplitude. Knowing this decay rate tells you how many harmonics you need for a given approximation accuracy.

Key Formula

For the sawtooth wave x(t) = 2At/T for -T/2 < t < T/2 with amplitude A and period T, the Fourier series is x(t) = (2A/π) · Σ [(-1)^(k+1)/k] · sin(kω0t), summed over k = 1, 2, 3, ... The fundamental angular frequency is ω0 = 2π/T. Coefficients: a_0 = 0, a_k = 0 (odd symmetry), b_k = (2A/π) · (-1)^(k+1) / k.

Given: x(t) = 2At/T for -T/2 < t < T/2. Period T. Amplitude A.

Step 1: Verify symmetry.
         x(-t) = 2A(-t)/T = -x(t)  → odd symmetry.
         Therefore a_0 = 0 and all a_k = 0.

Step 2: Compute b_k using odd-symmetry formula.
         b_k = (4/T) ∫₀^(T/2) x(t) sin(kω0t) dt
             = (4/T) ∫₀^(T/2) (2At/T) sin(kω0t) dt
             = (8A/T²) ∫₀^(T/2) t sin(kω0t) dt

Step 3: Integrate by parts. Let u = t, dv = sin(kω0t) dt.
         ∫ t sin(kω0t) dt = -t cos(kω0t)/(kω0) + sin(kω0t)/(kω0)²

Step 4: Evaluate from 0 to T/2.
         = -[T/2 · cos(kπ)]/(kω0) + [sin(kπ)]/(kω0)²
         = -(T/2)(-1)^k / (kω0)   [since sin(kπ) = 0]

Step 5: Substitute ω0 = 2π/T.
         b_k = (8A/T²) · (T/2)(-1)^(k+1) / (k · 2π/T)
             = (8A/T²) · T²(-1)^(k+1) / (4kπ)
             = 2A(-1)^(k+1) / (kπ)

Final Answer:
  x(t) = (2A/π)[sin(ω0t) - sin(2ω0t)/2 + sin(3ω0t)/3 - ...]

Exam Tip: The alternating sign (-1)^(k+1) in the sawtooth series means positive amplitude for k=1, negative for k=2, positive for k=3, and so on. Do not confuse this with the square wave series where even terms simply vanish. Also, if the sawtooth ramps down instead of up (x(t) = -2At/T), all b_k change sign but the magnitudes stay the same.

Properties Summary

• Odd symmetry: DC and all cosine coefficients are zero; only sine terms survive.
• All harmonics present: both even and odd k appear, unlike the square wave.
• Amplitude decay: b_k = 2A(-1)^(k+1)/(kπ), falling as 1/k due to one discontinuity per period.
• Alternating sign: b_k alternates positive and negative as k increases.
• Average power: P = A²/3, computed directly from x(t) = 2At/T and verified by Parseval.
• Gibbs overshoot: sawtooth also shows 9% overshoot near the vertical drop because of the jump discontinuity.

Quick Revision

• Sawtooth is odd → only sine harmonics; b_k = 2A(-1)^(k+1)/(kπ).
• All harmonics present: k = 1, 2, 3, ... no even harmonics vanish.
• Amplitude falls as 1/k — same rate as square wave — one discontinuity per period causes this.
• Triangle wave has no discontinuities and its amplitude falls as 1/k², which is faster.
• Reversing the ramp direction changes the sign of all b_k but not their magnitude.
• DC term is zero because the average value of the centred sawtooth over one period is zero.
• Sawtooth average power = A²/3, check this against Parseval sum Σ b_k²/2.
• Exam trap: writing b_k = A(-1)^(k+1)/(kπ) instead of 2A(-1)^(k+1)/(kπ) — missing the factor of 2 from the integration limits.