Fourier Series Symmetry

Symmetry properties of a periodic signal let you predict which Fourier coefficients vanish before doing any integration. Recognising even, odd, half-wave, or quarter-wave symmetry cuts calculation time in half — or more — on exam problems and filter design tasks.

Core Concept

When a signal is even — meaning x(-t) = x(t) — its graph is a mirror image about the vertical axis. The Fourier series of an even function contains only cosine terms. All the sine coefficients b_k are exactly zero because the integral of an odd function over a symmetric interval is always zero.

An odd signal satisfies x(-t) = -x(t). Its graph has rotational symmetry about the origin. The Fourier series of an odd function contains only sine terms. All cosine coefficients a_k, including the DC term a_0, vanish. Any signal can be split into its even part x_e(t) = [x(t)+x(-t)]/2 and odd part x_o(t) = [x(t)-x(-t)]/2.

Half-wave symmetry means x(t + T/2) = -x(t): the second half of each period is a flipped copy of the first half. This forces all even-indexed harmonics to zero. Only odd harmonics (k = 1, 3, 5, ...) survive. Quarter-wave symmetry combines half-wave with either even or odd symmetry, leaving only odd cosines or odd sines.

Key Formula

For even symmetry: b_k = 0 for all k. The a_k integral simplifies to a_k = (4/T) ∫₀^(T/2) x(t) cos(kω0t) dt. For odd symmetry: a_k = 0 for all k including a_0. b_k = (4/T) ∫₀^(T/2) x(t) sin(kω0t) dt. For half-wave symmetry: a_k = b_k = 0 for even k. For odd k: a_k = (4/T) ∫₀^(T/2) x(t) cos(kω0t) dt and similarly for b_k.

Given: x(t) = t for -T/2 < t < T/2, period T (sawtooth-like, odd signal)

Step 1: Check symmetry.
         x(-t) = -t = -x(t)  → odd symmetry confirmed.

Step 2: Conclude a_k = 0 for all k (including DC a_0 = 0).
         Only b_k terms survive.

Step 3: Use simplified odd-symmetry formula.
         b_k = (4/T) ∫₀^(T/2) t · sin(kω0t) dt

Step 4: Integrate by parts (u = t, dv = sin(kω0t) dt).
         b_k = (4/T) · [T/(2kπ) · cos(kπ) - ... ]
         b_k = (2/kπ) · (-1)^(k+1)
              = 2(-1)^(k+1) / (kπ)

Final Answer:
  a_0 = 0, a_k = 0 (odd symmetry)
  b_k = 2(-1)^(k+1) / (kπ)  for k = 1,2,3,...

Exam Tip: Half-wave symmetry is the most frequently tested. Remember: only odd harmonics exist, even harmonics are zero. For a signal with both half-wave and even symmetry, only odd cosine terms remain; for half-wave plus odd symmetry, only odd sine terms remain. Always check for half-wave symmetry first — it halves your integration work even before you look for even or odd symmetry.

Properties Summary

• Even symmetry x(t) = x(-t): all b_k = 0, Fourier series is a pure cosine series.
• Odd symmetry x(t) = -x(-t): a_0 = 0 and all a_k = 0, Fourier series is a pure sine series.
• Half-wave symmetry x(t+T/2) = -x(t): all even-indexed coefficients are zero, only odd k survive.
• Quarter-wave even: half-wave plus even symmetry, only odd cosine terms; b_k = 0 for all k.
• Quarter-wave odd: half-wave plus odd symmetry, only odd sine terms; a_k = 0 for all k.
• Symmetry speeds integration: even/odd symmetry lets you integrate over half the period and double; reduces integral from T to T/2.
• DC component a_0 = 0 whenever x(t) is odd or has half-wave symmetry — the signal's average value is zero.

Quick Revision

• Even signal → only cosines in Fourier series; odd signal → only sines.
• Half-wave symmetry → even harmonics vanish; odd harmonics only.
• Any signal = even part + odd part; decompose before testing symmetry.
• Quarter-wave symmetry is stricter: signal must satisfy both half-wave AND even or odd conditions.
• Symmetry does not change the Fourier series result; it just tells you which terms are zero without computing them.
• Even symmetry: integrate from 0 to T/2 and multiply by 4/T for a_k.
• Square wave and triangle wave both have half-wave symmetry — a common exam verification task.
• Exam trap: confusing half-wave symmetry x(t+T/2) = -x(t) with time-reversal x(-t) = -x(t), which is odd symmetry.