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Exponential FS coefficients; Gibbs phenomenon

Short answer

The coefficient is a_k = (1/T) ∫_T x(t) e^{-j k ω0 t} dt. Gibbs phenomenon: truncating the series for a signal with a jump causes ripples near the discontinuity with a fixed overshoot of about 9% of the jump height, which does not vanish as more terms are added, though the ripples narrow.

Derive the expression to compute Fourier series coefficients of exponential Fourier series. Explain Gibbs phenomenon.

Exponential Fourier series coefficients

A periodic signal x(t)x(t) with period TT and ω0=2π/T\omega_0 = 2\pi/T is written as

x(t)=∑k=−∞∞akejkω0t(synthesis)x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t} \qquad \text{(synthesis)}

Multiply both sides by e−jnω0te^{-jn\omega_0 t} and integrate over one period TT:

∫0Tx(t)e−jnω0tdt=∑k=−∞∞ak∫0Tej(k−n)ω0tdt\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt

For k≠nk \ne n:

∫0Tej(k−n)ω0tdt=ej(k−n)ω0T−1j(k−n)ω0=ej2π(k−n)−1j(k−n)ω0=0\int_0^T e^{j(k-n)\omega_0 t}dt = \frac{e^{j(k-n)\omega_0 T} - 1}{j(k-n)\omega_0} = \frac{e^{j2\pi(k-n)} - 1}{j(k-n)\omega_0} = 0

For k=nk = n the integral is ∫0T1 dt=T\int_0^T 1\,dt = T. Therefore

∫0Tx(t)e−jnω0tdt=anT⇒ak=1T∫Tx(t)e−jkω0tdt(analysis)\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T \quad\Rightarrow\quad a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \qquad \text{(analysis)}
  • a0=1T∫Tx(t)dta_0 = \frac{1}{T}\int_T x(t)dt is the dc value.
  • aka_k is complex: ∣ak∣|a_k| is the magnitude spectrum and ∠ak\angle a_k the phase spectrum.
  • For real x(t)x(t), a−k=ak∗a_{-k} = a_k^*.

Example: a square wave of height 1 for ∣t∣<T1|t| < T_1 in period TT gives a0=2T1/Ta_0 = 2T_1/T and ak=sin⁡(kω0T1)kπa_k = \frac{\sin(k\omega_0 T_1)}{k\pi}.

Gibbs phenomenon

When a signal with a jump discontinuity (e.g. a square wave) is approximated by a finite number of Fourier terms,

xN(t)=∑k=−NNakejkω0t,x_N(t) = \sum_{k=-N}^{N}a_k e^{jk\omega_0 t},

the partial sum shows ripples near the discontinuity, with an overshoot of about 9% of the jump height on each side of it.

Key features:

  • As NN increases, the ripples become narrower and crowd closer to the discontinuity, but the peak overshoot does not decrease; it stays at about 9% of the jump (about 1.09 times for a unit jump).
  • At the discontinuity itself, the series converges to the midpoint 12[x(t−)+x(t+)]\frac{1}{2}[x(t^-) + x(t^+)].
  • The energy of the error ∫T∣x(t)−xN(t)∣2dt→0\int_T|x(t) - x_N(t)|^2dt \to 0 as N→∞N \to \infty (mean-square convergence), because the area under the ripples shrinks.
  • It was explained by J. W. Gibbs (1899) after Michelson noticed the overshoot in a mechanical harmonic analyser.
 x_N(t) near a jump (square wave)
       overshoot ~9%
        _/\_/\/\/\________
       |
       |   finite N ripples
 ______|
  \/\/\
       ^ jump at t = T1

Practical meaning: truncating a Fourier series (or ideal filtering, which cuts off high harmonics) produces ringing near sharp edges; window functions are used to reduce it.

✓ Answer checked by Claude Opus 5.5 (high effort)

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