---
title: "Exponential FS coefficients; Gibbs phenomenon"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:exponential-fs-gibbs"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/exponential-fs-gibbs
site: BCT Engineering
---

# Exponential FS coefficients; Gibbs phenomenon

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2076 Bhadra, 8 marks
- Video: [Gibbs Phenomenon by Ms. P Saritha](https://www.youtube.com/watch?v=ctQwsDdU6P4) — Institute of Aeronautical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/exponential-fs-gibbs
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

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

## Answer

#### Exponential Fourier series coefficients
A periodic signal $x(t)$ with period $T$ and $\omega_0 = 2\pi/T$ is written as

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

Multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period $T$:

$$
\int_0^T x(t)e^{-jn\omega_0 t}dt = \sum_{k=-\infty}^{\infty}a_k\int_0^T e^{j(k-n)\omega_0 t}dt
$$

For $k \ne n$:

$$
\int_0^T e^{j(k-n)\omega_0 t}dt = \frac{e^{j(k-n)\omega_0 T} - 1}{j(k-n)\omega_0} = \frac{e^{j2\pi(k-n)} - 1}{j(k-n)\omega_0} = 0
$$

For $k = n$ the integral is $\int_0^T 1\,dt = T$. Therefore

$$
\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T \quad\Rightarrow\quad a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt \qquad \text{(analysis)}
$$

- $a_0 = \frac{1}{T}\int_T x(t)dt$ is the dc value.
- $a_k$ is complex: $|a_k|$ is the magnitude spectrum and $\angle a_k$ the phase spectrum.
- For real $x(t)$, $a_{-k} = a_k^*$.

Example: a square wave of height 1 for $|t| < T_1$ in period $T$ gives $a_0 = 2T_1/T$ and $a_k = \frac{\sin(k\omega_0 T_1)}{k\pi}$.

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

$$
x_N(t) = \sum_{k=-N}^{N}a_k e^{jk\omega_0 t},
$$

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 $N$ 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 $\frac{1}{2}[x(t^-) + x(t^+)]$.
- The energy of the error $\int_T|x(t) - x_N(t)|^2dt \to 0$ as $N \to \infty$ (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.

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In deriving a_k, which fact removes all terms except k = n?
   A. The integral of e^{j(k-n)w0 t} over one period is 0 for k not equal to n and T for k = n
   B. x(t) is zero outside one period
   C. a_k is real
   D. The series has finitely many terms
   Answer: A. Orthogonality of the exponentials leaves only a_n T. (Key point: Orthogonality)

2. For a real x(t), the coefficients satisfy:
   A. a_{-k} = a_k*
   B. a_{-k} = a_k
   C. a_{-k} = -a_k*
   D. a_k = 0 for k < 0
   Answer: A. Real signals give conjugate-symmetric coefficients, a_{-k} = a_k*. (Key point: Coefficients of real signals)

3. As more harmonics N are added to approximate a square wave, the peak overshoot near the jump:
   A. Disappears completely
   B. Grows without limit
   C. Stays at about 9% of the jump height
   D. Falls in proportion to 1/N
   Answer: C. Ripples get narrower and move closer to the jump, but the peak stays near 9% of the jump. (Key point: Gibbs overshoot)

4. Even with Gibbs ripples, why does the series still converge in the mean-square sense?
   A. Because the overshoot shrinks to zero
   B. Because the ripples only occur at t = 0
   C. Because the jump is removed
   D. Because the ripples get narrower so the error energy tends to zero as N grows
   Answer: D. The area under the ripples shrinks, so the error energy goes to 0 even though the peak overshoot does not. (Key point: Convergence in energy)

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