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

# Exponential FS coefficients and existence

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2079 Chaitra, 5+3 marks
- Video: [Complex Exponential Fourier Series](https://www.youtube.com/watch?v=oXrbcRyXj84) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-exponential-existence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Derive the expression of Fourier Series Coefficient in exponential form for Continuous time periodic signal. Describe the condition for existence of Fourier series with suitable examples.

## Short answer

The exponential Fourier series coefficient is a_k = (1/T) ∫_T x(t) e^{-j k ω0 t} dt. Existence requires Dirichlet conditions: absolute integrability, finite maxima/minima, and finite discontinuities per period. For example, 1/t violates absolute integrability, while a square wave satisfies all conditions.

## Answer

#### Exponential Fourier series coefficients
Let $x(t)$ be periodic with period $T$, $\omega_0 = 2\pi/T$. It is expressed as a sum of harmonically related complex exponentials (synthesis equation):

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

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

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

The right-hand integral:

$$
\int_0^T e^{j(k-n)\omega_0 t}dt = \left[\frac{e^{j(k-n)\omega_0 t}}{j(k-n)\omega_0}\right]_0^T = \frac{e^{j(k-n)2\pi} - 1}{j(k-n)\omega_0} = 0 \quad (k \ne n)
$$

and equals $T$ when $k = n$. Only the $k = n$ term remains:

$$
\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T
$$

$$
a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt
$$

This is the analysis equation. $a_0 = \frac{1}{T}\int_T x(t)dt$ is the average value, and $|a_k|$, $\angle a_k$ give the magnitude and phase spectra.

#### Conditions for existence (Dirichlet conditions)
Over one period:
1. **Absolutely integrable:** $\int_T|x(t)|dt < \infty$.
   - Violated by $x(t) = 1/t$ on $0 < t \le 1$ (period 1): $\int_0^1 dt/t = \infty$.
2. **Finite number of maxima and minima** in a period.
   - Violated by $x(t) = \sin(2\pi/t)$ on $0 < t \le 1$: it oscillates infinitely often near $t = 0$.
3. **Finite number of discontinuities**, each of finite size, in a period.
   - Violated by a staircase that steps down by half on $[0, \frac12), [\frac12, \frac34), [\frac34, \frac78), \dots$: infinitely many jumps in one period.

**A signal that satisfies them:** a square wave has two finite jumps per period, is bounded and has no oscillations, so its Fourier series exists; it converges to the midpoint value at each jump.

## Check yourself

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

1. What is the integral over one period of e^{j(k-n)w0 t} when k is not equal to n?
   A. T
   B. 1
   C. 0
   D. Infinity
   Answer: C. The integral evaluates to (e^{j(k-n)2 pi} - 1)/(j(k-n)w0) = 0. (Key point: Orthogonality integral)

2. After the orthogonality step, the integral of x(t) e^{-jn w0 t} over a period equals a_n T. So a_k is:
   A. a_k = T integral of x(t) dt
   B. a_k = (1/T) integral over T of x(t) e^{-jk w0 t} dt
   C. a_k = (1/T) integral of x(t) e^{jk w0 t} dt
   D. a_k = integral of x(t) e^{-jk w0 t} dt with no scaling
   Answer: B. Dividing both sides by T gives the analysis equation with a 1/T factor and a negative exponent. (Key point: Analysis equation)

3. What does a_0 represent?
   A. The fundamental frequency
   B. The average (dc) value of x(t) over a period
   C. The peak value of x(t)
   D. The phase of the first harmonic
   Answer: B. a_0 = (1/T) integral of x(t) dt, the average value. (Key point: Meaning of a_0)

4. Which signal violates the absolute integrability condition?
   A. A square wave
   B. x(t) = 1/t on 0 < t <= 1, repeated with period 1
   C. A sine wave
   D. x(t) = sin(2 pi/t) on 0 < t <= 1
   Answer: B. The integral of 1/t from 0 to 1 is infinite. sin(2 pi/t) instead violates the finite maxima/minima condition. (Key point: Violating example)

## Related in Signals and Systems

- [Derive exponential Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-exponential-derivation) — Signals and Systems, chapter 2, asked 15×
- [Derive the discrete-time Fourier series](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-derivation) — Signals and Systems, chapter 2, asked 15×
- [Exponential FS coefficients; Gibbs phenomenon](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aexponential-fs-gibbs) — Signals and Systems, chapter 2
- [Periodic signal whose Fourier series fails](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-existence-examples) — Signals and Systems, chapter 2
- [FS coefficients of periodic square wave](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asquare-wave-fourier-coefficients) — Signals and Systems, chapter 2
- [FS coefficients; time scaling and conjugation](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-significance-scaling-conjugation) — Signals and Systems, chapter 2

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