---
title: "FS equations and existence conditions"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:fourier-series-equations-conditions"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/fourier-series-equations-conditions
site: BCT Engineering
---

# FS equations and existence conditions

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2078 Poush, 5+3 marks
- Video: [Conditions for Existence of Fourier Series (Dirichlet Conditions)](https://www.youtube.com/watch?v=blS_OImUJ-c) — 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/fourier-series-equations-conditions
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Derive the formula for synthesis equation and analysis equation of Fourier series. Mention the three conditions that guarantee the existence of Fourier series.

## Short answer

Synthesis: x(t)=∑ a_k e^{jkω0t}. Multiply by e^{-jnω0t} and integrate over one period; orthogonality yields a_k = (1/T)∫_T x(t)e^{-jkω0t} dt. Dirichlet conditions: absolute integrability, finite extrema, and finite discontinuities per period.

## Answer

A periodic signal $x(t)$ with period $T$ and fundamental frequency $\omega_0 = 2\pi/T$ can be written as a sum of harmonically related complex exponentials (the exponential Fourier series).

#### Synthesis equation
The functions $e^{jk\omega_0 t}$, $k = 0, \pm1, \pm2, \dots$, all repeat every $T$, so

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

is periodic with period $T$. This is the synthesis equation; $a_k$ are the Fourier series coefficients.

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

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

Orthogonality of the exponentials:

$$
\int_0^T e^{j(k-n)\omega_0 t}dt = \begin{cases} T, & k = n \\ 0, & k \ne n \end{cases}
$$

(for $k \ne n$, the cosine and sine parts complete a whole number of cycles and integrate to zero). Hence

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

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

This is the analysis equation; $a_0 = \frac{1}{T}\int_T x(t)dt$ is the dc value.

| Equation | Formula |
|---|---|
| Synthesis | $x(t) = \sum_k a_k e^{jk\omega_0 t}$ |
| Analysis | $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ |

#### Conditions for existence (Dirichlet conditions)
Over any one period:
1. **Absolutely integrable:** $\int_T |x(t)|\,dt < \infty$, which guarantees every $|a_k|$ is finite.
2. **Finite number of maxima and minima** (bounded variation) in one period.
3. **Finite number of discontinuities** in one period, and each discontinuity is finite.

When these hold, the series converges to $x(t)$ at every point where $x(t)$ is continuous, and to the average $\frac{1}{2}[x(t^-) + x(t^+)]$ at a discontinuity. (A weaker condition, finite energy over one period, $\int_T|x(t)|^2dt < \infty$, guarantees convergence in the mean-square sense.)

## Check yourself

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

1. What is the synthesis equation of the exponential Fourier series of a periodic x(t) with fundamental frequency w0?
   A. x(t) = sum over k of a_k e^{jk w0 t}
   B. x(t) = (1/T) integral of a_k e^{-jk w0 t} dt
   C. x(t) = sum over k of a_k e^{-jk w0 t} / k
   D. x(t) = a_0 + a_1 e^{j w0 t} only
   Answer: A. x(t) is built as a sum of harmonically related complex exponentials e^{jk w0 t} weighted by the coefficients a_k. (Key point: Synthesis equation)

2. When x(t) is multiplied by e^{-jn w0 t} and integrated over one period, why does only the k = n term survive?
   A. Because the integral of e^{j(k-n)w0 t} over a period is T for k = n and 0 for k not equal to n
   B. Because all a_k except a_n are defined to be zero
   C. Because e^{j(k-n)w0 t} equals 1 for every k
   D. Because the integral of x(t) over a period is always zero
   Answer: A. The exponentials are orthogonal over one period: the integral is T when k = n and 0 otherwise, leaving a_n T. (Key point: Orthogonality in the derivation)

3. Which expression gives the coefficient a_k?
   A. a_k = T integral of x(t) e^{jk w0 t} dt
   B. a_k = (1/T) integral over T of x(t) e^{-jk w0 t} dt
   C. a_k = (1/T) integral over T of x(t) e^{jk w0 t} dt
   D. a_k = integral over T of x(t) dt
   Answer: B. Dividing a_n T by T gives a_k = (1/T) times the integral of x(t) e^{-jk w0 t} over one period. (Key point: Analysis equation)

4. Which set lists the three Dirichlet conditions over one period?
   A. Even symmetry, finite energy, zero mean
   B. Absolutely integrable, finite number of maxima and minima, finite number of finite discontinuities
   C. Bounded amplitude, continuous, and differentiable everywhere
   D. Absolutely integrable, infinite harmonics, and odd symmetry
   Answer: B. The conditions are absolute integrability, a finite number of maxima/minima, and a finite number of finite discontinuities per period. (Key point: Dirichlet conditions)

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