---
title: "Derive trigonometric Fourier series (CT)"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:ctfs-trigonometric-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-trigonometric-derivation
site: BCT Engineering
---

# Derive trigonometric Fourier series (CT)

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2081 Chaitra, 5 marks; 2079 Asoj, 5 marks; 2078 Chaitra, 5 marks (asked 3 times)
- Video: [Trigonometric Fourier Series](https://www.youtube.com/watch?v=0d-xE2SvsSo) — 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-trigonometric-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Derive the trigonometric Fourier series expansion of a continuous-time periodic signal x(t).

## Short answer

Trigonometric series: x(t)=a0+∑(a_n cos nω0t + b_n sin nω0t). Coefficients found via orthogonality: a0=(1/T)∫ x dt, a_n=(2/T)∫ x cos nω0t dt, b_n=(2/T)∫ x sin nω0t dt. Even signals have only cosine terms; odd only sine terms.

## Answer

A periodic signal $x(t)$ with period $T$ ($\omega_0 = 2\pi/T$) that meets the Dirichlet conditions can be written as a dc term plus sums of cosines and sines at multiples of $\omega_0$. This is the trigonometric Fourier series:

$$
x(t) = a_0 + \sum_{n=1}^{\infty}\left[a_n\cos n\omega_0 t + b_n \sin n\omega_0 t\right]
$$

#### Orthogonality relations (over one period $T$)
For integers $m, n \ge 1$:

$$
\begin{aligned}
&\int_T \cos m\omega_0 t\,dt = \int_T \sin m\omega_0 t\,dt = 0 \\
&\int_T \cos m\omega_0 t\,\cos n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\
&\int_T \sin m\omega_0 t\,\sin n\omega_0 t\,dt = \begin{cases} T/2, & m = n \\ 0, & m \ne n \end{cases} \\
&\int_T \sin m\omega_0 t\,\cos n\omega_0 t\,dt = 0 \quad \text{for all } m, n
\end{aligned}
$$

#### Finding $a_0$
Integrate both sides over one period. Every sine and cosine term integrates to zero:

$$
\int_T x(t)\,dt = a_0 T \quad\Rightarrow\quad a_0 = \frac{1}{T}\int_T x(t)\,dt
$$

#### Finding $a_n$
Multiply both sides by $\cos m\omega_0 t$ and integrate over $T$. Only the term with $n = m$ in the cosine sum survives:

$$
\int_T x(t)\cos m\omega_0 t\,dt = a_m\frac{T}{2}
\quad\Rightarrow\quad
a_n = \frac{2}{T}\int_T x(t)\cos n\omega_0 t\,dt
$$

#### Finding $b_n$
Multiply by $\sin m\omega_0 t$ and integrate; only the $n = m$ sine term survives:

$$
\int_T x(t)\sin m\omega_0 t\,dt = b_m\frac{T}{2}
\quad\Rightarrow\quad
b_n = \frac{2}{T}\int_T x(t)\sin n\omega_0 t\,dt
$$

#### Summary and notes
| Coefficient | Formula |
|---|---|
| $a_0$ (dc) | $\frac{1}{T}\int_T x(t)\,dt$ |
| $a_n$ | $\frac{2}{T}\int_T x(t)\cos n\omega_0 t\,dt$ |
| $b_n$ | $\frac{2}{T}\int_T x(t)\sin n\omega_0 t\,dt$ |

- Even $x(t)$: $b_n = 0$ (cosine terms only). Odd $x(t)$: $a_0 = a_n = 0$ (sine terms only).
- **Compact (polar) form:** $x(t) = C_0 + \sum_{n=1}^{\infty}C_n\cos(n\omega_0 t + \theta_n)$ with $C_0 = a_0$, $C_n = \sqrt{a_n^2 + b_n^2}$, $\theta_n = -\tan^{-1}(b_n/a_n)$.
- Link to exponential form: $c_0 = a_0$, $c_n = \frac{a_n - jb_n}{2}$, $c_{-n} = \frac{a_n + jb_n}{2}$.

## Check yourself

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

1. What is the trigonometric Fourier series?
   A. a_0 + sum of a_n cos(n w0 t) + b_n sin(n w0 t)
   B. sum of a_n e^(jn w0 t) only
   C. a_0 + sum of a_n cos(n t) only
   D. sum of b_n sin(n w0 t) for n >= 0 only
   Answer: A. It is a dc term plus cosines and sines at multiples of w0. (Key point: Form of the series)

2. What is the integral over T of cos(m w0 t) cos(n w0 t) dt?
   A. T for m = n
   B. 0 for m = n
   C. T/2 for m = n and 0 for m not equal to n
   D. T/2 for all m, n
   Answer: C. Cosines of different harmonics are orthogonal; for m = n the integral is T/2. (Key point: Orthogonality of cosines)

3. What is a_n (n >= 1)?
   A. (1/T) integral x(t) cos(n w0 t) dt
   B. (2/T) integral x(t) cos(n w0 t) dt
   C. (2/T) integral x(t) sin(n w0 t) dt
   D. (T/2) integral x(t) cos(n w0 t) dt
   Answer: B. The integral equals a_n T/2, so a_n = (2/T) times the integral. (Key point: Formula for a_n)

4. What is true for an odd signal x(t)?
   A. b_n = 0
   B. a_0 and all a_n are 0; only sine terms remain
   C. Only a_0 is non-zero
   D. All coefficients are non-zero
   Answer: B. Odd functions have zero average and no cosine terms. (Key point: Symmetry)

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