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

# Derive exponential Fourier series (CT)

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2082 Chaitra, 5 marks; 2082 Kartik, 6 marks; 2083 Bhadra (new course), 4 marks; 2082 Bhadra (new course), 4 marks; 2075 Baisakh, 5 marks; 2074 Bhadra, 5 marks; 2072 Asoj, 5 marks; 2071 Bhadra, 7 marks; 2069 Bhadra, 8 marks (asked 9 times)
- 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-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Derive the exponential form of the Fourier series of a continuous-time periodic signal x(t) (synthesis and analysis equations).

## Short answer

Synthesis: x(t)=∑ a_k e^(jkω0 t). Analysis: a_k = (1/T)∫_T x(t) e^(-jkω0 t) dt. Derived by multiplying synthesis by e^(-jnω0 t), integrating over one period T, and applying orthogonality: the integral of e^(j(k-n)ω0 t) is T for k=n, 0 otherwise.

## Answer

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

#### Synthesis equation
The set $\phi_k(t) = e^{jk\omega_0 t}$, $k = 0, \pm1, \pm2, \dots$, contains signals that are all periodic with period $T$ (the $k$th one has frequency $k\omega_0$). A linear combination of them is also periodic with period $T$:

$$
x(t) = \sum_{k=-\infty}^{\infty} a_k\, e^{jk\omega_0 t} = \sum_{k=-\infty}^{\infty} a_k\, e^{jk(2\pi/T)t}
$$

This is the **synthesis equation**. $a_0$ is the dc term, $k = \pm1$ are the fundamental components, and $k = \pm N$ are the $N$th harmonics.

#### Analysis equation
To find $a_k$, multiply both sides by $e^{-jn\omega_0 t}$ and integrate over one period $T$:

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

Evaluate the integral on the right (orthogonality):

$$
\int_0^T e^{j(k-n)\omega_0 t}dt = \int_0^T \cos((k-n)\omega_0 t)\,dt + j\int_0^T \sin((k-n)\omega_0 t)\,dt
$$

- For $k \ne n$, both integrals cover a whole number $(k-n)$ of periods, so they are 0.
- For $k = n$, the integrand is 1 and the integral is $T$.

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

So only the $k = n$ term survives:

$$
\int_0^T x(t)e^{-jn\omega_0 t}dt = a_n T
\quad\Rightarrow\quad
a_n = \frac{1}{T}\int_0^T x(t)e^{-jn\omega_0 t}dt
$$

Since any interval of length $T$ gives the same result, the **analysis equation** is

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

#### The Fourier series pair
$$
x(t) \overset{FS}{\longleftrightarrow} a_k
$$

| Equation | Formula |
|---|---|
| Synthesis | $x(t) = \sum_k a_k e^{jk\omega_0 t}$ |
| Analysis | $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$ |
| DC value | $a_0 = \frac{1}{T}\int_T x(t)\,dt$ (average value) |

Notes:
- The $a_k$ are the **Fourier series coefficients** or **spectral coefficients**; they are complex in general, $a_k = |a_k|e^{j\angle a_k}$, giving the magnitude and phase spectra.
- For real $x(t)$, $a_{-k} = a_k^*$, so $|a_k|$ is even and $\angle a_k$ is odd in $k$.

Example: $x(t) = \cos\omega_0 t = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$ gives $a_1 = a_{-1} = 1/2$ and all other $a_k = 0$.

## Check yourself

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

1. What are the basis functions of the exponential Fourier series of a signal with period T?
   A. cos(kt) and sin(kt)
   B. e^(jk w0 t) with w0 = 2pi/T
   C. e^(-k w0 t)
   D. e^(j w0 t/k)
   Answer: B. Harmonically related complex exponentials e^(jk w0 t) are all periodic with period T. (Key point: Basis functions)

2. What is the integral over one period of e^(j(k-n) w0 t) dt?
   A. T for every k and n
   B. 0 for every k and n
   C. T when k = n and 0 when k is not n
   D. T/2 when k = n and 0 otherwise
   Answer: C. For k not equal to n it covers whole cycles and gives 0; for k = n the integrand is 1, giving T. (Key point: Orthogonality)

3. Which is the analysis equation for a_k?
   A. a_k = (1/T) integral over T of x(t) e^(-jk w0 t) dt
   B. a_k = integral over T of x(t) e^(jk w0 t) dt
   C. a_k = (2/T) integral over T of x(t) e^(-jk w0 t) dt
   D. a_k = (1/T) integral over T of x(t) cos(k w0 t) dt
   Answer: A. Multiplying by e^(-jn w0 t) and integrating gives a_n T, so a_n = (1/T) times the integral. (Key point: Analysis equation)

4. What does a_0 represent?
   A. The fundamental amplitude
   B. The average (dc) value (1/T) integral of x(t) dt
   C. The peak value of x(t)
   D. The total power
   Answer: B. With k = 0 the exponential is 1, so a_0 is the average value over one period. (Key point: DC coefficient)

## Close variants set in other exams

- [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×

## Related in Signals and Systems

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