---
title: "How Fourier series coefficients are calculated"
subject: "Digital Signal Analysis and Processing"
chapter: 3
key: "semester_7/digital_signal_analysis_and_processing:fourier-series-coefficients"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/fourier-series-coefficients
site: BCT Engineering
---

# How Fourier series coefficients are calculated

Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain

- Asked in: 2072 Chaitra, 4 marks; 2073 Shrawan, 3 marks (asked 2 times)
- Video: [Introduction to DT Fourier Series](https://www.youtube.com/watch?v=OMdUYrKMZjE) — Darryl Morrell
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/fourier-series-coefficients
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3

## Question

Explain the process of calculating Fourier series coefficients.

## Short answer

Multiply the periodic signal x(t) by e^{-j k ω_0 t}, integrate over one period T, divide by T: c_k = (1/T)∫_T x(t) e^{-j k ω_0 t} dt. Orthogonality keeps only the k-th harmonic. c_0 gives DC; magnitude and phase spectra are |c_k| and ∠c_k.

## Answer

A periodic signal x(t) with period T (ω₀ = 2π/T) can be written as a sum of harmonically related sinusoids. The Fourier series coefficients tell how much of each harmonic is present.

#### Exponential Fourier series
x(t) = Σ(k = −∞ to ∞) cₖ e^(jkω₀t)

**Process to calculate cₖ:**
1. Find the period T and ω₀ = 2π/T. Check that x(t) satisfies the Dirichlet conditions (finite number of maxima, minima and discontinuities per period, and absolutely integrable).
2. Multiply both sides by e^(−jmω₀t) and integrate over one period. Because ∫ e^(j(k−m)ω₀t) dt over T equals T for k = m and 0 otherwise (orthogonality), only one term survives.
3. This gives the analysis formula:

   cₖ = (1/T) ∫(over one period) x(t) e^(−jkω₀t) dt

4. c₀ = (1/T)∫x(t)dt is the average (DC) value; |cₖ| and ∠cₖ give the magnitude and phase spectra.

#### Trigonometric form
x(t) = a₀ + Σ(n=1 to ∞) [aₙ cos(nω₀t) + bₙ sin(nω₀t)]

- a₀ = (1/T)∫x(t)dt
- aₙ = (2/T)∫x(t) cos(nω₀t) dt
- bₙ = (2/T)∫x(t) sin(nω₀t) dt

with cₖ = (aₖ − jbₖ)/2 for k > 0 and c₋ₖ = cₖ* for real x(t). Symmetry simplifies the work: even x(t) has bₙ = 0; odd x(t) has a₀ = aₙ = 0.

#### Discrete-time (DTFS) for a sequence of period N
cₖ = (1/N) Σ(n = 0 to N−1) x[n] e^(−j2πkn/N), with x[n] = Σ(k=0 to N−1) cₖ e^(j2πkn/N). The cₖ repeat with period N.

#### Example
x(t) = +1 for 0 < t < T/2 and −1 for T/2 < t < T. Then

cₖ = (1/T)[∫₀^(T/2) e^(−jkω₀t)dt − ∫_(T/2)^T e^(−jkω₀t)dt] = (1 − (−1)ᵏ)/(jπk)

So cₖ = 2/(jπk) for odd k and 0 for even k, c₀ = 0. In trigonometric form, bₙ = 4/(πn) for odd n: x(t) = (4/π)[sin ω₀t + (1/3) sin 3ω₀t + (1/5) sin 5ω₀t + …].

## Check yourself

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

1. Why does multiplying by e^(-jmw0t) and integrating over one period isolate a single coefficient?
   A. Because the integral equals 1 for every k
   B. Because e^(-jmw0t) is an even function so the other terms cancel
   C. Because the integral of e^(j(k-m)w0t) over T is T for k = m and 0 otherwise
   D. Because x(t) is zero outside one period
   Answer: C. The harmonics are orthogonal over a period, so only the k = m term survives and gives T*c_m. (Key point: Orthogonality)

2. Which expression gives the exponential Fourier coefficient c_k?
   A. (2/T) times the integral over one period of x(t) e^(jkw0t) dt
   B. The integral over one period of x(t) e^(-jkw0t) dt, with no 1/T
   C. (1/T) times the sum of x(t) over one period
   D. (1/T) times the integral over one period of x(t) e^(-jkw0t) dt
   Answer: D. The analysis formula is c_k = (1/T) integral of x(t) e^(-jkw0t) dt. (Key point: Analysis formula)

3. For a real, odd x(t), which trigonometric coefficients are nonzero?
   A. Only b_n
   B. Only a0 and a_n
   C. Both a_n and b_n
   D. Only a0
   Answer: A. An odd signal has a0 = a_n = 0, so only the sine terms b_n remain. (Key point: Trig relation and symmetry)

4. What does c0 represent?
   A. The amplitude of the fundamental
   B. The average (DC) value of x(t) over one period
   C. The phase of the first harmonic
   D. The total energy in one period
   Answer: B. c0 = (1/T) integral of x(t) dt, the mean value. (Key point: DC term)

## Related in other subjects

- [Fourier series coefficients of a square wave](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-coefficients-square-wave) — Data Communication, chapter 3
- [Fourier series coefficients and plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-coefficients-plot) — Signals and Systems, chapter 2
- [Rayleigh energy theorem; power from Fourier series](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arayleigh-periodic-power-fourier) — Communication Systems, chapter 1
- [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×