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Signal Processing Tool

Fourier Series Calculator

Decompose periodic functions into harmonic components using Fourier series analysis. Our signal processing calculator supports coefficient calculation, convergence analysis, and comprehensive frequency domain studies.

Last updated: December 15, 2024

Fourier coefficient calculation
Harmonic analysis
Convergence properties

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Fourier Series Calculator
Decompose periodic functions into harmonic components with Fourier analysis

Use x as variable. Supported: +, -, *, /, ^, sin, cos, tan, ln, sqrt, abs, exp

Fourier Analysis

Fourier Series:

f(x) ≈ 0.0000 + 0.6366sin(1πx/1.0) - 0.3183sin(2πx/1.0) + 0.2122sin(3πx/1.0) - 0.1591sin(4πx/1.0) + 0.1273sin(5πx/1.0)

DC Component:

a₀/2 = 0.0000

Series Type:

Complete Fourier series

Cosine Coefficients:

a₍1₎ = -0.0000
a₍2₎ = 0.0000
a₍3₎ = -0.0000
...

Sine Coefficients:

b₍1₎ = 0.6366
b₍2₎ = -0.3183
b₍3₎ = 0.2122
...

Analysis:

The Fourier series representation of f(x) with period 2 contains 5 harmonic terms. The DC component (a₀/2) is 0.0000, representing the average value.

Calculation Steps:

  1. Function: f(x) = x
  2. Period: T = 2, Half-period: L = 1
  3. DC component: a₀/2 = 0.0000
  4. Fourier series with 5 terms calculated
  5. Series: f(x) ≈ 0.0000 + 0.6366sin(1πx/1.0) - 0.3183sin(2πx/1.0) + 0.2122sin(3πx/1.0) - 0.1591sin(4πx/1.0) + 0.1273sin(5πx/1.0)

Convergence Info:

Energy in first 5 terms: 0.2966

Fourier Series Properties:

  • a₀: DC component (average value)
  • aₙ: Cosine coefficients (even symmetry)
  • bₙ: Sine coefficients (odd symmetry)
  • Convergence: More terms → better approximation

Quick Example Result

For f(x) = x with period T = 2:

f(x) ≈ 0.6366sin(πx) - 0.3183sin(2πx) + 0.2122sin(3πx) - ...

Sawtooth wave decomposition with odd harmonic dominance

How This Calculator Works

Our Fourier series calculator applies advanced signal processing principles to decompose periodic functions into harmonic components. The calculator uses Fourier analysisto compute coefficients and provides comprehensive frequency domain representation.

Fourier Series Formulas

General Form:
f(x) = a₀/2 + Σ(aₙcos(nωx) + bₙsin(nωx))
DC Component:
a₀ = (2/T) ∫ f(x) dx
Fourier Coefficients:
aₙ = (2/T) ∫ f(x)cos(nωx) dx
bₙ = (2/T) ∫ f(x)sin(nωx) dx

The Fourier series decomposes any periodic function into its fundamental frequency and harmonics. Each coefficient represents the amplitude of a specific frequency component, revealing the spectral content of the signal.

📊 Frequency Spectrum Visualization

Shows harmonic components and their amplitudes in frequency domain

Signal Processing Foundation

Fourier series analysis is fundamental to signal processing and engineering applications. By decomposing complex periodic signals into simple sinusoidal components, we can understand the frequency content, design filters, analyze system responses, and solve differential equations in the frequency domain.

  • DC component (a₀/2) represents the average value over one period
  • Fundamental frequency (n=1) determines the basic repetition rate
  • Higher harmonics (n>1) add detail and shape to the waveform
  • Coefficient magnitudes indicate the strength of each frequency component

Sources & References

  • Signals and Systems - Alan V. Oppenheim, Alan S. Willsky (2nd Edition)Comprehensive treatment of Fourier series and signal processing
  • IEEE Signal Processing Society - Professional Standards and EducationLeading organization for signal processing research and education
  • MIT OpenCourseWare - Signals and SystemsEducational resources for Fourier analysis and applications

Need help with other signal processing calculations? Check out our Fourier transform calculator and convolution calculator.

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Example Analysis

Square Wave Decomposition
Analyzing the Fourier series of a square wave signal

Signal Properties:

  • Function: Square wave, ±1
  • Period: T = 2π
  • Symmetry: Odd function
  • Application: Digital signal processing

Fourier Analysis:

  1. DC component: a₀ = 0 (zero average)
  2. Cosine terms: aₙ = 0 (odd symmetry)
  3. Sine terms: bₙ = 4/(nπ) for odd n
  4. Result: Only odd harmonics present

Result: f(x) = (4/π)[sin(x) + sin(3x)/3 + sin(5x)/5 + ...]

The square wave contains only odd harmonics with amplitudes decreasing as 1/n. This is a classic example in signal processing, showing how sharp transitions require many high-frequency components. The Gibbs phenomenon causes overshoot near discontinuities when truncating the series.

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