FOURIER SERIES (Prerequisite for Fourier Transform)

1. PERIODIC FUNCTIONS

A periodic function is a function that repeats its values in regular intervals or periods. Mathematically:
f(t) = f(t + T)
where:
- f(t) is the function.
- T is the period, the smallest positive number such that the function repeats every T units.

Examples:

• sin(t), cos(t) have period 2π.
• Square waves, sawtooth waves, and triangular waves are all periodic.

Importance: Any periodic function can be expressed as a sum of sine and cosine functions (Fourier series). This helps us analyze signals in frequency domain.

2. TRIGONOMETRIC FOURIER SERIES

We can represent any reasonable periodic function f(t) (with period 2π) as:
f(t) = a0/2 + Σ [ an cos(n t) + bn sin(n t) ]
where the summation is from n = 1 to ∞.

Where:

• a0 is the average (DC component)
• an and bn are Fourier coefficients

Formulas for coefficients (for period 2π):

• a0 = (1/π) ∫-ππ f(t) dt
• an = (1/π) ∫-ππ f(t) cos(n t) dt
• bn = (1/π) ∫-ππ f(t) sin(n t) dt

Example 1: f(t) = t on [-π, π]
This is an odd function:
- a0 = 0
- an = 0
- bn = 2 * (-1)n+1 / n

So:
f(t) = Σ [2(-1)n+1/n * sin(n t)]

3. EXPONENTIAL FOURIER SERIES

We can also write the Fourier series using complex exponentials:
f(t) = Σ cn ej n t, where the sum is from n = -∞ to ∞

Where:

Relation with Trigonometric Form:

• an = cn + c-n
• bn = j(cn - c-n)

4. EVEN AND ODD FUNCTION SYMMETRY

Even Functions:

f(-t) = f(t)
- Symmetrical about the y-axis
- Only cosine terms appear (bn = 0)
Example: f(t) = |t|

Odd Functions:

f(-t) = -f(t)
- Symmetrical about the origin
- Only sine terms appear (an = 0)
Example: f(t) = t on [-π, π]

Why this helps: It reduces computation. Knowing the symmetry lets you skip sine or cosine calculations.

5. HALF-RANGE EXPANSIONS

Used when a function is defined on [0, L] instead of [-L, L]. You extend the function to [-L, L] by:

• Even extension: use only cosine terms
• Odd extension: use only sine terms

Formulas (for period 2L):

• a0 = (1/L) ∫0L f(t) dt
• an = (1/L) ∫0L f(t) cos(nπt/L) dt
• bn = (1/L) ∫0L f(t) sin(nπt/L) dt

Example 2: f(t) = t on [0, π]
- Use an odd extension → sine-only Fourier series

DEEP COMPETENT UNDERSTANDING

The idea behind Fourier series is breaking complex, periodic signals into basic building blocks of sines and cosines. It's like using LEGO bricks to recreate any shape.

Once you understand:

• How periodicity influences frequency components
• What each Fourier coefficient tells us
• Why symmetry saves effort
• How half-range expansions mimic real-world conditions

Then you’re not just solving problems — you're analyzing wave behavior, signal compression, electrical systems, acoustics, and image processing.

With deep understanding, you may propose better basis functions than sinusoids — that’s how wavelets, short-time Fourier transforms, and even machine-learned signal bases were born.

20 Solved Examples on Fourier Series

Master periodic functions, Fourier coefficients, symmetry, and half-range expansions with these detailed examples.

Example 1: Verify if f(t) = sin(t) is periodic, and find its period.

Solution:

A function f(t) is periodic if f(t) = f(t + T) for some smallest positive T.

Here, f(t) = sin(t).
Since sin(t + 2π) = sin(t), the period T = 2π.

Answer: Period of sin(t) is 2π.

Example 2: Find a₀ for f(t) = t on [-π, π].

Recall formula: a₀ = (1/π) ∫-ππ f(t) dt

Where:

• a₀ is the average (DC) component of the function over one period.

Calculate:

a₀ = (1/π) ∫-ππ t dt
   = (1/π) [t²/2] from -π to π
   = (1/π) (π²/2 - (−π)²/2)
   = (1/π) (π²/2 - π²/2) = 0
  

Answer: a₀ = 0

Example 3: Find Fourier cosine coefficients aₙ for even function f(t) = |t| on [-π, π].

Recall formula: aₙ = (1/π) ∫-ππ f(t) cos(n t) dt

Since |t| is even: aₙ = (2/π) ∫0π t cos(n t) dt

We calculate:

aₙ = (2/π) ∫0π t cos(n t) dt

Use integration by parts:
Let u = t ⇒ du = dt
Let dv = cos(n t) dt ⇒ v = sin(n t)/n

aₙ = (2/π) [t * sin(n t)/n]_0^π - (2/π) ∫_0^π sin(n t)/n dt
   = (2/π) [π * sin(n π)/n - 0] - (2/π) * (1/n) * [-cos(n t)/n]_0^π
   = (2/π) * (π/n) * sin(n π) + (2/π) * (1/n²) [cos(n π) - 1]

Since sin(n π) = 0 for all integers n,

aₙ = (2/π) * (1/n²) [cos(n π) - 1]
   = (2/π n²) [(-1)^n - 1]
  

Explanation:

• Used integration by parts to solve integral of t cos(n t).
• cos(n π) = (-1)^n alternates between 1 and -1.

Answer: aₙ = (2/(π n²)) * [(-1)^n - 1]

Example 4: Find sine coefficients bₙ for odd function f(t) = t on [-π, π].

Recall formula: bₙ = (1/π) ∫-ππ f(t) sin(n t) dt

Since t is odd: bₙ = (2/π) ∫0π t sin(n t) dt

Calculate by integration by parts:

Let u = t ⇒ du = dt
Let dv = sin(n t) dt ⇒ v = -cos(n t)/n

bₙ = (2/π) [ -t * cos(n t)/n ]_0^π + (2/π) ∫_0^π cos(n t)/n dt
   = (2/π) [ -π * cos(n π)/n - 0 ] + (2/π n) [ sin(n t)/n ]_0^π
   = (-2 π / (π n)) * (-1)^n + 0
   = (2 / n) * (-1)^{n+1}
  

Answer: bₙ = (2 / n) * (-1)^{n+1}

Example 5: Write full Fourier series for f(t) = t on [-π, π].

Since f(t) is odd: no cosine terms (aₙ = 0) and no constant term (a₀ = 0).

Therefore,

f(t) = Σn=1^∞ bₙ sin(n t)
     = Σn=1^∞ (2 / n) * (-1)^{n+1} sin(n t)
  

This series represents the sawtooth wave.

Example 6: Find exponential Fourier coefficient cₙ for f(t) = cos(t) on [-π, π].

Recall formula:

cₙ = (1 / 2π) ∫-ππ cos(t) * e^{-j n t} dt
  

Rewrite cosine using Euler's formula:

cos(t) = (e^{j t} + e^{-j t}) / 2
  

Then:

cₙ = (1 / 2π) ∫-ππ [ (e^{j t} + e^{-j t}) / 2 ] * e^{-j n t} dt
    = (1 / 4π) ∫-ππ (e^{j (1-n) t} + e^{-j (1+n) t}) dt
  

Integral of complex exponentials:

∫ e^{j m t} dt = (e^{j m t}) / (j m)
  

For m ≠ 0 the integral over one full period is zero. So only terms with m=0 survive:

• When 1 - n = 0 ⇒ n = 1
• When -(1 + n) = 0 ⇒ n = -1

Evaluating:

c₁ = 1/2, c_{-1} = 1/2, otherwise 0
  

Answer: cₙ = 1/2 if n = ±1, else 0.

Example 7: Show that Fourier series of even function has zero sine coefficients.

Recall: For sine coefficients:

bₙ = (1/π) ∫-ππ f(t) sin(n t) dt
  

If f(t) is even and sin(n t) is odd, their product is odd.
Integral of an odd function over symmetric limits is zero.
Hence, bₙ = 0.

Example 8: Find half-range sine series for f(t) = t on [0, π].

Recall formula for sine coefficients:

bₙ = (2 / π) ∫_0^π f(t) sin(n t) dt
  

Calculate:

bₙ = (2/π) ∫_0^π t sin(n t) dt

Use integration by parts:

u = t ⇒ du = dt
dv = sin(n t) dt ⇒ v = -cos(n t)/n

bₙ = (2/π) [ -t cos(n t)/n ]_0^π + (2/π) ∫_0^π cos(n t)/n dt
   = (2/π) [ -π cos(n π)/n + 0 ] + (2/π n) [ sin(n t)/n ]_0^π
   = (-2 π / (π n)) * (-1)^n + 0
   = (2 / n) * (-1)^{n+1}
  

Answer: bₙ = (2 / n) * (-1)^{n+1}, same as full-range odd function.

Example 9: Find Fourier cosine series for f(t) = t² on [−π, π].

Since t² is even: only cosine terms and constant term.

Calculate constant term:

a₀ = (1/π) ∫_−π^π t² dt
   = (1/π) * [t³/3]_−π^π
   = (1/π) * (π³/3 − (−π)³/3)
   = (1/π) * (2 π³ / 3)
   = (2 π²) / 3
  

Calculate cosine coefficients:

aₙ = (1/π) ∫_−π^π t² cos(n t) dt
Since t² and cos(n t) are even, integrate from 0 to π and multiply by 2:

aₙ = (2/π) ∫_0^π t² cos(n t) dt

Use integration by parts twice (omitted for brevity):

aₙ = (−4 / n²) * (−1)^n
  

Answer:

f(t) = π²/3 + Σ (−4 / n²) (−1)^n cos(n t)
  

Example 10: Use symmetry to find Fourier series of f(t) = sin²(t).

Note: sin²(t) is even because sin(−t) = −sin(t) but squared makes it positive.

Use identity:

sin²(t) = (1 − cos(2 t)) / 2
  

Fourier series is just constant plus cosine term:

f(t) = 1/2 − (1/2) cos(2 t)
  

Example 11: Calculate Fourier series of a square wave defined by:

f(t) = {
  1 for 0 < t < π
  -1 for -π < t < 0
}
  

Observations: This is an odd function with period 2π.

Sine coefficients:

bₙ = (2/π) ∫_0^π 1 * sin(n t) dt = (2/π) * [−cos(n t)/n]_0^π
    = (2/π) * (1 − cos(n π))/n
    = (4/π n) for odd n, 0 for even n
  

Fourier series:

f(t) = Σn=1,3,5... (4 / (π n)) sin(n t)
  

Example 12: Compute half-range cosine expansion for f(t) = t on [0, π].

Use even extension:

a₀ = (1/π) ∫_0^π t dt = π/2

aₙ = (2/π) ∫_0^π t cos(n t) dt
Using integration by parts (similar to Example 3):

aₙ = (2/π) * [ (−1)^{n+1} / n² ] * π
   = (2 (−1)^{n+1}) / n²
  

Fourier cosine series:

f(t) = π/2 + Σn=1 (2 (−1)^{n+1} / n²) cos(n t)
  

Example 13: Show that the Fourier series of a constant function f(t) = c is just the constant.

a₀ = (1/π) ∫_−π^π c dt = (1/π) * (2π c) = 2 c

aₙ = bₙ = 0 because integrals of sine or cosine times constant over full period are zero.

Fourier series = a₀/2 = c
  

Example 14: Find the Fourier series for a sawtooth wave defined by f(t) = t/π on [-π, π].

Note: Odd function with period 2π.

bₙ = (2/π) ∫_0^π (t/π) sin(n t) dt
Use integration by parts:

bₙ = (2 / (π²)) * [−t cos(n t)/n + ∫ cos(n t)/n dt]_0^π
    = (2 / (π²)) * ( (−π cos(n π)/n) + (sin(n π)/n²) - 0 )
Since sin(n π) = 0:

bₙ = (−2 / (π n)) (−1)^n

Fourier series:

f(t) = Σ (−2 / (π n)) (−1)^{n+1} sin(n t)
  

Example 15: Calculate Parseval's theorem for f(t) = t on [-π, π].

Parseval's theorem states:

(1/π) ∫_−π^π |f(t)|² dt = (a₀²)/2 + Σ (aₙ² + bₙ²)
  

For odd function f(t) = t, a₀ = aₙ = 0, so:

(1/π) ∫_−π^π t² dt = Σ bₙ²
Calculate LHS:

(1/π) * ∫_−π^π t² dt = (1/π) * (2 π³ / 3) = (2 π²) / 3

Calculate RHS using bₙ = 2(-1)^{n+1}/n:

Σ bₙ² = Σ (4 / n²) = 4 Σ (1/n²)
Using known series Σ 1/n² = π² / 6

So Σ bₙ² = 4 * (π² / 6) = (2 π²) / 3

Both sides equal, confirming Parseval's theorem.
  

Example 16: Find Fourier series of f(t) = e^{−|t|} with period 2π.

This is an even function, so only cosine terms and constant term appear.

Constant term:

a₀ = (1/π) ∫_−π^π e^{−|t|} dt
    = (2/π) ∫_0^π e^{−t} dt
    = (2/π) [−e^{−t}]_0^π
    = (2/π)(1 - e^{−π})
  

Cosine coefficients:

aₙ = (2/π) ∫_0^π e^{−t} cos(n t) dt

Use formula for ∫ e^{−α t} cos(β t) dt:
= (1/(α² + β²)) [1 − e^{−α π} (α cos(β π) + β sin(β π))]

Set α=1, β=n:

aₙ = (2/π) * (1/(1 + n²)) * [1 − e^{−π} (cos(n π) + n sin(n π))]

Since sin(n π) = 0:

aₙ = (2/π) * (1/(1 + n²)) * [1 − e^{−π} (−1)^n]
  

Example 17: Explain time-shifting property of Fourier series.

If f(t) has Fourier series with coefficients aₙ, bₙ, then f(t − t₀) has coefficients:

aₙ' = aₙ cos(n t₀) + bₙ sin(n t₀)

bₙ' = bₙ cos(n t₀) − aₙ sin(n t₀)
  

Explanation: Shifting in time domain corresponds to phase shifts in frequency domain.

Example 18: Compute Fourier series of periodic extension of f(t) = t² on [0, L] with half-range cosine expansion.

a₀ = (1/L) ∫_0^L t² dt = L² / 3

aₙ = (2/L) ∫_0^L t² cos(n π t / L) dt

Use integration by parts twice to solve aₙ (details omitted):

aₙ = (4 L² / (n² π²)) (−1)^n
  

Example 19: Explain why Fourier series converges faster for smooth functions.

Because Fourier coefficients decay faster for smooth functions. For infinitely differentiable functions, coefficients decay faster than any power of 1/n.

Sharp corners or discontinuities cause slow decay and Gibbs phenomenon.

Example 20: Discuss how Fourier series relate to Fourier transform.

Fourier series represent periodic signals by discrete frequencies. Fourier transform generalizes this to non-periodic signals, representing them as continuous frequency spectra.

Fourier series coefficients correspond to samples of the Fourier transform at discrete frequencies.

End of examples. With these, you have mastered Fourier Series from basics to advanced understanding.