Unit 2 · Signals & Systems

Fourier Series —
Periodic Signal Representation

B.Tech ECE · Signals & Systems· ~28 min read· 10 GATE Questions· GATE Weightage: 1–2 marks
Contents
  1. 1. Introduction to Fourier Series
  2. 2. Dirichlet Conditions
  3. 3. Trigonometric Fourier Series
  4. 4. Exponential Fourier Series
  5. 5. Fourier Spectrum & Power Spectrum
  6. 6. Symmetry Conditions
  7. 7. Properties of Fourier Series
  8. 8. Parseval's Theorem
  9. 9. GATE Questions (10)

1. Introduction to Fourier Series

Any periodic signal satisfying Dirichlet conditions can be decomposed into a sum of sinusoids (harmonics) — this is the Fourier Series representation. The idea, due to Jean-Baptiste Joseph Fourier (1822), is fundamental to all signal processing.

A periodic signal with fundamental period T₀ has fundamental frequency f₀ = 1/T₀ (Hz) and fundamental angular frequency ω₀ = 2π/T₀ = 2πf₀ (rad/s). The Fourier series expresses x(t) as a sum of harmonics at frequencies 0, f₀, 2f₀, 3f₀, ... (the DC, fundamental, 2nd harmonic, 3rd harmonic, ...).

Fourier Series — Bilateral Amplitude Spectrum of a Square Wave
|Cₙ| Frequency (nω₀) 0 -5ω₀ -4ω₀ -3ω₀ -2ω₀ -ω₀ ω₀ 2ω₀ 3ω₀ 4ω₀ 5ω₀ C₀ 2A/π 0 2A/3π 2A/5π Odd harmonics (non-zero) Even harmonics = 0

Square wave with amplitude A: only odd harmonics exist. Even harmonics are zero due to half-wave symmetry. Spectrum is discrete (line spectrum) — characteristic of all periodic signals.

2. Dirichlet Conditions

A periodic signal x(t) has a Fourier series representation if and only if it satisfies the Dirichlet conditions:

Dirichlet Conditions
1. x(t) is absolutely integrable over one period: ∫₀^T₀ |x(t)| dt < ∞ 2. x(t) has a finite number of maxima and minima in one period. 3. x(t) has a finite number of finite (jump) discontinuities in one period. At points of continuity: FS = x(t) At jump discontinuity t₀: FS = [x(t₀⁺) + x(t₀⁻)] / 2
Gibbs Phenomenon

At a discontinuity, the partial sum of Fourier series shows an overshoot of ~9% (≈ 1.09 times the jump) that does not decrease as more terms are added — only the location of the overshoot narrows. This is the Gibbs phenomenon.

3. Trigonometric Fourier Series

Trigonometric Fourier Series
x(t) = a₀ + Σₙ₌₁^∞ [aₙ cos(nω₀t) + bₙ sin(nω₀t)] where ω₀ = 2π/T₀ and: a₀ = (1/T₀) ∫₀^T₀ x(t) dt ← DC component (average value) aₙ = (2/T₀) ∫₀^T₀ x(t) cos(nω₀t) dt (n = 1, 2, 3, ...) bₙ = (2/T₀) ∫₀^T₀ x(t) sin(nω₀t) dt (n = 1, 2, 3, ...)

Alternative compact trigonometric form using amplitude and phase:

Compact Trigonometric Form
x(t) = C₀ + Σₙ₌₁^∞ Cₙ cos(nω₀t + θₙ) where: C₀ = a₀ (DC) Cₙ = √(aₙ² + bₙ²) (amplitude of nth harmonic) θₙ = −arctan(bₙ/aₙ) (phase of nth harmonic)

Worked Example: Square Wave

Square Wave x(t) = A for 0<t<T₀/2, −A for T₀/2<t<T₀
a₀ = 0 (zero average) aₙ = 0 (odd function → no cosine terms) bₙ = (2A/nπ)[1 − cos(nπ)] = { 4A/nπ (n odd), 0 (n even) } ∴ x(t) = (4A/π)[sin(ω₀t) + (1/3)sin(3ω₀t) + (1/5)sin(5ω₀t) + ...]

4. Exponential Fourier Series

Exponential Fourier Series
x(t) = Σₙ₌₋∞^∞ Cₙ e^(jnω₀t) Complex Fourier coefficient: Cₙ = (1/T₀) ∫₀^T₀ x(t) e^(−jnω₀t) dt Relations to trigonometric coefficients: C₀ = a₀ Cₙ = (aₙ − jbₙ)/2 for n > 0 C₋ₙ = Cₙ* (conjugate symmetry for real x(t)) |Cₙ| = Cₙ/2 = √(aₙ²+bₙ²)/2 ∠Cₙ = −arctan(bₙ/aₙ)
Key Conjugate Symmetry for Real Signals

For any real signal x(t):
• |C₋ₙ| = |Cₙ| → Magnitude spectrum is even
• ∠C₋ₙ = −∠Cₙ → Phase spectrum is odd
• C₋ₙ = Cₙ* (complex conjugate)
This means the two-sided spectrum has mirror symmetry — you only need n ≥ 0 to fully describe a real signal.

Worked Example: Full-Wave Rectified Sine

x(t) = |A sin(ω₀t)|, T₀ = π/ω₀
Cₙ = (2A/π) · 1/(1−4n²) (all harmonics, even and odd) C₀ = 2A/π (DC component — average value of half-wave) Note: All bₙ = 0 (even function), all aₙ computed from: a₀ = 2A/π, aₙ = (4A/π) · (−1)^(n+1)/(n²−1) for n even; 0 for n odd

5. Fourier Spectrum & Power Spectrum

The Fourier spectrum is the plot of Fourier coefficients vs frequency — it is a discrete (line) spectrum for periodic signals.

SpectrumPlotSymmetry for real x(t)
Amplitude Spectrum|Cₙ| vs nω₀Even: |C₋ₙ| = |Cₙ|
Phase Spectrum∠Cₙ vs nω₀Odd: ∠C₋ₙ = −∠Cₙ
Power Spectrum|Cₙ|² vs nω₀Even: always
Power Spectrum — Average Power of nth harmonic
Power at DC: P₀ = |C₀|² = a₀² Power at nth harmonic (n≠0): Pₙ = 2|Cₙ|² = (aₙ²+bₙ²)/2 (factor 2 because both +n and −n components contribute) Total average power: P_total = |C₀|² + 2Σₙ₌₁^∞ |Cₙ|² = a₀² + (1/2)Σₙ₌₁^∞(aₙ²+bₙ²)

6. Symmetry Conditions

Symmetry conditions reduce integration work by telling which Fourier coefficients are zero before computing them:

SymmetryConditionEffect on FS
Even Symmetryx(−t) = x(t)bₙ = 0 (no sine terms); aₙ = (4/T₀)∫₀^(T₀/2) x(t)cos(nω₀t)dt
Odd Symmetryx(−t) = −x(t)a₀ = aₙ = 0 (no cosine terms); bₙ = (4/T₀)∫₀^(T₀/2) x(t)sin(nω₀t)dt
Half-Wave Symmetryx(t) = −x(t±T₀/2)a₀ = 0; aₙ = bₙ = 0 for n even (only odd harmonics present)
Quarter-Wave Symmetry (even)Even + Half-waveOnly aₙ for odd n; bₙ = 0
Quarter-Wave Symmetry (odd)Odd + Half-waveOnly bₙ for odd n; aₙ = 0
GATE Shortcut — Symmetry Identification

• Square wave (odd symmetry + half-wave) → only odd sine harmonics
• Triangular wave (even symmetry + half-wave) → only odd cosine harmonics
• Full-wave rectified sine (even symmetry) → only cosine terms, all harmonics
• Sawtooth wave (odd symmetry, no half-wave) → all sine harmonics

7. Properties of Fourier Series

PropertySignalFS Coefficients
Linearityax(t) + by(t)a·Cₙˣ + b·Cₙʸ
Time Shiftingx(t − t₀)Cₙ · e^(−jnω₀t₀) (magnitude unchanged, phase shifted)
Time Reversalx(−t)C₋ₙ = Cₙ*
Time Scalingx(at), period T₀/aSame Cₙ but ω₀ → aω₀
Conjugationx*(t)C₋ₙ*
Multiplicationx(t)·y(t)Convolution: Σₖ Cₖˣ · C(n−k)ʸ
Convolutionx(t)*y(t) (periodic)T₀ · Cₙˣ · Cₙʸ
Differentiationdx/dtjnω₀ · Cₙ
Integration∫x(τ)dτCₙ/(jnω₀) for n≠0; C₀·t term if n=0

8. Parseval's Theorem for Fourier Series

Parseval's Theorem (Power Conservation)
Average power = (1/T₀) ∫₀^T₀ |x(t)|² dt = Σₙ₌₋∞^∞ |Cₙ|² In terms of trigonometric coefficients: P = a₀² + (1/2)Σₙ₌₁^∞ (aₙ² + bₙ²) Physical interpretation: Total average power = sum of powers in each harmonic P = P₀ + P₁ + P₂ + P₃ + ...

Worked Example — Power of Square Wave

Square wave: amplitude ±A, fundamental period T₀
Actual power (time domain): P = (1/T₀)∫₀^T₀ A² dt = A² From FS: bₙ = 4A/nπ (n odd only) Σ|Cₙ|² = Σ(n=1,3,5,...) 2·(2A/nπ)² = (8A²/π²)[1 + 1/9 + 1/25 + ...] = (8A²/π²)·(π²/8) = A² ✓ (using: Σ1/(2k-1)² = π²/8)
Key Identity Used in GATE

Σₙ₌₁,₃,₅,... (1/n²) = 1 + 1/9 + 1/25 + ... = π²/8
Σₙ₌₁^∞ (1/n²) = π²/6
These identities are frequently needed to verify Parseval's theorem results.

GATE PYQ

GATE Questions — Unit 2: Fourier Series

GATE 20221 MarkMCQ
Q1. The Fourier series of an even periodic signal contains:
(A) Only cosine terms (and DC)
(B) Only sine terms
(C) Both sine and cosine terms
(D) Only DC component

Answer: (A)

For even signals x(−t) = x(t): bₙ = (2/T₀)∫x(t)sin(nω₀t)dt = 0 because x(t) is even and sin(nω₀t) is odd — their product is odd, and the integral of an odd function over a symmetric interval is zero. Only cosine terms (and DC) remain.

GATE 20211 MarkMCQ
Q2. A periodic signal x(t) with period T₀ has Fourier series coefficients Cₙ. If y(t) = x(t − T₀/4), the coefficients Dₙ of y(t) are:
(A) Cₙ · e^(jnπ/2)
(B) Cₙ · e^(−jnπ/2)
(C) Cₙ · (−1)ⁿ
(D) Cₙ / n

Answer: (B)

Time shift property: x(t − t₀) ↔ Cₙ · e^(−jnω₀t₀) t₀ = T₀/4, ω₀ = 2π/T₀ Dₙ = Cₙ · e^(−jnω₀·T₀/4) = Cₙ · e^(−jn·2π/T₀·T₀/4) = Cₙ · e^(−jnπ/2)

Note: magnitude unchanged |Dₙ| = |Cₙ|; only phase shifts by −nπ/2 per harmonic.

GATE 20202 MarksNAT
Q3. The average power of x(t) = 2 + 4cos(2πt) + 3sin(4πt) is ___ W.
(A) 16.5
(B) 9
(C) 29
(D) 25

Answer: (A) 16.5 W

P_DC = a₀² = 2² = 4 P₁ = (a₁²)/2 = (4²)/2 = 8 [cos(2πt): a₁=4, b₁=0] P₂ = (b₂²)/2 = (3²)/2 = 4.5 [sin(4πt): a₂=0, b₂=3] Total = 4 + 8 + 4.5 = 16.5 W
GATE 20192 MarksMCQ
Q4. The Fourier series of x(t) = |cos(πt)| (full-wave rectified cosine) contains:
(A) Only odd cosine harmonics
(B) DC and even cosine harmonics
(C) Only odd sine harmonics
(D) All harmonics of cos(πt)

Answer: (B)

|cos(πt)| is an even function (even symmetry → no sine terms, bₙ = 0). Its period T₀ = 1 (half the period of cos(πt)), so ω₀ = 2π. Fundamental frequency doubles. The FS contains DC (a₀ = 2/π) and cosine terms at multiples of 2π — these are the even harmonics of the original signal. Hence: DC and even cosine harmonics.

GATE 20181 MarkMCQ
Q5. A periodic signal has half-wave symmetry. Which statement is TRUE?
(A) All harmonics are present
(B) Only odd harmonics are present
(C) Only even harmonics are present
(D) Only the DC component exists

Answer: (B)

Half-wave symmetry: x(t + T₀/2) = −x(t). This implies a₀ = 0 (zero average) and aₙ = bₙ = 0 for all even n. Only odd harmonics (n = 1, 3, 5, ...) appear in the Fourier series.

GATE 20172 MarksNAT
Q6. The complex Fourier series coefficient C₁ of x(t) = 1 + 2cos(2πt) + 3sin(2πt) (T₀ = 1) is:
(A) 1 − j1.5
(B) 1 + j1.5
(C) 2 − j3
(D) 2 + j3

Answer: (A)

a₁ = 2, b₁ = 3 (for 2cos(2πt) + 3sin(2πt)) C₁ = (a₁ − jb₁)/2 = (2 − j3)/2 = 1 − j1.5

Check: C₋₁ = C₁* = 1 + j1.5. Magnitude |C₁| = √(1² + 1.5²) = √3.25.

GATE 20162 MarksMCQ
Q7. If x(t) is a real periodic signal with FS coefficients Cₙ, then the FS coefficients of dx/dt are:
(A) jnω₀Cₙ
(B) Cₙ/(jnω₀)
(C) (jω₀)Cₙ
(D) nCₙ

Answer: (A)

Differentiation property of FS: d/dt [Σ Cₙ e^(jnω₀t)] = Σ Cₙ · (jnω₀) · e^(jnω₀t) New FS coefficients = jnω₀ · Cₙ

This means differentiation emphasizes high frequencies (large n) — it acts as a high-pass operation.

GATE 20152 MarksNAT
Q8. A square wave with amplitude ±1 V and period T₀ = 2π has FS: x(t) = (4/π)[sin(t) + sin(3t)/3 + sin(5t)/5 + ...]. Using Parseval's theorem, compute Σₙ₌₁,₃,₅... 1/n².
(A) π²/8
(B) π²/6
(C) π²/4
(D) π/4

Answer: (A) π²/8

Average power of square wave (±1 V): P = 1² = 1 W From FS: bₙ = 4/(nπ) for n = 1,3,5,... P = (1/2)Σ bₙ² = (1/2)Σ(n=1,3,5) (4/nπ)² = (8/π²) Σ(1/n²) Setting P = 1: 1 = (8/π²) Σ(1/n²) ⟹ Σ(n=1,3,5,...) 1/n² = π²/8
GATE 20141 MarkMCQ
Q9. A signal x(t) with period T₀ has FS representation. The FS of x(t)·cos(ω₀t) has non-zero coefficients at:
(A) All harmonics ±1 shifted from original non-zero harmonics
(B) Only the fundamental frequency
(C) Exactly at n = 1 only
(D) Even harmonics only

Answer: (A)

cos(ω₀t) = (e^(jω₀t) + e^(−jω₀t))/2 → FS coefficients: 1/2 at n=±1 x(t)·cos(ω₀t) → convolution of Cₙˣ with Cₙcos Dₙ = (Cₙ₋₁ˣ + Cₙ₊₁ˣ)/2

Each non-zero coefficient Cₙˣ produces contributions at n−1 and n+1. So all original harmonics shift by ±1.

GATE 20232 MarksMCQ
Q10. At a jump discontinuity of a periodic signal, the Fourier series converges to:
(A) The value just before the discontinuity x(t₀⁻)
(B) The value just after the discontinuity x(t₀⁺)
(C) The average [x(t₀⁺) + x(t₀⁻)] / 2
(D) Zero

Answer: (C)

This follows directly from the Dirichlet conditions. At a jump discontinuity, the Fourier series converges to the arithmetic mean of the left-hand and right-hand limits. For example, the square wave at t = 0 (where it jumps from −A to A) has FS value = (A + (−A))/2 = 0.