Unit 3 · Signals & Systems

Fourier Transform —
Aperiodic Signal Analysis

B.Tech ECE · Signals & Systems· ~32 min read· 12 GATE Questions· GATE Weightage: 2–4 marks ⭐
Contents
  1. 1. From Fourier Series to Fourier Transform
  2. 2. Fourier Transform — Definition & Inverse
  3. 3. Standard Fourier Transform Pairs
  4. 4. Properties of Fourier Transform
  5. 5. Convolution Theorem & LTI Systems
  6. 6. Energy Spectral Density (ESD)
  7. 7. Parseval's Theorem for FT
  8. 8. LTI System Frequency Response H(jω)
  9. 9. GATE Questions (12)
Fourier Transform — Time Domain ↔ Frequency Domain
Time Domain x(t) = e⁻ᵃᵗu(t) t x 0 e⁻ᵃᵗ, a > 0 Causal, decays to 0 FT / IFT X(jω)=∫x(t)e⁻ʲωᵗdt x(t)=(1/2π)∫X(jω)eʲωᵗdω Frequency Domain |X(jω)| = 1/√(a²+ω²) ω |X| 0 1/a -a +a Continuous, even symmetric spectrum

For an aperiodic signal, the spectrum is continuous (not a line spectrum like periodic signals). The FT maps time-domain signal ↔ frequency-domain spectrum. For real-valued x(t): amplitude spectrum |X(jω)| is even, phase spectrum ∠X(jω) is odd.

1. From Fourier Series to Fourier Transform

The Fourier Series represents periodic signals as a discrete sum of harmonics. To handle aperiodic signals, we let the period T₀ → ∞ — the discrete harmonics merge into a continuum, and the sum becomes an integral. This limiting process yields the Fourier Transform (FT).

Intuition: FS → FT (as T₀ → ∞)
Fourier Series: x(t) = Σ Cₙ · e^(jnω₀t), Cₙ = (1/T₀)∫x(t)e^(-jnω₀t)dt As T₀ → ∞: nω₀ → ω (continuous), T₀·Cₙ → X(jω) Result: x(t) = (1/2π)∫X(jω)e^(jωt)dω [Inverse FT]
Key Difference: FS vs FT

Fourier Series: Periodic signals → discrete line spectrum (Cₙ at nω₀)
Fourier Transform: Aperiodic signals → continuous spectrum X(jω)
The FT can also represent periodic signals using impulse functions δ(ω).

2. Fourier Transform — Definition & Inverse

Fourier Transform (Analysis Equation)
X(jω) = ∫₋∞^∞ x(t) e^(-jωt) dt Also written as: X(f) = ∫₋∞^∞ x(t) e^(-j2πft) dt [using f in Hz] Note: X(jω) = X(f)|_{f=ω/2π} · (1/2π) — be careful about 2π factors in problems!
Inverse Fourier Transform (Synthesis Equation)
x(t) = (1/2π) ∫₋∞^∞ X(jω) e^(jωt) dω Notation: x(t) ↔ X(jω) means x(t) and X(jω) are a FT pair

Existence Conditions for FT

The FT of x(t) exists if any of these is satisfied:

Sufficient Conditions (Dirichlet for FT)
1. x(t) is absolutely integrable: ∫₋∞^∞ |x(t)| dt < ∞ → Guarantees existence of FT as a regular function 2. x(t) has finite energy: ∫₋∞^∞ |x(t)|² dt < ∞ → FT exists in mean-square sense (Plancherel theorem) 3. For periodic / power signals: FT exists as a distribution with δ(ω) impulses Example: cos(ω₀t) ↔ π[δ(ω−ω₀) + δ(ω+ω₀)]

3. Standard Fourier Transform Pairs

These pairs must be memorized for GATE — especially the exponential, sinc, rect, Gaussian, and signum functions.

x(t)X(jω)Notes / Conditions
δ(t)1Impulse → flat spectrum (all frequencies equally)
1 (DC)2πδ(ω)Constant → impulse at DC
u(t) (unit step)πδ(ω) + 1/(jω)Non-absolutely integrable, distributional FT
e^(−at)u(t)1/(a + jω)a > 0; |X| = 1/√(a²+ω²)
e^(−a|t|)2a/(a² + ω²)a > 0; even, real, Lorentzian spectrum
te^(−at)u(t)1/(a + jω)²a > 0
e^(jω₀t)2πδ(ω − ω₀)Complex exponential → impulse at ω₀
cos(ω₀t)π[δ(ω−ω₀) + δ(ω+ω₀)]Two impulses at ±ω₀
sin(ω₀t)(π/j)[δ(ω−ω₀) − δ(ω+ω₀)]Two impulses, imaginary
rect(t/τ)τ·sinc(ωτ/2) = τ·sin(ωτ/2)/(ωτ/2)rect pulse → sinc spectrum
sinc(Wt) = sin(Wt)/(Wt)(π/W)·rect(ω/2W)sinc in time → rect in freq (by duality)
sgn(t) = ±12/(jω)Signum function
e^(−t²/2σ²)σ√(2π)·e^(−σ²ω²/2)Gaussian → Gaussian (self-dual)
δ(t − t₀)e^(−jωt₀)Time shift → phase shift in frequency
rect(t/τ) ↔ τ·sinc(ωτ/2) — Classic FT Pair & Duality
x(t) = rect(t/τ) -τ/2 +τ/2 A=1 Width = τ in time FT X(jω) = τ·sinc(ωτ/2) -2π/τ +2π/τ Wider τ → narrower sinc (uncertainty principle)

Time-Bandwidth product: Narrower pulse (small τ) → wider spectrum; wider pulse (large τ) → narrower spectrum. This is the time-frequency uncertainty principle.

4. Properties of Fourier Transform

All 10 properties below are GATE-critical. Master these to solve FT problems without computing integrals from scratch.

PropertyTime DomainFrequency Domain
Linearityax(t) + by(t)aX(jω) + bY(jω)
Time Shiftingx(t − t₀)e^(−jωt₀) · X(jω) [magnitude unchanged, phase changes]
Frequency Shifting (Modulation)x(t) · e^(jω₀t)X(j(ω − ω₀)) [spectrum shifted by ω₀]
Time Scalingx(at), a ≠ 0(1/|a|) X(jω/a) [compress time → expand freq]
Time Reversalx(−t)X(−jω) = X*(jω) [for real x(t)]
DualityX(jt) [replace ω by t]2π x(−ω) [key: swap time and freq]
Differentiation (time)d^n x(t)/dt^n(jω)^n X(jω) [nth derivative → multiply by (jω)ⁿ]
Integration∫₋∞^t x(τ)dτX(jω)/(jω) + π X(0)δ(ω) [divides by jω + DC term]
Convolution in timex(t) * h(t)X(jω) · H(jω) [convolution → multiplication]
Multiplication (modulation)x(t) · g(t)(1/2π) X(jω) * G(jω) [multiplication → convolution/2π]
GATE Trap — Time Shifting Phase

Time delay x(t−t₀) adds phase −ωt₀ to X(jω): the magnitude |X(jω)| is unchanged, only the phase angle changes by −ωt₀. Many GATE questions check if you know the magnitude spectrum is invariant under time shifts.

Duality Shortcut — Derive New Pairs

Known: rect(t/τ) ↔ τ·sinc(ωτ/2)
Apply duality (X(jt) ↔ 2πx(−ω)):
τ·sinc(tτ/2) ↔ 2π·rect(−ω/τ) = 2π·rect(ω/τ) [since rect is even]
→ sinc(Wt) ↔ (π/W)·rect(ω/2W) where W = τ/2

4.1 Symmetry Properties (for real x(t))

Symmetry of X(jω) for Real Signals
If x(t) is real: X(−jω) = X*(jω) [Hermitian symmetry] |X(−jω)| = |X(jω)| [Amplitude spectrum is EVEN] ∠X(−jω) = −∠X(jω) [Phase spectrum is ODD] If x(t) is real AND even: X(jω) is real and even If x(t) is real AND odd: X(jω) is purely imaginary and odd

5. Convolution Theorem & LTI Systems

The convolution theorem is the most important result in signal processing — it converts a time-domain convolution (complex integral) to simple multiplication in the frequency domain.

Convolution Theorem
Time domain: y(t) = x(t) * h(t) = ∫x(τ)h(t−τ)dτ ↕ FT Frequency domain: Y(jω) = X(jω) · H(jω) Dual (multiplication in time): y(t) = x(t) · g(t) ↔ Y(jω) = (1/2π) X(jω) * G(jω)
LTI System — Time Domain vs Frequency Domain
TIME DOMAIN x(t) LTI System h(t) — impulse resp. y(t) = x(t)*h(t) Convolution ↓ FT ↓ FT ↓ FT FREQUENCY DOMAIN X(jω) H(jω) = FT{h(t)} Frequency response Y(jω)=X(jω)·H(jω) Multiplication!

The convolution theorem is the power of the Fourier Transform. Instead of computing ∫x(τ)h(t−τ)dτ, simply multiply X(jω)·H(jω) and take the inverse FT.

6. Energy Spectral Density (ESD)

For energy signals (finite energy), the ESD describes how energy is distributed across frequencies.

Energy Spectral Density (ESD)
ESD: S_xx(ω) = |X(jω)|² [units: joules/Hz or J·s/rad] Total Energy: E = ∫₋∞^∞ |x(t)|² dt = (1/2π) ∫₋∞^∞ |X(jω)|² dω [Parseval's theorem] For bilateral (two-sided) representation, energy in band [ω₁, ω₂]: E_band = (1/2π) ∫_{ω₁}^{ω₂} |X(jω)|² dω + (1/2π) ∫_{-ω₂}^{-ω₁} |X(jω)|² dω = (1/π) ∫_{ω₁}^{ω₂} |X(jω)|² dω [since |X|² is even]

ESD of e^(−at)u(t)

Worked Example
x(t) = e^(-at)u(t), a > 0 X(jω) = 1/(a + jω) |X(jω)|² = 1/(a² + ω²) Total energy: E = (1/2π)∫₋∞^∞ 1/(a²+ω²) dω = (1/2π) · (π/a) = 1/(2a) Check via time domain: E = ∫₀^∞ e^(-2at) dt = 1/(2a) ✓

7. Parseval's Theorem for FT

Parseval's Theorem (Energy Conservation)
∫₋∞^∞ |x(t)|² dt = (1/2π) ∫₋∞^∞ |X(jω)|² dω General form (cross-energy): ∫₋∞^∞ x(t)y*(t) dt = (1/2π) ∫₋∞^∞ X(jω)Y*(jω) dω In Hz (using f): ∫₋∞^∞ |x(t)|² dt = ∫₋∞^∞ |X(f)|² df [no 1/2π factor in Hz form]
GATE Application: Compute Energy via FT

∫₀^∞ [sin(ωt)/(ωt)]² dt — use Parseval's theorem. x(t) = sinc(t/π) whose FT is π·rect(ω/2). Then E = (1/2π)∫|π·rect|² dω = (1/2π)·π²·(2ω_c) where ω_c=1, giving E = π/2. Parseval's is consistently tested as a shortcut to avoid direct integration.

8. LTI System Frequency Response H(jω)

Frequency Response of LTI System
H(jω) = Y(jω)/X(jω) = FT{h(t)} Output for sinusoidal input x(t) = A cos(ω₀t + φ): y(t) = A|H(jω₀)| cos(ω₀t + φ + ∠H(jω₀)) Magnitude response: |H(jω)| — scales amplitude at each frequency Phase response: ∠H(jω) — phase shift at each frequency

Ideal Filters

FilterH(jω) = 1 forH(jω) = 0 forh(t)
Ideal LPF (cutoff ωc)|ω| ≤ ωc|ω| > ωcωc/π · sinc(ωct/π) — non-causal, unrealizable
Ideal HPF|ω| > ωc|ω| ≤ ωcδ(t) − LPF impulse response
Ideal BPF (ω₁ to ω₂)ω₁ ≤ |ω| ≤ ω₂OtherwiseLPF(ω₂) − LPF(ω₁) impulse
Ideal BSF (band-stop)|ω| < ω₁ or |ω| > ω₂ω₁ ≤ |ω| ≤ ω₂δ(t) − BPF impulse response
Ideal Filters Are Non-Causal

Ideal LPF: h(t) = (ωc/π)sinc(ωct/π) — non-zero for t < 0, hence non-causal. Real filters (Butterworth, Chebyshev) approximate this with causal, realizable transfer functions.

GATE PYQs

GATE Questions — Fourier Transform (12)

GATE 2023 2 Marks MCQ
The Fourier Transform of x(t) = e^(−2|t|) is:
(A) 4/(4 + ω²)
(B) 4/(4 + ω²)  [same as A — corrected below]  → Answer: 4/(4+ω²)
(C) 2/(2 + jω)
(D) 1/(2 + jω)²

x(t) = e^(−2|t|) is a two-sided decaying exponential.

Using the FT pair: e^(−a|t|) ↔ 2a/(a² + ω²) with a = 2:

X(jω) = 2(2)/(2² + ω²) = 4/(4 + ω²)

Answer: X(jω) = 4/(4 + ω²)

GATE 2022 2 Marks MCQ
If x(t) ↔ X(jω), then the FT of x(t−3) is:
(A) X(jω) e^(j3ω)
(B) X(jω) e^(−j3ω)
(C) X(j(ω−3))
(D) 3·X(jω)

Time shifting property: x(t − t₀) ↔ e^(−jωt₀) X(jω)

With t₀ = 3: FT{x(t−3)} = e^(−j3ω) X(jω)

Note: magnitude spectrum unchanged, phase gets −3ω added.

Answer: (B) X(jω) e^(−j3ω)

GATE 2021 2 Marks MCQ
The FT of x(2t) is (given x(t) ↔ X(jω)):
(A) 2·X(2jω)
(B) (1/2)·X(jω/2)
(C) X(2jω)
(D) (1/2)·X(j2ω)

Time scaling property: x(at) ↔ (1/|a|)·X(jω/a)

With a = 2: FT{x(2t)} = (1/2)·X(jω/2)

Interpretation: compressing time (a=2 > 1) → stretching frequency by 2, amplitude scaled by 1/2.

Answer: (B) (1/2)·X(jω/2)

GATE 2020 2 Marks Numerical
The energy of the signal x(t) = e^(−3t)u(t) is:
(A) 1/6
(B) 1/3
(C) 1/9
(D) 3

Method 1 (time domain): E = ∫₀^∞ |e^(−3t)|² dt = ∫₀^∞ e^(−6t) dt = 1/6

Method 2 (Parseval's): X(jω) = 1/(3+jω), |X|² = 1/(9+ω²)

E = (1/2π)∫₋∞^∞ 1/(9+ω²) dω = (1/2π)·(π/3) = 1/6

Answer: (A) 1/6

GATE 2019 2 Marks MCQ
The Fourier Transform of the signal x(t) = rect(t − 0.5) [unit rectangle pulse of width 1 centered at t=0.5] is:
(A) sinc(ω/2π)
(B) e^(−jω/2)·sinc(ω/2π)
(C) e^(jω/2)·sinc(ω/2π)
(D) sinc(ω/2π)·cos(ω/2)

rect(t) centered at 0: FT = sinc(ω/2π) [using normalized sinc]

Time shift by +0.5 → multiply by e^(−jω·0.5) = e^(−jω/2)

FT{rect(t−0.5)} = e^(−jω/2)·sinc(ω/2π)

Magnitude |X(jω)| = |sinc(ω/2π)| [unchanged by shift], phase gains −ω/2.

Answer: (B) e^(−jω/2)·sinc(ω/2π)

GATE 2018 2 Marks MCQ
By duality, if sinc(t) ↔ rect(ω/2), then rect(t) ↔ ?
(A) sinc(ω)
(B) sinc(ω/2)
(C) 2·sinc(ω/2π)
(D) (1/2)·sinc(ω/2)

Duality: If x(t) ↔ X(jω), then X(jt) ↔ 2πx(−ω)

Known pair: rect(t) ↔ sinc(ω/2) [derived from rect↔sinc definition with proper normalization]

By duality from sinc(t) ↔ rect(ω/2): applying duality to this pair gives rect(t) ↔ 2π·sinc(−ω/2)/2π = sinc(ω/2) [since sinc is even]

Answer: (B) sinc(ω/2)

GATE 2017 1 Mark MCQ
The magnitude spectrum |X(jω)| of x(t) = e^(−3t)u(t) is:
(A) 1/√(9 + ω²)
(B) 1/(9 + ω²)
(C) 1/(3 + ω²)
(D) 3/√(9 + ω²)

X(jω) = 1/(3 + jω) = 1/|3 + jω| · e^(−j∠(3+jω))

|X(jω)| = 1/|3 + jω| = 1/√(3² + ω²) = 1/√(9 + ω²)

Answer: (A) 1/√(9 + ω²)

GATE 2016 2 Marks MCQ
An LTI system has impulse response h(t) = e^(−2t)u(t). If the input is x(t) = e^(−t)u(t), then the output Y(jω) in frequency domain is:
(A) 1/[(1+jω)(2+jω)]
(B) 1/(1+jω) + 1/(2+jω)
(C) e^(−3t)u(t)
(D) 1/(3+jω)

By convolution theorem: Y(jω) = X(jω)·H(jω)

X(jω) = 1/(1+jω), H(jω) = 1/(2+jω) Y(jω) = X·H = 1/[(1+jω)(2+jω)]

Note: y(t) = e^(−t)u(t) * e^(−2t)u(t) = (e^(−t) − e^(−2t))u(t) [by partial fractions in time]

Answer: (A) 1/[(1+jω)(2+jω)]

GATE 2015 2 Marks MCQ
If X(jω) = 2πδ(ω − 5), then x(t) is:
(A) δ(t − 5)
(B) e^(j5t)
(C) cos(5t)
(D) u(t − 5)

Known FT pair: e^(jω₀t) ↔ 2πδ(ω − ω₀)

With ω₀ = 5: X(jω) = 2πδ(ω − 5) → x(t) = e^(j5t)

Intuition: a complex exponential at frequency ω₀ has all its energy at exactly ω₀.

Answer: (B) e^(j5t)

GATE 2014 2 Marks MCQ
The FT of the derivative dx(t)/dt of x(t) = e^(−|t|) is:
(A) 2jω/(1 + ω²)
(B) −2jω/(1 + ω²)
(C) 2/(1 + ω²)
(D) jω·(2/(1+ω²))

x(t) = e^(−|t|) → X(jω) = 2/(1 + ω²) [standard pair with a=1]

Differentiation property: FT{dx/dt} = jω·X(jω)

FT{x'(t)} = jω · 2/(1+ω²) = 2jω/(1+ω²)

Note: x(t) = e^(−|t|) has a kink at t=0, derivative is a sgn(t) correction: x'(t) = −sgn(t)e^(−|t|). The FT result = 2jω/(1+ω²). Answer choice B shows −2jω/(1+ω²) — check sign conventions used in the original exam. Most editions give 2jω/(1+ω²).

Answer: FT{x'(t)} = 2jω/(1+ω²) = jω·X(jω) [differentiation property]

GATE 2016 1 Mark MCQ
For a real-valued signal x(t), which of the following is always true?
(A) X(jω) is real-valued
(B) |X(jω)| = |X(−jω)| and ∠X(jω) = ∠X(−jω)
(C) |X(jω)| = |X(−jω)| and ∠X(jω) = −∠X(−jω)
(D) X(jω) = X*(jω)

For real x(t): Hermitian symmetry holds → X(−jω) = X*(jω)

This means: |X(−jω)| = |X*(jω)| = |X(jω)| [magnitude is EVEN]

And: ∠X(−jω) = ∠X*(jω) = −∠X(jω) [phase is ODD]

Answer: (C)

GATE 2018 2 Marks Numerical
Using Parseval's theorem, the value of ∫₋∞^∞ sinc²(t) dt (where sinc(t) = sin(πt)/(πt)) is:
(A) π
(B) 1
(C) 1/π
(D) 2

sinc(t) = sin(πt)/(πt) has FT: X(jω) = rect(ω/2π) [width 2π, height 1, using normalized sinc convention]

Equivalently: sinc(t) ↔ rect(f) in Hz, and ∫|sinc(t)|² dt = ∫|rect(f)|² df

∫₋∞^∞ sinc²(t) dt = ∫₋∞^∞ |rect(f)|² df = ∫₋₁/₂^{1/2} 1² df = 1

rect(f)=1 for |f| ≤ 1/2 and 0 elsewhere, so the integral is just 1 × (width 1) = 1.

Answer: (B) 1