Unit 6 · Signals & Systems

Sampling Theory —
Nyquist, Aliasing & Reconstruction

B.Tech ECE · Signals & Systems· ~25 min read· 8 GATE Questions· GATE Weightage: 1–2 marks
Contents
  1. 1. Why Sampling? CT → DT
  2. 2. Nyquist-Shannon Sampling Theorem
  3. 3. Spectrum of Sampled Signal
  4. 4. Aliasing — Cause & Effect
  5. 5. Ideal Reconstruction
  6. 6. Practical Sampling — ZOH, Natural, Flat-Top
  7. 7. Sampling Rates in Real Systems
  8. 8. GATE Questions (8)
Real-World Context — Why Sampling Matters Everywhere

CD Audio (1980s): Audio bandwidth ≈ 20 kHz. Nyquist rate = 40 kHz. CD standard: fs = 44.1 kHz (10% above Nyquist for filter rolloff). 16-bit samples → 1.4 Mbps data rate.

Medical ECG: Heart rate signal bandwidth ≈ 150 Hz. Standard ECG sampling: 250–1000 Hz (well above Nyquist). Anti-aliasing LPF at 75 Hz before sampling to prevent muscle artifact aliasing.

Phone calls (GSM/4G): Voice bandwidth = 300–3400 Hz (telephony). Nyquist rate = 6800 Hz. Sampled at 8 kHz → 64 kbps (PCM voice). Your phone's codec applies a 3.4 kHz anti-aliasing filter before sampling every 125 μs.

Stroboscopic effect (aliasing in video): A fan spinning at 30 Hz recorded at 30 fps appears stationary — it completes one full rotation per frame. This is spatial/temporal aliasing. The same effect makes car wheels appear to spin backward in movies.

Sampling & Reconstruction — Complete System Block Diagram
Analog Signal x(t) fₘ Hz BW Anti-aliasing LPF cutoff = fₘ Sampler T = 1/fs fs ≥ 2fₘ Clock (fs) Discrete x[n]=x(nT) digital data Ideal LPF Reconstruct cutoff = fs/2 Analog Output x̂(t) ≡ x(t) if fs≥2fₘ ← Transmitter/ADC side ——————— Receiver/DAC side → 1 2 3 4 5 6

Complete ADC→DAC chain: (1) Analog signal → (2) Anti-aliasing LPF removes components above fₘ → (3) Sampler creates x[n]=x(nT) at rate fs≥2fₘ → (4) Digital data → (5) Reconstruction LPF → (6) Perfect analog output x̂(t)=x(t).

1. Why Sampling? CT → DT

Digital systems can only process discrete sequences, not continuous-time waveforms. Sampling converts a CT signal x(t) into a DT sequence x[n] = x(nT) by measuring x(t) at uniform intervals T seconds apart. The key question: How fast must we sample to avoid losing information?

2. Nyquist-Shannon Sampling Theorem

Nyquist-Shannon Sampling Theorem
A band-limited signal x(t) with maximum frequency fₘ Hz can be PERFECTLY reconstructed from its uniform samples x[n] = x(nT) if: Sampling frequency: fs ≥ 2fₘ [Nyquist criterion] Nyquist rate: fₛ_Nyquist = 2fₘ [minimum sampling rate] Nyquist interval: T_Nyquist = 1/(2fₘ) [maximum sampling period] If fs < 2fₘ → aliasing occurs → perfect reconstruction IMPOSSIBLE. If fs = 2fₘ → theoretically just sufficient (practical systems use fs > 2fₘ).
GATE Shortcut — Nyquist Rate of Product/Sum of Signals

x₁(t) has BW = f₁ Hz, x₂(t) has BW = f₂ Hz:
x₁(t) + x₂(t): Nyquist rate = 2·max(f₁, f₂)
x₁(t) · x₂(t): Bandwidth = f₁ + f₂ → Nyquist rate = 2(f₁+f₂) [multiplication = convolution in freq]
x(t²): time compression by 2 → BW doubles → Nyquist rate doubles

3. Spectrum of Sampled Signal

Sampling in time domain is multiplication by an impulse train. In frequency domain, this causes periodic replication of the spectrum.

Sampling = Multiplication by Impulse Train
Impulse train: p(t) = Σₙ δ(t − nT) ↔ P(jω) = (2π/T) Σₖ δ(ω − kωs) Sampled signal: xs(t) = x(t) · p(t) = Σₙ x(nT)δ(t − nT) Spectrum: Xs(jω) = (1/2π) X(jω) * P(jω) = (1/T) Σₖ X(j(ω − kωs)) → Spectrum of xs(t) is X(jω) repeated at multiples of ωs = 2π/T = 2πfs → Called PERIODIC EXTENSION (spectral replicas)
Nyquist Sampling vs Aliasing — Frequency Spectrum Comparison
Original Spectrum X(jω) — Bandwidth fₘ X(jω) -fₘ +fₘ 0 Case 1: fs > 2fₘ — NO Aliasing (Nyquist satisfied) original gap gap gap Replicas are separate — LPF recovers X(jω) perfectly ✓ LPF selects this Case 2: fs < 2fₘ — ALIASING (Nyquist violated!) OVERLAP! Replicas overlap → components mix → irrecoverable aliasing ✗ High-freq components appear as low-freq aliases — cannot be separated by LPF ⚠ Reduce fs → more severe aliasing; increase fs above 2fₘ to prevent it

Adequate sampling keeps spectral copies separated (top case). Under-sampling causes overlap (aliasing) — the reconstruction LPF cannot separate them, so higher-frequency components appear at wrong frequencies.

4. Aliasing — Cause & Effect

Aliasing occurs when the sampling rate is below the Nyquist rate, causing spectral replicas to overlap. A signal at frequency f appears at an alias frequency |f − kfs| after sampling.

Alias Frequency Formula
When sampling signal at frequency f with sampling rate fs (fs < 2f): Alias frequency = |f − round(f/fs)·fs| Simple rule: alias of f = |f mod fs| if |f mod fs| ≤ fs/2 = fs − |f mod fs| otherwise Example: Signal at f = 7 kHz, sampled at fs = 8 kHz Alias = |7000 − 8000| = 1000 Hz → 7 kHz signal appears as 1 kHz alias!
Real-Life Example — Aliasing in Telephone Systems

Telephone bandwidth is limited to 3.4 kHz for a reason: The public switched telephone network (PSTN) uses 8 kHz sampling. Nyquist says maximum frequency = fs/2 = 4 kHz. A practical anti-aliasing LPF with rolloff limits to 3.4 kHz (not 4 kHz), leaving 600 Hz guard band for the filter transition.

If a voice signal at 5 kHz were allowed through: alias = |5000 − 8000| = 3000 Hz. The 5 kHz sound would masquerade as 3 kHz — wrong pitch! This is why high-quality phone calls (HD Voice, VoLTE) use 16 kHz sampling and 7 kHz bandwidth — much better voice quality.

5. Ideal Reconstruction

Ideal Reconstruction from Samples
Step 1: Convert samples to impulse train: xs(t) = Σₙ x[n] · δ(t − nT) Step 2: Pass through ideal LPF: H_r(jω) = T for |ω| ≤ π/T = ωs/2 0 otherwise [cutoff at Nyquist freq ωs/2 = πfs] Result: x̂(t) = xs(t) * h_r(t) where h_r(t) = sinc(πt/T) x̂(t) = Σₙ x[n] · sinc(π(t−nT)/T) [Whittaker-Shannon interpolation] If fs ≥ 2fₘ: x̂(t) = x(t) exactly (perfect reconstruction)
Reconstruction = Sinc Interpolation

Each sample x[nT] is multiplied by a shifted sinc function and the results are summed. The sinc functions have zeros at all other sample points, so they interpolate perfectly between samples. This is the theoretical basis — practical systems use ZOH (staircase) or polynomial interpolation instead of the ideal sinc.

6. Practical Sampling — ZOH, Natural, Flat-Top

Sampling MethodDescriptionSpectrum EffectApplication
Ideal (Impulse) SamplingMultiply by impulse train. Instantaneous samples.Periodic replicas, no distortion within replicaTheoretical analysis
Natural SamplingGate pulses follow signal shape during sampling interval τ.Replicas with sinc (due to pulse width) envelope: sinc(nτ/T)Analog multiplexer switches
Flat-Top (Sample & Hold)Sample held constant (flat top) during pulse interval τ. Rectangular pulse shape.Spectrum distorted by sinc(ωτ/2): high-freq roll-offADC input stage (S&H circuit)
ZOH (Zero-Order Hold)Each sample held for full period T (staircase output).H_ZOH(jω) = T·sinc(ωT/2)·e^(−jωT/2): sinc distortion + linear phaseDAC output stage

ZOH Frequency Response

Zero-Order Hold (ZOH) Transfer Function
Pulse shape: rect pulse of width T → FT: T·sinc(ωT/2)·e^(-jωT/2) H_ZOH(jω) = T · sinc(ωT/2) · e^(-jωT/2) = T · sin(ωT/2)/(ωT/2) · e^(-jωT/2) Magnitude: |H_ZOH| = T · |sinc(ωT/2)| — rolls off at high freq Phase: ∠H_ZOH = −ωT/2 — linear phase (T/2 group delay) To compensate: apply post-reconstruction equalizer with 1/sinc response. In practice: the sinc roll-off is small at ω ≪ ωs/2 (within signal band) and often acceptable without equalization.

7. Sampling Rates in Real Systems

ApplicationSignal BW (fₘ)Nyquist RateActual fsReason for higher fs
Telephone (PSTN)3.4 kHz6.8 kHz8 kHzFilter transition band margin
CD Audio20 kHz40 kHz44.1 kHz10% margin + historical reasons
DAT / Studio Audio20 kHz40 kHz48 kHzProfessional standard
Hi-Res Audio40 kHz80 kHz96/192 kHzHigher fidelity, easier anti-aliasing
Medical ECG150 Hz300 Hz500–1000 HzHigh oversampling for accuracy
Seismic monitoring100 Hz200 Hz500 HzCapture all relevant frequencies
GATE PYQs

GATE Questions — Sampling Theory (8)

GATE 2023 1 Mark MCQ
A signal x(t) = cos(2000πt) + sin(4000πt) is sampled at fs = 3000 Hz. What aliasing occurs?
(A) No aliasing
(B) The 2000 Hz component aliases
(C) The 2000 Hz component aliases to 1000 Hz
(D) The 1000 Hz component aliases

x(t) has components at f₁ = 1000 Hz and f₂ = 2000 Hz. fs = 3000 Hz.

Nyquist rate = 2 × 2000 = 4000 Hz. Since fs = 3000 < 4000 → aliasing!

f₁ = 1000 Hz: |1000| < fs/2 = 1500 → No aliasing for 1000 Hz ✓

f₂ = 2000 Hz: 2000 > fs/2 = 1500 → Alias = |2000 − 3000| = 1000 Hz

Answer: (C) — 2000 Hz component aliases to 1000 Hz

GATE 2022 1 Mark MCQ
The Nyquist sampling rate for the signal x(t) = sinc(100πt)·sinc(200πt) is:
(A) 100 Hz
(B) 200 Hz
(C) 300 Hz
(D) 400 Hz

sinc(2Wt) ↔ (1/2W)rect(f/2W): bandwidth of sinc(2Wt) = W Hz.

sinc(100πt) = sinc(2·50·πt): bandwidth = 50 Hz

sinc(200πt) = sinc(2·100·πt): bandwidth = 100 Hz

Multiplication in time → convolution in frequency: BW = 50 + 100 = 150 Hz

Nyquist rate = 2 × 150 = 300 Hz

Answer: (C) 300 Hz

GATE 2021 2 Marks MCQ
A 5 kHz signal is sampled at 8 kHz. What is the alias frequency that appears in the reconstructed signal?
(A) 5 kHz
(B) 8 kHz
(C) 3 kHz
(D) 13 kHz

Signal at 5 kHz, sampled at fs = 8 kHz. Nyquist limit = 4 kHz.

5 kHz > 4 kHz → aliasing. Alias = |f − fs| = |5000 − 8000| = 3000 Hz = 3 kHz

Answer: (C) 3 kHz

GATE 2020 1 Mark MCQ
The Nyquist rate for x(t) = 10cos(50πt)cos(300πt) is:
(A) 150 Hz
(B) 300 Hz
(C) 350 Hz
(D) 700 Hz

cos(50πt) has frequency 25 Hz. cos(300πt) has frequency 150 Hz.

Product: x(t) = 10cos(50πt)cos(300πt) = 5[cos(250πt) + cos(350πt)]

Using: 2cosA·cosB = cos(A+B) + cos(A-B):

x(t) = 5cos((300-50)πt) + 5cos((300+50)πt) = 5cos(250πt) + 5cos(350πt) → frequencies: 125 Hz and 175 Hz

Maximum frequency = 175 Hz → Nyquist rate = 2 × 175 = 350 Hz

Answer: (C) 350 Hz

GATE 2019 2 Marks MCQ
The ideal reconstruction filter for a signal sampled at fs Hz is:
(A) LPF with cutoff fₘ and gain 1
(B) LPF with cutoff fs/2 and gain T = 1/fs
(C) HPF with cutoff fs/2 and gain T
(D) BPF centered at fs with gain 1

After ideal sampling (impulse train), spectrum is (1/T)·X(jω) repeated at multiples of ωs.

Reconstruction filter: select the baseband copy → LPF with cutoff at ωs/2 = πfs (or fs/2 in Hz), gain = T.

The gain T compensates for the 1/T factor in the sampled spectrum.

Answer: (B) LPF with cutoff fs/2 and gain T = 1/fs

GATE 2018 2 Marks MCQ
In flat-top sampling, the distortion compared to ideal sampling is:
(A) Linear distortion only
(B) Nonlinear distortion that cannot be corrected
(C) Aperture effect (high-frequency attenuation) that can be equalized
(D) No distortion at all

Flat-top sampling: each sample is held at constant amplitude for pulse duration τ.

This introduces a sinc(ωτ/2) envelope distortion — high frequencies are attenuated (aperture effect).

It is a LINEAR (amplitude/phase) distortion → can be corrected by an equalizer with response 1/sinc(ωτ/2).

Answer: (C) Aperture effect (linear, correctable)

GATE 2016 2 Marks MCQ
The Nyquist sampling rate for x(t) = sinc²(100t) is:
(A) 100 Hz
(B) 200 Hz
(C) 400 Hz
(D) 50 Hz

sinc(100t) = sin(100πt)/(100πt). This is sinc(2W·πt) with 2W = 100 → W = 50 Hz? Let's use the standard: sinc(Wt) = sin(Wt)/(Wt) has FT = (π/W)rect(ω/2W). Bandwidth = W/2π Hz when using ω, or W/(2π) Hz.

sinc(100t) — bandwidth using convention: FT = (1/100)rect(f/100)? Depends on convention. For GATE purposes: sinc²(100t) → bandwidth = 100/π Hz... Let's use the simpler approach.

x(t) = sinc²(100t). In freq domain: X(f) = triangle(f/100) [since sinc² ↔ triangle]. Bandwidth = 100 Hz (max non-zero frequency at ±100)? Actually for sinc²(Wt) where sinc(Wt)=sin(Wt)/(Wt): bandwidth is W/π. But GATE typically uses: sinc(100t) has bandwidth 100/(2π)... This is convention-dependent. Most GATE solutions use BW of sinc²(100t) = 100/π ≈ 31.8 Hz → Nyquist rate ≈ 200/π Hz. Most accepted GATE answer is 200 Hz based on treating the bandwidth as 100 Hz for the function sinc(100πt).

Using the standard GATE convention where sinc(2πfₘt) has BW = fₘ, and sin(100πt)/(100πt) = sinc(100πt) → fₘ = 50 Hz → sinc²(100πt) has BW = 100 Hz → Nyquist rate = 200 Hz.

Answer: (B) 200 Hz

GATE 2014 1 Mark MCQ
A band-limited signal with fₘ = 5 kHz is sampled at exactly fs = 2fₘ = 10 kHz (Nyquist rate). Perfect reconstruction is:
(A) Always possible
(B) Never possible
(C) Theoretically possible but practically difficult
(D) Possible only for periodic signals

At exactly fs = 2fₘ, spectral replicas just touch at ωs/2 = ωₘ — they neither overlap nor have a gap.

Theoretically, an ideal brick-wall LPF with exactly unity gain for |f|≤fₘ and zero for |f|>fₘ can separate them.

Practically: an ideal brick-wall LPF is non-causal (sinc impulse response extending to −∞). Practical filters have finite transition bands and cannot achieve perfect brick-wall cutoff → reconstruction error at the boundary.

This is why practical systems use fs = 1.1 to 2× the theoretical Nyquist rate.

Answer: (C) Theoretically possible but practically requires an ideal brick-wall filter (non-realizable)