--- title: "How Nyquist criterion prevents aliasing" subject: "Signals and Systems" chapter: 4 key: "semester_4/signals_and_systems:nyquist-prevents-aliasing" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/nyquist-prevents-aliasing site: BCT Engineering --- # How Nyquist criterion prevents aliasing Signals and Systems · Chapter 4: Sampling - Asked in: 2080 Chaitra, 4 marks - Video: [Sampling, Aliasing & Nyquist Theorem](https://www.youtube.com/watch?v=yWqrx08UeUs) — 0612 TV w/ NERDfirst - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/nyquist-prevents-aliasing - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4 ## Question How does Nyquist criteria prevent aliasing? Explain with diagram. ## Short answer The Nyquist criterion (f_s ≥ 2f_m) prevents aliasing by ensuring that spectral copies created by sampling do not overlap; the original band −f_m to f_m stays separated from the first copy at f_s−f_m, so an ideal low-pass filter can recover the signal without fold-over distortion. ## Answer The **Nyquist criterion** says a signal band-limited to $f_m$ must be sampled at $f_s \ge 2f_m$. It prevents aliasing because it keeps the spectral copies created by sampling from overlapping. **How it works:** Sampling at rate $f_s$ makes the spectrum periodic: $$ X_s(f) = f_s\sum_{k=-\infty}^{\infty}X(f - kf_s) $$ The original band occupies $-f_m$ to $f_m$. The first copy starts at $f_s - f_m$. There is no overlap when $$ f_s - f_m \ge f_m \;\Rightarrow\; f_s \ge 2f_m $$ ```text (a) fs > 2fm : copies separate, guard band ___ ___ ___ / \ / \ / \ --+--+--+-----+--+--+-----+--+--+--> f -fm 0 fm fs-fm fs 2fs |<->| guard band [ LPF cutoff fs/2 recovers centre copy ] (b) fs = 2fm : copies just touch (ideal LPF needed) ___ ___ ___ / V V \ --+---+---+---+--> f 0 fs 2fs (c) fs < 2fm : copies overlap -> aliasing ____ ____ / X \ --+---+---+----> f 0 fs ``` **Explanation:** - In (a), the original spectrum stays separate, so an ideal low-pass filter with cutoff $f_s/2$ removes the copies and gives back $x(t)$ exactly. - In (b), recovery is possible only with an ideal brick-wall filter. - In (c), the tail of the copy at $f_s$ folds into the band $0$ to $f_s/2$. Those components appear at false frequencies $|f - f_s|$ and cannot be separated by any filter. This is aliasing. **Example:** A 3 kHz tone needs $f_s \ge 6$ kHz. At $f_s = 8$ kHz it is recovered correctly; at $f_s = 4$ kHz it appears as $|3 - 4| = 1$ kHz. In practice, an anti-aliasing low-pass filter (cutoff $f_s/2$) is placed before the sampler to make sure the signal really is band-limited. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does fs >= 2fm avoid overlap? A. Copies move closer together B. The signal amplitude is reduced C. The first copy starts at fs - fm, which is >= fm D. The copies are filtered by the sampler Answer: C. fs - fm >= fm gives fs >= 2fm. (Key point: Condition) 2. When fs < 2fm A. Signal recovery is still exact with a sharp filter B. Only DC is lost C. The spectrum becomes narrower D. Tail of the copy at fs folds into the band and cannot be separated Answer: D. The overlap produces false frequencies. (Key point: fs < 2fm) 3. When fs = 2fm exactly A. Copies just touch and an ideal brick-wall filter is needed B. Aliasing is severe C. No filter is needed D. Copies are separated by a wide guard band Answer: A. No guard band exists. (Key point: fs = 2fm) 4. A 3 kHz tone sampled at 4 kHz appears at A. 3 kHz B. 1 kHz C. 7 kHz D. 4 kHz Answer: B. |3 - 4| = 1 kHz. 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