--- title: "Nyquist criteria, aliasing; 30sin100πt + …" subject: "Signals and Systems" chapter: 4 key: "semester_4/signals_and_systems:nyquist-criteria-aliasing-example" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/nyquist-criteria-aliasing-example site: BCT Engineering --- # Nyquist criteria, aliasing; 30sin100πt + … Signals and Systems · Chapter 4: Sampling - Asked in: 2081 Asoj, 4+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-criteria-aliasing-example - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4 ## Question Explain the Nyquist Criteria for sampling and aliasing effect. Find the Nyquist rate for the signal x(t) = 30 sin(100πt) + 40 cos(400πt) sin(300πt). ## Short answer Nyquist criterion: for a signal band-limited to fm, choose fs ≥ 2fm; otherwise spectral copies overlap and aliasing makes frequency f appear at |f−kfs|. Use sinAcosB to simplify x(t) = 10sin100πt + 20sin700πt; fm = 350 Hz. Nyquist rate = 700 samples/s, interval ≈ 1.43 ms. ## Answer #### Nyquist criterion For a signal band-limited to $f_m$ Hz, the sampling frequency must satisfy $$ f_s \ge 2f_m $$ The minimum value $f_N = 2f_m$ is the **Nyquist rate** and $T_N = 1/(2f_m)$ is the **Nyquist interval**. When this holds, the spectral copies produced by sampling, $X_s(f) = f_s\sum_k X(f - kf_s)$, do not overlap, and an ideal low-pass filter of cutoff $f_m$ to $f_s - f_m$ recovers the original signal exactly. #### Aliasing effect If $f_s < 2f_m$, the copies at $0$ and $\pm f_s$ overlap. A component of frequency $f > f_s/2$ then appears at the alias frequency $|f - kf_s|$ (the nearest one in $0$ to $f_s/2$). Example: 900 Hz sampled at 1200 Hz looks like 300 Hz. The overlap cannot be undone after sampling. It is prevented by an anti-aliasing low-pass filter before the sampler and by choosing $f_s > 2f_m$. ```text fs >= 2fm: /\ /\ /\ separate copies -+--+---+--+---+--+--> f 0 fs 2fs fs < 2fm: /\/\/\/\/\ copies overlap -+----+----+---> f 0 fs 2fs ``` #### Nyquist rate of $x(t) = 30\sin(100\pi t) + 40\cos(400\pi t)\sin(300\pi t)$ Convert the product using $\sin A\cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ with $A = 300\pi t$, $B = 400\pi t$: $$ \begin{aligned} 40\cos(400\pi t)\sin(300\pi t) &= 20\left[\sin(700\pi t) + \sin(-100\pi t)\right] \\ &= 20\sin(700\pi t) - 20\sin(100\pi t) \end{aligned} $$ So $$ x(t) = 10\sin(100\pi t) + 20\sin(700\pi t) $$ | Component | $\omega$ (rad/s) | $f$ (Hz) | |---|---|---| | $10\sin(100\pi t)$ | $100\pi$ | 50 | | $20\sin(700\pi t)$ | $700\pi$ | 350 | Highest frequency $f_m = 350$ Hz. $$ f_N = 2f_m = 700\ \text{Hz}, \qquad T_N = \frac{1}{700} = 1.43\ \text{ms} $$ **Answer:** Nyquist rate $= 700$ samples/s ($\omega_N = 1400\pi$ rad/s); Nyquist interval $\approx 1.43$ ms. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Nyquist criterion requires A. fs <= 2fm B. fs = fm C. fs >= 2fm D. fs >= fm/2 Answer: C. Spectral copies then do not overlap. (Key point: Criterion) 2. 900 Hz sampled at 1200 Hz looks like A. 900 Hz B. 2100 Hz C. 600 Hz D. 300 Hz Answer: D. |900 - 1200| = 300 Hz. (Key point: Alias frequency) 3. 40cos(400πt)sin(300πt) equals A. 20sin(700πt) - 20sin(100πt) B. 40sin(700πt) C. 20sin(100πt) D. 20sin(700πt) + 20sin(100πt) Answer: A. sinAcosB = 1/2[sin(A+B)+sin(A-B)] with A-B = -100πt. (Key point: Product conversion) 4. x(t) simplifies to A. 30sin(100πt) + 20sin(700πt) B. 10sin(100πt) + 20sin(700πt) C. 50sin(100πt) D. 10sin(100πt) + 40sin(700πt) Answer: B. 30-20 = 10 for the 100πt term. (Key point: Total signal) 5. Nyquist rate of this signal A. 800 Hz B. 350 Hz C. 700 Hz D. 1400 Hz Answer: C. fm = 350 Hz so 2fm = 700 Hz. 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