--- title: "Nyquist rate and interval for 6cos50πt + 20sin300πt − 10cos100πt" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:nyquist-rate-6cos50-512" source: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-rate-6cos50-512 site: BCT Engineering --- # Nyquist rate and interval for 6cos50πt + 20sin300πt − 10cos100πt Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2072 Chaitra (CS II), 2+5 marks; 2071 Chaitra (CS II), 2+5 marks (asked 2 times) - Video: [Nyquist Rate (Solved Problem 1)](https://www.youtube.com/watch?v=hRH9edUACcI) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-rate-6cos50-512 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question State Nyquist sampling theory. Determine the Nyquist rate and Nyquist interval for a continuous time signal x(t) = 6cos50πt + 20sin300πt − 10cos100πt is to be sampled and quantize using 512 levels. ## Short answer Nyquist theorem: bandlimited signal with max frequency fm can be recovered from samples at fs ≥ 2fm. For x(t)=6cos50πt+20sin300πt-10cos100πt, max frequency 150 Hz, Nyquist rate = 300 Hz, Nyquist interval = 3.33 ms. With 512 levels (9 bits), bit rate = 2700 bps, bandwidth = 1350 Hz. ## Answer #### Nyquist sampling theorem A band-limited signal with no frequency components above $f_m$ Hz is **completely described by, and can be exactly recovered from**, its samples taken uniformly at a rate $$ f_s \ge 2f_m $$ The minimum rate $f_s = 2f_m$ is the **Nyquist rate** and $T_s = 1/(2f_m)$ is the **Nyquist interval**. Recovery is done by passing the samples through an ideal low-pass filter of cut-off $f_m$. If $f_s < 2f_m$, spectral copies overlap (**aliasing**) and recovery is impossible. #### Nyquist rate and interval of the given signal $$ x(t) = 6\cos 50\pi t + 20\sin 300\pi t - 10\cos 100\pi t $$ Frequencies of the components ($\omega = 2\pi f$): | Term | $\omega$ (rad/s) | $f$ (Hz) | |---|---|---| | $6\cos 50\pi t$ | $50\pi$ | 25 | | $20\sin 300\pi t$ | $300\pi$ | 150 | | $10\cos 100\pi t$ | $100\pi$ | 50 | Highest frequency: $f_m = 150$ Hz. $$ \text{Nyquist rate } f_N = 2f_m = 2 \times 150 = 300\ \text{Hz (samples/s)} $$ $$ \text{Nyquist interval } T_N = \frac{1}{f_N} = \frac{1}{300} = 3.33\ \text{ms} $$ #### Effect of quantizing with 512 levels $$ n = \log_2 512 = 9\ \text{bits/sample} $$ At the Nyquist rate the bit rate is $$ R_b = n f_N = 9 \times 300 = 2700\ \text{bits/s} $$ and the minimum transmission bandwidth is $R_b/2 = 1350$ Hz. **Answer:** Nyquist rate = 300 Hz, Nyquist interval = 3.33 ms (9 bits/sample, 2.7 kbps). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the Nyquist theorem say? A. Sampling must be exactly 4 fm B. A signal band-limited to fm is recoverable from samples at fs >= 2 fm C. Sampling at fs >= fm suffices D. Any signal can be recovered at any rate Answer: B. Sampling at least twice the highest frequency avoids aliasing. (Key point: Theorem) 2. What is the frequency of 20 sin(300 pi t)? A. 600 Hz B. 150 Hz C. 75 Hz D. 300 Hz Answer: B. omega = 2 pi f, so f = 300 pi/(2 pi) = 150 Hz. (Key point: Frequencies) 3. What is the Nyquist rate and interval for the signal? A. 150 Hz and 6.67 ms B. 100 Hz and 10 ms C. 300 Hz and 300 ms D. 300 Hz and 3.33 ms Answer: D. fN = 2 x 150 = 300; interval = 1/300 = 3.33 ms. (Key point: Nyquist rate) 4. With 512 levels, what is the bit rate at the Nyquist rate? A. 2700 bits/s B. 512 x 300 bits/s C. 8 x 300 = 2400 bits/s D. 1350 bits/s Answer: A. n = log2 512 = 9; 9 x 300 = 2700 b/s. 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