---
title: "Sampling; Nyquist of 1 + cos200πt + sin4000πt"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:sampling-nyquist-200pi"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-nyquist-200pi
site: BCT Engineering
---

# Sampling; Nyquist of 1 + cos200πt + sin4000πt

Signals and Systems · Chapter 4: Sampling

- Asked in: 2073 Bhadra, 2+3 marks
- Video: [Examples of Nyquist rate and Sampling Theory in Digital Communication by Engineering Funda](https://www.youtube.com/watch?v=6a8pmz-HPH4) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-nyquist-200pi
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

What is sampling? Determine the Nyquist rate for the following signal: x(t) = 1 + cos(200πt) + sin(4000πt)

## Short answer

Sampling converts a continuous-time signal to a discrete sequence x[n]=x(nT) at intervals T. For x(t)=1+cos(200πt)+sin(4000πt), the frequencies are 0 Hz (DC), 100 Hz, and 2000 Hz. The highest frequency is 2000 Hz, so Nyquist rate = 2×2000 = 4000 Hz.

## Answer

#### Sampling
**Sampling** is the conversion of a continuous-time signal $x(t)$ into a discrete-time sequence $x[n] = x(nT)$ by taking its values at regular intervals $T$ (sampling period). The sampling frequency is $f_s = 1/T$. By the sampling theorem, a band-limited signal can be recovered from its samples if $f_s > 2f_M$, where $f_M$ is its highest frequency. Sampling is the first step of analog-to-digital conversion.

#### Nyquist rate of $x(t) = 1 + \cos(200\pi t) + \sin(4000\pi t)$

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $1$ | $0$ | $0$ (DC) |
| $\cos(200\pi t)$ | $200\pi$ | $100$ |
| $\sin(4000\pi t)$ | $4000\pi$ | $2000$ |

$f_M = 2000$ Hz.
$$
f_N = 2 f_M = 4000\ \text{Hz}
$$
**Answer: Nyquist rate = 4000 Hz (4000 samples/s).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Sampling gives x[n] = x(nT), where fs equals
   A. T
   B. 2T
   C. 1/T
   D. 1/(2T)
   Answer: C. fs is the reciprocal of the sampling period. (Key point: Sampling)

2. sin(4000πt) has frequency
   A. 4000 Hz
   B. 200 Hz
   C. 1000 Hz
   D. 2000 Hz
   Answer: D. 4000π/2π = 2000. (Key point: Frequency of sin4000πt)

3. Nyquist rate of 1 + cos200πt + sin4000πt
   A. 4000 Hz
   B. 200 Hz
   C. 2000 Hz
   D. 8000 Hz
   Answer: A. 2 x 2000 = 4000. (Key point: Nyquist rate)

## Close variants set in other exams

- [Nyquist rate of 5cos3000πt·sin1000πt](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Anyquist-rate-cos-sin-product) — Signals and Systems, chapter 4, asked 2×

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