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

# Sampling theorem; Nyquist of 1 + cos + sin

Signals and Systems · Chapter 4: Sampling

- Asked in: 2079 Chaitra, 2+4 marks
- 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_4/signals_and_systems/sampling-theorem-nyquist-1-cos-sin
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

Define sampling theorem. Determine the Nyquist rate for the continuous time signal x(t) = 1 + cos(2000πt) + sin(4000πt).

## Short answer

The signal contains DC (0 Hz), a 1000 Hz component from cos(2000πt), and a 2000 Hz component from sin(4000πt). The highest frequency is 2000 Hz, so the Nyquist rate is 4000 Hz (or 8000π rad/s).

## Answer

#### Sampling theorem
**Sampling theorem:** A continuous-time signal $x(t)$ that is band-limited to $\omega_M$ (that is, $X(j\omega)=0$ for $|\omega|>\omega_M$) is uniquely determined by its samples $x(nT)$ if the sampling frequency satisfies
$$
\omega_s = \frac{2\pi}{T} > 2\omega_M \quad\text{or}\quad f_s > 2f_M
$$
The minimum rate $2f_M$ is called the **Nyquist rate**, and $T_N = 1/(2f_M)$ is the **Nyquist interval**.

The original signal is recovered by passing the samples through an ideal low-pass filter of cutoff between $\omega_M$ and $\omega_s-\omega_M$ and gain $T$.

#### Nyquist rate of $x(t) = 1 + \cos(2000\pi t) + \sin(4000\pi t)$
Write each term as $\cos(2\pi f t)$ or $\sin(2\pi f t)$ and read its frequency:

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $1$ (DC) | $0$ | $0$ |
| $\cos(2000\pi t)$ | $2000\pi$ | $1000$ |
| $\sin(4000\pi t)$ | $4000\pi$ | $2000$ |

The highest frequency present is $f_M = 2000$ Hz, so
$$
\begin{aligned}
f_N &= 2 f_M = 2 \times 2000 = 4000\ \text{Hz} \\
\omega_N &= 2\pi f_N = 8000\pi\ \text{rad/s}
\end{aligned}
$$
**Answer: Nyquist rate = 4000 Hz (4000 samples/s, or $8000\pi$ rad/s).**

## Check yourself

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

1. The sampling theorem requires
   A. ωs < 2ωM
   B. ωs = ωM
   C. ωs > 2ωM for a band-limited signal
   D. ωs > ωM/2
   Answer: C. Samples then uniquely determine the signal. (Key point: Statement)

2. Recovery is done by
   A. HPF with gain 1
   B. LPF with cutoff 2ωs
   C. Differentiator
   D. Ideal LPF with gain T and cutoff between ωM and ωs - ωM
   Answer: D. It passes only the baseband copy. (Key point: Recovery)

3. The highest frequency of 1 + cos2000πt + sin4000πt is
   A. 2000 Hz
   B. 4000 Hz
   C. 1000 Hz
   D. 0 Hz
   Answer: A. 4000π/2π = 2000 Hz. (Key point: Highest frequency)

4. Nyquist rate in rad/s is
   A. 4000π
   B. 8000π
   C. 2000π
   D. 4000
   Answer: B. 2π x 4000 = 8000π. (Key point: Nyquist rate in rad/s)

## Close variants set in other exams

- [Aliasing; Nyquist rate of 1 + cos + sin](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-nyquist-1-cos-sin) — Signals and Systems, chapter 4, asked 4×
- [Nyquist rate of (1/2π)cos4000πt·cos1000πt](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-theorem-nyquist-product) — Signals and Systems, chapter 4, asked 2×

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## Related in other subjects

- [Flat top vs natural sampling; Nyquist rate of sin(800πt)/πt + 10cos2000t](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aflat-top-vs-natural-nyquist-rate) — Communication Systems, chapter 4
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