---
title: "Nyquist rate of 5cos3000πt·sin1000πt"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:nyquist-rate-cos-sin-product"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/nyquist-rate-cos-sin-product
site: BCT Engineering
---

# Nyquist rate of 5cos3000πt·sin1000πt

Signals and Systems · Chapter 4: Sampling

- Asked in: 2082 Bhadra (new course), 3 marks
- Video: [Nyquist Rate (Solved Problem 2)](https://www.youtube.com/watch?v=-AOGX_95yAQ) — 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/nyquist-rate-cos-sin-product
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

Determine the Nyquist rate for the signal x(t) = 5cos(3,000πt)sin(1,000πt).

## Short answer

x(t) = 5cos(3000πt)sin(1000πt) = 2.5 sin(4000πt) − 2.5 sin(2000πt), giving frequencies 2000 Hz and 1000 Hz. Maximum frequency is 2000 Hz, so Nyquist rate = 4000 Hz.

## Answer

The product must be written as a sum of sinusoids. Use $\sin A\cos B = \tfrac12[\sin(A+B) + \sin(A-B)]$ with $A = 1000\pi t$, $B = 3000\pi t$:
$$
\begin{aligned}
x(t) &= 5\cos(3000\pi t)\sin(1000\pi t) \\
&= \frac52\big[\sin(4000\pi t) + \sin(-2000\pi t)\big] \\
&= 2.5\sin(4000\pi t) - 2.5\sin(2000\pi t)
\end{aligned}
$$

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $2.5\sin(4000\pi t)$ | $4000\pi$ | $2000$ |
| $-2.5\sin(2000\pi t)$ | $2000\pi$ | $1000$ |

$f_M = 2000$ Hz, so
$$
f_N = 2 f_M = 4000\ \text{Hz}
$$
**Answer: Nyquist rate = 4000 Hz (4000 samples/s, $\omega_N = 8000\pi$ rad/s).** Note it is not twice 1500 Hz: multiplication creates sum and difference frequencies.

## Check yourself

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

1. Why must 5cos(3000πt)sin(1000πt) be rewritten before finding the Nyquist rate?
   A. Multiplication creates sum and difference frequencies
   B. It has no frequency
   C. Product terms have the same frequency as each factor
   D. Only sin terms have a Nyquist rate
   Answer: A. The product equals a sum of sinusoids at the sum and difference frequencies. (Key point: Product to sum)

2. What frequencies are present in x(t)?
   A. 1500 Hz and 500 Hz
   B. 3000 Hz and 1000 Hz
   C. 2000 Hz and 1000 Hz
   D. 4000 Hz and 2000 Hz
   Answer: C. x(t) = 2.5sin4000πt - 2.5sin2000πt, giving 2000 Hz and 1000 Hz. (Key point: Resulting frequencies)

3. What is the Nyquist rate?
   A. 2000 Hz
   B. 3000 Hz
   C. 8000 Hz
   D. 4000 Hz
   Answer: D. fM = 2000 Hz, so the Nyquist rate is 4000 Hz. (Key point: Nyquist rate)

## Close variants set in other exams

- [Sampling; Nyquist of 1 + cos200πt + sin4000πt](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-nyquist-200pi) — Signals and Systems, chapter 4, asked 2×

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