---
title: "Nyquist rate of (1/2π)cos4000πt·cos1000πt"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:sampling-theorem-nyquist-product"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-theorem-nyquist-product
site: BCT Engineering
---

# Nyquist rate of (1/2π)cos4000πt·cos1000πt

Signals and Systems · Chapter 4: Sampling

- Asked in: 2082 Chaitra, 2+4 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/sampling-theorem-nyquist-product
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

Define sampling theorem. Determine the Nyquist rate and Nyquist interval for the given continuous time signal. x(t) = (1/2π) cos(4000πt) cos(1000πt)

## Short answer

Sampling theorem says a band-limited signal with highest frequency fm is exactly recoverable if sampled at fs ≥ 2fm. Convert the product: x(t) = (1/4π)cos(5000πt) + (1/4π)cos(3000πt), giving 2500 Hz and 1500 Hz; fm = 2500 Hz. Nyquist rate = 5000 samples/s, interval = 0.2 ms.

## Answer

#### Sampling theorem

A band-limited signal with no frequency components above $f_m$ Hz can be completely represented by, and exactly recovered from, its uniformly spaced samples if the sampling frequency satisfies $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate; recovery is done with an ideal low-pass filter.

#### Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(4000\pi t)\cos(1000\pi t)$

**Step 1: convert the product to a sum** using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$
\begin{aligned}
x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(5000\pi t) + \cos(3000\pi t)\right] \\
&= \frac{1}{4\pi}\cos(5000\pi t) + \frac{1}{4\pi}\cos(3000\pi t)
\end{aligned}
$$

**Step 2: frequencies.**

| Component | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $\cos(5000\pi t)$ | $5000\pi$ | 2500 |
| $\cos(3000\pi t)$ | $3000\pi$ | 1500 |

Highest frequency $f_m = 2500$ Hz.

**Step 3: Nyquist rate and interval.**

$$
\begin{aligned}
f_N &= 2f_m = 2\times2500 = 5000\ \text{Hz} \\
T_N &= \frac{1}{f_N} = \frac{1}{5000} = 0.2\ \text{ms}
\end{aligned}
$$

**Answer:** Nyquist rate $= 5000$ samples/s (5 kHz, $\omega_N = 10000\pi$ rad/s); Nyquist interval $= 0.2$ ms $= 200\ \mu$s.

Note: the highest frequency is not 2000 Hz (from $\cos 4000\pi t$ alone); multiplication creates the sum frequency 2500 Hz.

## Check yourself

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

1. Why must the product cos(4000πt)cos(1000πt) be converted before finding fm?
   A. It makes the signal non-periodic
   B. The product has frequency 0
   C. The product creates sum and difference frequencies
   D. Multiplication halves frequencies
   Answer: C. cosAcosB = 1/2[cos(A+B)+cos(A-B)]. (Key point: Why convert)

2. Which frequencies are present?
   A. 2000 Hz and 500 Hz
   B. 4000 Hz and 1000 Hz
   C. 5000 Hz and 3000 Hz
   D. 2500 Hz and 1500 Hz
   Answer: D. 5000π/2π = 2500 and 3000π/2π = 1500. (Key point: Components)

3. What is the Nyquist rate?
   A. 5000 Hz
   B. 4000 Hz
   C. 3000 Hz
   D. 10000 Hz
   Answer: A. 2 x 2500 = 5000 Hz. (Key point: Nyquist rate)

4. What is the Nyquist interval?
   A. 0.25 ms
   B. 0.2 ms
   C. 0.5 ms
   D. 2 ms
   Answer: B. 1/5000 = 0.2 ms. (Key point: Nyquist interval)

## Close variants set in other exams

- [Sampling theorem; Nyquist of 1 + cos + sin](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-theorem-nyquist-1-cos-sin) — Signals and Systems, chapter 4, asked 5×

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