---
title: "Nyquist rate of (1/2π)cos50πt·cos500πt"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:nyquist-theorem-product-50-500"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/nyquist-theorem-product-50-500
site: BCT Engineering
---

# Nyquist rate of (1/2π)cos50πt·cos500πt

Signals and Systems · Chapter 4: Sampling

- Asked in: 2079 Jestha, 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/nyquist-theorem-product-50-500
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

State Nyquist's sampling theorem. Find the Nyquist rate for the given CT signal x(t) = (1/2π) cos(50πt) cos(500πt)

## Short answer

Nyquist sampling theorem: exact recovery from samples requires fs ≥ 2fm; the minimum rate is the Nyquist rate. Use cosAcosB to get (1/4π)cos550πt + (1/4π)cos450πt, so fm = 275 Hz. Nyquist rate = 550 samples/s, interval ≈ 1.82 ms.

## Answer

#### Nyquist sampling theorem

A continuous-time signal band-limited to $f_m$ Hz can be exactly recovered from its uniformly spaced samples if the sampling rate is at least twice the highest frequency, $f_s \ge 2f_m$. The minimum rate $2f_m$ is the Nyquist rate, and $1/(2f_m)$ is the Nyquist interval. Sampling below this rate causes aliasing.

#### Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(50\pi t)\cos(500\pi t)$

Using $\cos A\cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$:

$$
\begin{aligned}
x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\left[\cos(550\pi t) + \cos(450\pi t)\right] \\
&= \frac{1}{4\pi}\cos(550\pi t) + \frac{1}{4\pi}\cos(450\pi t)
\end{aligned}
$$

| Component | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $\cos(550\pi t)$ | $550\pi$ | 275 |
| $\cos(450\pi t)$ | $450\pi$ | 225 |

Highest frequency $f_m = 275$ Hz.

$$
\begin{aligned}
f_N &= 2f_m = 2\times275 = 550\ \text{Hz} \\
T_N &= \frac{1}{550} = 1.82\ \text{ms}
\end{aligned}
$$

**Answer:** Nyquist rate $= 550$ samples/s ($\omega_N = 1100\pi$ rad/s); Nyquist interval $\approx 1.82$ ms.

## Check yourself

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

1. cos(50πt)cos(500πt) is a sum of cosines with ω =
   A. 500π and 50π
   B. 25π and 250π
   C. 1000π and 100π
   D. 550π and 450π
   Answer: D. A+B = 550π, A-B = 450π (cos is even). (Key point: Product to sum)

2. The highest frequency is
   A. 275 Hz
   B. 250 Hz
   C. 550 Hz
   D. 225 Hz
   Answer: A. 550π/2π = 275 Hz. (Key point: Highest frequency)

3. Nyquist rate
   A. 500 Hz
   B. 550 Hz
   C. 275 Hz
   D. 450 Hz
   Answer: B. 2 x 275 = 550 Hz. (Key point: Nyquist rate)

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