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

# Sampling requirement; Nyquist of product

Signals and Systems · Chapter 4: Sampling

- Asked in: 2075 Bhadra, 2+4 marks
- Video: [Nyquist Rate & Nyquist Interval](https://www.youtube.com/watch?v=clDgsheEb9g) — 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-requirement-nyquist-product
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

What do you mean by sampling, state the requirement of sampling frequency? Determine Nyquist rate for x(t) = (1/2π) cos(100πt) cos(300πt)

## Short answer

Sampling converts a continuous signal to discrete values at interval T, requiring f_s > 2f_M for perfect reconstruction. The signal (1/2π)cos(100πt)cos(300πt) = (1/(4π))[cos(200πt)+cos(400πt)]; the highest frequency is 200 Hz, so Nyquist rate is 400 Hz.

## Answer

#### Sampling and the required sampling frequency
**Sampling** is the process of converting a continuous-time signal $x(t)$ into a discrete-time signal $x[n] = x(nT)$ by taking its values at equally spaced instants $t = nT$. $T$ is the sampling interval and $f_s = 1/T$ the sampling frequency.

**Requirement:** for a signal band-limited to $f_M$, the sampling frequency must be more than twice the highest frequency:
$$
f_s > 2 f_M
$$
$2f_M$ is the Nyquist rate. If this holds, $x(t)$ can be reconstructed exactly from its samples by an ideal low-pass filter; if not, aliasing occurs.

#### Nyquist rate of $x(t) = \frac{1}{2\pi}\cos(100\pi t)\cos(300\pi t)$
The product must first be written as a sum, using $\cos A\cos B = \tfrac12[\cos(A-B) + \cos(A+B)]$:
$$
\begin{aligned}
x(t) &= \frac{1}{2\pi}\cdot\frac{1}{2}\big[\cos(300\pi t - 100\pi t) + \cos(300\pi t + 100\pi t)\big] \\
&= \frac{1}{4\pi}\big[\cos(200\pi t) + \cos(400\pi t)\big]
\end{aligned}
$$

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $\cos(200\pi t)$ | $200\pi$ | $100$ |
| $\cos(400\pi t)$ | $400\pi$ | $200$ |

$f_M = 200$ Hz, so
$$
f_N = 2 f_M = 400\ \text{Hz}
$$
**Answer: Nyquist rate = 400 Hz (400 samples/s, $\omega_N = 800\pi$ rad/s).**

## Check yourself

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

1. Sampling converts
   A. x[n] to x(t)
   B. x(t) to its derivative
   C. x(t) to x[n] = x(nT)
   D. A DT signal to its FT
   Answer: C. Values are taken at t = nT. (Key point: Sampling)

2. The sampling frequency must satisfy
   A. fs < 2fM
   B. fs = fM
   C. fs > fM/2
   D. fs > 2fM
   Answer: D. Otherwise aliasing occurs. (Key point: Requirement)

3. (1/2π)cos100πt cos300πt =
   A. (1/4π)[cos200πt + cos400πt]
   B. (1/2π)cos400πt
   C. (1/4π)cos300πt
   D. (1/π)[cos200πt+cos400πt]
   Answer: A. Half-sum and half-difference angles with factor 1/2. (Key point: Product to sum)

4. Nyquist rate
   A. 200 Hz
   B. 400 Hz
   C. 300 Hz
   D. 800 Hz
   Answer: B. fM = 200 Hz. (Key point: Nyquist rate)

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## Related in other subjects

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