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

# Sampling theorem; aliasing cause; Nyquist

Signals and Systems · Chapter 4: Sampling

- Asked in: 2083 Baisakh (new course), 2+2+2 marks
- Video: [Aliasing or Effect of Under Sampling | Generation of Aliasing | Effect and Solution of Aliasing](https://www.youtube.com/watch?v=lT8wKrLfcFU) — Engineering Funda
- 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-aliasing-cause-nyquist
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

Define sampling theorem. Explain the aliasing effect and its cause. Find the Nyquist rate and Nyquist time interval of the following signal. x(t) = 5cos(150πt) + 10sin(350πt) + cos(150πt)

## Short answer

Sampling theorem: a band-limited signal with max frequency f_M is recoverable from its samples if f_s > 2f_M. Aliasing, caused by under-sampling, makes high frequencies appear as lower ones due to spectral overlap. For x(t)=5cos(150πt)+10sin(350πt)+cos(150πt), f_M=175 Hz; Nyquist rate=350 Hz, interval=1/350 s≈2.86 ms.

## 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**.

#### Aliasing effect and its cause
**Aliasing** is the appearance of a high-frequency component as a false lower frequency after sampling.

**Cause:** sampling repeats the spectrum at every multiple of $\omega_s$: $X_p(j\omega) = \frac1T\sum_k X(j(\omega - k\omega_s))$. If $\omega_s < 2\omega_M$ (under-sampling), or if the signal is not band-limited, neighbouring copies overlap. Frequencies above $f_s/2$ fold back into $0$ to $f_s/2$ and cannot be separated. Example: 700 Hz sampled at 1 kHz looks like 300 Hz.

#### Nyquist rate and interval of $x(t) = 5\cos(150\pi t) + 10\sin(350\pi t) + \cos(150\pi t)$
The first and last terms have the same frequency, so $x(t) = 6\cos(150\pi t) + 10\sin(350\pi t)$.

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $6\cos(150\pi t)$ | $150\pi$ | $75$ |
| $10\sin(350\pi t)$ | $350\pi$ | $175$ |

$f_M = 175$ Hz.
$$
\begin{aligned}
f_N &= 2 f_M = 350\ \text{Hz} \\
T_N &= \frac{1}{350} = 2.857\ \text{ms}
\end{aligned}
$$
**Answer: Nyquist rate = 350 Hz; Nyquist interval = 1/350 s ≈ 2.86 ms.**

## Check yourself

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

1. The Nyquist rate for a signal with maximum frequency fM is:
   A. fM
   B. fM/2
   C. 2fM
   D. 4fM
   Answer: C. The theorem requires fs > 2fM, and 2fM is the Nyquist rate. (Key point: Sampling theorem)

2. Aliasing is caused by:
   A. Overlap of spectral copies due to fs < 2fM or a non-band-limited signal
   B. Using too high a sampling rate
   C. Adding a low-pass filter before sampling
   D. Quantization error
   Answer: A. Spectral copies at multiples of ωs overlap when under-sampling. (Key point: Cause of aliasing)

3. In x(t) = 5cos150πt + 10sin350πt + cos150πt, the cos terms combine to:
   A. 5cos150πt
   B. 6cos150πt
   C. cos300πt
   D. 50cos150πt
   Answer: B. Same frequency terms add amplitudes: 5 + 1 = 6. (Key point: Combining terms)

4. What are the Nyquist rate and interval?
   A. 150 Hz and 1/150 s
   B. 175 Hz and 1/175 s
   C. 350 Hz and 1/175 s
   D. 350 Hz and 1/350 s
   Answer: D. fM = 175 Hz, so f_N = 350 Hz and T_N = 1/350 s ≈ 2.86 ms. (Key point: Nyquist rate result)

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