---
title: "Band-limited signals; aliasing; Nyquist rate"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:bandlimited-aliasing-nyquist"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/bandlimited-aliasing-nyquist
site: BCT Engineering
---

# Band-limited signals; aliasing; Nyquist rate

Signals and Systems · Chapter 4: Sampling

- Asked in: 2083 Bhadra (new course), 1+3+3 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/bandlimited-aliasing-nyquist
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

What do you mean by band limited signals? Explain aliasing effect with examples. Find the Nyquist rate and Nyquist time interval of the following signal. x(t) = 3sin(230πt) + 50cos(225πt) + sin(200πt)

## Short answer

A band-limited signal has zero spectrum beyond a maximum frequency f_M. Aliasing occurs when sampling below 2f_M, causing frequency overlap (e.g., 50 Hz and 10 Hz indistinguishable at 40 Hz sampling). For x(t) = 3sin(230πt)+50cos(225πt)+sin(200πt), f_M=115 Hz; Nyquist rate = 230 Hz, Nyquist interval = 1/230 s ≈ 4.35 ms.

## Answer

#### Band-limited signal
A signal $x(t)$ is **band-limited** if its Fourier transform is zero above some finite frequency $\omega_M$: $X(j\omega) = 0$ for $|\omega| > \omega_M$. Example: a sum of sinusoids, or speech after a 3.4 kHz low-pass filter.

#### Aliasing effect with example
When a band-limited signal is sampled at $f_s < 2f_M$, the shifted copies of its spectrum overlap. A component of frequency $f_0 > f_s/2$ then gives the same samples as a lower frequency $|f_0 - kf_s|$ and is mistaken for it. This is **aliasing**.

**Example:** $x_1(t) = \cos(2\pi\cdot 10t)$ and $x_2(t) = \cos(2\pi\cdot 50t)$, sampled at $f_s = 40$ Hz:
$$
x_2[n] = \cos\!\left(2\pi\frac{50}{40}n\right) = \cos\!\left(2\pi n + \frac{2\pi n}{4}\right) = \cos\!\left(2\pi\frac{10}{40}n\right) = x_1[n]
$$
The samples are identical, so 50 Hz is an alias of 10 Hz. Aliasing is avoided by sampling above the Nyquist rate and using an anti-aliasing filter.

#### Nyquist rate and interval of $x(t) = 3\sin(230\pi t) + 50\cos(225\pi t) + \sin(200\pi t)$

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $3\sin(230\pi t)$ | $230\pi$ | $115$ |
| $50\cos(225\pi t)$ | $225\pi$ | $112.5$ |
| $\sin(200\pi t)$ | $200\pi$ | $100$ |

$f_M = 115$ Hz.
$$
\begin{aligned}
f_N &= 2f_M = 2 \times 115 = 230\ \text{Hz} \\
T_N &= \frac{1}{f_N} = \frac{1}{230} = 4.348\ \text{ms}
\end{aligned}
$$
**Answer: Nyquist rate = 230 Hz; Nyquist interval = 1/230 s ≈ 4.35 ms.**

## Check yourself

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

1. A signal is band-limited if:
   A. x(t) = 0 for |t| > T
   B. X(jω) = 0 for |ω| > ωM for some finite ωM
   C. X(jω) is constant
   D. x(t) is periodic
   Answer: B. Band-limited means the spectrum has finite extent. (Key point: Band-limited definition)

2. For x(t) = 3sin230πt + 50cos225πt + sin200πt, what is the Nyquist rate?
   A. 230 Hz
   B. 115 Hz
   C. 450 Hz
   D. 225 Hz
   Answer: A. Highest frequency is 230π/2π = 115 Hz, so Nyquist rate = 230 Hz. (Key point: Nyquist rate)

3. What is the Nyquist interval for this signal?
   A. 1/115 s ≈ 8.7 ms
   B. 1/230 s ≈ 4.35 ms
   C. 1/460 s
   D. 230 s
   Answer: B. T_N = 1/f_N = 1/230 s. (Key point: Nyquist interval)

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