---
title: "Aliasing in a band-limited signal"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:aliasing-bandlimited-w"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/aliasing-bandlimited-w
site: BCT Engineering
---

# Aliasing in a band-limited signal

Signals and Systems · Chapter 4: Sampling

- Asked in: 2071 Bhadra, 5 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/aliasing-bandlimited-w
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

Explain aliasing that may occur in any arbitrary band-limited signal x(t) with band-width 'W'.

## Short answer

When a signal band-limited to W (rad/s) is sampled at ω_s, its spectrum repeats every ω_s. If ω_s < 2W, shifted copies overlap: the baseband [−W,W] and the first copy [ω_s−W, ω_s+W] overlap from ω_s−W to W. High frequencies alias as ω_s−ω_0, and the original spectrum cannot be separated, making exact reconstruction impossible.

## Answer

Let $x(t)$ be band-limited with bandwidth $W$, i.e. $X(j\omega) = 0$ for $|\omega| > W$ (in rad/s). It is sampled with an impulse train of period $T$, sampling frequency $\omega_s = 2\pi/T$.

#### Spectrum of the sampled signal
$$
X_p(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big)
$$
The baseband copy ($k=0$) occupies $[-W, W]$. The copy for $k=1$ occupies $[\omega_s - W,\ \omega_s + W]$.

#### Three cases
| Case | Condition | Result |
|---|---|---|
| Over-sampling | $\omega_s > 2W$ | Gap of $\omega_s - 2W$ between copies; no aliasing |
| Critical | $\omega_s = 2W$ | Copies just touch; ideal LPF needed |
| Under-sampling | $\omega_s < 2W$ | Copies overlap from $\omega_s - W$ to $W$: **aliasing** |

```text
ws < 2W :
       k=0           k=1
     _______       _______
    /       \     /       \
---/----------\-/----------\---> w
  -W     ws-W  W  ws       ws+W
          |<-->|
         overlap
```

#### What aliasing does
- In the overlap band $\omega_s - W < |\omega| < W$, $X_p$ is the sum of the original spectrum and the tail of the neighbouring copy. A component at $\omega_0 > \omega_s/2$ appears at the alias frequency $\omega_s - \omega_0$.
- A low-pass filter cannot separate the two, so the reconstructed signal differs from $x(t)$: high frequencies are lost and false low frequencies appear.
- Example: $W = 2\pi(1000)$, $\omega_s = 2\pi(1500)$. A 900 Hz component appears at $1500 - 900 = 600$ Hz.

**Avoiding it:** sample at $\omega_s > 2W$ and place an anti-aliasing low-pass filter of cutoff $\omega_s/2$ before the sampler.

## Check yourself

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

1. What is the spectrum of an impulse-train sampled signal?
   A. X(jω) multiplied by ωs
   B. (1/T) times the sum of copies of X(jω) shifted by multiples of ωs
   C. A single copy of X(jω) with no repetition
   D. The sum of copies shifted by multiples of W
   Answer: B. Xp(jω) = (1/T) Σ X(j(ω - kωs)): copies repeat at every multiple of ωs. (Key point: Sampled spectrum)

2. For a signal band-limited to W, when do the k=0 and k=1 copies overlap?
   A. When ωs > 2W
   B. When ωs = 4W only
   C. When ωs < 2W
   D. Only when W is infinite
   Answer: C. The k=1 copy begins at ωs - W, which is below W when ωs < 2W. (Key point: Overlap condition)

3. What does an LPF do about the overlapped region?
   A. It cannot separate the original from the neighbouring copy, so reconstruction is distorted
   B. It removes the overlap perfectly
   C. It amplifies the overlap to restore the signal
   D. It shifts the overlap to higher frequency
   Answer: A. Overlapped spectra add, so no filter can recover x(t) exactly. (Key point: Effect of aliasing)

4. W = 2π(1000) and ωs = 2π(1500). Where does a 900 Hz component appear?
   A. 900 Hz
   B. 1500 Hz
   C. 2400 Hz
   D. 600 Hz
   Answer: D. Alias = fs - f0 = 1500 - 900 = 600 Hz. (Key point: Alias example)

## Related in Signals and Systems

- [Band-limited signals; aliasing; Nyquist rate](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Abandlimited-aliasing-nyquist) — Signals and Systems, chapter 4
- [Impulse train sampling, aliasing, reconstruction](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aimpulse-train-aliasing-reconstruction) — Signals and Systems, chapter 4

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