---
title: "Aliasing; Nyquist of 6cos50πt + …"
subject: "Signals and Systems"
chapter: 4
key: "semester_4/signals_and_systems:aliasing-overcome-50pi"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/aliasing-overcome-50pi
site: BCT Engineering
---

# Aliasing; Nyquist of 6cos50πt + …

Signals and Systems · Chapter 4: Sampling

- Asked in: 2071 Magh, 3+3 marks
- Video: [Sampling, Aliasing & Nyquist Theorem](https://www.youtube.com/watch?v=yWqrx08UeUs) — 0612 TV w/ NERDfirst
- 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-overcome-50pi
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4

## Question

What is aliasing effect and how can we overcome it? Determine the Nyquist rate for a continuous time signal x(t) = 6cos50πt + 20sin300πt − 10cos100πt.

## Short answer

Aliasing distorts a sampled signal when f_s < 2f_max, making high frequencies appear as lower ones. It is overcome with an anti-aliasing low-pass filter before sampling and sampling at f_s > 2f_max. For x(t)=6cos50πt+20sin300πt−10cos100πt, frequencies are 25, 150, 50 Hz; f_max=150 Hz. Nyquist rate = 300 Hz; sampling must be >300 Hz to avoid zeroing the sine term.

## Answer

#### Aliasing effect
**Aliasing** is the distortion that occurs when a signal is sampled at a rate lower than its Nyquist rate ($f_s < 2f_M$). The shifted copies of the spectrum in the sampled signal overlap, so a high-frequency component "folds back" and appears as a lower frequency (its *alias*). After overlap the original spectrum cannot be separated by any filter, so the signal cannot be recovered exactly.

Example: a 700 Hz tone sampled at $f_s = 1000$ Hz gives the same samples as a $1000-700 = 300$ Hz tone.

**How to overcome aliasing:**
1. **Sample fast enough:** choose $f_s > 2f_M$ (in practice $f_s$ is kept 2.5 to 5 times $f_M$ to leave a guard band).
2. **Anti-aliasing filter:** pass the signal through an analog low-pass filter with cutoff $f_s/2$ *before* sampling, so that no component above $f_s/2$ reaches the sampler. Real signals and noise are never strictly band-limited, so this filter is always used in practice.
3. **Oversampling:** sample at a much higher rate, then low-pass filter and decimate digitally. This relaxes the sharpness needed in the analog anti-aliasing filter.

#### Nyquist rate of $x(t) = 6\cos 50\pi t + 20\sin 300\pi t - 10\cos 100\pi t$

| Term | $\omega$ (rad/s) | $f$ (Hz) |
|---|---|---|
| $6\cos 50\pi t$ | $50\pi$ | $25$ |
| $20\sin 300\pi t$ | $300\pi$ | $150$ |
| $-10\cos 100\pi t$ | $100\pi$ | $50$ |

$f_M = 150$ Hz.
$$
f_N = 2 f_M = 2\times150 = 300\ \text{Hz}
$$
**Answer: Nyquist rate = 300 Hz (300 samples/s, $\omega_N = 600\pi$ rad/s).** (Sampling at exactly 300 Hz would make every sample of $\sin 300\pi t$ zero, so in practice $f_s$ must be strictly greater than 300 Hz.)

## Check yourself

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

1. When does aliasing occur?
   A. Sampling rate is below the Nyquist rate (fs < 2fM)
   B. Sampling rate is above 2fM
   C. The signal has only one frequency
   D. The reconstruction filter has too high a gain
   Answer: A. Aliasing happens when fs < 2fM so the spectral copies overlap. (Key point: Cause of aliasing)

2. A 700 Hz tone is sampled at 1000 Hz. At what frequency does it appear?
   A. 700 Hz
   B. 1700 Hz
   C. 300 Hz
   D. 500 Hz
   Answer: C. The tone folds back to fs - f0 = 1000 - 700 = 300 Hz. (Key point: Alias frequency)

3. Where is the anti-aliasing filter placed and what cutoff does it have?
   A. After the sampler, digital high-pass with cutoff fs
   B. Before the sampler, analog low-pass with cutoff fs/2
   C. Before the sampler, high-pass with cutoff fs/2
   D. After the DAC, low-pass with cutoff 2fs
   Answer: B. It removes components above fs/2 before sampling so nothing can fold back. (Key point: Anti-aliasing filter)

4. For x(t) = 6cos50πt + 20sin300πt - 10cos100πt, what is the Nyquist rate?
   A. 150 Hz
   B. 600 Hz
   C. 100 Hz
   D. 300 Hz
   Answer: D. Highest frequency is 300π/2π = 150 Hz, so the Nyquist rate is 2 x 150 = 300 Hz. (Key point: Nyquist rate calculation)

## Close variants set in other exams

- [Aliasing; Nyquist of 2sin40πt + …](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aaliasing-overcome-40pi) — Signals and Systems, chapter 4, asked 2×

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## Related in other subjects

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