--- title: "Sampling theorem; aliasing and aperture effect remedies" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:sampling-theorem-aliasing-aperture-remedy" source: https://www.bctengineer.com/q/semester_6/communication_systems/sampling-theorem-aliasing-aperture-remedy site: BCT Engineering --- # Sampling theorem; aliasing and aperture effect remedies Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2074 Asoj (CS II), 4+6 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_6/communication_systems/sampling-theorem-aliasing-aperture-remedy - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question State Sampling theorem in terms of transmitter and receiver. Explain aliasing and aperture effect with remedy solutions. ## Short answer Sampling theorem: transmitter samples bandlimited signal (max W Hz) at fs ≥ 2W; receiver reconstructs via ideal LPF of bandwidth W. Aliasing occurs when fs < 2W, causing spectral overlap and false low frequencies. Remedies: anti-aliasing filter before sampling, and sampling above Nyquist for guard band. Aperture effect in flat-top sampling: finite pulse width τ multiplies spectrum by sinc(fτ), attenuating high frequencies. Remedies: small τ ≪ Ts, and equalizer with response 1/sinc(fτ) after reconstruction filter. ## Answer #### Sampling theorem **At the transmitter:** a band-limited signal of finite energy with no frequency components above $W$ Hz is completely described by its sample values taken at uniform intervals of $T_s \le 1/(2W)$ seconds, i.e. at a rate $f_s \ge 2W$. **At the receiver:** such a signal can be completely recovered from its samples taken at the rate of $2W$ samples per second (or more) by passing them through an ideal low-pass filter of bandwidth $W$. The rate $2W$ is the Nyquist rate. Proof sketch: ideal sampling gives $G_\delta(f) = f_s\sum_n G(f - nf_s)$; the copies separate when $f_s \ge 2W$, so a LPF extracts $G(f)$. #### Aliasing When $f_s < 2W$, or the signal is not strictly band-limited, the spectral copies overlap. High-frequency components fold back and appear as false low frequencies $|f - f_s|$. Example: a 7 kHz tone sampled at 10 kHz appears as 3 kHz after reconstruction. ```text copy n=0 copy n=1 /‾‾‾‾‾‾\ /‾‾‾‾‾‾\ / \ / \ / X \ overlap = aliasing -W 0 fs/2 fs ``` **Remedies** 1. Use an **anti-aliasing (pre-alias) low-pass filter** before the sampler to strictly band-limit the signal to $W$. 2. Sample **slightly above** the Nyquist rate ($f_s > 2W$) to create a guard band, so a practical reconstruction filter with finite roll-off can be used. Example: speech 3.4 kHz sampled at 8 kHz. #### Aperture effect In flat-top sampling each sample is held for width $\tau$. The sampled spectrum becomes $f_sH(f)\sum_n G(f - nf_s)$ with $H(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}$. Because $|H(f)|$ falls with frequency, high-frequency parts of the message are attenuated and slightly delayed: this amplitude distortion is the **aperture effect**. **Remedies** 1. Keep the pulse width small, $\tau \ll T_s$ (duty cycle $\tau/T_s \le 0.1$ makes the loss under about 0.5%). 2. Use an **equalizer** after the reconstruction LPF with response $$ H_{eq}(f) = \frac{1}{H(f)} = \frac{1}{\tau\,\text{sinc}(f\tau)}\ \text{for}\ |f| \le W $$ which boosts high frequencies to cancel the sinc loss. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At the receiver, a signal band-limited to W is recovered from its samples by: A. A high-pass filter B. An ideal low-pass filter of bandwidth W C. A quantizer D. A differentiator Answer: B. An ideal LPF extracts the n = 0 copy of the spectrum. (Key point: Receiver side) 2. A 7 kHz tone sampled at 10 kHz appears after reconstruction as: A. 7 kHz B. 3 kHz C. 17 kHz D. 5 kHz Answer: B. |f - fs| = |7 - 10| = 3 kHz. (Key point: Aliasing example) 3. Which two measures remedy aliasing? A. Equalizer and small tau B. Anti-aliasing LPF and sampling slightly above 2W C. Companding and expanding D. Larger pulse width and lower fs Answer: B. Pre-filtering band-limits the signal; the guard band permits a practical reconstruction filter. (Key point: Aliasing remedies) 4. Which measures reduce the aperture effect? A. Making tau much smaller than Ts and using a 1/H(f) equalizer B. Raising tau to equal Ts C. Using an anti-aliasing filter only D. Lowering fs Answer: A. A narrow pulse makes sinc(f tau) nearly flat and the equalizer boosts the lost high frequencies. 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