---
title: "Aperture effect correction; bandpass sampling 80–115 kHz"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:aperture-correction-bandpass-80-115"
source: https://www.bctengineer.com/q/semester_6/communication_systems/aperture-correction-bandpass-80-115
site: BCT Engineering
---

# Aperture effect correction; bandpass sampling 80–115 kHz

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2069 Chaitra (CS II), 5+3 marks
- Video: [Sampling Types in digital communication - Ideal, Natural, Flat top Sampling - Aperture effect](https://www.youtube.com/watch?v=sFiCgEVpTWs) — Electronics Subjectified
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/aperture-correction-bandpass-80-115
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

What do you mean by aperture effect in Sampling? How can it be corrected? A band pass signal with the spectrum in the range of (80–115) kHz is to be digitized. Calculate minimum sampling frequency required for the signal.

## Answer

#### Aperture effect
In flat-top sampling each sample is held for a finite pulse width $\tau$. This is equivalent to passing ideal samples through a filter with impulse response $h(t) = \text{rect}(t/\tau)$:
$$
S(f) = f_sH(f)\sum_n G(f - nf_s), \qquad H(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}
$$
Since $|H(f)|$ falls as $f$ rises, the higher message frequencies are attenuated (and delayed by $\tau/2$). This amplitude distortion is the **aperture effect**; the larger $\tau$, the worse it is.

#### Correction
1. **Equalizer:** after the reconstruction LPF, use a filter with
$$
H_{eq}(f) = \frac{1}{H(f)} = \frac{1}{\tau\,\text{sinc}(f\tau)}, \quad |f| \le W
$$
which boosts high frequencies to cancel the sinc droop.
2. **Small duty cycle:** keep $\tau \ll T_s$; for $\tau/T_s \le 0.1$ the droop at the band edge is under about 0.5%, so equalization is often unnecessary.

#### Band-pass sampling of 80–115 kHz
$f_L = 80$ kHz, $f_H = 115$ kHz, $B = f_H - f_L = 35$ kHz.

Band-pass sampling condition:
$$
\frac{2f_H}{k} \le f_s \le \frac{2f_L}{k - 1}, \qquad k_{max} = \left\lfloor\frac{f_H}{B}\right\rfloor
$$
$$
\begin{aligned}
k_{max} &= \left\lfloor\frac{115}{35}\right\rfloor = \lfloor 3.29 \rfloor = 3 \\
f_{s,min} &= \frac{2f_H}{k} = \frac{2 \times 115}{3} = 76.67\ \text{kHz} \\
\text{upper limit} &= \frac{2f_L}{k - 1} = \frac{2 \times 80}{2} = 80\ \text{kHz}
\end{aligned}
$$
So any $f_s$ from 76.67 to 80 kHz works, far below the low-pass Nyquist rate $2f_H = 230$ kHz.

**Answer: minimum sampling frequency ≈ 76.67 kHz.** (If treated as a low-pass signal, 230 kHz would be needed.)

## Check yourself

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

1. Aperture effect is attenuation of high frequencies because:
   A. The spectrum is multiplied by tau sinc(f tau)
   B. fs is too low
   C. Quantization is coarse
   D. Carriers drift
   Answer: A. The finite hold time acts as a rectangular filter. (Key point: Aperture effect)

2. Aperture effect is corrected by:
   A. Equalizer with 1/H(f) after the reconstruction LPF
   B. Raising the pulse width
   C. Lowering fs
   D. Using a high-pass filter
   Answer: A. The equalizer boosts high frequencies to cancel the sinc droop. (Key point: Correction)

3. For 80-115 kHz, B = 35 kHz and kmax is:
   A. 2
   B. 3
   C. 4
   D. 1
   Answer: B. floor(115/35) = 3. (Key point: kmax)

4. Minimum band-pass sampling frequency is:
   A. 230 kHz
   B. 80 kHz
   C. 76.67 kHz
   D. 35 kHz
   Answer: C. 2 fH / k = 230/3 = 76.67 kHz, valid up to 2fL/(k-1) = 80 kHz. (Key point: Min fs)

## Related in Communication Systems

- [Sampling theorem; aliasing and aperture effect remedies](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asampling-theorem-aliasing-aperture-remedy) — Communication Systems, chapter 4
- [Sub-sampling theory and aperture effect](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asubsampling-aperture-effect) — Communication Systems, chapter 4
- [Prove sampling theorem; aliasing and aperture effect](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asampling-prove-aliasing-aperture) — Communication Systems, chapter 4
- [Delta modulator numerical (3 kHz, 10 kHz sampling)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-3khz-10khz-numerical) — Communication Systems, chapter 4, asked 2×
- [Aperture effect, DPCM and DM slope overload numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aaperture-effect-dpcm-dm-slope) — Communication Systems, chapter 4
- [Nyquist-Kotelnikov sampling theorem; aliasing and aperture](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-kotelnikov-aliasing-aperture) — Communication Systems, chapter 4

## Related in other subjects

- [Band pass (10 kHz, 4 kHz) from normalized LPF TF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-10khz-from-transfer-function) — Filter Design, chapter 3
- [Practical sampling; band-pass sampling](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Apractical-bandpass-sampling) — Signals and Systems, chapter 4
- [Why band-limit a signal before sampling](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Asampling-bandlimit) — Data Communication, chapter 5
- [Importance of frequency transformation; obtaining band pass](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afrequency-transformation-importance-bandpass) — Filter Design, chapter 3, asked 3×