---
title: "Aperture effect, DPCM and DM slope overload numerical"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:aperture-effect-dpcm-dm-slope"
source: https://www.bctengineer.com/q/semester_6/communication_systems/aperture-effect-dpcm-dm-slope
site: BCT Engineering
---

# Aperture effect, DPCM and DM slope overload numerical

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2078 Chaitra, 2+3+4 marks
- Video: [SLOPE OVERLOAD DISTORTION AND GRANULAR NOISE - Types of noise in delta modulation - Hindi](https://www.youtube.com/watch?v=jGXfuw54Q0I) — 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-effect-dpcm-dm-slope
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Explain the aperture effect during flat-topped sampling. Illustrate the DPCM scheme that overcomes the disadvantages of PCM. A delta modulator system is designed to operate at 5 times the Nyquist rate for a signal having a bandwidth equal to 3 kHz bandwidth. Calculate the maximum amplitude of a 2 kHz sinusoidal for which the delta modulator does not have slope overload. The given step size is 250 mV.

## Short answer

Flat-top sampling multiplies the spectrum by T sinc(fT), causing high-frequency attenuation (aperture effect), corrected by an equalizer. DPCM sends quantized prediction error to reduce bits. For DM: fs = 5×2×3 kHz = 30 kHz; maximum amplitude = (Δ fs)/(2π fm) = (0.25×30000)/(2π×2000) = 0.597 V.

## Answer

#### Aperture effect in flat-top sampling

In flat-top sampling each sample is held for a pulse width $T$. The flat-top signal is the ideally sampled signal convolved with a rectangular pulse $h(t)$ of width $T$:

$$
S(f) = f_s\sum_k M(f - kf_s)\,H(f), \qquad H(f) = T\,\text{sinc}(fT)\,e^{-j\pi fT}
$$

The message spectrum is multiplied by $\text{sinc}(fT)$, so high frequencies are attenuated. This amplitude distortion (plus a delay $T/2$) is the **aperture effect**. It is reduced by keeping $T \ll T_s$ (duty ratio ≤ 0.1) or by an **equalizer** after the reconstruction filter with response $1/H(f)$.

#### DPCM scheme

PCM sends full samples although adjacent samples are highly correlated. DPCM sends only the quantized **prediction error**:

```text
m[n]->(+)-e[n]->[Quant.]--+--eq[n]-->[Encoder]--> out
       ^-                 |
     m^[n]               (+)<-+
       |                  |   |
       +--[Predictor]<-mq[n]  |
              +---------------+
Receiver: eq[n]->(+)->mq[n]->[LPF]; feedback via
          the same predictor gives m^[n]
```

$e[n] = m[n] - \hat m[n]$, $m_q[n] = \hat m[n] + e_q[n] = m[n] + q[n]$. Since $e[n]$ is small, fewer bits are needed, or SQNR improves by the prediction gain.

#### Numerical: maximum amplitude without slope overload

**Given:** $W = 3$ kHz, $f_s = 5 \times$ Nyquist rate, $f_m = 2$ kHz, $\Delta = 250$ mV.

$$
f_s = 5 \times 2W = 5 \times 6\ \text{kHz} = 30\ \text{kHz}
$$

No slope overload requires $2\pi f_m A_m \le \Delta f_s$:

$$
A_{m,max} = \frac{\Delta f_s}{2\pi f_m} = \frac{0.25 \times 30000}{2\pi \times 2000} = \frac{7500}{12566.4} = 0.597\ \text{V}
$$

**Answer:** maximum amplitude $\approx 0.597$ V (597 mV).

## Check yourself

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

1. What causes the aperture effect in flat-top sampling?
   A. The message spectrum is multiplied by sinc(fT), attenuating high frequencies
   B. Quantization noise
   C. Aliasing from low sampling rate
   D. Clock jitter in the ADC
   Answer: A. The holding pulse acts as a sinc low-pass response. (Key point: Aperture effect)

2. How is the aperture effect reduced?
   A. Lower the sampling rate
   B. Small T/Ts (duty <= 0.1) or an equalizer with response 1/H(f)
   C. Use a wider anti-aliasing filter only
   D. Increase the pulse width
   Answer: B. Narrow pulses make sinc flat in band; an equalizer corrects the rest. (Key point: Remedy)

3. With W = 3 kHz and sampling at 5 times Nyquist rate, what is fs?
   A. 30 kHz
   B. 15 kHz
   C. 6 kHz
   D. 18 kHz
   Answer: A. Nyquist rate 6 kHz; times 5 = 30 kHz. (Key point: Sampling rate)

4. What is the maximum amplitude without slope overload (Delta = 250 mV, fm = 2 kHz)?
   A. 1.19 V
   B. 0.149 V
   C. 0.597 V
   D. 0.25 V
   Answer: C. Am = Delta fs/(2 pi fm) = 0.25 x 30000/(2 pi x 2000) = 0.597 V. (Key point: Max amplitude)

## Related in Communication Systems

- [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×
- [Flat top vs natural sampling; Nyquist rate of sin(800πt)/πt + 10cos2000t](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aflat-top-vs-natural-nyquist-rate) — Communication Systems, chapter 4
- [PCM, DPCM and DM comparison; SQNR of PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-dpcm-dm-compare-sqnr) — Communication Systems, chapter 4
- [Natural and flat top sampling; Nyquist rate of sin²(400πt)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anatural-flat-top-sin2-400pit) — Communication Systems, chapter 4
- [PCM vs DM SQNR for 6cos(5000πt) with 128 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-dm-6cos5000pit-128) — Communication Systems, chapter 4
- [Delta modulation encoder/decoder; PCM vs DM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-encoder-decoder-pcm-vs-dm) — Communication Systems, chapter 4

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