--- title: "Merits and demerits of flat-top sampling" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:flat-top-sampling-merits" source: https://www.bctengineer.com/q/semester_6/communication_systems/flat-top-sampling-merits site: BCT Engineering --- # Merits and demerits of flat-top sampling Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2081 Baisakh (CS II), 7 marks - Video: [Flat Top Sampling or PAM - Pulse Amplitude Modulation in Digital Communication by Engineering Funda](https://www.youtube.com/watch?v=PdIYPU1AfTc) — 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/flat-top-sampling-merits - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question Explain the merits and demerits of flat-top sampling of a continuous time signal using relevant mathematical expressions. ## Short answer Flat-top sampling uses sample-and-hold; flat tops ease ADC quantization and detection. The sinc(fT) aperture effect attenuates higher message frequencies (e.g. 3.9 dB at band edge) and requires an equalizer. Wide pulses also increase crosstalk in TDM. ## Answer **Flat-top sampling** produces pulses whose amplitude equals the instantaneous sample value $m(nT_s)$ and stays **constant** (flat) for the pulse duration $T$. It is produced by a **sample-and-hold** circuit. #### Mathematical model The flat-top signal is $$ s(t) = \sum_{n=-\infty}^{\infty} m(nT_s)\,h(t - nT_s), \qquad h(t) = \begin{cases} 1, & 0 < t < T \\ 0, & \text{otherwise} \end{cases} $$ This equals the ideally sampled signal convolved with $h(t)$: $$ s(t) = \left[m(t)\sum_n \delta(t - nT_s)\right] * h(t) = m_\delta(t) * h(t) $$ Taking the Fourier transform: $$ S(f) = M_\delta(f)\,H(f) = f_s\sum_{k=-\infty}^{\infty} M(f - kf_s)\;H(f) $$ where $$ H(f) = T\,\text{sinc}(fT)\,e^{-j\pi fT} $$ ```text |H(f)| T |\___ | \__ sinc envelope | \___ attenuates higher +-----------\----->f message frequencies 0 W 1/T ``` #### Merits 1. **Easy to generate** with a sample-and-hold circuit; the constant level gives the ADC time to quantize, so it is the usual practical method before PCM encoding. 2. Pulse tops are flat, so amplitude detection is simple and noise riding on the top is easier to handle. 3. Since the amplitude is constant during the pulse, the energy per sample is well defined, giving a better SNR than very narrow pulses. #### Demerits 1. **Aperture effect:** each spectral copy is multiplied by $\text{sinc}(fT)$, so higher message frequencies are attenuated: amplitude distortion. Example: at $f = W$ with $T = T_s$ and $f_s = 2W$, $\text{sinc}(0.5) = 0.637$, i.e. a 3.9 dB loss. 2. **Delay distortion:** the term $e^{-j\pi fT}$ introduces a delay of $T/2$. 3. An **equalizer** with response $\frac{1}{H(f)} = \frac{\pi f}{\sin(\pi fT)}e^{j\pi fT}$ (within the message band) is needed after the reconstruction LPF. The distortion is negligible if $T/T_s \le 0.1$. 4. Wide pulses leave less guard time between samples, which increases crosstalk between channels in TDM. Compared with natural sampling (where each copy of $M(f)$ keeps its shape), flat-top sampling distorts the spectrum, but it is preferred in practice because of the simple hold-circuit implementation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is flat-top sampling produced? A. Differentiation B. Multiplying by pulse train C. PLL D. Sample-and-hold Answer: D. The held level gives the ADC time to quantize. (Key point: Flat-top generation) 2. What is S(f) of flat-top sampling? A. fs sum M(f - k fs) times H(f) = T sinc(fT) e^(-j pi fT) B. fs M(f) C. M(f) only D. Sum of Cn M(f - n fs) Answer: A. Ideal sampled spectrum times pulse transform. (Key point: Spectrum) 3. What is the main demerit? A. Aperture effect: high message frequencies attenuated by sinc(fT) B. Aliasing always C. No hold needed D. Needs infinite bandwidth Answer: A. Plus a T/2 delay. (Key point: Aperture loss) 4. What corrects it? A. A high-pass filter B. Higher quantizer levels C. Equalizer with response 1/H(f) after the LPF D. Larger T Answer: C. Distortion is negligible if T/Ts <= 0.1. 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