--- title: "Flat top vs natural sampling; Nyquist rate of sin(800πt)/πt + 10cos2000t" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:flat-top-vs-natural-nyquist-rate" source: https://www.bctengineer.com/q/semester_6/communication_systems/flat-top-vs-natural-nyquist-rate site: BCT Engineering --- # Flat top vs natural sampling; Nyquist rate of sin(800πt)/πt + 10cos2000t Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2079 Bhadra (CS II), 5+5 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/flat-top-vs-natural-nyquist-rate - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question Briefly explain flat top sampling and compare its advantage over natural sampling. Find the Nyquist rate and the Nyquist interval for the signal x(t) = sin(800πt)/πt + 10 cos 2000t. ## Short answer Flat top sampling holds each sample constant, easy for digitization but introduces aperture effect equalized by 1/H(f). Nyquist rate is 800 Hz, Nyquist interval is 1.25 ms for the given signal. ## Answer #### Flat-top sampling In flat-top sampling, each sample value $g(nT_s)$ is held constant for a pulse width $\tau$ (sample-and-hold). The PAM wave is $$ s(t) = \sum_n g(nT_s)\,h(t - nT_s), \qquad S(f) = f_s H(f)\sum_n G(f - nf_s) $$ where $h(t)$ is a rectangular pulse of width $\tau$ and $H(f) = \tau\,\text{sinc}(f\tau)\,e^{-j\pi f\tau}$. The spectrum copies are shaped by $H(f)$, so high message frequencies are slightly attenuated (**aperture effect**), corrected by an equalizer $1/H(f)$ at the receiver. **Advantages of flat-top over natural sampling** | Point | Natural sampling | Flat-top sampling | |---|---|---| | Pulse top | Follows the signal | Constant (flat) | | Circuit | Analog switch/multiplier | Simple sample-and-hold | | Quantization/ADC | Hard (value changes during pulse) | Easy (value constant) | | Noise | Top varies, harder to detect amplitude | Fixed amplitude, easier detection and less noise interference | | Distortion | No aperture distortion | Small aperture effect, easily equalized | So flat-top sampling is preferred in practical PCM/PAM systems because the held value can be digitized and transmitted easily. #### Nyquist rate and interval $$ x(t) = \frac{\sin(800\pi t)}{\pi t} + 10\cos(2000t) $$ - First term: $\sin(2\pi Wt)/(\pi t)$ is an ideal low-pass (sinc) signal with flat spectrum up to $W$. Here $2\pi W = 800\pi$, so $W_1 = 400$ Hz. - Second term: $\omega = 2000$ rad/s, so $f_2 = 2000/(2\pi) = 318.31$ Hz (the argument has no $\pi$). - Highest frequency: $f_{max} = 400$ Hz. $$ \begin{aligned} f_{Nyquist} &= 2f_{max} = 2 \times 400 = 800\ \text{Hz} \\ T_{Nyquist} &= \frac{1}{800} = 1.25\ \text{ms} \end{aligned} $$ **Answer: Nyquist rate = 800 Hz (samples/s); Nyquist interval = 1.25 ms.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is flat-top sampling preferred over natural sampling in practice? A. It has no aperture distortion B. The held constant value is easy to quantize and detect C. It needs a higher sampling rate D. Its spectrum copies keep the exact shape of G(f) Answer: B. Sample-and-hold gives a constant amplitude during the pulse, which makes quantizing easy. The price is a small aperture effect. (Key point: Flat-top advantage) 2. The aperture effect of flat-top sampling is corrected at the receiver by: A. A high-pass filter B. An equalizer with response 1/H(f) C. A wider sampling pulse D. A compressor Answer: B. H(f) = tau sinc(f tau) attenuates high frequencies, so 1/H(f) restores them. (Key point: Aperture effect fix) 3. sin(800 pi t)/(pi t) has bandwidth W equal to: A. 800 Hz B. 400 Hz C. 1600 Hz D. 200 Hz Answer: B. sin(2 pi W t)/(pi t) with 2 pi W = 800 pi gives W = 400 Hz. (Key point: Frequency of first term) 4. For x(t) = sin(800 pi t)/(pi t) + 10 cos(2000t), the Nyquist rate and interval are: A. 800 Hz and 1.25 ms B. 637 Hz and 1.57 ms C. 2000 Hz and 0.5 ms D. 400 Hz and 2.5 ms Answer: A. The cosine is only 2000/(2 pi) = 318 Hz, so fmax = 400 Hz, giving 2 x 400 = 800 Hz and T = 1.25 ms. 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