--- title: "Sub-sampling; sampling sinc(5πt) at 10 Hz" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:subsampling-sinc5pit-10hz" source: https://www.bctengineer.com/q/semester_6/communication_systems/subsampling-sinc5pit-10hz site: BCT Engineering --- # Sub-sampling; sampling sinc(5πt) at 10 Hz Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2072 Kartik (CS II), 2+2+2+1 marks - Video: [037. Sub-sampling](https://www.youtube.com/watch?v=ryJPVHrj0rE) — Ali Hajimiri (Academy) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/subsampling-sinc5pit-10hz - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question Explain Sub-sampling theorem. A signal x(t) = sinc(5πt) is sampled (using uniformly spaced impulses) at a rate of 10 Hz. i) Sketch the sampled signal (not to scale); ii) Sketch the spectrum of the sampled signal for the range |f| < 30 Hz; iii) Explain whether you can recover the signal x(t) from the sampled signal. ## Short answer Sub-sampling theorem allows a bandpass signal of bandwidth B to be sampled at a rate as low as 2B provided the aliased band occupies baseband without overlap. x(t) = sinc(5πt) has bandwidth 2.5 Hz; sampled at 10 Hz, the spectrum consists of 5 Hz wide rectangles at multiples of 10 Hz. Recovery is possible with an ideal LPF cut-off between 2.5 and 7.5 Hz. ## Answer #### Sub-sampling theorem A **band-pass** signal occupying $f_L$ to $f_H$ (bandwidth $B = f_H - f_L$) can be sampled at a rate much lower than $2f_H$, roughly $2B$, without loss of information. The valid rates are $$ \frac{2f_H}{k} \le f_s \le \frac{2f_L}{k - 1}, \qquad 1 \le k \le \left\lfloor\frac{f_H}{B}\right\rfloor $$ Here aliasing is used on purpose: the band is shifted down to baseband without overlap. This is used in IF-sampling receivers. #### Numerical: $x(t) = \text{sinc}(5\pi t)$, $f_s = 10$ Hz Using $\text{sinc}(\theta) = \sin\theta/\theta$: $$ x(t) = \frac{\sin 5\pi t}{5\pi t} \;\Longleftrightarrow\; X(f) = \frac{1}{5}\,\text{rect}\!\left(\frac{f}{5}\right) $$ so $X(f) = 0.2$ for $|f| < 2.5$ Hz and zero elsewhere. Bandwidth $W = 2.5$ Hz, Nyquist rate $= 5$ Hz. **i) Sampled signal** ($T_s = 0.1$ s): $x(nT_s) = \dfrac{\sin(n\pi/2)}{n\pi/2}$ | $n$ | 0 | ±1 | ±2 | ±3 | ±4 | ±5 | |---|---|---|---|---|---|---| | $x(n/10)$ | 1 | 0.637 | 0 | −0.212 | 0 | 0.127 | ```text 1 ^ 0.637 | 0.637 ^ | ^ 0.127 | | | 0.127 ^ | | | ^ ---+----+----+----+----+----+----+----+----+----+----+-------> t (s) | | v v -0.212 -0.212 -0.5 -0.3 -0.1 0 0.1 0.3 0.5 ``` Impulses occur every 0.1 s, with heights following the sinc envelope; the samples at $t = \pm 0.2, \pm 0.4$ s are zero. **ii) Spectrum of the sampled signal** $$ X_s(f) = f_s\sum_n X(f - nf_s) = 2\sum_n \text{rect}\!\left(\frac{f - 10n}{5}\right) $$ Rectangles of height $10 \times 0.2 = 2$ and width 5 Hz, centred at $0, \pm 10, \pm 20, \pm 30$ Hz: ```text 2 __ ___ ___ ___ ___ ___ __ | | | | | | | | | | | | | | ----+-+---+---+---+---+---+---+---+---+---+---+---+-+--> f (Hz) -30 -20 -10 0 10 20 30 ``` Occupied bands for $|f| < 30$ Hz: $|f| < 2.5$, $7.5$–$12.5$, $17.5$–$22.5$ and $27.5$–$30$ Hz (and their mirror images at negative frequencies). **iii) Recovery** Yes. $f_s = 10$ Hz is greater than the Nyquist rate $2W = 5$ Hz, so the copies are separated by 5 Hz gaps (2.5 to 7.5 Hz). An ideal LPF with gain $1/10$ and cut-off anywhere between 2.5 and 7.5 Hz (e.g. 5 Hz) recovers $x(t)$ exactly. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the sub-sampling (band-pass sampling) theorem allow? A. Sampling any signal below its Nyquist rate B. Sampling a band-pass signal at about 2B rather than 2fH without loss C. Sampling above 4fH D. Recovery without filters Answer: B. A band-pass signal of bandwidth B can be sampled near 2B with the band shifted to baseband without overlap. (Key point: Sub-sampling theorem) 2. What is the bandwidth of x(t) = sinc(5 pi t)? A. 5 Hz B. 10 Hz C. 2.5 Hz D. 5 pi Hz Answer: C. Its spectrum is a rectangle of width 5 Hz, i.e. |f| < 2.5 Hz. (Key point: Bandwidth of sinc(5 pi t)) 3. At fs = 10 Hz, the sampled spectrum consists of: A. Rectangles of width 5 Hz and height 2 repeated every 10 Hz B. Rectangles overlapping each other C. A single rectangle at 0 D. Impulses at multiples of 10 Hz Answer: A. Xs(f) = fs sum X(f - n fs): copies of height 10 x 0.2 = 2 centred at multiples of 10 Hz. (Key point: Sampled spectrum) 4. Can x(t) be recovered from samples at 10 Hz? A. No, because fs < 2W B. No, because the signal is not band-limited C. Yes, with an ideal LPF cut-off between 2.5 and 7.5 Hz, since fs exceeds 5 Hz D. Only with a high-pass filter Answer: C. fs = 10 Hz exceeds the Nyquist rate of 5 Hz, leaving a gap from 2.5 to 7.5 Hz. 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