--- title: "State and prove Nyquist sampling theorem" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:nyquist-sampling-prove" source: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-sampling-prove site: BCT Engineering --- # State and prove Nyquist sampling theorem Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2080 Bhadra (CS II), 8 marks - Video: [Sampling Theorem](https://www.youtube.com/watch?v=XyZerq7-IvA) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/nyquist-sampling-prove - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question State and prove Nyquist sampling theorem. ## Short answer A band-limited signal with max frequency W is uniquely determined by samples taken at fs ≥ 2W. Ideal sampling repeats the spectrum every fs; if fs ≥ 2W, copies do not overlap, and an ideal LPF recovers the original. Aliasing occurs if fs < 2W. ## Answer #### Statement A band-limited signal $m(t)$ with no frequency components above $W$ Hz ($M(f) = 0$ for $|f| > W$) is **uniquely determined by its samples** taken at uniform intervals $T_s \le \frac{1}{2W}$, i.e. at a rate $f_s \ge 2W$. It can be recovered exactly from the samples by an ideal low-pass filter of bandwidth $W$. The minimum rate $2W$ is the **Nyquist rate**; $1/(2W)$ is the **Nyquist interval**. #### Proof **1. Ideal sampling.** Multiply $m(t)$ by a periodic impulse train: $$ m_\delta(t) = m(t)\sum_{n=-\infty}^{\infty}\delta(t - nT_s) = \sum_{n=-\infty}^{\infty} m(nT_s)\,\delta(t - nT_s) $$ **2. Spectrum of the impulse train.** It is periodic, with Fourier series coefficients $1/T_s$: $$ \sum_n \delta(t - nT_s) = \frac{1}{T_s}\sum_{k=-\infty}^{\infty} e^{j2\pi k f_s t} \;\longleftrightarrow\; f_s\sum_{k}\delta(f - kf_s) $$ **3. Spectrum of the sampled signal.** Multiplication in time is convolution in frequency: $$ M_\delta(f) = M(f) * f_s\sum_k\delta(f - kf_s) = f_s\sum_{k=-\infty}^{\infty} M(f - kf_s) $$ So $M_\delta(f)$ is $M(f)$ repeated every $f_s$ Hz, scaled by $f_s$. ```text f_s >= 2W (no overlap): /\ /\ /\ / \ / \ / \ --/----\----/----\----/----\---> f -W 0 W fs-W fs 2fs f_s < 2W: copies overlap -> aliasing ``` **4. Condition for no overlap.** The copy at $f_s$ occupies $f_s - W$ to $f_s + W$. It does not overlap the baseband copy ($-W$ to $W$) if $$ f_s - W \ge W \;\Rightarrow\; f_s \ge 2W $$ **5. Recovery.** If $f_s \ge 2W$, pass $m_\delta(t)$ through an ideal LPF with $H(f) = T_s$ for $|f| < W$ (any cut-off between $W$ and $f_s - W$): $$ M_\delta(f)\,H(f) = M(f) $$ so $m(t)$ is recovered exactly. **6. Interpolation formula.** With $f_s = 2W$, the LPF impulse response is $h(t) = \text{sinc}(2Wt)$, so $$ m(t) = \sum_{n=-\infty}^{\infty} m\!\left(\frac{n}{2W}\right)\text{sinc}(2Wt - n) $$ Each sample is replaced by a sinc pulse and the sum rebuilds $m(t)$ at all instants. This proves the theorem. If $f_s < 2W$, the copies overlap (**aliasing**) and $m(t)$ cannot be recovered; an anti-aliasing filter is therefore used before sampling. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The Nyquist condition is: A. fs = W B. fs >= 2W C. fs >= W/2 D. fs <= 2W Answer: B. Sampling at least twice the highest frequency. (Key point: Statement) 2. What is the spectrum of the ideally sampled signal? A. M(f) only B. fs sum M(f - k fs) C. fs M(f/ k) D. sum M(f) fs Answer: B. M(f) repeated every fs. (Key point: Sampled spectrum) 3. What condition prevents overlap of copies? A. fs <= 2W B. fs = 0 C. fs - W >= W D. fs + W >= W Answer: C. Gives fs >= 2W. (Key point: No overlap) 4. How is m(t) recovered? A. Differentiator B. Squarer C. Ideal LPF with gain Ts and cutoff between W and fs - W D. High-pass filter Answer: C. Passes only the baseband copy. (Key point: Recovery) ## Related in Communication Systems - [Nyquist sampling theorem; flat top sampling](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-sampling-flat-top) — Communication Systems, chapter 4 - [Nyquist sampling theorem; natural sampling](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-natural-sampling) — 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 - [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 - [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 - [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 ## Related in other subjects - [Prove sampling theorem; Nyquist of product](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aprove-sampling-theorem-nyquist) — Signals and Systems, chapter 4, asked 2× - [Sampling theorem; Nyquist of 1 + cos + sin](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-theorem-nyquist-1-cos-sin) — Signals and Systems, chapter 4, asked 5× - [Sampling theorem; aliasing cause; Nyquist](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asampling-aliasing-cause-nyquist) — Signals and Systems, chapter 4 - [Sampling theorem, Nyquist criterion and aliasing](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_i%3Asampling-nyquist-aliasing) — Instrumentation I, chapter 6