--- title: "Prove sampling theorem for low pass signals" subject: "Signals and Systems" chapter: 4 key: "semester_4/signals_and_systems:sampling-theorem-lowpass" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-theorem-lowpass site: BCT Engineering --- # Prove sampling theorem for low pass signals Signals and Systems · Chapter 4: Sampling - Asked in: 2070 Magh, 6 marks - Video: [Sampling Theorem for Low Pass Signals || Part 1 ||](https://www.youtube.com/watch?v=312h2T6bV-s) — Dr. Pallavi Ingale - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/sampling-theorem-lowpass - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4 ## Question State and prove sampling theorem for low pass signals. ## Short answer If a continuous-time signal is band-limited to ω_M, sampling at ω_s > 2ω_M uniquely recovers it. Sampling with an impulse train yields spectrum copies; no overlap when ω_s - ω_M > ω_M, and an ideal low-pass filter with gain T reconstructs exactly. ## Answer **Statement (low-pass sampling theorem):** If a continuous-time signal $x(t)$ is band-limited with $X(j\omega) = 0$ for $|\omega| > \omega_M$, then $x(t)$ is uniquely determined by its samples $x(nT)$, $n = 0, \pm1, \dots$, provided $$ \omega_s = \frac{2\pi}{T} > 2\omega_M $$ $2\omega_M$ is the Nyquist rate. **Proof.** Let $x(t)$ be band-limited: $X(j\omega) = 0$ for $|\omega| > \omega_M$. **Step 1 – sampling model.** Sample with the impulse train $p(t) = \sum_n \delta(t-nT)$: $$ x_p(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\delta(t-nT) $$ **Step 2 – transform of $p(t)$.** $p(t)$ is periodic with period $T$; its Fourier series coefficients are all $1/T$, so $$ P(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s), \qquad \omega_s = \frac{2\pi}{T} $$ **Step 3 – multiplication property.** $$ \begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big) \end{aligned} $$ **Step 4 – no-overlap condition.** The $k=0$ copy lies in $[-\omega_M, \omega_M]$ and the $k=1$ copy starts at $\omega_s - \omega_M$. They do not overlap if $$ \omega_s - \omega_M > \omega_M \;\Rightarrow\; \omega_s > 2\omega_M $$ **Step 5 – recovery.** Pass $x_p(t)$ through an ideal LPF $H(j\omega) = T$ for $|\omega| < \omega_c$, 0 otherwise, with $\omega_M < \omega_c < \omega_s - \omega_M$. Only the $k=0$ term survives: $$ X_r(j\omega) = H(j\omega)X_p(j\omega) = T\cdot\frac{1}{T}X(j\omega) = X(j\omega) $$ Hence $x_r(t) = x(t)$: the samples determine $x(t)$ uniquely. $\blacksquare$ ```text X(jw) Xp(jw), ws > 2wM LPF H(jw) /\ /\ /\ /\ ______ / \ / \ / \ / \ | | -wM wM -ws 0 ws -wc wc ``` In time domain the recovered signal is the sinc interpolation $x(t) = \sum_n x(nT)\,\dfrac{\sin(\omega_s(t-nT)/2)}{\omega_s(t-nT)/2}$ (with $\omega_c = \omega_s/2$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition lets a band-limited x(t) be uniquely determined by its samples? A. ωs < 2ωM B. ωs = ωM C. ωs > 2ωM D. T > 2π/ωM only Answer: C. Sampling above twice the highest frequency avoids overlap and ensures unique recovery. (Key point: Theorem statement) 2. What is the Fourier transform of p(t) = Σδ(t - nT)? A. (2π/T) Σ δ(ω - kωs) B. Σ δ(ω - kT) C. T Σ δ(ω - kωs) D. A single impulse at ω = 0 Answer: A. p(t) is periodic with Fourier coefficients 1/T, giving an impulse train of strength 2π/T at multiples of ωs. (Key point: Impulse train transform) 3. By the multiplication property, Xp(jω) equals: A. T Σ X(j(ω - kωs)) B. (1/T) Σ X(j(ω - kωs)) C. X(jω) P(jω) D. (1/T) X(jω) Answer: B. Multiplication in time is (1/2π) convolution in frequency, giving scaled shifted copies of X. (Key point: Sampled spectrum) 4. What ideal LPF recovers x(t) from xp(t)? A. Gain 1/T with cutoff above ωs B. Gain T with cutoff below ωM C. Gain 1 with cutoff exactly 2ωs D. Gain T with cutoff ωc between ωM and ωs - ωM Answer: D. Gain T cancels the 1/T factor and the cutoff selects only the k=0 copy. 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