--- title: "Impulse train sampling and reconstruction" subject: "Signals and Systems" chapter: 4 key: "semester_4/signals_and_systems:impulse-train-sampling-reconstruction" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/impulse-train-sampling-reconstruction site: BCT Engineering --- # Impulse train sampling and reconstruction Signals and Systems · Chapter 4: Sampling - Asked in: 2078 Baisakh, 2+6 marks - Video: [Sampling , Aliasing and Reconstruction | Solved Problem | Nyquist Criteria | Impulse Train](https://www.youtube.com/watch?v=KYZLjXIE6JY) — Fakhar STEM Sphere - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/impulse-train-sampling-reconstruction - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-4 ## Question What is impulse train sampling? How do you reconstruct original signal from its sample? ## Short answer Impulse train sampling multiplies the signal by a periodic impulse train, producing impulse-weighted samples. Reconstruction passes these samples through an ideal low-pass filter with gain T and cutoff between ω_m and ω_s−ω_m, provided ω_s > 2ω_m; in time domain this corresponds to sinc interpolation of the sample values. ## Answer #### Impulse train sampling **Impulse train (ideal) sampling** multiplies the continuous signal $x(t)$ by a periodic train of unit impulses spaced $T$ seconds apart. The result is a train of impulses whose strengths are the sample values: $$ x_p(t) = x(t)\sum_{n=-\infty}^{\infty}\delta(t - nT) = \sum_{n=-\infty}^{\infty}x(nT)\,\delta(t - nT) $$ $T$ is the sampling period and $\omega_s = 2\pi/T$ the sampling frequency. #### Spectrum of the sampled signal $p(t) = \sum\delta(t - nT) \leftrightarrow P(j\omega) = \frac{2\pi}{T}\sum_k\delta(\omega - k\omega_s)$. Multiplication in time is convolution in frequency: $$ X_p(j\omega) = \frac{1}{2\pi}X(j\omega) * P(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X\big(j(\omega - k\omega_s)\big) $$ So $X_p(j\omega)$ is $X(j\omega)$ repeated at every multiple of $\omega_s$, scaled by $1/T$. ```text X(jw) Xp(jw), ws > 2wm /\ /\ /\ /\ / \ / \ / \ / \ -+-+--+--> w --+--+--+-+--+--+-+--+--> w -wm 0 wm -ws 0 wm ws ``` #### Reconstruction from samples 1. **Condition:** $x(t)$ must be band-limited to $\omega_m$, and $\omega_s > 2\omega_m$ (Nyquist). Then the copies do not overlap. 2. **Ideal low-pass filter:** pass $x_p(t)$ through an LPF with gain $T$ and cutoff $\omega_c$, where $\omega_m < \omega_c < \omega_s - \omega_m$ (usually $\omega_c = \omega_s/2$): $$ H(j\omega) = \begin{cases} T, & |\omega| < \omega_c \\ 0, & \text{otherwise}\end{cases} \quad\Rightarrow\quad X_r(j\omega) = X_p(j\omega)H(j\omega) = X(j\omega) $$ 3. **Time-domain view (interpolation):** the filter's impulse response is $h(t) = \frac{T\sin(\omega_c t)}{\pi t}$. Convolving it with the impulse train gives $$ x_r(t) = \sum_{n=-\infty}^{\infty}x(nT)\,\frac{T\omega_c}{\pi}\,\frac{\sin\big(\omega_c(t - nT)\big)}{\omega_c(t - nT)} $$ With $\omega_c = \omega_s/2$ this becomes $x_r(t) = \sum x(nT)\,\text{sinc}\left(\frac{t - nT}{T}\right)$: each sample is replaced by a sinc pulse, and the sum of these pulses fills in the values between samples exactly. ```text x(t) --(x)--> xp(t) --> [ LPF: gain T, cutoff wc ] --> xr(t) = x(t) ^ p(t) = sum delta(t - nT) ``` **Practical note:** an ideal LPF is non-causal, so real systems use a zero-order hold (D/A converter) followed by a smoothing filter, and sample a little above the Nyquist rate to leave a guard band. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Impulse train sampling gives x_p(t) = A. x(t) convolved with the impulse train B. Sum of δ(t - nT) with unit areas only C. Sum of x(nT) δ(t - nT), impulses weighted by the sample values D. x(t) + sum δ(t - nT) Answer: C. Each impulse carries the sample value as its area. (Key point: Impulse train sampling) 2. X_p(jω) is A. T sum X(j(ω - kωs)) B. X(jω) only C. (1/T) X(jω) P(jω) D. (1/T) sum X(j(ω - kωs)) Answer: D. Original spectrum repeated at multiples of ωs, scaled by 1/T. (Key point: Spectrum) 3. Reconstruction uses an ideal LPF with A. Gain T and cutoff ωm < ωc < ωs - ωm B. Gain 1 and cutoff above ωs C. Gain T and cutoff below ωm D. Gain 1/T and cutoff 2ωs Answer: A. It passes the k = 0 copy and restores gain to 1. (Key point: Reconstruction filter) 4. In time, the reconstruction is A. Straight lines between samples B. Sum of sinc pulses centred at each sample C. Staircase of constant values D. Sum of exponentials Answer: B. x_r(t) = sum x(nT) sinc((t - nT)/T) for ωc = ωs/2. 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