--- title: "Practical factors in sampling; sampling interval for X1(t)X2(t)" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:sampling-practical-x1x2" source: https://www.bctengineer.com/q/semester_6/communication_systems/sampling-practical-x1x2 site: BCT Engineering --- # Practical factors in sampling; sampling interval for X1(t)X2(t) Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2070 Chaitra (CS II), 6+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/sampling-practical-x1x2 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question What are the practical factors to be considered while sampling? Explain. If two band limited signals X1[t] and X2[t] have bandwidths of W1 and W2 Hertz respectively, estimate the maximum sampling interval required for the signal given by Y[t] = X1[t] X2[t]. ## Short answer Practical sampling considers anti-aliasing filters, guard bands, aperture effect, jitter, and clock stability. For Y(t)=X1(t)X2(t), bandwidth is W1+W2, so maximum sampling interval is 1/(2(W1+W2)) seconds. ## Answer #### Practical factors in sampling 1. **Signals are not strictly band-limited.** Real signals (and noise) have some energy above $W$, which would alias. An **anti-aliasing LPF** is used before the sampler. 2. **Non-ideal filters.** Ideal brick-wall filters do not exist, so the sampling rate is chosen **above** $2W$ to leave a **guard band** for the filter's transition region (speech 3.4 kHz sampled at 8 kHz instead of 6.8 kHz). 3. **Finite pulse width (aperture effect).** Practical samples are pulses of width $\tau$, not impulses. Flat-top sampling multiplies the spectrum by $\tau\,\text{sinc}(f\tau)$, attenuating high frequencies; fixed by small $\tau/T_s$ and an **equalizer** $1/H(f)$. 4. **Sample-and-hold limits.** Acquisition time, droop of the hold capacitor and **aperture jitter** (uncertainty in sampling instant) cause errors, especially for fast signals. 5. **Clock stability.** Jitter in $f_s$ produces timing errors, so stable crystal clocks are used. 6. **Choice of rate.** Higher $f_s$ eases filtering but raises bit rate, storage and bandwidth; a compromise is chosen. 7. **Noise and quantization.** The sampled value is followed by quantization; ADC resolution and noise must suit the required SNR. ```text m(t)->[Anti-alias LPF]->[S/H, fs>2W]->[ADC]->... ...->[DAC]->[Recon. LPF]->[Equalizer]->m(t) ``` #### Maximum sampling interval for $Y(t) = X_1(t)X_2(t)$ Multiplication in time is convolution in frequency: $$ Y(f) = X_1(f) * X_2(f) $$ If $X_1$ occupies $|f| \le W_1$ and $X_2$ occupies $|f| \le W_2$, the convolution occupies $|f| \le W_1 + W_2$. So the bandwidth of $Y$ is $W_1 + W_2$. $$ \begin{aligned} f_{s,min} &= 2(W_1 + W_2) \\ T_{s,max} &= \frac{1}{2(W_1 + W_2)} \end{aligned} $$ **Answer: maximum sampling interval $= \dfrac{1}{2(W_1 + W_2)}$ seconds.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is an anti-aliasing LPF needed in practice? A. Real signals and noise are not strictly band-limited B. Samples are impulses C. Quantizer is non-linear D. To reduce pulse width Answer: A. Energy above W would alias into the band. (Key point: Anti-alias filter need) 2. Why is speech (3.4 kHz) sampled at 8 kHz rather than 6.8 kHz? A. To reduce aliasing from clock only B. To leave a guard band for the filter transition region C. To increase quantization levels D. To avoid companding Answer: B. Non-ideal filters need transition space. (Key point: Guard band) 3. Aperture effect due to finite pulse width is reduced by: A. Small tau/Ts and an equalizer 1/H(f) B. Higher quantization levels C. Larger tau D. Removing the LPF Answer: A. Flat-top sampling multiplies the spectrum by tau sinc(f tau), compensated by equalization. (Key point: Aperture fix) 4. Y(t) = X1(t)X2(t) has bandwidth: A. max(W1, W2) B. W1 - W2 C. W1 + W2 D. W1 W2 Answer: C. Multiplication convolves the spectra, giving width W1 + W2. (Key point: Bandwidth of product) 5. The maximum sampling interval for Y(t) is: A. 1/(W1+W2) B. 1/(2(W1+W2)) C. 2/(W1+W2) D. 1/(2 max(W1,W2)) Answer: B. fs,min = 2(W1+W2), so Ts,max = 1/(2(W1+W2)). 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