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Practical factors in sampling; sampling interval for 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.

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].

Practical factors in sampling

  1. Signals are not strictly band-limited. Real signals (and noise) have some energy above WW, 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 2W2W 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 τ\tau, not impulses. Flat-top sampling multiplies the spectrum by τ sinc(fτ)\tau\,\text{sinc}(f\tau), attenuating high frequencies; fixed by small τ/Ts\tau/T_s and an equalizer 1/H(f)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 fsf_s produces timing errors, so stable crystal clocks are used.
  6. Choice of rate. Higher fsf_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.
m(t)->[Anti-alias LPF]->[S/H, fs>2W]->[ADC]->...
...->[DAC]->[Recon. LPF]->[Equalizer]->m(t)

Maximum sampling interval for Y(t)=X1(t)X2(t)Y(t) = X_1(t)X_2(t)

Multiplication in time is convolution in frequency:

Y(f)=X1(f)∗X2(f)Y(f) = X_1(f) * X_2(f)

If X1X_1 occupies ∣f∣≤W1|f| \le W_1 and X2X_2 occupies ∣f∣≤W2|f| \le W_2, the convolution occupies ∣f∣≤W1+W2|f| \le W_1 + W_2. So the bandwidth of YY is W1+W2W_1 + W_2.

fs,min=2(W1+W2)Ts,max=12(W1+W2)\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 =12(W1+W2)= \dfrac{1}{2(W_1 + W_2)} seconds.

✓ Answer checked by Claude Opus 5.5 (high effort)

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