JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B. Tech III DIGITAL COMMUNICATIONS Previous Question Paper (August - 2022 - R16)

Ideal sampling, or impulse sampling, involves multiplying a continuous-time signal x(t) by a periodic impulse train s(t) with period T_s. Mathematically, the sampled signal is x_s(t) = x(t) ∑ δ(t - nT_s). In the frequency domain, this results in the replication of the spectrum of x(t) at integer multiples of the sampling frequency f_s = 1/T_s. According to the Nyquist-Shannon sampling theorem, for perfect reconstruction, the sampling frequency must satisfy f_s ≥ 2f_m, where f_m is the highest frequency component of the signal.

Reconstruction is achieved by passing the sampled signal through an ideal low-pass filter with a cutoff frequency f_c such that f_m ≤ f_c < f_s - f_m. This filter removes the spectral replicas, leaving only the original signal spectrum scaled by 1/T_s, thereby recovering x(t) exactly.

Using the trigonometric identity cos(A)cos(B) = 0.5[cos(A-B) + cos(A+B)], the signal is expressed as: x(t) = 5[cos(6000πt) + cos(10000πt)]. The frequency components are f₁ = 3000 Hz and f₂ = 5000 Hz.

i) The highest frequency component is f_max = 5000 Hz. According to the low-pass uniform sampling theorem, the minimum sampling rate is f_s = 2f_max = 2 × 5000 = 10,000 Hz.

ii) For band-pass sampling, the signal occupies the band [3000, 5000] Hz. The bandwidth B = 5000 - 3000 = 2000 Hz. The minimum sampling rate is given by f_s = 2B(1 + 1/k), where k = floor(f_H/B). Here, f_H = 5000 and B = 2000, so k = floor(5000/2000) = 2. Thus, f_s = 2 × 2000 × (1 + 1/2) = 4000 × 1.5 = 6000 Hz.