A real continuous-time signal x(t) = sin(2 pi(4 x 103)t) + sin (2 pi (8 x 103)t)

Question

A real continuous-time signal x(t) = sin(2 pi(4 x 103)t) + sin (2 pi (8 x 103)t) is presented to the system shown in Figure 1. a- What is the system's Nyquist sampling rate (Sa/s)? f(Nyquist Rate): b- The Shannon interpolator (filter), shown in Figure 1, maps x[k] into y(t) = sin(2 pi f1t) + sin(2 pi f2 t). if fs = 6kHz. What are the baseband frequency values for f1, and f2? f1: f2: c- What is the lowest sampling frequency fs 6kHz, that will result in a reconstructed signal y(t) = 0 for all t? (Hint: What is required to make y(t) = 0?) Figure 1: Interpolation System

Explanation / Answer

a) Nyquist rate represents the twice the highest frequency of message signal.
In given signal, the highest frequency is 8x103 Hz

Nyquist rate=2x8x103 =16x103 Sa/s

c)given fs =6kHz

here the nyquist rate must be greater than or equal to the sampling rate to reconstruct the signal

given that the baseband signal frequency as f1. so nyquist rate = 2f1

now to reconstruct the signal, 2f1=6x103

hence f1=3kHz. Below this frequency, y(t)=0

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