Question

H(o) s(t) ﹁ | y(t) | lyst) Impulse sample at rate o -B 0 B c) (5 pts) Using your value of B from part b, what is minimum value of the sampling rate co, that will allow the filter output y(t) to be perfectly recovered from its impulse sampled version ys(0)? d) (5 pts) What is the purpose of the filter H()? (One sentence answer please.) e) (10 pts) Suppose the sampling rate o, is double your answer in part (c). (This is called 2x over-sampling.) Carefully sketch the Fourier Transform Y (00) of the impulse sampled output ys(t), showing all significant amplitudes and frequencies. f (5 pts) Now suppose that the sampled signal ys(t) from part e) is the input to a recovery filter with frequency response H(o). Carefully sketch H4o) so that the output of the recovery filter is the signal y(t), showing all significant amplitudes and frequencies

Answer 1

a. We cannot recover from $s_\delta(t)$ since $S(\omega)$ is non zero for  $\omega \quad \epsilon \quad (-\infty,\infty)$ , i.e. S(w) is not a low/band pass signal so Nyquist sampling with a finite sampling frequency $\omega_s$ is not possible.

b.

Such

c. B = 3 Hz, the minimum sampling frequency, $\omega_s = 2B = 6$ Hz.

d. Filter H(w) is applied for making the signal s(t) into band limited/lowpass so that it can be sampled at finite sampling rate.

e.

3-223

f )