Question

1. (50pt) NOTE: To get full mark, you are required to: (1) Plot sampled signals and fitered signals in the frequency domain, (e) Provide solution in time domain and (3 Provide the reasoning. HINT: Take notice on the height of the low pass filter for correct solutions A signal r(t)-cos(3rt) is sampled at a rate of f, samples per second. The sampled signal is then passed through an ideal low pass filter (LPF) with a cutoff frequency at 2 Hz as shown in Fig. 1. Determine the signal at the output of the LPF if sampling rate f, is 1/f Hif) Figure I: LPF in Problem 1.

Answer 1

X(t) = cos(3 pi t) wm = 3 pi rad/sec 2 pi middot fm = 3 pi fm = 1.5 Hz \r\n Block diagram Now y_z (f) = 1/T_s sigma^n = infinity_n = -infinity X(f - n f_s) where f_s is sampling frequency \r\n So frequency component present in sampled signal n f_s plusminus f_m for n = 0, 1, 2, 3 f_s = sampling frequency. < i > f_s = 4 Hz, f_m = 1.5 Hz so 4n plusminus 1.5 (sampled frequency) n = 0 plusminus 1.5 Hz n = 1 2.5 Hz, 5.5 Hz n = 2 5.5 Hz, 9.5 Hz \r\n So Y(f) has frequency component at plusminus 1.5 Hz Y(f) = delta(f + 1.5) + delta(f - 1.5) Taking inverse fourier Transform y(f) = 2 cos 2 pi middot (1.5)t y(t) = 2 cos(3 pi t) \r\n < b > 3 Hz 3n plusminus 1.5 Hz n = 0 plusminus 1.5 Hz n = 1 3 - 1.5, 3