EE422

Solution to Homework #17

1a) Taking the Fourier Transform we find that X(f)=2d (f)+2/2d (f-4)+2/2d (f+4). Plot:

(Ideal) Sampling this signal at f_s = 16 Hz:

Obviously, we can recover the original signal if we put the sampled signal through an ideal low-pass filter with a gain of 1/16.

b) f_s = 8 Hz:

We CANNOT recover the original signal because of aliasing. If we use a low pass filter with a gain of 1/f_s,= 1/8 and a bandwidth of 4 Hz, then we would obtain the following recovered spectrum:

We immediately recognize this spectrum to be the Fourier transform of x_recovered(t) = 2 + 4cos(8p t) which is NOT the original signal!

2a-i) f_s = 80 Hz

ii) f_s = 40 Hz

iii) f_s = 20 Hz

b) Sample rates of f_s ³ 2*f_h = 2*20=40 Hz are acceptable, so part ii) with f_s = 40 Hz is barely acceptable (almost aliasing) and part i) with f_s = 80 Hz is excellent! In Lab 2, we will see a signal which CANNOT be sampled at exactly f_s = 2*f_h but must be sampled at f_s > 2*f_h!!!