Question

1. Consider the system shown in Figure 1 below. Ideal Ideal r(t) Sample period Sample period T Figure 1. System for Problem 1. The input to the above system is x(), the output is y(), and the sampling period for both the ideal A/D and ideal D/A is T>0. For your answers to the various pa be sure to include appropriate units, where applicable. rts below., Suppose that x(1) Justify your answer. a) 5. Determine all numerical values of T such that y(t) x(1). b) Now, suppose that x()-2cos(2 ft)+4cos(2t f,), where f,-100Hz and = 500 Hz. Determine all numerical values of Y such that y(1)-x(t). Justify your Finally, suppose that x(t)-2cosCrnl) + 4 cos(2nh1), fi-100Hz,左-500 Hz, y(t) = A cos(2r/), that the output y) satisfies the above form, and then determine the positive numerical answer. c) and output y() has the form 0 is the output peak value. Firs, determine the numerical 7 value such l T value such peak output value A. Show your work.

Answer 1

(a) x(t) = 5 T can have any values as the input x(t) is a constant value. Any value of T will satisfy y(t) = x(t). (b) x(t) = 2 cos(2 pi f_1) + 4 cos(f_2 t) f_1 = 100 Hz f_2 = 500 Hz. ie, x(t) = 2 cos(2 pi 100 t) + 4 cos (2 pi 500 t) Here the highest frequency = 500 Hz. To satisfy sampling theorem f_s greaterthanorequalto 2 f_m f_s greaterthanorequalto 2 times 500 Hz ie, f_s greaterthanorequalto 1000 Hz. To get y(t) = x(t), sampling theorem must be satisfied. ie, f_s = 1 k Hz T = 1/f_s = 1/1 k Hz = 1 ms. T lessthanorequalto 1 ms \r\n (c) x(t) = 2 cos(2 pi f_1 t) + 4 cos(2 pi f_2 t) f_1 = 100 Hz and f_2 = 500 Hz. y(t) = A cos(2 pi f_1 t) Here f_s greaterthanorequalto 2 times 100 Hz. (ie, 2 f_1) f_s greaterthanorequalto 200 Hz. T lessthanorequalto 1/200 s lessthanorequalto 5 ms
​​​​​​