Figure 1 shows the overall system for filtering a continuous time signal using a

Question

Figure 1 shows the overall system for filtering a continuous time signal using a discrete time filter. Suppose that x_c(t) = cos(10^4 pi t) + sin(2 middot 1064 pi t)/pi t, and H(e^j omega) is as follows: H(e^j ohm) = {1 if |ohm| lessthanorequalto pi/4 0 if |ohm| elementof (pi/4, pi) (H(e^I ohm): 2 pi-periodic) 1. Suppose that the sampling period T_s is 2.5 middot 106-5 second. Sketch X_c(j omega), X_delta(j omega), X(e^j ohm), Y(e^i ohm), Y_delta(j omega), and Y_c(j omega). What is y_c(t)/2. Obtain the Nyquist sampling rate for x_c(t). 3. Suppose that omega_s is greater than the Nyquist rate obtained in part 2 such that aliasing does not occur. Sketch X_c(j omega), X_delta(j omega), and X (e^j ohm). 4. If we choose T_s properly, we can make y_c(t) equal to x_c(t) (to see this, observe the plots you sketched in part 3.) What is the largest T_s that makes this happen?

Explanation / Answer

2 ans:) we have two signals here first one is cos signal and second one is sinc signal.in cos signal w1=10^4*pi.from this 2*pi*fm1=10^4*pi so fm1=5khz.and second signal sinc have w2=2*10^4*pi.from this 2*pi*fm2=2*10^4*pi then we get fm2=10khz.from two frequencies fm2 is higest one.so nyquist rate fs=2*fm2 i.e., 20khz is maximum rate beyond which if we increase frequency then aliasing may occur.

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