Suppose that r(t) is a band-limited signal with the bandwidth W. Suppose that we sampled this...

 

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Suppose that r(t) is a band-limited signal with the bandwidth W. Suppose that we sampled this signal with the sampling interval T, to generate the sample sequence r[n]; suppose that 27/T, is larger than the Nyquist rate 2W. Given z[n], we reconstructed a continuous time signal z, () using the zero-order-hold method. In other words, Ir(t) = z[n] for t€ [nTs, (n+1)Ts). In the last lecture, we derived that Ir (l) = Is(l) * h(1) where s(l), as usual, denotes the continuous time representation of the samples, and 1 if L € [0, T₂) 0 otherwise If we consider this relation in the frequency domain, we have h(1) Now, select a correct statement. X, (jw) = Xs (jw) - H (jw). (3) Note that h(t) = rect (7/2). By applying the time scaling and time delay properties of Fourier transform (FT) to the FT of rect(), we can show that H(jw) = 23 sin(@T,/2) (a) X, (jw) has bandwidth 2W, i.e., X, (jw) = 0 for w with |w| > 2W. (b) X, (jw) has bandwidth w, +W, i.e., X, (jw)=0 for w with |w|>w+W. (d) None of the above is correct. (c) Let z(t) denote the output signal of an LTI system with frequency response H₁(jw) = 2 when the input is r(t). Then, (t-T/2) is equivalent to z, (l). (1) sin(wT,/2) لیا (4) Suppose that r(t) is a band-limited signal with the bandwidth W. Suppose that we sampled this signal with the sampling interval T, to generate the sample sequence r[n]; suppose that 27/T, is larger than the Nyquist rate 2W. Given z[n], we reconstructed a continuous time signal z, () using the zero-order-hold method. In other words, Ir(t) = z[n] for t€ [nTs, (n+1)Ts). In the last lecture, we derived that Ir (l) = Is(l) * h(1) where s(l), as usual, denotes the continuous time representation of the samples, and 1 if L € [0, T₂) 0 otherwise If we consider this relation in the frequency domain, we have h(1) Now, select a correct statement. X, (jw) = Xs (jw) - H (jw). (3) Note that h(t) = rect (7/2). By applying the time scaling and time delay properties of Fourier transform (FT) to the FT of rect(), we can show that H(jw) = 23 sin(@T,/2) (a) X, (jw) has bandwidth 2W, i.e., X, (jw) = 0 for w with |w| > 2W. (b) X, (jw) has bandwidth w, +W, i.e., X, (jw)=0 for w with |w|>w+W. (d) None of the above is correct. (c) Let z(t) denote the output signal of an LTI system with frequency response H₁(jw) = 2 when the input is r(t). Then, (t-T/2) is equivalent to z, (l). (1) sin(wT,/2) لیا (4)

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