Question

g A signal x(????) with Fourier transform ????(???? ) undergoes impulse-train sampling to generate 4 where For each of the following sets of constraints on x(????) or ????(???? ), does the sampling guarantee that x(????) can be recovered from xxpp(????????)? Justify your answer and show your work. Simple yes/no answers are not acceptable. a) ????????(???????????? )=0 for |????????|>5,000???????? b) ????????(???????????? )=0 for |????????|>15,000???????? c) ???????????? {????????(???????????? )}=0 for |????????|>5,000???????? d) xx(????????) is real and ????????(???????????? )=0 for |????????|>15,000???????? [hint: if xx(????????) is real-valued then ????????(???????????? )=????????∗(−???????????? )] e) ????????(???????????? )∗????????(???????????? )=0 for |????????|>15,000???????? f) |????????(???????????? )|=0 for ????????>5,000???????? Problem

Answer 1

(a) The Nyquist rate for the given signal is 2 × 5000π = 10000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

10000π = 2 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax, x(t) can be recovered from xp(t).

(b) The Nyquist rate for the given signal is 2 × 15000π = 30000π. Therefore, in order to be able

to recover x(t) from xp(t), the sampling period must at most be Tmax =

2π

30000π = 0.66×10−4

sec. Since the sampling period used is T = 10−4 > Tmax, x(t) cannot be recovered from

xp(t).

(c) Here, Im{X(jω)} is not specified. Therefore, the Nyquist rate for the signal x(t) is indeterminate. This implies that one cannot guarantee that x(t) would be recoverable from xp(t).

(d) Since x(t) is real, we may conclude that X(jω) = 0 for |ω| > 5000. Therefore, the answer to

this part is identical to that of part (a).

(e) Since x(t) is real, X(jω) = 0 for |ω| > 15000π. Therefore, the answer to this part is identical

to that of part (b).

(f) If X(jω) = 0 for |ω| > ω1, then X(jω) ∗ X(jω) = 0 for |ω| > 2ω1. Therefore, in this

part, X(jω) = 0 for |ω| > 7500π. The Nyquist rate for this signal is 2 × 7500π = 15000π.

Therefore, in order to be able to recover x(t) from xp(t), the sampling period must at most

be Tmax =

2π

15000π = 1.33 × 10−4

sec. Since the sampling period used is T = 10−4 < Tmax,

x(t) can be recovered from xp(t).

(g) If |X(jω)| = 0 for |ω| > 5000π, then X(jω) = 0 for |ω| > 5000π. Therefore, the answer to

this part is identical to the answer of part (a).