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

Question

Mathematics

Rolando Villanueva

15 March, 19:51

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

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Answers (1)

Know the Answer?

Darius Osborn · 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).