Question is ⇒ The minimum step size required for a Delta modulator operating at 36 samples/sec to track the signal (here u(t) is the unit step function) x(t) = 125t (u(t) - u(t - 1)) + (250 - 125t) (u(t - 1) - u(t - 2)) So that slope overload is avoided, would be, Options are ⇒ (A) 2-10, (B) 2-8, (C) 2-6, (D) 2-4, (E) , Leave your comments or Download question paper.

Question is ⇒ The minimum step size required for a Delta modulator operating at 36 samples/sec to track the signal (here u(t) is the unit step function) x(t) = 125t (u(t) - u(t - 1)) + (250 - 125t) (u(t - 1) - u(t - 2)) So that slope overload is avoided, would be, Options are ⇒ (A) 2-10, (B) 2-8, (C) 2-6, (D) 2-4, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	The minimum step size required for a Delta modulator operating at 36 samples/sec to track the signal (here u(t) is the unit step function) x(t) = 125t (u(t) - u(t - 1)) + (250 - 125t) (u(t - 1) - u(t - 2)) So that slope overload is avoided, would be
A.	2^-10 [Wrong Answer]
B.	2^-8 [Correct Answer]
C.	2^-6 [Wrong Answer]
D.	2^-4 [Wrong Answer]
View Answer 1 -4 Explanation:- Answer : B Discuss it below : !! OOPS Login [Click here] is required to post your answer/result Help other students, write article, leave your comments

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