Question is ⇒ If h1[n] = 3δ[n] + δ[n - 1], h2[n] = 2δ[n] + δ[n, Options are ⇒ (A) {6, 18, 7, 4, 0, - 6, 9}, (B) {9, 0, - 6, 21, 7, 6, 1}, (C) {7, - 1, 3, 1, 7, 0, 6}, (D) {9, - 6, 0, 1, 21, 7, 18, 6}, (E) , Leave your comments or Download question paper.

Question is ⇒ If h1[n] = 3δ[n] + δ[n - 1], h2[n] = 2δ[n] + δ[n, Options are ⇒ (A) {6, 18, 7, 4, 0, - 6, 9}, (B) {9, 0, - 6, 21, 7, 6, 1}, (C) {7, - 1, 3, 1, 7, 0, 6}, (D) {9, - 6, 0, 1, 21, 7, 18, 6}, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	If h₁[n] = 3δ[n] + δ[n - 1], h₂[n] = 2δ[n] + δ[n
A.	{6, 18, 7, 4, 0, - 6, 9} [Wrong Answer]
B.	{9, 0, - 6, 21, 7, 6, 1} [Wrong Answer]
C.	{7, - 1, 3, 1, 7, 0, 6} [Wrong Answer]
D.	{9, - 6, 0, 1, 21, 7, 18, 6} [Correct Answer]
View Answer 1 -1 Explanation:- Answer : D 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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