‫حل سوال ‪)1‬‬
‫حل سوال ‪)2‬‬
‫حل سوال ‪)3‬‬
‫‪1/2‬‬
‫‪1‬‬
‫‪1‬‬
‫𝑎 = )𝑎 ‪= (𝑥[1] + 𝑥[−1]) = (𝑎 +‬‬
‫‪2‬‬
‫‪2‬‬
‫حل سوال ‪)4‬‬
‫حل سوال ‪)5‬‬
‫حل سوال ‪)6‬‬
‫‪1‬‬
‫‪1‬‬
‫𝑎 = )𝑎 ‪= (𝑥[1] + 𝑥[−1]) = (𝑎 +‬‬
‫‪2‬‬
‫‪2‬‬
)7 ‫حل سوال‬
(b) From part (a) we know that h[n] is length 3 with even symmetry around h[1]. Let h[1] = b and
h[0] = h[2] = a from (iv) we have:
ℎ[𝑛] =
𝑛=0
1 𝜋
1 𝜋
∫ 𝐻(𝑒 𝑗𝜔 )𝑒 𝑗𝜔𝑛 𝑑𝜔 ⇒ ℎ[0] =
∫ 𝐻(𝑒 𝑗𝜔 )𝑑𝜔 ≝ 2 ⟹ ℎ[0] = 𝑎 = 2
2𝜋 −𝜋
2𝜋 −𝜋
From (v) we also have
2
𝑗𝜋
𝐻(𝑒 ) = ∑ ℎ[𝑛]𝑒 −𝑗𝜋𝑛 = 𝑎 − 𝑏 + 𝑎 = 0 ⟹ 𝑏 = 4
𝑛=0
So we have h[0]=2, h[1]=4, h[2]=2.
‫حل سوال ‪)8‬‬