EE 511 Solutions to Problem Set 5

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CDf of X 0 F X0 (x) CDF of X 0.25 F X0.25 (x) CDF of X 0.25 F X0.5 (x) 1 0.5 0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 x 1 0.5 0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 x 1 0.5 0 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 x Figure 2: Another possible choice for Θ is: ⎧ ⎪⎨ Θ = ⎪⎩ 0 with prob. 1 12 π 4 with prob. 1 6 π 2 with prob. 1 12 3π 4 with prob. 1 6 π with prob. 1 12 5π 4 with prob. 1 6 3π 2 with prob. 1 12 7π 4 with prob. 1 6 A more general choice for f Θ (θ) can be made as follows: (i) Let us assume that the range of Θ is from 0 to 2π. (ii) The condition for mean to be constant can be obtained as follows: E[cos (2πf c t + Θ)] = ∫ 2π 0 cos (2πf c t + θ)f Θ (θ)dθ = ∫ π 0 cos (2πf ct + θ)f Θ (θ)dθ + ∫ 2π π cos (2πf ct + θ)f Θ (θ)dθ (using θ ′ = θ − π) = ∫ π 0 cos (2πf ct + θ)f Θ (θ)dθ + ∫ π 0 cos (2πf ct + θ ′ + π)f Θ (θ ′ + π)dθ ′ = ∫ π 0 cos (2πf ct + θ)f Θ (θ)dθ + ∫ π 0 [− cos (2πf ct + θ ′ )]f Θ (θ ′ + π)dθ ′ = ∫ π 0 cos (2πf ct + θ)[f Θ (θ) − f Θ (θ + π)]dθ Therefore, if f Θ (θ) = f Θ (θ + π) for θ in [0,π], then E[cos (2πf c t + Θ)] = 0. 2
(iii) The additional condition for the auto-correlation funtion to be a function of τ can be obtained as follows: E[cos (2πf c (2t + τ) + 2Θ)] = ∫ 2π 0 cos (2πf c (2t + τ) + 2θ)f Θ (θ)dθ (using φ = 2θ) = 1 2 = 1 2 (using φ ′ = φ − 2π) = 1 2 = 1 2 + 1 2 + 1 2 ∫ 4π 0 cos (2πf c (2t + τ) + φ)f Θ ( φ 2) dφ ∫ 2π 0 cos (2πf c (2t + τ) + φ)f Θ ( φ 2) dφ ∫ 4π 2π cos (2πf c(2t + τ) + φ)f Θ ( φ 2) dφ ∫ 2π 0 cos (2πf c (2t + τ) + φ)f Θ ( φ 2) dφ ( ) ∫ 2π 0 cos (2πf c (2t + τ) + φ ′ φ )f ′ Θ + π dφ ′ 2 [ ∫ ( ) )] 2π 0 cos (2πf c (2t + τ) + φ) f φ Θ 2 + (φ ′ fΘ + π dφ. 2 Assuming that we satisfy the condition from (ii) above, we get E[cos (2πf c (2t + τ) + 2Θ)] = Now, proceeding as in (ii), we need ) f Θ ( φ 2 ∫ 2π 0 cos (2πf c (2t + τ) + φ)f Θ ( φ 2 ( ) φ + π = f Θ 2 for φ/2 in [0,π]. Equivalently, we need ( f Θ (θ) = f Θ θ + π ) 2 ) dφ. for θ in [0,π/2]. (iv) Combining the conditions from (ii) and (iii), we get ( f Θ (θ) = f Θ θ + kπ ) 2 (1) for θ in [0,π/2] and k = 1, 2, 3. Therefore, we can choose any arbitrary f Θ (θ) for θ in [0,π/2] such that ∫ π 2 f Θ (θ)dθ = 1 4 . f Θ (θ) for θ in [π/2, 2π] can be set using (1). A sample pdf that fives a W.S.S. X t is shown in Figure 3. 0 3
Page 1: EE 511 Solutions to Problem Set 5 1
Page 5 and 6: 5. E [|X t+τ − X t | 2 ] = E [ X
Page 7: 13. (a) Y n = X n + X n−1 + X n

EE 511 Solutions to Problem Set 5

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