solution_set5

# solution_set5 - EE 511 Solutions to Problem Set 5 1. (a)...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE 511 Solutions to Problem Set 5 1. (a) The sample functions are shown in Figure 1.-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 1 2 t X t (0)-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1-1 1 t X t (1)-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1-1 1 t X t (2)-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1-1 1 t X t (3) Figure 1: (b) X = 0. X . 5 (0) = 1, X . 5 (1) = − 1, X . 5 (2) = 1, and X . 5 (3) = − 1. X . 25 (0) = 1, X . 25 (1) = 0, X . 25 (2) = − 1, and X . 25 (3) = 0. The marginal CDF’s of X , X . 25 , and X . 5 are shown in Figure 2. (c) Given that X . 5 = − 1, X . 25 = 0 with probability 1. (d) Given that X . 5 = 1, X . 25 = 1 with probability 0 . 5 and X . 25 = − 1 with proba- bility 0 . 5. 2. a) E [ X t ] = 0. E [ X t + τ X t ] = E [ A 2 cos (2 πf c t + Θ) cos (2 πf c ( t + τ ) + Θ)] = A 2 2 [cos 2 πf c τ + E [cos (2 πf c (2 t + τ ) + 2Θ)]] = A 2 2 cos 2 πf c τ b) We can choose any pdf for Θ as long as E [cos (2 πf c (2 t + τ ) + 2Θ)] = 0 and E [cos (2 πf c t + Θ)] = constant for any t , τ . Θ can be defined as follows: Θ =          with prob. 1 4 π 2 with prob. 1 4 π with prob. 1 4 3 π 2 with prob. 1 4 1-2-1.5-1-0.5 0.5 1 1.5 2 2.5 3 0.5 1 x CDf of X F X (x)-2-1.5-1-0.5 0.5 1 1.5 2 2.5 3 0.5 1 x CDF of X 0.25 F X 0.25 (x)-2-1.5-1-0.5 0.5 1 1.5 2 2.5 3 0.5 1 x CDF of X 0.25 F X 0.5 (x) Figure 2: Another possible choice for Θ is: Θ =                              with prob. 1 12 π 4 with prob. 1 6 π 2 with prob. 1 12 3 π 4 with prob. 1 6 π with prob. 1 12 5 π 4 with prob. 1 6 3 π 2 with prob. 1 12 7 π 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 2 π . (ii) The condition for mean to be constant can be obtained as follows: E [cos (2 πf c t + Θ)] = integraltext 2 π cos (2 πf c t + θ ) f Θ ( θ ) dθ = integraltext π cos (2 πf c t + θ ) f Θ ( θ ) dθ + integraltext 2 π π cos (2 πf c t + θ ) f Θ ( θ ) dθ (using θ ′ = θ − π ) = integraltext π cos (2 πf c t + θ ) f Θ ( θ ) dθ + integraltext π cos (2 πf c t + θ ′ + π ) f Θ ( θ ′ + π ) dθ ′ = integraltext π cos (2 πf c t...
View Full Document

## This note was uploaded on 10/26/2011 for the course ECE 171 taught by Professor Tu during the Spring '11 term at Aarhus Universitet.

### Page1 / 7

solution_set5 - EE 511 Solutions to Problem Set 5 1. (a)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online