# hwsol2 - EE 278 Handout#5 Statistical Signal Processing...

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Unformatted text preview: EE 278 Handout #5 Statistical Signal Processing Monday, July 20, 2009 Homework #2 Solutions 1. (15 points) The general formula for calculating the pdf of a differentiable function of a continuous random variable given in the lecture notes is f Y ( y ) = summationdisplay θ k : f Θ ( θ k )= y f Θ ( θ ) | dy/dθ | θ = θ k . We can apply that general formula to this special case. Here f Θ ( θ ) = 1 2 π , − π ≤ θ ≤ π . and dy dθ = cos( ωt + θ ) . For y ∈ [ − 1 , 1], there are two solutions to the equation y = sin( ωt + θ ) with θ ∈ [ − π,π ]. (Actually, there are two points where there is only one solution ( y = ± 1) and one point where there are three ( y = 0). We ignore these points since their probability is zero.) When y is plotted as a function of θ , it is apparent that the slopes at the two solutions do not depend on the phase shift ωt (also f Θ ( θ ) is uniform). Thus without loss of generality, we can take ωt = 0, which gives the two solutions: θ 1 = arcsin y and θ 2 = π − θ 1 . Thus, vextendsingle vextendsingle vextendsingle vextendsingle dy dθ vextendsingle vextendsingle vextendsingle vextendsingle θ 1 = vextendsingle vextendsingle vextendsingle vextendsingle dy dθ vextendsingle vextendsingle vextendsingle vextendsingle θ 2 = cos(arcsin y ) = radicalbig 1 − y 2 . Using the general formula for f Y ( y ), we get f Y ( y ) = 1 2 π parenleftBigg 1 radicalbig 1 − y 2 + 1 radicalbig 1 − y 2 parenrightBigg = braceleftBigg 1 π √ 1 − y 2 | y | ≤ 1 | y | > 1 . Note that f Y ( y ) does not depend on time t , that is, Y ( t ) is time invariant (or station- ary). (More on this later in the course.) 2. (15 points) (a) If k < 0 then p Y ( k ) = 0. If k ≥ 0 then p Y ( k ) = P { Y = k } = P { k ≤ X < k + 1 } = F X ( k + 1) − F X ( k ) = ( 1 − e − λ ( k +1) ) − ( 1 − e − λk ) = e − λk − e − λ ( k +1) = e − λk ( 1 − e − λ ) . In other words, Y , the quantization of exponential random variable X , is Geom( ǫ − λ ), where ǫ − λ is the probability of success in one time unit. 1 (b) Note that P { X < x } = P { X ≤ x } = F X ( x ) . because the probability of any given point is 0. Since ≤ Z = X − Y = X − ⌊ X ⌋ < 1 , the quantization error pdf f Z ( z ) is 0 if z < 0 or z ≥ 1. If 0 ≤ z < 1 then f Z ( z ) = 1 Δ z f Z ( z )Δ z ≈ 1 Δ z P { z ≤ Z < z + Δ z } = 1 Δ z P { z ≤ X − ⌊ X ⌋ < z + Δ z } = 1 Δ z ∞ summationdisplay k =0 P { k + z ≤ X < k + z + Δ z } ≈ 1 Δ z ∞ summationdisplay k =0 f X ( k + z )Δ z = ∞ summationdisplay k =0 f X ( k + z ) = ∞ summationdisplay k =0 λe − λ ( k + z ) = λe − λz ∞ summationdisplay k =0 e − λk = λe − λz 1 − e − λ ....
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hwsol2 - EE 278 Handout#5 Statistical Signal Processing...

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