Engelberg Random Signals and Noise 1st Edition Solutions Manual & Answers Instant Download ISBN: 9780849375545)

All 11ChaptersCovered
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SOLUTIONS

,Solutions Manual w




S UMMARY: In this chapter we present complete solution to the exercis
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es set in the text.
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Chapter 1 w




1. Problem 1. As defined in the problem, A B—
w is composed of the elements in
w w w w w w w w w w w w w w




A that are not in B. Thus, the items to be noted are true. Making use of
w w w w w w w w w w w w w w w w w




the properties of the probability function, we find that:
w w w w w w w w




P(A ∪ B) = P (A) + P (B — A)
w w w w w w w w w w w




and that:w




P(B) = P (B — A) + P (A ∩ B).
w w w w w w w w w w w




Combining the two results, we find that: w w w w w w




P (A ∪ B) = P (A) + P (B) — P (A ∩ B).
w w w w w w w w w w w w w w




2. Problem 2. w




(a) It is clear that fX (α)
w w w w w ≥
0. Thus, we need only check that the i w w w w w w w w




ntegral of the PDF is equal to 1. We find that: w w w w w w w w w w



∫∞
∫ ∞
w
w




fX (α) dα = 0.5 e−|α| dα w w w w w




−∞ −∞
∫0 ∫∞ w w w




= 0.5 α
e dα + e−α dα w w w w



−∞ 0
= 0.5(1 + 1)w w w




= 1. w




Thus fX(α) is indeed a PDF. w w w w w w




(b) Because fX(α) is even, its expected value must be zero. Addition-w w w w w w w w w w w




ally, because α 2fX(α) is an even function of α, we find that:
w w w w w w w w w w w w w




∫ ∞ ∫ ∞ w w



α2f X (α) dα = 2 α2f X (α) dα w w w w
w



−∞ 0


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1

,2 Random Signals and Noise: A Mathematical Introduction
w w w w w w




∫ ∞ w




= α2e−α dα w


0
∫ w
∞
by parts
=
w

(—α2e −α|∞
0 +2 w
w w
w w αe −α dα
∫∞
0 w
w w

by partsw
−α ∞ w −α
= 2(—αe |0 ) + 2
w
w w e dα
0
= 2.
Thus, E(X2 ) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
w w w w w w w w w w w w w w w w



√ X
2.

3. Problem 3. w




The expected value of the random variable
w
∫ ∞ is: w w w w w
w
w
w




E(X) = √ αe−(α− dα
1 µ)2/(2σ 2) w w




2πσ
∫ ∞ −∞ w


u=(α−µ)/σ 1 −u 2/2 w w w w




√
w


= (σu + µ)e dα. w w


2π −∞
2
Clearly the piece of the integral associated with ue−u /2 is zero. The
w w w w w w w w w w w w




remaining integral is just µ times the integral of the PDF of the standard no
w w w w w w w w w w w w w w




rmal RV—and must be equal to µ as advertised.
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Now let us consider the variance of the RV—
w w w — w w w w w




let us consider E((X µ)2). We find that:
w w ∫∞ w w w w w w
w




E((X — µ)2) = √ (α — µ)2e−(α−
w w dα w w


1 µ)2/(2σ 2) w w




2πσ ∫ −∞
∞ w



u=(α−µ)/σ 2 1 2 −u 2/2 ww w w
w w



= σ√ ue dα. w w


2π −∞

As this is just σ2 times the variance of a standard normal RV, we find th
w w w w w w w w w w w w w w w




at the variance here is σ2.
w w w w w




4. Problem 4. w




(a) Clearly (β —α)2 ≥ w




0. Expanding this and rearranging it a bit we find that: w w w w w w w w w w




β2 ≥ 2αβ — α2. w w w w




(b) Because β2 ≥ 2αβ — w




α2 and e−a is a decreasing function of a, the inequali w w w w w w w w w w




ty must hold. w w




(c)
∫ w
∞ w
2
∫∞ w
w
2

e − β /2 w
dβ ≤ w w e−(2αβ− α )/2
w w
dβ
α α

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, Solutions Manual w 3
The PDF Function
w w






 0


 1/2
 2





−2 2


 1/2
−2

 0





FIGURE 1.1 w




The PDF of Problem 6.
w w w w




∫ w
∞ w

2
=eα w w
w /2
e −2αβ/2 dβ
α
∞
2 e−αβ w
w


= eα w /2 w



—α α
w


w w


2
2 e−α
= eα /2 w
w




α
2
e−α
=
α
The final step is to plug this into the formula given at the beginning of
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the problem statement.
w w




5. Problem 5. w




If two random variables are independent, then their joint PDF must be th
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e product of their marginal PDFs. That is, fXY (α, β) = fX (α)fY (β). The r
w w w w w w w w w w w w w w w w




egions in which the joint PDF are non-
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zero must be the intersection of regions in which both marginal PDFs are
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non-
zero. As these regions are strips in the α, β plains, their intersections ar
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e rectangles in that plain. (Note that for our purposes an infinite region all o
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f whose borders are right angles to one another is also considered a recta
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ngle.)
6. Problem 6. w




Consider the PDF given in Figure 1.1. It is the union of two rectangu-
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lar regions. Thus, it is at least possible that the two random variables ar
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e independent. In order for the random variables to actually be in-
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dependent it is necessary that f XY (α, β) = fX(α)fY (β) at all points.
w w w w w w w w w w w w w w w



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