signal detection and estimation (mourad barkat, solution manual)

Chapter 1

Probability Concepts 1.1 The given sets are:

A = {1,2,3,4} B = {0,1,2,3,4,5,6,7,8}

C = {x | x real and 1≤ x <3} D = {2,4,7} E = {4,7,8,9,10}.

We observe that:

A is finite and countable. D is finite and countable. B is finite and countable. E is finite and countable. C is infinite and uncountable.

1.2 By inspection,

(a) BAI = A = {1,2,3,4}.

(b) EDBA UUU = {1,2,3,4,5,6,7,8,9,10}.

(c) DEB IU )( = D = {2,4,7}.

(d) EB − = {1,2,3,5,6}.

(e) EDBA III ={4}.

1.3 The universal set is U = {0,1,2,3,4,5,6,7,8,9,10,11,12}. The subsets are A = {0,1,4,6,7,9}, B = {2,4,6,8,10,12} and C = {1,3,5,7,9,11}. By inspection,

(a) BAI = {4,6}. (b) ( BAU ) CI = CAI = {1,7,9}.

1

Signal Detection and Estimation 2

(c) CBU = {0}. (d) AB − = {0,1,7,9}.

(e) )()( CABA UIU = )( CBA IU = A = {0,1,4,6,7,9}.

(f ) CAI = {0,4,6}. (g) −B =C ∅.

(h) CBI = B = {2,4,6,8,10,12}.

1.4 Applying the definitions, we have

U A B

(a) BA− (b) CBA IU )(

UA B

C

U A B

(e) A∩B

U A B C D

A

U

Ad )(DCBAc III )(

Probability Concepts 3

1.5 This is easily verified by Venn diagrams. Hence, we have

BA ⊂ and CB ⊂ , then CA ⊂

1.6 By inspection, B and C are mutually exclusive. 1.7 Let R, W and B denote red ball drawn, white ball drawn and blue ball drawn, respectively

(a) 5.021

731010

balls ofnumber Totalballs red ofNumber )( ==

++==RP .

(b) 15.0203)( ==WP . (c) 35.0

207)( ==BP .

(d) 5.021)(1)( ==−= RPRP . (e) 65.0

2013

20310)( ==

+=WRP U .

1.8 Let B1 ≡ first ball drawn is blue. W2 ≡ 2nd ball drawn is white. R3 ≡ 3rd ball drawn is red.

(a) The ball is replaced before the next draw ⇒ the events are independent and hence,

)|()|()()( 213121321 WBRPBWPBPRWBP III = )()()( 321 RPWPBP=

02625.08000210

2010

203

207

===

A B

C

U

B C A B

10R , 3W,

7B


(b) Since the ball is not replaced, the sample size changes and thus, the events are dependent. Hence,

)|()|()()( 213121321 WBRPBWPBPRWBP III =

0307.01810

193

207

==

1.9

Let R1 and R2 denote draw a red ball from box B1 and B2 respectively, and let

W1 and W2 denote also draw a white ball from B1 and B2.

(a) )()()|()()( 2112121 RPRPRRPRPRRP ==I since the events are independent. Hence,

111.091

92

2010)( 21 ===RRP I .

(b) Similarly, 1.096

203)()()( 2121 === WPWPWWP I

(c) Since we can have a different color from each box separately, then

25.0207

96

91

203)()()( 1221 =+=+= BWPBWPBWP III .

1.10 Let B1 and B2 denote Box 1 and 2 respectively. Let B denote drawing a black ball and W a white ball. Then ,

10R , 3W

7B

2R , 6W

1B

B1 B2

4W , 2B

3W , 5B

B1 B2


Let B2 be the larger box, then P(B2) = 2P(B1). Since 1)()( 12 =+ BPBP , we

obtain 32)(and

31)( 21 == BPBP .

(a) P(1B | B2) = 625.085= .

(b) P(1B | B1) = 3333.062= .

(c) This is the total probability of drawing a black ball. Hence

.5278.031

62

32

85

)()|1()()|1()1( 1122

=+=

+=

BPBBPBPBBPBP

(d) Similarly, the probability of drawing a white ball is

.4722.031

64

32

83

)()|1()()|1()1( 1122

=+=

+=

BPBWPBPBWPWP

1.11 In four tosses:__ __ __ __, we have three 1s and one is not 1. For example

1111 . Hence, the probability is 65

61

65

61

61

61 3

= but we have

34

ways of

obtaining this. Therefore, the probability of obtaining 3 ones in 4 tosses is

01543.065

61

!1!3!4 3

=

.

1.12 Let R, W and G represent drawing a red ball, a white ball, and a green ball respectively. Note that the probability of selecting Urn A is P(Urn A) = 0.6, Urn B is P(Urn B) = 0.2 and Urn C is P(Urn C) = 0.2 since

P(Urn A)+P(Urn B)+P(Urn C) =1.

(a) (P 1W | Urn B) = 3.010030

)(Urn)Urn1(

==BP

BWP

I .

(b) (P 1G | Urn B) = 4.010040

= .

(c) P(Urn C | R) =)(

)(UrnRP

RCP I . Also,


P(R | Urn C) = )(

)(Urn)Urn|()|(Urn)(Urn

(UrnRP

CPCRPRCPCP

RCP=⇒

) I .

We need to determine the total probability of drawing a red ball, which is

( ) ( ) ( ) 32.02.0100402.0

100306.0

10030

)(Urn)Urn|()(Urn)Urn|()(Urn)Urn|()(

=++=

++=

CPCRPBPBRPAPARPRP

Thus, 25.032.0

)2.0()4.0()|(Urn ==RCP .

1.13 In drawing k balls, the probability that the sample drawn does not contain a particular ball in the event Ei, i = 0, 1,2, … , 9, is

M

k

ji

k

i

EEP

EP

=

=

108)(

109)(

(a) P(A) = P(neither ball 0 nor ball1) = P(E0E1) = k

k

108 .

(b) P(B) = P( ball 1 does not appear but ball 2 does)

=k

kk

k

k

k

kEEPEP

1089

108

109)()( 211

−=−=− .

(c) P(AB) = )( 210 EEEP = =− )()( 21010 EEEPEEPk

kk

k

k

k

k

1078

107

108 −

=− .

(d) k

kkkABPBPAPBAP

10789)()()()( +−

=−+=U .

1.14 We have

<

≥−δ+=

−

0,0

0,)3(21

21

)(x

xxexf

x

X


(a)

∫∫∫∫∞∞

−∞

−∞

=+=−+=−+=0000

121

21)3(δ

21

21)]3(

21

21[)( dxx dxedxxδ edxxf xx

X .

Hence, )(xf X is a density function.

(a) P(X = 1) = 0 (the probability at a point of a continuous function is zero).

5.021)3( ===XP .

( ) 6839.0121

21

21)()1( 1

11

=+=+==≥ −−∞∞

∫∫ edxedxxfXP xX .

1.15

(a) The cumulative distribution function of X for all the range of x is,

∫∫−∞−

−≤≤−+===xx

XX x xduduufxF3

13for83

81

81)()( ,

and ∫−

≤≤−+=+x

xxdu1

11for 21

41

41

41 ,

and 31for85

881

43

1

≤≤+=+ ∫ x xdux

,

(1/2)

1/2

. . x 0 1 2 3

1/8

fX(x)

. . . . x

-3 -2 -1 0 1 2 3

1/4

fx(x)


and 3for1)( ≥= xxFX .

Thus,

(b) Calculating the area from the graph, we obtain 21

412)1( ==<XP .

1.16 The density function is as shown

(a) 22for21

41

41)()(

2

<≤−+===≤ ∫−

xxduxFxXPx

X

(b) 21

41)1(

1

1==≤ ∫

−

dxXP

(c) 34

412][,0][

2

0

222 =

==σ= ∫ dxxXEXE x .

(d) ωω

=ω

−==ω

ω−ωω 2sin

21

4][)(Φ

22

jeeeE

jjXj

x .

3,1

31,85

81

11,21

41

13,83

81

3,0

≥

<≤+

<≤−+

−<≤−+

−<

x

x x

x x

x x

x

FX (x) =

fX(x)

x -2 -1 0 2 1

1/4


1.17 The density function is shown below

(a) 75.043)2(

23

21 2/3

1

1

2/1

==−+=

<< ∫∫ dxxxdxXP .

(b) 1)2(][2

1

1

0

2 =−+= ∫∫ xdxxdxxXE as can be seen from the graph.

(c) The MGF of X is

)12(1)2(][)( 22

2

1

1

0

+−=−+== ∫∫ tttxtxtXx ete

tdxexxdxeeEtM .

(d) 3

2

4

222

02)4()1(2)12(2)(2)(

ttette

teteteet

dttdM tttttt

tx −−+−

=+−−−

==

Using L'hopital's rule, we obtain 1)0( ==′ tM x .

1.18 (a) 42

)(32][

1

0

2 β+

α=β+α== ∫ dxxxXE and

13

)()(1

0

2 =β

+α=β+α=∫ ∫+∞

∞−

dx xdxxf x .

Solving the 2 equations in 2 unknowns, we obtain

=β=α

23/1

(b) 511.045232

31][

1

0

222 ==

+= ∫ dxxxXE .

Then, the variance of X is ( ) 667.0453][][ 222 ==−=σ XEXEx .

fX (x)

1

x 0 1/2 1 3/2 2


1.19 (a) ∑=ji

jiji yxPyxXYE,

),(][

0]0[61)]1)(1()1)(1()1)(1()1)(1[(

121

=+−++−−+−++−−=

( )31

1241and

31

620,

31

1241

where,)(][

========−=

== ∑

) P(X) P(XXP

xXPxXEi

ii

Hence, the mean of X is 0311

310

311][ =

+

+

−=XE .

Similarly, 0311

310

311][ =

+

+

−=YE . Therefore, ][][][ YEXEXY E = .

(b) We observe that 121)1,1( ==−= YXP

91)1,1( =−=−=≠ YXP , thus X

and Y are not independent.

1.20 (a) ∫ ∫ ∫ ∫+∞

∞−

+∞

∞−

=⇒=+=2

0

2

0 811)(),( kdxdyyxkdxdyyxf XY .

(b) The marginal density functions of X and Y are:

∫ ≤≤+=+=2

020for

41

4)(

81)( xxdyyxxf X .

∫ ≤≤+=+=2

0

20for 41

4)(

81)( yydxyxyfY .

(c) P(X < 1 | Y < 1)=31

8/38/1

)41

41(

)(81

1

0

1

0

1

0 ==

+

+

∫

∫ ∫

dyy

dxdyyx.

(d) [ ] [ ]∫ ==+=2

0 67)1(

4YEdxxxXE .


To determine xyρ , we solve for

[ ] ∫ ∫ =+=2

0

2

0 34)(

8dxdyyxxyXYE .

611Thus,.

35][][ 22 =σ=σ== yxYEXE and the correlation coefficient is

[ ] [ ] [ ] 0909.011

1−=

−=

σσ−

=ρyx

YEXEXYE .

(e) We observe from (d) that X and Y are correlated and thus, they are not independent.

1.21 (a) ∫ ∫ ∫ ∫+∞

∞−

+∞

∞−

=⇒==4

0

5

1 9611),( kdydxkxydxdyyxf XY .

(b) ∫ ∫ ===≤≥2

0

5

3

09375.0323

96)2,3( dxdyxyYXP .

03906.0128

596

)32,21(3

2

2

1

===<<<< ∫ ∫ dxdyxyYXP .

(c)

∫=

<<<<<<

=<<<<3

2)(

128/5)32(

)32,21()32|21(

dyyfYP

YXPYXP

Y

where,

∫ <<==5

140

896)( y, ydxxyyfY . Therefore,

125081

16/5128/5)32|21( .YXP ===<<<<

(d) [ ] ∫ ∫ =====5

1

5

1444.3

931

12)(| dxxxdxxxfyYXE X .


1.22 (a) We first find the constant k. Hence, ∫ ∫ =⇒=2

1

3

1 611 kdydxkxy

(b) The marginal densities of X and Y are

∫ <<==2

1

31for46

1)( x xxydyxfX

and ∫ <<==3

121for

32

61)( yydydxxyyfY .

Since ⇒== ),(61)()( yxfxyyfxf xyYX X and Y are independent.

1.23 We first determine the marginal densities functions of X and Y to obtain

∫ ==1

033

816)(

xdy

xyxf X for x > 2.

and ∫∞

==2

32

16)( ydx

xyyfY for 0 < y < 1.

Then, the mean of X is [ ] ∫∞

==2

4)( dxxxfXE X ,

and the mean of Y is [ ] ∫ ==1

0

2

32

2 dyyYE .

4792.04823

61)3(

2

1

3

1

===<+ ∫ ∫− y

xydxdyYXP

yx −= 3

1

y

x

2

1 2

3

3


1.24 We first find the constant of k of )(yfY to be ∫∞

− =⇒=0

3 91 kdykye y .

(a) 321

0

1

0

32 149181)1(1)1( −−−

−− −=−=≤+−=>+ ∫ ∫ eedxdyyeYXPYXPy

yx .

(b) ∫ ∫∞

−− −==≥<<1

2

1

75 44),()1,21( eedxdyyxfYXP XY .

(c) ∫ −−− −==<<2

1

4222)21( eedxeXP x .

(d) ∫∞

−− ==≥1

33 49)1( edyyeYP y .

(e) 523

75

444

)1()1,21(

)1|21( −−−

−−

−=−

=≥

≥<<=≥<< ee

eee

YPYXPYXP .

1.25 (a) Using )(xf X , [ ] [ ] ∫ ∫+∞

∞−

∞− ====

0

2 122)()()( dxexdxxfxgXgEYE xX .

(b) We use the transformation of random variables (the fundamental theorem)

to find the density function of Y to be

yy

XY eeyfyf −−==

= 22

221

221)( .

Then, the mean of Y is [ ] ∫∞

− ==0

1dyyeYE y .

x + y = 1

y 1

x 0 1


Both results (a) and (b) agree.

1.26 (a) To find the constant k, we solve

∫ ∫ ∫ ∫+∞

∞−

+∞

∞−

=⇒==3

0

3

8181),(

yXY kdydxkxydxdyyxf .

(b) The marginal density function of X is

∫∞

≤≤==0

3 30for814

818)( xxxydyxfX .

(c) The marginal density function of Y is

∫ ≤≤−==3

3 30for )9(814

818)(

yY yyyxydxyf .

(d)

≤≤===

otherwise, 0

0, 2

814818

)(),(

)|( 2

3|

xyxy

x

xy

xfyxf

xyfX

XYXY

and

≤≤−==

otherwise, 0

3, 9

2

)(),(

)|( 2|

xyyx

yfyxf

YXfY

XYYX

1.27 The density function of YXZ += is the convolution of X and Y given by

∫+∞

∞−

−=∗= dxxzfxfYfxfzf YXYXZ )()()()()(

0 z - 4 zz-4 0 z


≥−

=

≤≤−

=

=−

−

−

−−

∫

∫

otherwize,0

4,4

)1(41

40,4

141

)(4

4

0

zeedxe

zedxe

zfz

xz

z

zzx

Z

1.28 The density function of YXZ += is ∫+∞

∞

−=_

)()()( dyyzfyfzf XYZ .

Graphically, we have

Hence, we have

fY (y)

0 0.3 0.5 0.2 0 0 0

Z fZ (z)

z = 0 0.4 0.2 0.4 0 0 0 0 0 0 0 0

z = 1 0.4 0.2 0.4 0 0 0 0 0 0 0

z = 2 0.4 0.2 0.4 0 0 0 0 0 0.12

z = 3 0 0.4 0.2 0.4 0 0 0 0 0.26

z = 4 0 0.4 0.2 0.4 0 0 0 0.30

z = 5 0 0 0.4 0.2 0.4 0 0 0.24

z = 6 0 0 0 0.4 0.2 0.4 0 0.08

z = 7 0 0 0 0 0.4 0.2 0.4 0

x 0 1 2 3

0.4 0.4 0.2

fX(x)

y 0 1 2 3

0.5 0.3 0.2

fY(y)

x -3 -2 -1 0

0.4 0.4 0.2

fX(-x)


The plot of )(zf Z is

Note that ∑ =++++=−

iiZ zzf 0.108.024.03.026.012.0)( as expected.

1.29 (a) ∫ ∫∞

+− ≥=≤=⇒=0

/

0

)( 0for)()(yz

yxZ zdxdyezXYPzFXYZ .

∫ ∫ ∫∫∞ ∞ ∞ −−

−−−− −=−=

⇒0 0 0

//

0

1)1( dyedyeedyedxe yzy

yyzyyz

x .

Therefore,

≥=

−== ∫∫

∞ −−∞ −−

otherwise, 0

0, 1)()(

0

)(

0

zdyeydye

dzdzF

dzdzf

yzy

yzy

ZZ .

(b) ∫+∞

∞−

−=⇒+= dyyzfyfzfYXZ XYZ )()()(

∫ −−−=z

yzy dyee0

)(

≥

=−

otherwise, 00, zze z

1.30 The density function of XYZ = is

∫ ∫+∞

∞−

<<−===1

10 forln1),(1)(z

XYZ zzdyy

dyyyzf

yzf .

z 0 1 2 3 4 5 6 7

0.3 0.26 0.24 0.12 0.08 0 0

fZ(z)

0 z


1.31 (a) The marginal density function of X is

∫β

α−α− ≥α=βα

=0

0for )( xedyexf xxX .

(b) The marginal density function of Y is

∫∞

α− ≤≤β

=βα

=0

β0 for 1)( ydxeyf xY .

(c) Since ⇒= ),()()( yxfyfxf XYYX X and Y are statistically independent.

(d) ∫+∞

∞−

−=∗=⇒+= dyyzfyfyfxfzfYXZ XYYXZ )()()()()( .

y β

β1

xe α−α

x

α

)(xf X( )yfY

For β<≤ y0

α

z-β 0 z

( )∫ α−α− −β

=βα

=z

zxZ edxezf

0

11)(


Then,

1.32 ( ) )()()(zxYPyzXPz

YXPzZPzF

YXZ Z ≥=≤=

≤=≤=⇒=

0,11αβ1

0 /

βα >

+==−∞ ∞

−−∫ ∫ z zα

βdxdyeezx

yx

Hence, the density function is

<

>

αβ

+αβ

==

00

01

)()(2

z ,

z ,

zzFdzdzf ZZ

1.33 (a) Solving the integral ∫ ∫ =⇒=2

1

3

12121 6

11 kdxdxxkx .

(b) The Jacobian of the transformation is

β

( )αβ−−β

e11

)(zfz

z 0

0 z-β z

α

For ∞<≤β y

( )[ ]∫β−

α−β−α−α− −β

=βα

=z

z

zzxZ eedxezf 1)(


.22

01),( 21

2121

2

2

1

2

2

1

1

1

21 xxxx x

xy

xy

xy

xy

xxJ ==

∂∂

∂∂

∂∂

∂∂

=

Hence,

∈

==otherwise,0

, 121

),(

),(),( 21

21

2121

21

21

D,xxxxJ

xxfYYf XX

YY

where D is the domain of definition.

Side 1 : 22211 1 xyxy =⇒== , then .41

4211

222

22 ≤≤⇒

=⇒==⇒=

yyxyx

Side 2 : 22211 33 xyxy =⇒== , then .123

12231

222

22 ≤≤⇒

=⇒==⇒=

yyxyx

Side 3 : 1122 442 yxyx ==⇒= , then

=⇒==⇒=

.12341

21

21

yxyx

Side 4 : 1122 1 yxyx ==⇒= . Therefore, D is as shown below

x2

3 11 xy = 2 2

212 xxy = 1 2 1 4 x1

0 1 2 3


1.34 (a) The marginal density functions of X1 and X2 are

∫+∞

∞−

α−+α−

≤>α

=α=0,00,)(

1

12

)(21

121

1 x xedxexf

xxx

X

and

∫+∞

∞−

α−+α−

≤>α

=α=0,00,)(

2

21

)(22

221

2 x xedxexf

xxx

X

Since ⇒= )()(),( 2121 2121xfxfxxf XXXX X1 and X2 are independent.

(b) The joint density function of ),( 21 YY is given by

( )

∈

=

. ,

D,x x, xxJ

xxfyyff

XX

YY

otherwise0,

),(),( 21

21

21

21

21

21

The Jacobian of the transformation is given by

D

+ + + y1 1 2 3

12 + + + + + + + 4 +

3 1

2y


.11

11),(

222

1

2

1

22

2

1

2

2

1

1

1

21 xxx

xx

x

xy

xy

xy

xy

xxJ −−=−

=

∂∂

∂∂

∂∂

∂∂

=

Hence,

222

1

)(2

211

),(21

21

xxxeyyf

xx

YY

−−

α=

+α−

, but 211 xxy += and 2212

12 xyx

xx

y =⇒= .

Thus, .),(21

212

211

21 xxx

eyyf yYY +

α= α− Also, 221112 xyyxyx −=−=

)1( 212 yyx +=⇒ .

Making the respective substitutions, we obtain

22

12

1

22

21

221

)1()1(

),( 1121 y

ye

yy

y

eyyf yyYY

+α=

+α= α−α− for 01 >y and 02 >y .

Chapter 2

Distributions 2.1 Let A = { seven appears} = {(1 ,6) , (2 , 5) , (3 , 4) , (4 , 3) , (5 , 2) , (6 , 1)}.

Then, ( ) .65

611and

61

366)( =−=== APAP

(a) This is Bernoulli trials with k = 2 successes and n = 6 trials. Hence,

2009065

61

!2!4!6

65

61

26

trials)6in successes2(4242

. nkP =

=

===

(b) 3349.065

65

61

06

trials)6in successes (no660

=

=

== nP

2.2 The number of ways of obtaining 4 white balls out of 10 is

410

. The other

number of different ways of obtaining 3 other balls (not white) is

39

. Hence, the

probability of obtaining the fourth white ball in the seventh trial is

3501.0

719

39

410

=

or, using the formula of the hypergeometric distribution without replacement, we have 19=N balls, 10=r and 4=k in 7=n trials. Hence,

22

Distributions 23

.3501.0

719

39

410

)4( =

==XP

2.3 The probability of success is 9.0=p while the probability of failure is 1.01 =−= pq .

(a) )9()8()7()6(zone)in land6least(at =+=+=+== XPXPXPXP P

)10( =+ XP

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) .. ...

......

9980901010

1090910

1090810

1090710

1090610

109

283746

=

+

+

+

+

=

(b) ( ) ( ) ( ) .1.01.09.0010

zone)in lands none(P 10100 =

=

(c) Probability that at least 70 0/0 land in zone is

======+= )10()9()8()7( XPXPXPXP

( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.987. 9.01010

1.09.0910

1.09.0810

1.09.0710 1092837 =

+

+

+

Hence, the program is successful.

2.4 Substitution for 0=k in the Poisson density function, we have λ−=== eXP 2.0)0( . Hence, λ = 1.609.

..

e.e.

XPXPXPXPXP

.

2190

]!2

)6091(60912.0[1

)]2()1()0([1)2(1)2(

60912

609.1

=

++−=

=+=+=−=≤−=>

−−

2.5 Let X represent the Poisson distribution of the incoming calls with hour12=λ .


(a) The probability of more than 15 calls per a given hour is

( )∑=

− =−=≤−=>15

0

12 1556.0!

121)15(1)15(k

k

keXPXP

(b) No calls in 15 minute ( 4/1 hour) break )0( =⇒ XP in 15 minutes. Hence,

0498.0)0( 3 === −eXP

2.6 X is Poisson distributed and )1()3/2()2( === XPXP . That is,

!132

!2

2 λ=

λ λ−λ− ee . Solving for λ we obtain 340

34

=λ⇒=

−λλ since 0=λ is

not a solution. Therefore,

2636.0)0( 3/4 === −eXP and ( ) 1041.0!33/4)3(

33/4 === −eXP

2.7 The lack of memory property is

( )( )

( )( )

( )( )2

1

21

1

121121

2

1

21

)|(

xXPee

e

xXPxxXP

xXPxXxxXP

xXxxXP

αxαx

xxα≥===

>+≥

=>

>+≥=>+≥

−−

+−

I

2.8 (a) In this case the parameter 1211

=λ

=β and the density function is

≥=

−

otherwise , 0

0 ,121

)(12

xexf

x

X

Hence, 232501211)15(1)15(

15

0

12 .eXPXPk

k

∑=

−=−=≤−=>

(b) 0833.0121)0( ===XP .

Distributions 25

2.9 X is the standard normal X ~ N (0. 1)

(a) )]1P(2[1)1(2)1()1()1( ≤−=>=>+−<=> XXPXPXPXP 3174.0)8413.02(1)]1(2[1 =−=−= I

(b) 15870)1(1)1(1)1( .IXPXP =−=≤−=> 2.10 X ~ N (0,1). Then, 00130331313 .)Q()I()P(X)P(X ==−=≤−=> 2.11 200/0 of 200 = 40 times. A success is when

X = 7 ≡ {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}.

Hence, we have success with .651and

61

366

=−=== pqp

(a) )40( time) theof20leastat (success 0

0 ≥= XPP

∑=

−

=

200

40

200

65

61200

k

kk

k= 0.1223.

(b) Using the central limit theorem to obtain the approximation to the normal distribution, we have

..).I(npq

X-np-PXPXP 142300711

361000

620039

1)39(1)40( =−=

−

≤≈≤−=≥

+1-1 x

fX(x)


2.12 10021 XXXXS k +++++= LL , Xk is Poisson distributed with λ=3.2

(a) ∑∑==

=−=≤−==≥4

1

23100

5

2.3 26020!)23(1)4(1

!)2.3()5(

k

k.

k

k..

k.eSP

keSP

(b) Using the central limit theorem, S becomes normal with mean 3.2 and variance 3.2. That is S ~ N(0.032× 100, 0.032× 100) . Hence,

3264.0)45.0(450123234

23231)4(1)5( ==−≅

−≤−≈≤−=≥ Q).I(

..

..S-PSPSP

2.13 X ~ N(1, 2)

(a) From the tables,

−≤

−−≈≤−=>

212

211212 XP) P(X)P(X

2399070701 .).I( =−=

(b)

−≤

−≤

−=≤≤

2222

22

2261)2261( .X.P .X.P

..).I().I(.XP.X P 16602801402802

21402

2=−−=

−≤

−−

≤

−=

2.14

≤≤

=otherwise,0

61,51

)(

xxf X

Using the fundamental theorem, we have

22

and11 xdydxdx

xdy

xy =−=⇒= . Hence,

+ + + + + + x 1 2 3 4 5 6

fX(x)

51

Distributions 27

.161for

51

51)()()( 2

2 ≤≤==== y y

xdydxxf

dxdy

xfyf XX

Y

2.15 (a) We found in Example 1.19 for 2XY = that

≤

>−−=

0,00,)()()(

y y yFyFyF XX

Y

and

[ ]

≤

>−−=

0,0

0,)()(2

1)(

y

y yfyfyyf XX

Y

For X uniformly distributed between 0 and 1, we obtain

≤

<<

≤

=

1,110,

0,0

)( y

y y

y

yFY

and

≤<=

otherwise,0

10,2

1)(

y yyfY

1

FY(y)

y 0 1

y 1

fY(y)

1/2

≈

+ + + + + + y

61

62

63

64

65 1

51

fY(y)

536


(b) For eZ X ,=

)()( zZPz F Z ≤=

>=≤=≤

≤=

0,)(ln)ln()(

0,0

z zFzXPzeP

z

XX

Hence,

≤>

=0,00,ln

z zz) ( F

(z) F XZ . The density function is

≤

>==

0,0

0,)(ln1)()(

z

z zfz

dzzdF

zf XZZ

Substituting for z, we obtain

≥<≤

<=

ezezz

zzFZ

,11,ln

1,0)( and

≤≤

=otherwise,0

1,1)(

ez zzf Z

2.16 X and Y are standard normal, that is 2

2

21)(

x

X exf−

π= and

2

2

21)(

y

Y eyf−

π= .

(a)

<−

>==

0,

0

YYX

Y,YX

YXZ

The distribution function of Z is

<≤−+

>≤=≤= 00)()( Yz

YXPYz

YXPzZPzFZ

)yyzXP()yyzXP( 00 <−≤+>≤= with the regions shown below.

Distributions 29

( ) ( )∫∫

∫∫

∫ ∫∫ ∫

∞−

−+∞

−−

∞

∞∞−

−

∞−

∞

∞−

−=

−−=

+=

022

0

22

0

0

0

0

2222

21

21

),(

),(),()(

dyeyeπ

dyeyeπ

yz,y)dy(yfdyyyzyf

dxdyyxfdxdyyxfzF

yyzyyz

-XYXY

yz

XY

y

XYZ

Using Leibniz’s rule, we obtain

z.z

zf Z allfor )1(

1)(2 +π

=

(b) For YX

W = , we have .YX

YXZ

YX Z ==⇒= Thus,

<−>

===00

Z , Z, ZZ

ZYX

W .

Using the fundamental theorem of transformation of random variables, we have

x < -yzy < 0

y

x

x=-yz y

x

x=yz

x<yz y>0

w

w

z z1 z2


)(')(

)(')(

)(2

2

1

1

zgzf

zgzf

wf ZZW += where wz −=1 and wz =2 .

Also, 1)(0101

)(' =⇒

<−>+

= zg'z ,z ,

zg .

Substituting, we obtain

( ) ]1[

1]1[

1)()()(2221+

++−

=+=wπwπ

zfzfwf ZZW

Therefore, .w wπ

wfW ∞<≤+

= 0for)1(

2)(2

2.17 The joint density function of ),( 21 XX is

( )2

221

212

2212

1),( σ

+−

πσ=

xx

XX exxf with 2

12

22

211 and

XX YXXY =+= .

Solving 22121

22

21 and yxxyxx ==+ , we obtain

( )22

2112

2

12

21

22

22

21

22

22

22

1and

11

y

yy xy

yxyyxyxyx+

±=+

±=⇒=+⇒=+

By definition, 01 ≥y since 221 yxx = and hence, we have 2 solutions:

22

12

22

121

22

12

22

121

11

11

y

yx

y

yyx

y

yx

y

yyx

+−=

+−=

+=

+=

The Jacobian of the transformation is

Distributions 31

( ) ( )

1

22

11

22

2/122

21

2/122

21

22

21

22

1

2

22

21

2

22

21

1

21

111

111

),(

yy

yyy

xxxxxx

xx

x

xx

x

xx

x

xxJ

+−=−−=

+−

+−=

−

++=

Therefore,

+

−

+

−+

+++=

22

1

22

21

22

1

22

2122

121

1,

11,

11),(

21y

y

y

yyf

y

y

y

yyf

yy

yyf XYXYYY

Note that ),(),( 2121 2121xxfxxf XXXX −−= . Hence,

)()()(2

11

2),( 211

222

2

121 21

221

21yfyfyue

yy

yyf YYy

YY =πσ+

= σ−

where, )()( 12

1122

11

yuekyyf yY

σ−= . We determine the constant k to be

∫∞

σ−

σ=⇒=

021

21

1122

1 kdyeyk y . Thus, the density function of Y1 and Y2 are

respectively

)(2

1)( 12

21

122

11

yuey

yf yY

σ−

πσσ=

and

22

21

11)(2 y

yfY+π

=


2.18 X is a standard normal 22

21)( x

X exf −

π=⇒ .

Y is a chi-square distributed random variable with n degrees of freedom

( )0for

2/Γ21)( 2

12

2/>=⇒

−− yey

nyf

yn

nY

Let nY

XT/

= , then the cumulative distribution of T is

( )( ) ∫ ∫

∞

∞−

+−−

π=≤=≤=

0

/2

)(1

22/

2

22/Γ21/)()(

nyt xyn

nT dxdyeyn

nytXPtTPtF

since the region of integration is the xy-plane with nytx /≤ . Note that the joint density function of X and Y is just the product of the individual density functions since we assume X and Y independent. Making the change of variables

nyux /= , then dunydx /= and )/(22 nyux = . Substituting in the integral, we obtain

( )

( ) ∫ ∫

∫ ∫

−∞=

∞

=

+−−

∞

∞−

−−−

π=

π=

t

u y

nuyn

n

tn

yuyn

nT

dudyeyn

dudyeny

eyn

tF

0

122

1

2/

0

21

22/

2

2

22/Γ21

22/Γ21)(

Let dz

nu

dy z

nu

yzn

uy22

2

1

2and1

212

+

=

+

=⇒=

+ . The integral

becomes

Distributions 33

( ) ∫ ∫−∞=

∞

=+

−−+

+

π=

t

u zn

zn

n

n

T dudz

nu

ezn

tF0 2

12

21

2/

21

122/Γ2

2)(

( )

du

nu

nn

nt

un∫

−∞=+

+

π

+

=2

12

1

12/Γ

21Γ

since .nmnmmdzez zm

21with

21Γ)1(Γ!

0

−=

+

=+==∫∞

−

Taking the derivative of FT (t) using Leibniz’s rule, we obtain the required density function given by (2.171).

2.19 With α = 0, the Cauchy density function is given by .1)(22 x

xf X+βπ

β=

The moment generating function of X is then

∫∫∫∞

∞−

∞

∞−

∞

∞−

ωω

+β

ωπβ

++β

ωπβ

=+βπ

β==ω dx

xxjdx

xxdx

xeeE

xjXj

x 222222sincos][)(Φ

since xjxe xj ω+ω=ω sincos . Also, 0)(lim =∫−

∞→

p

ppdxxf when f(x) is an odd

function of x. Then,

∫∞

+β

ωπβ

=ω0

22cos2)(Φ dx

xx

x

since 22

cosxx

+β

ω is even. Using the tables of integrals, we obtain

. 0βand0ω )(Φ >>=ω βω− ,ex


2.20 (a) The mean value of Weibull distribution is given by

.][0

dxeabxXEbaxb∫

∞−=

Let b

bbb

au

bududx

xdxaxbdxabxduaxu

/11 )(

=⇒==⇒= − since

b

aux

/1

= .

Hence,

+==

= −

∞−−

∞− ∫∫ b

Γadueuaduau

buuebXE bubb

bu 111][ /1

0

/1/1

0

/1

.

(b) The variance is ( )222 ][][ XEXE −=σ . We need to determine the mean

square value, which is .][0

12 dxeabxXEbaxb∫

∞−+= Following the same approach as

in (a), we obtain

+==

−∞−−

∫ badu euaXE bubb 21Γ][

2

0

222 . Hence,

+−

+=σ

−22

2 11Γ21Γbb

a b

Chapter 3

Random Processes 3.1 (a) The mean and autocorrelation functions of )(tX are

∫π

π−

ωπ

=θπ

θ+ω=8/

8/00 cos244)cos()]([ tAdtAtXE

)22cos(2cos2

)]222[cos(2

cos2

)]cos()cos([),(

00

2

0

00

2

0

000

τω+ωπ

+τω=

θ+τω+ω+τω=

θ+ωθ+τω+ω=τ+

tAA

tEAA

AtAEttRxx

2

2

(b) E[X(t)] and Rxx (t + τ , t) are functions of time, then the process X(t) is not stationary. 3.2 (a) At )0(1)(00 XtsT ==⇒= , and at )1(0)(10 XtsT ==⇒= . Then, we have

(b) ]})()([{)]()([),( 0002012121 tTTtXTtXEEtXtXEttRxx =−−==

35

FX (x ; 0) fX (x ; 0)

1

1 x

x 1 0

21

21

21


)1()1(

21)()

21

)1()1()1()0()0()0(

2121

021021

−−+=

=−−+=−−=

tststss(t

TPtstsTPtsts

3.3 (a) The time average of )(tX is

)]([0)cos(21lim)( 0 tXEdtAT

txT

TT≠=θθ+ω=>< ∫

−∞→

≡ ensemble average.

Therefore, the process )(tX is not ergodic in the mean

(b) ∫−

∞→θ+ωθ+τω+ω=>τ+<

T

TTdttAtA

Ttxtx )cos()cos(

21lim)()( 000

⇒τ+≠τω= ),(cos2 0

2ttRA

xx The process is not ergodic in

the autocorrelation. 3.4 (a) )]()([),( 22 tXtXEttRyy τ+=τ+

)2cos(84

)]2cos()424[cos(84

)]22cos()222[cos(44

)22cos(21

21)222cos(

21

21

)](cos)(cos[

0

44

000

44

000

44

0004

022

0022

τω+=

τω+θ+τω+ω+=

θ+ωθ+τω+ω+=

θ+ω+

θ+τω+ω+=

θ+ωθ+τω+ω=

AA

tEAA

ttEAA

ttEA

tAtAE

t2

1/2

1/2 -1/2

-1/2

3/2

3/2

Height 21

t1

Random Processes

37

(b) )](cos[)]([)]([ 0222 θ+ω== tAEtXEtYE

2

)]22[cos(22

2

0

22 AtEAA=θ+ω+= = constant. Therefore, Y(t)

is wide-sense stationary.

3.5 (a) ][][][)]([ )Θ()Θ( +ω+ω == tjtj eEAEAeEtXE , where

∫∞

σ−

σπ

=σ

=0

22

22

2

][a

daeaAEa

and

.0)]([021][][

2

0

Θ)Θ( =⇒=θπ

== ∫π

θωω+ω tXEdeeeEeeE jtjjtjtj

(b) )(2)Θ()Θ(21

2121 ][][),( ttjtjtjxx eAEAeAeEttR −ω+ω−+ω == , where

∫∞

σ−

σ=σ

=0

222

32 2][ 2

2

daeaAEa

Let t1 = t + τ and t2 = t )(2),( 2 τ=σ=τ+⇒ ωτxx

jxx RettR . Therefore, )(tX is

wide-sense stationary. 3.6 (a) The autocorrelation function of Z(t) is

)()(][)]()()()([)]()([)( 22 ττ=τ+τ+=τ+=τ yyxxzz RRAEtYtYtXtXAEtZtZER ,

since A, X(t) and Y(t) are statistically independent.

.13)2(9][][ 2222 =+=+σ= AEAE a Therefore,

)9(cos26)( 32 τ−τ− +ωτ=τ eeRzz

(b) From(3.31), we have 0)]([)(lim 2 ==τ∞→τ

tZERzz . Therefore, the mean of

)(tZ

0)]([ =tZE


Since 0=zm , then )]([ 22 tZEz =σ . Hence,

.260)19(26)0()]([ 2 =+== zzRtZE

3.7 Let s(t) be the square wave with amplitude A± and without the shift t0. s(t)is periodic. From (3.40) we have

).()()()()(][),(0

0020

01

2

21 τ=τ+σ

=−−= ∫∫ xx

TT

xx RdttstsT

dtttsttsTAEttR

2

Two possible cases (i) 2

0 T≤τ≤ and (ii) 0

2≤τ≤−

T .

(i) For 2

0 T≤τ≤ , we have

(ii) For 02

≤τ≤−T , we have

τ−=

⇓

τ−−−+

τ−−=τ+∫

Tστ R

TTTdttsts

xx

T

41)(

22)1(

2)1()()(

2

022

s(t+τ)

t

t

T T/2

-1

+1

)2

( τ−T

s(t)

Random Processes

39

)41()( 2

TτστRxx += as shown below

A plot of the autocorrelation function is shown below 3.8 (a) As in the previous problem, s(t) is periodic and T0 is uniformly distributed over the period ⇒ X(t) is stationary in the wide-sense.

s(t+τ)

t

t

T T/2

-1

)2

( τ−T

s(t)

⇓

τ−−+−τ−−+τ−−++τ−−=τ+∫

)]2

()[1(]2

)2

[()1()](2

)[1())(1()()(0

TTTTTdttstsT

τ

RXX (τ)

T/2 -T/2 T/4 -3T/4

3T/4 -T/4

-σ2

σ2


(b) Consider one period only ⇒

A.x A

x xF

ATxTtTTtPtT

ATx

t P

TTtA

TxTTPA

TxTtTPxtXP

tX, P

Ax xtXPtXPxF

A x xF

tt

tX

tt

ttt

tttX

ttX

t

t

t

≤≤+=

+−≤<−+<≤−=

+<≤−+++≤<=≤<

==

≤≤<<+==

>=

0for 44

3)(Therefore,

].844

[]8

[

]484

[]8

[])(0[

and43]0)([Hence

.0for])(0[]0)([)(

and,for1)(

00

0000

x(t)

X(t)

A

T t

T0

40TT +

AtTxT

8)(

0 + AtTxTT

8)(

40 −+

xt A

3/4

1

)( ttX xF

Random Processes

41

(d) ∫ ∫∞

∞−

===84

)()]([0

AdxA

xdxxfxtXE

A

tt

ttXt t and

.19213

12)]([ 22

22 AAtXE

tx =σ⇒=

(e) ∫ ==><T Adttx

Ttx

0 8)(1)( and

12)(

22 Atx =>< .

3.9 (a) In fixing 2/11 =t and 2/32 =t , we obtain two independent random variables ( )2/1X and ( )2/3X with marginal density functions

=

2rect

21

21; xxf X and

=

2rect

21

23; xxf X . Therefore, the joint density

function is just the product of the marginal density functions to yield

41

21

21)5.1,5.0;0,0( == Xf .

(b) We observe that the duration of 1 second represents the length of a pulse

and thus, the samples spaced 1 second apart will be independent

41)5.1,5.0;0,0( =⇒ Yf as in (a).

3.10 )]()([)( tYτtYEτRyy +=

.τRτRτR

tXtXτtXτtXE

xxxxxx )1()1()(2)]}1()()][1()({[

−+++=−+−+++=

)( ttX xf

xt A 0

(3/4)

A41(c)

≤≤

=δ

=

otherwise,0

0,41

0,)(43

)(

Ax A

x x

xf t

tt

tX t


3.11 The autocorrelation function of the process )1()()( −−= tXtYtZ is

)]}1()()][1()({[)( −−−τ+−τ+=τ tXtYtXtYERzz

)()(

)()1()1()(

τ+τ=

τ+−τ−+τ−τ=

xxyy

xxxyyxyy

RR

RRRR

since Ryx = Rxy = 0 from orthogonality. Therefore, )(2)(2)( fSfSfS xxyyzz == as shown below. 3.12 )].([)]([ 62 tXEtYE = From Equation (2.80), we have

( )3263

66 1515

2!3!6)]([ σ=σ=

σ=

tXE where,

the variance .2

)( 0

00

02

α====σ ∫∫ ∫

∞α−

∞

∞−

∞

∞−

α− NdfeNdfe

NdffS ff

xx Therefore,

the mean square value .α

15)]([3

02

=

NtYE

+1 -1 +2 -2

2 2

1

+1 -1 +2 -2 τ

0 0

1

Ryy (τ)

τ

⇒

2

+1 -1

+ 1

f 0

Szz (f)

Random Processes

43

3.13 )2()1()()()()()( 21 −+−=∗+∗= tXtXthtXthtXtY . Thus,

)1()1()(2)1()1()()(

)]}2()1([)]2()1({[)(

−τ++τ+τ=−τ++τ+τ+τ=

−+−τ+−+τ+−=τ

xxxxxx

xxxxxxxx

yy

RRRRRRR

tXtXtXtXER

3.14 (a) )1()()( −+= tNtNtY or, we have with )1(δ)(δ)( −+= ttth . From (3.135), 2)()()( fHfSfS nnyy = where

fjefH π21)( −+= and thus, ( )( ) ( )feefH fjfj π+=++= π+π− 2cos1211)( 222 . Hence, the output power spectral density is ]2cos1[)rect(2)( fffS yy π+= .

τ 2 1 0-1 -2

1

)(τyyR

)(th )(tY)(tN

-1/2 1/2 0 f

)( fSyy

4


(b) )]}()()][()({[)]()([)( tNtVtNtUEtZtWERwz∗∗∗ +τ++τ+=τ+=τ

)()( τ+τ= nnuv RR Since )(tU and )(tN are statistically independent and zero mean. Hence,

)()()( fSfSfS nnuvwz += as shown below. 3.15 )()()()(

)(

τ−∗τ∗τ=ττ

hhRRg

xxyy 4434421

For 01 ≤τ≤− , we have

∫∫τ

−

τ−τ

−

−τ− +=+=τ11

)( )1()1()( dtetedtetg tt τ−−τ= 2e

For 10 ≤τ≤ , we have

eedtetdtetg tt τ−+−

−

τ−τ−−τ− −−+=−++=τ ∫ ∫ 2τ)2()1()1()( )1(τ

0

1 0

)()(

+1 -1 0 τ

+1 -1 0 τ

)( fSwz

f 1/2 -1/2

3/2

Random Processes

45

For 1≥τ , we have

eeedtetdtetg ttτ τ−−−+−

−

−−−− −+=−++=τ ∫ ∫ 2)1()1()( )1(τ)1(τ0

1

1

0

)(τ)(

Now, ).()()( τ∗τ=τ hgRyy In the same manner, we have: For 1−≤ τ ,

) eeee e

dteeee

dteetedteetR

tτttt

tτtttτtyy

2133

1

)1()1(

0

1

1

0

)1(2

32

31

31(

]2[

]2)2([)()(

−−−τ−

∞−−−−+−

−

−−+−−−

−++−−=

−++

−−++−=τ

∫

∫ ∫

For 01 ≤≤− τ ,

eeeeeτe

dteeee

dteetedteetτR

τττ

tτttt

tτtttτtyy

]32

31)1[(

]2[

)22()()(

2133

1

)1()1(

0 1

0

12

−−−−−

∞−−−−+−

τ

−−+−−−

−++−+=

−++

−−++−=

∫

∫ ∫

For 1 τ0 ≤≤ ,

eeeeτeeee

dteeeedteeteR

τττττ

tτttttτttyy

]21

32

32

21[

]2[]22[)(

213312

1

)1()1(1

)1(

−−−−−−−−

∞−−−−+−

τ

−−+−

−++−−+=

−++−−+=τ ∫∫

+1 -1 0 τ


For 1τ ≥ ,

.ee21e

21]2[)( τ1τ1τ)1()1( −+−−−

∞

τ

−τ−−−+− −+=−+=τ ∫ dteeeeR ttttyy

3.16 )()()( 2 fSfHfS xxyy = . The transfer function of the RC network is

RCfjfH

π+=

211)( . Hence,

2222

42)(

fdeefS fj

xx π+αα

=τ= τπ−∞

∞−

τα−∫

)41)(4(2)(

2222222 CRfffS yy

π+π+α

α=⇒

3.17 The transfer function of the RLC network is

RCjLCCj

LjR

cjfH

ω+ω−=

ω+ω+

ω=

211

1

1

)(

The mean of the output is 2)0()()( ==⇒= xyxy mmHtmtm . Also,

)()()( 2 fSfHfS xxyy = where 2244

4)(4)(f

ffS xxπ+

+δ=

Therefore,

π++δ

ω+ω−=

22222 11)(4

)()1(1)(

ff

RCLCfS yy

3.18 The spectrum of )( fSnn does not contain any impulse at f and thus,

0)]([)]([ =τ+= tNEtNE . The samples at t and τ+t are uncorrelated if 0),( =τ+ttCnn . It follows that the samples are uncorrelated provided

.0)( =τnnR Hence,

Random Processes

47

.2

2sin2

)()]()([)( 0202 ∫∫

−

τπ∞

∞−

τπ

τπτπ

===τ+=τB

B

fjfjnnnn B

BBNdfeN

dfefStNtNER

From the plot of )(τnnR ,

We observe that 0)( =τnnR for ...,2,1,2

±±==τ kBk . Therefore, the

sampling rates are ,21kBf s ==

τ K,3,2,1 =k .

3.19 (a) Nyquist rate .sec21

21

==cf

T

(b)

<−

=∗=otherwise,0

1,1)rect()rect()(

fffffS xx

sincsincsinc)( 2 τ. τττRxx =⋅=

⇓

or, { } )21()(])1[()( xxxx RTRTnXnTXE ==+ since sec

21

=T . Therefore,

2

2

2

2 2)2(

)2

(sin)

21(sinc)

21(

π

=π

π

==xxR and 22

=π

ρ since the process is zero

mean and stationary; that is, a shift in time does not change ρ.

)(τnnR

B21

B22

B23

B24

B21−

B22−

B23−

B24−

BN 0

τ 0


3.20 From (3.31), ( ) .2)]([4)(lim)]([ 2 ±=⇒=τ=∞→τ

tXERtXE xx The mean of

Y(t) is ∫ ∫ ⇒±=τ±=ττ=t t

tddXEtYE0 0

22)]([)]([ Y(t) is not stationary.

3.21 If ),(2),( 2121 ttttRxx −δ= then ∫ ∫ αββ−αδ=1 2

0 021 )(2),(

t t

yy ddttR .

We have 2 cases: 21 tt > and 21 tt < .

Case1: 21 tt > ⇒ ∫ ∫∫ =β=β

αβ−αδ=

2 21

0 02

021 .22)(2),(

t tt

yy tdddttR

Case2: 21 tt < ⇒ ∫ ∫∫ =α=α

ββ−αδ=

1 12

0 01

021 .22)(2),(

t tt

yy tdddttR

Therefore,

<<

==122

2112121 ,2

,2),min(2),(

tttttt

ttttRyy .

α

β α = β

t1 t2

t2

0

α

β

α = β

t1

t1

t2

0

Random Processes

49

3.22 (a) ∫=1

0

)( dttXI a . From (2.80),

σ=

odd,0

even,2)!2/(

!][ 2

n

nn

nXE n

n

n .

Hence, 2

44

2 !2σ !4][ =aIE . The variance of Ia is ][][ 222

aai IEIEa

−=σ with

∫ ==1

0

0)]([][ dttXEIE a and .32][ 2 =aIE Hence, .

32][σ 22 == aIE After

substitution, we obtain .34

4 !232 !4

][

4

4 =

=aIE

(b) 0][][][ == baba IEIEIIE since 0][ =aIE and the random variable Ib is

obtained independently from Ib.

(c) The mean of Ic is ∫∫ ==

=

TT

c dttXEdttXEIE00

.0)]([)(][ Hence,

].[]var[ 2cc IEI = Using (3.203), the variance of Ic is

∫ ∫ ∫− − −

=ττ≈ττ

τ−=τττ−=

T

T

T

Txxxxxxc TdRTdR

TTdRTI

1

1.)()(1)()(]var[

or, ∫∫∫∫ ττ−τ−ττ−=τττ−=τττ−=−

1

0

1

00

)1(2)1(2)()(2)()(]var[ ddTdRTdRTIT

xx

T

Txxc

TT ≈−=31 for .1>>T

3.23 (a) We first compute the mean of Y(t) to obtain

∫ ∫ ττ=ττ=t t

dXEdXEtYE0 0

)]([])([)]([

But,

1)]([1)(lim)]([2 ±=⇒=τ=∞→τ

tXERtXE xx


Therefore, tdtYEt

±=τ±= ∫0

)1()]([ , which is function of time )(tYt ⇒ is not

stationary.

(b) ∫ ∫∫∫ βαβα=

ββαα==

1 221

0 0002121 ),()()()]()([),(

t t

xx

tt

yy ddRdXdXEtYtYEttR

∫ ∫ βαβ−α=1 2

0 0)(

t t

xx ddR

3.24 (a) )(tX and )(ˆ tX orthogonal .0)(ˆ =τ⇒ xxR From (3.225),

)(ˆ)(ˆ)(ˆ τ−=τ−=τ xxxxxx RRR which is not zero for all τ.⇒ (a) is False.

(b) j H jtX =)}(~{ H )](ˆ)(ˆ[)}(ˆ)({ tXtXjtXjtX +=+ , but )()(ˆ tXtX −= and

hence, j H jtX =)}(~{ ⇒=+ )(~)()(ˆ tXtXtX (b) is true.

(c) If tfjetXtX 0π21 )()( = is an analytic signal 0)(

11=⇒ fS xx for 0<f .

.)(])()([)]()([)( 00011

)(11

τωω−∗τ+ω∗ τ=τ+=τ+=τ jxx

tjtjxx eRetXetXEtXtXER

The power spectral density of the process )(1 tX is then )()( 011ffSfS xxxx −= ,

which is zero if cff >0 so that all the spectrum will be shifted to the right.

(d) ∫ ∫∞

∞−

===cf

xxxxxx dffSdffSRtXE0

~~~~2 )(4)()0()](~[ , since from (3.235),

<>

=0,00,)(4

)(~~fffS

fS xxxx .

Hence, ∫∫ ⇒==cc f

xx

f

xx tXEdffSdffS0

2

0)]([2)(22)(4 (d ) is true.

Also,

)]0(ˆ)0([2)0()](~[ ~~2xxxxxx RjRRtXE +== from (3.233) ⇒ possibly true if

,0)0(ˆ =xxR but )(ˆ)(ˆ τ−=τ− xxxx RR from (3.225). At 0=τ , we have

Random Processes

51

)0(ˆ)0(ˆxxxx RR −=− and thus, ⇒==⇒= )]([2)0(2)](~[0)0(ˆ 22 tXERtXER xxxx

(d ) is true. 3.25 (a) The equivalent circuit using a noiseless resistor is

The transfer function RCjLCLjR

Cj

CjjH

nvvω+ω−

=ω++

ω

ω=ω

)1(1

1

1

)(20

. Hence,

the power spectral density of )(0 tv is

RCjLCkTRfSjHfS

nnn vvvvvvω+ω−

=ω=)1(

2)()()(2

2

000.

(b) The input impedance is

222

22

222 )()1()1(

)()1()(1

)(1

)(RCLC

CRLCLj

RCLCR

LjRCj

jLRCj

jZω+ω−

ω−ω−ω+

ω+ω−=

ω++ω

ω+ω

=ω

But Nyquist theorem says

kTSvv 2)( =ω ℜe222 )ω()ω1(

2)}ω({RCLC

kTRjZ+−

= which agrees with the result

obtained in (a).

R

L C v0(t)

+

_

±

Vn (t)

)( ω⇐ jZ


3.26 (a) The equivalent circuit with noiseless resistors is

22

21 )()()()()()()(

2211221100fHfSfHfSfSfSfS

eeee nnnnvvvvvv +=+=

Using superposition principle, the power spectral density at the terminal pairs for each source is

221

11)]([1

12)(11 RRC

RkTfS vv+ω+

= and2

2122

)]([112)(

22 RRCRkTfS vv

+ω+= .

Hence, the output power spectral density is

221

2211

)]([1)(2

)(00 RRC

RTRTkfS vv

+ω+

+=

(b) In order to determine the autocorrelation function, we rewrite )(00

fS vv as

222

21

21

21

2211

4])[(

1)(

12

)()(

)(00

fCRR

CRRCRRRTRTk

fS vvπ+

+

+

++

=

Hence,

+τ

−++

=τCRRCRR

RTRTkRvv )(

exp)(

)()(

2121

2211

(c) The mean square value is

CRRRTRTk

Rvv )()(

)0(21

2211

++

=

_

)( 11 TR

C

+ ±

)(1 tVn ±

)(2 tVn

)( 11 TR v0(t)

Random Processes

53

Substituting for the given values of CTTRR and ,,, 2121 , we obtain 1010457.0)0( −×=vvR . Therefore, the root mean square value is

. V76.6volts1076.6 6 µ=− 3.27 The equivalent circuit using a noiseless resistor is

(a) The transfer function relating )(tI to )(tVn is LjR

jHnin ω+

=ω1)( .

Therefore, the power spectral density of )(tI is

22

2

)(2)()()(

LRkTRSjHS

nnn vviviiω+

=ω=ω=ω .

(b) From (3.244), we need to determine the power spectral density of the

short-circuit current. Hence, we have, for the circuit below,

2222 )ω(ω

)ω(ω11

LRLj

LRR

LjRZY

inin

+−

+=

+==

Therefore, the power spectral density of the short-circuit current is

kTSii 2)( =ω ℜe22 )ω(

2}{LR

RkTYin+

=

R

Yin L

R

I(t)L

±

)(tVn


3.28 (a) ∫∞

∞−

−= α)α()α()( dhtXtY . The mean of Y(t) is

∫∞

∞−

=ααα−= )0()]()([)]([ HmdhtXEtYE x

where, .1α)0(0

α == ∫∞

− deH Hence .xy mm =

(b) Since ,0)]([ == xmtXE the mean of Y(t) is 0)]([ =tYE and the variance is

∫ ∫∞

∞−

∞

∞−

βαβαα−β= .)()()()]([ 2 ddhhRtYE xx

The autocorrelation function of the input is )()( τδ=τ kRxx , where k is a constant. Therefore,

∫ ∫∫ ∫∞

∞−

∞α−

∞

∞−

∞

∞−

=α=αα=βαβαα−βδ= .2

)()()()()]([0

222 kdekdhkddhhktYE

3.29 Since )( fSnn does not have an impulse at 0=f , 0)]([ =tNE and the mean of the output of the linear filter is .0)0()]([)]([ == HtNEtYE Hence, the variance of Y(t) is

∫∞

∞−

===σ dffSRtYE yyyyy )()0()]([ 22

where, 202 )(2

)()()( fHN

fHfSfS nnyy == .

The system function is given by

>

≤

−

=

Bf

BfBf

KfH

,0

,1)(

Therefore,

Random Processes

55

∫∫∫

−=

−+

+=σ−

BB

By df

Bf

KNdfBf

KN

dfBf

KN

0

22

0

2

0

020

02 .112

12

3

20BKN

=

Chapter 4

Discrete Time Random Processes

4.1 (a) Using 0131111322

)det( =λ−−

λ−−λ−

=λ− IA , we obtain

31 =λ , 12 =λ and 23 −=λ .

Then,

=

−

−⇒λ=

cba

cba

3131

111322

111 xAx

Solving for a, b and c, we obtain 1=== cba and thus,

=

111

1x

Similarly,

−=⇒λ=

111

2222 xxAx and

−−

=⇒λ=1071.0786.0

3323 xxAx .

The modal matrix is then

−−−

=111071.011786.011

M ,

−−−=−

93.093.0033.083.05.04.01.05.0

1M

The Jordan form is

56

Discrete Time Random Processes 57

AMMJ 1−=

−−−

−

−

−−−=

111071.011786.011

131111322

93.093.0033.083.05.04.01.05.0

−=

200010003

(b) 6and

3,30

600021024

3

21

=λ−=λ+=λ

⇒=λ−

λ−−λ−

=−jj

λIA

Solving 0and

5.05.0,1)3(

600021024

111 =−==

⇒

+=

−⇒λ=

cjba

cba

jcba

xAx

Thus,

−=

021

21

1

1 jx and

+=

021

21

1

2 jx .

Again, solving 333 xAx λ= we obtain

=

100

3x

The modal matrix is then

−+

−=⇒

+−= −

10005.05.005.05.0

10005.05.05.05.0011

1 jjjj

jj MM

and

−

+== −

600030003

1 jj

AMMJ

Signal detection and estimation 58

(c) Similarly, we solve

40240

160124

321 =λ=λ=λ⇒=λ−−

λ−λ−

=λ− IA

Note that we have an algebraic multiplicity of 3=r . Solving, ⇒=λ−=λ

12IA degeneracy 213 =−=q ; that is, we have two eigenvectors and one generalized eigenvector. Thus,

⇒= 11 4xAx

=

001

1x or

−=

211

2x

Solving for the generalized eigenvector, we have

=⇒=−

001

)4( 22222 xxxIA

Therefore, the modal matrix is [ ]

−==

120010011

2221 xxxM

−=⇒ −

120010011

1M and the Jordan form is

== −

400

140

0041 AMMJ

4.2 (a) Solving for the determinant of A, we have ⇒≠−= 06)det(A the matrix is of full rank rA = 3.

(b) Solving

−=−=−=

⇒=−−−

−−−−−

⇒=−4142.3λ

3λ5858.0λ

0λ510

3 λ1120λ1

0)λdet(

3

2

1

IA


We observe that all 0<λ i , for i = 1, 2, 3, and thus the matrix is negative definite.

(c) Solving vAv λ= , we obtain

−−−

=1511.06670.07296.0

1v ,

−

−=

4082.08165.0 4082.0

2v and

−=

4879.0 7737.04042.0

3v .

4.3 The characteristic equation is

)λ3()λ2(

λ10011λ31100λ20100λ3

λ 3 −−=

−−−

−−

=− IA

==λ==λ

⇒1mty multiplici algebraicwith 3

3tymultiplici algebraicwith 2

2

11

2

m

Note that the rank of r==λ− 21IA . Thus, 2241 =−=−= rnq . Thus, for

21 =λ , we have 2 eigenvectors and 1 generalized eigenvector since .31 =m

−=

−

=⇒=⇒=λ

01

10

and

1001

22 31 x xxxA . The generalized eigenvector is

−=⇒=−

11

00

)2( 12112 xxxIA

For

−=⇒==λ

01

00

3,3 444 xxAx


Hence, the modal matrix is

−−−−

==

00111110

01000001

][ 43121 xxxxM

Note that

== −

3000020000200012

1 AMMΛ

4.4 Let the M roots of iλ , i = 1, 2, …, M, be the eigenvalues of R, then 0)λdet( =− IR . Also,

0)det()det()](det[)det( 11 =λ−=λ−=λ− −− RIRRIRIR

Since the correlation matrix R is nonsingular )0)(det( ≠R , then

01det0)det( 11 =λ

−

λ

==λ− −− mRIRI

The eigenvalues are non-zero for the non trivial solution )0( ≠λ and thus,

0λ1det 1 =

−− IR , which means that Mi

i,,2,1,1

L=λ

, are eigenvalues of

.1−R 4.5 From (4.121), two eigenvectors iv and jv are orthogonal if ,0=j

Hi vv i ≠ j.

From the definition,

iii vRv λ= (1)

and

jjj vRv λ= (2)

Premultiplying both sides of (1) by Hjv , the Hermitian vector of jv ,we obtain


iHiii

Hj vvRvv λ= (3)

Since the correlation matrix R is Hermitian, RR =H . Taking the Hermitian of (2), we have

Hjj

Hj vRv λ= (4)

since iλ is real. Postmultiplying (4) by iv yields

iHjji

Hj vvRvv λ= (5)

Subtracting (5) from (3), we obtain

0)( =λ−λ iHjji vv (6)

which yields 0=iHj vv sine ji λ≠λ . Therefore, the eigenvectors iv and jv are

ortogonal. 4.6 Let Mvvv ,...,, 21 be the eigenvectors corresponding to M eigenvalues of the correlation matrix R. From (4.120), the eigenvectors are linearly independent if

=+++ nnaaa vvv L2211 0 (1)

for 021 ==== naaa L . Let ii λ−= IRT , then =iivT 0 and

jijji vvT )( λ−λ= if i ≠ j. Multiplying (1) by 1T gives

=λ−λ++λ−λ+λ−λ nnnaaa vvv )()()( 131332121 L 0 (2)

Similarly multiplying (2) by 2T and then 3T and so on until 1−nT , we obtain

=λ−λλ−λ++λ−λλ−λ − nnnnaa vv ))(())(( 211323133 L 0 (3) M

LL ))(()())(( 2112121111 λ−λλ−λ+λ−λλ−λλ−λ −−−−−− nnnnnnnnn aa v =λ−λ − nnn v)( 2 0 (4)

=λ−λλ−λλ−λλ−λ −− nnnnnnnna v))(())(( 1221 L 0 (5)


From (5), since 0)( ≠λ−λ in for i ≠ n. ⇒ an = 0. Using (5) and (4), we see again 01 =−na , and so on going backward until Equation (1). Hence,

===== − nn aaaa 121 L 0

and thus, the eigenvectors are linearly independent. 4.7 From (4.121), since the matrix is symmetric, the normalized eigenvectors x1

and x2 corresponding to the eigenvalues 1λ and 2λ are orthogonal and A has the

form

=

2212

1211

aaaa

A since it is symmetric.

Let 21 xxX y x += . Then 22121 λλ xxxAxAAX y x y x +=+= since

ii xAx λ= . Also,

1λλ)λλ)(( 22

12

221121 =+=++= yxy x y x xxxxAXX T (1)

The equation of the ellipse has the form

12

2

2

2=+

by

ax

Therefore, (1) represents an ellipse for 1/1 λ=a and 2/1 λ=b . Assuming

21 λ>λ , then a is the minor axis and b is the major axes and the ellipse is a shown below.

2

1λ

1

1λ

y

x1 x2

x


(b) For

=

5335

A , we solve for the eigenvalues⇒

max12 8)2)(8(9)5(

5335

det)det( λ==λ⇒−λ−λ=−−λ=

−λ−−−λ

=−λ AI and

min2 λ==λ . Solving for the eigenvectors, we have ⇒=− 01)8( xIA

=

11

21

1x , and

−

=⇒=−11

210)2( 22

xxIA .

Note that x1 and x2 are orthogonal. From (a), the semi-major axis is ( ) ( ) 707.02/1/1 min ==λ and the semi-minor axis is

( ) ( ) 354.08/1/1 max ==λ . The ellipse is shown below 4.8 (a) The second order difference equation of the AR process is.

)()2()1()( 21 nenXanXanX +−−−−=

and thus, the characteristic equation is

01 22

11 =++ −− ZaZa

y

x2

x1

x

0.354

0.707


(b) Solving for the roots of the second order equation, we obtain

−−−= 2

2111 4

21 aaaP and

−+−= 2

2112 4

21 aaaP

For stability of the system, the poles 1P and 2P must be inside the unit circle, that is 11 <P and 12 <P . Applying these two conditions, we obtain

11

21

21

−≥−−≥+

aaaa

and

11 2 ≤≤− a

4.9 (a) The Yule-Walker equations for the AR(2) process are rR =ω or raR = . Applying this to our system, we have

=

ωω

−

−)2()1(

)0()1()1()0(

2

1

rr

rrrr

For a real-valued stationary process )1()1( rr =− , and thus solving the two equations in two unknowns, we obtain

)1()0()]2()0()[1(

2211rr

rrra−

−=−=ω

Z -1

Z -1

)(nX

)1( −nX

)2( −nX

)(ne ∑

-a1

-a2


)1()0()1()2()0(

22

2

22rr

rrra

−

−=−=ω

where 2)0( xr σ= .

(b) Note that r(1) and r(2) may be expressed in terms of parameters of the systems as in (4.184) and (4.186) to obtain

21

2

2

1

1)1( xxa

ar σρ=σ

+−

= with 2

11 1 a

a+−

=ρ

and 22

22

2

21

1)2( xxa

aa

r σρ=σ

−

+= with 2

2

21

2 1a

aa

−+

=ρ

4.10 The state diagram is We have S1 and S2: irreducible ergodic. S3: aperiodic and transient. S4: absorbing. 4.11 Let S1, S2 and S3 represent symbols 1, 2 and 3, respectively. Then, the state diagram is

S1 S2

1/3

1/2

1/2 2/3

S3 S4

1/4

1/2

1

1/4


(b) The n-step transition matrix is

==

3400.02700.03900.03200.02800.04000.03000.02700.04300.0

)2( 2PP

==

3220.02730.04050.03200.02720.04080.03140.02730.04130.0

)3( 3PP

==

3290.02727.04083.03284.02728.04088.03174.02727.04099.0

)4( 4PP

==

3283.02727.04089.03282.02727.04090.03180.02727.04093.0

)5( 5PP

M

==

3282.02727.04091.03282.02727.04091.03182.02727.04091.0

)6( 6PP

0.4

S1 S2

S3

0.3

0.3

0.4 0.2 0.3

0.4

0.5 0.2


=

3282.02727.04091.03282.02727.04091.03182.02727.04091.0

)20(P

(c) The state probabilities are given by nTn PpP =)( . Thus,

]3400.02700.03900.0[)1( == PppT T

with [ ]4.03.03.0)0( == pP T .

]3220.02730.04050.0[)2( 2 == PppT T

]3150.02727.04083.0[)3( 3 == PppT T

]3183.02727.04089.0[)4( 4 == PppT T

M

]3182.02727.04091.0[)5( 4 == PppT T

]3182.02727.04091.0[)20( 20 == PppT T

4.12 (a)

0.5

S1 ≡ R S2 ≡ N

S3 ≡ S

0.25

0.25

0.5 0.25

0.25

0.5

0.5


(b)

=

500.0250.0250.0500.0000.0500.0250.0250.0500.0

SnowNiceRain

)1(

SnowNiceRain

P

=

438.0188.0375.0375.0250.0375.0375.0188.0438.0

SnowNiceRain

)2(

SnowNiceRain

P

=

406.0203.0391.0406.0188.0406.0391.0203.0406.0

SnowNiceRain

)3(

SnowNiceRain

P

=

402.0199.0398.0398.0203.0398.0398.0199.0402.0

SnowNiceRain

)4(

SnowNiceRain

P

=

400.0200.0399.0400.0199.0400.0399.0200.0400.0

SnowNiceRain

)5(

SnowNiceRain

P

=

400.0200.0400.0400.0200.0400.0400.0200.0400.0

SnowNiceRain

)6(

SnowNiceRain

P

We observe that after 6 days of weather predictions, we have probability of Rain = 0.4, probability of Nice = 0.2 and probability of Snow = 0.4 no matter where the chain started. Therefore, this chain is a regular Markov chain.


(c) Using nTn PpP =)( , we have

PppT T=)1( with [ ]1.02.07.0)0( == pP T .

Therefore, ]325.0200.0475.0[)1( =Tp .

]381.0200.0419.0[)2( 2 == PppT T

]395.0200.0404.0[)3( 3 == PppT T

]399.0200.0401.0[)4( 4 == PppT T

M

]400.0200.0400.0[)5( 5 == PppT T

]400.0200.0400.0[)20( 20 == PppT T

Hence, the steady state distribution vector is

=

ωωω

=4.02.04.0

3

2

1

P

4.13 (a) This is a two-state Markov chain as shown below

(b) To verify that it is true by induction, we must verify that it is true for 1=n first, then assuming it is true for n yields it is true for 1+n . That is,

)1()1()1( nn PPP =+ must be verified. Since )1()( nn PP = , for 1=n , we have

S0 S1

a

b

1-b 1- a

Chapter 5

Statistical Decision Theory

5.1 (a) The LRT is η<>=

0

1

0|

1|

)|(

)|()(Λ

0

1

H

H

Hyf

Hyfy

HY

HY

We observe that for )2ln(2/120

2

1

2

1

η<>⇒η<

>⇒≤≤−

H

H

y

H

H

ey

y, while for 2>y , we

always decide H0.

(b) For minimum probability of error criterion 01100 ==⇒ CC and 11001 == CC

(i) ⇒=⇒=η⇒= 693.0)2ln(121

0P choose H0 for 693.00 ≤≤ y ;

otherwise choose H1. The minimum probability of error is

355.021)(

693.0

00

2

693.01 =+=ε ∫∫ − dyPdyePP y

72

)|( 00| Hyf HY

)|( 11| Hyf HY21

0.693 1 2

1

y


73

(ii) Similarly, ⇒=32

1P choose H1 for 239.1 ≤≤ y and 308.0)( =εP .

(iii) 31

1 =P , ⇒<> 0

0

1

H

H

y always decide H1 and 288.0)( =εP .

5.2 (a) ⇒η<>=

0

1

0|

1|

)|(

)|()(Λ

0

1

H

H

Hyf

Hyfy

HY

HY

(i) 21

<η ,

⇒η>)(Λ y always decide H1

(ii) 21

>η ,

⇒η<≤≤ )(Λ,10 yy decide H0 ⇒η>≤≤ )(Λ,21 yy decide H1

21

1 2

∞)(Λ y

21

1 2 y

)(Λ y

η

21

y

)(Λ y

η

1 2


(iii) ,21

=η

decide H1 or H0 in at the range 10 ≤≤ y and decide H1 for 21 ≤< y .

(b) (i) 21

<η , the probability of false alarm is

11)|(1

00|

1 0=== ∫∫ dydyHyfP

Z HYF .

The probability of detection is

01121)|(

2

01|

1 1=−=⇒=== ∫∫ DMZ HYD PPdydyHyfP

(ii) 21

>η , 00)|(2

10|

1 0=== ∫∫ dydyHyfP

Z HYF and

21

21

21)|(

2

11|

1 1=⇒=== ∫∫ MZ HYD PdydyHyfP

5.3 Minimum probability of error criterion 1and0 10011100 ====⇒ CCCC .

(a) The conditional density functions are

( )

σ

+−

σπ=

2

2

0|2

exp21)|(

0

AyHyf HY

( )

σ

−−

σπ=

2

2

1|2

exp21)|(

1

AyHyf HY

21

=η

y

)(Λ y

1 2


75

( ) ( ) ( )

1

02

0

1

1

0

0

1

2

2

2

2

1

0

0

1

22

22

|

|

ln2

ln22

lnΛ

]2/)(exp[]2/)(exp[

)Λ(0

1

PP

AH

H

y

PP

H

HAyAy

y

PP

H

H

AyAy

f

fy

HY

HY

σ<>⇒

<>

σ

++

σ

−=⇒

=η<>

σ+−

σ−−==

(b) AA

H

H

yP

P22

0

1

01

σ549.03ln2σ

3=<

>⇒=

0

0

1

01

H

H

yPP <>⇒=

AA

H

H

yP

P22

0

1

01

σ256σ405.03

5 −=<

>⇒=

As P1 increases ⇒ DP increases and FP increases, but FP increases at a faster rate. 5.4 The received signals under each hypothesis are

NAYHNYHNAYH

+==

+−=

:::

2

0

1

-A

H1 γ H0

A 0

)|( 00| Hyf HY )|( 11| Hyf HY

y


77

σ−

−+

σ−

σπ= ∫ ∫

∞

γ

γ

∞−

dyAydyy2

2

2

2

2)(exp

2exp

21

32

Now, ( )2

02

exp2

exp02

2

2

2 AAPe =γ⇒=

σ

−γ−+

σ

γ−−⇒=

γ∂∂

(b) Substituting for the value of 2A

=γ and solving the integrals we obtain

σ=

σ=

234

22erfc

32(error) AQAP

5.5 (a)

≤≤+−

≤≤−

−≤≤−+

=∗=

otherwise,0

31,83

81

11,41

13,83

81

)()()|( 1| 1

yy

y

yy

nfsfHyf NSHY

as shown below

The LRT is then

<<∞

≤≤+−

≤≤−

−≤≤−+

−≤≤−∞

=

32,

21,23

21

11,1

12,23

21

23,

)(Λ

y

yy

y

yy

y

y

1/4

1 2 3 -1 -2 -3 y

)|( 11| Hyf HY


(i) 41

=η , we have

Λ(y) > η ⇒ always decide H1.

(ii) η=1

2 cases: decide H1 when

≤<≤<−−<≤−≤<−<≤−

⇒η=32and11and23when

21and12when)(Λ

1

0

yyyHyyH

y

or, decide H0 when

≤<−<≤−<≤−

⇒η=32and23when

22when)(Λ

1

0

yyHyH

y

(iii) 2=η

decide H0 when 22 ≤≤− y since 2)(Λ =η<y

decide H1 when 23 −≤≤− y and 32 ≤≤ y since η>)(Λ y

1/2

1 2 3-1 -2 -3 y

1

∞ ∞ )(Λ y

4/1=η

1/2

1 2 3-1 -2 -3 y

1

∞ ∞

η=1

1/2

1 2 3-1 -2 -3 y

1

∞ ∞ η=2

)(Λ y


79

(b) ∫=1 0

)|( 0|Z HYF dyHyfP and ∫=1 1

)|( 1|Z HYD dyHyfP

(i) 141

==⇒=η DF PP

(ii) 625.0and211 ==⇒=η DF PP or, 125.0and0 == DF PP

(iii) 125.0and02 ==⇒=η DF PP

(c) The ROC is shown below

5.6 (a) 0allfor),()(00

00 0≥α=α=

α== α−α−α−

∞

∞−∫∫∫ sednedne

Ndnnsfsf s

Ns

Ns

SNS

000 0

0,1)( NnN

dseN

nf sN ≤≤=

α= α−

∞

∫

(b) ⇒= )()(),( nfsfnsf NSSN S and N are statistically independent.

(c) ∫∞

λλ−λ=∗=0

)()()()()( dyffnfsfyf SNNSY

Solving the convolution as shown in Chapter 2 in detail, we obtain

PD

PF 1

1

1/2

1/2

0.625

0.162

21

=η

121

<η<

1>η


[ ]{ }

∞<≤α−−α−

≤≤−

=

α−

yNyNyN

NyeN

yf

y

Y

000

00

,)exp()(exp1

0,)1(1

)(

5.7 (a) The LRT is

γ=

η

π<>−=⇒η<

>

π

=−

−

2ln21)(

21

21

)(Λ

0

1

2

0

1

2

2

H

H

yyyT

H

H

e

ey

y

y

Solving, γ+±=⇒=γ−−= 211021)( 2 yyyyT as shown below

To determine the decisions, we observe that we have 3 cases:

(i) 21

−≤γ , (ii) 021

<γ<− and (iii) 0>γ

T(y)

y

-1/2

0 1 2-1-2

fY (y)

yN0

)1(1 0

0

NeN

α−−


81

(i) e22

1 π≤η⇒−≤γ

⇒γ>)(yT always decide H1

(ii) 22

021 π

≤η<π

⇒<γ<−e

γ+−−= 2111y , γ++−= 2112y , γ+−= 2113y and γ++= 2114y

Decide H1 when 1yy ≤ , 32 yyy ≤≤ and 4yy ≥ . Decide H0 when 21 yyy << and 43 yyy << .

(iii) 2

0 π>η⇒>γ

y

-1/2

1 2-1-2

γ

y1 y2 y3 y4

H1 H1 H1

H0 H0

T(y)

y

-1/2

0 1 2-1 -2

γ

y

-1/2

0 1 2-1-2

γ

y1 y2

H1 H1 H0


γ+−−= 2111y and γ++= 2112y . Decide H1 when 1yy ≤ and 2yy ≥ Decide H0 when 21 yyy <<

(b) 47.0232

1

00 ≈γ⇒==η⇒=

PP

P

02.0)393.2(221

21

)|()|(

211

2211

2

0|01

22

1 0

==π

+π

=

==

∫∫

∫∞

γ++

−γ+−−

∞−

−Qdyedye

dyHyfHHPP

yy

Z HYF

09.021

21)|()|(

211

211

1|111 1

=+=== ∫∫∫∞

γ++

−γ+−−

∞−

+ dyedyedyHyfHHPP yyZ HYD

(c) )211(2 γ++= QPF

)211exp(11)]211(exp[ γ+−−−=−=⇒γ++−= DMD PPP

The optimum threshold optγ is obtained when MF PP = , or

)211(2)211exp(1 optopt Q γ++=γ+−−−

(d) 2.021

21

1

21

2

22 =π

+π

= ∫∫∞

γ

−γ

∞−

−dyedyeP

yy

F

or, 022.02.02

erf1.0)(2.0)(2 11

11 >≅γ⇒=

γ⇒=γ⇒=γ QQ ; that is the

decision regions as given in (a) part (iii).


83

5.8

−−

π=

2)1(

exp21)|(

2

1| 1

yHyf HY

−

π=

2exp

21)|(

2

0| 0

yHyf HY

(a) The LRT is 211

2exp

21

2)1(

exp21

)(Λ

0

1

0

1

2

2

H

H

y

H

H

y

y

y <>⇒<

>

−

π

−−

π=

(b) 005.021)|( 2

0|

2

1 0=

π== ∫∫

∞

α

−dyedyHyfP

y

Z HYF

581.2005.02

erf121)( ≈α⇒=

α−=α⇒ Q

(c) ∫∫∞

−α

−∞

=α

≈=π

=

−−

π=

1

2

581.2

2013.0)581.1(

21

2)1(

exp21

2

Qdxedyy

Px

D

5.9 The LRT is

η<>

σ−

π

σ

−−

π==

∏

∏

=

=

0

1

12

2

12

2

0|

1|

2exp

21

2)(

exp21

)|(

)|()(Λ

0

1

H

H

y

my

Hf

Hfy

K

k

k

K

k

k

H

H

yy

Y

Y

44 344 21321γ

=+η

σ<>⇒ ∑ 2

ln2

0

1

)(1

Kmm

H

H

y

yT

K

kk as given in Example 5.2. Hence, γ<

>

0

1

)(

H

H

yT .


5.10 ∏=

σ−

σπ=

K

k

kH

yHf

120

2

00|

2exp

21)|(

0yY

∏=

σ−

σπ=

K

k

kH

yHf

121

2

11|

2exp

21)|(

1yY

∑=

γ<>=⇒

K

kk

H

H

yT1

0

1

2)( y where,

σσ

−ησ−σ

σσ=γ

1

020

21

21

20 lnln

2K from Example 5.9.

5.11 (a) The probability of false alarm is

σγ

=σπ

== ∫∫∞

γ

σ−

0

2

00|

0

2

1 0 21)|( QdyedHfP

y

Z HF yyY

where, 1andlnln2

1

020

21

21

20 =

σσ

−ησ−σ

σσ=γ K .

σγ

−=⇒

σγ

=σπ

=−= ∫∞

γ

σ−

11

2

1

1211 1

2

QPQdyePP M

y

MD

(b) The ROC is PD versus PF. For 22 20

21 =σ=σ , we have

η=

2)2ln(4

QPD and )]2ln(4[ η= QPF for various values of η. Hence,

1/2

1/2

PD

PF


85

(c) The minimax criterion when 01100 == CC and 11001 == CC yields

MF PP = . Hence,

σ

γ=

σ

γ−

011 optopt QQ .

5.12 (a)

σ−

σπ=

2

2

0|2

exp21)|(

0

yHyf HY

σ

+−

σπ=

2

2

1|2

)(exp

21)|(

1

myHyf HY

σ

−−

σπ=

2

2

2|2

)(exp

21)|(

2

myHyf HY

The receiver based on the minimum probability of error selects the hypothesis having the largest a posteriori probability )|( yHP j , where

)(

)()|()|(

|

yf

HPHyfyHP

Y

jjHYj

j=

31)( =jHP and )(yfY is common to all a posteriori probabilities ⇒ We choose

Hj for which )|(| jHY Hyfj

is largest. This is equivalent to choosing Hj for which

)( jmy − is smallest. Hence, we have

(b) The minimum probability of error is

∑∑==

ε=ε=ε3

1

3

1)|(

31)|()()(

jj

jjj HPHPHPP where,

H0 H2 H1

-m m

y -m/2 m/2

0


σ

=π

=

σ

−−

σπ=

−>=ε

∫

∫

∞

σ

−

∞

−

221

2)(

exp21

2)|(

2/

2

2/2

21

11

2

mQdxe

dymy

HmyPHP

m

x

m

By symmetry, )|()|( 31 HPHP ε=ε and

dyedye

HmYmYPHmYPHP

m

ym y

∫∫∞

σ−−

∞−

σ−

σπ+

σπ=

><=

>=ε

2/

22/

2

000

2

2

2

2

21

21

2and

22)|(

σ

=ε⇒

σ

=π

= ∫∞

σ

−

234)(

22

212

2/

2

2

mQPmQdxem

x

(c) The conditional density functions become

20|

2

0 21)|(

y

HY eHyf−

π=

−−

π=

2)1(

exp21)|(

2

1| 1

yHyf HY

82|

2

2 221)|(

y

HY eHyf−

π=

The boundary between H0 and H1 is 21

=y , while the boundary between H0 and

H2 is obtained from

36.1)|()|( 822|0|

22

20±≈⇒=⇒=

−−yeeHyfHyf

yy

HYHY

For the boundary between H1 and H2, we have


87

18.0and85.221)|()|( 21

22)1(

2|1|

22

21−≈≈⇒=⇒=

−−

−yyeeHyfHyf

yy

HYHY

)|()()(3

1j

jj HPHPP ε=ε ∑

= where,

527.0)0()36.1(21

21)|()|(

0

236.1

20|0

22

21 0

=+=π

+π

==ε ∫∫∫∞ −−

∞−

−

QQ

dyedyedyHyfHPyy

ZZ HYU

29.0)85.1(21

21

21

21

21)|()|(

85.1

22/1

2

85.2

2)1(2/1

2)1(

1|1

22

22

20 1

=+

=

π+

π=

π+

π==ε

∫∫

∫∫∫

∞ −−

∞−

−

∞ −−

∞−

−−

QQ

dxedxe

dyedyedyHyfHP

xx

yy

ZZ HYU

H0 H2 H1

y 2.851.36 1 0.5 0 -0.18-1.36

H2

-1 2

2.851.36 1 0.5y

0 -0.18-1.36

)|( 11| Hyf HY)|( 00| Hyf HY )|( 22| Hyf HY


1)7.5()72.2(1

)7.5()72.2(21

21

21

221)|()|(

7.5

2

72.2

2

7.5

72.2

285.2

36.1

82|2

22

22

10 2

≈−−=

−−=π

−π

=

π=

π==ε

∫∫

∫∫∫

∞ −∞

−

−

−

−

−

−

QQ

QQdxedxe

dxedyedyHyfHP

xx

xy

ZZ HYU

6.0)]|()|()|([31)( 210 ≈ε+ε+ε=ε⇒ HPHPHPP

5.13 ∏=

σ−

σπ=

K

k

kH

yHf

12

2

0|2

exp21)|(

0yY

∏=

σ+σ−

σ+σ=

K

k m

k

m

Hy

Hf1

22

2

221|

)(2exp1)|(

1yY

1)(2

exp)(Λ)|(

)|(

0

1

222

22/

22

2

0|

1|

0

1

H

H

Hf

Hf

m

mTK

mH

H

<>

σ+σσ

σ

σ+σ

σ⇒= yyy

yy

Y

Y

Taking the logarithm on both sides and rearranging terms, we obtain the decision rule

γ≡

σ

σ+σ

σ

σ+σσ<>

2

22

2

222

0

1

ln)( m

m

mT K

H

H

yy

or,

γ<>∑

=

0

1

1

2

H

H

yK

kk


89

5.14 The conditional density functions are

[ ]

σ+−

σ+π= yyyY

T

mK

m

H Hf)1(2

1exp)1(2

1)|(22/2

1| 1

where [ ]TKyyy L21=y and

−

π= yyyY

TKH Hf

21exp

)2(1)|(

2/0| 0

The LRT is then

η<>

σ+

σ

σ+=

0

1

2

22/

2 )1(2exp

11)(Λ

H

H

m

mTK

m

yyy

Taking logarithm η<>

σ++

σ+

σ⇒ ln

11ln

2)1(20

1

22

2

H

HK

mm

mT yy

12

2

2

0

1

)1ln(2

)1(2γ≡+σ+η

σ

σ+<>⇒ m

m

mT K

H

H

yy

or,

1

0

1

1

2 γ<>∑

=H

H

yK

kk

We observe that the LRT does not require knowledge of 2mσ to make a decision.

Therefore, a UMP test exists.


5.15 ∏=

−

π=

K

k

kH

yHf

1

2

0| 2exp

21)|(

0yY

∏=

−−

π=

K

k

kH

myHf

1

2

1| 2)(

exp21)|(

1yY

(a)

∑ −=⇒η<>=

=

K

kk

H

H mKym

H

H

Hf

Hf

1

2

0

1

0|

1|

2exp)(Λ

)|(

)|()(Λ

0

1 yyy

yY

Y

or, γ≡+η

<>∑

= mKm

H

H

yK

kk 2

ln2 2

0

1

1. Therefore, a test can be conducted without

knowledge of m ⇒ A UMP test exists.

(b) ⇒= 05.0FP The test decides H0 when γ>= ∑=

K

kkyyT

1)( , where T is

Gaussian with mean zero and variance K under H0. Hence,

05.02

erf121)|( 0| 0

=

γ−=

γ== ∫

∞

γ KKQdtHtfP HTF

Using 9.0)|( 1| 1>= ∫

∞

γ

dtHtfP HTD where T is Gaussian with mean Km under H1,

we obtain from the table in the appendix 16≈K .

5.16 Since the observations are independent, the LRT becomes

η<>

θ

−θ

+++−

θθ

==

0

1

0121

1

0

0|

1| 11)(exp)|(

)|()(Λ

0

1

H

H

yyyHf

HfK

K

H

HL

yy

yY

Y


91

η<>

θθθ−θ

θθ

=⇒ ∑=

0

1

10

01

11

0 exp)(Λ

H

H

yK

kk

K

y

Taking the natural logarithm and simplifying the expression, we obtain

γ=

θθ

−ηθ−θθθ

<>= ∑

=

K

kk

H

H

yT1 0

1

01

10

0

1

lnln)( y

For a UMP test of level α, we need

05.0]|)([ 0 =γ>HTP Y or 95.0]|)([ 0 =γ≤HTP Y

We now determine the distribution of the test statistic )(YT using the characteristic function such that

[ ] [ ] [ ] [ ] [ ])(Φ)(Φ)(Φ

)(Φ

21

2121 )(

ωωω====ω ωωω+++ωω

K

KK

yyy

YjYjYjYYYjjt eEeEeEeEeE

L

LLY

since the Yks, Kk ,,2,1 L= are statistically independent. From (2.93), Kj −θω−=ω )1()(Φ t . Hence, from (2.102), )(YT is a gamma distribution

),( PKG with density function

>β=

β−−

otherwise,0

0,)(Γ

1)(

/1 tetKtf

tKK

T

Therefore, for 21=K , (see Table 9 page 456, Dudewicz1)

62.290062.2995.0]|)([0

0 =γ⇒=θγ

⇒=γ≤HTP Y

The test decides H1 (rejects H0) if 62.290)( >YT 1 Dudewicz, E. J., Introduction to Statistics and Probability, Holt, Rinehart and Winston, New York,

1976.


−

−=

+−−+

+++−

+++−

+−−+

=bb

aa

bababba

bababbb

baabaaa

baabaab

11

)1( 22

22

P

+−−+

+−−−

+−−−

+−−+

−

−==+

bababa

bababb

babaaa

babaab

bbaa

n nn

nn

n

)1()1(

)1()1(

11

)1()1()1( PPP

Let bax −−=1 , then

++

+−

+−

++

−

−=+

babxa

babxb

baaxa

baaxb

bbaa

n nn

nn

11

)1(P

)1(1

)1()1()1()1(1

1

11

11

2222

2222

+=

+−−+

+=

−−−−−−−−−−−+

+=

−−++−+−−+++++−−−+−−+

+=

++

++n

bxabxbaxaaxb

ba

babxababxbbaaxabaaxb

ba

xbabbxaabxabxbbbxbabxbabxaxaaaxaabxabxaabaxb

ba

nn

nn

nn

nn

nnnnnn

nnnnnn

P

and )1()1()1( nn PPP =+ is verified.

The limiting transition matrix is

++

++=

+

=∞→

baa

bab

baa

bab

abab

ban

n

1)(lim P

if 1<x . Note that bax −−= 1 and thus, 1<x requires 10 << a and 10 << b .


(c) For the special case, 0== ba , we have

)0(1001

)0()( PPP =

=

n

n

Also, for 1== ba , we have

n

n

=

0110

)0()( PP

=⇒=

1001

)1(1 Pn

)0(0110

)2(2 PP =

=⇒=n

Continuing for all values of n, we observe that

=

=oddfor,

0110

)2(

evenfor,)0()(

n

nnP

P

P

and thus, the limiting state probabilities do not exist.

Chapter 6

Parameter Estimation 6.1 kkk ZbxaY ++= . Yk is Gaussian with mean kbxa + and variance 2σ . Since KYYY ,,, 21 L are statistically independent, the likelihood function is

∏=

==K

kkY yfbaLf

k1

)(),()( yY

( )

+−σ

−σπ

=⇒ ∑=

K

kkkK bxaybaL

1

22 )]([

21exp

2

1),(

Taking the logarithm, we have

∑=

−−σ

−π−σ−=K

kkk bxayKKbaL

1

22

)(2

12ln2

ln),(ln

Hence,

0),(ln),(ln=

∂∂

=∂

∂b

baLa

baL

∑∑ ∑ ∑

∑∑∑

−

−

−=⇒ k

kkk

kkkk

k xxx

Kx

xyK

xy

Ky

Ka

1

111ˆ

2

92

Parameter Estimation

93

∑ ∑ ∑

∑ ∑∑

−

−

=

kkk

kkkk

xxK

x

xyK

xyb

1

1

ˆ

6.2 The conditional density function is

σ−

π=σσ 2

2

2exp

21)(

yyfY

(a) 2

2

2ln2ln

21)(ln

σ−σ−π−=σσ

yyfY . Hence,

)(11)(ln22

33

2σ−

σ=

σ+

σ−=

σ∂

σσy

yyfY, which cannot be written as

[ ]⇒σ−σσ )(ˆ)( yc No efficient estimate exists for σ .

(b) )(12

1)(

)(ln22

42

2

22σ−

σ=

σ+

σ−=

σ∂

σσy

yyfY ≡ [ ]222 )(ˆ)( σ−σσ yc

Therefore, an efficient estimate for 2σ exists.

Note that [ ] [ ] ⇒σ=σ=σσ 22222ˆ YEE The estimate is unbiased.

6.3 (a) The likelihood function is )()()()( 21 21

yfyffmL YY== yY since Y1 and

Y2 are statistically independent ( ) ( )[ ]

−+−−

π=⇒ 2

22

1 321exp

21)( mymymL .

( ) ( )222

1 321

212ln)(ln mymymL −−−−π−=

103ˆ0

)(ln 21 yym

mmL +

=⇒=∂

∂

The statistic is )3(101)(ˆ 21 YYm −=Y


(b) ≡+=+=+ )3(3][ 21212211 aammamaYaYaE m if unbiased. Thus, we must have 1)3( 21 =+ aa . 6.4 (a) The likelihood function is

θ

−θ

=θ

=θ ∑∏==

θ− K

kkK

K

k

y

yefk

11Θ

1exp11)( yY

Taking the logarithm ∑=θ

−θ−=θ⇒K

kkyKf

1Θ

1ln)(ln yY

Hence, ∑∑==

=θ⇒=θ

+θ

−=θ∂

θ∂ K

kkml

K

kk y

KyKf

112

Θ 1ˆ01)(ln yY

(b) [ ] θ=θ=

∑=θ=

KK

YK

EEK

kK

11Θ1

. Therefore, the estimator is unbiased.

(c) To determine the Cramer-Rao bound, we solve

2221

322

Θ2

22)(ln

θ−=

θ−

θ=

θ−

θ=

θ∂

θ∂∑=

KKKYKEf

EK

kk

yY

(d) The variance of mlθ is

[ ] [ ]

∑∑

∑∑∑

= =

= ==

=

=

=θθ−θ=θθ−θ

K

k

K

kkk

K

k

K

k

K

kk

YYCK

YYK

EYK

EE

1 12

1 12

2

1

2

),(1

11)ˆ()ˆ(varl

l

Since the observations are independent

21]ˆvar[ yml Kσ=θ


95

where 2222 ][][]var[ θ=−==σ YEYEYy . Hence, ≡θ

=θKml

2]ˆvar[ Cramer-Rao

bound. Hence, the estimator is consistent. 6.5 (a) Y is binomial npYE =⇒ ][ . An unbiased estimate for p is

Yn

YpnYEYE 1ˆ]ˆ[ =⇒=

= .

(b) ( )2

)ˆvar(ˆε

≤ε>−Y

pYP where [ ]n

pppYEY )1()ˆ(]ˆvar[ 2 −=−= . Thus,

( ) ∞→→ε

−≤ε>− n

npp

pYp as0)1(ˆ

2. Therefore, Y is consistent.

6.6 ( )∏ ∑

= =

−

σ−

πσ=

σ

−−

σπ=

K

k

K

kkK

k mymy

f1 1

222/22

2)(

21exp

2

12

)(exp

21)( yY

Let θ=σ2

∑=

−θ

−πθ−=θK

kk myKmL

1

2)(12ln2

),(

We need 0),(ln),(ln=

∂θ∂

=θ∂

θ∂mmLmL

Applying ∑=

=⇒=∂

θ∂ K

kky

Km

mmL

1

1ˆ0),(ln

and

∑=

−=θ⇒=θ∂

θ∂ K

kk my

KmL

1

2)(2ˆ0),(ln

where ∑=

=K

kky

Km

1

1ˆ

6.7 (a) 0)(

2)(

exp21)(

2

12

2=

∂

∂⇒

σ

−−

σπ=∏

= x

xfxyxf

X

k

kX

yy

YY yields

221 ≤+ yy . Consequently,


≤++

−≤+−≥+

=

2if,)(21

2if,12if,1

ˆ

2121

21

21

yyyy

yyyy

xml

(b) [ ] AxxYYExE ml =+=+= ][21][

21)(ˆ 21Y . Therefore, )(ˆ Ymlx is

unbiased.

6.8 (a) The likelihood function is given by

∏∏=

=

θ−θ−

∑θ

=θ

=θ=K

kK

kk

y

K

k

y

ye

yef

K

kk

k

1

1

Θ

!!

)(1

yY

Taking the logarithm, we have

−θ+θ−=θ ∏∑

==

K

kk

K

kk yyKf

11!lnln)(ln y

01)(ln

1=

θ+−=

θ∂

θ∂∑=

K

kkyK

f y

∑=

=θ⇒K

kkml y

K 1

1ˆ

(b) mlθ unbiased θ=θ=

=θ⇒ ∑

=)(11]ˆ[

1K

KYE

KE

K

kkml which is true, since

∑=

K

kkY

1 is also a Poisson with parameter θK .

We have,

θ

−

θ=

θ

θθ∂∂

= ∑=

2

1

2 1)(lnK

kk KYEfEJ y


97

( )θ

=

θ+θ

θ+θ

θ−=

θ+

θ−= ∑∑

==

KKKKKKE

YYKKEK

kk

K

kk

222

2

2

12

1

2

1)(2

12

Hence, [ ]θ

≥θθ−θK)ˆ(var is the Cramer-Rao bound.

6.9 (a) The conditional density function is given by

=θ≤≤θ−

θ=θotherwise,0

,,2,1,,21

)(ΘKky

yf kkYk

L

The likelihood function is

( )

=θ≤≤θ−θ=θ

otherwise,0

,,2,1,,2

1)(

KkyL kK

L

Maximizing )(θL is equivalent to selecting θ as the smallest possible value while )(θL is positive. Hence, ky≥θ and ky−≥θ for Kk ,,2,1 L= . Note that

1y−≥θ , 2y−≥θ , …, Ky−≥θ , 1y≥θ , 2y≥θ , …, Ky≥θ is written as ),,,,,,,( 2121 KK yyyyyy LL −−−≥θ which is true if and only if

( )Kyyy ,,, 21 L≥θ . Therefore, ( )Kml yyy ,,,maxˆ21 L=θ .

y

fY(y)

1/2θ

θ-θ 0


(b) From the MLE,

( )[ ] [ ] [ ] [ ]yyPyyPyyPyyyyPyP KK ≤≤≤=≤=≤θ LL 2121 ,,,max)ˆ(

[ ]

<

θ<≤

θ

θ≥

=≤=

0,0

0,

,1

x

yy

y

yYPn

n

θ<≤θ

=⇒−

θ ynyyf n

n0,)(

1

ˆ . Hence, 1

]ˆ[0

1

+θ

=θ

=θ ∫θ −

nndy

nyyE

n

n and thus,

the unbiased estimator is θ

+ ˆ1

nn .

6.10 (a) The likelihood function is

==−

=== ∏∏ =

−

= otherwise,0

,,2,1and1,0,)1()()()( 1

1

KkypppyfpfpL k

K

k

ykyK

kkP

kk LyY

KyKKy pp −−= )1( , since the Yks are i. i. d.

Taking the logarithm )1ln()(ln)(ln pKyKpKypL −−+=⇒ and

yppKyK

pKy

ppL

ml =⇒=−−

−⇒=∂

∂ ˆ01

)(0)(ln

(b) Solving for one sample, we have

=−

=

−

∂∂

==

∂∂

=

∂∂

0,)1(

1)1ln(

1,1ln)(ln

2

2

2

2

2

yp

pp

yp

pp

pfp

y

and


99

)1(1

111

)1(1)1(1)(ln

22 pppppp

PPpf

pE

−=

−+=

−−+

=

∂∂ y

Therefore, the Cramer-Roa bound for the K independent and identical observations is

KppY )1(]var[ −

≥

6.11 ∫∞

∞−

= dxxfxyfyf XXYY )()()(

[ ]

+

σ−+

−

σ−

σπ=

+δ+−δ

σ

−−

σπ= ∫

∞

∞−

22

22

2

2

)1(2

1exp)1(2

1exp221

)1()1(21

2)(exp

21

yy

dxxxxy

[ ]

+

σ−+

−

σ−

+δ+−δ

−

σ−

==⇒2

22

2

22

)1(2

1exp)1(2

1exp

)1()1()(2

1exp

)(

)()()(

yy

xxxy

yf

xfxyfyxf

Y

XXYYX

As in Example 6.5

<−≥+

=⇒0if,10if,1

ˆyy

xmap

[ ]dx

yy

xxxyxdxyxxfx YXms ∫∫

∞

∞−

∞

∞−

+

σ−+

−

σ−

+δ+−δ

−

σ−

==2

22

2

22

)1(2

1exp)1(2

1exp

)1()1()(2

1exp)(ˆ

2

2

22

22

2

2

1

1

σ−

σ−

σ−

σ

σ−

σ

+

−=

+

−=

y

y

yy

yy

e

e

ee

ee

Therefore, mapms xx ≠ˆ


6.12 (a)

<

≥α

=α−−

0,0

0,)(Γ)(

1

x

xexrxf

xrr

X

X is a Gamma distribution with mean α

=rXE ][ and variance

2]var[

α=

rX .

(b) (i) The marginal density function of Y is

∫∫∞

∞−

∞

∞−

== dxxfxyfdxxyfyf XXYYXY )()(),()(

4444 34444 211functiondensityGamma

0

)(1

1

0

1

)1()(

)(

)(Γ

⇒

∞+α−

+

+

∞α−−−

∫

∫

++α

+α

α=

α=

dxexrr

yy

r

dxexr

xe

xyrr

r

r

xrr

xy

≥

+α

α=⇒ +

otherwise,0

0,)()( 1

yy

ryf r

r

Y

Therefore, [ ]

⇒+

+α−+α==

+

44444 344444 21ondistributiGamma

1

)1(Γ)(exp)(

)(

)()()(

rxyyx

yf

xfyxfxyf

rr

Y

XYXXY

MMSE estimate of x is

yryXExms +α+

==1][ˆ

(ii) The variance of the estimate is ]ˆ[]ˆ[]ˆvar[ 22msmsms xExEx −= where,

α=

rxE ms ]ˆ[ and )2(

)1(]ˆ[

2

22

+α

+=

rrr

xE ms . Hence, 2

1]ˆvar[2 +α

=r

rxms


101

(c) ∏ ∑=

=

−

<

≥>==

K

kk

k

K

kk

xK

kXYX

y

yxyexxyfxf

k1

1

0,0

0,0,)()( yY

In order to obtain )( yxf YX , we need

4444444 34444444 21

L

1ondistributiGamma

10

11

1

exp)(Γ

)()2)(1(

)()(),()(

⇒

=

∞−+=

+

=

∞

∞−

∞

∞−

+α

+

+α

+α

α−+−+=

==

∑∫∑

∑

∫∫

dxxyxKr

y

y

rKrKr

dxxfxfdxxff

K

kk

Kr

K

kk

KrK

kk

r

XXX yyy YYY

≥

+α

α−+−+

=⇒+

=∑

otherwise,0

0,)()2)(1(

)(1

kKrK

kk

ry

y

rKrKr

f

L

yY

4444444 34444444 21ondistributiGamma

1

11 exp)(Γ)(

)()()(

+α−

+

+α==⇒ ∑

∑

=

−+= xyxKr

y

f

xfxfxf

K

kk

Kr

K

kk

XXX y

yy

Y

YY

⇒ MMSE estimate of X is

[ ]∑=

+α

+==

K

kk

ms

y

KrXEX

1

)(ˆ yy

(ii) The variance of the estimate is

]ˆ[]ˆ[]ˆvar[ 22msmsms XEXEX −= where,

α=

rXE ms ]ˆ[ and

)1()1)((

]ˆ[2

2

++α

++=

KrKrKr

XE ms)1(

]ˆvar[2 ++α

=⇒Kr

rKX ms .


(d) xyxKrKr

yxf

K

kk

K

kk

X

+α−−++

+

+α= ∑

∑

=

=

1

1 ln)1()(Γ

ln)(ln yY

The MAP estimate is

∑=

+α

−+=⇒=

∂

∂K

kk

mapX

y

Krx

x

xf

1

)1(ˆ0)(ln yY

. Therefore,

msmap xx ˆˆ ≠

6.13 (a)

≥

=−

otherwise,00,)( nenf

n

N

≤≤

=otherwise,0

10,1)(

xxf X

∫∞

∞−

=⇒ dxxfxyfyf XXYY )()()( where

>−−

=otherwise,0

ln,)]ln(exp[)(

xyxyxyf XY

Hence, the marginal density function of Y is

≤−−

≥−−

=

∫

∫yeY

ydxxy

ydxxy

yf

0

1

0

0,)]ln(exp[

0,)]ln(exp[

)(

and


103

<−−

≥−−

==

∫

∫∫∞

∞−yeXYms

ydxxyx

ydxxyx

xyfxx

0

1

0

0,)]ln(exp[

0,)]ln(exp[

)(ˆ

or,

<

≥=

0,3

2

0,32

ˆye

yx yms

(b) map

mapxx

XXY

xx

YX

xxf

x

xyf

x

yxf

ˆˆ

)(ln)(ln)(ln

==∂

∂+

∂

∂=

∂

∂

and

>≤≤−−

=otherwise,1

lnand10,)]ln(exp[)()(

xyxxyxfxyf XXY

The quantity is maximized when 0)ln( >− xy

or,

≥≤

=0,10,ˆ

yyex

y

map

(c)

≤≤≤

≥≤≤== − 0and0,2

0and10,2)(

)()()( 2 yexxe

yxxyf

xfxyfyxf yy

Y

XXYYX

{median

ˆ

ˆ

21)()( == ∫∫

∞

∞− abs

abs

xYX

x

YX dxyxfdxyxf

Hence, for 21ˆ2

210

ˆ=⇒∫=⇒≥

∞−abs

xxdxxy

abs


and

for 2

ˆ2210

ˆ

0

2y

abs

xy exdxxey

abs

=⇒=⇒≤ ∫ −

6.14

−−

σ−

σπ=−= 2

2)1(

21exp

21)()( xyxyfxyf NXY

≤≤

=otherwise,0

20,21

)(x

xf X

20for1ˆ0)(ln)(ln

≤≤−=⇒=∂

∂+

∂

∂xyx

xxf

x

xyfmap

XXY

6.15 Van Trees1 shows that the mean square estimation commutes over a linear transformation. To do so, consider a linear transformation on Θ given by

ΘΦ D=

where D is an KL× matrix. The cost function of the random vector Φ is

∑ =−==

L

i

Tii

1

2 )()(]Φ)(Φ[]),(ˆ[ yyyy ΦΦΦΦC

and )(~ yΦ is

−

−−

=

KK Φ)(Φ

Φ)(ΦΦ)(Φ

~ 22

11

y

yy

MΦ

Following the same procedure as we did in estimating )(ˆ ymsΦ , we obtain 1 Van Trees, H. L., Detection, Estimation, and Modulation Theory, Part I, John Wiley and Sons, New

York, 1968.


105

∫∞

∞−

== φφφ dfEms )|(]|[)(ˆ| yyy YΦΦΦ

substituting (6.92) in (6.94), we obtain

]|[]|[ yDy ΘΦ EE =

and thus,

)(ˆ)(ˆ yDy msms ΘΦ =

6.16 θ can be expressed as baY +=θ . Since 0and0][][ ==+=θ ymbYaEE ,

then 0=b

From (6.108), y

yaσσ

=ρ= θθ , and from (6.110), the conditional variance is given

by

[ ] )1()(]var[ 222 ρ−σ=−−θ=θ=ℜ θYbaYEYms

Note that for this Gaussian case, the linear mean square estimate is the mean square estimate. The conditional density function is then

−σ

−θ−

ρ−πσ=θ

θθθ

θ )1(2)(

exp)1(2

1)(22

2

22 yay

yfy

Y

6.17 The BLUE of θ is given by

][][ˆ 1yyyyblue mYCCE −+θ=θ −

θ

Using )(

))()(

yf

fyfyf

Y

YY

θθ=θ

θθ , the conditional density function

)( yf Y θθ is then


≤θ≤θ−

=θθotherwise,0

121,)820(

71

)( yf Y

We compute 121and

21 2 =σ== θθθθ Cm ,

92and

34 2 =σ== nnnn Cm and

611

=+= θ ny mmm .

Since Y and θ are statistically independent, then

361122 =σ+σ== θθθθ nyyy CandCC

Hence, after substitution, the best linear unbiased estimate is yblue 113ˆ =θ .

Chapter 7

Filtering 7.1 (a) The error is orthogonal to the data ⇒

ξ∀=

ξ

ααα−−− ∫∞

∞−

,0)()()()( 0 YdhtYttsE

ξ∀ααα−ξ−=−ξ−⇒ ∫∞

∞−

,)()()( 0 dhtRttR yysy

Let ⇒ξ−=τ t

∞<τ<∞−τ∗τ=

∞<τ<∞−ααα−τ=−τ ∫∞

∞−

for)()(

for)()()( 0

hR

dhRtR

yy

yysy

Taking the Fourier transform, we have

)()()( 02 fHfSefS yytfj

sy =π−

0

2

0

0

222

02

0

22

4)]/4([/4

)()()(

)()(

)(

tfj

tfj

nnss

ss

yy

tfjsy

efN

N

efSfS

fSfS

efSfH

π−

β

π−π−

π+α+α

α=

+==⇒

44 344 21

Taking the inverse Fourier transform, we obtain

107


0

2

0

4,2)( 0

Ne

Nth tt α

+α=ββα

= −β−

⇒

(b) The minimum mean square error is

−

ααα−−−= ∫∞

∞−

)()()()( 00 ttsdhtYttsEem

αα−α== ∫∞

∞−

dhtRR syss )()()0( 0

Using Equation (7.55) βα

=∫

−−=⇒

∞

∞−df

fSfSfS

fSeyy

sysyssm )(

)()()( from

Examples 7.3 and 7.4.

7.2 τ−=τ 5.0)( eRss and )()( τδ=τnnR

(a) From Equation (7.54) the transfer function of the optimum unrealizable filter is

)()()(

)(fSfS

fSfH

nnss

ss

+=

where, 22425.0

5.0)(f

fS ssπ+

= and 1)( =fSnn .

Hence, 2222 475.0

75.0275.0

25.0475.05.0)(

fffH

π+=

π+=

t t0

βα

0

2N

h(t)

Filtering 109

Taking the inverse Fourier transform, the impulse response is

τ−≈τ 75.029.0)( eh

(b) This is similar to Example 7.5 where 5.0=α and 12

0 =N

. Hence,

)(62.0)( 12.1 τ=τ τ− ueh

(c) The minimum mean-square error for the unrealizable filter is

∫∫∫∞

∞−

∞

∞−

∞

∞− π+==

+= df

fdffSfHdf

fSfSfSfS

e nnnnss

nnssm 22475.0

5.0)()()()(

)()(

Using ∫∞

∞−

≈⇒απ

=αα+

29.0122 med

x.

For the realizable filter, the minimum mean-square error is

62.0)62.0(1)()()0(0

12.15.0 ≈τ−=τττ== ∫∫∞

τ−τ−∞

∞−

deedhRRe syssm .

7.3 )()()(

2 tNtsdt

tds=+ and thus, we have

We see that 2222 41

2)(41

1)(22 f

fSf

fS nnssπ+

=π+

= and

22412)(

11 ffS nn

π+= . Thus,

222222 414

412

412)()()(

11 ffffSfSfS nnssyy

π+=

π++

π+=+=

121+π fj

)(22 fS nn

)( fSss


Using Laplace transform, we write

12

12

)1)(1(4)(

++

−−

=+−

−=

pppppS yy , fjp π= 2

Then, 1

2)(+

=+

ppS yy and

12)(−−

=−

ppS yy . Also,

)(1

1)()(

)(

)(pB

ppSpS

pS

pS

yy

ss

yy

sy +−−

=+

==

Therefore, the transfer function and the impulse response are

)(21)(

21

)()(

)( tthpSpB

pHyy

δ=⇒==+

+

7.4 fjpf

fSfSfS nnssyy π=+π+

=+= 2,21

411)()()(

22

)1(23

)1(23

)1(23

)(2

2

−

+

−

−=

+−

+−=⇒

pp

pp

pp

pS yy

212

1

122

1

21

12/

)(

)()()()(

−++

++=

−+=⇒==

−

′′

ppppp

pS

pSppSppSpS

yy

yssssyys

The transfer function 3

122

2)()(

)(++

==+

+

ppSpB

pHyy

and thus, the impulse

response is )(22

2)( 3 tueth t−

+= .

Filtering 111

7.5 224)4/1(

3/5)(f

fS ssπ+

= , 2241

3/7)(f

fSnnπ+

=

2413/20)(p

pSss−

=⇒ , 21

3/7)(p

pSnn−

=

and )1)(41(

169)()()(

22

2

ppp

pSpSpS nnssyy−−

−=+=

)1)(21(

43)1)(21(

43pp

ppp

p++

+−−

−=

)()( pSepS ssp

syα=⇒

Also,

+

+−

=++

−= αα

− ppe

pp

pe

pS

pS pp

yy

sy

212

433/2

)43)(21(

)1(3

20

)(

)(

Knowing, 2/

212 te

p−→

+ and

α+−→

+α )(

21exp

212 t

pe p , the transfer

function is

0,43

12

)()(

)( 2/ >α++

== α−+

+

pp

epSpB

pHyy

7.6 )4/1(1

2/1)(−

=n

ss nR ,

≠=

=0,00,2

)(nn

nRnn

Taking the Z-transform, we have )2)](2/1([

234)()(

−−−

==ZZ

ZZSZS sssy and

2)( =ZSnn . Hence,

34

)2()]2/1([)186.3)(314.0(

)2)](2/1([272

34)()()(

)()(

2

32143421ZSZS

nnssyy

yyyy

ZZZZ

ZZZZZSZSZS

−+

−−−−

=−−

+−=+=


Also,

4342143421)()(

186.3372.2

)2/1(372.0

)186.3)](2/1([2

)(

)(

ZBZByy

sy

ZZZZZ

ZS

ZS

−+

−−

+−

=−−

−=

−

The pulse transfer function is then

L,2,1,0,)314.0(372.0)(and314.0

372.0)()(

)( ==−

==+

+

nnhZZS

ZBZH n

yy

(b) The mean square error is given by

61.0314.01

1)372.0(34

34

)314.0(21)372.0(

34

34)()()0(

00

=−

−=

−=−= ∑∑

∞

=

∞

= n

n

nsyssm nhnRRe

7.7 )2)](2/1([

2)(2

1)(−−

−=⇒=

ZZZZSnR ssnss

1)(0,00,1

)( =⇒

≠=

= ZSnn

nR nnnn

Hence,

)2)](2/1([15.4)()()(

2

−−+−

=+=ZZ

ZZZSZSZS nnssyy

32143421)()(

2265.4

)2/1(234.0)(

ZSZS

yy

yyyy

ZZ

ZZZS

−+

−−

−−

=⇒

and 4342143421)()(

265.4265.2

)2/1(265.0

)265.4)](2/1([2

)(

)(

ZBZByy

sy

ZZZZZ

ZS

ZS

−+

−−

+−

=−−

−=

−

Filtering 113

Hence, L,2,1,)234.0(265.0)(234.0

265.0)()(

)( ==⇒−

==+

+

nnhZZS

ZBZH n

yy

(b) The mean-square error is

n

nnsyssm nhnRRe )]5.0)(235.0[(265.01)()()0(

00∑∑∞

=

∞

=−=−=

7.01175.011265.01 =

−−=

7.8 (a) From (7.113), the optimum weights are given by

ysyy RR 10

−=ω

Computing, we have

−

−=−

1456.15208.05208.01456.11

yyR and

−

=

−

−

−=

ωω

=7853.0

8360.04458.0

5272.01456.15208.05208.01450.1

02

010ω

That is, 8360.001 =ω and 7853.002 −=ω

(b) From (7.105), the minimum mean-square error is

00002σ ωωωω RRR T

ysTT

yssme +−−=

Substituting the values and computing, we obtain 1579.0=me .

Chapter 8

Representation of Signals

8.1 (a) We have, ∫ =

πTdt

Ttk

TT0

0cos21

111

0

=∫T

dtTT

and

≠=

=

ππ∫ jk

jkdt

Ttj

TTtk

T

T

,0,1

cos2cos2

0

Therefore,

π

Ttk

TTcos2,1 are orthonormal functions.

(b) Similarly, to verify that the set functions is orthonormal in the interval

]1,1[− , we do 121

21

=∫−

T

T

dtTT

0cos2

12cos121

0

=π

=π

∫∫−

TT

T

dtT

tkT

dtT

tkTT

and

114

Representation of Signals 115

kj

TT

T

dtT

tjT

tkT

dtT

tjTT

tkT

δ=ππ

=ππ

∫∫− 0

coscos2cos1cos1

Hence, the set is orthonormal on the interval ]1,1[− .

8.2 (a) We solve 0)()(1

1

1

121 == ∫∫

−−

tdtdttsts

22)(1

0

1

1

21 == ∫∫

−

dtdtts

and

322)(

1

0

21

1

21

1

22 ∫∫∫ ==

−−

dttdttdtts

Therefore, )(and)( 21 tsts are orthogonal.

(b) )(1 ts orthogonal to 30)1(1)(1

1

23 −=β⇒=β+α+⇒ ∫

−

dtttts

)(2 ts orthogonal to 00)1()(1

1

23 =α⇒=β+α+⇒ ∫

−

dttttts .

Therefore, 23 31)( tts −= .

8.3 Note that ⇒−= )(2)( 13 tsts We have 2 independent signals.

The energy of )(1 ts is thus,

TTTdtdtdttsET

T

TT=+=−+== ∫∫∫ 22

)1(1)(2/

22/

00

211


≤≤−

≤≤

==φTtT

T

TtT

E

tst

2,1

20,1

)()( 1

1

)()()( 12122 tststf φ−= where ∫ φ=T

dtttss0

1221 )()( . Then,

21)2(1)1(

2/

2/

021

TdtT

dtT

sT

T

T+=

−−+

−= ∫∫

Tttststf ≤≤−=φ−= 023)()()( 12122

and

TtT

dt

tT

≤≤−=

−

−=φ

∫

01

23

2/3)(

0

22

(b) )()( 11 tTts φ=

)(23)(

2)( 212 tTtTts φ+φ=

)(2)( 13 tTts φ−=

Thus, the signal constellation is

T23

2T TT2−

1φ

2φ

1s

2s

3s


[ ]0,1 T=s

= TT

23,

22s [ ]0,23 T−=s

8.4 )()()(

)()( 2

2

22t

tnt

dttd

tdt

tdt

tnt

dttd

tdtd

φ

−+

φ+

φ=φ

−+

φ

)()()()( 2

2

22 tnt

dttd

tdt

tdt φ−+

φ+

φ=

where,

[ ]∫π

π−

θθ−θθ−=φ

dtnjdt

td)sin(expsin

)(

[ ]∫π

π−

θθ−θθ=φ

dtnjdt

td)sin(expsin

)( 22

2

After substitution in the differential equation, we have

∫π

π−

θθ−θ−θ−θ=φ++φ′+φ ′′ dtnjnjttnttt )]sin(exp[)sincos()( 222222

but,

∫π

π−

θθ−θθ− dtnjjt )]sin(exp[sin ππ−

θ−θθ= )]sin(exp[cos tnjjt

∫π

π−

θθ−θθ−θ+ dtnjtnt )]sin(exp[)cos(cos

∫π

π−

θθ−θ−θ+= dtnjntt )]sin(exp[cos)cos(0 2

Thus,

0)]sin(exp[)cos()( 222 =−=θθ−θθ−−=φ−φ′+φ ′′ ∫∫π

π−

π

π−

n

n

juduendtnjtnnnttt

where θ−θ= sintnu .


8.5 Given the differential system 0)()( =λφ+φ ′′ tt , 0)1()0( =φ=φ′ , we first

integrate with respect to t 0)()0()(0

=φλ+φ′−φ′⇒ ∫t

duut .

⇓

0)()()0()0()(0

=φ−λ+φ′−φ−φ ∫t

duuuttt

Using ∫∫ φ−λ−φ−λ=φ⇒=φ=φ′t

duuutduuut0

1

0

)()()()1()(0)1()0(

since ∫ φ−λ−=φ1

0)()1()0( duuu ∫∫ φ−λ+φ−=φ⇒

1

0)()1()()1()(

t

tduuuduutt

Therefore, the kernel is

≤≤−≤≤−

=1,1

0,1),(

ututut

tuk

8.6 The integral equation can be reduced to the differential equation by taking the derivative with respect to t 0)()( =λφ+φ ′′⇒ tt with ( ) 02/and0)0( =πφ′=φ .

Let tjtj ecect λ−λ +=φ 21)( . Then, 21210)0( cccc −=⇒+==φ

tjtj ejcejct λ−λ λ−λ=φ′ 21)( and 02

221 =

+λ=

πφ′

πλ−

πλ jj

eejc .

01 =⇒ c trivial solution 22

02

cos π=

πλ⇒=

πλ⇒ or

L,2,1,0,)12(2

=+π

=λ kk

Therefore, ( ) L,2,1,0,)12sin(][ )12()12(1 =+=−=φ +−+ ktkceect tkjtkj and

L,1,0,)12( 2 =+=λ kkk .


8.7 Differentiating twice with respect to t, the integral equation reduces to the differential equation

0)()( =λφ+φ ′′ tt with 0)()0( =φ=φ′ T

Let tjtj ecect λ−λ +=φ 21)( . Then,

210)0( cc =⇒=φ′ and tjtj ececT λ−λ +==φ 210)(

or, 00cos ≠⇒=λ cTc and L,2,1,0,2

)12(2

=+π

=λ⇒π+π

=λ kTkkT

Therefore, the eigenfunctions are

L,2,1,0,2

)12(cos)( =π+

=φ ktT

kct

8.8 tnBtnAtjn ω+ω=φ⇒ω±=λ⇒=ω+λ cossin)(02

For tnAtnAtut ω+ω=φ⇒≤≤ cossin)(0 21

20)0( A==φ and tnAt ω=φ sin)( 1

For tnBtnBtTtu ω+ω=φ⇒≤≤ cossin)( 21

0cossin0)( 21 =ω+ω⇒=φ TnBTnBT

Also, )(tφ continuous ⇒

unBunBunAuu ω+ω=ω⇒+φ=−φ cossinsin)0()0( 211

and

1cossincos1)0()0( 212 =ωω+ωω−ωω⇒=+φ′−−φ′ unBnunnBunnBuu

Solving for the constants, we obtain


≤≤−ωωω

ω

≤≤ωωω−ω

−=φ

TtutTnTnn

un

uttnTnnTun

t,)(sin

sinsin

0,sinsin

)(sin

)(

8.9 For ut ≤ For ut ≥

21),( ctcutk += 43),( ctcutk +=

0)0( 2 == ck 0),( 43 =+= cTcuTk

),( utk continuous ⇒ At ut = , we have 431 cucuc +=

1),0( cuukt =− 3),0( cuukt =+

13331 11),0(),0( cucucuccuukuuk tt +=+⇒+=⇒=+−−⇒

Tucuc −==⇒ 34 , and

Tuc −=11

Therefore,

≤≤+−

≤≤−

=tuut

Tu

uttT

uT

utk0,

0,),(

8.10 Taking the integral of the second order integro-differential equation

)()(),(0

2

2tduuutk

dtd T

φ−=

φ∫

we have

∫∫∫

∫∫∫

∫∫∫∫

φ−

φ−φ−=

φ−φ+φ−φ−φ+φ−φ−=

φ−φ+φ+φ−

T

t

T

t

t

T

t

T

t

t

T

t

Tut

dttduuTuduu

Tu

dtd

duuTut

Tttduuttttduu

Tut

Ttt

dtd

duuTutduutduuuduu

Tut

dtd

)()()(

)()()()()()()(

)()()()(

0

0

000


Thus,

)()(),(0

2

2tduuutk

dtd T

φ−=

φ∫

For 0)()( =λφ+φ ′′ tt , 0)()0( =φ=φ T , we have

∫ φλ=φT

duuutkt0

)(),()( a solution since 0)()()()( =λφ+λφ−⇒λφ−=φ ′′ tttt as

expected.

8.11 For Problem 8.8, we have

≥ωω

−ωω

≤ωω

ω−ω

=ut

TnntTnun

utTnn

tnuTn

utk,

sin)(sinsin

,sin

sin)(sin

),(

and

≥ωω

−ωω

≤ωω

ω−ω

=tu

TnnuTntn

tuTnn

unuTn

tuk,

sin)(sinsin

,sin

sin)(sin

),(

We verify if unuTntnuTn ω−ω=ω−ω sin)(sinsin)(sin?

.

We know that ⇒+−−= )]cos()[cos(21sinsin bababa

)](cos)([cos21)](cos)([cos

21 utTnutTntuTntuTn +−ω−−−ω=+−ω−−−ω

Thus, they are equal and therefore ),(),( tukutk = .

For Problem 8.9, we have


≥+−

≤−

=utut

Tu

uttT

uT

utk,

,),(

and

≥+−

≤−

=tutu

Tt

tuuT

tT

tuk,

,),(

We observe that T

TtutuTtt

TuT ++−

=+−=−

Therefore, ),(),( tukutk = .

8.12 Here, we have two methods. We have ( ))(),,( tutkc nn φ , that is

∫∫∫π+−

+π−

=φ=T

u

uT

nn dtT

tnTT

uutdtT

tnT

tT

uTdttutkc sin2sin2)(),(00

Solving the integrals, we obtain the desired result T

unnT

Tcn

π

π= sin

)(2

2

2

Note that we can use the results of Problems 8.10 and 8.11, that is

∫ φ=∫ φ=T

n

T

nn duutukdttutkc00

)(),()(),( from Problem 8.11. Then, from Problem

8.11, ⇒=λ

φ=∫ φ n

nT

n ct

duutuk)(

)(),(0

L,2,1,sin2)( 2

2=

π

π= n

Tun

TnTcn

8.13 We have,

≥ωω

ω−ω

≤ωω

−ωω

=tu

TmmtmtTm

tuTmm

tTmum

uth,

sinsin)(sin

,sin

)(sinsin

),(


∫

∫∫

φωω

ω−ω+

φωω

−ωω=φ′

λ⇒φλ=φ

T

t

tT

duuTmm

tmuTm

duuTmm

uTmumtduuutht

)(sin

sin)(sin

)(sin

)(sinsin)(1)(),()(00

and

( )

( ) ∫

∫

φωω−ω

ωω−

φωω−ω

ωω−φωω

ω−ωω−

φωω

ω−ωω−=φ ′′

λ

T

t

t

duuTmmuTmtmm

tTmmtTm

tmmduuTmm

umtTmm

tTmm

tmtTmmt

)(sin

)(sinsin

)(sin

)(sincos)(

sinsin)(sin

)(sin

sin)(cos)(1

2

0

2

Simplifying the above equation, we have

∫

∫

φωω

ω−ωλ+

φωω

−ωωλ+ωλφ=φ ′′

λT

t

t

duuTmm

tmuTm

duuTmm

tTmumTmtt

)(sin

sin)(sin

)(sin

)(sinsinsin)()(1

2

0

2

From Problem 8.12, )(sinsinsin)(sin tTmumtmuTm −ωω=ω−ω

φλλ+λφ−=φλ+λφ−⇒ ∫∫

TTduuuthtduuutht

00

2 )(),()()(),()(

Thus, ∫ φλ=φT

duuutht0

)(),()( is a solution of

0)()0(,0)(])[()( 2 =φ=φ=φλ+ω+φ ′′ Ttmt

In the second part of the question, we use the integral equation to obtain )(ucn in

)()(),(1

tucuth nn

n φ= ∑∞

=. Here,


π

=φT

tnT

tn sin2)( and

ω−

π

=λ∈λ 22

)(mTn

n

This gives ( ) ∫∫ φ=φ=φ=T

n

T

nnn duutuhdttuthtuthc00

)(),()(),()(),,(( from Problem

8.12. Therefore, by Problem 8.10, we haveλ

φ=φ∫

)()(),(

0

tduutuh n

T

n where,

Ttn

Ttn

π=φ sin2)( and 2

2

)( ω−

π

=λ mTn and

L,2,1,sin)(2 22

=π

ω−

π

= nT

tnm

Tn

Tcn

8.16 Let tjtj ecect λ−λ +=φ 21)( , 12210)0( cccc −=⇒+==φ and thus,

tcecect tjtj λ=−=φ λ−λ sin)( 11

Let 0, 22 >ββ=λ , then L,2,1sin)( =β=φ ktct k

αβ

−=β⇒=ββ+βα⇒=φ′+αφ kkkkk tan0cossin0)1()1(

Therefore, L,2,1,sin)( =β=φ ktt k for positive roots of αβ

−=β kktan .

Case 1: Let 21)(0 ctct +=φ⇒=λ

tctc 12 )(and00)0( =φ=⇒=φ

100)1()1( 33 −=α⇒=+α⇒=φ′+αφ cc but α is positive and thus, 0=λ is not an eigenvalue.

Case 2: 0>λ such that 0and 22 >ββ=λ . Then,


tjtj ecect β−β +=φ 21)(

12210)0( cccc −=⇒+==φ

αβ

−=β⇒=φ′+αφ tan0)1()1(

when 00,0 >αβ

−⇒>β<α

Thus, )(sin)( tt kk β=φ is α solution where kβ are consecutive positive roots of

αβ

−=βtan .

+1

-1

β

tanh β

tanh β

-β/α

-β/α

tan β

β β1 β2


Case 3: If 0<λ , let )0(2 >γγ−=λ⇒γ=λ j .

Then γ=φ sinh)(t .

From 0)1()1( =φ′+αφ , we have 0coshsinh =γγ+γα

0>γ and 00 >αγ

−⇒<α . So tt 00 sinh)( β=φ is a solution

where 0tanh >αβ

−=β .

8.17 0)()( =λφ+φ ′′ tt , 0)()0( =φ′=φ′ T

Let tjtj ecect λ−λ +=φ 21)( . Then,

210)0( cc =⇒=φ′ and 0sin)( 1 =λ=

−λ=φ′ λ−λ TceejcT TjTj

0=c trivial solution L,2,1,0sin =π

=λ⇒π=λ⇒=λ⇒ kTkkTT

2

π

=λ⇒Tk and L,3,2,1,cos)( =

π=φ k

Ttkt

and

1)(0 =φ t when 00 =λ→=k

Chapter 9

The General Gaussian Problem

9.1 (a) We first diagonalize the matrix C.

5.12/1

012/1

2/11

2

1

=λ=λ

⇒=λ−

λ−=λ− IC

=

⇒=

ba

ba

21

12/12/11

λ 111 φφC2

2and22 −

==⇒ ba

Therefore,

−

=

−=φ

11

22

2/22/2

1 .

=

=⇒=

11

22

2/22/2λ 2222 φφφC

We form the modal matrix [ ]

−

=⇒=1111

22

21 MM φφ and

−=−

1111

221M .

Therefore, the observation vector y ′ in the new coordinate system is

127


)(22and)(

22

22

22

22

22

1212112

1 yyyyyyyy

−=′+=′⇒

−==′ Myy

The mean vector 1m ′ is

)(22and)(

22

22

22

22

22

11121212111112

11

12

11 mmmmmmmm

mm

−=′+=′⇒

−=

′′

101∆ mmmm ′=′−′=′ . The sufficient statistic is

))((31))((

5.1∆

2/1∆

λ∆

)(

121112211211

2121112

1

yymmyymm

ymymymT

k k

kk

−−+++=

′′+

′′=

′′=′ ∑

=y

or 11

11

0

1

21)( mCmy −+γ=γ<

>′ T

H

H

T .

(b) 1.19.0

11.01.01

2

1

=λ=λ

⇒

=C

Then,

−=

22

22

1φ ,

=

22

22

2φ ,

−

=

−= −

22

22

22

22

,

22

22

22

22

1MM

The General Gaussian Problem 129

and )(

22

)(22

22

22

22

22

121

211

2

1

yyy

yyy

yy

−=′

+=′⇒

−==′ Myy

)(22,)(

22

111212121111 mmmmmm −=′+=′ and 1∆ mm ′=′

The sufficient statistic is

))((45.0))((55.01.1

∆9.0

∆λ

∆)(

121112211211

2121112

1

yymmyymm

ymymymT

k k

kk

−−+++=

′′+

′′=

′′=′ ∑

=y

(c) 9.11.0

19.09.01

2

1

=λ=λ

⇒

=C

Then,

−=

22

22

1φ ,

=

22

22

2φ

))((26.0))((59.11.0λ

∆)(

121112211211

2121112

1

yymmyymm

ymymymT

k k

kk

−−+++=

′′+

′′=

′′=′⇒ ∑

=y

9.2 53.247.0

29.09.01

2

1

=λ=λ

⇒

=C

−

=51.0

86.01φ ,

=

86.051.0

2φ ,

−=⇒

−

= −

86.051.051.086.0

86.051.051.086.0 1MM


Then, 2112

1

2

1 51.086.086.051.051.086.0

yyyyy

yy

+=′⇒

−

=

′′

⇒=′ Myy and

122 51.086.0 yyy −=′

12111212111111 86.051.0and51.086.0 mmmmmm +−=′+=′⇒=′ Mmm

56.2)86.051.0)(86.051.0.(

47.0)51.086.0)(51.086.0(

)(

211211

211211

yymm

yymmT

+−+−+

++=′⇒ y

)34.02.0)(34.02.0()09.183.1)(09.183.1()(

211211

211211

yymmyymmT

+−+−+++=′⇒ y

9.3 Noise ∼N ),0( 2nσ

(a) 0==⇒==

= 011,02,1

,0]|[ mmjk

HYE jk

σσ

=σ==⇒= 2

22

00 00:n

nnnkk NYH ICC

σ+σσ+σ

=+=⇒+= 22

22

11 00:

ns

nsnskkk NSYH CCC , since IC 2

ss σ= .

From Equation (9.64), the LRT reduces to the following decision rule

∑=

γ<>

σ+σσ

σ=

2

12

0

1

2222

2

)()(

kk

nsn

s

H

H

yT y

where ( )

−+η=γ 012 lnln

21ln2 CC


or, 2

2

12

222

3

0

1

2 )()( γ

σ

σ+σσ=γ<

>= ∑=k s

nsnk

H

H

yT y

(b) 21

01 == PP and minimum probability of error criterion 1=η⇒ ,

2

22

2 ln2n

ns

σ

σ+σ=γ and

2

22

2

222

3 ln)(

2n

ns

s

nsn

σ

σ+σ

σ

σ+σσ=γ

The density functions of the sufficient statistics under H1 and H0, from Equation (9.71) and (9.72), are

>σ=

σ−

otherwise,0

0,2

1)(

21

1

2/211

teHtf

t

HT

and

>σ=

σ−

otherwise,0

0,2

1)(

20

0

2/200

teHtf

t

HT

where 2221 ns σ+σ=σ and 22

0 nσ=σ . Consequently,

23

3

2 2/2/22

1nn edteP t

nF

σγ−∞

γ

σ− =σ

= ∫

and

)(2/2/2/21

223

213

3

21

21

sneedteP tD

σ+σγ−σγ−∞

γ

σ− ==σ

= ∫


9.4 (a)

σσ

σσ

==⇒=

2

2

2

2

0

000000000000

4

n

n

n

n

nK CC

σ+σσ+σ

σ+σσ+σ

=+=

22

22

22

22

1

000000000000

ns

ns

ns

ns

ns CCC

where IC 2ss σ= . Hence,

2

0

14

1

2222

2

)()( γ<

>σ+σσ

σ= ∑

=H

H

yTk

knsn

sy

or, 22

222

3

0

14

1

2222

2 )()(

)( γσ

σ+σσ=γ<

>σ+σσ

σ= ∑

= s

nsn

kk

nsn

s

H

H

yT y

The statistic is ∑=

=4

1

2)(k

kyT y .

(b) 2

22

2ln4

n

ns

σ

σ+σ=γ and

2

22

2

222

3 ln)(

n

ns

s

nsn

σ

σ+σ

σ

σ+σσ=γ .

The conditional density functions are then

>σ=

σ−

otherwise,0

0,8

1)(

20

0

2/400

tteHtf

t

HT

and


>σ=

σ−

otherwise,0

0,8

1)(

21

1

2/411

tteHtf

t

HT

where 220 nσ=σ and 222

1 ns σ+σ=σ . The probability of false alarm and detection are then

23

3

2 2/2

32/4 2

121

81

nn edttePn

t

nF

σγ−∞

γ

σ−

σ

γ+=

σ= ∫

213

3

21 2/

21

32/41 2

121

81 σγ−

∞

γ

σ−

σ

γ+=

σ= ∫ edtteP t

D

9.5 ROC of Problem 9.3 with 1=SNR , 2=SNR and 10=SNR .

0

0 0.2

0.2

0.4

0.4

0.6

0.6

0.8

0.8

1

1

0.1

0.3

0.5

0.9

0.7

PD

PF


9.6

σσ

=

σσ

= 2

2

2

2

00,

200

n

nn

s

ss CC

From (9.78), the LRT is

2

2

1

0

1

222

2

21)( γ<

>σ+σ

σ

σ= ∑

=kk

ns

s

nH

H

yTk

ky

or,

2

22222

2

0

1

222

122 )2)((

)(2)2(s

nsnsnnsns

H

H

yyσ

σ+σσ+σσγ<

>σ+σ+σ+σ

9.7 (a)

+

=ns

n

CCC

C0

00 and

+=

s

ns

CCC

C0

01

where

=

1001

nC and

=

2002

sC

=⇒

3000030000100001

0C and

=

1000010000300003

1C

From (9.88), the optimum test reduces to

3

0

14

3

22

1

2)( γ<>−= ∑∑

==H

H

yyTk

kk

ky

where 3γ is


22

222

3)(γ

σ

σ+σσ=γ

s

nsn and ( )12,lnln

21ln2 2

2

012=σ=σ

−+η=γ

n

sCC

(b) ⇒=γ 03 The test reduces to

∑∑==

<> 4

3

2

0

12

1

2

kk

kk y

H

H

y

From (9.94), (9.95) and (9.96), the probability of error is

=ε

=ε

=ε

∫ ∫

∫ ∫∞∞

∞

0101011

0 0100010

1

01

1

01

),,()|(

),,()|(

)(

tTT

t

TT

dtdtHttfHP

dtdtHttfHP

P

where, 6/1

11 18

1)( tT etf −= and 2/

00

0 21)( t

T etf −= . Therefore,

41

361)(

101

00

2/

0

6/1 ==ε ∫∫ −

∞−

ttt dteedtP .

9.8 (a)

=

11.05.01.019.05.09.01

C

0741.29153.00105.0

011.05.0

1.019.05.09.01

3

2

1

=λ=λ=λ

⇒=λ−

λ−λ−

=λ− IC

−−=⇒=

3009.06249.0

7204.0

1111 φφλφC ,

Similarly,


−−

=8750.04812.00519.0

2φ , and

=

3792.06148.06916.0

3φ

The modal matrix is

−−−−

=3792.08750.03009.06148.04812.06249.06916.00519.07204.0

M

and

3213

3212

3211

38.0875.03.0615.048.0625.0

69.0052.072.0

yyyyyyyy

yyyy

++−=′+−−=′

+−=′⇒=′ Myy

Similarly, 11 Mmm =′ and then we use

∑∑==

′′=

′′=

3

1

3

1 λλ∆

)(k k

kk

k k

kk ymymT y

(b)

=

18.06.02.08.018.06.06.08.018.02.06.08.01

C

In this case, 1394.01 =λ , 0682.02 =λ , 9318.2and8606.0 43 =λ=λ

whereas,

=

−−

=

−−

=

−

−

=

4445.05499.05499.04445.0

and

6768.02049.02049.06768.0

,

5499.04445.04445.0

5499.0

,

2049.06768.0

6768.02049.0

4321 φφφφ

and the modal matrix is


−−−−−−

=

44.068.055.02.055.02.044.068.055.02.044.068.044.068.055.02.0

M

Chapter 10

Detection and Parameter Estimation 10.1 (a) tts π= 2cos)(1

π

π−

π

π=

π

+π=3

2sin)2sin(3

2cos)2cos(3

22cos)(2 tttts

π

π+

π

π=

π

−π=3

2sin)2sin(3

2cos)2cos(3

22cos)(3 tttts

21

21

≤≤− t

Also, ⇒==π ∫∫−− 2

1)()2(cos2/1

2/1

21

2/1

2/1

2 dttsdtt21

21

,2sin22)(

,2cos22)(

2

1≤≤−

π=φ

π=φt

tt

tt

Therefore,

)(2)( 11 tts φ=

)(23)(

22)( 212 ttts φ−φ−=

)(23)(

22)( 213 ttts φ+φ−=

138

Detection and Parameter Estimation

139

(b) The decision space is

10.2 2112

21 for1)(],[ TtTTT

tfTTt T <<−

=⇒∈

)()(:)()()(:

0

1

tNtYHtNtstYH

=+=

A

t

s(t)

T1 t0 t0+T T2

1φ

2φ

1s

2s

3s

Decide

Decide

Decide

2

2/2−

2/3

2/3−

3s

2s

1s

Received signal

)(1 tφ

)(2 tφ

Choose largest

variable

∫−

2/1

2/1

∫−

2/1

2/1


where 02 /

0

1)( NnN e

Nnf −

π= .

The problem may be reduced to

Under H0, we have

∫==2

0

)()()( 1

T

tdttNtNtY

0)]([)]([2

0

1 ==⇒ ∫T

tdttNEtNE and

∫ ∫∫ ∫ =

=

2

0

2

0

2

0

2

0

2121221121 )]()([)()()]([

T

t

T

t

T

t

T

t

dtdttNtNEdttNdttNEtNE

where

≠

==

21

210

21,0

,2)](]([

tt

ttN

tNtNE

∫ ∫ ≡−=−δ=⇒2

0

2

0

)]([var)(2

)(2

)]([ 1020

212102

1

T

t

T

t

tNtTN

dtdtttN

tNE

Under H1, we have )()()( 012

2

0

tTAdtAdttstsT

t

T

t

−=== ∫∫ . Then,

tNtTAtYHtNtYH

()()(:)()(:

101

10

+−==

The LRT is

∫2

0

T

t

LRT )()()( tNtstY += )()()( 11 tNtstY +=


141

)(

)(),(

)(

)()(Λ

0

11,

0

1

0

11

0

1

Hyf

dtHtfHtyf

Hyf

Hyfy

HY

HTHTY

HY

HY∫

==

−−

−π

−

−−−

−−π

=∫

)(exp

)(1

1)(

)]([exp

)(1

020

2

020

12020

20

020

2

1

tTNy

tTN

dtTTtTN

ttAy

tTN

T

T

η<>

−−

−−−

−−

=∫

0

1

)(exp

)()]([

exp1

020

2

020

20

12

2

1

H

H

tTNy

dttTN

ttAyTT

T

T

[ ] η<>

+−

−−

−++

−

−= ∫

0

1

)1(2exp

)(exp

)()2(

exp1 2

1

022

020

2

020

002

12H

H

dttAtAtA

tTNy

tTNAtAty

TT

T

T

[ ]44444 344444 21

γ

∫ +−

−η<>

−++

−⇒2

1

)1(2exp

)(

0

1

)()2(2

exp

022

12

020

002

T

T

dttAtAtA

TT

H

H

tTNAtAty

Therefore, 2

)2(ln)( 00020

1

0

2 AtAttTN

H

H

y+−γ−

<> .

10.3 From (10.85), the probability of error is

α=ε

0

221)(

NQP where 0121 2 EEEE −+=α .

and


TAdttsET

2

0

211 )( == ∫

TAdttsET

2

0

200 )( == ∫

TATAdttstsEET

22

02121 2

1,2

)()( =α⇒=ρ==ρ ∫ and

=ε

0

2221)(

NTAQP

The optimum receiver is shown below

10.4 We have,

)()(:)()()()()()(

:

0

2

11

tWtYHtWtstYtWtstY

H

=

+=+=

Under H1, we have 110

111 )()]()([ WEdttstWtsYT

+=+= ∫

220

222 )()]()([ WEdttstWtsYT

+=+= ∫ .

∫T

0

∫T

0

0

0

1

1 Y

H

H

Y <>

H1

H0

Y(t) )(1 ts

)(2 ts

Y1

Y2


143

The problem reduces to:

+=

00

11

111

1

:::

HWHWHWE

Y

+=

00

12

122

2

:::

HWHWHWE

Y

Under H0, we have

.2,1,)()( 00

=≡== ∫ kWWdttstWY k

T

kk

The LRT is

)(

)(),()(),(

)(

)(

0

221,111,

0

1

0

2111

0

1

Hf

sPsHfsPsHf

Hf

Hf

H

SHSH

H

H

y

yy

y

y

Y

YY

Y

Y +=

where

−−

π=

0

211

0111,

)(exp1),(

111 NEy

NsHyf SHY

∫T

0

∫T

0

Y(t) )(1 ts

)(2 ts

Y1

Y2


−

π=

0

21

0211, exp1),(

211 Ny

NsHyf SHY

−

π=

0

22

0112, exp1),(

212 Ny

NsHyf SHY

−−

π=

0

222

0212,

)(exp1),(

212 NEy

NsHyf SHY

and

2,1,exp1)(0

2

000

=

−

π= k

Ny

NHyf k

kHYk

Therefore, the LRT becomes

0220

21

0210

22

//

0

0

222/

0

/

0

211

0

1

21)(

exp121)(

exp1

NyNy

NyNy

eeN

NEy

eN

eN

EyN

−−

−−

π

−−

π+

−−

π

= η<>

−−

−−

0

1

2222

011

21

0)2(1exp)2(1exp

21

H

H

EyEN

EyEN

When 1=η , the LRT becomes

2exp2

expexp2

exp

0

1

0

22

20

2

0

21

10

1

H

H

NE

yNE

NE

yNE

<>

−

−+

−

−

The optimum receiver may be


145

10.5 (a) The probability of error is given by

α=ε

0

221)(

NQP where

0101 2 EEEE ρ−+=α

49.0)1(21 4

2

0

21 =−== −−∫ edteE t

The signals are antipodal

=ε=α⇒−=ρ⇒

0

92.321)(and96.11

NQP .

(b) The block diagram is shown below with )(43.1)( 1 tst ≈φ .

10.6 At the receiver, we have

TttWtstYHTttWtstYH

≤≤+=≤≤+=

0,)()()(:0,)()()(:

22

11

∫T

0 1

0

0

1

PP

H

H

<>Y(t) y1

)(1 tφ

H1

H0

]exp[ ⋅

]exp[ ⋅

1Y

1Y

0

12NE

−0/2

1 NEe−

∫T

0

∫T

0

2

0

1

H

H

<>∑

Y(t) H1 )(1 ts

)(2 ts

0

22NE

− 0/22 NEe−

H0


221TEE == and )(and)(0)()( 21

02112 tstsdttsts

T⇒==ρ ∫ are uncorrelated.

The receiver is

where .2,1,)(2)()( ===φ kts

TE

tst k

k

kk

The observation variables Y1 and Y2 are then

=φ

+=φ

=

∫

∫

10

10

110

11

1

)()(:

)()(:

WdtttYH

WEdtttYH

YT

T

+=φ

=φ

=

∫

∫

220

10

20

11

2

)()(:

)()(:

WEdtttYH

WdtttYH

YT

T

This is the general binary detection case. Then,

=

=

=

22

212

12

111

2

1 and,ss

ss

YY

ssY

∫T

0

∫T

0

H1

H0

Y(t) )(1 tφ

)(2 tφ

Choose largest

Y1

Y2


147

The conditional means are

112

11111 0][ sYm =

=

==

ssEHE

222

21

222

0][ sYm =

=

==

ss

EHE

)(1 ts and )(2 ts uncorrelated ⇒ the covariance matrix is

CCC ==

= 2

0

01 2/0

02/N

N

and the probability of error is

=

α=ε

00 212

21)(

NTQ

NQP where EEE 221 =+=α

10.7 At the receiver, we have

TttWtsEtYH

TttWtsEtYH

≤≤+=

≤≤+=

0,)()()(:

0,)()()(:

222

111

∫T

0

∫T

0

Y(t) )(1 tφ

)(2 tφ

Y1

Y2


with ∫=φT

dtE

tst

0 1

11

)()( and ∫∫ ==φ

TTdttsdt

tst

02

0

22 )(2

2/1

)()(

Since the signals are orthogonal, we can have a correlation receiver with two orthogonal functions or with one orthonormal function )(∆ ts given by

−=

+

−= )(

21)(

23)()(

)( 2121

2211∆ tsts

EE

tsEtsEts

We obtain the sufficient statistic as follows

The conditional means are

32)(

21)(

32)]()([]|)([

02111 =

−+= ∫

TdttststWtsEHyTE

61)(

21)(

22)()(

21]|)([

02122 −=

−

+= ∫

TdttststWtsEHyTE

The noise variance is 2/1]|)(var[ 0 =HyT . Hence, the performance index is

2d ≜ { }3

2/1)6/13/2(

]|)(var[|)([]|)([ 2

0

201 =

+=

−HyT

HyTEHyTE

The probabilities of false alarm and detection are

=

=

23

2QdQPF

∫T

0

)(∆ tS

y(t) T(y)


149

−=

−=

23

2QdQPD

and thus, the achievable probability of error is

=

π=ε ∫

∞−

23

21)(

2/3

2/2QdxeP x

(b) In this case, the two signals will have the same energy E and thus,

EdEEd 242/1

22 =⇒==

From 2

1

23

23)(

2)(

=⇒==

=ε −QEEQdQP

10.8 We need to find the sufficient statistic. Since )(1 ts and )(2 ts are orthogonal, let

21

22111

)()()(

EE

tsEtsEt

+

−=φ

Then,

+

−

+

−+

=φ=

∫

∫∫

T

T

T

dtEE

tsEtsEtWH

dtEE

tsEtsEtWtsEH

dtttyY

0 21

22110

0 21

22111

011

)()()(:

)()()]()([:

)()(

Decision region

E1

E2

E

E


Y1 is Gaussian with conditional means

001 0]|[ mHYE ==

and

121

22

21

11

0

22

21

22

0

21

21

11

01222

0111111

)()(

)()()()(]|[

mEE

EP

EEE

P

dttsEE

EPdtts

EEEP

dtttsEPdtttsEPHYE

TT

TT

=+

−+

=

+−

+=

φ−φ=

∫∫

∫∫

The variance is 2/0N and thus,

−

π=

0

21

001| exp1)|(

01 Ny

NHyf HY

−−

π=

0

211

011|

)(exp1)|(

11 Nmy

NHyf HY

Applying the likelihood ratio test, taking the natural logarithm and rearranging terms, we obtain

22ln 1

1

0

0

1

1m

mN

H

H

y +η

<>

For minimum probability of error, 1=η and the decision rule becomes

21

2211

0

1

122 EE

EPEPm

H

H

y+

−=<

>

The optimum receiver is


151

10.9 (a) The energy BdttdttdttAdttsETTTT

k +

φ+φ+φ== ∫∫∫∫

0

23

0

22

0

21

2

0

2 )()()()(

where B is the sum involving terms of the form

kjdtttT

jj ≠φφ∫ ,)()(0

But the φs are orthonormal 0=⇒ B and thus,3

3 2 EAAE =⇒= .

(b) The signals 7,,1,0),( L=ktsk , span a 3-dimentional space. The coefficients are

kk

T

kk

T

kk

WsdtttWts

kdtttyy

+=φ+′=

=φ=

∫

∫

0

0

)()]()([

3,2,1,)()(

such that

=

3

2

1

yyy

y ,

=

3

2

1

WWW

W and

=

3

2

1

k

k

k

k

sss

s

Hence,

=

111

30Es ,

−=

111

31Es ,

−=11

1

32Es ,

−−=

11

1

33Es ,

−=

111

34Es ,

∫T

02

1

0

1

m

H

H

<>y(t) y1

)(1 tφ

H1

H0


−

−=

111

35Es ,

−−

=111

36Es ,

−−−

=111

37Es .

Since the criterion is minimum probability of error, the receiver is then a "minimum distance" receiver.

The receiver evaluates the sufficient statistic

7,,1,0,)]()([0

22L=−=−= ∫ kdttstyT

T

kkj sy

and chooses the hypothesis for which jT is smallest.

Since the transmitted signals have equal energy, the minimum probability of error receiver can also be implemented as a "largest of " receiver. The receiver computes the sufficient statistic

7,,1,0,)()(0

L=== ∫ kdttytsTT

kTkj ys

and chooses the hypothesis for which jT is largest.

(c)

1φ

2φ

1s

2s

3s4s

5s

6s

7s3φ 0s


153

Using "minimum distance" or "nearest neighbor", the decision regions are

0,0,00,0,00,0,00,0,00,0,00,0,00,0,00,0,0

3217

3216

3215

3214

3213

3212

3211

3210

<<<><<<><>><<<>><><>>>>>

yyyHyyyHyyyHyyyHyyyHyyyHyyyHyyyH

(d) The probability of error is

)()()()( 0

7

00

7

0HPPHPHPPP

jj

jjj ε=ε=ε=ε ∑∑

==

1Y , 2Y and 3Y are independent Gaussian random variables with conditional means

3][][][ 030201

EHYEHYEHYE ===

and conditional variances

2]var[]var[]var[ 0

030201N

HYHYHY ===

Therefore, ]0,0,0[1)|()( 3210 >>>−=ε=ε YYYPHPP

3

0

3

0 0

2

0

321

321

)3/(exp11

)0()0()0(1

−−=

−−

π−=

>>>−=

∫∞

NEQ

dyNEy

N

YPYPYP


10.10 (a) We observe that the dimension of the space is 2 and that we have 4

signal levels per axis ⇒ Basis functions { }21 ,φφ such that 0)()(0

21 =φφ∫T

dttt

and

1)()(0

22

0

21 =φ=φ ∫∫

TTdttdtt

The receiver is then

with

tfT

t

tfT

t

02

01

2sin2)(

2cos2)(

π=φ

π=φ

∫T

0

∫T

0

Y(t) )(1 tφ

)(2 tφ

1 Threshold

Threshold 4-level signal

4-level signal 2

3

4


155

(c) From (b), we observe that the probability of a correct decision is

)along()along()alongdecisioncorrectandalongdecision(correct)(

21

21

φφ=φφ=

cPcPPcP

where, )along( 1φcP is, from the figure below, given by

[ ]

−=−+−+−+−=

′=φ ∑=

qqqqq

PcPk

k

461)1()21()21()1(

41

)decisioncorrect(41)along(

4

11 s

where,

=

02NdQq .

1φ

2φ

1s′ 2s′ 4s′1φ

d d

3s′


Similarly,

−=φ qcP

461)along( 2 . Therefore, the probability of a correct

decision is

2

461)(

−= qcP

and the probability of error is

2

49

93)(1)( qqcPP −=−=ε

10.11 From (10.104), we have

Mj

T

Tjj

jTj

Tj

Tj

Tj

Tjj

,,2,12

2))(()(22

22

L=−+=

+−=−−=−=

yssy

ssysyysysysyy

For equal energy ⇒ 2R and 2

js are common to all hypotheses⇒ Minimizing

)(2 yjT is equivalent to maximizing ysTj . Therefore, the receiver computes the

sufficient statistic

MjdttytsTT

jTj

Tj ,,2,1,)()()(

0

L=== ∫ysy

and chooses the hypothesis having the largest dot product. The "Largest of " receiver is


157

"Largest of " receiver

10.12 We have

)()(:)()()(:

0

1

tWtYHtWtAstYH

=+=

where

∫=T

dttstWW0

1 )()(

AHYE =][ 11

2]var[

2]2[][ 0

1102

12

12

12

1N

HYN

AAWWAEHYE =⇒+=++=

A unknown ⇒ H1 is a Composite hypothesis and

∫T

0

)(ts

Y(t)

==+

110

111

::

YWHYWAH

∫T

0

∫T

0

y(t)

)(1 ts

)(tsM

Choose largest

decision variable

∫T

0

)(2 ts

Decision

T1

T2

TM


)(

),(max)(Λ

0

11,

0

111

Hyf

Hyfy

HY

HY

g

θ=

θθ

We need the estimate A of A such that ⇒=∂

∂0

)(ln

a

ayf AY the ML estimate is

YA =ˆ , the observation itself; i. e, where the distribution is maximum. Hence,

−

π

−−

π==

0

2

0

0

2

0

0

1,

exp2

1

)(exp

21

)(

),()(Λ

0

1

Ny

N

Nay

NHyf

Hayfy

HY

HAYg

η<>

−−+−=⇒

0

1

222

0)2(1exp)(Λ

H

H

yayayN

yg

but 1=η and 1exp)(Λˆ

0

1

0

2

H

H

Ny

yya g <>

−=⇒=

or 0

0

1

0

2

H

H

Ny

<> . Therefore, always decide H1 since 0/ 0

2 >Ny .

10.13

Y1 is a sufficient statistic and thus,

∫T

0

)(tφ

Y(t) ∫ φ=T

dtttYY0

1 )()(


159

111 )( YWtEY ⇒+φθ

= is Gaussian with mean θ/E and variance 2/0N .

The conditional density function becomes

θ−−

π=θ

0

21

01Θ

)]/([exp1)(

1 NEy

NyfY

Hence,

θ−

θ∂∂

=⇒=θ∂

θ∂ 2

10

1Θ 100)(ln

1 EyN

yfY

or θ

=Ey1 . Thus,

1

ˆyE

ml =θ and the optimum receiver is shown below.

10.14 The density function of θ is 22 2/

Θ2

1)( θσθ−

θσπ=θ ef . Hence, from the

MAP equation, we have

020)(ln)(ln

2210ˆˆ

Θ1Θ1 =

σ

θ−

θ

θ−−⇒=

θ∂θ∂

+θ∂

θ∂

θθ=θ

EEyN

fyf

map

Y

022

ˆ01

02

4=

−θ+

σ

θ⇒

θ=θθmap

NEEy

N

∫T

0

Ets /)(

y(t) Inverter

E

y1 1

1y

mlθ


As ∞→σθ , we have

022

ˆ01

0=−θ

θ=θmapN

EEyN

Therefore, mlmap yE

θ==θ∞→σθ

ˆˆlim1

2.

10.15 The ML equation is given by

0),(

)],()([2

00=

θ∂θ∂

θ−∫ dtts

tstyN

T

where, )cos(),( θ+ω=θ tAts c and )sin(),(

θ+ω−=θ∂θ∂

tAts

c .

Substituting into the ML equation, we have

0)sin()]cos()([2

00=θ+ωθ+ω−− ∫ dtttAty

NA

c

T

c

∫∫∫ θ+ω=θ+ωθ+ω=θ+ω⇒T

c

T

ccc

TdttAdtttAdttty

000

)](2sin[2

)sin()cos()sin()(

Assuming many cycles of the carrier within [0,T], the integral involving the double frequency terms is approximately zero. Hence,

∫ ≈ωθ+ωθT

cc dtttty0

0]cossinsin)[cos(

Therefore, ∫∫ ωθ−=ωθT

c

T

c dtttytdtty00

cos)(sinsin)(cos


161

∫

∫

ω

ω

−=θ⇒T

c

T

c

tdtty

tdtty

0

0

cos)(

sin)(tan

or,

ω

ω

−=θ

∫

∫−

T

c

T

c

ml

tdtty

tdtty

0

01

cos)(

sin)(tanˆ .

(b) Indeed, it can be shown that mlθ is unbiased and thus, we can apply the Cramer-Rao lower bound. The Cramer-Rao inequality is given by

∫

θ∂θ∂

≥θTml

dtts

N

0

20

),(2

]ˆvar[

with

2)22cos(

22)]22cos(1[

2

)(sin),(

)sin(),(

2

0

22

0

20

22

0

2

TAdttATAdttA

dttAdtts

tAts

T

c

T

c

T

c

T

c

∫∫

∫∫

≈θ+ω−=θ+ω−=

θ+ω=

θ∂θ∂

⇒θ+ω−=θ∂θ∂

Hence,

TAN

ml 20]ˆvar[ ≥θ then 1

0

2<<

NTA .

10.16 (a) The matched filters to )(1 ts and )(2 ts are )()( 11 tTsth −= and )()( 22 tTsth −= , respectively, as shown below.


(b) The filters outputs as a function of time when the signal matched to it is

the input are the resulting convolutions )(1 ty and )(2 ty as shown below.

t

t

s1(t)

h1(t)=s1(T-t)

0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7

t

y1(t)

0 1 2 3 4 5 6 7 8

1.0

1.0

1

2

t

h1(t)=s1(T-t)

0 1 2 3 4 5 6 7

t

h2(t)=s2(T-t)

0 1 2 3 4 5 6 7

0.5

-0.5

1.0


163

t

t

s2(t)

h2(t)=s2(T-t)

0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7

t

y2(t)

0 1 2 3 4 5 6 7 8

0.5

-0.5

0.5

-0.5

1.75

-0.5 9 10

11 12 13 14 10


(c) The output of the filter matched to )(2 ts when the input is )(1 ts is )()()( 21 thtsty ∗= as shown below.

10.17 (a) The signals )(1 ts and )(2 ts are orthonormal

Hence,

≤≤

=−=otherwise,0

2/,/2)()( 11TtTTtTsth

and

≤≤=−=

otherwise,02/0,/2)()( 22

TtTtTsth .

0 1 2 3 4 5 6 7 8 9 10 11

0.5

-0.5

1.0

-1.0

t

)()()( 21 thtsty ∗=

2/T T t

)(1 ts

2/T

2/T Tt

)(2 ts

2/T


165

(b) The noise free output of the matched filters is 2,1,)()()( =∗= kthtsty kkk . Hence,

Note that we sample at Tt = and thus, 1)(1 =Ty and 0)(2 =Ty .

(c) The SNR at the output of the matched filter is

000

2

022

2/ NNE

NdSNR === since 1=E .

10.18 tts cω= cos)(1 , then the signal energy is 2/TE = , and the first basis

function is tTt cω=φ cos/2)(1 . Consequently, the first coefficient in the Karhunen-Loève expansion of )(tY is

φ

φ+θ+ω

=φ=

∫

∫∫ T

T

cT

dtttWH

dtttWtAH

dtttYY

010

011

011

)()(:

)()]()cos([:

)()(

Then, we select a suitable set of functions L,3,2),( =φ ktk , orthogonal to )(1 tφ . We observe that for 2>k , we always obtain kW independently of the hypothesis. Only 1Y depends on which hypothesis is true. Thus, 1Y is a sufficient statistic.

2/T T t

)(1 ty

1

2/T Tt

)(2 ty

2/3T

1

0

2/T T t

)(1 th

2/T

2/T Tt

)(2 th

2/T


1Y is a Gaussian random variable with conditional means

θ=

θ=θ coscos

2],,[ 11 EaTaEHaYE

0][],,[ 101 ==θ WEHaYE

and variances

2],,var[],,var[ 0

0111N

HaYHaY =θ=θ

The conditional likelihood ratio is given by

θ−

θ=

θ

θ=θ 22

01

001,Θ,

11,Θ,cos1expcos2exp

),,(

),,(],)([Λ

01

11 EN

EyNHayf

Hayfaty

HAY

HAY

)()(),( ΘΘ, θ=θ fafaf AA since A and Θ are independent. Hence,

∫ ∫ θ θθ=A

dadatyty ],)([Λ)]([Λ

Substituting for ],)([Λ θaty and ),(Θ, θaf A into the above integral, the decision rule reduces to

η<>

+σ

σ

+σ=

0

1

21

02

0

2

02

0

)2(2

exp2

)]([Λ

H

H

yNNN

Nty

a

a

a

or,

γ<>

0

1

21

H

H

y

with


167

0

02

20

20 )2(

ln2

)2(N

NNN a

a

a +ση

σ

+σ=γ

(b) The receiver can be implemented as follows

10.19 Under hypothesis H0, no signal is present and the conditional density function was derived in Example 10.7 to be

σ

+−

πσ=

2

22

202

exp2

1),(0

scscHYY

yyHyyf

sc

Using the transformations θ= cosrYc and θ= sinrYs then,

σ−

σ=

2

2

202

exp)(0

rrHrf HR

and the probability of false alarm is

γ−=

σ

γ−=

σ−

σ= ∫

∞

γ 022

2

2exp

2exp

2exp

NdrrrPF

The probability of detection is

∫= A ADD daafaPP )()(

where,

∫∞

γ

σ

+−

σ= dr

ATrraPD 2

22

2 2)2/(

exp)(

∫T

0

γ<>

0

1

H

HY(t)

tT cωcos2

Squarer H1

H0


Solving for the expressions of )(aPD and )(af A , and solving the integral, we obtain

σ+

γ−=

)(2exp

20

2

aD

TNTP

Expressing DP in terms of FP , we obtain

( ) TNN

FD aPP 20

0

σ+=

10.20 (a) 0)()]([)()(][00

=φ=

φ= ∫∫

T

k

T

kk dtttNEdtttNENE

and

kkkN

NEN λ+==2

][]var[ 02

(b) 2/0N is the variance of the white noise process. kλ may be considered as the variance of the colored noise. That is, we assume that the variance is composed of the white noise variance and the colored noise variance. The white noise coefficients are independent; the others are Karhunen-Loève coefficients, which are Gaussian and uncorrelated ⇒ Independent.

(c) ≡−δ=′′ )(2

),( 0 utN

utc nn white Gaussian noise

0][ =′⇒ kNE and 2

][]var[ 02 NNEN kk =′=′

10.21 (a) WtN =)(1 has one eigenfunction. 0N is the component to filter ⇒ It cannot be whitened since the process has no contribution in any other direction in the signal space.

(b) In this case, the noise )()()( 21 tNtNtN += can be whitened by:


169

That is, the whitening is performed by an amplifier and a dc canceller.

Delay T

20

2

)2/( w

w

N σ+σ

)(tN )(tN ′

0/2 N

T/1

∑−

+

signal detection and estimation (mourad barkat, solution manual)

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