signal detection and estimation (mourad barkat, solution manual)
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Signal Detection and Estimation, Second Edition, Mourad Barkat, Solution ManualTRANSCRIPT
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
Signal Detection and Estimation 4
(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
Probability Concepts 5
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,
Signal Detection and Estimation 6
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
Probability Concepts 7
(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)
Signal Detection and Estimation 8
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
Probability Concepts 9
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
Signal Detection and Estimation 10
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 .
Probability Concepts 11
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 .
Signal Detection and Estimation 12
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
Probability Concepts 13
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
Signal Detection and Estimation 14
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
Probability Concepts 15
≥−
=
≤≤−
=
=−
−
−
−−
∫
∫
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)
Signal Detection and Estimation 16
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
Probability Concepts 17
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)(
Signal Detection and Estimation 18
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)(
Probability Concepts 19
.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
Signal Detection and Estimation 20
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
Probability Concepts 21
.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 .
Signal Detection and Estimation 24
(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)
Signal Detection and Estimation 26
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
Signal Detection and Estimation 28
(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
Signal Detection and Estimation 30
)(')(
)(')(
)(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
=
Signal Detection and Estimation 32
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
Signal Detection and Estimation 34
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
Signal Detection and Estimation 36
)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
Signal Detection and Estimation 38
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
Signal Detection and Estimation 40
(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
Signal Detection and Estimation 42
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
Signal Detection and Estimation 44
(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
Signal Detection and Estimation 46
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
Signal Detection and Estimation 48
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
Signal Detection and Estimation 50
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
Signal Detection and Estimation 52
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
Signal Detection and Estimation 54
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
Discrete Time Random Processes 59
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
Signal detection and estimation 60
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
Discrete Time Random Processes 61
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)
Signal detection and estimation 62
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
Discrete Time Random Processes 63
(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
Signal detection and estimation 64
(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
Discrete Time Random Processes 65
)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
Signal detection and estimation 66
(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
Discrete Time Random Processes 67
=
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
Signal detection and estimation 68
(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.
Discrete Time Random Processes 69
(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
Statistical Decision Theory
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
Signal detection and estimation 74
(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
Statistical Decision Theory
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
Signal detection and estimation 76
(a) By symmetry, we observe that the thresholds are –γ and γ, and
)|(error)|(error 21 HPHP =
( )∫∞
γ−
σ
+−
σπ= dy
AyHP
2
2
12
exp21)|(error
( )∫γ
∞−
σ
−−
σπ= dy
AyHP
2
2
22
exp21)|(error
∫
∫∫∞
γ
∞
γ
γ−
∞−
σ−
σπ=
σ−
σπ+
σ−
σπ=
dyy
dyy
dyy
HP
2
2
2
2
2
2
0
2exp
212
2exp
21
2exp
21)|(error
But )|(error)|(error 21 HPHP = and hence,
σ
−−+
σ−+
σ
+−
σπ=
∫
∫∫
γ
∞−
∞
γ
∞
γ−
dyAy
dyy
dyAy
P
2
2
2
2
2
2
2)(
exp
2exp2
2)(
exp21
31(error)
-A γ
H0
A 0 -γ
H1 H2
y
)|( 00| Hyf HY )|( 22| Hyf HY)|( 11| Hyf HY
Statistical Decision Theory
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
Signal detection and estimation 78
(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
Statistical Decision Theory
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
Signal detection and estimation 80
[ ]{ }
∞<≤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−−
Statistical Decision Theory
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
Signal detection and estimation 82
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).
Statistical Decision Theory
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 .
Signal detection and estimation 84
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
Statistical Decision Theory
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
Signal detection and estimation 86
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
Statistical Decision Theory
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 ∫∫∫∞ −−
∞−
−
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
∫∫
∫∫∫
∞ −−
∞−
−
∞ −−
∞−
−−
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
Signal detection and estimation 88
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
∫∫
∫∫∫
∞ −∞
−
−
−
−
−
−
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
Statistical Decision Theory
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.
Signal detection and estimation 90
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
Statistical Decision Theory
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.
Signal detection and estimation 70
−
−=
+−−+
+++−
+++−
+−−+
=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 .
Discrete Time Random Processes 71
(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
Signal detection and estimation 94
(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
Parameter Estimation
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,
Signal detection and estimation 96
≤++
−≤+−≥+
=
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
Parameter Estimation
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
Signal detection and estimation 98
(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
Parameter Estimation
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
Signal detection and estimation 100
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
Parameter Estimation
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 .
Signal detection and estimation 102
(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
Parameter Estimation
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
Signal detection and estimation 104
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.
Parameter Estimation
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
Signal detection and estimation 106
≤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
Signal detection and estimation 108
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
Signal detection and estimation 110
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
−+
−−−−
=−−
+−=+=
Signal detection and estimation 112
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
Signal detection and estimation 116
≤≤−
≤≤
==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
Representation of Signals 117
[ ]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 .
Signal detection and estimation 118
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 .
Representation of Signals 119
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
Signal detection and estimation 120
≤≤−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
Representation of Signals 121
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
Signal detection and estimation 122
≥+−
≤−
=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
),(
Representation of Signals 123
∫
∫∫
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,
Signal detection and estimation 124
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,
Representation of Signals 125
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
Signal detection and estimation 126
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
Signal detection and estimation 128
)(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
Signal detection and estimation 130
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
The General Gaussian Problem 131
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
= ∫
Signal detection and estimation 132
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
The General Gaussian Problem 133
>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
Signal detection and estimation 134
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
The General Gaussian Problem 135
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,
Signal detection and estimation 136
−−
=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
The General Gaussian Problem 137
−−−−−−
=
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
Signal detection and estimation 140
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 +=
Detection and Parameter Estimation
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
Signal detection and estimation 142
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
Detection and Parameter Estimation
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
Signal detection and estimation 144
−
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
Detection and Parameter Estimation
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
Signal detection and estimation 146
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
Detection and Parameter Estimation
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
Signal detection and estimation 148
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)
Detection and Parameter Estimation
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
Signal detection and estimation 150
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
Detection and Parameter Estimation
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
Signal detection and estimation 152
−
−=
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
Detection and Parameter Estimation
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
Signal detection and estimation 154
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
Detection and Parameter Estimation
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′
Signal detection and estimation 156
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
Detection and Parameter Estimation
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
Signal detection and estimation 158
)(
),(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 )()(
Detection and Parameter Estimation
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
Signal detection and estimation 160
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
Detection and Parameter Estimation
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.
Signal detection and estimation 162
(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
Detection and Parameter Estimation
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
Signal detection and estimation 164
(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
Detection and Parameter Estimation
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
Signal detection and estimation 166
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
Detection and Parameter Estimation
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
Signal detection and estimation 168
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:
Detection and Parameter Estimation
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
∑−
+