Signal Detection and Estimation - Solution Manual

Chapter 1 Probability Concepts 1.1 The given sets are: A = {1,2,3,4}

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

C = {x | x real and 1≤ x <3}

D = {2,4,7}

E = {4,7,8,9,10}.

We observe that: A is finite and countable. B is finite and countable. C is infinite and uncountable.

D is finite and countable. E is finite and countable.

1.2 By inspection, (a) A I B = A = {1,2,3,4}. (b) A U B U D U E = {1,2,3,4,5,6,7,8,9,10}. (c) ( B U E ) I D = D = {2,4,7}. (d) B − E = {1,2,3,5,6}. (e) A I B I D I E ={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) A I B = {4,6}.

(b) ( A U B ) I C = A I C = {1,7,9}.

1

Signal Detection and Estimation

2

(d) B − A = {0,1,7,9}.

(c) B U C = {0}.

(e) ( A U B) I ( A U C ) = A U ( B I C ) = A = {0,1,4,6,7,9}. (f )

(g) B − C = ∅.

A I C = {0,4,6}.

(h) B I C = B = {2,4,6,8,10,12}. 1.4 Applying the definitions, we have

(a) A − B U

A

(b) ( A U B) I C

U

B

B

A C

(c ) A I B I C I D

(d ) A

U A

B

U A

C

D

(e) A∩B

U A

B

Probability Concepts

3

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

B ⊂ C , then

and

A⊂C U

B

B C

A

B

A C

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

10R , 3W, 7B

Number of red balls 10 1 = = = 0.5 . Total number of balls 10 + 3 + 7 2 3 7 = 0.15 . = 0.35 . (c) P( B) = (b) P(W ) = 20 20 10 + 3 13 1 (e) P( R U W ) = = = 0.65 . (d) P( R ) = 1 − P ( R) = = 0.5 . 20 20 2

(a) P( R) =

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, P( B1 I W 2 I R3 ) = P( B1 )P (W 2 | B1 ) P( R3 | B1 I W 2 ) = P ( B1 ) P(W 2 ) P ( R3 )

=

7 3 10 210 = = 0.02625 20 20 20 8000


4

(b) Since the ball is not replaced, the sample size changes and thus, the events are dependent. Hence, P( B1 I W 2 I R3 ) = P ( B1 ) P(W 2 | B1 ) P( R3 | B1 I W 2 ) =

7 3 10 = 0.0307 20 19 18

1.9

B1 10R

,

B2 3W

2R

,

7B

6W

1B

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) P( R1 I R 2 ) = P ( R1 ) P( R 2 | R1 ) = P ( R1 ) P( R 2 ) independent. Hence,

P( R1 I R 2 ) =

since the events are

10 2 1 = = 0.111 . 20 9 9

3 6 = 0.1 20 9 (c) Since we can have a different color from each box separately, then

(b) Similarly, P(W1 I W2 ) = P (W1 ) P (W2 ) =

P(W I B ) = P(W1 I B 2 ) + P (W 2 I B1 ) =

3 1 6 7 + = 0.25 . 20 9 9 20

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 ,

B1 4W ,

B2 2B

3W ,

5B


5

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

1 2 and P( B2 ) = . 3 3

5 = 0.625 . 8 2 (b) P(1B | B1) = = 0.3333 . 6 (c) This is the total probability of drawing a black ball. Hence

(a) P(1B | B2) =

P(1B) = P(1B | B 2 ) P( B 2 ) + P (1B | B1 ) P( B1 ) =

52 21 + = 0.5278. 83 63

(d) Similarly, the probability of drawing a white ball is P(1W ) = P (1W | B 2 ) P( B 2 ) + P(1W | B1 ) P( B1 ) =

32 41 + = 0.4722. 83 63

1.11 In four tosses:__ __ __ __, we have three 1s and one is not 1. For example 3  4 1 1 1 5 1 5 111 1 . Hence, the probability is =  but we have   ways of 6 6 6 6 6 6 3  obtaining this. Therefore, the probability of obtaining 3 ones in 4 tosses is 3

4!  1  5 = 0.01543 .   3! 1!  6  6

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. P(1W I Urn B) 30 = = 0.3 . P(Urn B) 100 40 (b) P( 1G | Urn B) = = 0.4 . 100 P (Urn C I R ) (c) P(Urn C | R) = . Also, P( R)

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


6 P(R | Urn C) =

P(Urn C I R ) P( R | Urn C ) P(Urn C ) ⇒ P(Urn C | R) = . P(Urn C ) P( R)

We need to determine the total probability of drawing a red ball, which is P( R) = P( R | Urn A) P(Urn A) + P( R | Urn B) P(Urn B) + P( R | Urn C ) P(Urn C ) 30 (0.6) + 30 (0.2) + 40 (0.2) = 0.32 100 100 100 (0.4) (0.2) = 0.25 . Thus, P(Urn C | R) = 0.32 =

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 9 P( E i ) =    10 

k

8 P( E i E j ) =    10  M

k

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

8k

.

10 k (b) P(B) = P( ball 1 does not appear but ball 2 does) = P( E1 ) − P( E1 E 2 ) =

9k 10 k

−

8k 10 k

=

9k − 8k 10 k

.

(c) P(AB) = P( E 0 E1 E 2 ) = P( E 0 E1 ) − P ( E 0 E1 E 2 ) = (d) P( A U B) = P( A) + P( B) − P( AB) =

9k − 8k + 7 k 10 k

8k 10 k .

1.14 We have 1 −x 1  e + δ( x − 3) , x ≥ 0 f X ( x) =  2 2 0 , x<0

−

7k 10 k

=

8k − 7 k 10 k

.


7

(a) fx(x) 1/2 (1/2)

0

∞

∫ 0

.

.

1

2

x 3

∞ ∞ ∞ 1 1 1 1 1 1 f X ( x) dx = ∫ [ e − x + δ ( x − 3)]dx = ∫ e − x dx + ∫ δ ( x − 3)dx = + = 1 . 2 2 2 2 2 2 0 0 0

Hence, f X (x) is a density function. (a) P(X = 1) = 0 (the probability at a point of a continuous function is zero). P( X = 3) =

P( X ≥ 1) =

1 = 0.5 . 2 ∞

1

∞

1

∫ f X ( x)dx = 2 + ∫ 2 e 1

−x

dx =

1

(

)

1 1 + e −1 = 0.6839 . 2

1.15 fX(x) 1/4

-3

1/8

.

.

-2

-1

0

.

.

1

2

x 3

(a) The cumulative distribution function of X for all the range of x is, x x 1 1 3 FX ( x) = ∫ f X (u ) du = ∫ du = x + for − 3 ≤ x ≤ −1 , 8 8 8 −∞ −3 x

and

1

1

1

1

∫ 4 du + 4 = 4 x + 2

for − 1 ≤ x ≤ 1 ,

x 5 3 x1 + ∫ du = + 4 18 8 8

for 1 ≤ x ≤ 3 ,

−1

and


8

and F X ( x) = 1 for x ≥ 3 . Thus, 0 1 3 x+ 8 8 1 1 x+ 4 2 1 5 x+ 8 8 1

FX (x) =

,

x < −3

,

− 3 ≤ x < −1

,

−1 ≤ x < 1

,

1≤ x < 3 x≥3

,

(b) Calculating the area from the graph, we obtain P( X < 1) = 2

1.16 The density function is as shown

fX(x) 1/4

-2

-1

(a) P( X ≤ x) = F X ( x) =

x

1

0

1

1

∫ 4 du = 4 x + 2

x

2

1

for

−2≤ x < 2

−2

(b) P( X ≤ 1) =

1

1

1

∫ 4 dx = 2

−1

2 1  4 (c) E[ X ] = 0 , σ 2x = E[ X 2 ] = 2 ∫ x 2 dx  = .  4  3 0 

(d) Φ x (ω) = E[e jωX ] =

1 e jω 2 − e − jω 2 = sin 2ω . 4 jω 2ω

1 1 = . 4 2


9

1.17 The density function is shown below fX (x) 1

x 1/2

0

1

3/2

2

3/ 2 3 3 1 1 (a) P < X <  = ∫ xdx + ∫ (2 − x) dx = = 0.75 . 4 2  1/ 2 2 1 1

2

0

1

(b) E[ X ] = ∫ x 2 dx + ∫ (2 − x) xdx = 1 as can be seen from the graph. (c) The MGF of X is 1

2

M x (t ) = E[e tX ] = ∫ e tx xdx + ∫ ( 2 − x)e tx dx = 0

(d)

dM x (t ) dt

t =0

=

1

2t 2 (e 2t − e t ) − 2t (te 2t − 2e t + 1)

t

4

1 t

=

2

(te 2t − 2e t + 1) .

2te 2t (t − 1) + e t (4 − t ) − 2

Using L'hopital's rule, we obtain M ′x (t = 0) = 1 .

1.18 (a) E[ X ] = +∞

∫

−∞

α β 2 1 = x(α + β x 2 ) dx = + and 3 ∫0 2 4 1

f x ( x) dx = ∫ (α + β x 2 ) dx = α + 0

β = 1. 3

α = 1 / 3 Solving the 2 equations in 2 unknowns, we obtain  β = 2 1 23  1 = 0.511 . (b) E[ X 2 ] = ∫ x 2  + 2 x 2 dx = 3 45   0

Then, the variance of X is σ 2x = E[ X 2 ] − (E[ X ])2 =

3 = 0.667 . 45

t3


10

1.19 (a) E[ XY ] = ∑ x i y j P( x i , y j ) i, j

1 1 = [(−1)(−1) + (+1)(−1) + (−1)(−1) + (+1)(−1)] + [0] = 0 12 6 E[ X ] = ∑ x i P( X = x i )

where,

i

P( X = −1) =

4 1 2 1 4 1 = , P(X = 0 ) = = and P(X = 1 ) = = 12 3 6 3 12 3

1 1 1 Hence, the mean of X is E[ X ] = −1  + 0  + 1  = 0 . 3 3 3 1 1 1 Similarly, E[Y ] = −1  + 0  + 1  = 0 . Therefore, E[ XY ] = E[ X ]E[Y ] .  3 3 3

(b) We observe that P( X = −1, Y = 1) =

1 1 ≠ P( X = −1, Y = −1) = , thus X 9 12

and Y are not independent. +∞ +∞

1.20 (a)

∫ ∫

−∞ −∞

22

f XY ( x, y ) dxdy = ∫ ∫ k ( x + y )dxdy = 1 ⇒ k = 00

1 . 8

(b) The marginal density functions of X and Y are: 2 1 x 1 f X ( x) = ∫ ( x + y ) dy = + for 0 ≤ x ≤ 2 . 8 4 4 0 2 y 1 1 f Y ( y ) = ∫ ( x + y )dx = + 8 4 4 0 11

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

1

∫ ∫ 8 ( x + y)dxdy 00 1

1

1

∫ ( 4 y + 4 )dy 0

2

(d) E [X ] = ∫ 0

for 0 ≤ y ≤ 2 .

x 7 ( x + 1)dx = = E [Y ] . 4 6

=

1/ 8 1 = . 3/8 3


11

To determine ρ xy , we solve for 22

E [XY ] = ∫ ∫ 00

xy 4 ( x + y )dxdy = . 8 3

5 11 . Thus, σ x = σ y = 3 6 E [XY ] − E [X ]E [Y ] − 1 ρ= = = −0.0909 . σ xσ y 11 E[ X 2 ] = E[Y 2 ] =

and the correlation coefficient is

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

∫ ∫

1.21 (a)

−∞ −∞

45

f XY ( x, y )dxdy = ∫ ∫ kxy dx dy = 1 ⇒ k = 01

1 . 96

25

xy 3 dxdy = = 0.09375 . 96 32 03

(b) P( X ≥ 3, Y ≤ 2) = ∫ ∫

32

xy 5 dxdy = = 0.03906 . 96 128 21

P(1 < X < 2, 2 < Y < 3) = ∫ ∫

(c) P(1 < X < 2 | 2 < Y < 3) =

P(1 < X < 2, 2 < Y < 3) = P( 2 < Y < 3)

5 / 128 3

∫ f Y ( y)dy 2

5

xy y f Y ( y ) = ∫ dx = , 0 < y < 4 . Therefore, 96 8 1

P(1 < X < 2 | 2 < Y < 3) =

5 / 128 1 = = 0.125 5 / 16 8

5

5

1

1

(d) E [X | Y = y ] = ∫ xf X ( x)dx = ∫ x

x 31 dx = = 3.444 . 12 9

where,


12

23

1.22 (a) We first find the constant k. Hence,

1

∫ ∫ kxy dx dy = 1 ⇒ k = 6 11

y

x =3− y

3

2

3− y

1

1

P( X + Y < 3) = ∫

∫

1 23 xydxdy = = 0.4792 6 48

2

1

1

2

3

(b) The marginal densities of X and Y are 2

f X ( x) = ∫ 1

1 x xydy = 6 4

for 1 < x < 3

3

1 2 xy dx dy = y for 1 < y < 2 . 6 3 1

and f Y ( y ) = ∫

Since f X ( x) f Y ( y ) =

1 xy = f xy ( x, y ) ⇒ X and Y are independent. 6

1.23 We first determine the marginal densities functions of X and Y to obtain 1

f X ( x) = ∫ 0

and

f Y ( y) =

∞

∫ 2

16 y x3

dx = 2 y

16 y x3

dy =

for

∞

8 x3

0 < y < 1.

Then, the mean of X is E [X ] = ∫ xf X ( x)dx = 4 , 2

1

and the mean of Y is E [Y ] = ∫ 2 y 2 dy = 0

2 . 3

for x > 2.

x


13 ∞

1.24 We first find the constant of k of f Y ( y ) to be

∫ kye

−3 y

dy = 1 ⇒ k = 9 .

0

1 y −1

(a) P( X + Y > 1) = 1 − P( X + Y ≤ 1) = 1 − ∫

∫ 18 ye

− 2 x −3 y

dxdy = 9e − 2 − 14e −3 .

0 0

y x+y=1 1

x 0

(b) P(1 < X < 2, Y ≥ 1) =

1

∞2

∫ ∫ f XY ( x, y)dxdy = 4e

−5

− 4e −7 .

11

2

(c) P(1 < X < 2) = ∫ 2e − 2 x dx = e − 2 − e − 4 . 1

∞

(d) P(Y ≥ 1) = ∫ 9 ye −3 y dy = 4e −3 . 1

(e) P(1 < X < 2 | Y ≥ 1) =

P(1 < X < 2, Y ≥ 1) 4e −5 − 4e −7 = = e − 2 − e −5 . P (Y ≥ 1) 4e − 3

1.25 (a) Using f X (x) , E [Y ] = E [g ( X )] =

+∞

∫

−∞

∞

g ( x) f X ( x)dx = ∫ 2 x 2e − 2 x dx = 1 . 0

(b) We use the transformation of random variables (the fundamental theorem) to find the density function of Y to be f Y ( y) = ∞

1  y 1 − f X   = 2e 2 2 2

Then, the mean of Y is E [Y ] = ∫ ye − y dy = 1 . 0

2y 2

= e−y .


14

Both results (a) and (b) agree. 1.26 (a) To find the constant k, we solve +∞ +∞

∫ ∫

−∞ −∞

33

f XY ( x, y ) dxdy = ∫ ∫ kxy dx dy = 1 ⇒ k = 0 y

8 . 81

(b) The marginal density function of X is f X ( x) =

∞

8

4

∫ 81 xydy = 81 x

3

for 0 ≤ x ≤ 3 .

0

(c) The marginal density function of Y is 3

8 4 xydx = (9 y − y 3 ) 81 81 y

fY ( y ) = ∫

(d)

8 xy 2 y f XY ( x, y ) 81  = =  x2 f Y | X ( y | x) = 4 3  f X ( x) x 0 81

for 0 ≤ y ≤ 3 .

, 0≤ y≤ x , otherwise

and  2x f XY ( x, y )  f X |Y ( X | Y ) = = 9 − y 2 f Y ( y) 0 

,

y≤ x≤3

, otherwise

1.27 The density function of Z = X + Y is the convolution of X and Y given by f Z ( z ) = f X ( x) ∗ f Y (Y ) =

+∞

∫ f X ( x) f Y ( z − x)dx

−∞

z-4

0

z

0

z-4

z


15

z −x 1 1− e −z dx = ∫ e 4 4 0  z e − z (e 4 − 1) 1  f Z ( z ) =  ∫ e − x dx = 4  z −4 4  0   

1.28 The density function of Z = X + Y

, 0≤ z≤4 z≥4

,

, otherwize

f Z ( z) =

is

+∞

∫ f Y ( y ) f X ( z − y )dy .

_∞

Graphically, we have fY(y)

fX(x)

fX(-x) 0.5

0.4

0.4

0.4

0.4 0.3

0.2

0.2

0.2

x 0

1

2

y

3

0

1

2

x

-3 -2

3

-1

0

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

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

0


16 The plot of f Z (z ) is fZ(z)

0.3 0.26

0.24

0.12

0.08

0

0 z

0

Note that

1

2

3

4

5

6

7

∑ f Z ( z − z i ) = 0.12 + 0.26 + 0.3 + 0.24 + 0.08 = 1.0

as expected.

i

1.29 (a) Z = XY ⇒ FZ ( z ) = P( XY ≤ z ) =

∞z/ y

∫ ∫e

−( x + y )

dxdy for z ≥ 0 .

0 0

z

∞ z / y

∞ ∞ − y−   y ⇒ ∫  ∫ e − x dx e − y dy = ∫ (1 − e − z / y )e − y dy = 1 − ∫ e dy .   0 0 0 0 

Therefore, z ∞ z 1 (− y − y )   d d  ∞ − y − y   ∫ e dy f Z ( z) = FZ ( z ) = dy  =  y − ∫ e 0 dz dz  0    0

(b) Z = X + Y ⇒ f Z ( z ) =

,

z≥0

, otherwise

+∞

∫ f Y ( y) f X ( z − y)dy

−∞ z

= ∫ e − y e − ( z − y ) dy 0

 ze − z = 0

, z≥0 , otherwise

0

z

1.30 The density function of Z = XY is f Z ( z) =

+∞

∫

−∞

1 z 1 1 f XY ( , y )dy = ∫ dy = − ln z y y y z

for 0 < z < 1 .

.


17

1.31 (a) The marginal density function of X is β

α − αx e dy = αe −αx β

f X ( x) = ∫ 0

for x ≥ 0 .

(b) The marginal density function of Y is fY ( y ) =

∞

α

∫βe

− αx

0

dx =

1 for 0 ≤ y ≤ β . β

(c) Since f X ( x) f Y ( y ) = f XY ( x, y ) ⇒ X and Y are statistically independent. (d) Z = X + Y ⇒ f Z ( z ) = f X ( x) ∗ f Y ( y ) =

+∞

∫ f Y ( y) f X ( z − y)dy .

−∞

fY ( y )

f X (x) α e − αx

α

1 β y

x

β

For 0 ≤ y < β z

f Z ( z) = ∫

α

z-β

0

0

z

(

α − αx 1 e dx = 1 − e − αz β β

)


18 For β ≤ y < ∞ α

f Z ( z) =

z-β

0

[

z

α − αx 1 e dx = e − α ( z −β ) − e −αz β β z −β

∫

z

Then, f z (z )

(

1 1 − e −αβ β

)

1.32 Z =

z

β

0

X x X  ⇒ FZ ( z ) = P ( Z ≤ z ) = P ≤ z  = P( X ≤ yz ) = P(Y ≥ ) Y z Y  =

∞ ∞

∫ ∫ αβ e 0 x/z

− α x −β y

e

β 1  dxdy = 1 +  α z 

−1

,

Hence, the density function is 1 β α 2 d β   f Z ( z) = FZ ( z ) =   z +  dz α   0  23

1.33 (a) Solving the integral

, z>0

, z<0 1

∫ ∫ kx1 x 2 dx1dx 2 = 1 ⇒ k = 6 . 11

(b) The Jacobian of the transformation is

z>0

]


J ( x1 , x 2 ) =

Hence, f Y1Y2 (Y1 , Y2 ) =

∂y1 ∂x1

∂y1 ∂x 2

∂y 2 ∂x1

∂y 2 ∂x 2

f X 1 X 2 ( x1 , x 2 ) J ( x1 , x 2 )

=

1 x12

1  = 12 0

19

0 2 x1 x 2

= 2 x1 x 2 .

, x1 ,x 2 ∈ D , otherwise

where D is the domain of definition. x2

y1 = x1 y2 = x1 x22

3 2

1

2

1 4 0

1

2

3

x1

x2 = 1 ⇒ y 2 = 1 Side 1 : y1 = x1 = 1 ⇒ y 2 = x22 , then  ⇒ 1 ≤ y 2 ≤ 4. x2 = 2 ⇒ y 2 = 4 x2 = 1 ⇒ y 2 = 3 Side 2 : y1 = x1 = 3 ⇒ y 2 = 3 x22 , then  ⇒ 3 ≤ y 2 ≤ 12.  x 2 = 2 ⇒ y 2 = 12  x1 = 1 ⇒ y 2 = 4 Side 3 : x2 = 2 ⇒ y 2 = 4 x1 = 4 y1 , then   x1 = 3 ⇒ y 2 = 12.

Side 4 : x 2 = 1 ⇒ y 2 = x1 = y1 . Therefore, D is as shown below


20 y2 12 + + + + + + + 4+

D

3 1

+ 1

+ 2

+ 3

y1

1.34 (a) The marginal density functions of X1 and X2 are f X 1 ( x1 ) =

+∞

αe − αx1 α 2 e −α ( x1 + x2 ) dx 2 =   0 −∞

∫

,

x1 > 0

,

x1 ≤ 0

,

x2 > 0

,

x2 ≤ 0

and f X 2 (x2 ) =

+∞

αe − αx2 α 2 e −α ( x1 + x2 ) dx1 =   0 −∞

∫

Since f X 1 X 2 ( x1 , x 2 ) = f X 1 ( x1 ) f X 2 ( x 2 ) ⇒ X1 and X2 are independent. (b) The joint density function of (Y1 , Y2 ) is given by  f X 1 X 2 ( x1 , x 2 )  f Y1 f Y2 ( y1 , y 2 ) =  J (x1 , x 2 )  0  The Jacobian of the transformation is given by

,

x1 ,x 2 ∈ D

,

otherwise .


J ( x1 , x 2 ) =

∂y1 ∂x1

∂y1 ∂x 2

∂y 2 ∂x1

∂y 2 ∂x 2

21

1 = 1 x2

1

−

x 1 . x1 = − 12 − x x2 2 x2

Hence, f Y1Y2 ( y1 , y 2 ) =

α 2 e − α ( x1 + x2 ) −

Thus,

x1 x 22

−

1 x2

, but y1 = x1 + x 2 and y 2 =

f Y1Y2 ( y1 , y 2 ) = α 2 e −αy1

x12 . x1 + x 2

Also,

x1 ⇒ x1 = y 2 x 2 . x2

x 2 = y 1 − x1 = y1 − y 2 x 2

⇒ x 2 = y1 (1 + y 2 ) .

Making the respective substitutions, we obtain y12 f Y1Y2 ( y1 , y 2 ) = α 2 e − αy1

(1 + y 2 ) 2 y1 = α 2 e −αy1 for y1 > 0 and y 2 > 0 . y1 (1 + y 2 ) 2

Chapter 2 Distributions 2.1 Let A = { seven appears} = {(1 ,6) , (2 , 5) , (3 , 4) , (4 , 3) , (5 , 2) , (6 , 1)}. 6 1 1 5 = and P A = 1 − = . Then, P( A) = 36 6 6 6

( )

(a) This is Bernoulli trials with k = 2 successes and n = 6 trials. Hence, 2 4 2 4  6  1   5  6!  1   5  P(k = 2 successes in n = 6 trials) =      =     = 0.2009 4! 2!  6   6   2  6   6  0

6

6

 6  1   5  5 (b) P(no successes in n = 6 trials) =      =   = 0.3349 0 6 6 6     

10  2.2 The number of ways of obtaining 4 white balls out of 10 is   . The other 4  9 number of different ways of obtaining 3 other balls (not white) is   . Hence, the 3 probability of obtaining the fourth white ball in the seventh trial is 10  9      4  3  = 0.3501 19    7 

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

22

Distributions

23

10  9     4 3 P( X = 4) =    = 0.3501. 19    7 

2.3 The probability of success is p = 0.9 while the probability of failure is q = 1 − p = 0.1 .

(a) P (at least 6 land in zone) = P ( X = 6) + P ( X = 7) + P ( X = 8) + P ( X = 9) + P( X = 10) 10  10  10  =  (0.9 )6 (0.1)4 +  (0.9 )7 (0.1)3 +  (0.9 )8 (0.1)2 6 7     8  10  10  +  (0.9 )9 (0.1) +  (0.9 )10 = 0.998. 9  10 

10  (b) P(none lands in zone) =  (0.9)0 (0.1)10 = (0.1)10 . 0 

(c) Probability that at least 70 0/0 land in zone is P( X = 7) + P( X = 8) = P( X = 9) = P( X = 10) =

10  10  10  10   (0.9 )7 (0.1)3 +  (0.9)8 (0.1)2 +  (0.9)9 (0.1) +  (0.9 )10 = 0.987. 7 8 9 10        Hence, the program is successful. 2.4 Substitution for k = 0 in the Poisson density function, we have P ( X = 0) = 0.2 = e − λ . Hence, λ = 1.609. P( X > 2) = 1 − P( X ≤ 2) = 1 − [ P( X = 0) + P( X = 1) + P( X = 2)] = 1 − [0.2 + 1.609e −1.609 +

(1.609) 2 −1.609 e ] 2!

= 0.219.

2.5 Let X represent the Poisson distribution of the incoming calls with λ = 12 hour .


24

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

(12)k

k =0

k!

P( X > 15) = 1 − P( X ≤ 15) = 1 − ∑ e −12

= 0.1556

(b) No calls in 15 minute ( 1 / 4 hour) break ⇒ P( X = 0) in 15 minutes. Hence, P( X = 0) = e −3 = 0.0498

2.6 X is Poisson distributed and

P( X = 2) = ( 2 / 3) P( X = 1) .

That is,

2

2 λ λ 4 4  = e − λ . Solving for λ we obtain λ λ −  = 0 ⇒ λ = since λ = 0 is 2! 3 1! 3 3   not a solution. Therefore, e −λ

P( X = 0) = e −4 / 3 = 0.2636 and P( X = 3) = e − 4 / 3

(4 / 3)3 3!

= 0.1041

2.7 The lack of memory property is P( X ≥ x1 + x 2 | X > x1 ) =

P( X ≥ x1 + x 2 I X > x1 ) P( X ≥ x1 + x 2 ) = P( X > x1 ) P( X > x1 ) =

2.8 (a) In this case the parameter β =

e − α ( x1 + x2 ) e −αx1

= e − αx2 = P( X ≥ x 2 )

1 1 = and the density function is λ 12

1 −x 12 ,  f X ( x) = 12 e 0 ,  15

x≥0 otherwise k

1 − 12 e = 0.2325 k = 0 12

Hence, P( X > 15) = 1 − P( X ≤ 15) = 1 − ∑

(b) P( X = 0) =

1 = 0.0833 . 12

Distributions

25

2.9 X is the standard normal X ~ N (0. 1) (a) P ( X > 1) = P ( X < −1) + P ( X > 1) = 2 P ( X > 1) = 2[1 − P( X ≤ 1)] = 2[1 − I (1)] = 2(1 − 0.8413) = 0.3174

fX(x)

-1

x

+1

(b) P( X > 1) = 1 − P( X ≤ 1) = 1 − I (1) = 0.1587

2.10 X ~ N (0,1). Then, P(X > 3 ) = 1 − P(X ≤ 3 ) = 1 − I( 3 ) = Q( 3 ) = 0.0013 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 p =

6 1 = 36 6

and q = 1 − p =

5 . 6

(a) P(success at least 20 0 0 of the time) = P( X ≥ 40) =

k

 200  1   5  ∑  k  6   6      k = 40  200

200 − k

= 0.1223.

(b) Using the central limit theorem to obtain the approximation to the normal distribution, we have  200   39 − X-np  6  = 1 − I( 1.07 ) = 0.1423. ≤ P( X ≥ 40) = 1 − P ( X ≤ 39) ≈ 1-P  npq 1000     36  


26

2.12

S = X 1 + X 2 + L + X k + L + X 100 , Xk is Poisson distributed with λ=3.2 (a) P( S ≥ 5) =

100

∑ e3.2

k =5

4 (3.2) k (3.2) k = 1 −P( S ≤ 4) = 1 − e3.2 ∑ = 0.2602. k! k =1 k!

(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,  S-3.2 4 − 3.2   ≅ 1 − I( 0.45 ) = Q(0.45) = 0.3264 P( S ≥ 5) = 1 − P ( S ≤ 4) ≈ 1 − P ≤  3.2   3.2

2.13 X ~ N(1, 2)  X −1 2 −1   (a) From the tables, P(X > 2 ) = 1 − P(X ≤ 2 ) ≈ 1 − P ≤  2   2 = 1 − I( 0.707 ) = 0.2399

 1.6 − 2 X − 2 2.2 − 2   ≤ ≤ (b) P(1.6 ≤ X ≤ 2.2) = P  2 2   2  X −2   X −2  = P ≤ 0.14  − P ≤ −0.28  = I( 0.14 ) − I( − 0.28 ) = 0.166.  2   2 

2.14

1 , 1≤ x ≤ 6  ( ) = f X x 5 0 , otherwise  fX(x) 1 5

+

+

+

+

+

+

1

2

3

4

5

6

Using the fundamental theorem, we have y=

1 1 ⇒ dy = − 2 dx x x

and

dx = x 2 . Hence, dy

x

Distributions

fY ( y ) =

27

f X ( x) dx 1 2 1 = f X ( x) = x = 2 dy dy 5 5y dx

for

1 ≤ y ≤ 1. 6

fY(y) 36 5

≈ 1 5

+

+

+

+

+

+

1 6

2 6

3 6

4 6

5 6

1

y

2.15 (a) We found in Example 1.19 for Y = X 2 that  F ( y ) − FX (− y ) FY ( y ) =  X 0

,

y>0

,

y≤0

]

,

y>0

,

y≤0

and

[

 1 f X ( y ) − f X (− y )  fY ( y ) =  2 y  0 

For X uniformly distributed between 0 and 1, we obtain 0  FY ( y ) =  y 1 

,

y≤0

,

0 < y < 1 and

,

y ≤1

 1  f Y ( y) =  2 y 0 

, 0 < y ≤1 , otherwise

fY(y) FY(y) 1 1/2 y 0

1

y 1


28 (b) For Z = e X , F

Z

, z≤0 0 (z) = P (Z ≤ z) =  X  P(e ≤ z ) = P( X ≤ ln z ) = F X (ln z ) , z > 0

Hence,  F ( ln z) , z > 0 F Z (z) =  X . , z ≤ 0 0 1 dF ( z )  f X (ln z ) , z > 0 = z f Z ( z) = Z dz 0 , z≤0

The

density

function

is

Substituting for z, we obtain 0  FZ ( z ) = ln z 1 

, z <1 , 1 ≤ z < e and , z≥e

1  f Z ( z) =  z 0

2.16 X and Y are standard normal, that is fY ( y) =

1 2π

e

−

y2 2

, 1≤ z ≤ e , otherwise

f X ( x) =

1 2π

e

−

x2 2

and

.

X X  Y = (a) Z = Y  X −  Y

, Y >0 , Y <0

The distribution function of Z is X   X  FZ ( z ) = P( Z ≤ z ) = P ≤ z Y > 0  + P − ≤ z Y < 0  Y Y     = P( X ≤ yz y > 0 ) + P( X ≤ − yz y < 0 ) with the regions shown

below.

Distributions

29 x=-yz

y

y

x=yz

x0 x

x x < -yz y<0

FZ ( z ) =

∞ y

∫∫

0 −∞ ∞ 0

1 2π

∫ ∫ f XY ( x, y)dxdy

−∞ −∞

= ∫ yf XY ( yz, y )dy − =

0 − yz

f XY ( x, y )dxdy +

∞

− ( yz ) ∫ ye

2

0

∫ yf XY ( − yz,y)dy

-∞

2 − y2 2

e

dy −

0

1 2π

0

∫ ye

+ ( yz )2 2 − y 2 2

e

dy

−∞

Using Leibniz’s rule, we obtain f Z ( z) =

(b) W=

For

W =

X Y

,

we

1 2

π( z + 1)

for all z.

Z=

have

X X X ⇒ Z = = . Y Y Y

Thus,

Z , Z > 0 . = Z = Y − Z , Z < 0 X

w w

z1

z2

z

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


30

f W ( w) =

f Z ( z1 ) f Z ( z 2 ) + g ' ( z1 ) g ' ( z 2 )

where

z1 = − w and z 2 = w .

+ 1 , z > 0 Also, g ' ( z ) =  ⇒ g' ( z ) = 1 . − 1 , z < 0 Substituting, we obtain

f W ( w) = f Z ( z1 ) + f Z ( z 2 ) =

Therefore, f W ( w) =

2

1

π[(− w) + 1] 2

+

1

π[ w 2 + 1]

for 0 ≤ w < ∞.

π ( w 2 + 1)

2.17 The joint density function of ( X 1 , X 2 ) is

f X 1 X 2 ( x1 , x 2 ) =

1 2πσ

2

Solving x12 + x 22 = y12

e

−

( x1 + x2 )2 2σ 2

and

(

with Y1 =

X 12 + X 22

and

Y2 =

x1 = x 2 y 2 , we obtain

)

x22 y22 + x22 = y12 ⇒ x22 1 + y22 = y12 ⇒ x2 = ±

y1 1+

y22

and x1 = ±

By definition, y1 ≥ 0 since x1 = x 2 y 2 and hence, we have 2 solutions: x1 = x1 = −

X1 . X2

y 2 y1 1+

y 22

y 2 y1 1+

y 22

The Jacobian of the transformation is

x2 = x2 = −

y1 1 + y 22 y1 1 + y 22

y1 y2 1 + y22

Distributions

x2

x1 J ( x1 , x 2 ) =

x12

+

31

x 22

x12

1 x2

+ x 22 −

=−

x1

x12 x 22

x 22 = − y 22

(

x12

1 +

)

1/ 2 x 22

−

(

x12

1 + x 22

)

1/ 2

1 + y 22 1 1 − =− y1 y1 y1

Therefore,

f Y1Y2 ( y1 , y 2 ) =

   f  y1 y 2 , y1 XY   1+ y 2 1 + y 22  1 + y 22 2  

  − y1  − y1 y 2  + , f XY    1+ y 2  1 + y 22 2  

y1

    

Note that f X 1 X 2 ( x1 , x 2 ) = f X 1 X 2 (− x1 ,− x 2 ) . Hence, f Y1Y2 ( y1 , y 2 ) = 2

f Y1 ( y1 ) = ky1e − y1

where, ∞

2

k ∫ y1 e − y1

2σ 2

0

2 y1 1+

2σ 2

dy1 = 1 ⇒ k =

1

y 22

2πσ

2

2

e − y1

2σ 2

u ( y1 ) = f Y1 ( y1 ) f Y2 ( y 2 )

u ( y1 ) . We determine the constant k to be

1 σ2

.

Thus, the density function of Y1 and Y2 are

respectively f Y1 ( y1 ) =

2 y1 1 e − y1 σ 2πσ 2

2σ 2

and

f Y2 ( y 2 ) =

1 1 π 1 + y 22

u ( y1 )


32

1

2.18 X is a standard normal ⇒ f X ( x) =

2π

e −x

2

2

.

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

1

⇒ f Y ( y) =

2 n / 2 Γ(n / 2 )

Let T =

X Y /n

−1 −

y2 e

y 2

for y > 0

, then the cumulative distribution of T is

(

)

FT (t ) = P (T ≤ t ) = P X ≤ t y / n =

1 2 n / 2 Γ(n / 2 ) 2π

∞ t y/n

∫ ∫ 0

n ( y+ x2 ) −1 − 2 y e 2 dxdy

−∞

since the region of integration is the xy-plane with x ≤ t y / n . 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 x = u y / n , then dx =

y / n du

x 2 = u 2 ( y / n) . Substituting in the

and

integral, we obtain FT (t ) =

1 2 n / 2 Γ(n / 2 ) 2π

∞ t

∫ ∫

0 −∞

n y −1 − y2 e 2

y − e n

u2 y n dudy

 ∞ n −1 − y  1+ u 2     2  n  2 = dy  du ∫  ∫y e n/2 2 Γ(n / 2) 2π u = −∞  y = 0    1

Let becomes

y  u 2 1+ 2  n

t

  = z ⇒ y = 2 z and dy = 2 dz .  u2 u2  1+ 1+ n n

The

integral

Distributions

33

    n +1 n −1  t  ∞ −z 2 2 2 z e   du FT (t ) = dz ∫ ∫ n +1  2 n / 2 Γ(n / 2) 2π u = −∞  z =0   u2  2    1 +   n      n +1 Γ  t 1  2  = du ∫ n +1 nπ Γ(n / 2) u = −∞ 2 2  u  1 +   n   ∞

since

∫z 0

n −1  n +1 e dz = m! = Γ(m + 1) = Γ .  with m = 2  2 

m −z

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 f X ( x) =

β 1 . π β2 + x 2

The moment generating function of X is then Φ x (ω) = E[e jωX ] =

β ∞ e jωx β ∞ cos ωx β ∞ sin ωx dx = dx + j dx ∫ ∫ π −∞ β 2 + x 2 π −∞ β 2 + x 2 π −∫∞ β 2 + x 2

since e jωx = cos ωx + j sin ωx . Also, lim

p →∞

p

∫ f ( x)dx = 0

when f(x) is an odd

−p

function of x. Then, Φ x (ω) =

since

cos ωx β2 + x 2

2β ∞ cos ωx dx π ∫0 β 2 + x 2

is even. Using the tables of integrals, we obtain Φ x (ω) = e −βω , ω > 0 and β > 0.


34 2.20 (a)

The

mean

∞

value

of

Weibull

distribution

is

given

by

b

E[ X ] = ∫ abx b e − ax dx. 0

Let u = ax b ⇒ du = abx b −1 dx = b(ax b ) ∞

Hence, E[ X ] = b ∫ ue −u 0

1 u   bu  a 

1/ b

dx du  u  ⇒ dx =   x bu  a 

1/ b

u since x =   a

1/ b

.

∞

 1 du = a −1 / b ∫ u 1 / b e −u du = a −1 / b Γ 1 +  .  b 0

(b) The variance is σ2 = E[ X 2 ] − (E[ X ])2 . We need to determine the mean ∞

b

square value, which is E[ X 2 ] = ∫ abx b +1e − ax dx. Following the same approach as 0

in (a), we obtain E[ X 2 ] = a

−

2∞ 2 b u b e −u du

∫

=a

−

2 b

0

σ2 = a

−

 2 Γ1 +  . Hence,  b 2 b

  2    1  2   Γ 1 +  −  Γ 1 +      b    b  

Chapter 3 Random Processes 3.1 (a) The mean and autocorrelation functions of X (t ) are E[ X (t )] =

π/8

∫

A cos(ω 0 t + θ)

−π / 8

4 4 2A dθ = cos ω 0 t π π

R xx (t + τ, t ) = E[ A cos(ω 0 t + ω 0 τ + θ) A cos(ω 0 + θ)] A2 A2 cos ω 0 τ + E[cos(2ω 0 t + 2ω 0 τ + 2θ)] 2 2 A2 2A2 = cos ω 0 τ + cos(2ω 0 t + 2ω 0 τ) 2 π

=

(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 T0 = 0 ⇒ s (t ) = 1 = X (0) , and at T0 = 1 ⇒ s (t ) = 0 = X (1) . Then, we have fX (x ; 0)

FX (x ; 0)

1 2 0

1 2 1

1 1 2

x

1

(b) R xx (t1 , t 2 ) = E[ X (t1 ) X (t 2 )] = E{E[ X (t1 − T0 ) X (t 2 − T0 ) T0 = t 0 ]}

35

x


36

= s (t1 − 0) s (t 2 − 0) P(T0 = 0) + s (t1 − 1) s (t 2 − 1) P (T0 = 1) =

1 1 s(t1 ) s (t 2 ) + s (t1 − 1) s (t 2 − 1) 2 2 t2

Height

1 2

3/2

1/2 t1

-1/2

1/2

3/2

-1/2

3.3 (a)

The

< x(t ) > = lim

T →∞

1 2T

time

average

of

X (t )

is

T

∫ A cos(ω 0 t + θ)dθ = 0 ≠ E[ X (t )] ≡

ensemble

average.

−T

Therefore, the process X (t ) is not ergodic in the mean 1 T →∞ 2T

(b) < x(t + τ) x(t ) > = lim =

T

∫ A cos(ω 0 t + ω 0 τ + θ) A cos(ω 0 t + θ)dt

−T

A2 cos ω 0 τ ≠ R xx (t + τ, t ) ⇒ The process is not ergodic in 2

the autocorrelation. 3.4 (a) R yy (t + τ, t ) = E[ X 2 (t + τ) X 2 (t )] = E[ A2 cos 2 (ω0t + ω0 τ + θ) A2 cos 2 (ω0t + θ)]  1 1  1 1  = A4 E  + cos(2ω0t + 2ω0 τ + 2θ)  + cos(2ω0t + 2θ)   2 2 2 2      A4 A4 E[cos(2ω0t + 2ω0 τ + 2θ) cos(2ω0t + 2θ)] + 4 4 A4 A4 E[cos(4ω0t + 2ω0 τ + 4θ) + cos(2ω0 τ)] = + 4 8 A4 A4 = + cos(2ω0 τ) 4 8

=

Random Processes

37

(b) E[Y (t )] = E[ X 2 (t )] = E[ A 2 cos 2 (ω 0 t + θ)] A2 A2 A2 = constant. Therefore, Y(t) + E[cos(2ω 0 t + 2θ)] = 2 2 2 is wide-sense stationary. =

3.5 (a) E[ X (t )] = E[ Ae j (ωt + Θ ) ] = E[ A]E[e j (ωt + Θ ) ] , where

E[ A] =

∞

a2

∫ σ2 e

−

a2 2σ 2

da =

0

E[e j (ωt + Θ) ] = e jωt E[e jΘ ] = e jωt

π σ and a

1 2 π jθ e dθ = 0 ⇒ E[ X (t )] = 0. 2π ∫0

(b) R xx (t1 , t 2 ) = E[ Ae j (ωt1 + Θ ) Ae − j (ωt 2 + Θ ) ] = E[ A 2 ]e jω(t1 −t 2 ) , where

2

E[ A ] =

∞

a3

∫ σ2 e

−

a2 2σ 2

da = 2σ 2

0

Let t1 = t + τ and t2 = t ⇒ R xx (t + τ, t ) = 2σ 2 e jωτ = R xx (τ) . Therefore, X (t ) is wide-sense stationary. 3.6 (a) The autocorrelation function of Z(t) is R zz ( τ) = E[ Z (t + τ) Z (t )] = E[ A 2 X (t + τ) X (t )Y (t + τ)Y (t )] = E[ A 2 ]R xx (τ) R yy ( τ) ,

since A, X(t) and Y(t) are statistically independent. E[ A 2 ] = σ 2a + E 2 [ A] = 9 + (2) 2 = 13. Therefore,

R zz (τ) = 26e

−2 τ

cos ωτ(9 + e

−3 τ

)

(b) From(3.31), we have lim R zz (τ) = E 2 [ Z (t )] = 0 . Therefore, the mean of τ→∞

Z (t ) E[ Z (t )] = 0


38

Since m z = 0 , then σ 2z = E[ Z 2 (t )] . Hence, E[ Z 2 (t )] = R zz (0) = 26(9 + 1) = 260.

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 R xx (t1 , t 2 ) =

E[ A 2 ] T σ2 s (t1 − t 0 )s (t 2 − t 0 ) dt 0 = ∫ T 0 T

Two possible cases (i) 0 ≤ τ ≤ (i) For 0 ≤ τ ≤

T

∫ s(t + τ)s(t )dt = R xx (τ). 0

T T and (ii) − ≤ τ ≤ 0 . 2 2

T , we have 2

s(t)

T/2

-1

T

t

T

∫0

s(t+τ)

T  T T   s (t ) s (t + τ) dt = ( −1) 2  − τ  + ( −1) 2  −  − τ  2   2  2 ⇓

+1 t

(

(ii) For −

T − τ) 2

T ≤ τ ≤ 0 , we have 2

 4τ  R xx ( τ ) = σ 2 1 −  T  

Random Processes

T

∫0

s (t ) s (t + τ) dt = (−1)(− τ) + (+1)[

39

T T T T − (− τ)] + ( −1)[( − τ) − ] + (1)[T − ( − τ)] 2 2 2 2

⇓

R xx ( τ ) = σ 2 (1 +

4τ ) as shown below T

s (t)

T/2

T

t

-1

s(t+τ)

t T ( − τ) 2

A plot of the autocorrelation function is shown below

RXX (τ) σ2

-3T/4

-T/2

T/4

-T/4

T/2

τ 3T/4

-σ2

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.


40

(b) Consider one period only ⇒ X(t)

A x(t) T T0

T0 +

Tx(t ) 8A

t

T T0 + 4 T Tx (t ) T0 + − 4 8A

F X t ( x t ) = 1 for x t > A , and F X t ( x t ) = P[ X (t ) = 0] + P[0 < X (t ) < x t ] Hence , P[ X (t ) = 0] =

3 4

for 0 ≤ x t ≤ A.

and

Tx t T Tx T ] + P[T0 + − t ≤ t < T0 + ] 8A 4 8A 4 Tx t T T Tx t = P[t − ≤ T0 < t ] + P[t − < T0 ≤ t − + ]. 8A 4 4 8A x 3 Therefore, F X t ( x t ) = + t for 0 ≤ x t ≤ A. 4 4A P[0 < X (t ) ≤ x t ] = P[T0 < t ≤ T0 +

FX t ( xt )

1 3/4

A

xt

Random Processes

3  4 δ( x t )   1 (c) f X t ( x t ) =  4A 0  

(d) E[ X (t )] =

∞

∫

−∞

E[ X 2 (t )] =

41 f X t ( xt )

,

xt = 0

,

0 ≤ xt ≤ A

,

otherwise

1 4A

(3/4)

0

A

x t f X t ( x t )dx t = ∫ 0

A

xt

xt A dx t = and 4A 8

2

13 2 A ⇒ σ 2xt = A . 12 192

(e) < x(t ) > =

1T A x(t ) dt = 8 T ∫0

and < x 2 (t ) > =

A2 . 12

3.9 (a) In fixing t1 = 1 / 2 and t 2 = 3 / 2 , we obtain two independent random X (1 / 2 ) and X (3 / 2) with marginal density functions variables  1 1  3 1 x x f X  x;  = rect  and f X  x;  = rect  . Therefore, the joint density  2 2  2 2 2 2 function is just the product of the marginal density functions to yield 11 1 f X (0, 0; 0.5, 1.5) = = . 22 4 (b) We observe that the duration of 1 second represents the length of a pulse thus, the samples spaced 1 second apart will be independent 1 ⇒ f Y (0, 0; 0.5, 1.5) = as in (a). 4

and

3.10

R yy ( τ ) = E[Y (t + τ )Y (t )] = E{[ X (t + τ ) + X (t + τ − 1)][ X (t ) + X (t − 1)]} = 2 R xx ( τ ) + R xx ( τ + 1) + R xx ( τ − 1).


42

Ryy (τ) 2

2

⇒

1

-2

3.11

0

-1

1

+1

τ

+2

-2

-1

0

+1

+2

τ

The autocorrelation function of the process Z (t ) = Y (t ) − X (t − 1) is R zz ( τ) = E{[Y (t + τ) − X (t + τ − 1)][Y (t ) − X (t − 1)]} = R yy (τ) − R yx (τ + 1) − R xy (τ − 1) + R xx ( τ) = R yy (τ) + R xx ( τ)

since Ryx = Rxy = 0 from orthogonality. Therefore, S zz ( f ) = 2 S yy ( f ) = 2 S xx ( f ) as shown below. Szz (f)

2

+1 f 0

-1

3.12

+1

E[Y 2 (t )] = E[ X 6 (t )]. From Equation (2.80), we have

E[ X 6 (t )] =

6! σ 6 3

3! 2 ∞ N the variance σ 2 = ∫ S xx ( f )df = 0 2 −∞

( )

= 15 σ 6 = 15 σ 2 ∞

∫e

−α f

−∞

∞

N  the mean square value E[Y 2 (t )] = 15 0  .  α 

where,

df = N 0 ∫ e − α f df = 0

3

3

N0 . Therefore, α

Random Processes

43

3.13 Y (t ) = X (t ) ∗ h1 (t ) + X (t ) ∗ h2 (t ) = X (t − 1) + X (t − 2) . Thus, R yy (τ) = E {[ X (t − 1 + τ) + X (t − 2 + τ)] [ X (t − 1) + X (t − 2)]} = R xx ( τ) + R xx (τ) + R xx (τ + 1) + R xx (τ − 1) = 2 R xx ( τ) + R xx ( τ + 1) + R xx (τ − 1)

R yy (τ)

1

-2

-1

0

1

τ

2

3.14 (a) Y (t ) = N (t ) + N (t − 1) or, we have h(t )

N (t )

with

h(t ) = δ(t ) + δ(t − 1) .

From (3.135),

H ( f ) = 1 + e − j 2 π f and thus, H ( f )

2

(

Y (t )

S yy ( f ) = S nn ( f ) H ( f )

)(

2

where

)

= 1 + e − j 2 π f 1 + e + j 2 π f = 2(1 + cos 2π f ) .

Hence, the output power spectral density is S yy ( f ) = 2 rect( f ) [1 + cos 2π f ] .

S yy ( f ) 4

-1/2

0

1/2

f


44

(b) R wz ( τ) = E[W (t + τ) Z ∗ (t )] = E{[U (t + τ) + N (t + τ)][V ∗ (t ) + N ∗ (t )]} = Ruv (τ) + R nn ( τ) Since U (t ) and N (t ) are statistically independent and zero mean. S wz ( f ) = Suv ( f ) + Snn ( f ) as shown below. S wz ( f ) 3/2

-1/2

3.15

f

1/2

R yy (τ) = R xx (τ) ∗ h(τ) ∗ h(−τ) 142 4 43 4 g ( τ)

For −1 ≤ τ ≤ 0 , we have

-1

τ

0

+1

τ

τ

−1

−1

g ( τ) = ∫ (1 + t )e − ( τ −t ) dt = e − τ ∫ (1 + t )e t dt = τ − e −2 τ

For 0 ≤ τ ≤ 1 , we have

-1

0

τ

−1

0

0

τ

+1

g ( τ) = ∫ (1 + t )e − ( τ −t ) dt + ∫ (1 − t )e −( τ −t ) dt = e − (τ +1) + (2 − τ) − 2e − τ

Hence,

Random Processes

45

For τ ≥ 1 , we have

-1 0

1

−1

0

0

+1

τ

g ( τ) = ∫ (1 + t )e − ( τ −t ) dt + ∫ (1 − t )e − (τ −t ) dt = e − (τ +1) + e −(τ −1) − 2e − τ

Now, R yy (τ) = g (τ) ∗ h(τ). In the same manner, we have: For τ ≤ −1 , 0

1

R yy ( τ) = ∫ (t − e − 2t )e τ −t dt + ∫ [e −(t +1) + ( 2 − t ) − 2e −t ]e τ −t dt −1

0

∞

+ ∫ [e − (t +1) + e − (t −1) − 2e −t ]e τ −t dt 1

= e − τ (−

1 1 3 2 −3 − e + e + e −1 − e − 2 ) 3 3 3

For −1 ≤ τ ≤ 0 , 0

1

R yy ( τ ) = ∫ (t − e − 2t )e τ −t dt + ∫ (e −t +1 + 2 − t − 2e −t ) e τ −t dt τ

0

∞

+ ∫ [e − (t +1) + e − (t −1) − 2e −t ]e τ −t dt 1

1 2 = e τ [( τ + 1)e − τ − e −3τ + e −3 + e −1 − e − 2 ] 3 3

For 0 ≤ τ ≤ 1 , ∞

1

R yy ( τ) = ∫ [e − (t +1) + 2 − t − 2e −t ]e τ −t dt + ∫ [e − (t +1) + e − (t −1) − 2e −t ]e τ −t dt τ

1

1 2 2 1 = e [ e − 2 τ −1 + e − τ − τe − τ − e −3τ + e −3 + e −1 − e − 2 ] 2 3 3 2 τ


46 For τ ≥ 1 , ∞

1 − τ −1 1 − τ +1 − τ e + e −e . 2 2

Ryy (τ) = ∫ [e − (t +1) + e− (t −1) − 2e −t ]eτ −t dt = τ

3.16

2

S yy ( f ) = H ( f ) S xx ( f ) .

H( f ) =

The transfer function of the RC network is

1 . Hence, 1 + j 2π f RC S xx ( f ) =

∞

∫e

−α τ − j 2π f τ

e

dτ =

−∞

⇒ S yy ( f ) =

2α α 2 + 4π2 f 2

2α 2

2

2

(α + 4π f )(1 + 4π 2 f 2 R 2 C 2 )

3.17 The transfer function of the RLC network is

H( f ) =

1 jωc 1 R + jωL + jωC

=

1 2

1 − ω LC + jωRC

The mean of the output is m y (t ) = m x (t ) H (0) ⇒ m y = m x = 2 . Also, 4

2

S yy ( f ) = H ( f ) S xx ( f ) where S xx ( f ) = 4δ( f ) +

Therefore, S yy ( f ) =

4 + 4π 2 f

 1 4δ( f ) + (1 − ω LC ) + (ωRC )  1+ π2 f 1

2

2

2

2

2

  

3.18 The spectrum of S nn ( f ) does not contain any impulse at f and thus, E[ N (t )] = E[ N (t + τ)] = 0 . The samples at t and t + τ are uncorrelated if C nn (t , t + τ) = 0 . It follows that the samples are uncorrelated provided R nn (τ) = 0. Hence,

Random Processes

R nn (τ) = E[ N (t + τ) N (t )] =

∞

∫ S nn ( f )e

j 2π f τ

df =

−∞

47 B

∫

−B

N 0 j 2π f τ sin 2πBτ e df = N 0 B . 2 2πBτ

From the plot of R nn (τ) , Rnn (τ)

N0B

τ −4 2B

−3 2B

−2 2B

−1 2B

We observe that R nn (τ) = 0 for τ = sampling rates are

1 2B = fs = , τ k

3.19 (a) Nyquist rate T =

1 2B

0

2 2B

3 2B

k , k = ±1, ± 2, ... . 2B

4 2B

Therefore, the

k = 1, 2, 3, K .

1 1 = sec . 2 fc 2

1 − f (b) S xx ( f ) = rect( f ) ∗ rect( f ) =  0 ⇓

,

f <1

, otherwise

R xx ( τ ) = sinc τ ⋅ sinc τ = sinc 2 τ. 1 1 E{X (nT ) X [(n + 1)T ]} = R xx (T ) = R xx ( ) since T = sec . Therefore, 2 2 π sin 2 ( ) 2 2 1 2 1 2 =  2  and ρ =  2  since the process is zero R xx ( ) = sinc ( ) =   2 2 ( π 2) 2 π π mean and stationary; that is, a shift in time does not change ρ.

or,


48

3.20 From (3.31), (E[ X (t )] )2 = lim R xx ( τ) = 4 ⇒ E[ X (t )] = ±2. The mean of τ→∞

t

t

0

0

Y(t) is E[Y (t )] = ∫ E[ X ( τ)]dτ = ± ∫ 2dτ = ±2t ⇒ Y(t) is not stationary. t1 t 2

3.21 If R xx (t1 , t 2 ) = 2δ(t1 − t 2 ), then R yy (t1 , t 2 ) = 2 ∫ ∫ δ(α − β) dβ dα . 0 0

We have 2 cases: t1 > t 2 and t1 < t 2 . t 2 t1 t2  Case1: t1 > t 2 ⇒ R yy (t1 , t 2 ) = ∫  ∫ 2δ(α − β)dα  dβ = ∫ 2dβ = 2t 2 .  0 0 0

β

α=β

t2

t2

0

α

t1

t1 t 2 t1  Case2: t1 < t 2 ⇒ R yy (t1 , t 2 ) = ∫  ∫ 2δ(α − β)dβ dα = ∫ 2dα = 2t1 .  0 0 0

β t2

α=β

t1

0

t1

α

2t1 , t1 < t2 Therefore, Ryy (t1, t2 ) = 2 min(t1, t2 ) =  . 2t2 , t2 < t1

Random Processes

49

 n! σ n  3.22 (a) I a = ∫ X (t )dt . From (2.80), E[ X ] =  (n / 2)! 2 n 2  0  0 1

, n even

n

Hence, E[ I a4 ] =

4! σ

4

2! 2 2

.

.

, n odd

The variance of Ia is σ i2a = E[ I a2 ] − E 2 [ I a ] with

1

E[ I a ] = ∫ E[ X (t )]dt = 0 and E[ I a2 ] = 0

2 . 3

Hence, σ 2 = E[ I a2 ] =

2 . 3

After

4

2 4!   4 3 = . substitution, we obtain E[ I a4 ] = 2! 4 3

(b) E[ I a I b ] = E[ I a ]E[ I b ] = 0 since E[ I a ] = 0 and the random variable Ib is obtained independently from Ib. (c)

T  T The mean of Ic is E[ I c ] = E  ∫ X (t )dt  = ∫ E[ X (t )]dt = 0. Hence,  0  0

var[ I c ] = E[ I c2 ]. Using (3.203), the variance of Ic is

var[ I c ] =

or, var[ I c ] =

T

T



τ

−T

−T



1

∫ (T − τ ) R xx (τ)dτ = T ∫ 1 − T  R xx (τ)dτ ≈ T ∫ R xx (τ)dτ = T . −1

T

T

1

1

−T

0

0

0

∫ (T − τ ) R xx (τ)dτ =2 ∫ (T − τ) R xx (τ)dτ = 2∫ T (1 − τ)dτ − 2∫ τ(1 − τ)dτ

1 =T − ≈T 3

for T >> 1.

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

t

0

0

E[Y (t )] = E[ ∫ X (τ)dτ] = ∫ E[ X (τ)]dτ

But, E 2 [ X (t )] = lim R xx ( τ) = 1 ⇒ E[ X (t )] = ±1 τ→∞


50 t

Therefore, E[Y (t )] = ∫ (±1)dτ = ±t , which is function of time t ⇒ Y (t ) is not 0

stationary. t2 t1  t1 t 2 (b) R yy (t1 , t 2 ) = E[Y (t1 )Y (t 2 )] = E  ∫ X (α)dα ∫ X (β)dβ = ∫ ∫ R xx (α, β)dαdβ  0  0 0 0 t1 t 2

= ∫ ∫ Rxx (α − β) dαdβ 0 0

and Xˆ (t ) orthogonal ⇒ R xxˆ ( τ) = 0. From (3.225), R xxˆ ( τ) = Rˆ xx (− τ) = − Rˆ xx (τ) which is not zero for all τ. ⇒ (a) is False.

3.24 (a)

X (t )

~ ˆ ˆ (b) j H { X (t )} = j H { X (t ) + jXˆ (t )} = j[ Xˆ (t ) + Xˆ (t )] , but Xˆ (t ) = − X (t ) and ~ ~ hence, j H { X (t )} = j Xˆ (t ) + X (t ) = X (t ) ⇒ (b) is true.

(c) If X 1 (t ) = X (t )e j 2 πf 0t is an analytic signal ⇒ S x1 x1 ( f ) = 0 for f < 0 . R x1 x1 (τ) = E[ X 1 (t + τ) X 1∗ (t )] = E[ X (t + τ)e jω0 (t + τ) X ∗ (t )e − jω0t ] = R xx (τ)e jω0 τ .

The power spectral density of the process X 1 (t ) is then S x1 x1 ( f ) = S xx ( f − f 0 ) , which is zero if f 0 > f c so that all the spectrum will be shifted to the right. ~ (d) E[ X 2 (t )] = R ~x ~x (0) =

∞

∫

−∞

4 S ( f ) S ~x ~x ( f ) =  xx  0 fc

Hence,

∫ 4S xx ( f )df 0

fc

S ~x ~x ( f )df = ∫ 4 S xx ( f )df , since from (3.235), 0

, f >0 , f <0

.

fc

= 2 ∫ 2S xx ( f )df = 2 E[ X 2 (t )] ⇒ (d ) is true. 0

Also, ~ E[ X 2 (t )] = R ~x ~x (0) = 2[ R xx (0) + jRˆ xx (0)] from (3.233) ⇒ Rˆ xx (0) = 0, but Rˆ xx (− τ) = − Rˆ xx (τ) from (3.225). At

possibly true if

τ = 0 , we have

Random Processes

51

~ Rˆ xx (−0) = − Rˆ xx (0) and thus, Rˆ xx (0) = 0 ⇒ E[ X 2 (t )] = 2 R xx (0) = 2 E[ X 2 (t )] ⇒ (d ) is true.

3.25 (a) The equivalent circuit using a noiseless resistor is Vn (t)

±

+ R

L

C

v0(t)

⇐ Z ( jω) _

1 jωC

The transfer function H v0vn ( jω) =

1 + R + jωL jωC the power spectral density of v 0 (t ) is

=

1 (1 − ω 2 LC ) + jωRC

2kTR

2

S v0v0 ( f ) = H v0vn ( jω) S vn vn ( f ) =

2

(1 − ω LC ) + jωRC

. Hence,

.

(b) The input impedance is 1 ( R + jLω) jωC ωL(1 − ω 2 LC ) − ωR 2 C R Z ( jω) = = + j 2 2 2 1 (1 − ω 2 LC ) 2 + (ωRC ) 2 + ( R + jωL) (1 − ω LC ) + (ωRC ) jωC

But Nyquist theorem says S vv (ω) = 2kT ℜe {Z ( jω)} =

obtained in (a).

2kTR (1 − ω 2 LC ) 2 + (ωRC ) 2

which agrees with the result


52

3.26 (a) The equivalent circuit with noiseless resistors is Vn2 (t )

±

+

R1 (T1 )

R1 (T1 )

Vn1 (t )

v0(t)

C

± _ 2

S v0v0 ( f ) = S v1v1 ( f ) + S v2v2 ( f ) = S ne ne ( f ) H 1 ( f ) + S ne 1

1

n 2 e2

( f ) H2( f )

2

Using superposition principle, the power spectral density at the terminal pairs for each source is S v1v1 ( f ) = 2kT1 R1

1 1 + [ωC ( R1 + R 2 )] 2

and S v2v2 ( f ) = 2kT2 R 2

1 1 + [ωC ( R1 + R 2 )] 2

.

Hence, the output power spectral density is S v0 v0 ( f ) =

2k (T1 R1 + T2 R 2 ) 1 + [ωC ( R1 + R 2 )] 2

(b) In order to determine the autocorrelation function, we rewrite S v0 v0 ( f ) as

k (T1 R1 + T2 R 2 ) S v0 v0 ( f ) = ( R1 + R 2 )C

  1 2  R R C ( + ) 2  1  1 + 4π 2 f [( R1 + R 2 )C ] 2

Hence, Rvv (τ) =

  k (T1 R1 + T2 R 2 ) τ exp −  R R C ( R1 + R 2 )C ( ) + 1 2  

(c) The mean square value is Rvv (0) =

k (T1 R1 + T2 R 2 ) ( R1 + R 2 )C

2

Random Processes

R1 , R 2 , T1 , T2 and C , we obtain

Substituting for the given values of Rvv (0) = 0.457 × 10 6.76 10

−6

−10

.

Therefore,

53

the

root

mean

square

value

is

volts = 6.76 µ V .

3.27 The equivalent circuit using a noiseless resistor is Vn (t )

R

± I(t)

L

(a) The transfer function relating I (t ) to V n (t ) is H inn ( jω) = Therefore,

the

S ii (ω) = H ivn ( jω)

power 2

spectral

= S vn vn (ω) =

2kTR 2

R + ( ωL ) 2

density

of

1 . R + j ωL I (t )

is

.

(b) From (3.244), we need to determine the power spectral density of the short-circuit current. Hence, we have, for the circuit below, R L

Yin =

Yin

1 1 R ωL = = −j 2 Z in R + jωL R 2 + (ωL) 2 R + (ωL) 2

Therefore, the power spectral density of the short-circuit current is S ii (ω) = 2kT ℜe {Yin } = 2kT

R R 2 + (ωL) 2


54 3.28 (a) Y (t ) =

∞

∫ X (t − α)h(α)dα .

The mean of Y(t) is

−∞

E[Y (t )] =

∞

∫ E[ X (t − α)h(α)]dα = m x H (0)

−∞ ∞

where, H (0) = ∫ e − α dα = 1. Hence m y = m x . 0

(b) Since E[ X (t )] = m x = 0, the mean of Y(t) is E[Y (t )] = 0 and the variance is E[Y 2 (t )] =

∞ ∞

∫ ∫ R xx (β − α)h(α)h(β)dαdβ.

−∞ −∞

The autocorrelation function of the input is R xx (τ) = kδ(τ) , where k is a constant. Therefore, E[Y 2 (t )] =

∞ ∞

∞

−∞ −∞

−∞

∫

∫ kδ(β − α)h(α)h(β)dαdβ = k

∫

∞

h 2 (α)dα = k ∫ e − 2α dα = 0

k . 2

3.29 Since S nn ( f ) does not have an impulse at f = 0 , E[ N (t )] = 0 and the mean of the output of the linear filter is E[Y (t )] = E[ N (t )]H (0) = 0. Hence, the variance of Y(t) is ∞

σ 2y = E[Y 2 (t )] = R yy (0) =

∫ S yy ( f )df

−∞

where, S yy ( f ) = S nn ( f ) H ( f )

2

=

N0 2 H( f ) . 2

  f   K 1 −  , B  The system function is given by H ( f ) =    ,  0

Therefore,

f ≤B f >B

Random Processes

σ 2y =

0

∫

−B

=

N0 2

2

55 2

2

B B   N0   f  f  f  2  K 1 df K 1 df N K + + − =     1 −  df . 0     ∫ ∫ 2   B  B   B  0 0

N 0 BK 2 3

Chapter 4 Discrete Time Random Processes 4.1 (a) Using det( A − λI ) =

2−λ

−2

1

1− λ

1

1

3

−1− λ

3 = 0 , we obtain

λ 1 = 3 , λ 2 = 1 and λ 3 = −2 .

a  2 − 2 3  a  Then, Ax1 = λ 1 x1 ⇒ 1 1 1  b  = 3b   c  1 3 − 1  c 

Solving for a, b and c, we obtain a = b = c = 1 and thus, 1 x1 = 1 1 − 1 − 0.786   Similarly, Ax 2 = λ 2 x 2 ⇒ x 2 =  1  and Ax 3 = λ 2 x 3 ⇒ x 3 =  − 0.071 .  1   1  0.1 0.4  1 − 1 − 0.786  0.5    −1 The modal matrix is then M = 1 1 − 0.071 , M = − 0.5 0.83 − 0.33 1 1  0 − 0.93 0.93  1  The Jordan form is

56

Discrete Time Random Processes

J=M

−1

0.1 0.4  2 − 2 3  1 − 1 − 0.786  0.5  AM = − 0.5 0.83 − 0.33 1 1 1  1 1 − 0.071  0 − 0.93 0.93  1 3 − 1 1 1 1  3 0 0  = 0 1 0  0 0 − 2 4−λ

(b) A − I λ =

1 0

−2

0

2−λ 0 =0 ⇒ 0 6−λ

λ 1 = 3 + j, λ 2 = 3 − j and λ 3 = 6

a   4 − 2 0  a  a = 1, b = 0.5 − j 0.5     Solving Ax1 = λ 1 x1 ⇒ 1 2 0 b  = (3 + j ) b  ⇒ and c = 0  c  0 0 6  c 

 1 1 Thus, x1 =  − j 2 0 

  1  1 1 1  and x 2 =  + j  . 2 2  0 2   

0 Again, solving Ax 3 = λ 3 x 3 we obtain x 3 = 0 1

The modal matrix is then 1 1 0  0.5 − j 0.5 j 0   −1 M = 0.5 − j 0.5 0.5 + j 0.5 0 ⇒ M = 0.5 + j 0.5 − j 0   0 0 1 0 0 1

and J = M

−1

0 0 3 + j  3 − j 0 AM =  0  0 0 6

57

Signal detection and estimation

58 (c) Similarly, we solve

A− I λ =

4−λ 2 1 0 6−λ 1 = 0 ⇒ λ1 = λ 2 = λ 3 = 4 0 −4 2−λ

Note that we have an algebraic multiplicity of r = 3 . Solving, A − Iλ λ = 2 = 1 ⇒ degeneracy q = 3 − 1 = 2 ; that is, we have two eigenvectors and one generalized eigenvector. Thus, 1  1 Ax1 = 4 x1 ⇒ x1 = 0 or x 2 =  1  0 − 2

Solving for the generalized eigenvector, we have

( A − 4 I ) x 22 = x 2 ⇒ x 22

Therefore, the modal matrix is M = [x1

⇒ M −1

x2

1  = 0 0

1 1 0  x 22 ] = 0 1 0 0 − 2 1

4 1 − 1 0     = 0 1 0 and the Jordan form is J = M −1 AM = 0  0 2 1 0

0 4 0

0  1  4

4.2 (a) Solving for the determinant of A, we have det( A) = −6 ≠ 0 ⇒ the matrix is of full rank rA = 3. −1 − λ

(b) Solving det( A − Iλ ) = 0 ⇒

1 0

0 −2 λ 1 = −0.5858  −1 − λ 3 = 0 ⇒  λ 2 = −3 λ = −3.4142 −1 −5−λ  3


59

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

(c) Solving Av = λv , we obtain − 0.7296 v1 = − 0.6670 ,  − 0.1511

 − 0.4082 v 2 =  0.8165  − 0.4082

 0.4042 and v 3 = − 0.7737  .  0.4879

4.3 The characteristic equation is 3− λ 0 0 1 0 2−λ 0 0 A − Iλ = = (2 − λ ) 3 (3 − λ ) 1 1 3− λ 1 −1 0 0 1− λ

λ = 2 with algebraic multiplicity m1 = 3 ⇒ 1 λ 2 = 3 with algebraic multiplicity m 2 = 1

Note that the rank of A − Iλ 1 = 2 = r . Thus, q1 = n − r = 4 − 2 = 2 . Thus, for λ1 = 2 , we have 2 eigenvectors and 1 generalized eigenvector since m1 = 3.

0 1 1 0 λ = 2 ⇒ Ax = 2 x ⇒ x1 =   and x3 =   . The generalized eigenvector is − 1 0     0 − 1

( A − 2 I ) x12 = x1 ⇒ x12

0 0 For λ = 3, Ax 4 = 3 x 4 ⇒ x 4 =   − 1   0

0 0 =  − 1   1


60

Hence, the modal matrix is M = [ x1

2 0 Note that Λ = M −1 AM =  0  0

1 2 0 0

0 0 2 0

x12

x3

1 0 0 0 0 0 1 0  x4 ] =   0 − 1 − 1 − 1   − 1 1 0 0 

0 0 0  3

4.4 Let the M roots of λ i , i = 1, 2, …, M, be the eigenvalues of R, then det( R − Iλ ) = 0 . Also, det( R − Iλ) = det[ R( I − R −1 λ)] = det( R) det( I − R −1λ ) = 0

Since the correlation matrix R is nonsingular (det( R) ≠ 0) , then   1 det( I − R −1 λ) = 0 = det  I   − R −1  λm = 0   λ

The eigenvalues are non-zero for the non trivial solution (λ ≠ 0) and thus,  1  1  det  R −1 − I   = 0 , which means that , i = 1, 2, L , M , are eigenvalues of λi  λ   R −1 .

4.5 From (4.121), two eigenvectors v i and v j are orthogonal if v iH v j = 0, i ≠ j.

From the definition, Rv i = λ i v i

(1)

Rv j = λ j v j

(2)

and

Premultiplying both sides of (1) by v Hj , the Hermitian vector of v j ,we obtain


61

v Hj Rv i = λ i v iH v i

(3)

Since the correlation matrix R is Hermitian, R H = R . Taking the Hermitian of (2), we have v Hj R = λ j v Hj

(4)

since λ i is real. Postmultiplying (4) by v i yields v Hj Rv i = λ j v Hj v i

(5)

(λ i − λ j )v Hj v i = 0

(6)

Subtracting (5) from (3), we obtain

which yields v Hj v i = 0 sine λ i ≠ λ j . Therefore, the eigenvectors v i and v j are ortogonal. 4.6 Let v1 , v 2 ,..., v M be the eigenvectors corresponding to M eigenvalues of the correlation matrix R. From (4.120), the eigenvectors are linearly independent if a1 v1 + a 2 v 2 + L + a n v n = 0 a1 = a 2 = L = a n = 0 . Let Ti = R − I λ i , for Ti v j = (λ j − λ i )v j if i ≠ j. Multiplying (1) by T1 gives

(1) then

Ti v i = 0

a1 (λ 2 − λ 1 )v 2 + a 3 (λ 3 − λ 1 )v 3 + L + a n (λ n − λ 1 )v n = 0

and

(2)

Similarly multiplying (2) by T2 and then T3 and so on until Tn −1 , we obtain a 3 (λ 3 − λ 1 )(λ 3 − λ 2 )v 3 + L + a n (λ n − λ 1 )(λ n −1 − λ 2 )v n = 0

(3)

M a n −1 (λ n −1 − λ 1 )(λ n −1 − λ 2 ) L (λ n −1 − λ n − 2 )v n −1 + a n (λ n − λ 1 )(λ n − λ 2 ) L (λ n − λ n − 2 )v n = 0 (4) a n (λ n − λ 1 )(λ n − λ 2 ) L (λ n − λ n − 2 )(λ n − λ n −1 )v n = 0

(5)


62

From (5), since (λ n − λ i ) ≠ 0 for i ≠ n. ⇒ an = 0. Using (5) and (4), we see again a n −1 = 0 , and so on going backward until Equation (1). Hence, a1 = a 2 = L = a n −1 = a n = 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 λ 1 and λ 2 are orthogonal and A has the a form A =  11 a12

a12  since it is symmetric. a 22 

Let X = x x1 + y x 2 . Then AX = x A x1 + y A x 2 = x λ x1 + y λ 2 x 2 since Ax i = λx i . Also, X T AX = ( x x1 + y x 2 )( x λ 1 x1 + y λ 2 x 2 ) = x 2 λ 1 + y 2 λ 2 = 1

(1)

The equation of the ellipse has the form x2 a

2

+

y2 b2

=1

Therefore, (1) represents an ellipse for a = 1 / λ 1 and b = 1 / λ 2 . Assuming λ 1 > λ 2 , then a is the minor axis and b is the major axes and the ellipse is a shown below. y

x1

x2

x

1 λ2

1 λ1


63

5 3 (b) For A =   , we solve for the eigenvalues ⇒ 3 5 λ − 5 − 3  2 det(λI − A) = det   = (λ − 5) − 9 = (λ − 8)(λ − 2) ⇒ λ 1 = 8 = λ max and  − 3 λ − 5 λ = 2 = λ min . Solving for the eigenvectors, we have ( A − 8 I ) x1 = 0 ⇒ x1 =

1 1 1  1 1 , and ( A − 2 I ) x 2 = 0 ⇒ x 2 =  . 2 2  − 1

Note that x1 and x2 are orthogonal.

(1 / (1 /

From (a), the semi-major axis is

) ( ) and the semi-minor ) = (1 / 8 ) = 0.354 . The ellipse is shown below

λ min = 1 / 2 = 0.707 λ max

y 0.354

x1

x

0.707

x2

4.8 (a) The second order difference equation of the AR process is. X (n) = − a1 X (n − 1) − a 2 X (n − 2) + e( n)

and thus, the characteristic equation is 1 + a1 Z −1 + a 2 Z −2 = 0

axis

is


64

e( n )

X ( n)

∑ Z -1 -a1

X (n − 1) Z -1

-a2

X ( n − 2)

(b) Solving for the roots of the second order equation, we obtain P1 =

1 1 2 2    − a1 − a1 − 4a 2  and P2 =  − a1 + a1 − 4a 2    2 2

For stability of the system, the poles P1 and P2 must be inside the unit circle, that is P1 < 1 and P2 < 1 . Applying these two conditions, we obtain a1 + a 2 ≥ −1 a1 − a 2 ≥ −1

and −1 ≤ a 2 ≤ 1

4.9 (a) The Yule-Walker equations for the AR(2) process are R ω = r or R a = r . Applying this to our system, we have  r (0) r (−1)  ω1   r (1)  r ( −1) r (0)  ω  = r ( 2)   2   

For a real-valued stationary process r (−1) = r (1) , and thus solving the two equations in two unknowns, we obtain ω1 = − a1 =

r (1)[r (0) − r (2)] r 2 (0) − r 2 (1)


ω 2 = −a 2 =

65

r (0)r ( 2) − r 2 (1) r 2 (0) − r 2 (1)

where r (0) = σ 2x . (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 r (1) =

−a1 2 −a1 σ x = ρ1 σ 2x with ρ1 = 1+ a2 1+ a2

 a2  a2 and r (2) =  1 − a 2 σ 2x = ρ 2 σ 2x with ρ 2 = 1 − a 2  1+ a2  1+ a2  

4.10 The state diagram is

1/3

2/3

S1

1/2

1/2

S2

1

1/4

S3

S4 1/2

1/4

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


66

0.3 0.5

0.2 0.4

S1

S2

0.3

0.3 0.2

0.4

S3

0.4

(b) The n-step transition matrix is 0.4300 0.2700 0.3000 P (2) = P = 0.4000 0.2800 0.3200 0.3900 0.2700 0.3400 2

0.4130 0.2730 0.3140 P (3) = P = 0.4080 0.2720 0.3200 0.4050 0.2730 0.3220 3

0.4099 0.2727 0.3174 P (4) = P = 0.4088 0.2728 0.3284 0.4083 0.2727 0.3290 4

0.4093 0.2727 0.3180 P (5) = P 5 = 0.4090 0.2727 0.3282 0.4089 0.2727 0.3283

0.4091 0.2727 0.3182 P (6) = P 6 = 0.4091 0.2727 0.3282 0.4091 0.2727 0.3282 M


67

0.4091 0.2727 0.3182 P (20) = 0.4091 0.2727 0.3282 0.4091 0.2727 0.3282

(c) The state probabilities are given by P (n) = p T P n . Thus, p T (1) = p T P = [0.3900 0.2700 0.3400]

with P T = p(0) = [0.3 0.3 0.4] . p T (2) = p T P 2 = [0.4050 0.2730 0.3220] p T (3) = p T P 3 = [0.4083 0.2727 0.3150]

p T (4) = p T P 4 = [0.4089 0.2727 0.3183] p T (5) = p T P 4 = [0.4091 0.2727 0.3182] M p T (20) = p T P 20 = [0.4091 0.2727 0.3182]

4.12 (a)

0.25 0.5

S1 ≡ R

S2 ≡ N

0.5

0.25

0.25 0.25

0.5

S3 ≡ S

0.5


68 (b)

Rain

Nice Snow

Rain 0.500 0.250 0.250 P (1) = Nice 0.500 0.000 0.500 Snow 0.250 0.250 0.500 Rain

Nice Snow


Nice Snow

Rain 0.406 0.203 0.391 P (3) = Nice 0.406 0.188 0.406 Snow  0.391 0.203 0.406 Rain

Nice Snow


Nice Snow


Nice Snow

Rain 0.400 0.200 0.400 P (6) = Nice 0.400 0.200 0.400 Snow 0.400 0.200 0.400

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.


69

(c) Using P (n) = p T P n , we have p T (1) = p T P with P T = p(0) = [0.7 0.2 0.1] .

Therefore, p T (1) = [0.475 0.200 0.325] . p T (2) = p T P 2 = [0.419 0.200 0.381] p T (3) = p T P 3 = [0.404 0.200 0.395] p T (4) = p T P 4 = [0.401 0.200 0.399]

p T (5) = p T P 5 = [0.400 0.200 0.400] M

p T (20) = p T P 20 = [0.400 0.200 0.400]

Hence, the steady state distribution vector is  ω1  0.4 P = ω 2  = 0.2  ω 3  0.4

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

1- a

S0

1-b

S1 b

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

Chapter 5 Statistical Decision Theory H1 f Y | H1 ( y | H 1 ) > η 5.1 (a) The LRT is Λ( y ) = f Y |H ( y | H 0 ) < 0

1

H0

fY | H 0 ( y | H 0 )

1 2

f Y | H1 ( y | H1 ) y

0.693

1

2

H1 H1 1/ 2 > > ln(2η) , while for y > 2 , we η⇒ y We observe that for 0 ≤ y ≤ 2 ⇒ − y < < e H2 H2 always decide H0.

(b) For minimum probability of error criterion ⇒ C 00 = C11 = 0 and C01 = C10 = 1 1 ⇒ η = 1 ⇒ ln(2) = 0.693 ⇒ choose H0 2 otherwise choose H1. The minimum probability of error is

(i)

P0 =

P(ε) = P1

2

∫e

−y

dy + P0

0.693

0.693

∫ 0

72

1 dy = 0.355 2

for

0 ≤ y ≤ 0.693 ;

Statistical Decision Theory

(ii) Similarly, P1 =

73

2 ⇒ choose H1 for 1.39 ≤ y ≤ 2 and P(ε) = 0.308 . 3

H1 1 > (iii) P1 = , y 0 ⇒ always decide H1 and P(ε) = 0.288 . < 3 H0 Λ( y )

∞

H1

5.2 (a) Λ( y ) =

f Y | H1 ( y | H 1 ) > η⇒ f Y |H 0 ( y | H 0 ) < H0

1 2

1

2

Λ( y )

(i)

η<

1 , 2 η

1 2 y 1

2

Λ( y ) > η ⇒ always decide H1 Λ( y )

(ii) η >

1 , 2

η 1 2 y 1

0 ≤ y ≤ 1 , Λ( y ) < η ⇒ decide H0

1 ≤ y ≤ 2 , Λ( y ) > η ⇒ decide H1

2


74

Λ( y )

(iii) η =

1 , 2 η=

1 2 y 2

1

decide H1 or H0 in at the range 0 ≤ y ≤ 1 and decide H1 for 1 < y ≤ 2 . (b) (i) η < PF = ∫

1 , 2

the

probability

of

false

alarm

is

1

Z1

f Y | H 0 ( y | H 0 )dy = ∫ 1 dy = 1 . 0

The probability of detection is PD = ∫

(ii) η >

2

Z1

1 dy = 1 ⇒ PM = 1 − PD = 0 2 0

f Y | H1 ( y | H 1 )dy = ∫

2 1 , PF = ∫ f Y | H 0 ( y | H 0 ) dy = ∫ 0 dy = 0 and Z1 2 1

PD = ∫

2

Z1

1 1 1 dy = ⇒ PM = 2 2 2 1

f Y | H1 ( y | H 1 ) dy = ∫

5.3 Minimum probability of error criterion ⇒ C 00 = C11 = 0 and C 01 = C10 = 1 . (a) The conditional density functions are f Y |H 0 ( y | H 0 ) =

 ( y + A)2  exp −  2σ 2  2πσ 

f Y | H1 ( y | H 1 ) =

 ( y − A)2  exp −  2σ 2  2πσ 

1

1


Λ( y ) =

f Y | H1 f Y |H 0

=

2

2

75 H1

P exp[−( y − A) / 2σ ] > η= 0 P1 exp[−( y + A) 2 / 2σ 2 ] < H0

( y − A)2 + ( y + A) 2 ⇒ lnΛ( y ) = 2σ

2

2σ

2

H1 P > ln 0 < P1 H0

H1

⇒y

2 P > σ ln 0 < 2 A P1 H0

H1 2 P0 0.549σ 2 > σ ⇒y ln 3 = (b) P1 = < 2A A 3 H0

P1 = P0 ⇒ y

> 0 < H0

-A H0

H1

P1 =

fY | H1 ( y | H1 )

fY | H 0 ( y | H 0 )

H1

y

A

0 γ

H1

2 5 P0 − 256σ 2 > 0.405σ ⇒y = < A A 3 H0

As P1 increases ⇒ PD increases and PF increases, but PF increases at a faster rate.

5.4 The received signals under each hypothesis are H1 : Y = − A + N H0 :Y =

N

H2 :Y = A+ N


76

fY | H 0 ( y | H 0 )

fY | H1 ( y | H1 )

-A

0

-γ

H1

fY | H 2 ( y | H 2 )

y

A

γ

H0

H2

(a) By symmetry, we observe that the thresholds are –γ and γ, and P(error | H 1 ) = P (error | H 2 ) P(error | H 1 ) =

∞

∫

−γ

γ

P(error | H 2 ) =

∫

−∞

−γ

P(error | H 0 ) =

∫

−∞ ∞

= 2∫ γ

 ( y + A)2  exp −  dy 2σ 2  2πσ  1

 ( y − A)2  exp −  dy 2σ 2  2π σ  1

1

 y2 exp − 2  2σ 2πσ 

∞ 2   dy + ∫ 1 exp − y   2σ 2 γ 2πσ  

 y2 exp − 2  2σ 2πσ 

 dy  

1

 dy  

But P(error | H 1 ) = P (error | H 2 ) and hence, P(error) =

1 3

∞  ∞  ( y + A) 2   y2 − + 2 exp dy   ∫ exp − ∫  2σ 2 2σ 2  2πσ − γ  γ 

1

 ( y − A) 2   exp −  dy  2σ 2   −∞  γ

+

 dy  

∫


=

Now,

2 3

77

γ ∞  y2   ( y − A) 2    dy   ∫ exp − 2 dy + ∫ exp − 2σ 2   2πσ  γ  2σ   −∞

1

 γ2 ∂Pe = 0 ⇒ − exp − 2  2σ ∂γ 

2    + exp − (γ − A)  = 0 ⇒ γ = A  2 2σ 2   

(b) Substituting for the value of γ =

P(error) =

A and solving the integrals we obtain 2

 A  4  A  2  = Q  erfc    3  2 2σ  3  2 σ 

5.5 (a) 3 1 8 y + 8   1  f Y | H1 ( y | H 1 ) = f S ( s ) ∗ f N ( n) =  4  1 3 − y + 8 8   0

, − 3 ≤ y ≤ −1 , −1 ≤ y ≤ 1 , 1≤ y ≤ 3 , otherwise

as shown below fY | H1 ( y | H1 ) 1/4

-3

-2

-1

 ∞ 1  y+ 3 2 2  The LRT is then Λ( y ) =  1  1 3 − y + 2  2  ∞ 

1

2

, − 3 ≤ y ≤ −2 , − 2 ≤ y ≤ −1 , −1 ≤ y ≤ 1 , 1≤ y ≤ 2 , 2< y<3

3

y


78 (i) η =

1 , we have 4

Λ( y)

∞

∞

1 1/2

-3

1

-1

-2

η = 1/ 4 y

3

2

Λ(y) > η ⇒ always decide H1. ∞

(ii) η=1

∞ η=1

1 1/2

-3

1

-1

-2

2

y

3

2 cases: decide H1 when  H 0 when − 2 ≤ y < −1 and 1 < y ≤ 2 Λ( y ) = η ⇒   H 1 when − 3 ≤ y < −2 and − 1 < y ≤ 1 and 2 < y ≤ 3  H 0 when − 2 ≤ y < 2 or, decide H0 when Λ( y ) = η ⇒   H 1 when − 3 ≤ y < −2 and 2 < y ≤ 3

(iii) η = 2

Λ( y )

∞

∞ η=2

1 1/2

-3

-2

-1

1

decide H0 when −2 ≤ y ≤ 2 since Λ( y ) < η = 2 decide H1 when −3 ≤ y ≤ −2 and 2 ≤ y ≤ 3 since Λ( y ) > η

2

3

y


(b) PF = ∫ (i) η =

Z1

f Y | H 0 ( y | H 0 )dy and PD = ∫

Z1

79

f Y | H1 ( y | H 1 )dy

1 ⇒ PF = PD = 1 4

(ii) η = 1 ⇒ PF =

1 and PD = 0.625 or, PF = 0 and PD = 0.125 2

(iii) η = 2 ⇒ PF = 0 and PD = 0.125 (c) The ROC is shown below PD

η=

1

1 < η <1 2

0.625 1/2

η >1

1 2

0.162 1/2

5.6 (a) f S ( s ) =

∞

N0

−∞

0

∫ f SN ( s, n)dn = ∫ f N ( n) =

∞

α

∫N 0

α − αs e dn = αe − αs N0

e −αs ds =

0

PF

1 N0

∫ dn = αe

− αs

for all s ≥ 0

0

1 , 0 ≤ n ≤ N0 N0

(b)

f SN ( s, n) = f S ( s ) f N (n) ⇒ S and N are statistically independent.

(c)

f Y ( y ) = f S ( s ) ∗ f N ( n) =

∞

∫ f N (λ) f S ( y − λ)dλ 0

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


80

 1 − αy , 0 ≤ y ≤ N0  N (1 − e )  0 f Y ( y) =   1 {exp[− α( y − N )] exp(−αy )} , N ≤ y < ∞ 0 0  N 0 fY (y) 1 (1 − e − αN 0 ) N0

y N0

5.7 (a) The LRT is

Λ( y) =

1 −y e 2 1 2π

Solving, T ( y ) =

e

−

y2 2

H1 H1 1 > >  2  η ⇒ T ( y) = y 2 − y η =γ ln < <  π  2   H0 H0

1 2 y − y − γ = 0 ⇒ y = 1 ± 1 + 2 γ as shown below 2 T(y)

-2

-1

0

1

2

-1/2

To determine the decisions, we observe that we have 3 cases: (i) γ ≤ −

1 1 , (ii) − < γ < 0 and (iii) γ > 0 2 2

y


(i) γ ≤ −

81

π 2e

1 ⇒η≤ 2

T(y)

T ( y ) > γ ⇒ always decide H1 -2

-1

1

0

2 y

-1/2

(ii) −

γ

π <η≤ 2e

1 <γ<0⇒ 2

π 2

H0 H1 -2

y1

-1

H0

H1 y2

H1 y3

1

y4

2

y γ

-1/2

y1 = −1 − 1 + 2 γ , y 2 = −1 + 1 + 2γ , y 3 = 1 − 1 + 2γ and y 4 = 1 + 1 + 2 γ Decide H1 when y ≤ y1 , y 2 ≤ y ≤ y 3 and y ≥ y 4 . Decide H0 when y1 < y < y 2 and y 3 < y < y 4 . (iii) γ > 0 ⇒ η >

π 2

H1

H0 -2

-1

0

y1

γ 1

2

y y2

-1/2

H1


82

y1 = −1 − 1 + 2 γ and y 2 = 1 + 1 + 2γ . Decide H1 when y ≤ y1 and y ≥ y 2 Decide H0 when y1 < y < y 2 (b) P0 =

P 2 ⇒ η = 0 = 2 ⇒ γ ≈ 0.47 3 P1

PF = P( H 1 | H 0 ) = ∫ −1− 1+ 2 γ

=

1

∫

2π

−∞

PD = P ( H 1 | H 1 ) = ∫

e

Z1 −

f Y | H 0 ( y | H 0 )dy

y2 2

∞

1

∫

dy +

2π

1+ 1+ 2 γ

e

−1− 1+ 2 γ Z1

∫

f Y | H1 ( y | H 1 )dy =

−∞

−

y2 2

dy = 2Q( 2.393) = 0.02

1 +y e dy + 2 1+

∞

∫

1+ 2 γ

1 −y e dy = 0.09 2

(c) PF = 2Q(1 + 1 + 2γ ) PD = exp[−(1 + 1 + 2 γ )] ⇒ PM = 1 − PD = 1 − exp(−1 − 1 + 2γ )

The optimum threshold γ opt is obtained when PF = PM , or 1 − exp(−1 − 1 + 2 γ opt ) = 2Q(1 + 1 + 2γ opt )

(d) PF =

γ1

∫

−∞

1 2π

e

−

y2 2

dy +

∞

∫

γ1

1 2π

e

−

y2 2

dy = 0.2

 γ  or, 2Q ( γ 1 ) = 0.2 ⇒ Q( γ 1 ) = 0.1 ⇒ erf  1  = 0.2 ⇒ γ 1 ≅ 0.22 > 0 ; that is the  2 decision regions as given in (a) part (iii).


5.8 f Y | H1 ( y | H 1 ) =

f Y |H 0 ( y | H 0 ) =

83

 ( y − 1) 2  exp −  2  2π 

1

 y2   exp −  2  2π  

1

 ( y − 1) 2  H exp −  1 2  > 2π  1 ⇒ <  y2  1 exp −  H 0 2π  2 

1

(a) The LRT is Λ( y ) =

(b) PF = ∫

Z1

f Y | H 0 ( y | H 0 )dy =

⇒ Q (α ) =

∞

(c) PD =

∫

α = 2.581

∞

1

∫

2π

α

e

−

y2 2

H1 > 1 y < 2 H0

dy = 0.005

 α  1  = 0.005 ⇒ α ≈ 2.581 1 − erf   2   2  2

x ∞  ( y − 1) 2  1 −2 exp − e dx = Q (1.581) ≈ 0.013 dy = ∫ 2  2π α −1 2π 

1

5.9 The LRT is  ( y − m) 2  exp − k 2  f Y | H1 ( y | H 1 ) k =1 2π 2σ   Λ( y ) = = 2 K f Y |H 0 ( y | H 0 )  y  1 exp − k2  ∏  2σ  k =1 2π   K

∏

1

H1 > η < H0

H1 H1 2 Km > σ > ⇒ ∑ yk ln η + γ. as given in Example 5.2. Hence, T ( y ) < < m 2 13 k1 =2 144244 3 γ H0 T ( y) H 0 K


84

5.10

K

f Y |H 0 ( y | H 0 ) = ∏

k =1 K

f Y | H1 ( y | H 1 ) = ∏

k =1

 y2  exp − k2   2σ  2πσ 0 0   2  y  1 exp − k2   2σ  2πσ 1 1   1

H1 2σ 2 σ 2 > ⇒ T ( y ) = ∑ y k2 γ where, γ = 2 0 12 < σ1 − σ 0 k =1 H0 K

 σ  ln η − K ln 0 σ1 

  from Example 5.9. 

5.11 (a) The probability of false alarm is

PF = ∫

where, γ =

Z1

f Y | H 0 ( y | H 0 ) dy =

σ 2σ 02 σ12   ln η − ln 0 2 2  σ1 σ1 − σ 0 

PD = 1 − PM =

∞

∫ γ

1 2π σ 1

∞

∫ γ

1 2π σ 0

e

−

y2 2σ0

 γ dy = Q  σ0

   

  and K = 1 . 

e

−

y2 2 σ1

 γ   γ  dy = Q  ⇒ PM = 1 − Q   σ1   σ1 

(b) The ROC is PD versus PF.

For

σ12 = 2σ 02 = 2 , we have

 4 ln(η 2 )  PD = Q   and PF = Q[4 ln(η 2 )] for various values of η. Hence, 2   PD

1/2

1/2

PF


85

(c) The minimax criterion when C 00 = C11 = 0 and C01 = C10 = 1 yields PF = PM . Hence,  γ opt 1 − Q  σ1

5.12 (a) f Y | H 0 ( y | H 0 ) =

 γ  = Q opt   σ   0

 y2 exp − 2  2σ 2πσ  1

 .  

   

f Y | H1 ( y | H 1 ) =

 ( y + m) 2  exp −  2σ 2  2πσ 

f Y |H 2 ( y | H 2 ) =

 ( y − m) 2  exp −  2σ 2  2πσ 

1

1

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

f Y |H j ( y | H j ) P( H j ) f Y ( y)

1 and f Y ( y ) is common to all a posteriori probabilities ⇒ We choose 3 Hj for which f Y | H j ( y | H j ) is largest. This is equivalent to choosing Hj for which P( H j ) =

( y − m j ) is smallest. Hence, we have H1

H0 -m/2

-m

H2 m/2

0

y m

(b) The minimum probability of error is 3

P ( ε ) = ∑ P ( H j ) P (ε | H j ) = j =1

1 3 ∑ P(ε | H j ) where, 3 j =1


86

 ( y − m1 ) 2  exp − dy 2σ 2  2πσ − m / 2 

  m P(ε | H 1 ) = P y > − H 1  = 2   =

∞

∫

1 2π

m / 2σ

e

−

x2 2

∞

1

∫

 m dx = Q   2σ 

By symmetry, P(ε | H 1 ) = P (ε | H 3 ) and     m m m P (ε | H 0 ) = P Y > H 0  = P Y < and Y > H 0  2 2 2     =

=2

1 2πσ

1 2π

2 −m / 2 − y 2 σ 2 e dy

∫

+

∞ −x e 2

∫

m / 2σ

∫

2πσ m / 2

−∞ 2

2 ∞ − y 2 σ 2 e dy

1

 m dx = 2Q  ⇒  2σ 

P (ε ) =

4  m Q  3  2σ 

(c) The conditional density functions become 1

f Y |H 0 ( y | H 0 ) =

f Y | H1 ( y | H 1 ) =

2π

e

−

y2 2

 ( y − 1) 2  exp −  2  2π 

1

f Y |H 2 ( y | H 2 ) =

The boundary between H0 and H1 is y =

1 2 2π

e

−

y2 8

1 , while the boundary between H0 and 2

H2 is obtained from f Y |H 0 ( y | H 0 ) = f Y |H 2 ( y | H 2 ) ⇒ e

For the boundary between H1 and H2, we have

−

y2 2

=e

−

y2 8

⇒ y ≈ ±1.36


f Y | H1 ( y | H 1 ) = f Y | H 2 ( y | H 2 ) ⇒ e

−

( y −1) 2 2

1 − = e 2

y2 2

87

⇒ y1 ≈ 2.85 and y 2 ≈ −0.18

f Y | H 1 ( y | H1 )

fY | H 0 ( y | H 0 ) fY | H 2 ( y | H 2 )

-0.18

-1.36

H2

y

0

1 1.36

0.5

H1

H0 -1.36

-0.18

-1

2.85

H2 y

0

1 1.36

0.5

2

2.85

3

P(ε) = ∑ P( H j )P(ε | H j ) where, j =1

P (ε | H 0 ) = ∫

f (y Z1 U Z 2 Y | H 0

| H 0 ) dy =

−1.36

1

∫

−∞

2π

1

−

e

−

y2 2 dy

∞

+∫ 0

1 2π

e

−

y2 2 dy

−

( y −1) 2 2

= Q(1.36) + Q (0) = 0.527

P (ε | H 1 ) = ∫ =

Z0 U Z2

−1 / 2

∫

−∞

f Y | H1 ( y | H 1 )dy =

1 2π

e

−

x2 2

dx +

∞

∫

1.85

1 = Q  + Q(1.85) = 0.29 2

1/ 2

∫

2π

−∞

1 2π

e

−

x2 2

e

dx

( y −1) 2 2

dy +

∞

∫

2.85

1 2π

e

dy


88

P (ε | H 2 ) = ∫

Z 0 U Z1 ∞

−

1

x2 2

∞

1

∫

2 2π

−1.36

1

e dx − ∫ e 2π 5.7 2π = 1 − Q(2.72) − Q(5.7) ≈ 1 =

∫

f Y | H 2 ( y | H 2 )dy =

2.85

−

x2 2

e

−

y2 8

5. 7

dy =

∫

− 2.72

1 2π

e

−

x2 2

dx

dx = Q( −2.72) − Q(5.7)

− 2.72

1 ⇒ P(ε) = [ P (ε | H 0 ) + P (ε | H 1 ) + P(ε | H 2 )] ≈ 0.6 3

5.13

K

f Y |H 0 ( y | H 0 ) = ∏

k =1

K

f Y | H1 ( y | H 1 ) = ∏

k =1

 y2 exp − k2  2σ 2πσ  1

1 σ 2 + σ 2m

   

  y k2 exp −  2 2  2(σ + σ m ) 

 σ2 = Λ( y ) ⇒  2  σ + σ2 f Y |H 0 ( y | H 0 ) m  f Y | H1 ( y | H 1 )

   

K /2

H1 2  T  > σm exp  y y 1  2 2σ 2 ( σ 2 + σ m )  <  H0

Taking the logarithm on both sides and rearranging terms, we obtain the decision rule H1

yT y

2 2 2 2  σ2 + σm > σ (σ + σ m ) K ln 2 2  σ < σm  H0

or, H1 K

∑ y k2

k =1

> γ < H0

 ≡γ  


89

5.14 The conditional density functions are

f Y | H1 ( y | H 1 ) = where y = [ y1

y2

1

[2π(1 + σ )] 2 m

K /2

 1 yT exp − 2  2(1 + σ m )

 y 

L y K ]T and f Y |H 0 ( y | H 0 ) =

1 ( 2π) K / 2

 1  exp − y T y   2 

The LRT is then  1 Λ( y) =  1+ σ2 m 

   

K /2

H1 2  >    σ m η exp  y T y   2 2 ( 1 )   < + σm   H0

K  1 + ln Taking logarithm ⇒ y T y 2 2 2(1 + σ m ) 2  1 + σ m 2 σm

H1  >  ln η  <  H0

H1

T

⇒y y

2 > 2(1 + σ m ) K ln(η + σ 2m + 1) ≡ γ 1 2 < 2 σm H0

or, H1 K

∑ y k2

k =1

> γ < 1 H0

We observe that the LRT does not require knowledge of σ 2m to make a decision. Therefore, a UMP test exists.


90 K

f Y |H 0 ( y | H 0 ) = ∏

5.15

k =1

K

f Y | H1 ( y | H 1 ) = ∏

k =1

 y2  exp − k   2  2π  

1

 ( y − m) 2  exp − k  2 2π  

1

H1 f Y | H1 ( y | H 1 ) > K  K  η ⇒ Λ( y ) = exp m ∑ y k − m 2  (a) Λ( y ) = 2 f Y |H 0 ( y | H 0 ) <  k =1  H0 H1

or,

K

∑ yk

k =1

2 > 2 ln η + Km ≡γ. < 2m

Therefore, a test can be conducted without

H0 knowledge of m ⇒ A UMP test exists.

(b)

K

PF = 0.05 ⇒ The test decides H0 when T ( y ) = ∑ y k > γ , where T is k =1

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

∞

∫ f T |H γ

Using PD =

∞

∫ f T |H γ

1

0

 γ  1  γ  = 1 − erf  (t | H 0 ) dt = Q  2    K  2K

  = 0.05  

(t | H 1 ) dt > 0.9 where T is Gaussian with mean Km under H1,

we obtain from the table in the appendix K ≈ 16 . 5.16 Since the observations are independent, the LRT becomes

f Y | H1 ( y | H 1 )

θ =  0 Λ( y ) = f Y | H 0 ( y | H 0 )  θ1

K

  1  1  exp − ( y1 + y 2 + L + y K ) − θ    1 θ0

H1  >   < η  H0


θ ⇒ Λ( y ) =  0  θ1

K

91

K  θ − θ0   exp  ∑ y k  1   k =1  θ 0 θ1 

H1  >   < η  H0

Taking the natural logarithm and simplifying the expression, we obtain H1

K

T ( y) = ∑ y k k =1

> θ 0 θ1 < θ1 − θ 0 H0

 θ  ln η − ln 1  θ0 

 =γ  

For a UMP test of level α, we need P[T (Y ) | H 0 > γ ] = 0.05 or P[T (Y ) | H 0 ≤ γ ] = 0.95

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

] [

[

] [

][

]

[

Φ t (ω) = E e jωY = E e jω(Y1 +Y2 +L+YK ) = E e jωY1 E e jωY2 L E e jωYK

]

= Φ y1 (ω)Φ y2 (ω) L Φ y K (ω)

since the Yks, k = 1, 2, L , K Φ t (ω) = (1 − jθω)

−K

.

are statistically independent.

From (2.93),

Hence, from (2.102), T (Y ) is a gamma distribution

G ( K , P) with density function

1  t K −1 e −t / β  f T (t ) =  Γ( K )β K  0 

, t >0 , otherwise

Therefore, for K = 21 , (see Table 9 page 456, Dudewicz1) P[T (Y ) | H 0 ≤ γ ] = 0.95 ⇒

γ = 29.062 ⇒ γ = 290.62 θ0

The test decides H1 (rejects H0) if T (Y ) > 290.62 1 Dudewicz, E. J., Introduction to Statistics and Probability, Holt, Rinehart and Winston, New York, 1976.


70

 b + a − a 2 − ab  a+b P (1) =  2  b − b + ab + b a+b 

a − a + a 2 + ab   1 − a a  a+b =  2 a + b − ab − b   b 1 − b a+b 

 b + a (1 − a − b) n a  1 − a a+b P (n + 1) = P (1) P n (1) =   n  b 1 − b  b − b(1 − a − b)  a+b

a − a (1 − a − b) n   a+b  n a + b(1 − a − b)   a+b

Let x = 1 − a − b , then  b + ax n a  a + b 1 − a P (n + 1) =   n  b 1 − b  b − bx  a + b

a − ax n   a+b  n a + bx  a + b 

=

1  b + ax n − ab − a 2 x n + ab − abx n  a + b b 2 + abx n + b − bx n − b 2 + b 2 x n

a − ax n − a 2 + a 2 x n + a 2 + abx n   ab − abx n + a + bx n − ab − b 2 x n 

=

1 b + ax n (1 − a − b) a − ax n (1 − a − b)   a + b b − bx n (1 − a − b) a − bx n (1 − a − b) 

=

1 b + ax n +1  a + b b − bx n +1

a − ax n +1   = P (n + 1) a + bx n +1 

and P (n + 1) = P (1) P n (1) is verified. The limiting transition matrix is  b 1 b a   a + b lim P ( n) =  = n →∞ a + b b a   b a + b

if x < 1 . 0 < b <1.

Note that x = 1 − a − b

a  a +b a   a +b

and thus, x < 1 requires 0 < a < 1 and


(c) For the special case, a = b = 0 , we have n

1 0 P (n) = P (0)   = P ( 0) 0 1  Also, for a = b = 1 , we have 0 1  P (n) = P (0)   1 0

n

1 0 n = 1 ⇒ P (1) =   0 1  0 1  n = 2 ⇒ P ( 2) =   = P ( 0) 1 0

Continuing for all values of n, we observe that

, for n even  P (0)  P ( n) =  0 1    P (2) = 1 0 , for n odd    and thus, the limiting state probabilities do not exist.

71

Chapter 6 Parameter Estimation 6.1 Yk = a + bx k + Z k . Yk is Gaussian with mean a + bx k and variance σ 2 . Since Y1 , Y2 , L , Y K are statistically independent, the likelihood function is K

f Y ( y ) = L(a, b) = ∏ f Yk ( y k ) k =1

⇒ L ( a, b) =

(

1 2πσ

)

K

 1 exp− 2  2σ

K



k =1



∑ [ yk − (a + bxk )]2 

Taking the logarithm, we have ln L(a, b) = − K ln σ −

K 1 ln 2π − 2 2 2σ

K

∑ ( y k − a − bx k ) 2

k =1

Hence,

∂ ln L(a, b) ∂ ln L(a, b) = =0 ∂a ∂b 1

⇒ aˆ =

1 K

1

∑ yk − K

∑ y k xk − K ∑ y k xk ∑

x k2

92

1 − K

∑ xk ∑ xk

∑ xk

Parameter Estimation

93

1

bˆ =

∑ yk xk − K ∑ yk ∑ xk 1

∑ xk − K ∑ xk ∑ xk 6.2 The conditional density function is f Y σ ( y σ) =

(a) ln f Y σ ( y σ) = − ln f Y σ ( y σ)

y2 1 ln 2π − ln σ − 2 . Hence, 2 2σ

1 y2 1 + = ( y 2 − σ 2 ) , which cannot ∂σ σ σ3 σ3 c(σ)[σˆ ( y ) − σ] ⇒ No efficient estimate exists for σ .

(b)

=−

ln f Y σ ( y σ) 2

∂ (σ )

   

 y2 exp − 2  2σ 2π 

1

=−

1 2σ

2

+

y2 σ

2

=

1 σ4

(y 2 − σ2 )

≡

be

written

as

[

c(σ 2 ) σˆ 2 ( y ) − σ 2

]

Therefore, an efficient estimate for σ 2 exists.

[

] [

]

Note that E σˆ 2 σ 2 = E Y 2 σ 2 = σ 2 ⇒ The estimate is unbiased. 6.3 (a) The likelihood function is L( m) = f Y ( y ) = f Y1 ( y1 ) f Y2 ( y 2 ) since Y1 and

Y2 are statistically independent ⇒ L(m) =

ln L(m) = − ln 2π −

[

1 ( y1 − m )2 − 1 ( y 2 − 3m )2 2 2

y + 3y2 ∂ ln L(m) = 0 ⇒ mˆ = 1 ∂m 10

The statistic is mˆ (Y ) =

1 (Y1 − 3Y2 ) 10

]

1   1 exp− ( y1 − m )2 + ( y 2 − 3m )2  . 2π   2


94

(b) E[a1Y1 + a 2 Y2 ] = a1 m + a 2 3m = m (a1 + 3a 2 ) ≡ m if unbiased. Thus, we must have (a1 + 3a 2 ) = 1 . 6.4 (a) The likelihood function is fY

Taking the logarithm ⇒ ln f Y

Hence,

∂ ln f Y

Θ(y

y

k  1 K 1 −θ 1 = K exp − ∑ y k e θ k =1 θ  θ k =1

K

Θ ( y θ) = ∏

θ)

∂θ

=−

Θ(y

K 1 + θ θ2

θ) = − K ln θ −

K

∑ yk

k =1

  

1 K ∑ yk θ k =1

1 = 0 ⇒ θˆ ml = K

K

∑ yk

k =1

[ ]

K ˆ θ = E  1 ∑ Y  = 1 Kθ = θ . Therefore, the estimator is unbiased. (b) E Θ  K k =1 K  K  

(c) To determine the Cramer-Rao bound, we solve  ∂ 2 ln f Y Θ ( y θ)   = E  K − 2 E 2 2   ∂θ θ3 θ  

K



K



2

∑ Yk  =

k =1

θ

−

2K θ

2

=−

K θ2

(d) The variance of θˆ ml is

] [

]

 1 var (θˆ − θ) θ = E (θˆ − θ) 2 θ = E   K 

[

 ∑ Yk  k =1  K

2

Since the observations are independent var[θˆ ml ] =

1 2 σy K

 1 K K   = E Y Y 2 ∑∑ k l  K k =1 l =1    K K 1 = 2 ∑ ∑ C (Yk , Yk ) K k =1 k =1


95

where σ 2y = var[Y ] = E[Y 2 ] − E 2 [Y ] = θ 2 . Hence, var[θˆ ml ] =

θ2 ≡ Cramer-Rao K

bound. Hence, the estimator is consistent. 6.5 (a)

Y is binomial ⇒ E[Y ] = np .

An unbiased estimate for p is

1 Y  E[Yˆ ] = E   = p ⇒ Yˆ = Y . n n

)

(

[

]

var(Yˆ ) p(1 − p) . Thus, P Yˆ − p > ε ≤ where var[Yˆ ] = E (Yˆ − p) 2 = 2 n ε p (1 − p ) → 0 as n → ∞ . Therefore, Yˆ is consistent. p Yˆ − p > ε ≤ nε 2 (b)

(

)

K

6.6 f Y ( y ) = ∏

k =1

 ( y − m) 2  1 exp − k 2  = σ 2 2π σ   2πσ 2 1

(

)

K /2

 K 1  exp − ∑ 2 ( y k − m) 2   k =1 2σ 

Let σ 2 = θ L(m, θ) = −

K K 1 ln 2πθ − ∑ ( y k − m) 2 2 θ k =1

We need

∂ ln L(m, θ) ∂ ln L(m, θ) = =0 ∂θ ∂m

Applying

∂ ln L(m, θ) 1 = 0 ⇒ mˆ = ∂m K

K

∑ yk

k =1

and ∂ ln L(m, θ) 2 = 0 ⇒ θˆ = ∂θ K

6.7 (a)

fY

2

X ( y x) = ∏

k =1

y1 + y 2 ≤ 2 . Consequently,

K

∑ ( y k − m) 2

k =1

where mˆ =

1 K

K

∑ yk

k =1

 ( y − x) 2  ∂f Y X ( y x) exp − k 2  ⇒ =0 ∂x 2σ 2πσ   1

yields


96

xˆ ml

(b)

E [xˆ ml (Y )] =

 1  = − 1 1  ( y1 + y 2 ) 2

, if y1 + y 2 ≥ 2 , if y1 + y 2 ≤ −2 , if y1 + y 2 ≤ 2

1 1 E[Y1 + Y2 ] = [ x + x] = A . 2 2

Therefore,

xˆ ml (Y )

is

unbiased. 6.8 (a) The likelihood function is given by K

fY

Θ(y

K

θ) = ∏ e − θ k =1

∑ yk

yk

θ θ k =1 = e − Kθ K yk ! ∏ yk ! k =1

Taking the logarithm, we have K  K  ln f ( y θ) = − Kθ + ∑ y k ln θ − ln ∏ y k ! k =1  k =1 

∂ ln f ( y θ) ∂θ

= −K +

1 ⇒ θˆ ml = K

1 K ∑ yk = 0 θ k =1

K

∑ yk

k =1

 1 1 K (b) θˆ ml unbiased ⇒ E[θˆ ml ] = E  ∑ Yk  = ( Kθ) = θ which is true, since K  k =1  K K

∑ Yk

is also a Poisson with parameter Kθ .

k =1

We have, 2 2  K   ∂   1     J = E  ln f ( y θ) θ = E  ∑ Yk − K  θ  θ k =1     ∂θ   


 2K K 1 = E K 2 − ∑ Yk + 2 θ k =1  θ  2K 1  = E K 2 − ( Kθ) + 2 θ θ 

[

]

Hence, var (θˆ − θ) θ ≥

K  ∑ Yk   k =1

(K

97    

2

  

)

 K θ + Kθ  =  θ

2 2

K is the Cramer-Rao bound. θ

6.9 (a) The conditional density function is given by

f Yk

1 , − θ ≤ y k ≤ θ, k = 1, 2, L , K  ( ) θ = y  2θ k Θ  0 , otherwise  fY(y)

1/2θ

-θ

0

y

θ

The likelihood function is  1  L(θ) =  (2θ) K  0 

, − θ ≤ y k ≤ θ, k = 1, 2, L , K , otherwise

Maximizing L(θ) is equivalent to selecting θ as the smallest possible value while L(θ) is positive. Hence, θ ≥ y k and θ ≥ − y k for k = 1, 2, L , K . Note that θ ≥ − y1 , θ ≥ − y 2 , …, θ ≥ − y K , θ ≥ y1 , θ ≥ y 2 , …, θ ≥ y K is written as θ ≥ ( − y1 ,− y 2 , L ,− y K , y1 , y 2 , L , y K ) which is true if and only if θ ≥ ( y , y , L , y ) . Therefore, θˆ = max( y , y , L , y ) . 1

2

K

ml

1

2

K


98 (b) From the MLE,

[

P(θˆ ≤ y ) = P max( y1 , y 2 , L , y K

[

=PY ≤y

⇒ f θˆ ( y ) =

ny n −1 θn

]n

 1  n  y  =    θ   0 

) ≤ y ] = P [ y1

][

]

[

≤ y P y2 ≤ y L P yK ≤ y

]

y≥θ

,

, 0≤ y<θ x<0

,

, 0≤ y<θ.

θ ny n −1 nθ Hence, E[θˆ ] = ∫ y n dy = and thus, n +1 θ 0

 n +1 ˆ the unbiased estimator is  θ.  n  6.10 (a) The likelihood function is

L( p ) = f Y

P(y

K

p) = ∏ f ( y k k =1

 K yk k−y ∏ p (1 − p) k , y k = 0, 1 and k = 1,2, L , K p) =  k =1  0 , otherwise 

= p Ky (1 − p) K − Ky , since the Yks are i. i. d.

Taking the logarithm ⇒ ln L( p) = Ky ln p + ( K − Ky ) ln(1 − p) and ∂ ln L( p) Ky ( K − Ky ) =0⇒ − = 0 ⇒ pˆ ml = y ∂p p 1− p

(b) Solving for one sample, we have 2  ∂  1    ln p  = 2 2  ∂ p  p ∂    ln f ( y p ) =  2 ∂ p    ∂  1  ln(1 − p) =  ∂p (1 − p) 2 

and

,

y =1

,

y=0


99

 1  1 ∂  1 1  1  E  ln f ( y p)  = P 2  + (1 − p )  = + = 2  p 1 p p ( 1 ∂ p − − p) P   (1 − p)   

Therefore, the Cramer-Roa bound for the K independent and identical observations is var[Y ] ≥

6.11

f Y ( y) =

∞

∫ fY

X

−∞

=

∞

⇒ fX

Y

( y x) f X ( x)dx

 ( y − x) 2   1  exp −   [δ( x − 1) + δ( x + 1)] dx 2 2σ  2π σ   2  1

∫

−∞

=

p(1 − p) K

  1  1 2 2  exp − 2 ( y − 1)  + exp − 2 ( y + 1)   2 2πσ   2σ   2σ  1

( x y) =

fY

X ( y x) f X ( x)

f Y ( y)

 1  exp − 2 ( y − x) 2 [δ( x − 1) + δ( x + 1)]  2σ  =  1   1  exp − 2 ( y − 1) 2  + exp − 2 ( y + 1) 2   2σ   2σ 

+ 1 , if y ≥ 0 As in Example 6.5 ⇒ xˆ map =   − 1 , if y < 0

xˆ ms =

∞

∫ xf X

−∞

 1  exp − 2 ( y − x) 2 [δ( x − 1) + δ( x + 1)]  2σ  dx Y ( x y ) dx = ∫ x − ∞ exp  − 1 ( y − 1) 2  + exp  − 1 ( y + 1) 2       2σ 2   2σ 2  ∞

y 2

=

eσ −e y 2

eσ +e Therefore, xˆ ms ≠ x map

− −

y σ2 y σ

2

=

1− e 1+ e

− −

2y σ2

2y σ2


100

 α r r −1 − αx x e  6.12 (a) f X ( x) =  Γ(r )  0 

x≥0

, ,

x<0

X is a Gamma distribution with mean E[ X ] =

r r and variance var[ X ] = 2 . α α

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

∫

f Y ( y) =

−∞ ∞

∫ fY

−∞

= ∫ xe − xy 0

=

∞

f YX ( y, x)dx =

( y x) f X ( x)dx

α r r −1 −αx x e dx Γ( r ) ∞

rα r (α + y )

X

r +1

(α + y ) r +1 r − (α + y ) x x e dx r ( r + 1) 0 14444244443

∫

Gamma density function ⇒1

 rα r  ⇒ f Y ( y ) =  (α + y ) r +1  0  Therefore, f Y

X ( y x) =

fX

Y

( x y ) f X ( x) f Y ( y)

=

,

y≥0

, otherwise x r (α + y ) r +1 exp[− (α + y ) x ] ⇒ Γ(r + 1) 144444244444 3 Gamma distribution

MMSE estimate of x is xˆ ms = E[ X y ] = (ii) E[ xˆ ms ] =

r +1 α+ y

2 ] − E 2 [ xˆ ms ] where, The variance of the estimate is var[ xˆ ms ] = E[ xˆ ms

r ( r + 1) 2 r 1 r 2 ]= 2 . Hence, var[ xˆ ms ] = 2 and E[ xˆ ms α α (r + 2) α r+2


K

(c) f Y

X

( y x) = ∏ f Yk k =1

In order to obtain f X f Y ( y) =

∞

∫

X

Y

( yk

101

 K −x K x e ∑ y k x) =  k =1  0 

, x > 0, y k ≥ 0 yk < 0

,

( x y ) , we need ∞

f YX ( y, x)dx =

−∞

∫ f Y X ( y x) f X ( x)dx

−∞

K

=

(r + K − 1)(r + K − 2) L (r )α K   α + ∑ y k k =1 

  

r ∞

α + ∑ yk

K   r + K −1  α + x exp y k  x dx ∑ ∫ Γ( r + K )  k =1 0 444444424 1 444 4443

r+K

k =1

Gamma distribution ⇒1

 (r + K − 1)(r + K − 2) L (r )α r  r+K K    ⇒ f Y ( y) =  α + ∑ yk     k =1    0 

yk ≥ 0

,

, otherwise

K

⇒ fX

Y

( x y) =

fY

X

( y x) f X ( x) f Y ( y)

=

α + ∑ yk

K     x r + K −1 exp −  α + ∑ y k  x  Γ( r + K )   k =1 3 1444444424 444444 k =1

Gamma distribution

⇒ MMSE estimate of X is

[

]

Xˆ ms ( y ) = E X y =

r+K K

α + ∑ yk k =1

(ii) The variance of the estimate is 2 var[ Xˆ ms ] = E[ Xˆ ms ] − E 2 [ Xˆ ms ]

(r + K )(r + 1) K 2 ]= 2 E[ Xˆ ms α ( r + K + 1)

⇒

E[ Xˆ ms ] =

where, var[ Xˆ ms ] =

rK 2

α (r + K + 1)

.

r α

and


102

(d) ln f X

K   α + ∑ yk k =1  Y ( x y ) = ln  Γ( r + K )  

The MAP estimate is

∂ ln f X

Y

  K  + (r + K − 1) ln x −  α + ∑ y k   k =1   

( x y)

∂x

= 0 ⇒ xˆ map =

(r + K − 1) K

α + ∑ yk k =1

xˆ map ≠ xˆ ms

e − n 6.13 (a) f N (n) =   0

, n≥0 , otherwise

 1 , 0 ≤ x ≤1 f X ( x) =   0 , otherwise ⇒ f Y ( y) =

∞

∫ fY

−∞

X

( y x) f X ( x) dx where

fY

X

exp[−( y − ln x)] , y > ln x ( y x) =  0 , otherwise 

Hence, the marginal density function of Y is 1 ∫ exp[−( y − ln x)]dx , y ≥ 0 0 f Y ( y) =  y e   ∫ exp[−( y − ln x)]dx , y ≤ 0 0

and

.

  x 

Therefore,


xˆ ms =

∞

∫x

−∞

fY

103

1 ∫ x exp[−( y − ln x)]dx , y ≥ 0 0 X ( y x ) = e y   ∫ x exp[−( y − ln x)]dx , y < 0 0

or,

xˆ ms

(b)

∂ ln f X

Y

( x y)

∂x

=

 2  3 = y  2e  3 ∂ ln f Y

X

,

y≥0

,

y<0

( y x)

∂x

x = xˆ map

+

∂ ln f X ( x) ∂x

x = xˆ map

and fY

X

exp[−( y − ln x)] , 0 ≤ x ≤ 1 and y > ln x ( y x) f X ( x) =  1 , otherwise 

The quantity is maximized when ( y − ln x) > 0  e y or, xˆ map =   1

(c) f X

Y

, y≤0 , y≥0

( x y) =

fY

−∞

Hence, for y ≥ 0 ⇒

( y x) f X ( x) f Y ( y)

xˆ abs

∫

X

fX

Y ( x y ) dx =

2 x = 2 xe − 2 y

∞

∫ fX

xˆ abs

1 xˆ abs 1 = ∫ 2 x dx ⇒ xˆ abs = 2 −∞ 2

Y

, 0 ≤ x ≤ 1 and y ≥ 0 , 0 ≤ x ≤ e y and y ≤ 0

( x y )dx =

1 2 {

median


104 and

1 = 2

for y ≤ 0 ⇒

6.14

fY

X

xˆ abs

−2 y ∫ 2 xe dx ⇒ xˆ abs = 0

( y x) = f N ( y − x) =

ey 2

  1 exp − 2 ( y − x − 1) 2  2πσ   2σ 1

1 , 0≤ x≤2  f X ( x) =  2  0 , otherwise

∂ ln f Y

X

∂x

( y x)

+

∂ ln f X ( x) = 0 ⇒ xˆ map = y − 1 for ∂x

0≤ x≤2

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 Φ = DΘ

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

ˆ ( y ), Φ] = ∑ [Φ ˆ ( y ) − Φ ] 2 = Φ T ( y )Φ( y ) C [Φ i i i =1

~ and Φ( y ) is ˆ ( y) − Φ ˆ  Φ 1 1  ˆ ˆ  − Φ ( y ) Φ  2 2  ~ Φ=  M     ˆ ˆ Φ K ( y ) − Φ K  ˆ ( y ) , we obtain Following the same procedure as we did in estimating Φ ms

1

Van Trees, H. L., Detection, Estimation, and Modulation Theory, Part I, John Wiley and Sons, New York, 1968.


ˆ ( y ) = E[Φ | y ] = Φ ms

105

∞

∫ φ f Φ|Y (φ | y)dφ

−∞

substituting (6.92) in (6.94), we obtain E[Φ | y ] = D E[Θ | y ]

and thus, ˆ ( y) ˆ ( y) = D Θ Φ ms ms

6.16 θ can be expressed as θ = aY + b . Since E[θ] = aE[Y ] + b = 0 and m y = 0 ,

then b = 0

From (6.108), a = ρ θy =

σθ , and from (6.110), the conditional variance is given σy

by

[

]

ℜ ms = var[θ Y ] = E (θ − aY − b) 2 Y = σ θ2 (1 − ρ 2 )

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

θ (y

θ) =

 (θ − ay ) 2  exp − 2  2 2πσ θ2 (1 − ρ θ2y )  2σ θ (1 − y )  1

6.17 The BLUE of θ is given by −1 θˆ blue = E[θ] + C θy C yy [Y − m y ]

Using fθ

Y

fθ

Y

(θ y ) =

(θ y ) is then

fY

θ (y

θ) f θ)

f Y ( y)

,

the

conditional

density

function


106

We compute m θ = m y = mθ + mn =

1  ( 20 − 8θ) (θ y ) =  7  0 

fθ

Y

1 2

and C θθ = σ θ2 =

1 ≤ θ ≤1 2 , otherwise

,

4 1 , mn = 12 3

and C nn = σ 2n =

11 . 6

Since Y and θ are statistically independent, then C θy = C θθ

and

C yy = σ θ2 + σ 2n =

11 36

3 Hence, after substitution, the best linear unbiased estimate is θˆ blue = y. 11

2 and 9

Chapter 7 Filtering 7.1 (a) The error is orthogonal to the data ⇒ ∞    E s (t − t 0 ) − ∫ Y (t − α) h(α) dα Y (ξ) = 0 , ∀ξ    −∞ ∞

⇒ R sy (t − ξ − t 0 ) =

∫ R yy (t − ξ − α) h(α) dα

, ∀ξ

−∞

Let τ = t − ξ ⇒ R sy ( τ − t 0 ) =

∞

∫ R yy (τ − α) h(α) dα

for

−∞ < τ < ∞

−∞

= R yy (τ) ∗ h(τ)

for

−∞ < τ < ∞

Taking the Fourier transform, we have S sy ( f )e − j 2 π f t0 = S yy ( f ) H ( f )

⇒ H( f ) =

S sy ( f )e − j 2 π f t0 S yy ( f )

= =

S ss ( f ) e − j 2 π f t0 S ss ( f ) + S nn ( f ) 4α / N 0 2

2

[α + (4α / N 0 )] + 4π f 1442443 β2

Taking the inverse Fourier transform, we obtain 107

2

e − j 2 π f t0


108

h(t ) =

2α −β t −t0 e N 0β

, β = α2 +

4α N0

⇒ h(t) 2α N 0β

t

t0

(b) The minimum mean square error is ∞    e m = E s (t − t 0 ) − ∫ Y (t − α) h(α) dα s (t − t 0 )  −∞   ∞

∫ R sy (α − t 0 ) h(α) dα

= R ss (0) =

−∞

Using

Equation

(7.55)

∞  S sy (− f ) S sy ( f )  α ⇒ e m = ∫  S ss ( f ) −  df = ( ) β S f −∞   yy 

from

Examples 7.3 and 7.4. 7.2 R ss (τ) = e

−0.5 τ

and R nn (τ) = δ(τ)

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

where, S ss ( f ) =

Hence, H ( f ) =

0.5 0.25 + 4π 2 f 0.5

0.75 + 4π 2 f

2

2

S ss ( f ) S ss ( f ) + S nn ( f )

and S nn ( f ) = 1 .

=

0.25

2 0.75

0.75 0.75 + 4π 2 f

2

Filtering

109

Taking the inverse Fourier transform, the impulse response is h(τ) ≈ 0.29e

−0.75 τ

(b) This is similar to Example 7.5 where α = 0.5 and

N0 = 1 . Hence, 2

h(τ) = 0.62e −1.12 τ u (τ)

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

∞

∫

−∞ ∞

Using

∫

−∞

∞ ∞ S ss ( f ) S nn ( f ) 0.5 df = ∫ H ( f ) S nn ( f )df = ∫ 2 S ss ( f ) + S nn ( f ) −∞ − ∞ 0.75 + 4π f

1 x2 + α2

dα =

df

2

π ⇒ e m ≈ 0.29 . α

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

7.3

∞

∞

−∞

0

∫ R sy (τ) h(τ) dτ = 1 − ∫ e

e m = R ss (0) =

(0.62)e −1.12 τ dτ ≈ 0.62 .

ds(t ) + s(t ) = N 2 (t ) and thus, we have dt S n 2 n2 ( f )

We

− 0.5 τ

see

S n1n1 ( f ) =

that 2

1 + 4π 2 f

2

1 j 2π f + 1

S ss ( f ) =

S ss ( f )

1 2

1 + 4π f

2

S n2 n 2 ( f ) =

2 1 + 4π 2 f

and

2

. Thus,

S yy ( f ) = S ss ( f ) + S n1n1 ( f ) =

2 1 + 4π 2 f

2

+

2 1 + 4π 2 f

2

=

4 1 + 4π 2 f

2


110

Using Laplace transform, we write S yy ( p) =

+ Then, S yy ( p) =

−4 −2 2 = + , p = j 2π f ( p − 1)( p + 1) p − 1 p + 1

2 −2 − and S yy . Also, ( p) = p +1 p −1

S sy ( p) − S yy ( p)

=

S ss ( p)

=

− S yy ( p)

1 = B + ( p) p +1

Therefore, the transfer function and the impulse response are B + ( p)

H ( p) =

7.4 S yy ( f ) = S ss ( f ) + S nn ( f ) =

⇒ S yy ( p ) =

=

+ S yy ( p)

1 1 ⇒ h(t ) = δ(t ) 2 2

1 2

2

1 + 4π f

−3+ p2 2( −1 + p 2 )

=

+

1 2

,

p = j 2π f

p− 3

p+ 3

2 ( p − 1)

2 ( p − 1)

1 1 p/ 2 1 2+ 2 2 +1 S s′y ( p) = pS sy ( p) = pS ss ( p) ⇒ − = = + p +1 p − 2 p +1 S yy ( p ) p− 2 S s′y ( p )

The transfer function H ( p ) = response is h(t ) =

2 2+ 2

e−

3t

B + ( p) + S yy ( p)

u (t ) .

=

2

1

2+ 2 p+ 3

and thus, the impulse

Filtering

7.5 S ss ( f ) =

5/3 2

(1 / 4) + 4π f ⇒ S ss ( p ) =

S nn ( f ) =

,

2

20 / 3 1− 4 p

and S yy ( p ) = S ss ( p ) + S nn ( p ) = =

111

1 + 4π 2 f

S nn ( p) =

,

2

7/3 2

7/3 1− p 2

9 − 16 p 2 (1 − 4 p 2 )(1 − p 2 ) 3− 4p 3+ 4p (1 − 2 p )(1 − p ) (1 + 2 p )(1 + p )

⇒ S sy ( p ) = e αp S ss ( p )

Also, S sy ( p ) − S yy ( p)

Knowing,

=e

αp

20 (1 − p )  2/3 2  3  = e αp  + (1 + 2 p )(3 + 4 p ) 3 − 4 p 1 + 2 p  

2 → e −t / 2 1+ 2 p

and

e αp

2  1  → exp − (t + α) , the transfer 1+ 2 p  2 

function is H ( p) =

B + ( p) + ( p) S yy

n

7.6 R ss (n) =

1/ 2 1 − (1 / 4)

,

= 2e −α / 2

1+ p 3+ 4p

, α>0

2 , n = 0 R nn (n) =  0 , n ≠ 0

Taking the Z-transform, we have S sy ( Z ) = S ss ( Z ) =

4 −2 Z and 3 [ Z − (1 / 2)]( Z − 2)

S nn ( Z ) = 2 . Hence, S yy ( Z ) = S ss ( Z ) + S nn ( Z ) =

( Z − 0.314)( Z − 3.186) 4 4 2Z 2 − 7 Z + 2 = 3 [ Z − (1 / 2)]( Z − 2) [ Z − (1 / 2)]( Z − 2) 3 14243 123 + S yy (Z )

− S yy (Z )


112 Also, S sy ( Z ) − S yy (Z )

=

− 2Z 0.372 − 2.372 = + [ Z − (1 / 2)]( Z − 3.186) Z − (1 / 2) 1 Z4 −2 3.186 4 3 1424 3 − B (Z )

B+ (Z )

The pulse transfer function is then H (Z ) =

B + (Z ) + S yy (Z )

=

0.372 and h(n) = 0.372(0.314) n Z − 0.314

, n = 0, 1, 2, L

(b) The mean square error is given by ∞

e m = R ss (0) − ∑ R sy (n) h( n) = n =0

∞ 4 4 − (0.372) ∑ 3 3 n =0

1   2 (0.314)   

4 4 1 = − (0.372) = 0.61 3 3 1 − 0.314

7.7 Rss (n) =

1 2

n

⇒ S ss ( Z ) =

−2Z [ Z − (1 / 2)]( Z − 2)

1 , n = 0 R nn (n) =  ⇒ S nn ( Z ) = 1 0 , n ≠ 0 Hence, S yy ( Z ) = S ss ( Z ) + S nn ( Z ) =

⇒ S yy ( Z ) =

Z 2 − 4.5Z + 1 [ Z − (1 / 2)]( Z − 2)

Z − 0.234 Z − 4.265 Z − (1 / 2) Z −32 2 1424 3 1 − + S yy (Z )

and

S sy ( Z ) − S yy (Z )

=

S yy ( Z )

− 2Z 0.265 − 2.265 = + [ Z − (1 / 2)]( Z − 4.265) Z − (1 / 2) 1 Z4 −2 4.265 4 3 1424 3 − B+ (Z )

B (Z )

n

Filtering

Hence, H ( Z ) =

B + (Z ) + S yy (Z )

=

113

0.265 ⇒ h(n) = 0.265(0.234) n Z − 0.234

, n = 1, 2, L

(b) The mean-square error is ∞

∞

n =0

n=0

e m = R ss (0) − ∑ R sy ( n) h(n) = 1 − ∑ 0.265[(0.235)(0.5)] n

1 = 1 − 0.265 = 0.7 1 − 0.1175 7.8 (a) From (7.113), the optimum weights are given by −1 ω 0 = R yy R ys

 1.1456 − 0.5208 −1 Computing, we have R yy =  and − 0.5208 1.1456   ω   1.1450 − 0.5208  0.5272   0.8360  ω 0 =  01  =   =  ω 02  − 0.5208 1.1456  − 0.4458 − 0.7853

That is, ω 01 = 0.8360 and ω 02 = −0.7853 (b) From (7.105), the minimum mean-square error is e m = σ 2s − R Tys ω 0 − ω T0 R ys + ω T0 Rω 0 Substituting the values and computing, we obtain e m = 0.1579 .

Chapter 8 Representation of Signals T

 1

∫ 

8.1 (a) We have,

T

0

T

∫ 0

kπt  2 cos dt = 0 T T 

1

1

T

T

dt = 1

and  2 kπ t cos T T 0

T

∫ 

1 , k = j jπ t  2 cos dt =  T T  0 , k ≠ j

 1 kπ t  2 , cos Therefore,   are orthonormal functions. T T   T

(b) Similarly, to verify that the set functions is orthonormal in the interval T 1 1 [−1, 1] , we do ∫ dt = 1 2 T 2 T −T T

∫

−T

1

1

2T

T

cos

T kπt kπt 1 dt = 2 ∫ cos dt = 0 T T 0T 2

and

114

Representation of Signals T

∫

−T

1

cos

T

115

kπt 1 j πt kπt jπ t 2T cos dt = ∫ cos cos dt = δ kj T T T T 0 T T

Hence, the set is orthonormal on the interval [−1, 1] . 8.2 (a) We solve

1

1

−1

−1

∫ s1 (t )s 2 (t )dt = ∫ tdt = 0 1

1

∫ s1 (t )dt = 2∫ dt = 2 2

−1

0

and 1

1

1

−1

−1

0

2 2 2 ∫ s 2 (t )dt = ∫ t dt = 2∫ t dt

2 3

Therefore, s1 (t ) and s 2 (t ) are orthogonal. 1

(b) s1 (t ) orthogonal to s 3 (t ) ⇒ ∫ 1 (1 + αt + β t 2 )dt = 0 ⇒ β = −3 −1

1

s 2 (t ) orthogonal to s 3 (t ) ⇒ ∫ t (1 + αt + βt 2 ) dt = 0 ⇒ α = 0 . −1

Therefore, s 3 (t ) = 1 − 3t 2 . 8.3 Note that s 3 (t ) = −2s1 (t ) ⇒ We have 2 independent signals.

The energy of s1 (t ) is thus, T

T /2

T

0

0

T /2

E1 = ∫ s12 (t )dt =

∫ 1dt + ∫ (−1)

2

dt =

T T + =T 2 2


116

 1 s (t )  T φ1 (t ) = 1 =  E − 1  T

, 0≤t≤

T 2

T ≤t ≤T 2

,

T

f 2 (t ) = s 2 (t ) − s 21 φ1 (t ) where s 21 = ∫ s 2 (t )φ1 (t )dt . Then, 0

s 21 =

T /2

∫ 0

T  1   1  T dt + ∫ ( −2) − dt = + (−1)    2 T T     T /2

f 2 (t ) = s 2 (t ) − s 21φ1 (t ) = −

3 2

0≤t ≤T

and φ 2 (t ) =

−3 / 2 T

2

=−

 3 ∫  − 2  dt 0

1

0≤t ≤T

T

(b) s1 (t ) = T φ1 (t ) s 2 (t ) =

T 3 φ1 (t ) + T φ 2 (t ) 2 2

s 3 (t ) = −2 T φ1 (t )

Thus, the signal constellation is φ2 3 T 2

s2

s3 −2 T

s1 T 2

T

φ1

Representation of Signals

s1 =

8.4

[T

, 0

]

d  dφ(t )   n 2 t + t − t dt  dt  

 T s2 =   2

 3 T 2 

,

117

[

s3 = − 2 T

 dφ(t ) d 2 φ(t )  n 2 φ(t ) = +t + t−   dt t dt 2  

= t2

d 2 φ(t ) dt

2

+t

, 0

]

 φ(t )  

dφ(t ) + (t − n 2 )φ(t ) dt

where, π dφ(t ) = − ∫ sin θ exp[ j (nθ − t sin θ)] dθ dt −π

d 2 φ(t )

=

dt 2

π

∫ sin

2

θ exp[ j (nθ − t sin θ)] dθ

−π

After substitution in the differential equation, we have π

t 2 φ ′′ + tφ ′ + (t 2 + n 2 )φ =

∫ (t

2

cos 2 θ − jt sin θ − n 2 ) exp[ j (nθ − t sin θ)]dθ

−π

but, π

π

− jt ∫ sin θ exp[ j (nθ − t sin θ)]dθ = jt cos θ exp[ j (nθ − t sin θ)] − π −π

π

+ ∫ t cos θ(n − t cos θ) exp[ j ( nθ − t sin θ)]dθ −π

π

= 0 + ∫ (t 2 cos θ − nt cos) exp[ j (nθ − t sin θ)]dθ −π

Thus, π

nπ

−π

− nπ

t 2 φ′′ + tφ′(t 2 − n 2 )φ = −n ∫ (n − t cos θ) exp[ j (nθ − t sin θ)]dθ = −n

where u = nθ − t sin θ .

∫e

ju

du = 0


118

8.5 Given the differential system φ ′′(t ) + λφ(t ) = 0 ,

φ ′(0) = φ(1) = 0 , we first

t

integrate with respect to t ⇒ φ ′(t ) − φ ′(0) + λ ∫ φ(u ) du = 0 . 0

⇓ t

φ(t ) − φ(0) − tφ ′(0) + λ ∫ (t − u )φ(u )du = 0 0

1

t

0

0

Using φ ′(0) = φ(1) = 0 ⇒ φ(t ) = λ ∫ (1 − u )φ(u )du − λ ∫ (t − u )φ(u ) du 1

t

1

0

0

t

since φ(0) = −λ ∫ (1 − u )φ(u ) du ⇒ φ(t ) = ∫ (1 − t )φ(u ) du + λ ∫ (1 − u )φ(u )du Therefore, the kernel is 1 − t , 0 ≤ u ≤ t k (u, t ) =  1 − u , t ≤ u ≤ 1 8.6 The integral equation can be reduced to the differential equation by taking the derivative with respect to t ⇒ φ ′′(t ) + λφ(t ) = 0 with φ(0) = 0 and φ ′(π / 2) = 0 .

Let φ(t ) = c1 e j

λt

+ c2 e − j

φ ′(t ) = c1 j λ e j

λt

λt

. Then, φ(0) = 0 = c1 + c 2 ⇒ c1 = −c 2

− c2 j λe − j

λt

⇒ c1 = 0 trivial solution ⇒ cos λ

λ= Therefore, λ k = (2k + 1)

 j π and φ ′  = c1 j λ  e  2 

π 2

+e

−j λ

π 2

 =0.  

π π π = 0 ⇒ λ = or 2 2 2

π (2k + 1) , k = 0, 1, 2, L 2

φ(t ) = c1 [e j ( 2 k +1)t − e − j ( 2 k +1)t ] = c sin(2k + 1)t 2

λ

, k = 0, 1, L .

, k = 0, 1, 2, L

and


119

8.7 Differentiating twice with respect to t, the integral equation reduces to the differential equation φ ′′(t ) + λφ(t ) = 0 with φ ′(0) = φ(T ) = 0

Let φ(t ) = c1 e j

λt

+ c2 e − j

λt

. Then,

φ ′(0) = 0 ⇒ c1 = c 2 and φ(T ) = 0 = c1 e j

or, c cos λT = 0 ⇒ c ≠ 0 and

λT =

λt

+ c2 e − j

π(2k + 1) π + kπ ⇒ λ = 2 2T

λt

, k = 0, 1, 2, L

Therefore, the eigenfunctions are φ(t ) = c cos

(2k + 1)π t 2T

, k = 0, 1, 2, L

8.8 λ2 + ω = 0 ⇒ λ = ± jnω ⇒ φ(t ) = A sin nωt + B cos nωt For 0 ≤ t ≤ u ⇒ φ(t ) = A1 sin nωt + A2 cos nωt φ(0) = 0 = A2 and φ(t ) = A1 sin nωt

For u ≤ t ≤ T ⇒ φ(t ) = B1 sin nωt + B 2 cos nωt φ(T ) = 0 ⇒ B1 sin nωT + B 2 cos nωT = 0 Also, φ(t ) continuous ⇒ φ(u − 0) = φ(u + 0) ⇒ A1 sin nωu = B1 sin nωu + B 2 cos nωu

and φ ′(u − 0) − φ ′(u + 0) = 1 ⇒ B 2 nω cos nωu − B1 nω sin nωu + nωB 2 cos nωu = 1

Solving for the constants, we obtain


120

 sin nω(u − T ) , 0≤t≤u − nω sin nωT sin nωt φ(t ) =   sin nωu sin nω(T − t ) , u ≤ t ≤ T  nω sin nωT

8.9 For t ≤ u

For t ≥ u

k (t , u ) = c1t + c 2

k (t , u ) = c 3 t + c 4

k ( 0) = c 2 = 0

k (T , u ) = c 3T + c 4 = 0

k (t , u ) continuous ⇒ At t = u , we have c1u = c 3 u + c 4

k t (u − 0, u ) = c1

k t (u + 0, u ) = c 3

⇒ k t (u − 0, u ) − k t (u + 0, u ) = 1 ⇒ c1 = 1 + c 3 ⇒ u + c 3 u = c 3 u + c1 ⇒ c 4 = u, c3 = −

u u and c1 = 1 − T T

T − u  T t Therefore, k (t , u ) =  − u t + u  T

, 0≤t ≤u , 0≤u≤t

8.10 Taking the integral of the second order integro-differential equation  d 2 T  k (t , u )φ(u ) du  = −φ(t ) 2 ∫ dt  0 

we have u T T  d  tu u − t ∫ φ(u )du + ∫ uφ(u )du + t ∫ φ(u )du − t ∫ φ(u ) du  dt  0 T T t 0 0 

=

t T T  d  t u t u − t φ(t ) − ∫ φ(u )du + tφ(t ) − tφ(t ) − ∫ φ(u )du + t φ(t ) − ∫ φ(u )du  dt  T T T T t t 0 

=

d dt

T  tu  T u − ∫ φ(u ) du − ∫ φ(u ) du  − ∫ φ(t )dt T  0 T  t t


121

Thus,  d 2 T  k (t , u )φ(u ) du  = −φ(t ) 2 ∫ dt  0 

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

φ(t ) = λ ∫ k (t , u )φ(u ) du a solution since φ ′′(t ) = −λφ(t ) ⇒ −λφ(t ) + λφ(t ) = 0 as 0

expected.

8.11 For Problem 8.8, we have  sin nω(T − u ) sin nωt  nω sin nωT k (t , u ) =  sin n u sin nω(T − t ) ω   nω sin nωT

, t≤u , t≥u

and  sin nω(T − u ) sin nωu , u≤t  nω sin nωT k (u, t ) =   sin nωt sin nω(T − u ) , u ≥ t  nω sin nωT ?

We verify if sin nω(T − u ) sin nωt = sin nω(T − u ) sin nωu . We know that sin a sin b =

1 [cos(a − b) − cos(a + b)] ⇒ 2

1 1 [cos nω(T − u − t ) − cos nω(T − u + t )] = [cos nω(T − t − u ) − cos nω(T − t + u )] 2 2

Thus, they are equal and therefore k (t , u ) = k (u , t ) . For Problem 8.9, we have


122

T − u  T t k (t , u ) =  − u t + u  T

, t ≤u , t≥u

and T − t  T u k (u, t ) =  − t u + t  T

We observe that T −

, u≤t , u≥t

u t −u + t + T t = − u +t = T T T

Therefore, k (t , u ) = k (u, t ) .

8.12 Here, we have two methods. We have c n (k (t , u ), φ n (t ) ) , that is T

u

T T −u 2 nπt − ut + u t sin dt + ∫ T T T T u 0

c n = ∫ k (t , u )φ n (t ) dt = ∫ 0

Solving the integrals, we obtain the desired result c n =

2 nπt sin dt T T

nπu 2 T2 sin 2 T ( nπ) T

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

T

0

0

c n = ∫ k (t , u )φ n (t ) dt = ∫ k (u, t )φ n (u )du from Problem 8.11. Then, from Problem T

8.11, ∫ k (u , t )φ n (u )du = 0

φ n (t ) T2 = cn ⇒ cn = λ ( n π) 2

2 nπu sin T T

8.13 We have,  sin mωu sin mω(T − t ) , u≤t  mω sin mωT h(t , u ) =   sin mω(T − t ) sin mωt , u ≥ t  mω sin mωT

, n = 1, 2, L

Representation of Signals T

φ(t ) = λ ∫ h(t , u )φ(u )du ⇒ 0

123

t sin mωu sin mω(T − u ) 1 φ ′(t ) = ∫ φ(u ) du λ mω sin mωT 0 T

+∫ t

sin mω(T − u ) sin mωt φ(u )du mω sin mωT

and 1 sin mωt φ(t ) φ ′′(t ) = − mω cos mω(T − t ) mω sin mωT λ t

− (mω)2 sin mω(T − t ) ∫ 0

sin mω(T − t ) sin mωu φ(u )du − mω cos mωt φ(t ) mω sin mωT mω sin mωT T

− (mω)2 sin mωt ∫ t

sin mω(T − u ) φ(u )du mω sin mωT

Simplifying the above equation, we have t sin mωu sin mω(T − t ) 1 φ ′′(t ) = λφ(t ) sin mωT + λ2 ∫ φ(u )du λ mω sin mωT 0 T

+ λ2 ∫ t

sin mω(T − u ) sin mωt φ(u )du mω sin mωT

From Problem 8.12, sin mω(T − u ) sin mωt = sin mωu sin mω(T − t ) T  T  ⇒ −λφ(t ) + λ2 ∫ h(t , u )φ(u )du = −λφ(t ) + λ λ ∫ h(t , u )φ(u )du   0  0 T

Thus, φ(t ) = λ ∫ h(t , u )φ(u )du is a solution of 0

φ ′′(t ) + [(mω) 2 + λ ]φ(t ) = 0, φ(0) = φ(T ) = 0

In the second part of the question, we use the integral equation to obtain c n (u ) in ∞

h(t , u ) = ∑ c n (u ) φ n (t ) . Here, n =1


124

2    nπt  2  nπ  2 sin  and λ ∈ λ n =   − (mω)  φ n (t ) = T T   T    T

T

This gives c n = ((h(t , u ), φ n (t ) ) = ∫ h(t , u )φ n (t )dt = ∫ h(u, t )φ n (u )du from Problem 0

0

T

8.12. Therefore, by Problem 8.10, we have ∫ h(u , t )φ n (u )du = 0

φ n (t ) where, λ

2

φ n (t ) =

cn =

8.16 Let φ(t ) = c1 e j

n πt 2  nπ  sin and λ =   − ( mω) 2 and T T T 

2  nπ t 2  nπ    − (mω) 2  sin T  T  T 

λt

+ c2e − j

φ(t ) = c1 e j

λt

, n = 1, 2, L

, φ(0) = 0 = c1 + c 2 ⇒ c 2 = −c1 and thus,

λt

− c1 e − j

Let λ = β 2 , β 2 > 0 , then φ(t ) = c sin β k t

λt

= c sin λ t

k = 1, 2, L

αφ(1) + φ ′(1) = 0 ⇒ α sin β k + β k cos β k = 0 ⇒ tan β k = −

Therefore, φ(t ) = sin β k t

βk α

, k = 1, 2, L for positive roots of tan β k = −

βk . α

Case 1: Let λ = 0 ⇒ φ(t ) = c1t + c 2 φ(0) = 0 ⇒ c 2 = 0 and φ(t ) = c1t αφ(1) + φ ′(1) = 0 ⇒ αc 3 + c 3 = 0 ⇒ α = −1 but α is positive and thus, λ = 0 is not an eigenvalue.

Case 2: λ > 0 such that λ = β 2

and

β2 > 0

. Then,


125

φ(t ) = c1 e jβt + c 2 e − jβt φ(0) = 0 = c1 + c 2 ⇒ c 2 = −c1

αφ(1) + φ ′(1) = 0 ⇒ tan β = −

when α < 0, β > 0 ⇒ −

β α

β >0 α tanh β +1

- β /α tanh β

β

-1

Thus, φ k (t ) = sin β k (t ) is α solution where β k are consecutive positive roots of tan β = −

β . α tan β

-β/α

β1

β2

β


126 Case 3: If λ < 0 , let

λ = jγ ⇒ λ = − γ 2 ( γ > 0) .

Then φ(t ) = sinh γ . From αφ(1) + φ ′(1) = 0 , we have α sinh γ + γ cosh γ = 0 γ > 0 and α < 0 ⇒ −

where tanh β = − 8.17

γ > 0 . So φ 0 (t ) = sinh β 0 t is a solution α

β >0. α

φ ′′(t ) + λφ(t ) = 0 , φ ′(0) = φ ′(T ) = 0

Let φ(t ) = c1 e j

λt

+ c2e − j

λt

. Then,

φ ′(0) = 0 ⇒ c1 = c 2 and φ ′(T ) = c1 j λ e j 

λT

− e− j

λT

 = c sin λT = 0 

c = 0 trivial solution ⇒ sin λT = 0 ⇒ λT = kπ ⇒ λ =  kπ  ⇒λ=  T 

2

and φ(t ) = cos

kπt T

kπ T

, k = 1, 2, L

, k = 1, 2, 3, L

and φ 0 (t ) = 1 when k = 0 → λ = 0

Chapter 9 The General Gaussian Problem 9.1 (a) We first diagonalize the matrix C. C−Iλ =

λ = 1/ 2 1− λ 1/ 2 =0⇒ 1 λ 2 = 1.5 1/ 2 1− λ

 1 1 / 2  a  1  a  2 =   ⇒a= C φ1 = λ 1 φ1 ⇒     2 1 / 2 1  b  2 b 

and b =

− 2 2

 2 /2  2 1 Therefore, φ1 =  = − 1 . 2 2 / 2 −      2 / 2 2 1 C φ2 = λ2 φ2 ⇒ φ2 =  = 1 2 2 / 2   

We

form

M −1 =

the

modal

matrix

M = [φ1

φ2 ]⇒ M =

2 2

 1 1 − 1 1  

2 1 − 1  . 2 1 1 

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

127

and


128  2  y ′ = My =  2 2  − 2

2  2   y1  ⇒ y ′ = 2 ( y + y ) and 1 1 2   2 2  y2   2 

y1′ =

2 ( y 2 − y1 ) 2

The mean vector m1′ is  ′    m11 m ′  =   12  − 

2 2 2 2

2  2   m11  ⇒ m ′ = 2 (m + m ) and m ′ = 2 ( m − m ) 11 11 12 12 12 11   2 2 2  m12  2 

∆m ′ = m1′ − m 0′ = m1′ . The sufficient statistic is ′ y1′ ∆m12 ′ y 2′ ∆m k′ y k′ ∆m11 = + λk 1/ 2 1.5 k =1 2

T ( y ′) = ∑

1 = (m11 + m12 )( y1 + y 2 ) + (m12 − m11 )( y 2 − y1 ) 3 H1

or T ( y ′)

1 > γ = γ + m1T C −1 m1 . < 1 2 H0

 1 0.1 λ 1 = 0.9 (b) C =  ⇒ 0.1 1  λ 2 = 1.1   Then, φ1 =   −

2   2  2 2    M =  −

,

2 2 2 2

  φ2 =    2  2  , 2 2 

2  2 , 2 2 

M

−1

  =  

2 2 2 2

−

2  2  2  2 

The General Gaussian Problem

  and y ′ = My =   − ′ = m11

2  2   y1  ⇒   2  y2   2 

2 2 2 2

2 ( y1 + y 2 ) 2 2 y1′ = ( y 2 − y1 ) 2

y1′ =

2 2 ′ = (m11 + m12 ) , m12 (m12 − m11 ) and ∆m ′ = m1′ 2 2

The sufficient statistic is ′ y1′ ∆m12 ′ y ′2 ∆m ′k y ′k ∆m11 = + λk 0.9 1.1 k =1 2

T ( y ′) = ∑

= 0.55(m11 + m12 )( y1 + y 2 ) + 0.45( m12 − m11 )( y 2 − y1 )

λ 1 = 0.1  1 0.9 (c) C =   ⇒ λ = 1.9 0 . 9 1 2    2    Then, φ1 =  2  2  − 2 

  φ2 =   

,

2  2  2 2 

′ y1′ m12 ′ y ′2 ∆m k′ y k′ m11 = + λk 0.1 1.9 k =1 2

⇒ T ( y ′) = ∑

= 5(m11 + m12 )( y1 + y 2 ) + 0.26( m12 − m11 )( y 2 − y1 )

λ 1 = 0.47  1 0.9 9.2 C =   ⇒ λ = 2.53 0 . 9 2 2    0.86  φ1 =   − 0.51

,

 0.86 0.51 M =  ⇒ − 0.51 0.86

 0.51 φ2 =  , 0.86

0.86 − 0.51 M −1 =    0.51 0.86 

129


130

 y ′   0.86 0.51  y1  y ′ = My ⇒  1  =     ⇒ y1′ = 0.86 y1 + 0.51 y 2  y 2′  − 0.51 0.86  y 2  y 2′ = 0.86 y 2 − 0.51 y1

Then,

′ = 0.86m11 + 0.51m12 m1′ = Mm1 ⇒ m11

⇒ T ( y ′) =

and

′ = −0.51m11 + 0.86m12 m12

and

(0.86m11 + 0.51m12 )(0.86 y1 + 0.51 y 2 ) 0.47 (−.0.51m11 + 0.86m12 )(−0.51 y1 + 0.86 y 2 ) + 2.56

⇒ T ( y ′) = (1.83m11 + 1.09m12 )(1.83 y1 + 1.09 y 2 ) + (−0.2m11 + 0.34m12 )(−0.2 y1 + 0.34 y 2 )

9.3 Noise ∼N (0, σ 2n ) (a) E[Yk | H j ] = 0 ,

k = 1, 2 j = 0, 1

⇒ m1 = m 0 = 0

σ 2 H 0 : Yk = N k ⇒ C n = C 0 = σ 2n I =  n  0

0  σ 2n 

σ 2 + σ 2n H 1 : Yk = S k + N k ⇒ C 1 = C s + C n =  s  0

0

σ 2s



 + σ 2n 

, since C s = σ 2s I .

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

T ( y) =

σ 2s σ 2n (σ 2s

H1 2

∑

+ σ 2n ) k =1

1   where γ 2 = 2ln η + (ln C 1 − ln C 0 ) 2  

y k2

> γ < 2

H0


131

H1

σ 2 (σ 2 + σ 2 ) > γ3 = n s 2 n γ2 < σ

2

or, T ( y ) = ∑ y k2 k =1

s

H0

(b) P1 = P0 = γ 2 = 2 ln

σ 2s + σ n2 σ 2n

1 2

and minimum probability of error criterion ⇒ η = 1 ,

and γ 3 = 2

σ 2n (σ 2s + σ 2n ) σ 2s

ln

σ 2s + σ 2n σ 2n

The density functions of the sufficient statistics under H1 and H0, from Equation (9.71) and (9.72), are  1 −t / 2σ12 e  f T H1 (t H 1 ) =  2σ12  0 

, t>0 , otherwise

and  1 −t / 2σ 02 e  f T H 0 (t H 0 ) =  2σ 02  0 

, t>0 , otherwise

where σ12 = σ 2s + σ 2n and σ 02 = σ 2n . Consequently, PF =

1

∞

∫e

2σ 2n γ 3

−t / 2σ 2n

2

dt = e − γ 3 / 2σ n

and PD =

1

∞

∫e

2σ12 γ 3

−t / 2σ12

2

2

2

dt = e − γ 3 / 2σ1 = e − γ 3 / 2(σ n + σ s )


132

σ 2n  0 9.4 (a) K = 4 ⇒ C 0 = C n =  0   0 σ 2s  C1 = C s + C n =    

0 σ 2n 0 0

+ σ 2n 0

0  0 0  σ 2n 

0 0 σ 2n 0 0 σ 2s

0

+ σ 2n

  0  0   σ 2s + σ 2n  0

0

0

0

σ 2s + σ 2n

0

0

0

where C s = σ 2s I . Hence,

T ( y) =

or, T ( y ) =

σ 2s

4

∑

σ n2 (σ 2s + σ n2 ) k =1

y k2

4 σ 2s y k2 2 2 2 σ n (σ s + σ n ) k =1

∑

H1 > γ < 2 H0

H1 σ 2 (σ 2 + σ 2 ) > γ3 = n s 2 n γ2 < σs H0

4

The statistic is T ( y ) = ∑ y k2 . k =1

(b) γ 2 =

4 ln σ 2s + σ 2n σ 2n

and γ 3 =

σ n2 (σ 2s + σ n2 ) σ 2s

ln

σ 2s + σ n2 σ 2n

The conditional density functions are then  1 −t / 2σ 02  4 te f T H 0 (t H 0 ) =  8σ 0  0 

and

, t>0 , otherwise

.


 1 − t / 2 σ12  4 te f T H1 (t H 1 ) =  8σ1  0 

133

, t >0 , otherwise

where σ 02 = σ 2n and σ12 = σ 2s + σ 2n . The probability of false alarm and detection are then ∞

1

∫ 8σ 4

PF =

γ3

PD =

∞

n

1

∫ 8σ 4 te

γ3

2

te − t / 2σ n dt =

−t / 2σ12

1

dt =

γ 1  1 + 32  2 2σ n

 − γ / 2σ 2 e 3 n  

γ 1  1+ 3 2  2σ12

 − γ / 2σ 2 e 3 1  

9.5 ROC of Problem 9.3 with SNR = 1 , SNR = 2 and SNR = 10 . 1 0.9 0.8 0.7 0.6 PD

0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.6

0.4 PF

0.8

1


134 σ 2 9.6 C s =  s  0

σ 2n 0  = , C   n 2σ 2s   0

0  σ 2n 

From (9.78), the LRT is

T ( y) =

1 σ n2

H1 > y k2 γ2 2 < + σn H0

σ 2sk

2

∑

2 k =1 σ sk

or,

( 2σ 2s

+ σ 2n ) y1

C n 9.7 (a) C 0 =   0

+ 2(σ 2s

H1 σ 2 (σ 2 + σ 2n )(2σ 2s + σ 2n ) > γ2 n s < σ 2s H0

+ σ 2n ) y 2

0  C s + C n 

and

C + C n C1 =  s  0

0 C s 

 2 0 1 0 and C s =  where C n =     0 2 0 1  1 0 ⇒ C0 =  0  0

0 1 0 0

0 0 3 0

0 0 0  3

and

3 0 C1 =  0  0

From (9.88), the optimum test reduces to H1 > T ( y ) = ∑ y k2 − ∑ y k2 γ3 < k =1 k =3 2

4

H0

where γ 3 is

0 3 0 0

0 0 1 0

0 0 0  1


γ3 =

σ n2 (σ 2s + σ n2 ) σ 2s

1 σ 2s = 2   γ 2 and γ 2 = 2ln η + (ln C 1 − ln C 0 ) , 2 σ 2n = 1  

(b) γ 3 = 0 ⇒ The test reduces to H1 > ∑ y k2 < k =1 2

4

∑ y k2

k =3

H0

From (9.94), (9.95) and (9.96), the probability of error is ∞ t1   P (ε | H 0 ) = ∫ ∫ f T1T0 (t1 , t 0 , H 0 )dt 0 dt1  00 P (ε ) =  ∞∞  P (ε | H ) = 1 ∫ ∫ f T1T0 (t1 , t 0 , H 1 )dt 0 dt1  0 t1 

where, f T1 (t1 ) =

1 1 −t1 / 6 e and f T0 (t 0 ) = e −t0 / 2 . Therefore, 18 2 1 1 ∞ 1 dt1 e −t1 / 6 ∫ e −t0 / 2 dt 0 = . ∫ 36 0 4 0

t

P (ε ) =

 1 0.9 0.5 9.8 (a) C = 0.9 1 0.1 0.5 0.1 1  1 − λ 0.9 0.5 λ 1 = 0.0105 C − I λ = 0.9 1 − λ 0.1 = 0 ⇒ λ 2 = 0.9153 0.5 0.1 1 − λ λ 3 = 2.0741  0.7204  Cφ1 = λ 1 φ1 ⇒ φ1 = − 0.6249 , − 0.3009

Similarly,

135


136

− 0.0519 0.6916   φ 2 = − 0.4812 , and φ 3 = 0.6148  0.8750  0.3792

The modal matrix is  0.7204 − 0.0519 0.6916 M = − 0.6249 − 0.4812 0.6148 − 0.3009 0.8750 0.3792

and y ′ = My ⇒ y1′ = 0.72 y1 − 0.052 y 2 + 0.69 y 3 y 2′ = −0.625 y1 − 0.48 y 2 + 0.615 y 3 y 3′ = −0.3 y1 + 0.875 y 2 + 0.38 y 3

Similarly, m1′ = Mm1 and then we use 3 m′ y ′ ∆m k′ y ′k =∑ k k λk k =1 k =1 λ k 3

T ( y) = ∑  1 0.8 0.6 0.2 0.8 1 0.8 0.6  (b) C =  0.6 0.8 1 0.8   0.2 0.6 0.8 1 

In this case, λ 1 = 0.1394 , λ 2 = 0.0682 , λ 3 = 0.8606 and λ 4 = 2.9318 whereas,  0.5499  − 0.2049    0.6768   , φ = − 0.4445 φ1 =  2 − 0.4445  − 0.6768      0.5499   0.2049 

and the modal matrix is

− 0.6768 − 0.2049  , φ3 =   0.2049     0.6768 

0.4445 0.5499  and φ 4 =  0.5499   0.4445


0.55 − 0.68  − 0.2  0.68 − 0.44 − 0.2 M = − 0.68 − 0.44 0.2  0.55 0.68  0.2

0.44 0.55 0.55  0.44

137

Chapter 10 Detection and Parameter Estimation 10.1 (a) s1 (t ) = cos 2πt 2π    2π   2π  s 2 (t ) = cos 2π t +  = cos(2π t ) cos  − sin(2π t ) sin   3 3      3  2π    2π   2π  s 3 (t ) = cos 2π t −  = cos(2π t ) cos  + sin( 2π t ) sin   3    3   3  1 1 − ≤t≤ 2 2 1/ 2

Also,

∫ (cos 2πt )

−1 / 2

2

dt =

1/ 2

∫

s12 (t ) dt −1 / 2

2 φ1 (t ) = cos 2π t , 1 1 1 2 = ⇒ − ≤t ≤ 2 2 2 2 φ 2 (t ) = sin 2π t , 2

Therefore, s1 (t ) = 2φ1 (t ) s 2 (t ) = −

2 3 φ1 (t ) − φ 2 (t ) 2 2

s3 (t ) = −

2 3 φ1 (t ) + φ 2 (t ) 2 2

138

Detection and Parameter Estimation

139

1/ 2

∫

−1 / 2

φ1 (t )

Received signal

Choose largest variable 1/ 2

∫

−1 / 2

φ 2 (t )

(b) The decision space is φ2 s3

3/ 2 Decide s3

s1

− 2 /2

Decide s1

φ1

2

Decide s2

− 3/ 2

s2

10.2

t ∈ [T1 , T2 ] ⇒ f T (t ) =

1 T2 − T1

for T1 < t < T2

s(t)

A t t0

T1

t0+T

T2

H 1 : Y (t ) = s (t ) + N (t ) H 0 : Y (t ) =

N (t )


140

where

1

f N ( n) =

πN 0

e −n

2

/ N0

.

The problem may be reduced to Y (t ) = s (t ) + N (t )

T2

∫

Y (t ) = s1 (t ) + N1 (t )

LRT

t0

Under H0, we have Y (t ) = N 1 (t ) =

T2

∫ N (t )dt

t0 T2

⇒ E[ N 1 (t )] = ∫ E[ N (t )]dt = 0 and t0

T2 T2  T2 T2 E[ N 12 (t )] = E  ∫ ∫ N (t1 ) dt1 N (t 2 )dt 2  = ∫ ∫ E[ N (t1 ) N (t 2 )]dt1 dt 2  t0 t0  t0 t0  N0  where E[ N (t1 ]N (t 2 )] =  2  0  ⇒ E[ N 12 (t )] =

T2 T2

∫∫

t0 t0

, t1 = t 2 , t1 ≠ t 2

N0 N δ(t1 − t 2 )dt1 dt 2 = 0 (T2 − t 0 ) ≡ var[ N 1 (t )] 2 2 T2

T

t0

t2

Under H1, we have s1 (t ) = ∫ s (t )dt = A ∫ dt = A(T − t 0 ) . Then, H 0 : Y (t ) =

N 1 (t )

H 1 : Y (t ) = A(T − t 0 ) + N 1 (t The LRT is


Λ( y) =

fY

H1 ( y

fY

H0

T , H1 ( y

=

fY

t, H1 ) f T H0

H1 (t

H 1 )dt

(y H0 )

1

∫

=

(y H0 )

∫ fY

 [ y − A(t − t 0 )] 2  1 dt exp−  N 0 (T2 − t 0 )  T2 − T1  πN 0 (T2 − t 0 )   y2 1 exp −  πN 0 (T2 − t 0 )  N 0 (T2 − t 0 ) 

T2

=

H1 )

141

T1

1 T2 − T1

T2

 [ y − A(t − t 0 )] 2  dt N 0 (T2 − t 0 )  

∫ exp−

T1

  y exp −   N 0 (T2 − t 0 )  2

H1

> < η

H0

 y 2 + At 0 (2 + At 0 )  H1 exp −  N 0 (T2 − t 0 )  T2 >  1 = exp A 2 t 2 − 2 A( At 0 + 1)t dt < η ∫ 2 T2 − T1   y T1 exp −  H0  N 0 (T2 − t 0 ) 

[

H1  2 y 2 + At 0 ( 2 + At 0 )  > ⇒ exp −  < N 0 (T2 − t 0 )   H0

]

η(T2 − T1 )

∫ exp[A

T2

]

2 2

t − 2 A( At 0 + 1)t dt

T1

1444442444443 γ

H0 > N 0 (T2 − t 0 ) ln γ − At 0 ( 2 + At 0 ) Therefore, y . < 2 2

H1

10.3 From (10.85), the probability of error is 1 P (ε ) = Q 2 

and

2α N0

  where α = E1 + E 2 − 2 E1 E 0 .  


142

T

E1 = ∫ s12 (t )dt = A 2 T 0

T

E 0 = ∫ s 02 (t ) dt = A 2 T 0

T

ρ E1 E 2 = ∫ s1 (t ) s 2 (t ) dt = 0

A 2T 2

1 ⇒ α = A 2 T and 2

, ρ=

1 P (ε) = Q  2 

2 A 2 T  N0  

The optimum receiver is shown below T

∫

Y1

0

H1

s1 (t )

Y(t)

Y1

> Y < 0 H0

T

∫ 0

Y2

s2 (t )

10.4 We have, Y (t ) = s1 (t ) + W (t ) H1 :  Y (t ) = s 2 (t ) + W (t ) H 0 : Y (t ) =

W (t )

T

Under H1, we have Y1 = ∫ [ s1 (t ) + W (t )]s1 (t )dt = E1 + W1 0 T

Y2 = ∫ [ s 2 (t ) + W (t )]s 2 (t )dt = E 2 + W 2 . 0

H1 H0


143

The problem reduces to: T

∫

Y1

0

s1 (t )

Y(t)

T

∫

Y2

0

s2 (t )

 E1 + W1 : H 1  Y1 =  W1 : H 1  W0 : H 0  E 2 + W2 : H 1  Y2 =  W2 : H 1  W0 : H 0 

Under H0, we have T

Yk = ∫ W (t ) s k (t )dt = W k ≡ W0

, k = 1, 2.

0

The LRT is fY

H1 ( y

fY

H0

H1 )

(y H0)

=

fY

H1 , S1 ( y

H 1 , s1 ) P ( s1 ) + f Y fY

H0

H1 , S 2

( y H 1 , s 2 ) P( s 2 )

(y H0)

where f Y1

H1 , S1 ( y1

H 1 , s1 ) =

 ( y − E1 ) 2  exp − 1  N0 πN 0   1


144

f Y2

f Y1

H1 , S 2

( y1 H 1 , s 2 ) =

 y2  exp − 1   N0  πN 0  

f Y2

H1 , S 2 ( y 2 H 1 , s1 ) =

 y2  exp − 2   N0  πN 0  

H1 , S 2

( y2 H1 , s2 ) =

1

1

 ( y − E2 ) 2  exp − 2  N0 πN 0   1

and f Yk

H0

( yk H 0 ) =

 y2  exp − k  , k = 1, 2  N0  πN 0   1

Therefore, the LRT becomes 1 πN 0

  ( y − E ) 2  − y 2 / N  1   ( y − E 2 ) 2   1  1  − y12 / N 0 1 1 exp − 2  e 2 0   +    exp − e N0 N0    2  πN 0      2  1 − y12 / N 0 − y22 / N 0 e e πN 0 H1     1   1 1  > ( E12 − 2 y1 E1 ) exp − ( E 22 − 2 y 2 E 2 )  η = exp − 2   N 0   N0   < H0

When η = 1 , the LRT becomes H1  E12   E 22  >  2 E1   2E 2   + exp −    y1  exp − exp −  N y 2  exp − N  < 2  N0  0 0   N0       H0

The optimum receiver may be


T

∫

Y1

145

exp[ ⋅ ]

0

Y (t )

s1 (t )

−

2 E1 N0

e

H1

−E 2 / N0 1

> 2 <

∑

H0 T

∫

Y1

exp[ ⋅ ]

0

s2 (t )

−

2 E2 N0

e

−E 2 / N0 2

2α N0

  where  

1 The signals are antipodal ⇒ ρ = −1 ⇒ α = 1.96 and P(ε) = Q 2 

3.92  . N 0 

1 10.5 (a) The probability of error is given by P(ε) = Q 2  α = E1 + E 0 − 2ρ E1 E 0 2

E1 = ∫ e − 2t dt = 0

1 (1 − e − 4 ) = 0.49 2

(b) The block diagram is shown below with φ(t ) ≈ 1.43s1 (t ) .

Y(t)

T

∫ 0

y1

H1 > P0 < P1 H0

φ1 (t )

10.6 At the receiver, we have H 1 : Y (t ) = s1 (t ) + W (t ) , 0 ≤ t ≤ T H 2 : Y (t ) = s 2 (t ) + W (t ) , 0 ≤ t ≤ T

H1 H0

H1

H0


146

E1 = E 2 =

T T and ρ12 = ∫ s1 (t ) s 2 (t ) dt = 0 ⇒ s1 (t ) and s 2 (t ) are uncorrelated. 2 0

The receiver is T

∫

Y1

0

Y(t)

φ1 (t )

Choose largest

H1 H0

T

∫

Y2

0

φ 2 (t )

where φ k (t ) =

s k (t ) Ek

=

2 s k (t ) , k = 1, 2. T

The observation variables Y1 and Y2 are then T   H 1 : ∫ Y (t )φ1 (t )dt = E1 + W1  0 Y1 =  T  H : W1  0 ∫ Y (t )φ1 (t ) dt = 0  T  W2  H 1 : ∫ Y (t )φ1 (t )dt =  0 Y2 =  T   H 0 : ∫ Y (t )φ1 (t )dt = E 2 + W 2 0 

This is the general binary detection case. Then, Y  Y =  1 , Y2 

s  s1 =  11   s12 

and

s  s 2 =  21   s 22 


147

The conditional means are  E  s  m1 = E[Y H 1 ] =  1  =  11  = s1  0   s12   0   s 21  m 2 = E[Y H 2 ] =   =   = s2  E 2   s 22 

s1 (t ) and s 2 (t ) uncorrelated ⇒ the covariance matrix is 0  N / 2 C1 =  0 = C2 = C N 0 / 2  0

and the probability of error is 1 P (ε ) = Q 2 

2α N0

   = Q 1 2   

T N0

  where α = E1 + E 2 = 2 E  

10.7 At the receiver, we have H 1 : Y (t ) = E1 s1 (t ) + W (t ) , 0 ≤ t ≤ T H 2 : Y (t ) = E 2 s 2 (t ) + W (t ) , 0 ≤ t ≤ T

T

∫

Y1

0

Y(t)

φ1 (t )

T

∫ 0

φ 2 (t )

Y2


148 T

with φ1 (t ) = ∫

s1 (t ) E1

0

dt and φ 2 (t ) =

T

s 2 (t )

0

1/ 2

∫

T

2 ∫ s 2 (t )dt

dt =

0

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

E1 s1 (t ) − E 2 s 2 (t ) E1 + E 2

=

 3 1 s 2 (t )  s1 (t ) − 2  2 

We obtain the sufficient statistic as follows

y(t)

T

∫

T(y)

0

S ∆ (t )

The conditional means are T  2     1 2 E[T ( y ) | H 1 ] = E  ∫ [ s1 (t ) + W (t )] s 2 (t ) dt  =  s1 (t ) − 2 3 0  3      T  1      2  1 1 E[T ( y ) | H 2 ] = E  ∫  s 2 (t ) + W (t )  s 2 (t )  dt  = −  s1 (t ) − 2 6  0  2      2  

The noise variance is var[T ( y ) | H 0 ] = 1 / 2 . Hence, the performance index is d2≜

{E[T ( y ) | H 1 ] − E[T ( y ) | H 0 }2 var[T ( y ) | H 0 ]

=

( 2 / 3 + 1/ 6)2 =3 1/ 2

The probabilities of false alarm and detection are d  3 PF = Q  = Q  2   2


149

 3   d PD = Q −  = Q −   2 2    

and thus, the achievable probability of error is P (ε) =

∞

∫

1

3/2

2π

e −x

2

/2

 3  dx = Q  2   

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

d2 =

2E = 4E ⇒ d = 2 E 1/ 2

d  From P(ε) = Q  = Q( E ) = 2

  3  3  ⇒ E = Q −1   2  2   

2

E2 E

Decision region

E1 E

10.8 We need to find the sufficient statistic. orthogonal, let

φ1 (t ) =

Since s1 (t ) and s 2 (t ) are

E1 s1 (t ) − E 2 s 2 (t ) E1 + E 2

T   E s (t ) − E 2 s 2 (t )   H 1 : ∫ [ E s (t ) + W (t )] 1 1  dt T    E1 + E 2 0  Then, Y1 = ∫ y (t )φ1 (t ) dt =  T  E1 s1 (t ) − E 2 s 2 (t )   0  dt  H 0 : ∫ W (t )    E1 + E 2 0 


150

Y1 is Gaussian with conditional means E[Y1 | H 0 ] = 0 = m 0

and T

T

E[Y1 | H 1 ] = P1 ∫ E1 s1 (t )φ1 (t )dt − P2 ∫ E 2 s 2 (t )φ1 (t )dt 0

=

0

P1 E1 E1 + E 2

= P1

T

∫ s1 (t )dt − 2

0

E1 E1 + E 2

− P2

T

P2 E 2 E1 + E 2 E2

E1 + E 2

∫ s 2 (t )dt 2

0

= m1

The variance is N 0 / 2 and thus, f Y1 | H 0 ( y1 | H 0 ) =

f Y1 | H1 ( y1 | H 1 ) =

 y2  exp − 1   N0  πN 0   1

 ( y − m1 ) 2  exp − 1  N0 πN 0   1

Applying the likelihood ratio test, taking the natural logarithm and rearranging terms, we obtain H1 > N 0 ln η m1 y1 + < 2m1 2 H0

For minimum probability of error, η = 1 and the decision rule becomes H1 > m P1 E1 − P2 E 2 = y1 < 2 2 E1 + E 2 H0

The optimum receiver is


y(t)

y1

T

∫ 0

151

H1

H1

> m1 < 2 H0

H0

φ1 (t )

T T T  T 10.9 (a) The energy E = ∫ s k2 (t ) dt = A 2  ∫ φ12 (t )dt + ∫ φ 22 (t ) dt + ∫ φ 32 (t )dt  + B   0 0 0 0 where B is the sum involving terms of the form T

∫ φ j (t )φ j (t )dt

,

j≠k

0

But the φs are orthonormal ⇒ B = 0 and thus, E = 3 A 2 ⇒ A =

E . 3

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

y k = ∫ y (t )φ k (t )dt

The

, k = 1, 2, 3

0 T

= ∫ [ s k′ (t ) + W (t )]φ k (t )dt = s k + W k 0

 y1  such that y =  y 2   y 3 

,

sk  W1   1   W = W 2  and s k =  s k 2  s  W3   k3 

Hence,

s0 =

1 E 1 , s1 = 3 1

1 E  1 , s2 = 3  − 1

1 E  −1 , s3 = 3   1 

1 E  −1 , s4 = 3  − 1

− 1 E  1 , 3   1 


152

s5 =

− 1 E  1 , s6 = 3  − 1

− 1 E  −1 , s7 = 3   1 

− 1 E  −1 . 3  − 1

Since the criterion is minimum probability of error, the receiver is then a "minimum distance" receiver. The receiver evaluates the sufficient statistic T j = y − sk

2

T

= ∫ [ y (t ) − s k (t )] 2 dt

, k = 0, 1, L , 7

0

and chooses the hypothesis for which T j 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 T

T j = s kT y = ∫ s k (t ) y (t )dt

, k = 0, 1, L , 7

0

and chooses the hypothesis for which T j is largest. (c) φ2 s6

s2

s1

s5

φ1

s3

s4

φ3

s7

s0


153

Using "minimum distance" or "nearest neighbor", the decision regions are H0

y1 > 0,

y 2 > 0,

y3 > 0

H1

y1 > 0,

y 2 > 0,

y3 < 0

H2

y1 > 0,

y 2 < 0,

y3 > 0

H3

y1 > 0,

y 2 < 0,

y3 < 0

H4

y1 < 0,

y 2 > 0,

y3 > 0

H5

y1 < 0,

y 2 > 0,

y3 < 0

H6

y1 < 0,

y 2 < 0,

y3 > 0

H7

y1 < 0,

y 2 < 0,

y3 < 0

(d) The probability of error is P (ε ) =

7

∑ P j P (ε j =0

7

H j ) =P(ε H 0 ) ∑ P j = P (ε H 0 ) j =0

Y1 , Y2 and Y3 are independent Gaussian random variables with conditional means E 3

E[Y1 H 0 ] = E[Y2 H 0 ] = E[Y3 H 0 ] = and conditional variances var[Y1 H 0 ] = var[Y2 H 0 ] = var[Y3 H 0 ] =

Therefore, P(ε) = P (ε | H 0 ) = 1 − P[Y1 > 0, Y2 > 0, Y3 > 0] = 1 − P (Y1 > 0) P(Y2 > 0) P (Y3 > 0) ∞ 1  ( y − E / 3 ) 2   = 1− ∫ exp −  dy  N0     0 πN 0   2E = 1 − Q −  3 N0  

   

3

3

N0 2


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 {φ1 , φ 2 } such that

T

∫ φ1 (t )φ 2 (t )dt = 0 0

and T

T

∫ φ1 (t )dt = ∫ φ 2 (t )dt = 1 2

0

2

0

The receiver is then

T

∫

Threshold

0

Y(t)

4-level signal

1

4-level signal

3

2

φ1 (t )

T

∫

Threshold

0

φ 2 (t )

with φ1 (t ) =

2 cos 2π f 0 t T

φ 2 (t ) =

2 sin 2π f 0 t T

4


155

φ2

φ1

(c) From (b), we observe that the probability of a correct decision is P(c) = P(correct decision along φ1 and correct decision along φ 2 ) = P (c along φ1 ) P (c along φ 2 ) where, P (c along φ1 ) is, from the figure below, given by

d

s1′

s′2

P (c along φ1 ) =

s3′

φ1

s′4

1 4 ∑ P(correct decision s k′ ) 4 k =1

=  d  . where, q = Q  2N  0  

d

1 [(1 − q) + (1 − 2q) + (1 − 2q) + (1 − q)] = 1 − 6 q  4  4 


156

 6  Similarly, P (c along φ 2 ) = 1 − q  . Therefore, the probability of a correct  4  decision is  6  P (c ) = 1 − q   4 

2

and the probability of error is P (ε ) = 1 − P (c ) =

3 9 q − q2 9 4

10.11 From (10.104), we have T j2 ( y ) = y − s j 2

2

= y + sj

For equal energy ⇒ R

2

= ( yT − s Tj )( y − s j ) = yT y − 2 s Tj y + s Tj s j 2

− 2 s Tj y

and s j

2

j = 1, 2, L , M

are common to all hypotheses ⇒ Minimizing

T j2 ( y ) is equivalent to maximizing s Tj y . Therefore, the receiver computes the sufficient statistic T

T jT ( y ) = s Tj y = ∫ s j (t ) y (t )dt ,

j = 1, 2, L , M

0

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


157

T1

T

∫ 0

s1 ( t )

T2

T

∫

y(t)

Choose largest decision variable

0

s 2 (t )

T

Decision

TM

∫ 0

s M (t )

"Largest of " receiver 10.12 We have

H 1 : Y (t ) = As(t ) + W (t ) H 0 : Y (t ) =

Y(t)

W (t )

T

∫ 0

 H1 : A + W1 = Y1  W1 = Y1 H 0 :

s (t )

where T

W1 = ∫ W (t ) s (t )dt 0

E[Y1 H 1 ] = A

E[Y12 H 1 ] = E[ A 2 + W12 + 2 AW1 ] = A 2 + A unknown ⇒ H1 is a Composite hypothesis and

N0 N ⇒ var[Y1 H 1 ] = 0 2 2


158

Λ g ( y) =

max f Y H ,θ ( y H 1 , θ1 ) 1 1 θ 1

fY H ( y H 0 ) 0

We need the estimate Aˆ of A such that

∂ ln f Y A ( y a) ∂a

= 0 ⇒ the ML estimate is

Aˆ = Y , the observation itself; i. e, where the distribution is maximum. Hence,  ( y − a) 2  exp −  N 0  2πN 0  1

Λ g ( y) =

fY

A, H 1 ( y a , H 1 )

fY

H0

(y H0 )

=

 y2  exp −  2πN 0  N 0  1

H1  1  > ⇒ Λ g ( y ) = exp − ( y 2 + a 2 − 2ay − y 2 ) η N 0   < H0 H1  y2  >  but η = 1 and aˆ = y ⇒ Λ g ( y ) = exp − 1  N  < 0   H0 H1 y2 > or 0 . Therefore, always decide H1 since y 2 / N 0 > 0 . N0 < H0

10.13 Y(t)

T

∫ 0

φ (t )

Y1 is a sufficient statistic and thus,

T

∫

Y1 = Y (t )φ(t ) dt 0


Y1 =

E φ(t ) + W1 ⇒ Y1 is Gaussian with mean θ

159

E / θ and variance N 0 / 2 .

The conditional density function becomes f Y1 Θ ( y1 θ) =

 [ y − ( E / θ)] 2  exp− 1  N0   πN 0 1

Hence, ∂ ln f Y1 Θ ( y1 θ) ∂θ

or y1 =

2  E   ∂  1  y1 =0⇒0= − ∂θ  N 0  θ    

E E . Thus, θˆ ml = and the optimum receiver is shown below. θ y1

y(t)

T

∫

1 y1

y1 Inverter

θˆ ml

0

E

s (t ) / E

10.14 The density function of θ is f Θ (θ) =

1 2πσ θ

e −θ

2

/ 2 σ θ2

. Hence, from the

MAP equation, we have ∂ ln f Y1 Θ ( y1 θ) ∂θ

+

∂ ln f Θ (θ) ∂θ

θˆ = θˆ map

 2     y1 − E  E − θ  = 0 = 0 ⇒ − θ  θ 2 σ θ2   N 0 

 θ4 2 2 E  y1 E θ − =0 ⇒ 2 + σ N 0  ˆ  θ N0 θ = θ map


160 As σ θ → ∞ , we have

2 2E y1 E θ − N0 N0

Therefore, lim θˆ map = 2 σ θ →∞

=0 θˆ map = θ

E ˆ = θ ml . y1

10.15 The ML equation is given by ∂s (t , θ) 2 T [ y (t ) − s (t , θ)] dt = 0 N 0 ∫0 ∂θ

where, s (t , θ) = A cos(ω c t + θ) and

∂s (t , θ) = − A sin(ω c t + θ) . ∂θ

Substituting into the ML equation, we have −

2A T [ y (t ) − A cos(ω c t + θ)] sin(ω c t + θ)dt = 0 N 0 ∫0

T

T

0

0

⇒ ∫ y (t ) sin(ω c t + θ)dt = A∫ cos(ω c t + θ) sin(ω c t + θ)dt =

AT sin[2(ω c t + θ)]dt 2 ∫0

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

∫ y(t )[cos θ sin ω c t + sin θ cos ω c t ]dt ≈ 0 0

T

T

0

0

Therefore, cos θ ∫ y (t ) sin ω c tdt = − sin θ ∫ y (t ) cos ω c t dt


161

T

∫ y(t ) sin ω c tdt

⇒ tan θ = − T0

∫ y(t ) cos ω c tdt 0

or,

θˆ ml

T   ∫ y (t ) sin ω c tdt    = − tan −1  T0 .  y (t ) cos ω c tdt   ∫0 

(b) Indeed, it can be shown that θˆ ml is unbiased and thus, we can apply the Cramer-Rao lower bound. The Cramer-Rao inequality is given by var[θˆ ml ] ≥

N0 2

T

 ∂s (t , θ)  2∫  dt ∂θ  0

with 2

T T ∂s (t , θ)  ∂s (t , θ)  dt = A 2 ∫ sin 2 (ω c t + θ) dt = − A sin(ω c t + θ) ⇒ ∫   ∂θ ∂θ  0 0

=

A2 T A 2T A 2 − ω + θ = − t dt [ 1 cos( 2 2 )] c 2 ∫0 2 2

T

∫ cos(2ω c t + 2θ)dt ≈ 0

A 2T 2

Hence, var[θˆ ml ] ≥

N0 A 2T

then

A 2T << 1 . N0

10.16 (a) The matched filters to s1 (t ) and s 2 (t ) are h1 (t ) = s1 (T − t ) and h2 (t ) = s 2 (T − t ) , respectively, as shown below.


162 h1(t)=s1(T-t) 1.0

0

1

2

3

4

5

6

7

5

6

7

t

h2(t)=s2(T-t) 0.5

0

1

3

2

4

t

-0.5

(b) The filters outputs as a function of time when the signal matched to it is the input are the resulting convolutions y1 (t ) and y 2 (t ) as shown below. s1(t) 1.0

0

t 1

2

3

4

5

6

7

1

2

3

4

5

6

7

1

2

3

4

5

6

7

h1(t)=s1(T-t) 1.0

0

t

y1(t)

2 1

0

t 8


163

s2(t) 0.5

0

1

2

3

4

5

t

6

7

-0.5

h2(t)=s2(T-t) 0.5

0

1

3

2

4

t

5

6

7

5

6

7

8

9

10

10

11

12

13

14

-0.5

y2(t)

1.75

0

1

2

3

4

-0.5

t


164

(c) The output of the filter matched to s 2 (t ) when the input is s1 (t ) is y (t ) = s1 (t ) ∗ h2 (t ) as shown below. y (t ) = s1 (t ) ∗ h2 (t ) 1.0 0.5

t 0

1

2

3

4

5

6

7

8

9

10

-0.5 -1.0

10.17 (a) The signals s1 (t ) and s 2 (t ) are orthonormal s1 (t )

s2 (t )

T /2

T /2 t T /2

t T /2

T

 2 / T Hence, h1 (t ) = s1 (T − t ) =   0

, T /2≤t ≤T , otherwise

and  2 / T h2 (t ) = s 2 (T − t ) =   0

, 0≤t ≤T /2 , otherwise

.

T

11


h1 (t )

165

h2 (t )

T /2

T /2 t T /2

t T /2

T

(b) The noise free output y k (t ) = s k (t ) ∗ hk (t ) , k = 1, 2 . Hence,

of

y1 (t )

the

T

matched

filters

is

y2 (t )

1

1 t T /2

T

3T / 2

0

t T /2

T

Note that we sample at t = T and thus, y1 (T ) = 1 and y 2 (T ) = 0 . (c)

The SNR at the output d2 2E 2 SNR 0 = = = since E = 1 . N0 / 2 N0 N0

of

the

matched

filter

is

10.18 s1 (t ) = cos ω c t , then the signal energy is E = T / 2 , and the first basis

function is φ1 (t ) = 2 / T cos ω c t . Consequently, the first coefficient in the Karhunen-Loève expansion of Y (t ) is T  H 1 : ∫ [ A cos(ω c t + θ) + W (t )]φ1 (t )dt  T  0 Y1 = ∫ Y (t )φ1 (t )dt =  T  0 H : 0 ∫ W (t )φ1 (t )dt  0 

Then, we select a suitable set of functions φ k (t ), k = 2, 3, L , orthogonal to φ1 (t ) . We observe that for k > 2 , we always obtain W k independently of the hypothesis. Only Y1 depends on which hypothesis is true. Thus, Y1 is a sufficient statistic.


166

Y1 is a Gaussian random variable with conditional means  T  E[Y1 a, θ, H 1 ] = E a cos θ = a E cos θ  2  E[Y1 a, θ, H 0 ] = E[W1 ] = 0

and variances var[Y1 a, θ, H 1 ] = var[Y1 a, θ, H 0 ] =

N0 2

The conditional likelihood ratio is given by Λ[ y (t ) a, θ] =

f Y1 A,Θ, H1 ( y1 a, θ, H 1 ) f Y1 A,Θ, H 0 ( y1 a, θ, H 0 )

 2   1 2 y1 E cos θ  exp − E cos 2 = exp N N 0  0  

 θ  

f A,Θ (a, θ) = f A (a ) f Θ (θ) since A and Θ are independent. Hence, Λ[ y (t )] = ∫

∫

A θ

Λ[ y (t ) a, θ]dadθ

Substituting for Λ[ y (t ) a, θ] and f A,Θ (a, θ) into the above integral, the decision rule reduces to H1   2σ a2 > y12  η Λ[ y (t )] = exp  2 2σ a2 + N 0  N 0 (2σ a + N 0 )  < H0 N0

or, H1 y12

> γ < H0

with


γ=

N 0 (2σ 2a + N 0 )

ln η

2σ 2a

167

( 2σ 2a + N 0 ) N0

(b) The receiver can be implemented as follows

Y(t)

H1

T

∫

> γ <

Squarer

0

H0

H1 H0

2 cos ωc t T

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

H0 ( yc , y s H 0 ) =

 y c2 + y s2  exp −  2πσ 2 2σ 2   1

Using the transformations Yc = r cos θ and Ys = r sin θ then, f R H 0 (r H 0 ) =

 r2 exp − 2  2σ σ  r

2

   

and the probability of false alarm is PF =

∞

∫

γ

 r2 − exp  2σ 2 σ2  r

   dr = exp − γ  = exp − γ    2   2σ   N0  

The probability of detection is PD = ∫ PD (a ) f A (a ) da A

where, PD (a ) =

∞

 r 2 + ( AT / 2) 2  exp  dr ∫ σ 2 − 2σ 2  γ  r


168

Solving for the expressions of PD (a) and f A (a) , and solving the integral, we obtain   2γ 2 PD = exp −  2  T ( N 0 + Tσ a ) 

Expressing PD in terms of PF , we obtain N0

PD = (PF ) N 0 + σ 2aT T  T 10.20 (a) E[ N k ] = E  ∫ N (t )φ k (t )dt  = ∫ E[ N (t )]φ k (t )dt = 0  0  0

and var[ N k ] = E[ N k2 ] =

N0 + λk 2

(b) N 0 / 2 is the variance of the white noise process. λ k 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) c n′n′ (t , u ) =

N0 δ(t − u ) ≡ white Gaussian noise 2

⇒ E[ N k′ ] = 0 and var[ N k′ ] = E[ N k′ 2 ] =

N0 2

10.21 (a) N 1 (t ) = W has one eigenfunction. N 0 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 N (t ) = N 1 (t ) + N 2 (t ) can be whitened by:


N (t )

Delay T

+

∑

169

N′(t )

−

2 / N0 σ 2w ( N 0 / 2) + σ 2w 1/ T

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

Signal Detection and Estimation - Solution Manual

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