Chapter 8

Representation of Signals


8.1 (a) We have,

=
|
|
.
|

\
|

T
dt
T
t k
T
T
0
0 cos
2 1

1
1 1
0
=

T
dt
T T

and

=
=
|
|
.
|

\
|

j k
j k
dt
T
t j
T T
t k
T
T
, 0
, 1
cos
2
cos
2
0

Therefore,

T
t k
T
T
cos
2
,
1
are orthonormal functions.
(b) Similarly, to verify that the set functions is orthonormal in the interval
] 1 , 1 [ , we do 1
2
1
2
1
=

T
T
dt
T T

0 cos
2
1
2 cos
1
2
1
0
=

T T
T
dt
T
t k
T
dt
T
t k
T T

and


114
Representation of Signals
115
kj
T T
T
dt
T
t j
T
t k
T
dt
T
t j
T
T
t k
T
=

=


0
cos cos
2
cos
1
cos
1

Hence, the set is orthonormal on the interval ] 1 , 1 [ .
8.2 (a) We solve 0 ) ( ) (
1
1
1
1
2 1
= =


tdt dt t s t s
2 2 ) (
1
0
1
1
2
1
= =

dt dt t s
and
3
2
2 ) (
1
0
2
1
1
2
1
1
2
2
= =

dt t dt t dt t s
Therefore, ) ( and ) (
2 1
t s t s are orthogonal.
(b) ) (
1
t s orthogonal to 3 0 ) 1 ( 1 ) (
1
1
2
3
= = + +

dt t t t s
) (
2
t s orthogonal to 0 0 ) 1 ( ) (
1
1
2
3
= = + +

dt t t t t s .
Therefore,
2
3
3 1 ) ( t t s = .
8.3 Note that = ) ( 2 ) (
1 3
t s t s We have 2 independent signals.
The energy of ) (
1
t s is thus,
T
T T
dt dt dt t s E
T
T
T T
= + = + = =

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

Signal detection and estimation
116

= =
T t
T
T
T
t
T
E
t s
t
2
,
1
2
0 ,
1
) (
) (
1
1

) ( ) ( ) (
1 21 2 2
t s t s t f = where

=
T
dt t t s s
0
1 2 21
) ( ) ( . Then,
2
1
) 2 (
1
) 1 (
2 /
2 /
0
21
T
dt
T
dt
T
s
T
T
T
+ =
|
|
.
|

\
|
+
|
|
.
|

\
|
=


T t t s t s t f = = 0
2
3
) ( ) ( ) (
1 21 2 2

and
T t
T
dt
t
T
=
|
.
|

\
|

0
1
2
3
2 / 3
) (
0
2
2

(b) ) ( ) (
1 1
t T t s =
) (
2
3
) (
2
) (
2 1 2
t T t
T
t s + =
) ( 2 ) (
1 3
t T t s =
Thus, the signal constellation is

T
2
3
2
T T T 2
1

1
s
2
s
3
s
Representation of Signals
117
| | 0 ,
1
T = s
(
(

= T
T
2
3
,
2
2
s | | 0 , 2
3
T = s
8.4 ) (
) ( ) (
) (
) (
2
2
2 2
t
t
n
t
dt
t d
t
dt
t d
t
t
n
t
dt
t d
t
dt
d

|
|
.
|

\
|
+

=
|
|
.
|

\
|
+ |
.
|

\
|

) ( ) (
) ( ) (
2
2
2
2
t n t
dt
t d
t
dt
t d
t +

=
where,
| |

d t n j
dt
t d
) sin ( exp sin
) (

| |

d t n j
dt
t d
) sin ( exp sin
) (
2
2
2

After substitution in the differential equation, we have

= + + + d t n j n jt t n t t t )] sin ( exp[ ) sin cos ( ) (
2 2 2 2 2 2

but,

d t n j jt )] sin ( exp[ sin


= )] sin ( exp[ cos t n j jt

+ d t n j t n t )] sin ( exp[ ) cos ( cos

+ = d t n j nt t )] sin ( exp[ cos) cos ( 0
2
Thus,
0 )] sin ( exp[ ) cos ( ) (
2 2 2
= = = +

n
n
ju
du e n d t n j t n n n t t t
where = sin t n u .
Signal detection and estimation
118
8.5 Given the differential system 0 ) ( ) ( = + t t , 0 ) 1 ( ) 0 ( = = , we first
integrate with respect to t 0 ) ( ) 0 ( ) (
0
= +

t
du u t .

0 ) ( ) ( ) 0 ( ) 0 ( ) (
0
= +

t
du u u t t t
Using

= = =
t
du u u t du u u t
0
1
0
) ( ) ( ) ( ) 1 ( ) ( 0 ) 1 ( ) 0 (
since

=
1
0
) ( ) 1 ( ) 0 ( du u u

+ =
1
0
) ( ) 1 ( ) ( ) 1 ( ) (
t
t
du u u du u t t
Therefore, the kernel is

=
1 , 1
0 , 1
) , (
u t u
t u t
t u k
8.6 The integral equation can be reduced to the differential equation by taking the
derivative with respect to t 0 ) ( ) ( = + t t with ( ) 0 2 / and 0 ) 0 ( = = .
Let
t j t j
e c e c t

+ =
2 1
) ( . Then,
2 1 2 1
0 ) 0 ( c c c c = + = =
t j t j
e j c e j c t

=
2 1
) ( and 0
2
2 2
1
=
|
|
.
|

\
|
+ = |
.
|

\
|

j j
e e j c .
0
1
= c trivial solution
2 2
0
2
cos

=

or
L , 2 , 1 , 0 , ) 1 2 (
2
= +

= k k
Therefore, ( ) L , 2 , 1 , 0 , ) 1 2 sin( ] [
) 1 2 ( ) 1 2 (
1
= + = =
+ +
k t k c e e c t
t k j t k j
and
L , 1 , 0 , ) 1 2 (
2
= + = k k
k
.
Representation of Signals
119
8.7 Differentiating twice with respect to t, the integral equation reduces to the
differential equation
0 ) ( ) ( = + t t with 0 ) ( ) 0 ( = = T
Let
t j t j
e c e c t

+ =
2 1
) ( . Then,
2 1
0 ) 0 ( c c = = and
t j t j
e c e c T

+ = =
2 1
0 ) (
or, 0 0 cos = c T c and L , 2 , 1 , 0 ,
2
) 1 2 (
2
=
+
= +

= k
T
k
k T
Therefore, the eigenfunctions are
L , 2 , 1 , 0 ,
2
) 1 2 (
cos ) ( =
+
= k t
T
k
c t
8.8 t n B t n A t jn + = = = + cos sin ) ( 0
2

For t n A t n A t u t + = cos sin ) ( 0
2 1

2
0 ) 0 ( A = = and t n A t = sin ) (
1

For t n B t n B t T t u + = cos sin ) (
2 1

0 cos sin 0 ) (
2 1
= + = T n B T n B T
Also, ) (t continuous
u n B u n B u n A u u + = + = cos sin sin ) 0 ( ) 0 (
2 1 1

and
1 cos sin cos 1 ) 0 ( ) 0 (
2 1 2
= + = + u n B n u n n B u n n B u u
Solving for the constants, we obtain
Signal detection and estimation
120

=
T t u t T n
T n n
u n
u t t n
T n n
T u n
t
, ) ( sin
sin
sin
0 , sin
sin
) ( sin
) (
8.9 For u t For u t
2 1
) , ( c t c u t k + =
4 3
) , ( c t c u t k + =
0 ) 0 (
2
= = c k 0 ) , (
4 3
= + = c T c u T k
) , ( u t k continuous At u t = , we have
4 3 1
c u c u c + =
1
) , 0 ( c u u k
t
=
3
) , 0 ( c u u k
t
= +
1 3 3 3 1
1 1 ) , 0 ( ) , 0 ( c u c u c u c c u u k u u k
t t
+ = + + = = +
T
u
c u c = =
3 4
, and
T
u
c =1
1

Therefore,

=
t u u t
T
u
u t t
T
u T
u t k
0 ,
0 ,
) , (
8.10 Taking the integral of the second order integro-differential equation
) ( ) ( ) , (
0
2
2
t du u u t k
dt
d
T
=
(
(

we have




(
(

=
(
(

+ + =
(
(

+ +
T
t
T
t
t
T
t
T
t
t
T
t
T u t
dt t du u
T
u
du u
T
u
dt
d
du u
T
u
t
T
t
t du u t t t t du u
T
u
t
T
t
t
dt
d
du u
T
u
t du u t du u u du u
T
u
t
dt
d
) ( ) ( ) (
) ( ) ( ) ( ) ( ) ( ) ( ) (
) ( ) ( ) ( ) (
0
0
0 0 0

Representation of Signals
121
Thus,
) ( ) ( ) , (
0
2
2
t du u u t k
dt
d
T
=
(
(

For 0 ) ( ) ( = + t t , 0 ) ( ) 0 ( = = T , we have

=
T
du u u t k t
0
) ( ) , ( ) ( a solution since 0 ) ( ) ( ) ( ) ( = + = t t t t as
expected.
8.11 For Problem 8.8, we have

=
u t
T n n
t T n u n
u t
T n n
t n u T n
u t k
,
sin
) ( sin sin
,
sin
sin ) ( sin
) , (
and

=
t u
T n n
u T n t n
t u
T n n
u n u T n
t u k
,
sin
) ( sin sin
,
sin
sin ) ( sin
) , (
We verify if u n u T n t n u T n = sin ) ( sin sin ) ( sin
?
.
We know that + = )] cos( ) [cos(
2
1
sin sin b a b a b a
)] ( cos ) ( [cos
2
1
)] ( cos ) ( [cos
2
1
u t T n u t T n t u T n t u T n + = +
Thus, they are equal and therefore ) , ( ) , ( t u k u t k = .
For Problem 8.9, we have
Signal detection and estimation
122

=
u t u t
T
u
u t t
T
u T
u t k
,
,
) , (
and

=
t u t u
T
t
t u u
T
t T
t u k
,
,
) , (
We observe that
T
T t u
t u
T
t
t
T
u
T
+ +
= + =
Therefore, ) , ( ) , ( t u k u t k = .
8.12 Here, we have two methods. We have ( ) ) ( ), , ( t u t k c
n n
, that is

+
+

= =
T
u
u T
n n
dt
T
t n
T T
u ut
dt
T
t n
T
t
T
u T
dt t u t k c sin
2
sin
2
) ( ) , (
0 0

Solving the integrals, we obtain the desired result
T
u n
n
T
T
c
n

= sin
) (
2
2
2

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

=
T
n
T
n n
du u t u k dt t u t k c
0 0
) ( ) , ( ) ( ) , ( from Problem 8.11. Then, from Problem
8.11, =

n
n
T
n
c
t
du u t u k
) (
) ( ) , (
0
L , 2 , 1 , sin
2
) (
2
2
=

= n
T
u n
T
n
T
c
n

8.13 We have,

=
t u
T m m
t m t T m
t u
T m m
t T m u m
u t h
,
sin
sin ) ( sin
,
sin
) ( sin sin
) , (
Representation of Signals
123

=
T
t
t T
du u
T m m
t m u T m
du u
T m m
u T m u m
t du u u t h t
) (
sin
sin ) ( sin
) (
sin
) ( sin sin
) (
1
) ( ) , ( ) (
0 0

and
( )
( )

T
t
t
du u
T m m
u T m
t m m
t
T m m
t T m
t m m du u
T m m
u m
t T m m
t
T m m
t m
t T m m t
) (
sin
) ( sin
sin
) (
sin
) ( sin
cos ) (
sin
sin
) ( sin
) (
sin
sin
) ( cos ) (
1
2
0
2

Simplifying the above equation, we have

+ =

T
t
t
du u
T m m
t m u T m
du u
T m m
t T m u m
T m t t
) (
sin
sin ) ( sin
) (
sin
) ( sin sin
sin ) ( ) (
1
2
0
2

From Problem 8.12, ) ( sin sin sin ) ( sin t T m u m t m u T m =
(
(

+ = +

T T
du u u t h t du u u t h t
0 0
2
) ( ) , ( ) ( ) ( ) , ( ) (
Thus,

=
T
du u u t h t
0
) ( ) , ( ) ( is a solution of
0 ) ( ) 0 ( , 0 ) ( ] ) [( ) (
2
= = = + + T t m t
In the second part of the question, we use the integral equation to obtain ) (u c
n
in
) ( ) ( ) , (
1
t u c u t h
n
n
n
=

=
. Here,
Signal detection and estimation
124

=
T
t n
T
t
n
sin
2
) ( and

|
.
|

\
|
=
2
2
) (m
T
n
n

This gives ( )

= = =
T
n
T
n n n
du u t u h dt t u t h t u t h c
0 0
) ( ) , ( ) ( ) , ( ) ( ), , ( ( from Problem
8.12. Therefore, by Problem 8.10, we have

) (
) ( ) , (
0
t
du u t u h
n
T
n
where,
T
t n
T
t
n

= sin
2
) ( and
2
2
) ( |
.
|

\
|
= m
T
n
and
L , 2 , 1 , sin ) (
2
2
2
=

(
(

|
.
|

\
|
= n
T
t n
m
T
n
T
c
n

8.16 Let
t j t j
e c e c t

+ =
2 1
) ( ,
1 2 2 1
0 ) 0 ( c c c c = + = = and thus,
t c e c e c t
t j t j
= =

sin ) (
1 1

Let 0 ,
2 2
> = , then L , 2 , 1 sin ) ( = = k t c t
k

= = + = +
k
k k k k
tan 0 cos sin 0 ) 1 ( ) 1 (
Therefore, L , 2 , 1 , sin ) ( = = k t t
k
for positive roots of

=
k
k
tan .
Case 1: Let
2 1
) ( 0 c t c t + = =
t c t c
1 2
) ( and 0 0 ) 0 ( = = =
1 0 0 ) 1 ( ) 1 (
3 3
= = + = + c c but is positive and thus, 0 = is not
an eigenvalue.
Case 2: 0 > such that 0 and
2 2
> = . Then,
Representation of Signals
125
t j t j
e c e c t

+ =
2 1
) (
1 2 2 1
0 ) 0 ( c c c c = + = =

= = + tan 0 ) 1 ( ) 1 (
when 0 0 , 0 >

> <

Thus, ) ( sin ) ( t t
k k
= is solution where
k
are consecutive positive roots of

= tan .

+1
-1

tanh
tanh
-/
-/
tan

1 2
Signal detection and estimation
126
Case 3: If 0 < , let ) 0 (
2
> = = j .
Then = sinh ) (t .
From 0 ) 1 ( ) 1 ( = + , we have 0 cosh sinh = +
0 > and 0 0 >

< . So t t
0 0
sinh ) ( = is a solution
where 0 tanh >

= .
8.17 0 ) ( ) ( = + t t , 0 ) ( ) 0 ( = = T
Let
t j t j
e c e c t

+ =
2 1
) ( . Then,
2 1
0 ) 0 ( c c = = and 0 sin ) (
1
= =
(

=

T c e e j c T
T j T j

0 = c trivial solution L , 2 , 1 , 0 sin =

= = = k
T
k
k T T
2
|
.
|

\
|
=
T
k
and L , 3 , 2 , 1 , cos ) ( =

= k
T
t k
t
and
1 ) (
0
= t when 0 0 = = k