Series representation of solistics lectr19.

1. 1
19. Series Representation of Stochastic Processes
Given information about a stochastic process X(t) in
can this continuous information be represented in terms of a countable
set of random variables whose relative importance decrease under
some arrangement?
To appreciate this question it is best to start with the notion of
a Mean-Square periodic process. A stochastic process X(t) is said to
be mean square (M.S) periodic, if for some T > 0
i.e
Suppose X(t) is a W.S.S process. Then
Proof: suppose X(t) is M.S. periodic. Then
,
0 T
t 

(19-1)
.
all
for
0
]
)
(
)
(
[
2
t
t
X
T
t
X
E 


( ) ( ) with 1 for all .
X t X t T probability t
 
)
(
PILLAI
( ) is mean-square perodic ( ) is periodic in the
ordinary sense, where
X t R 

*
( ) [ ( ) ( )]
R E X t X t T
  

2. 2
But from Schwarz’ inequality
Thus the left side equals
or
i.e.,
i.e., X(t) is mean square periodic.
.
period
with
periodic
is
)
( T
R 
2 2 2
*
1 2 2 1 2 2
0
[ ( ){ ( ) ( )} ] [ ( ) ] [ ( ) ( ) ]
E X t X t T X t E X t E X t T X t
    
* *
1 2 1 2 2 1 2 1
[ ( ) ( )] [ ( ) ( )] ( ) ( )
E X t X t T E X t X t R t t T R t t
      
Then
periodic.
is
)
(
Suppose
)
( 
R

0
)
(
)
(
)
0
(
2
]
|
)
(
)
(
[| *
2





 

 R
R
R
t
X
t
X
E
( ) ( ) for any
R T R
  
  
.
0
]
)
(
)
(
[
2


 t
X
T
t
X
E (19-2)
PILLAI
(19-3)
*
1 2 2
[ ( ){ ( ) ( )} ] 0
E X t X t T X t
  

3. 3
Thus if X(t) is mean square periodic, then is periodic and let
represent its Fourier series expansion. Here
In a similar manner define
Notice that are random variables, and
0
0
1
( ) .
T jn
n R e d
T
 
  

 
0
0
1
( )
T jk t
k
c X t e dt
T

 




k
ck,
(19-5)
PILLAI
(19-6)
)
(
R
0
0
2
( ) ,
jn
n
R e
T
  
  


 
 (19-4)
0 1 0 2
0 1 0 2
0 2 1 0 1
* *
1 1 2 2
2 0 0
2 1 1 2
2 0 0
( ) ( )
2 1 2 1 1
0 0
1
[ ] [ ( ) ( ) ]
1
( )
1 1
[ ( ) ( )]
m
T T
jk t jm t
k m
T T jk t jm t
T T jm t t j m k t
E c c E X t e dt X t e dt
T
R t t e e dt dt
T
R t t e d t t e dt
T T

 
 
 
 



   

 
  
 
 
 

4. 4
i.e., form a sequence of uncorrelated random variables,
and, further, consider the partial sum
We shall show that in the mean square sense as
i.e.,
Proof:
But
0 1
,
1 ( )
*
1
0
0,
[ ] { }
0 .
m k
T m
j m k t
k m m T
k m
E c c e dt
k m



    

  


 (19-7)




n
n
n
c }
{
0
( ) .
N
jk t
N k
K N
X t c e 


 
)
(
)
(
~
t
X
t
XN 
2 2 *
2
[ ( ) ( ) ] [ ( ) ] 2Re[ ( ( ) ( )]
[ ( ) ].
N N
N
E X t X t E X t E X t X t
E X t
  

(19-8)
.


N
.
as
0
]
)
(
~
)
(
[
2



 N
t
X
t
X
E N
(19-9)
(19-10)
PILLAI

5. 5
0
0
0
2
* *
( ) *
0
( )
0
[ ( ) ] (0) ,
[ ( ) ( )] [ ( )]
1
[ ( ) ( ) ]
1
[ ( ) ( )] .
k
k
k
N
jk t
N k
k N
N T jk t
k N
N N
T jk t
k
k N k N
E X t R
E X t X t E c e X t
E X e X t d
T
R t e d t
T

 
 


 
  




 

 
 
 


   


 
 

PILLAI
(19-12)
Similarly
i.e.,
0 0
2 ( ) ( )
* *
2
[ ( ) ] [ [ ] .
[ ( ) ( ) ] 2( ) 0 as
N
j k m t j k m t
N k m k m k
k m k m k N
N
N k k
k k N
E X t E c c e E c c e
E X t X t N
 

 
 


 
  
      
  
  (19-13)
0
( ) , .
jk t
k
k
X t c e t




    
 (19-14)
and

6. 6
Thus mean square periodic processes can be represented in the form
of a series as in (19-14). The stochastic information is contained in the
random variables Further these random variables
are uncorrelated and their variances
This follows by noticing that from (19-14)
Thus if the power P of the stochastic process is finite, then the positive
sequence converges, and hence This
implies that the random variables in (19-14) are of relatively less
importance as and a finite approximation of the series in
(19-14) is indeed meaningful.
The following natural question then arises: What about a general
stochastic process, that is not mean square periodic? Can it be
represented in a similar series fashion as in (19-14), if not in the whole
interval say in a finite support
Suppose that it is indeed possible to do so for any arbitrary process
X(t) in terms of a certain sequence of orthonormal functions.
*
,
( { } )
k m k k m
E c c  
 0 as .
k k
   
.
, 



k
ck
2
(0) [ ( ) ] .
k
k
R E X t P



    

k
k



 0 as .
k k
   
,
k  
,




 t 0 ?
t T
 
PILLAI

7. 7
i.e.,
where
and in the mean square sense
Further, as before, we would like the ck s to be uncorrelated random
variables. If that should be the case, then we must have
Now




1
)
(
)
(
~
n
k
k t
c
t
X  (19-15)
(19-16)
(19-17)
( ) ( ) in 0 .
X t X t t T
 
*
,
[ ] .
k m m k m
E c c  
 (19-18)
* * *
1 1 1 2 2 2
0 0
* *
1 1 2 2 2 1
0 0
*
1 1 2 2 2 1
0 0
[ ] [ ( ) ( ) ( ) ( ) ]
( ) { ( ) ( )} ( )
( ){ ( , ) ( ) }
T T
k m k m
T T
k m
T T
k XX m
E c c E X t t dt X t t dt
t E X t X t t dt dt
t R t t t dt dt
 
 
 



 
 
  (19-19)
PILLAI
*
0
*
,
0
( ) ( )
( ) ( ) ,
T
k k
T
k n k n
c X t t dt
t t dt

  






8. 8
and
Substituting (19-19) and (19-20) into (19-18), we get
Since (19-21) should be true for every we must have
or
i.e., the desired uncorrelated condition in (19-18) gets translated into the
integral equation in (19-22) and it is known as the Karhunen-Loeve or
K-L. integral equation.The functions are not arbitrary
and they must be obtained by solving the integral equation in (19-22).
They are known as the eigenvectors of the autocorrelation
*
1 1 2 2 2 1 1
0 0
( ){ ( , ) ( ) ( )} 0.
XX
T T
k m m m
t R t t t dt t dt
   
 
 
1 2 2 2 1
0
( , ) ( ) ( ) 0,
XX
T
m m m
R t t t dt t
  
 

( ), 1 ,
k t k
   
*
, 1 1 1
0
( ) ( ) .
T
m k m m k m
t t dt
    
  (19-20)
(19-21)
1 2 2 2 1 1
0
( , ) ( ) ( ), 0 , 1 .
XX
T
m m m
R t t t dt t t T m
  
     


1
)}
(
{ k
k t

(19-22)
PILLAI

9. 9
function of Similarly the set represent the eigenvalues
of the autocorrelation function. From (19-18), the eigenvalues
represent the variances of the uncorrelated random variables
This also follows from Mercer’s theorem which allows the
representation
where
Here and are known as the eigenfunctions
and eigenvalues of A direct substitution and
simplification of (19-23) into (19-22) shows that
Returning back to (19-15), once again the partial sum
1 2
( , ).
XX
R t t
*
1 2 1 2 1 2
1
( , ) ( ) ( ), 0 , ,
XX k k k
k
R t t t t t t T
  


  
 (19-23)
*
,
0
( ) ( ) .
T
k m k m
t t dt
  


1 2
( , ) respectively.
XX
R t t
)
(t
k
 ,
k

k
( ) ( ), , 1 .
k k k
t t k
   
    
1
( ) ( ) ( ), 0
N
N
k k N
k
X t c t X t t T
 

 
  

PILLAI
(19-24)
(19-25)
1
{ }
k k
 

k

,
k
c
1 .
k   
1
k   

10. 10
in the mean square sense. To see this, consider
We have
Also
Similarly
* * *
1
* *
0
1
*
0
1
*
1
[ ( ) ( )] ( ) ( )
[ ( ) ( )] ( ) ( )
( ( , ) ( ) ) ( )
( ) ( )= | (
N
N k k
k
N T
k k
k
N T
k k
k
N
k k k k k
k
E X t X t X t c t
E X t X t d
R t d t
t t

    
    
    










 
 2
1
) | .
N
k
t


(19-26)



N
k
k
k
N t
t
X
t
X
E
1
2
*
|
)
(
|
)]
(
~
)
(
[ 

PILLAI
2 2 *
* 2
[| ( ) ( ) | ] [| ( ) | ] [ ( ) ( )]
[ ( ) ( )] [| ( ) | ].
N N
N N
E X t X t E X t E X t X t
E X t X t E X t
  
 
2
[| ( ) | ] ( , ).
E X t R t t
 (19-27)
(19-28)
(19-29)

11. 11
and
Hence (19-26) simplifies into
i.e.,
where the random variables are uncorrelated and faithfully
represent the random process X(t) in provided
satisfy the K-L. integral equation.
Example 19.1: If X(t) is a w.s.s white noise process, determine the
sets in (19-22).
Solution: Here
2 * * 2
1
[| ( ) | ] [ ] ( ) ( ) | ( ) | .
N
N k m k m k k
k m k
E X t E c c t t t
   

 
 
.
as
0
|
)
(
|
)
,
(
]
|
)
(
~
)
(
[|
1
2
2





 

N
k
k
k
N t
t
t
R
t
X
t
X
E 

1
( ) ( ), 0 ,
k k
k
X t c t t T



 


1
}
{ k
k
c
0 ,
t T
  ( ),
k t

(19-30)
)
(
)
,
( 2
1
2
1 t
t
q
t
t
RXX 
 
PILLAI
(19-31)
(19-32)
1
{ , }
k k k
  

(19-33)
1 ,
k   

12. 12
and
can be arbitrary so long as they are orthonormal as in (19-17)
and Then the power of the process
and in that sense white noise processes are unrealizable. However, if
the received waveform is given by
and n(t) is a w.s.s white noise process, then since any set of
orthonormal functions is sufficient for the white noise process
representation, they can be chosen solely by considering the other
signal s(t). Thus, in (19-35)
2
1 1
[| ( ) | ] (0) k
k k
P E X t R q

 
 
     
 
)
(
)
(
)
( 2
1
2
1
2
1 t
t
q
t
t
R
t
t
R ss
rr 



 
.
1
, 


 k
q
k

( ) ( ) ( ), 0
r t s t n t t T
   
)
(t
k


PILLAI
(19-34)
(19-35)
(19-36)
1 2 2 1 1 2 2 1
0 0
1 1
( , ) ( ) ( ) ( )
( ) ( )
XX
T T
k k
k k k
R t t t dt q t t t dt
q t t
  
  
 
 
 


13. 13
and if
Then it follows that
Notice that the eigenvalues of get incremented by q.
Example19.2: X(t) is a Wiener process with
In that case Eq. (19-22) simplifies to
and using (19-39) this simplifies to
)
( 2
1 t
t
Rss 
*
1 2 1 2
1
( ) ( ) ( ) ( ).
rr k k k
k
R t t q t t
  


  

2 1 2
1 2 1 2
1 1 2
( , ) min( , ) , 0
XX
t t t
R t t t t
t t t

 



  



(19-39)
1
1
1 2 2 2 1 2 2 2
0 0
1 2 2 2 1
( , ) ( ) ( , ) ( )
( , ) ( ) ( ),
T t
XX k XX k
T
XX k k k
t
R t t t dt R t t t dt
R t t t dt t
 
  

 
 

PILLAI





1
2
*
1
2
1 )
(
)
(
)
(
k
k
k
k
ss t
t
t
t
R 

 (19-37)
(19-38)
0 1
t T
2
dt


1
1
2 2 2 1 2 2 1
0
( ) ( ) ( ).
t T
k k k k
t
t t dt t t dt t
     
 
  (19-40)

14. 14
Derivative with respect to t1 gives [see Eqs. (8-5)-(8-6), Lecture 8]
or
Once again, taking derivative with respect to t1, we obtain
or
and its solution is given by
But from (19-40)
1
1 1 1 1 2 2 1
( ) ( 1) ( ) ( ) ( )
T
k k k k k
t
t t t t t dt t
       
   

( ) cos sin .
k k
k k k
t A t B t
 
 
  
.
)
(
)
( 1
2
2
1
 
T
t k
k
k t
dt
t 


 
,
0
)
0
( 
k

PILLAI
(19-41)
(19-42)
1 1
( 1) ( ) ( )
k k k
t t
   
 
2
1
1
2
1
( )
( ) 0,
k
k
k
d t
t
dt
 


 
(19-43)

15. 15
(19-45)
(19-47)
(0) 0, 1 ,
( ) cos ,
k k
k k
k k
A k
t B t
 
 


    

PILLAI
and from (19-41)
This gives
and using (19-44) we obtain
Also
( ) 0.
k T
  (19-44)
 
2
2 2
1
2
( ) cos 0
2 1
2
, 1 .
( )
k k
k
k k
k
T B T
T k
T
k
k
 
 







 
  
    

(19-46)

16. 16
PILLAI
Further, orthonormalization gives
Hence
with as in (19-47) and as in (19-16),
( ) sin , 0 .
k
k k
t B t t T


    (19-48)
 
2 1 cos2
2 2 2
2
0 0 0
sin2 sin(2 1) 0
2 2 2
2 4
0
2
( ) sin
1
1
2 2 2
2/ .
k
k
k
k k
t
T T T
k k k
T
t k
k k k
k
T
t dt B t dt B dt
T T
B B B
B T




 
 





 
 
 
   
 
 
 
   
     
 
 
 
 
 
  
(19-49)
   
2 2
,
1
2
( ) sin sin
k
k T T
t
T
t t k



   
k
 k
c

17. 17
is the desired series representation.
Example 19.3: Given
find the orthonormal functions for the series representation of the
underlying stochastic process X(t) in 0 < t < T.
Solution: We need to solve the equation
Notice that (19-51) can be rewritten as,
Differentiating (19-52) once with respect to t1, we obtain
PILLAI




1
)
(
)
(
k
k
k t
c
t
X 
| |
( ) , 0,
XX
R e  
 

  (19-50)
0 1
t T
2
dt


2
dt


1 2
| |
2 2 1
0
( ) ( ).
T t t
n n n
e t dt t

  
 

 (19-51)
0 0
1
1 2 2 1
1
t ( ) ( )
2 2 2 2 1
0 t
( ) ( ) ( )
T
t t t t
n n n n
e t dt e t dt t
 
   
 
   
 
  (19-52)

18. 18
Differentiating (19-53) again with respect to t1, we get
or
1
1 2 2 1
1
1
1 2 2 1
1
( ) ( )
1 2 2 1 2 2
0
1
1
( ) ( ) 1
2 2 2 2
0
1
( ) ( ) ( ) ( ) ( )
( )
( )
( ) ( )
t T
t t t t
n n n n
t
n
n
t T
t t t t n n
n n
t
t e t dt t e t dt
d t
dt
d t
e t dt e t dt
dt
 
 
     


 
 

   
   
   

   
 
  (19-53)
1
1 2
2 1
1
( )
1 2 2
0
2
( ) 1
1 2 2 2
1
( ) ( ) ( )
( )
( ) ( )
t t t
n n
T t t n n
n n
t
t e t dt
d t
t e t dt
dt


  
 
  

 
 
  
  


1
1 2 2 1
1
1
( ) ( )
1 2 2 2 2
0
( ) {use (19-52)}
2
1
2
1
2 ( ) ( ) ( )
( )
n n
t T
t t t t
n n n
t
t
n n
t e t dt e t dt
d t
dt
 
 
   
 

   
 
  
 

 
PILLAI

19. 19
or
or
Eq.(19-54) represents a second order differential equation. The solution
for depends on the value of the constant on the
right side. We shall show that solutions exist in this case only if
or
In that case
Let
and (19-54) simplifies to
( )
n t

2
1
1 2
1
( )
( 2) ( ) n n
n n
d t
t
dt
 
 

 
2
1
1
2
1
( ) ( 2)
( ).
n n
n
n
d t
t
dt
  



 
  
 
(19-54)
(19-55)
PILLAI
( 2)/
n n
  

2,
n
 
( 2)/ 0.
n n
  
 
2
0 .
n


 
(19-56)
2
2
1
1
2
1
( )
( ).
n
n n
d t
t
dt

 
  (19-57)
2 (2 )
0,
n
n
n
 



 


20. 20
PILLAI
General solution of (19-57) is given by
From (19-52)
and
Similarly from (19-53)
and
Using (19-58) in (19-61) gives
1 1 1
( ) cos sin .
n n n n n
t A t B t
  
  (19-58)
2
2 2
0
1
(0) ( )
T t
n n
n
e t dt


 

  (19-59)
2
2 2
0
1
( ) ( ) .
T t
n n
n
T e T t dt


 

 
 (19-60)
2
1
1
2 2
0
1 0
( )
(0) ( ) (0)
T t
n
n n n
n
t
d t
e t dt
dt

 
  



  
 (19-61)
2
2 2
0
( ) ( ) ( ).
T t
n n n
n
T e T t dt T


  


    
 (19-62)

21. 21
PILLAI
or
and using (19-58) in (19-62), we have
or
Thus are obtained as the solution of the transcendental equation
,
n n
n
A
B



n n n
B A
 

(19-63)
(19-64)
sin cos ( cos sin ),
( )cos ( )sin
n n n n n n n n n n
n n n n n n n n
A T B T A T B T
A B T A B T
      
     
    
   
2
2 2 / 2 / 2( / )
tan .
( ) 1
1 1
n n n n n n
n
n
n n n n
n n n
n n
A A
B B
A A B A B
T
A B


 


    

 
   
 
 
2
2( / )
tan ,
( / ) 1
n
n
n
T
 

 


s
n


22. 22
which simplifies to
In terms of from (19-56) we get
Thus the eigenvalues are obtained as the solution of the transcendental
equation (19-65). (see Fig 19.1). For each such the
corresponding eigenvector is given by (19-58). Thus
since from (19-65)
and cn is a suitable normalization constant.
2
(or ),
n n
 
2
( ) cos sin
sin( ) sin ( ), 0
n n n n n
n n n n n
T
t A t B t
c t c t t T
  
  
 
     
(19-66)
2 2
2
0.
n
n


 
 

1 1
tan tan /2,
n n
n n
n
A
T
B

 

 
   
    
 
 
 
 
(19-68)
(19-67)
s
n

PILLAI
tan( / 2) .
n
nT



  (19-65)

23. 23
PILLAI
Fig 19.1 Solution for Eq.(19-65).
2

1

T


0
tan( / 2)
T

/
 




24. 24
PILLAI
Karhunen – Loeve Expansion for Rational Spectra
[The following exposition is based on Youla’s classic paper “The
solution of a Homogeneous Wiener-Hopf Integral Equation occurring
in the expansion of Second-order Stationary Random Functions,” IRE
Trans. on Information Theory, vol. 9, 1957, pp 187-193. Youla is
tough. Here is a friendlier version. Even this may be skipped on a first
reading. (Youla can be made only so much friendly.)]
Let X(t) represent a w.s.s zero mean real stochastic process with
autocorrelation function so that its power spectrum
is nonnegative and an even function. If is rational, then the
process X(t) is said to be rational as well. rational and even
implies
( ) ( )
XX XX
R R
 
 
( )
XX
S 
( )
XX
S 
0
( ) ( ) 2 ( )cos
XX XX XX
j t
S R e dt R d

    
 


 
  (19-69)
(19-70)
2
2
( )
( ) 0.
( )
XX
N
S
D



 

25. 25
PILLAI
The total power of the process is given by
and for P to be finite, we must have
(i) The degree of the denominator polynomial
must exceed the degree of the numerator polynomial
by at least two,
and
(ii) must not have any zeros on the real-frequency
axis.
The s-plane extension of is given by
Thus
and the Laplace inverse transform of is given by
2
2
( )
( )
1 1
2 2
( )
XX
N
D
P S d d


 
  
 
 
 
  (19-71)
( ) 2
D n
  2
( )
D 
2
( )
N 
( ) 2
N m
 
2
( )
D  ( )
s j

( )
XX
S 
(19-72)
2 2
( ) k
s  

2 2 2
( ) ( ) i
k
k
k
D s s 
  
 (19-73)
( )
s j
 
 
2
2
2
( )
( ) | ( ) .
( )
XX s j
N s
S S s
D s

 

 



26. 26
PILLAI
Let represent the roots of D(– s2) . Then
Let D+(s) and D–(s) represent the left half plane (LHP) and the right half
plane (RHP) products of these roots respectively. Thus
where
This gives
Notice that has poles only on the LHP and its inverse (for all t > 0)
converges only if the strip of convergence is to the right
1 2
, , , n
  
  
1 2
0 Re Re Re n
  
   
(19-74)
(19-75)
| |
2 2 1
1
1 ( 1) ( 2)! | |
( 1)! ( 1)!( )!
( ) (2 )
k k j
k
k k j
j
k j
e
k j k j
s
  
 


 

  

  


2
( ) ( ) ( ),
D s D s D s
 
  (19-76)
1 ( )
( )
C s
D s

*
0
( ) ( )( ) ( ).
n
k
k k k
k k
D s s s d s D s
 
 

     
  (19-77)
2
2 1 2
2
( ) ( )
( )
( )
( ) ( ) ( )
C s C s
N s
S s
D s D s D s
 

  

(19-78)

27. 27
PILLAI
of all its poles. Similarly
C2(s) /D–(s) has poles only on
the RHP and its inverse will
converge only if the strip is
to the left of all those poles. In
that case, the inverse exists for
t < 0. In the case of from
(19-78) its transform N(s2) /D(–s2)
is defined only for (see Fig 19.2). In particular,
for from the above discussion it follows that is given
by the inverse transform of C1(s) /D+(s). We need the solution to the
integral equation
that is valid only for 0 < t < T. (Notice that in (19-79) is the
reciprocal of the eigenvalues in (19-22)). On the other hand, the right
side (19-79) can be defined for every t. Thus, let
( ),
XX
R 
1 1
Re Re Re
s
 
  
0,
 

0
( ) ( ) ( ) , 0
XX
T
t R t d t T
     
   
 (19-79)
(19-80)
( )
XX
R 
Fig 19.2
 
Re n

 1
Re 
 Re n

1
Re 
strip of convergence
for ( )
XX
R 
s j

 
0
( ) ( ) ( ) ,
XX
T
g t R t d t
   
      



28. 28
PILLAI
and to confirm with the same limits, define
This gives
and let
Clearly
and for t > T
since RXX(t) is a sum of exponentials Hence it
follows that for t > T, the function f (t) must be a sum of exponentials
Similarly for t < 0
, for 0.
k t
k
k
a e t




.
kt
k
k
a e 


( ) 0
( ) .
0 otherwise
t t T
t


 

 

(19-81)
(19-82)
+
( ) ( ) ( )
XX
g t R t d
   


 

+
( ) ( ) ( ) ( ) ( ) ( ) .
XX
f t t g t t R t d
       


    
 (19-83)
( ) 0, 0
f t t T
   (19-84)
   
( ) 0
+
( ) { ( )} ( ) 0,
k
XX
D
d d
dt dt
D f t D R t d

    

 

 

   
 (19-85)

29. 29
PILLAI
and hence f (t) must be a sum of exponentials
Thus the overall Laplace transform of f (t) has the form
where P(s) and Q(s) are polynomials of degree n – 1 at most. Also from
(19-83), the bilateral Laplace transform of f (t) is given by
Equating (19-86) and (19-87) and simplifying, Youla obtains the key
identity
Youla argues as follows: The function is an entire
function of s, and hence it is free of poles on the entire
   
( ) 0
+
( ) { ( )} ( ) 0,
k
XX
D
d d
dt dt
D f t D R t d

    



 

   

, for 0.
k t
k
k
b e t



contributes to 0
contributions
in < 0
contributions in
( ) ( )
( )
( ) ( )
sT
t
t
t T
P s Q s
F s e
D s D s

 


 
(19-86)
2
2 1 1
( )
( )
( ) ( ) 1 , Re Re Re
N s
D s
F s s s
  


 
     
  (19-87)
(19-88)
2 2
( ) ( ) ( ) ( )
( ) .
( ) ( )
sT
P s D s e Q s D s
s
D s N s

  

 
  
0
( ) ( )
T st
s t e dt
 
  

30. 30
PILLAI
finite s-plane However, the denominator on the right
side of (19-88) is a polynomial and its roots contribute to poles of
Hence all such poles must be cancelled by the numerator. As a result
the numerator of in (19-88) must possess exactly the same set of
zeros as its denominator to the respective order at least.
Let be the (distinct) zeros of the
denominator polynomial Here we assume that
is an eigenvalue for which all are distinct. We have
These also represent the zeros of the numerator polynomial
Hence
and
which simplifies into
From (19-90) and (19-92) we get
).
Re
( 


 s
)
(s

1 2
( ), ( ), , ( )
n
     
  
2 2
( ) ( ).
D s N s

   
'
k s

'
k s

( ) ( ) ( ) ( ).
sT
P s D s e Q s D s
  

1 2
0 Re ( ) Re ( ) Re ( ) .
n
     
      (19-89)
( ) ( ) ( ) ( )
kT
k k k k
D P e D Q

   

 
 (19-90)
(19-91)
( ) ( ) ( ) ( )
kT
k k k k
D P e D Q

   
 
    
( ) ( ) ( ) ( ).
kT
k k k k
D P e D Q

   
 
   (19-92)
( ).
s


31. 31
PILLAI
i.e., the polynomial
which is at most of degree n – 1 in s2 vanishes at
(for n distinct values of s2). Hence
or
Using the linear relationship among the coefficients of P(s) and Q(s)
in (19-90)-(19-91) it follows that
are the only solutions that are consistent with each of those equations,
and together we obtain
( ) ( ) ( ) ( ), 1, 2, ,
k k k k
P P Q Q k n
   
    (19-93)
( ) ( ) ( ) ( ) ( )
L s P s P s Q s Q s
    (19-94)
2 2 2
1 2
, , , n
  
2
( ) 0
L s  (19-95)
( ) ( ) ( ) ( ).
P s P s Q s Q s
   (19-96)
( ) ( ) or ( ) ( )
P s Q s P s Q s
     (19-97)

32. 32
PILLAI
as the only solution satisfying both (19-90) and (19-91). Let
In that case (19-90)-(19-91) simplify to (use (19-98))
where
For a nontrivial solution to exist for in (19-100), we
must have
( ) ( )
P s Q s
   (19-98)
1
0
( ) .
n
i
i
i
P s p s


  (19-99)
0 1 1
, , , n
p p p 
1
0
( ) ( ) ( ) ( )
{1 ( 1) } 0, 1,2, ,
kT
k k k k
n
i i
k k i
i
P D e D P
a p k n

   


 



   
 (19-100)
( ) ( )
.
( ) ( )
k k
T T
k k
k
k k
D D
a e e
D D
 
 
 
 
 
 

  (19-101)

33. 33
PILLAI
The two determinant conditions in (19-102) must be solved together to
obtain the eigenvalues that are implicitly contained in the
and (Easily said than done!).
To further simplify (19-102), one may express ak in (19-101) as
so that
'
i s
 '
i
a s
'
i s

2
, 1, 2, ,
k
k
a e k n


 
1 1
1 1 1 1 1
1 1
2 2 2 2 2
1,2
1 1
(1 ) (1 ) (1 ( 1) )
(1 ) (1 ) (1 ( 1) )
0.
(1 ) (1 ) (1 ( 1) )
n n
n n
n n
n n n n n
a a a
a a a
a a a
 
 
 
 
 
 
 
 
  
 
(19-102)
(19-104)
(19-103)
/ 2 / 2
/ 2 / 2
1 ( ) ( )
tan
1 ( ) ( )
( ) ( )
( ) ( )
k
k k
k k k
k k
k k
T
k k k
k T
k k k
T T
k k
T T
k k
a D e D
e e
a
e e D e D
e D e D
e D e D

 
  
 
 
 

 
 
 

 

 
 

 

 
  

  

  
 

 
h

34. 34
PILLAI
Let
and substituting these known coefficients into (19-104) and simplifying
we get
and in terms of in (19-102) simplifies to
if n is even (if n is odd the last column in (19-107) is simply
Similarly in (19-102) can be obtained by
replacing with in (19-107).
0 1
( ) n
n
D s d d s d s

    (19-105)
2
tan ,
k
h  
2 3
0 2 1 3
2 3
0 2 1 3
( )tan ( / 2) ( )
tan
( ) ( )tan ( / 2)
k k k k
k
k k k k
d d T d d
d d d d T
   

   
    

    
(19-106)
1 1 1
1 2
[ , , , ] ).
n n n
n
T
  
  
1

cot k

h
tan k

h
2 3 1
1 1 1 1 1 1 1
2 3 1
2 2 2 2 2 2 2
1 tan tan tan
1 tan tan tan
1 tan
n
n
n
      
      



2 3 1
0
tan tan
n
n n n n n n
     

 (19-107)
h
h
h
h
h
h
h
h
h
h
h
h

35. 35
PILLAI
To summarize determine the roots with that satisfy
in terms of and for every such determine using (19-106).
Finally using these and in (19-107) and its companion
equation , the eigenvalues are determined. Once are
obtained, can be solved using (19-100), and using that can
be obtained from (19-88).
Thus
and
Since is an entire function in (19-110), the inverse Laplace
transform in (19-109) can be performed through any strip of
convergence in the s-plane, and in particular if we use the strip
,
 ,
k
 k

'
k s

2 2
( ) ( ) 0, 1, 2, ,
k k
D N k n
  
     (19-108)
k s
 tanh k s

k s
 k s

k
p s ( )
i s

( )
i s

2 2
( ) ( , ) ( ) ( , )
( )
( ) ( )
sT
i i
i
i
D s P s e D s Q s
s
D s N s
 

  

 
  
(19-109)
1
( ) { ( )}.
i i
t L s
 
  (19-110)
1

Re( ) 0
i
 

36. 36
PILLAI
then the two inverses
obtained from (19-109) will be causal. As a result
will be nonzero only for t > T and using this in (19-109)-(19-110) we
conclude that for 0 < t < T has contributions only from the first
term in (19-111). Together with (19-81), finally we obtain the desired
eigenfunctions to be
that are orthogonal by design. Notice that in general (19-112)
corresponds to a sum of modulated exponentials.
Re Re( ) (to the right of all Re( )),
n i
s  

1 1
2 2 2 2
( ) ( ) ( ) ( )
,
( ) ( ) ( ) ( )
D s P s D s Q s
L L
D s N s D s N s
 
 
 
   
   
     
   
(19-111)
(19-112)
 
2 2
1 ( ) ( )
( ) ( )
sT D s Q s
e
D s N s
L




  
( )
i t

1
2 2
( ) ( , )
( ) , 0 ,
( ) ( )
Re Re 0, 1,2, ,
k
k
k
n
D s P s
t L t T
D s N s
s k n





  
  
 
  
 
  

37. 37
PILLAI
Next, we shall illustrate this procedure through some examples. First,
we shall re-do Example 19.3 using the method described above.
Example 19.4: Given we have
This gives and P(s), Q(s) are constants
here. Moreover since n = 1, (19-102) reduces to
and from (19-101), satisfies
or is the solution of the s-plane equation
But |esT| >1 on the RHP, whereas on the RHP. Similarly
|esT| <1 on the LHP, whereas on the LHP.
| |
( ) ,
XX
R e  
 

( ) , ( )
D s s D s s
 
 
   
1 1
1 0, or 1
a a
   
1

1

sT s
e
s





2
2 2 2
2 ( )
( ) .
( )
XX
N
S
D
 

  
 

(19-113)
(19-114)
1 1 1
1
1
( )
( )
T D
e
D
   
 




 

1
s
s



 
1
s
s



 

38. 38
PILLAI
Thus in (19-114) the solution s must be purely imaginary, and hence
in (19-113) is purely imaginary. Thus with in (19-114)
we get
or
which agrees with the transcendental equation (19-65). Further from
(19-108), the satisfy
or
Notice that the in (19-66) is the inverse of (19-116) because as
noted earlier in (19-79) is the inverse of that in (19-22).
1
 1
s j


n

2 2
0.
2
n
n
 



  (19-116)
(19-115)
2 2 2 2
( ) ( ) 2 0
n
n n n
s j
D s N s 
   

      
1
1
tan( / 2)
T



 
1 1
1
j T j
e
j
  
 



s


39. 39
PILLAI
Finally from (19-112)
which agrees with the solution obtained in (19-67). We conclude this
section with a less trivial example.
Example 19.5
In this case
This gives With n = 2,
(19-107) and its companion determinant reduce to
1
2 2
( ) cos sin , 0
n n n n n
n
s
t L A t B t t T
s

  

  

    
 

 
(19-117)
| | | |
( ) .
XX
R e e
   
  
  (19-118)
2
2 2 2 2 2 2 2 2
2 2 2( )( )
( ) .
( )( )
XX
S
     

       
 
  
   
(19-119)
2
( ) ( )( ) ( ) .
D s s s s s
    

      
2 2 1 1
2 2 1 1
tan tan
cot cot
   
   


h h
h h

40. 40
PILLAI
or
From (19-106)
Finally can be parametrically expressed in terms of
using (19-108) and it simplifies to
This gives
and
(19-120)
(19-121)
2 2
1 2
and
 
2
2
1
( ) ( ) 4 ( )
2
b b c
  

 

1 2
tan tan .
 
 
h h
2
2
( )tan ( / 2) ( )
tan , 1,2
( ) ( ) tan ( / 2)
i i i
i
i i i
T
i
T
     

     
  
 
  
h
h
h

2 2 4 2 2 2
2 2
4 2
( ) ( ) ( 2 ( ))
2 ( )
0.
D s N s s s
s bs c
     
     
       
  
   


41. 41
PILLAI
and
and substituting these into (19-120)-(19-121) the corresponding
transcendental equation for can be obtained. Similarly the
eigenfunctions can be obtained from (19-112).
2
2 2 2
2 1
( ) ( ) 4 ( )
( ) 4 ( )
2
b b c
b c
  
   
 
   
i s
