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There is a close analogy between the theory of linear lters of section (11.7) and the

eigenvector decomposition of a matrix in linear algebra as treated in section (10.5). To

see this we will use the same notation as in section (11.7) and use the Fourier transform

(11.45) to write the output of the lter in the time domain as:

Z1

O(!)e;i!t d! :

o(t) = (11.70)

;1

Problem a: Use expression (11.58) to show that this can be written as

Z1

G(!)I(!)e;i!t d!

o(t) = 2 (11.71)

;1

and carry out an inverse Fourier transform of I(!) to nd the following expression

1

ZZ

G(!)e;i!t ei! i( )d!d :

o(t) = (11.72)

;1

In order to establish the connection with linear algebra we introduce by analogy with

(11.33) the following basis functions:

u! (t) 1 e;i!t (11.73)

p

2

and the inner product Z1

(f g) f (t)g(t)dt : (11.74)

;1

11.10. LINEAR FILTERS AND LINEAR ALGEBRA 149

Problem b: Show that these basisfunctions are orthonormal for this inner product in the

sense that ;u u 0 = (! !0) : (11.75)

!! ;

Problem c: These functions play the same role as the eigenvectors in section (10.5). To

which expression in section (10.5) does the above expression correspond?

Problem d: Show that equation (11.72) can be written as

Z1

o(t) = G(!)u! (t) (u! i) d! : (11.76)

;1

This expression should be compared with equation (10.60) of section (10.5)

X

N

Ap = nv

^ v(n) p

^

(n) (10:60) again :

n=1

The integration over frequency plays the same role as the summation over eigenvectors in

equation (10.60). Expression (11.76) can be seen as a description for the operator g(t) in

the time domain that maps the input function i(t) onto the output o(t).

Problem e: Use the equations (11.57), (11.74) and (11.76) to show that:

Z1

g(t ) = 2 G(!)u! (t)u! ( )d! : (11.77)

;1

;

There is a close analogy between this expression and the dyadic decompostion of a matrix

in its eigenvectors and eigenvalues derived in section (10.5).

Problem f: To see this connection show that equation (10.61) can be written in compo-

nent form as:

X (n) (n)T

N

Aij = n vi vj :

^ ^ (11.78)

n=1

The sum over eigenvalues in (11.78) corresponds with the integration over frequency

in (11.77). In section (10.5) linear algebra in a nite-dimensional vector space was

treated, in such a space there is a nite number of eigenvalues. In this section, a

function space with in nitely many degrees of freedom is analyzed it will be no

surprise that for this reason the sum over a nite number of eigenvules should be

replaced by an integration over the continuous variable !. The index i in (11.78)

corresponds with the variable t in (11.77) while the index j corresponds with the

variable .

Problem g: Establish the connection between all variables in the equations (11.77) and

(11.78). Show speci cally that G(!) plays the role of eigenvalue and u! plays the

role of eigenvector. Which operation in (11.77) corresponds to the transpose that is

taken of the second eigenvector in (11.78)?

CHAPTER 11. FOURIER ANALYSIS

150

You may wonder why the function u! (t) = exp (;i!t) = 2 de ned in (11.73) and not

p

some other function plays the role of eigenvector of the impulse respons operator g(t ).

;

To see this we have to understand what a linear lter actually does. Let us rst consider the

example of the reverberation lter of section (11.8). According to (11.59) the reverberation

lter is given by:

o(t) = i(t) r i(t T) + r2 i(t 2T ) + (11:59) again

; ; ;

It follows from this expression that what the lter really does is to take the input i(t),

translate it over a time nT to a new function i(t;nT), multiply each term with (;r)n and

sum over all values of n. This means that the lter is a combination of three operations,

(i) translation in time, (ii) multiplication and (iii) summation over n. The same conclusion

holds for any general time-invariant linear lter.

Problem h: Use a change of the integration variable to show that the action of a time-

invariant linear lter as given in (11.57) can be written as

Z1

o(t) = g( )i(t )d : (11.79)

;1

;

The function i(t; ) is the function i(t) translated over a time . This translated function

is multiplied with g( ) and an integration over all values of is carried out. This means

that in general the action of a linear lter can be seen as a combination of translation in

time, multiplication and integration over all translations . How can this be used to explain

that the correct eigenfunctions to be used are u! (t) = exp (;i!t) = 2 ? The answer does

p

not lie in the multiplication because any function is eigenfunction of the operator that

carries out multplication with a constant, i.e. af(t) = f(t) for every function f(t).

Problem i: What is the eigenvalue ?

This implies that the translation operator is the reason that the eigenfunctions are u! (t) =

exp (;i!t) = 2 . Let the operator that carries out a translation over a time be denoted

p

by T :

T f(t) f(t ) : (11.80)

;

Problem j: Show that the functions u! (t) de ned in (11.73) are the eigenfunctions of the

translation operator T , i.e. show that T u! (t) = u! (t). Express the eigenvalue

of the translation operator in the translation time .

Problem k: Compare this result with the shift property of the Fourier transform that

was derived in (11.61).

This means that the functions u! (t) are the eigenfunctions to be used for the eigenfunction

decomposition of a linear time-invariant lter, because these functions are eigenfunctions

of the translation operator.

Problem l: You identi ed in problem e the eigenvalues of the lter with G(!). Show

that this interpretation is correct, in other words show that when the lter g acts

on the function u! (t) the result can be written as G(!)u! (t). Hint: go back to

problem e of section (11.7).

11.10. LINEAR FILTERS AND LINEAR ALGEBRA 151

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