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### Transcript of The Fourier Transform â€¢ Introduction â€¢ Orthonormal bases for R

• The Fourier Transform

Introduction Orthonormal bases for Rn

Inner product Length Orthogonality Change of basis Matrix transpose

Complex vectors Orthonormal bases for Cn

Inner product Hermitian transpose

Orthonormal bases for 2 periodic functions Shah basis Harmonic signal basis Fourier series

Fourier transform

• Orthonormal bases for Rn

Let u = [u1,u2]T and v = [v1,v2]T be vectors inR

2. We define the inner product of u and v tobe

u,v = u1v1 +u2v2.We can use the inner product to define notionsof length and angle. The length of u is given bythe square root of the inner product of u withitself:

|u| = u,u12=

u21 +u22.

The angle between u and v can also be definedin terms of inner product:

u,v = |u||v|coswhere

= cos1(u,v|u||v|

)

.

• Orthogonality

An important special case occurs when

u,v = |u||v|cos = 0.When cos equals zero, = /2 = 90.

• Orthonormal bases for Rn

Any n orthogonal vectors which are of unit length

ui,u j ={

1 if i = j0 otherwise.

form an orthonormal basis for Rn. Any vectorin Rn can be expressed as a weighted sum of u1,u2, u3, ...,un:

v = w1u1 +w2u2 +w3u3 + ...+wnun.

Question How do we find w1, w2, w3, ...,wn? Answer Using inner product.

• Example

Consider two orthonormal bases. The first basis

is defined by the vectors u1 =[

10

]

and u2 =[

01

]

. It is easy to verify that these two vectors

form an orthonormal basis:[

10

]

,

[

01

]

= 1 0+0 1 = 0[

10

]

,

[

10

]

= 1 1+0 0 = 1[

01

]

,

[

01

]

= 0 0+1 1 = 1.

• Example (contd.)

The second, by the vectors u1 =[

cossin

]

and

u2 =[

sincos

]

. It is also easy to verify that

these two vectors form an orthonormal basis:[

cossin

]

,

[

sincos

]

= cossin+ cossin = 0

[

cossin

]

,

[

cossin

]

= cos2 + sin2 = 1

[

sincos

]

,

[

sincos

]

= cos2 + sin2 = 1.

• Example (contd.)

Let the coefficients of v in the first basis be w1and w2:

v = w1

[

10

]

+w2

[

01

]

.

What are the coefficients of v in the second ba-sis? Stated differently, what values of w1 andw2 satisfy:

v = w1

[

cossin

]

+w2

[

sincos

]

?

• Figure 1: Change of basis.

Example (contd.)

To find w1 and w2, we use inner product:

w1 =

[

cossin

]

,

[

w1w2

]

w2 =

[

sincos

]

,

[

w1w2

]

.

• Example (contd.)

The above can be written more economically inmatrix notation:

[

w1w2

]

=

[

cos sinsin cos

][

w1w2

]

w = Aw.

If the rows of A are orthonormal, then A isan orthonormal matrix. Multiplying by an or-thonormal matrix effects a change of basis. Achange of basis between two orthonormal basesis a rotation.

• Matrix transpose

If A rotates w by

A =[

cos sinsin cos

]

then A1 = AT rotates w by

AT =[

cos sinsin cos

]

.

In other words, AT undoes the action of A, i.e.,they are inverses:

AAT =[

cos2 + sin2 cossin sincoscossin sincos cos2 + sin2

]

=

[

1 00 1

]

.

For orthonormal matrices, multiplying by thetranspose undoes the change of basis.

• Complex vectors in C2

v = [a1ei1,a2ei2]T is a vector in C2.

Question Can we define length and angle inC

2 just like in R2?

Answer Yes, but we need to redefine innerproduct:

u,v = u1v1 +u2v2.

Note that this reduces to the inner product forR

2 when u and v are real. The norm of a com-plex vector is the square root of the sum of thesquares of the amplitudes. For example, forv C2:

|u| = u,u12=

u1u1 +u2u2.

• Orthonormal bases for Cn

Question How about orthonormal bases forC

n, do they exist?

Answer Yes. If ui,u j = 0 when i 6= j andui,u j = 1 when i = j, then the ui form anorthonormal basis for Cn.

Question Do complex orthonormal matricesexist?

Answer Yes, except they are called unitarymatrices and (A)T undoes the action of A.That is

A(A)T = I

where (A)T = AH is the Hermitian trans-pose of A.

• The space of 2 periodic functions

A function, f , is 2 periodic iff f (t) = f (t +2). We can think of two complex 2 periodicfunctions, e.g., f and g, as infinite dimensionalcomplex vectors. Length, angle, orthogonal-ity, and rotation (i.e., change of basis) still havemeaning. All that is required is that we gener-alize the definition of inner product:

f ,g =Z

f (t)g(t)dt.

The length (i.e., the norm) of a function is:

| f | = f , f 12 =

Z

f (t) f (t)dt.

Two functions, f and g, are orthogonal when

f ,g =Z

f (t)g(t)dt = 0.

• Scaling Property of the Impulse

The area of an impulse scales just like the areaof a pulse, i.e., contracting an impulse by a fac-tor of a changes its area by a factor of 1|a|:

Z

(at)dt =

1|a| = lim0

Z

(at)dt.

It follows that:

lim0

Z

|a|(at)dt = lim

0

Z

(t)dt = 1.

Since the impulse is defined by the above inte-gral property, we conclude that:

|a|(at) = (t).

• Shah basis

The Shah function is a train of impulses:

III(t) =

n=

(t n).

We can use the scaling property of the impulseto define a 2 periodic Shah function:

12

III( t

2

)

=1

2

n=

( t

2n

)

=22

n=

(

2( t

2n

))

=

n=

(t 2n) .

• Shah basis (contd.)

Consider the infinite set of 2 periodic Shahfunctions, 12III

(

t2

)

, for < . Because1

2III(

t2

)

= (t ) for t it followsthat

12

III

(

t 12

)

,1

2III

(

t 22

)

=Z

(t 1)(t 2)dt

equals 0 when 1 6= 2 and equalsR (t1)dt =

1 when 1 = 2. It follows that the infinite setof 2 periodic Shah functions, 12III

(

t2

)

, for < form an orthonormal basis for thespace of 2 periodic functions.

• . . . . . .

1

. . . . . .

Figure 2: Making a 2 periodic Shah function.

• Shah basis (contd.)

Question How do we find the coefficients,w(), representing f (t) in the Shah basis?How do we find w() such that

f (t) =1

2

Z

w()III

(

t 2

)

d?

Answer Take inner products of f with theinfinite set of 2 periodic Shah functions:

w() =

12

III

(

t 2

)

, f (t)

.

• Shah basis (contd.)

Because 12III(

t2

)

= (t ) for t itfollows that

w() =1

2

Z

f (t)III

(

t 2

)

dt

=Z

f (t)(t )dt

which by the sifting property of the impulse isjust:

w() = f ().We see that the coefficients of f in the Shahbasis are just f itself!

• Harmonic signal basis

Question How long is a harmonic signal? Answer The length of a harmonic signal is

|e j t| = e j t,e j t12

=

(

Z

e j te j tdt

)12

=

(

Z

dt

)12

=

2.

• Harmonic signal basis (contd.)

Question What is the angle between two har-monic signals with integer frequencies?

Answer The angle between two harmonicsignals with integer frequencies is

e j1t,e j2t =Z

e j1te j2tdt

=

[

e j(21)t

j(21)

]

.

Since this function is the same at and (forall integers 1 and 2), we conclude that

e j1t,e j2t = 0when 1 and 2 are integers and 1 6= 2.

• Fourier Series of 2 Periodic Functions

It follows that the infinite set of harmonic sig-nals, 1

2ej t for integer and

form an orthonormal basis for the space of 2periodic functions.

Question What are the coefficients of f inthe harmonic signal basis?

Answer Take inner products of f with theinfinite set of harmonic signals.

This is the analysis formula for Fourier series:

F() =

12

e j t, f

=12

Z

f (t)e j tdt

for integer frequency, .

• Fourier Series of 2 Periodic Functions (contd.)

The function can be reconstructed using the syn-thesis formula for Fourier series:

f (t) =12

=

F()e j t.

• Fourier Series Example

The Fourier series for the Shah basis function

f (t) =1

2III

( t2

)

is

F() =

12

e j t,1

2III

( t2

)

=12

Z

(t)e j tdt

=12

.

Consequently

f (t) =12

=

F()e j t

=1

2

=

e j t.

• Deep Thought

The analysis formula for Fourier series effects achange of basis. It is a rotation in the space of2 periodic functions. The synthesis formulaundoes the change of basis. It is the oppositerotation.

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