The Fourier Transform • Introduction • Orthonormal bases for R

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Transcript of The Fourier Transform • Introduction • 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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