Finite-Length Discrete Transforms

7/30/2019 Finite-Length Discrete Transforms

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Finite-Length Discrete

Transforms

1


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Orthogonal Transforms

Let x[n] denote a length-N time-

domain sequence with

denoting the coefficient of its N-

point orthogonal transform. The

general form of the orthogonal

transform pair is of the form

Referred to as the analysis and

synthesis equation respectively.

2

The above condition are said to be

orthogonal to each other

The verification of this equation in

book page 200

x [ k]= n=0

N 1

x [ n ] [ k, n ], 0 k N 1

x [ n ]=1

N n=0

N 1

x [n ] [ k, n ], 0 k N 1

n=0

N 1

[ k, n ] [ l,n] = {1, l=k0 l k}[k]


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The inverse discrete Fouriertransform (IDFT) is given by

As can be seen from the above

expression, the inverse DFT x[n] can

be a complex sequence even when

the DFT X[k] is real sequence.

4

The Discrete Transform

x [ n ]=1

Nk=0

N 1

X[ k] WN kn, 0 n N 1

X[k]

Example - Consider the length-Nsequence defined for 0 n N-1

Where is r is an integer in the range0 n N-1

Using a trigonometric identity, wecan write as

x [n ]= cos(2 rn N), 0 n N 1

x [ n ]= 12

(e j2rn /N+e j2rn /N)

1

2(WN

rn+WN

rn )


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5


The N-point DFT ofg[n] is thus given by

Making use of the identity

X[ k ]= n=0

N 1

g[n ] WNkn

1

2 (n=0N 1

WN ( r k)n+

n=0

N 1

WN( r+k)n)

0 k N 1

n=0

N 1

WN (k )n

={N,fork =rN,ran int eger

0,otherwise }G[ k]={

N/2, fork=r

N/2, fork=N r

0otherwis }0 k N 1

we

get

Matrix Relations

The DFT samples defined by

can be expressed in matrix form as

WhereX is the vector composed of

the N DFT samples and x is thevector ofNinput samples

X[ k]= n=0

N 1

x[ n]WN

k n, 0 k N 1

X=DNx

X=[ X[0 ] X[1 ] X[ N 1 ] ]t

x=[ x [0 ]x [1 ]x [ N 1 ] ]t


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6


and DNis the NNDFT matrix given

by

Likewise, the IDFT relation given by

DN=[1 1 1 1

1 WN1 W

N2 W

N(N 1)

1 WN

2 WN

4 WN

2( N 1)

1 WN

(N 1 ) WN

2(N 1) WN

(N 1)2 ]

can be expressed in matrix form as

where is the NNDFT matrix

Where

Note

DN

1

x=DN 1

X

DN 1

=

[1 1 1 1

1 WN

1 WN

2 WN

(N 1 )

1 WN

2 WN

4 WN

2(N 1)

1 WN

(N 1) WN

2(N 1) WN

(N 1)2

]D

N

1=1

N

DN

x [ n ]=1

Nk=0

N 1

X[K]WN kn, 0 n N 1


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The FT of length-Nsequence x[n]

limit the extent of the summation to

N points and evaluate the

continuous function of frequency at

Nequally spaced points

7

Relation with DTFT

X(ej)= n=

x [ n ]e jn

= n=0

N 1

x [ n ]e jn

X(ej)=

2

Nk

=X(k) =Xk

= n=0

N 1

x [ n ]e j2kn/N

Numerical Computation of the

DTFT using the DFT

A practical approach to the

numerical computation of the DTFT

of a finite-length sequence.

LetX(ej) be the DTFT of a length-N

sequencex[n]

We wish to evaluate X(ej) at a

dense grid of frequencies k=2k/M,0kM-1,where M>> N:

Relation Between the DTFT and the DFT and Their Inverses

ejn


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Define a new sequence

Then

X(ej

k)= n=0

N 1

x [ n ]e j

kn=

n=0

N 1

x [ n ]e j2kn/M

X(ej

k)= n=0

M 1

xe[ n ]e j2kn/M

xe[n ]= {x [n ],0 n N 10, N n M 1 }

Thus is essentially an M-point DFT Xe[k] of the length-Msequencexe[n]

The DFT Xe[k] can be computedvery efficiently using the FFTalgorithm ifMis an integer power of2

The function freqz employs this

approach to evaluate the frequencyresponse at a prescribed set offrequencies of a DTFT expressed asa rational function in e-j

X(ej

k)



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DTFT from DFT by Interpolation

The N-point DFT X[k] of a length-N

sequence x[n] is simply the

frequency samples of its DTFT

X(ej) evaluated at N uniformly

spaced frequency points

Given the N-point DFT X[k] of alength-N sequence x[n], its DTFT

X(ej) can be uniquely determined

fromX[k]

=k= 2k/N , 0 k N 1

Thus

X(ej

)= n= 0

N 1

x [n ]e jn

n= 0

N 1

[1

Nk=0

N 1

X[ k]WN kn ]e jn

1N k= 0

N 1

X[ k] n= 0

N 1

e j ( 2k/N)n

S



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To develop a compact expressionfor the sum S, let

Then

From the above

Or, equivalently,

r=e j( 2k/N)

S= rn

rS= rn= 1+ rn+rN 1

r

n

+r

N

1=S+rN

1

Hence

Therefore

S=1 rN

1 r=1 e j (N 2k)

1 e j [ (2k/N) ]

sin

(

N 2k

2 )sin(N 2k2N )

e j [ 2k/N)][(N 1)/2]

S rS=(1 r) S=1 rN

5.3 Relation Between the DTFT and the DFT and Their Inverses

X(ej)

1

Nk= 0

N 1

X[ k]

sin(N 2k2 )sin

(

N 2k

2N

)

e j [ 2k/N)][(N 1)/2 ]


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Sampling the Fourier Transform

Consider a sequencex[n] with a DTFT

X(ej)

We sampleX(ej) at Nequally spaced

points k=2k/N, 0kN-1 developingthe Nfrequency samples

These N frequency samples can be

considered as an N-point DFT Y[k]

whose N- point IDFT is a length-N

sequence y[n]

now

{X(ejk)}


X(ej

)=

x [ ]e j

Thus

An IDFT ofY[k] yields

Y[ k]=X(ejk)=X(e

j2k/N)

x [ ]e j2k/N=

x [ ]WNk

y [ n ]=1

Nk0

N 1

Y[k]WN kn

y [ n ]=1

Nk0

N 1

=

x[ ]WNkW

N kn

=

x [ ][1N k= 0N 1

WN k( n )]

1

N

n0

N 1

WN k( n r)= (

1, forr=n+mN

0, otherwise


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we arrive at the desired relation

Thus y[n] is obtained from x[n] by

adding an infinite number of shifted

replicas of x[n], with each replicashifted by an integer multiple of N

sampling instants, and observing the

sum only for the interval 0nN-1

y [n ]= m=

x [ n+mN],0 n N 1

To apply

To finite-length sequences, we

assume that the samples outside the

specified range are zeros

y [ n ]= m=

x [ n+mN], 0 n N 1


Thus ifx[n] is a length-M sequence

with MN, then y[n]= x[n] , for

0nN-1

Example 5.6

Let {x[n]}={0 1 2 3 4 5}

By sampling its DTFT X(ej) at

k=2k/4, 0k3 and then applying

a 4-point IDFT to these samples, we

arrive at the sequence y[n] given by

y[n]=x[n]+x[n+4]+x[n-4] 0n3

i.e. {y[n]}={4 6 2 3}

{x[n]} cannot be recovered from {y[n]}


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Circular Convolution

This operation is analogous to linear

convolution, but with a subtle

difference

Consider two length-N sequences,

g[n] and h[n], respectively

Their linear convolution results in a

length-(2N-1) sequence yL[n] given

by

yL

[n ]= m=0

N 1

g[ m ] h[ n m], 0 n 2N 2

In computing yL[n] we have assumedthat both length-N sequence havebeen zero-padded to extend theirlengths to 2N-1

The longer form ofyL[n] results fromthe time-reversal of the sequenceh[n] and its linear shift to the right

The first nonzero value ofyL[n] is

yL[0]= g[0]h[0] ,and the last nonzerovalue is

yL[2N-2]= g[N-1]h[N-1]


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Definition:

To develop a convolution-like

operation resulting in a length-N

sequence yC[n] , we need to define a

circular time-reversal, and then apply

a circular time-shift

Resulting operation, called a

circular convolution, is defined by

yC

[ n ]= m=0

N 1g[ m ] h [n m

N],

0 n N 1

Since the operation defined involves

two length-N sequences, it is often

referred to as an N-point circular

convolution, denoted as

The circular convolution is

commutative, i.e.

y[n] =g[n] h[n]

g[n] h[n] = h[n] g[n]



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The N-point circular convolution canbe written in matrix form as

Note: The elements of each diagonalof the NNmatrix are equal

Such a matrix is called a circulantmatrix

[yC[0 ]

yC[1 ]

yC[2 ]

yC[N 1 ]]=

[h[ 0 ] h [N 1 ] h[ N 2 ] h [1 ]

h[ 1 ] h[ 0 ] h [N 1] h[ 2 ]

h[ 2 ] h[ 1] h[ 0 ] h [3 ]

h [N 1 ] h[ N 2] h [N 3] h[ 0 ]][g[ 0 ]

g[1 ]

g[ 2 ]

g[N 1 ]]

5.4 Circular Convolution

Tabular Method

Consider the evaluation of y[n]= h[n]

g[n]

where {g[n]} and {h[n]} are length-4

sequences

First, the samples of the two

sequences are multiplied using theconventional multiplication method as

shown on the next slide


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The partial products generated in the 2nd, 3rd, and 4th rows are circularly shiftedto the left as indicated above



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The modified table after circular

shifting is shown below

The samples of the sequence {yc[n]}

are obtained by adding the 4 partial

products in the column above of eachsample

Thus

yc[0]=g[0]h[0]+ g[3]h[1]+ g[2]h[2]+ g[1]h[3]

yc[1]=g[1]h[0]+ g[0]h[1]+ g[3]h[2]+ g[2]h[3]

yc[2]=g[2]h[0]+ g[1]h[1]+ g[0]h[2]+ g[3]h[3]

yc[3]=g[3]h[0]+ g[2]h[1]+ g[1]h[2]+ g[0]h[3]

The definition of circular conjugate-symmetric sequence and circularconjugate-antisymmetric sequence.

Circular conjugate-symmetry

Circular conjugate-antisymmetry


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Classifications of Finite-Length Sequences

Classifications Based on Conjugate

Symmetry Any complex length-N sequence x[n]can be expressed as:

Where, Xcs[n] is its circularconjugate-symmetric part and Xca[n]is its circular conjugate-anti-symmetric part, defined by:

For a real sequence x[n], it can beexpressed as:

Where, Xev[n] is its circular evenpart and Xca[n] is its circular oddpart, defined by:


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Classifications Based on Geometric

Symmetry Two types of geometric symmetries

are usually defined:For a real

(1) Symmetric sequence

(2) Antisymmetric sequence

Since the length N of a sequencecan be either even or odd, four typesof geometric symmetry are defined:

Type1: Symmetric impulse responsewith odd length.

Type2: Symmetric impulse responsewith even length

Type1: Anti-symmetric impulseresponse with odd length.

Type1: Anti-symmetric impulseresponse with even length.


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Type 1 symmetric sequence, withN=9,

is

x[n]=x[0]+x[1]+x[2]+x[3]+x[4]+x[5]+

x[6]+x[7]+x[8]

The Fourier transform is

Type 1 Symmetry with Odd Length

Taking e-j4 as a common factor ineach group of.

Factoring out e-j4 in the right hand

side


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Notice that the quantity inside the

braces, { }, is a real function of andcan assume positive or negative values

in the range 0

The of the sequence is given by

() = 4 + where is either 0 or

, and hence the phase is a linear

function of

In general, for Type 1 linear-phase

sequence of length-N

Type 2 Symmetry with Even Length

Similarly, the Fourier transform ofType 2 symmetric sequence, with N=8,

can be written.

where the phase is given by

In general, for Type 2 linear-phase

sequence of length-N


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The Fourier transform of Type 3

antisymmetric sequence, with N=9, is

(notice that x[4]=0)

Now, x[0]=-x[8], x[1]=-x[7], x[2]=-x[6],

x[3]=-x[5] and x[4]=0

Type 3 Antisymmetry with Odd Length

Multiplying by j=ej/2and 2, we obtain

which results in

The phase is now

The antisymmetry introduces a phase

shift of/2


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In general, the Fourier transform of

Type 3 linear phase antisymmetric

sequence of odd length-N is

Similarly, the Fourier transform of

Type 4 linear phase antisymmetric

sequence of even length-N is

In both cases, j=ej/2 introduces a

phase shift of/2

Type 4 Antisymmetry with Even Length

Multiplying by j=ej/2and 2, we obtain

which results in

The phase is now

The antisymmetry introduces a phase

shift of/2


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DFT Symmetry Relations

In general, the DFT X[k] of a finite

sequence x[n] is a sequence of complex

numbers and can be expressed as

Real and imaginary parts of the DFT

sequence can be found as:

Assuming that the original time-domain

signal is complex

its DFT can be found as:

Therefore, real and imaginary parts ofthe DFT sequence are:

X[ k]= Xre

[ k]+ j Xim [ k]


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DFT Symmetry Relations

Symmetry Properties of the DFT of a

real sequence

Symmetry Properties of the DFT of a

complex sequence


26/44

Discrete Fourier Transform Theorems

The DFT satisfies a number of

properties that are useful in signalprocessing.

All time domain sequences are

assumed to be of length-N with N-point

DFT.

Linearity Theorem

Consider a sequence x[n] obtained by a

linear combination of g[n] and h[n]

Circular Time Shifting Theorem

The DFT of the circularly time shifting

sequence x[n] is given by

Circular Frequency Shifting Theorem

The inverse DFT of the circularly

frequency shifting DFT is given by

Duality Theorem

If the N-point DFT of the length-N

sequence g[n] is G[k], then

Circular Convolution Theorem

The N-point DFT Y[k] of the length N-sequence is given by


27/44

Computation of the DFT of Real Sequence

Modulation Theorem

The N-point DFT Y[k] of the length-Nproduct sequence is given by a

circulation of theirDFTs

Parsevals TheoremThe total energy of a length-N sequence

g[n] can be computed by summing the

square of the absolute values of the

DFT.

N-point DFTs of two Real Sequences

using a single N-point DFT

Let g[n] and h[n] be two length-N realsequences with G[k] and H[k] denoting

their respective N-point DFTs

These two N-point DFTs can be

computed efficiently using a single N-

point DFT

Define a complex length-N sequence

The inverse DFT of the circularly

frequency shifting DFT is given by

LetX[k] denote the N-point DFT of x[n]

Note that for 0 k N1,


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Computation of the DFT of Real Sequence

Thus

Linear Convolution using the DFT

Since a DFT can be efficiently

implemented using FFT algorithms, it is

of interest to develop methods for the

implementation of linear convolution

using the DFT

Let g[n] and h[n] be two finite-length

sequences of length N

and M, respectively

Define two length-L (L = N + M 1)sequences

Thus

2N-point DFT of a Real Sequence

using a single N-point DFTLet v[n] be a length-N real sequence

with a 2N-point DFT V[k]

Define two length-N real sequences

g[n] and h[n] as follows:

Let G[k] and H[k] denote their espective

N-point DFTs

Define a length-N complex sequence

with an N-point DFT X[k]


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Linear Convolution of a Finite-Length

Sequence with an infinite Length

Sequence

Overlap-Add Method

We first segment x[n], assumed to be a

causal sequence here without any loss

of generality, into a set of contiguous

finite-length subsequences of length N

each:

Where

Thus

where

The corresponding implementation

scheme is illustrated below

We next consider the DFT-basedimplementation of

where h[n] is a finite-length sequence oflength M and x[n] is an infinite length (or

a finite length sequence of length much

greater than M)


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The desired linear convolution y[n] =

h[n] x[n] is broken up into a sum of

infinite number of short-length linear

convolutions of length N + M 1 each:

ym[n] = h[n] xm[n]

Consider implementing the following

convolutions using the DFT-basedmethod, where now the DFTs (and the

IDFT) are computed on the basis of (N

+ M 1) points

In general, there will be an overlap ofM

1 samples between the samples of the

short convolutions h[n] xr-1[n]and h[n]

xm[n] for (r 1)N n rN + M 2


31/44


Therefore, y[n] obtained by a linear

convolution of x[n] and h[n] is given by

The above procedure is called the

overlap-add method

since the results of the short linear

convolutions overlap and the

overlapped portions are added to get

the correct final result

Overlap-Save Method

In implementing the overlap-add

method using the DFT, we need to

compute two (N + M1)-point DFTs and

one (N + M 1)-point IDFT for each

short linear convolution

It is possible to implement the overalllinear convolution by performing instead

circular convolution of length shorter

than (N + M 1)

To this end, it is necessary to

segment x[n] intooverlapping blocksxm[n] , keep the terms o f the circular

convolution of h[n] with that

corresponds to the terms obtained by a

linear convolution ofh[n] and xm[n], and

throw away the other parts of the

circular convolution


32/44


To understand the correspondence

between the linear and circular

convolutions, consider a length-4

sequencex[n] anda length-3 sequence

h[n]

Let yL[n] denote the result of a linear

convolution of x[n] with h[n]

The six samples ofyL[n] are given by

If we append h[n] with a single zero-

valued sample andconvert it into a

length-4 sequence he[n], the 4-point

circularconvolution yC[n] of he[n] and

x[n] is given by

Next form


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Or equivalently

Computing the above for m = 0, 1, 2, 3,

. . . , and substituting the values of xm[n]we arrive at

Then, we reject the first M 1 samples

of wm[n] and abutthe remaining M M

+ 1 samples of wm[n] to form yL[n], the

linear convolution ofh[n] and x[n]

Ifym[n] denotes the saved portion ofwm[n], i.e.,


34/44


The approach is called overlap-save

method since the input is segmented

into overlapping sections and parts of

the results of the circular convolutions

are saved and abutted to determine the

linear convolution result


35/44

Discrete Cosine Transform

In general, the N-point DFT X[k] oflength-N real sequence is a complex

sequence satisfying the symmetry

condition

For N even, the DFT samples X[0]

and X[N-2]/2 are real and distinct. The

remaining N-2 DFT samples are

complex and only half of these samples

are distinct

For N odd, the DFT samples x[0] is

real and the remaining N-1 DFT

samples are complex, of which only half

of these samples are distinct

The DFT of a real symmetric and anti-symmetric finite sequence is a product

of a linear phase term and a real

amplitude function.

Orthogonal transform is based onconverting an arbitrary sequence into

either a symmetric or an anti-symmetric

sequence and then extracting the real

orthogonal transform coefficients from

the DFT, the transform develop via this

approach is called discrete cosine

transform (DCT)

kX=X[k]*


36/44


To develop the expression for theDCT for symmetric periodic sequence,

consider type 1 DCT.

then the extracted periodic is given by

To develop the Type-2 DCT, extracty[n] of the symmetric periodic sequence

let x[n] be a length-N sequencedefined for 0nN-1. First, x[n] is

extended to a length-2N sequence by

zero padding.

Next, type-2 symmetric sequence y[n]

of length 2N is formed from xe[n]

according to

{}{ cba=x[n]

{}{ cdcba=y[n]

{}{ bcddcba=y[n]

2,010],[][

nN

Nnnxnxe

][nxe

120

]12[][][

Nn

nNxnxn ee

2],12[

0],[

nNnNx

nnx


37/44


The generated sequence y[n]satisfies the symmetry property.

the 2N-point DFT Y[k] of the length-2N sequence y[n] is thus

rewrite

By making a change of variables, the

DFT Y[k] can be expressed as

The Type-2 N point DCT,

The samples of XDCT[k] are real for

real sequence x[n]

]12[][ nNyn

20,][][12

0

2 nWnykYN

n

kn

N

120

,]12[,][

,][][][

12

0

2

1

0

2

1

0

12

0

22

Nk

WnNxWnx

WnyWnykY

N

n

kn

N

N

n

kn

N

N

n

N

n

kn

N

kn

N

120

,2

)12(cos][2

][

,][][][

1

0

2/

2

1

0

2/

22

2/

222

1

0

1

0

)12(

222

Nk

N

nknxW

WWWWnxW

WWnxWnxkY

N

n

k

N

N

n

k

N

kn

N

k

N

kn

N

kn

N

N

n

N

n

Nk

N

kn

N

kn

N

XDCT

[ k]= n= 0

N 1

2x[n ] cos(k(2n + 1)2N ),0 k 2N 1


38/44


The inverse discrete cosine transform

(IDCT) of an N-point DCT XDCT[k] is

given by

where

a DCT pair may often be denoted as

it can be shown that

In other words, the basis sequence

are orthogonal to each other.

To verify that x[n] is indeed the IDCT

of XDCT[k]

The IDCT of an N-point DCT XDCT[k]

can also be computed using the DFT

120

2

)12(cos][][

1][

1

0

Nk

N

nkkXk

Nnx

N

n

DCT

1102/1][k

kk

x [ k] DCT

XDCT[ k]

,,0

,0,2/1

,0,1

2

)12(cos

2

)12(cos

1 1

0

mk

mk

mk

N

nm

N

nk

N

N

n

Nnk2

)12(cos

10

2

)12(cos

2

)12(cos

1][][2

2

)12(cos

2

)12(cos][][

2][

1

0

1

0

1

0

1

0

Nk

N

nk

N

nl

NlXl

N

nk

N

nllXl

NkX

N

l

N

l

DCT

N

n

N

l

DCTDCT

21],2[

,,0

10][

][

2/

2

2/

2

kNkNXW

Nk

NkkXW

kY

DCT

k

N

DCT

k

N


39/44


The 2N-point IDFT y[n] of Y[k] is

given by

We get

The length-N IDCT x[n] of the N-point

DCT

The DCT satisfies a number ofproperties that are useful in certain

application. Assume all time domain

sequences to be length N with an N-

point DCT.

The DCT XDCT[k] of a sequence

obtained by a linear combination of two

sequences, g[n] and h[n]. Where and arbitrary constants.

20,][2

1][

12

0

2 nWkyNnN

k

kn

N

12

1

(

2

1

0

)2

1(

2 ]2[

2

1][

2

1][

N

Nk

DC

N

k

kn

NDCT kX

N

WkX

N

n

1

1

2)(2

1(

2

1

0

)2

1(

2 ][2

1][

2

1 N

k

n

NDC

N

k

kn

NDCTkX

NWkX

N

1

1

(

2

12

0

)2

1(

2 ][2

1][

2

1 N

k

DC

N

k

kn

NDCTkX

NWkX

N

120

2

)12(cos][

1

2

]0[ 1

1

Nn

nkkX

NN

XN

k

DCT

DCT

10][][ Nnnynx

DCT Properties

][][

][][

kHnh

kGng

DCT

DCT

DCT

DCT

Linearity Property

[][][][ kGnhng DDCTDCT


40/44

The Haar Transform

The DCT of the conjugate sequenceis given by

The energy preservation property of

the DCT is similar to the persevals

relation for the DFT.

Symmetry Properties

Energy Preservation Property

][][ ** kGng DCTDCT

1

0

1

0

2|][|][

2

1|][|

N

k

D

N

n

GkN

ng

The discrete-time Haar transform isderived by sampling the continouse-

time Haar function.

The set of Haar function hl(t) contains

N numbers, with N a power-of-2 positive

integer: that is N=2v+1, where v 0.

In defining the Haar function, the

integer subscript l is uniquely

represented as a function of two non

negative integer variable, r and s.

Where variables r and s have ranges

0 r v; 0 s 2r

The Haar Transform

12 sl r


41/44

The Haar Transform

The NN discrete-time Haar

transform matrix HN is obtained by

discretizing the Haar functions at

discrete values of t, given by t = n/N,

0 n N-1

Definition

To derive the high order Haartransform

Where denotes the kroneckerproduct and IK is the KK identity.

The N-point transform XHaar of length

N sequence x[n] is given by

where l= 2r+s 1

10,1)()( 0,00 tthth

00

22

5.0

2

,2

5.0

2

1,2

)()(

2/

2/

,

tforotherwise

s

t

s

st

s

thth rrr

rr

r

srl

.1,112

11

2

2/

2

12v

I

HH

v

v

vv

,.xHhNHaar

HaHaarHaarHaar XXXh []...1[]0[

Nxxxx ]1[]...1[]0[


42/44

The Haar Transform

The inverse Haar transform is giveby

2 2 normalized Haar ransform

matrix is given by

Higher order normalized Haar

transform is given by

Haar Transform Properties

The Haar transform matrix is

orthogonal and hence

Haar transform expression reduced to

If denote the (k, l)-th element of HN ashN(k,l) then

t

NN HNH

11

10

[),(1

][1

0

Nn

XlkhN

nxN

k

HaN

HaarNXHx1

2

1

2

1

2

1

2

1

x

.0,

2

1

2

1

2

1

2

1

2

,2

21v

I

H

H

v

v

v

n

Orthogonality Property

HaartNXH

Nx1


43/44

The Haar Transform

The expression is similar to theParsevals relation for the DFT also

exists for the Haar transform

If L samples of the transform with

indices in the range R are set to zero

with L < < N and if x(m) [n] denotes the

inverse of the modified transform, then

a measure of the energy compaction

property

Energy conservation property Consider the energy compaction

proper of the DFT, the DCT and theHaar transform.

N-point DFT, the high frequency

indices around

The Discrete Fourier Transform

12

1],[

2

1

2

1

,0

2

10],[

][)(

NkLN

kX

N

k

LN

LNkkX

kX

DFT

DFT

m

DFT

1

0

1

0

2|][|

1|][|

N

k

Ha

N

n

kHN

nx

Energy Compaction Properties

1

0

)(|][][|

1)(N

k

mxnx

NL


44/44

The Energy Compaction Properties

The original time domain sequence

x[n] is obtained by computing the IDFT

The corresponding approximation

error is given by

The high frequency samples havehigh indices and thus the modified DCT

obtained

The original time domain sequence

x[n] is obtained by computing the IDCT

Hence the approximation error is

The modified Haar transform obtained

The x[n] is obtained by computing theIDCT

1

0

)()(][

1][

N

k

m

DFT

m

DFT kXNn

1

0

)(|][][|

1)(N

k

m

DXnx

NL

The Discrete Cosine Transform

10][][)(

NkkXkDCTm

DCT

LN

k

mDCT

mDCT kkXkNn

1

0

)()(

22(co][][1][

1

0

)( |][][|1)(N

k

m

DCxnxNL

The Haar Transform

0

10][][)(

kLN

NkkXkHaarm

Haar

LN

k

H

m

Haar XnkhNnx

1

0

)( [],[1][

1)( |][][|

1)(N

m

HaxnxL

Documents

Finite-Length Discrete Transforms