chapter 5chapter 5 finite-length discrete transformscwlin/courses/dsp/notes/ch5_mitra_dsp_c.pdf ·...

Chapter 5Chapter 5

Finite-Length Discrete TransformsTransforms

清大電機系林嘉文

[email protected] 5731152

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-1

03-5731152

Digital Fourier TransformDigital Fourier Transform• Definition - The simplest relation between a length-N e t o e s p est e at o bet ee a e gt

sequence x[n], defined for 0 ≤ n ≤ N −1, and its DTFT X(ejω) is obtained by uniformly sampling X(ejω) on the ω-axis between 0 ≤ ω < 2π at ωk = 2πk/ N, 0 ≤ k ≤ N −1

• From the definition of the DTFT we thus have

0 ≤ k ≤ N −1

• Note: X[k] is also a length-N sequence in the frequency domain

• The sequence X[k] is called the Discrete Fourier Transform (DFT) of the sequence x[n]


Digital Fourier TransformDigital Fourier Transform• Using the notation WN = e−j2π/N the DFT is usually Us g t e otat o N e t e s usua y

expressed as:

• The Inverse Discrete Fourier Transform (IDFT) is ( )given by

• To verify the above expression we multiply both sides of the above equation by and sum the result from n = 0 to n = N −1


Digital Fourier TransformDigital Fourier Transform• This results in:s esu ts

• Making use of the identity

we observe the RHS of the last equation is equal to X[l]• Hence


Digital Fourier TransformDigital Fourier Transform• Example - Consider the length-N sequencea p e Co s de t e e gt seque ce

• Its N-point DFT is given by

0 ≤ k ≤ N 1

• Example - Consider the length-N sequence

, 0 ≤ k ≤ N −1

• Example - Consider the length-N sequence

• Its N-point DFT is given by

0 ≤ k ≤ N −1


, 0 ≤ k ≤ N 1

Digital Fourier TransformDigital Fourier Transform• Example - Consider the length-N sequence defined for

0 ≤ n ≤ N −1g[n] = cos(2πrn/ N), 0 ≤ r ≤ N −1

• Using a trigonometric identity we can write

• The N-point DFT of g[n] is thus given by

0 ≤ k ≤ N −1


0 ≤ k ≤ N 1


• We get , 0 ≤ k ≤ N −1

Matrix RelationMatrix Relation• The DFT samples defined by

can be expressed in matrix form ascan be expressed in matrix form asX = DNx

whereX = [X[0] X[1] ..... X[N −1]]Tx = [x[0] x[1] ..... x[N −1]]T

and DN is the N × N DFT matrix given by


Matrix RelationMatrix Relation• Likewise the IDFT relation given by• Likewise, the IDFT relation given by

can be expressed in matrix form as

where is the IDFT matrixwhere is the IDFT matrix

• Note:


DFT Computation Using MATLABDFT Computation Using MATLAB• The functions to compute the DFT and the IDFT are fft and

ifft• These functions make use of FFT algorithms which are

computationally highly efficient compared to the direct computation

• The DFT and DTFT of the following function is shown below• The DFT and DTFT of the following function is shown below

x[n] = cos(6πn/16), 0 ≤ n ≤ 15


DTFT from DFT by InterpolationDTFT from DFT by Interpolation• The DFT X[k] of a length-N sequence x[n] is simply the freq.

samples of its DTFT X(ejω) evaluated at N uniformly spaced frequency points ω = ωk = 2πk/N, 0 ≤ k ≤ N −1

• Compared to the direct computation, these functions are computationally highly efficient due to the use of FFTGi th N i t DFT X[k] f l th N [ ] it• Given the N-point DFT X[k] of a length-N sequence x[n], its DTFT X(ejω) can be uniquely determined from X[k]


DTFT from DFT by InterpolationDTFT from DFT by Interpolation• To develop a compact expression for the sum S, let

r = e−j(ω−2πk/N)

• Then

• Or, equivalentlyS− rS = (1− r)S =1− rNS rS = (1 r)S =1 r

• Hence


DTFT from DFT by InterpolationDTFT from DFT by Interpolation• Therefore


Sampling the DTFTSampling the DTFT• Consider a sequence x[n] with a DTFT X(ejω)• We sample X(ejω) at N equally spaced points ωk = 2πk/N, 0 ≤

k ≤ N −1 developing the N frequency 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

• Taking IDFT of Y[k] yields


Sampling the DTFTSampling the DTFT• That is


we arrive at the desired relationwe arrive at the desired relation


Sampling the DTFTSampling the DTFT• Thus y[n] is obtained from x[n] by adding an infinite number

of shifted replicas of x[n], with each replica shifted by an integer multiple of N sampling instants, and observing the

l f th i t lsum only for the interval• To apply

t fi it l th th t th lto finite-length sequences, we assume that the samples outside the specified range are zerosThus if x[n] is a length M sequence with M ≤ N then y[n] =• Thus if x[n] is a length-M sequence with M ≤ N then y[n] = x[n] for 0 ≤ n ≤ N −1


Sampling the DTFTSampling the DTFT• Example – Let {x[n]} = {0 1 2 3 4 5}• By sampling its DTFT X(ejω) at ωk = 2πk/4, 0 ≤ k ≤ 3 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] 0 ≤ n ≤ 3i.e.,

{y[n]} = {4 6 2 3}⇒ {x[n]} cannot be recovered from {y[n]}


Numerical Computation of the DTFT Using DFT

• A practical approach to the numerical computation of the• A practical approach to the numerical computation of the DTFT of a finite-length sequence

• Let X(ejω) be the DTFT of a length-N sequence x[n]Let X(e ) be the DTFT of a length N sequence x[n]• We wish to evaluate X(ejω) at a dense grid of frequencies ωk = 2πk/M, 0 ≤ k ≤ M −1, where M >> N:k , ,

• Define a new sequence

• Then


Numerical Computation of the DTFT Using DFT

• Thus X(ejω) is essentially an M-point DFT X [k] of the• Thus X(ej ) is essentially an M-point DFT Xe[k] of the length-M sequence xe[n]

• The DFT Xe[k] can be computed very efficiently using theThe DFT Xe[k] can be computed very efficiently using the FFT algorithm if M is an integer power of 2

• The function freqz employs this approach to evaluate the q p y ppfrequency response at a prescribed set of frequencies of a DTFT expressed as a rational function in e−jω


DFT PropertiesDFT Properties• Like the DTFT, the DFT also satisfies a number of

properties that are useful in signal processing applications • Some of these properties are essentially identical to those

of the DTFT, while some others are somewhat different

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-19x[n] is a complex sequence

Symmetry Relations of DFTSymmetry Relations of DFT

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-20x[n] is a real sequence

General Properties of DFTGeneral Properties of DFT


Circular Shift of a SequenceCircular Shift of a Sequence• This property is analogous to the time-shifting property of

the DTFT, but with a subtle difference• Consider length-N sequences defined for 0 ≤ n ≤ N −1.

Sample values of such sequences are equal to zero for values of n < 0 and n ≥ NIf [ ] i h th f bit i t• If x[n] is such a sequence, then for any arbitrary integer , the shifted sequence

x [n] = x[n n ]x1[n] = x[n − no]is no longer defined for the range 0 ≤ n ≤ N −1We thus need to define another type of a shift that will• We thus need to define another type of a shift that will always keep the shifted sequence in the range 0 ≤ n ≤ N −1


Circular Shift of a SequenceCircular Shift of a Sequence• The desired shift, called the circular shift, is defined using

a modulo operation:xc[n] = x[ n − no N]

• For no > 0 (right circular shift), the above equation implies

x[ n − 1 ] x[ n − 4 ]


x[ n 1 6]= x[ n + 5 6]

x[ n 4 6]= x[ n + 2 6]

Circular ConvolutionCircular Convolution• This operation is analogous to linear convolution, but with a

subtle difference • Consider two length-N sequences, g[n] and h[n]. Their linear

convolution results in a length-(2N−1) sequence yL[n]:

• In computing yL[n] we assume that both length-N sequences h b dd d t t d th i l th t 2N 1have been zero-padded to extend their lengths to 2N−1

• The longer form of yL[n] results from the time-reversal of the sequence h[n] and its linear shift to the rightsequence h[n] and its linear shift to the right

• The first nonzero value of yL[n] is yL[0] = g[0]h[0] , and the last nonzero value is y [2N−2] = g[N−1]h[N−1]


last nonzero value is yL[2N−2] = g[N−1]h[N−1]

Circular ConvolutionCircular Convolution• 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 R lti ti ll d i l l ti i• Resulting operation, called a circular convolution, is defined by:

• Since the operation defined involves two length-N p gsequences, it is often referred to as an N-point circular convolution, denoted as

• The circular convolution is commutative, i.e.


Circular ConvolutionCircular Convolution• Example - Determine the 4-point circular convolution of the

t l th 4two length-4 sequences {g[n]} = {1 2 0 1}, {h[n]} = {2 2 1 1}

• The result is a length-4 sequence yC[n] given by

• From above, we observe


Circular ConvolutionCircular Convolution• Likewise

yC[1] = g[0]h[1] + g[1]h[0] + g[2]h[3] + g[3]h[2] = 7

yC[2] = g[0]h[2] + g[1]h[1] + g[2]h[0] + g[3]h[3] = 6yC[2] g[0]h[2] g[1]h[1] g[2]h[0] g[3]h[3] 6

yC[3] = g[0]h[3] + g[1]h[2] + g[2]h[1] + g[3]h[0] = 5


Circular ConvolutionCircular Convolution• Example - Consider the two length-4 sequences repeated

b l f ibelow for convenience:

• The 4-point DFT G[k] of g[n] is given byG[k] = g[0] + g[1] e−j2πk/4 + g[2] e−j4πk/4 + g[3] e−j6πk/4

= 1 + 2e−jπk/2 + e−j3πk/2 , 0 ≤ k ≤ 3• Therefore G[0] = 4, G[1] = 1− j, G[2] = −2, G[3] = 1+ j• Likely

H[k] = 2 + 2e−jπk/2 + e−jπk + e−j3πk/2 , 0 ≤ k ≤ 3


Circular ConvolutionCircular Convolution• The two 4-point DFTs can also be computed using the

t i l ti i limatrix relation given earlier

• If YC[k] denotes the 4-point DFT of yC[n], YC[k] = G[k]H[k]


Circular ConvolutionCircular Convolution• A 4-point IDFT of YC[k] yields


Circular ConvolutionCircular Convolution• Example - Extend the two length-4 sequences to length 7 by

di h ith th l d l iappending each with three zero-valued samples, i.e.

• We next determine the 7-point circular convolution of ge[n]and he[n]

• From the above, we can obtain y[0] ~ y[6]y[ ] y[ ]• y[n] is precisely the sequence yL[n] obtained

by a linear convolution of g[n] and h[n]


Circular ConvolutionCircular Convolution• The N-point circular convolution can be written in matrix

fform as

• Note: The elements of each diagonal of the matrix are equal• Such a matrix is called a circulant matrix


Circular Convolution Using Tabular Method

• Consider the evaluation of y[n] = h[n] g[n] where {g[n]} Co s de e e a ua o o y[n] h[n] g[ ] e e {g[ ]}and {h[n]} are length-4 sequences

• First, the samples of the two sequences are multiplied using p q p gthe conventional multiplication method as shown below

• The partial products generated in the 2nd, 3rd, and 4th rows


p p gare circularly shifted to the left as indicated above

Circular Convolution Using Tabular Method

• The modified table after circular shifting is shown belowe od ed tab e a te c cu a s t g s s o be o

• The samples of the sequence yC[n] are obtained by adding the 4 partial products in the column above of each sample

yC[0] = g[0]h[0] + g[3]h[1] + g[2]h[2] + g[1]h[3] = 7yC[1] = g[1]h[0] + g[0]h[1] + g[3]h[2] + g[2]h[3] = 6yC[2] = g[2]h[0] + g[1]h[1] + g[0]h[2] + g[3]h[3] = 5


yC[3] = g[3]h[0] + g[2]h[1] + g[1]h[2] + g[0]h[3] = 5

N-Point DFTs of Two Length-N Real Sequences

• Let g[n] and h[n] be two length-N real sequences with G[k] et g[ ] a d [ ] be t o e gt ea seque ces t G[ ]and H[k] denoting their respective N-point DFTs

• These two N-point DFTs can be computed efficiently using a p p y gsingle N-point DFT

• Define a complex length-N sequencex[n] = g[n] + j h[n]

• Let X[k] denote the N-point DFT of x[n]

• Note that for 0 ≤ k ≤ N −1,


N-Point DFTs of Two Length-N Real Sequences

• Example - We compute the 4-point DFTs of the two real a p e e co pute t e po t s o t e t o easequences g[n] and h[n] given below

{g[n]} = {1 2 0 1}, {h[n]} = {2 2 1 1}{g[ ]} { } { [ ]} { }• Then {x[n]} = {g[n]}+ j{h[n]} = {1+j2 2+j2 j 1+j}• Its DFT X[k] is[ ]

• From the aboveX*[k] = [4−j6 2 −2 −j2], X*[ 4−k 4] = [4−j6 −j2 −2 2]

• Therefore


{G[k]} = {4 1−j −2 1+j}, {H[k]} = {6 1−j 0 1+j}

2N-Point DFTs of a Real Sequences Using an N-Point DFT

• Let v[n] be a length-N real sequence with a 2N-point DFT V[k]et [ ] be a e gt ea seque ce t a po t [ ]• Define two length-N real sequences g[n] and h[n] as follows:

g[n] = v[2n], h[n] = v[2n +1], 0 ≤ n ≤ Ng[n] v[2n], h[n] v[2n 1], 0 ≤ n ≤ N• Let G[k] and H[k] denote their respective N-point DFTs• Define a length-N complex sequenceDefine a length N complex sequence

{x[n]} = {g[n]}+ j{h[n]}with an N-point DFT X[k]with an N point DFT X[k]

• Now

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-37• That is

2N-Point DFTs of a Real Sequences Using an N-Point DFT

• Example - Let us determine the 8-point DFT V[k] of the a p e et us dete e t e 8 po t [ ] o t elength-8 real sequence

{v[n]} = {1 2 2 2 0 1 1 1}{ [ ]} { }• We form two length-4 real sequences as follows

g[n] = v[2n] = {1 2 0 1}, h[n] = v[2n +1] = {2 2 1 1}g[ ] [ ] { }, [ ] [ ] { }• Now

• Substituting the values of the 4-point DFTs G[k] and H[k], we can obtain V[k]we can obtain V[k]


Linear Convolution Using DFTLinear Convolution Using DFT• Since a DFT can be efficiently implemented using FFT S ce a ca be e c e t y p e e ted us g

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

• Then


Linear Convolution Using DFTLinear Convolution Using DFT• The corresponding implementation scheme is illustrated e co espo d g p e e tat o sc e e s ust ated

below

• We next consider the DFT-based implementation ofp

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


g ( g q ggreater than M)

Overlap Add MethodOverlap-Add Method• We first segment x[n], assumed to be a causal sequence

h ith t l f lit i t t f tihere without any loss of generality, into a set of contiguous finite-length subsequences of length N each:

wherewhere

• Thus we can writeThus we can write

wherewhere

• Since h[n] is of length M and is of length N, ym[n] is of length


[ ] g g ym[ ] gN + M −1

Overlap Add MethodOverlap-Add Method• The desired linear convolution y[n] = h[n] x[n] is broken up

i t f i fi it b f h t l th liinto 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 follo ing con ol tions sing the• Consider implementing the following convolutions using the DFT-based method, where now the DFTs (and the IDFT) are computed on the basis of (N + M −1) pointsare computed on the basis of (N M 1) points

Th fi l i i h b [ ] h[ ] [ ] i• The first convolution in the above sum, y0[n] = h[n] x0[n], is of length N + M −1 and is defined for 0 ≤ n ≤ N + M − 2Th d h t l ti [ ] h[ ] [ ] i l f• The second short convolution y1[n] = h[n] x1[n], is also of length N + M −1 but is defined for N ≤ n ≤ 3N + M − 2There is an overlap of samples between these two short


There is an overlap of samples between these two short linear convolutions

Overlap Add MethodOverlap-Add Method• In general, there will be an overlap of M −1 samples

b t th l f th h t l ti h[ ] [ ]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


Overlap Add MethodOverlap-Add Method

• Therefore, y[n] obtained by a linear convolution of x[n] and h[n] is given by


Overlap Add MethodOverlap-Add Method• The above procedure is called the overlap-add method

i th lt f th h t li l ti l dsince the results of the short linear convolutions overlap and the overlapped portions are added to get the correct final resultresult

• The MATLAB function fftfilt can be used to implement the above methodabove method

• The following illustrates an example of filtering of a noise-corrupted signal using a length-3 moving average filter:p g g g g g


Overlap Save MethodOverlap-Save Method• In implementing the overlap-add method using the DFT, we

d t t t (N M 1) i t DFT d (N Mneed to compute two (N + M −1)-point DFTs and one (N + M−1)-point IDFT for each short linear convolutionIt is possible to implement the o erall linear con ol tion b• It is possible to implement the overall linear convolution by performing instead circular convolution of length shorter than (N + M −1)(N M 1)

• To this end, it is necessary to segment x[n] into overlapping blocks xm[n], keep the terms of the circular pp g m[ ], pconvolution of h[n] with that corresponds to the terms obtained by a linear convolution of h[n] and xm[n], and throw

faway the other parts of the circular convolution


Overlap Save MethodOverlap-Save Method• To understand the correspondence between the linear and

i l l ti id l th 4 [ ] dcircular convolutions, consider a length-4 sequence x[n] and a length-3 sequence h[n]Let [n] denote the res lt of a linear con ol tion of [n] ith• Let yL[n] denote the result of a linear convolution of x[n] with h[n]

• The six samples of y [n] are given by• The six samples of yL[n] are given byyL[0] = h[0]x[0]y [1] = h[0]x[1] + h[1]x[0]yL[1] = h[0]x[1] + h[1]x[0]yL[2] = h[0]x[2] + h[1]x[1] + h[2]x[0]y [3] = h[0]x[3] + h[1]x[2] + h[2]x[1]yL[3] = h[0]x[3] + h[1]x[2] + h[2]x[1]yL[4] = h[1]x[3] + h[2]x[2]

[5] h[2] [3]


yL[5] = h[2]x[3]

Overlap Save MethodOverlap-Save Method• If we append h[n] with a single zero-valued sample and

t it i t l th 4 h [ ] th 4 i t i lconvert it into a length-4 sequence he[n], the 4-point circular convolution yC[n] of he[n] and x[n] is given by

y [0] = h[0]x[0] + h[1]x[3] + h[2]x[2]yC[0] = h[0]x[0] + h[1]x[3] + h[2]x[2]yC[1] = h[0]x[1] + h[1]x[0] + h[2]x[3]y [2] = h[0]x[2] + h[1]x[1] + h[2]x[0]yC[2] = h[0]x[2] + h[1]x[1] + h[2]x[0]yC[3] = h[0]x[3] + h[1]x[2] + h[2]x[1]

• If we compare the expressions for the samples yL[n] of withIf we compare the expressions for the samples yL[n] of with those of yC[n], we observe that the first 2 terms of yC[n] do not correspond to the first 2 terms of yL[n], whereas the last 2 Lterms of yC[n] are precisely the same as the 3rd and 4th terms of yL[n], i.e.


yL[0] ≠ yC[0], yL[1] ≠ yC[1], yL[2] = yC[2], yL[3] = yC[3]

Overlap Save MethodOverlap-Save Method• General case: N-point circular convolution of a length-M

h[ ] ith l th N [ ] ith N Msequence h[n] with a length-N sequence x[n] with N > M• First M − 1 samples of the circular convolution are incorrect

and are rejectedand are rejected• Remaining N − M + 1 samples correspond to the correct

samples of the linear convolution of h[n] with x[n]samples of the linear convolution of h[n] with x[n]• Now, consider an infinitely long or very long sequence x[n]• Break it up as a collection of smaller length (length 4)• Break it up as a collection of smaller length (length-4)

overlapping sequences xm[n] as xm[n] = x[n + 2m], 0 ≤ n ≤ 3, 0 ≤ m ≤ ∞0 m

• Next, formwm[n] = h[n] xm[n]


wm[n] h[n] xm[n]

Overlap Save MethodOverlap-Save Method• Or, equivalently,

wm[0] = h[0]xm[0] + h[1]xm[3] + h[2]xm[2]wm[1] = h[0]xm[1] + h[1]xm[0] + h[2]xm[3]

[2] h[0] [2] h[1] [1] h[2] [0]wm[2] = h[0]xm[2] + h[1]xm[1] + h[2]xm[0]wm[3] = h[0]xm[3] + h[1]xm[2] + h[2]xm[1]

C ti th b f 0 1 2 3 d b tit ti• Computing the above for m = 0, 1, 2, 3, . . . , and substituting the values of xm[n] we arrive at

w [0] = h[0]x[0] + h[1]x[3] + h[2]x[2] Rejectw0[0] = h[0]x[0] + h[1]x[3] + h[2]x[2] Rejectw0[1] = h[0]x[1] + h[1]x[0] + h[2]x[3] Rejectw0[2] = h[0]x[2] + h[1]x[1] + h[2]x[0] = y[2] Savew0[2] h[0]x[2] + h[1]x[1] + h[2]x[0] y[2] Savew0[3] = h[0]x[3] + h[1]x[2] + h[2]x[1] = y[3] Save


Overlap Save MethodOverlap-Save Methodw1[0] = h[0]x[2] + h[1]x[5] + h[2]x[4] Rejectw1[1] = h[0]x[3] + h[1]x[2] + h[2]x[5] Rejectw1[2] = h[0]x[4] + h[1]x[3] + h[2]x[2] = y[4] Save

[3] h[0] [5] h[1] [4] h[2] [3] [5] Sw1[3] = h[0]x[5] + h[1]x[4] + h[2]x[3] = y[5] Save

w [0] = h[0]x[4] + h[1]x[5] + h[2]x[6] Rejectw2[0] = h[0]x[4] + h[1]x[5] + h[2]x[6] Rejectw2[1] = h[0]x[5] + h[1]x[4] + h[2]x[7] Rejectw2[2] = h[0]x[6] + h[1]x[5] + h[2]x[4] = y[6] Savew2[2] = h[0]x[6] + h[1]x[5] + h[2]x[4] = y[6] Savew2[3] = h[0]x[7] + h[1]x[6] + h[2]x[5] = y[7] Save

• It should be noted that to determine y[0] and y[1] we need toIt should be noted that to determine y[0] and y[1], we need to form x−1[n]: x−1[0] = 0, x−1[1] = 0, x−1[2] = x[0] , x−1[3] = x[1]and compute w−1[n] = h[n] x−1[n] for 0 ≤ n ≤ 3, reject w−1[0]


a d co pu e −1[ ] [ ] −1[ ] o 0 3, ejec −1[0]and w−1[1], and save w−1[2] = y[0], and w−1[3] = y[1]

Overlap Save MethodOverlap-Save Method• General Case: Let h[n] be a length-N sequence• Let xm[n] denote the m-th section of an infinitely long

sequence x[n] of length N and defined byxm[n] = x[n + m(N − M + 1)], 0 ≤ n ≤ N − 1 with M < N

• Let wm[n] = h[n] xm[n]• Then, we reject the first M − 1 samples of wm[n] and “abut”

the remaining M − M + 1 samples of wm[n] to form yL[n], the linear convolution of h[n] and x[n]linear convolution of h[n] and x[n]

• If ym[n] denotes the saved portion of wm[n], i.e.,

Th [ + (N M + 1)] [ ] M 1 ≤ ≤ N 1

1


• Then yL[n + m(N − M + 1)] = ym[n], M − 1 ≤ n ≤ N − 1

Overlap Save MethodOverlap-Save Method• The approach is called overlap-save method since the

i t i t d i t l i ti d t f thinput is segmented into overlapping sections and parts of the results of the circular convolutions are saved and abutted to determine the linear convolution resultdetermine the linear convolution result


Overlap Save MethodOverlap-Save Method


Signal TransformSignal Transform• Motivation:Motivation:

– Represent a vector (e.g. a block of image samples) as the superposition of some typical vectors (block patterns)

+t1 t2 t3 t4


Transform Coding of a ImageTransform Coding of a Image


1 D 16 Pont DFT Basis Vectors1-D 16-Pont DFT Basis Vectors


Disadvantages of DFT in Signal Coding

• Fourier Transform of a real function results inou e a s o o a ea u ct o esu tscomplex numbers

• May result in artifacts due to discontinuityat the block boundary


Discontinuities (high freq. components)

From DFT to DCTFrom DFT to DCT• DFT of any real and symmetric sequence contains only

l ffi i t di t th i t freal coefficients corresponding to the cosine terms of the series

• Construct a new symmetric sequence y(n) of length 2N• Construct a new symmetric sequence y(n) of length 2N out of x(n) of length N

( ) ( ) 0 1= ≤ ≤ −= − − ≤ ≤ −

( ) ( ),0 1,( ) (2 1 ), 2 1.

y n x n n Ny n x N n N n N

• Y(n) is symmetrical about n = N - (1/2)

( ) ( ),y


From DFT to DCTFrom DFT to DCT• DCT has a higher compression ration than DFT

• DCT avoids the generation of spurious spectral components

No discontinuities


From DFT to DCTFrom DFT to DCT− +

= ∑12 1 ( )2( ) ( )

N n kY k y n W

=

− −+ +

=

= +

∑

∑ ∑

20

1 11 2 1( ) ( )2 2

( ) ( ) ,

( ) ( )

Nn

N Nn k n k

Y k y n W

y n W y n W= =

= +∑ ∑2 20

( ) ( )N Nn n N

y n W y n W

− −+ +

∑ ∑1 11 2 1( ) ( )N Nn k n k+ +

= =

− −+ − +

= + − −∑ ∑

∑ ∑

( ) ( )2 2

2 20

1 11 1( ) [ 2 ( ) ]

( ) ( 2 1 )n k n k

N Nn n N

N Nn k N n k

x n W x N n W

π

+ +

= =

−

= +

+

∑ ∑

∑

( ) [ 2 ( ) ]2 2

2 20 01

( ) ( )

( 2 1)

n k N n k

N Nn n

N

x n W x n W

n k

π

π=

−

+= ∑

02

( 2 1)2 ( ) c o s ,2n

j

n kx nN


≤ ≤ − = 220 2 1, j

NNk N a n d W e

1 D N Point DCT1-D N-Point DCTπ− +⎡ ⎤

⎢ ⎥∑1 ( 2 1)( ) ( ) ( ) c o s

N n kF k C k f n=

= ⎢ ⎥⎣ ⎦= −

∑0

( ) ( ) ( ) c o s ,2

0 ,1, , 1,n

F k C k f nN

k Nπ− +⎡ ⎤= ⎢ ⎥⎣ ⎦

∑1

0 ,1, , 1,( 2 1)( ) ( ) ( ) c o s ,

2

N

k Nn kf n C k F k

N= ⎢ ⎥⎣ ⎦= −

0 20 ,1, , 1,

k Nn N

1 2

where

= = = −1 2(0) , ( ) , 1,2, , 1C C k k NN N


• The constants are often defined differently

1 D N Point DCT1-D N-Point DCT

( )cos (2 1)0 /16n π+ ( )cos (2 1)2 /16n π+ ( )cos (2 1)4 /16n π+ ( )cos (2 1)6 /16n π+

( )cos (2 1)1 /16n π+ ( )cos (2 1)3 /16n π+ ( )cos (2 1)5 /16n π+ ( )cos (2 1)7 /16n π+


Example of 1 D DCTExample of 1-D DCTRow 256 of Lena 2500 00

160 00

200.00Row 256 of Lena

2000.00

2500.00

absolute DCT values of Lena row 256

120.00

160.00

1500.00

80.00 1000.00

40.00 500.00

0.00 200.00 400.00 600.00

0.00

0.00 200.00 400.00 600.00

0.00


Illustration of Image CodingUsing 2-D DCT


DCT-Based Image Coding with Different Quantization Levels

DCT di i h i i l i i bl k i 8 8• DCT coding with increasingly coarse quantization, block size 8x8

quantizer step-size for AC coefficient:




25 100 200

Image Coding with Different Numbers of DCT Coefficients

with 16/64 original coefficients

with 8/64 coefficients

with 4/64 coefficients


Comparison of Basis Vectors of Different Transform (1-D)


Comparison of Basis Vectors of Different Transform (2-D)

Cosine Sine

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-69Haar 69

DCT Based Coding: JPEGDCT-Based Coding: JPEG8x8 blocks Compressed

i dDC

Entropyencoder

image dataQ

DPCMZigzagscan

DCTAC

C

Quantizationtable

Tablespecification

Sourceimage data

specification

Compressed8x8 blocks

DC

Entropydecoder

image dataIQ

DPCMZigzagscanIDCT

AC

Quantization Table

Reconstructedimage data


table specification

chapter 5chapter 5 finite-length discrete transformscwlin/courses/dsp/notes/ch5_mitra_dsp_c.pdf ·...

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