A PROJECT REPORT ON

A VLSI PROGRESSIVE CODING FOR WAVELET-BASED IMAGE COMPRESSION

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

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This paper describes the hardware design flow of lifting based 2-D

Discrete Wavelet Transform (DWT) processor for JPEG 2000.

In order to build high quality image of JPEG 2000 codec, an effective

2-D DWT algorithm has been performed on input image file to get the

decomposed image coefficients.

The Lifting Scheme reduces the number of operations execution steps

to almost one-half of those needed with a conventional convolution

approach. Initially, the lifting based 2-D DWT algorithm has been

developed . The DWT modules were simulated using XPS(8.1i) design

tools.

A majority of today’s Internet bandwidth is estimated to be used for

images and video. Wavelet based techniques such as JPEG2000 for

image compression has a lot more to offer than conventional methods

in terms of compression ratio. The main aim of the project is to

implement and verify the image compression technique and to

investigate the possibility of hardware acceleration of DWT for signal

processing applications. A hardware design has to be provided to

achieve high performance, in comparison to the software

implementation of DWT. EDK Tool (XPS 8.1i), Matlab 7, Visual Basic

6.0, Xilinx Spartan-3E FPGA Kit

Languages Used: System C., JPEG2000 (SPHIT), FDWT

Fig : Block Diagram of image compression

Wavelet TransformWavelet TransformApplication of WT

Discrete Wavelet Transform

a

bt

at

ba 1

)(,

; a, b R1 and a>0

Function of DWTFunction of DWT

Fig: 1D WAVELET TRANSFORM

)(),(,,ttx

nmnmw ; m, n €Z

)},({),( jipDWTjic .

Where c (i, j) is the wavelet coefficients.

To illustrate how the SPIHT Coding works lets look at the following example

Fig: 2.4.4 a 4x4 matrix showing the pixel values of a digital image.The first step in SPIHT coding is the initialization of the sets LSP, LIS, LIP which is done as shown below

LIP (0, 0) = 26 (0, 1) = 6 (1, 0) = -7 (1, 1) = 7

LSPEMPTY

LIS(0, 1)= {13, 10, 6, 4}(1, 0)= {4,-4, 2,- 2}(1, 1)= {4,-3,-2, 0}

n = log2 (MAX COEFF)

n=log2(26) =4

After First Sorting Pass

Threshold To= 2n=24=16

•Process LIP

•S(0,0)=26>To, we transmit 1, since 26 is +ve ,we transmit 0; then move (0,0) to LSPT

(Temporary), then

•S(0,1)=6, S(1,0)= -7, S(1,1)=7 are all <To hence they are insignificant , therefore transmit

three 0,

Process LIS

•DS(0,0)=13, DS(0,1)=10, DS(1,0)=4 ,DS(1,1) are all less than To hence we transmit 0 for the

complete set S(0,1) similarly the sets S(1,0) and S(1,1) are insignificant hence transmit two

0’s

We need not process LSP since it is null

•Update LSPT to LSP

•The transmitted bit stream is 10000000(8 bits)

•LIP={(0,1),(1,0),(1,1)}

•LIS={D(0,1),D(1,0),D(1,1)}

•LSP={ (0,0)}

Fig: 2.4.6 BLOCK DIAGRAM AFTER FIRST SORTING PASS

n=4-1=3, Threshold T1= 2n=23=8

Process LIP S(0,1)=6, S(1,0)=-7, S(1,1),=7, are insignificant ,hence we transmit 3 0’S,

Process LIS DS (0, 0) =13, DS (0, 1) =10, these two are > T1 hence we transmit 1 for the set then we transmit 10 for 1 and again we transmit 10 for 10, then move (0, 2) and (0, 3) to LSPT DS(1,0)=6 and DS(1,1)=4 < T1 we transmit two O’S ,then move (1,2) and(1,3)

to LIP The sets D (1, 0) and D (1, 1) are insignificant hence we transmit two 0’SProcess LSP C (0, 0) =26= (11010)2 ---- TRANSMIT NTH MSB =1•Update LSPT to LSP•The transmitted bit stream is 0001101000001(13 bits)•LIP={(0,1),(1,0),(1,1),(1,2),(1,3)}•LIS={D(1,0),D(1,1)}•LSP={ (0,0),(0,2),(0,3) }

)},({),(ˆ jicIDWTjip

called as “progressive transmission”

DECODERAn example of decoding the above transmitted bit is as shown belowFirst Receive Get n=4, To=2n = 16

LIP = {(0, 0), (0, 1), (1, 0), (1, 1)} LIS ={ D(0,0),D(0,1),D(1,0),D(1,1)} LSP = { } The transmitted bit stream is 10000000(8 bits)Process LIP Get 1=S(0,0) is significant, next is zero hence +ve value; move S(0,0) to LSP, Then construct C(0,0)=(3/2)TO=(3/2)16 =24

Get three 0 = S (0, 1), S (1, 0), S (1, 1) are insignificantProcess LIS Get three 0 = DS (0, 1), DS (1, 0), DS (1, 1) are insignificant LIP = {(0, 1), (1, 0), (1, 1)} LIS ={ D(0,1),D(1,0),D(1,1)} LSP = {(0, 0)}

24 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

PIXELS AFTER FIRST RECIEVE

Second receive Get n=4-1=3, T1=2n = 8 LIP = {(0, 1), (1, 0), (1, 1)} LIS ={ D(0,1),D(1,0),D(1,1)} LSP = {(0, 0)} The transmitted bit stream is 0001101000001(13 bits) Process LIP Get 000 =S (0, 1), S (1, 0), S (1, 1) are insignificant Process LIS Get 1 = DS (0, 1) is significant Get 10 = C (0, 2) is a positive significance Move (0, 2) to LSP, then reconstruct C (0, 2) = + (3/2) T1 = (3/2)8 = 12

Get 10 = C (0, 3) is a positive significance

Move (0, 3) to LSP then reconstruct C (0, 3) = + (3/2) T1 = (3/2)8 =12

Get 00 = C (1, 2), C (1, 3) are insignificant move to LIP

Get 00 = DS (1, 0), DS (1, 1) are insignificant Process LSP Get 1, then add 2 n-1 to C (0, 0) = 24+2 n-1=24+2 2 =24+4 =28

28 0 12 12

0 0 0 0

0 0 0 0

0 0 0 0

PIXELS AFTER SECOND RECIEVE

inverse DWT is given as: m n

nmnm tWtx )()( ,,

EDK Tool (XPS 8.1i),

Matlab 7,

Visual Basic 6.0