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Application of Discrete Sine Transform in Image Processing


Call for Papers Engineering Journal, May 2019

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Application of Discrete Sine Transform in Image Processing

NCETRASECT-2015 Conference Proceedings

Nitesh Agarwal

Department of Computer Science Jodhpur Institute of Engineering & Technology

Jodhpur, India

Rahul Solanki Dr. A.M. Khan

Department of Mathematics Department Of Mathematics

Jodhpur Institute of Engineering & Technology Jodhpur Institute of Engineering & Technology Jodhpur, India Jodhpur, India

Abstract Digital devices & computational resources have limited communication & storage capabilities. Because of these limitation digital multimedia data need to compress. For example if there are multiple camera are connected in a network for communication then images captured by camera need to compress before communication to use proper bandwidth & for synchronized communication. Similarly local computer system has limited storage capacity & uncompressed multimedia data required high volume of storage to store. Hence multimedia data like images, videos need to be compress for many purpose. Image compression process use two technique to compress image lossless image compression & lossy image compression. Lossy Image compression needs some transformation like DCT, DFT, KLT, DST etc. Purpose of transformation is to convert the data into a form where compression is easier. This transformation will transform the pixels which are correlated into a representation where they are decorrelated. The new values are usually smaller on average than the original values. The net effect is to reduce the redundancy of representation. In Lossy image compression input image is divided in to 8*8 blocks & then each pixel is converted in to its equivalent frequency value

using various transformation like DCT, DST etc. The present paper deals with the study of transformation of an

converted from analog to digital form. Discrete Sine Transform (DST) converts this digital information into its equivalent frequency domain by partitioning image pixel matrix into blocks of size N*N, N depends upon the type of image. For example if we used a black & white image of 8 bit then all shading of black & white color can be expressed into 8 bit hence we use N=8, similarly for color image of 24 bit we can use N=24 but using block size N=24, time complexity may increase. Hence we operate DST on individual color component for a color image. Color image consist of 8 bit red

+ 8 bit green + 8 bit blue hence we apply DST on each color component (Red, Green, Blue) using block size N=8. DST give best result if block size is 8*8 or 16*16 hence block size is used in this way that it fulfill the constraint of DST as well constraint of pixel format of image.

    1. One-Dimensional DST

      If we have one-D sequence of signal value of length N then its equivalent DST can be expressed as

      su uN 1 f xsin 2x 1(u 1) 1

      x0 2N

      for u = 0,1,2,,N 1.

      & inverse transformation is defined as

      N 1 2x 1(u 1)

      8 bit (b/w) image into its frequency domain through Discrete Sine Transform.

      Key Words: DCT, DST, KLT, DFT

      f x ususin

      u0 2N

      2

      1. INTRODUCTION

        An image basically a 2D signal processed by the human visual system. The signal representing images are usually in analog form, but for processing, storage, transmission and computing by computer application analog images are

        Where f x is signal value at point x & u is

        transform coefficient for value u.

        1 for u N 1

        (u) N 3

      2. Apply FDSNTCE(FToRrAwSaErdCTD-i2s0c1r5etCeoSnifnereenTcreanPsrfoocremed)inogns each 8*8 block of pixel matrix to get equivalent 8*8 DST blocks.

        2

        N

        for u N 1

      3. To get Original image we apply IDST (Inverse Discrete Sine Transform) on each 8*8 block DST & get its equivalent 8*8 IDST block.

        It is clear from (1) for u=0,

        su 0 0 4

    2. Two Dimensional DST

An image is 2-D pixel matrix where each position (i,j)

  1. Using 8*8 IDST blocks we create modified pixel matrix to get modified image.

  2. Now we Find MSE (Mean Squared Error) & PSNR (Peak Signal To Noise Ratio) to determine quality of image obtain by IDST. MSE & PSNR calculated by following formulas

H 1 W 1

represents a color value for that particular point or position. Hence to transform an image into its equivalent DST matrix we use 2-D DST.

MSE

1 [o(x, y)m(x, y)]2 9

H *W x 0 y 0

2-D DST can be defined as

N 1 N 1 2x 1(u 1) 2y 1(v 1)

PSNR=20*log10 (MAX) – 10*log10 (MSE) (10)

Where H=Height of Image, W= Width of Image,

su,vu(v) f x, ysin

x 0 y 0

for u, v = 0,1,2,,N 1.

sin

2N 2N

5

variable MAX shows max value of a pixel for example if image is 8 bit then MAX=255.

Quality of image obtain by IDST is depend on MSE & PSNR value. If as the MSE value increases PSNR value decreases then we get a bad quality of image by

& inverse transformation is defined as

N 1 N 1 2x 1(u 1) 2y 1(v 1)

IDST & if as the MSE value decreases PSNR value increases we get a batter quality image hence a best

f x, y u(v)su,vsin

u 0 v0

sin

2N

6

2N

suitable transformation like DCT, DST, DFT is taken on the basis of this MSE & PSNR value.

Where su, v represents frequency value for u, v &

f x, y represents pixel color value at position ( x, y ).

1 for u N 1

N

2.1.1 Algorithm 1

Get_8*8_blocks (image)

{

n=8, k=0; width=width

(u)

2

N

for u N 1

7

of image;

height=height of image; for ( i=0;i < width/n; i++)

1 for v N 1

(v) N 8

2 for v N 1

N

2. MAIN RESULTS

    1. Implementation of DST

      This paper describe how a b/w image is convert into equivalent frequency domain using DST.

      Steps involved in this implementation

      1. Create pixel matrix of the image & divided it into blocks of size 8*8

        {

        for ( j=0; j < height/n; j++)

        {

        xpos = i * n; ypos = j * n;

        for ( a=0; a < n; a++)

        {

        for ( b=0; b < n; b++)

        { color = color at position(xpos+a, ypos+b); block[k][a][b]=color-128;

        } //end of for loop b

        k=k+1;

        } // end of for loop a }

        // end of for loop j

        } // end of for loop i }// end of Get_8*8_blocks

        N}C//ETenRdAoSfECfoTr-2lo0o1p5 Conference Proceedings

        y } // end of for loop x

        dst[i][j][k]=( (u) *(v) *sum);

        } // end of for loop v

        } // end of for loop u

        } // end of for loop i

        }//end of FDST

            1. Algorithm 2

              FDST (block [] [] [])

              { width=width of image, N=8; height=height of image;

              q=(width/8)*(height/8) for ( i=0;i < q; i++)

              { for ( u=0; u< N; u++)

              { for ( v=0; v < N; v++) { if (u==N-1) {

              (u) = 1

              N

              }

              else{

              2

              (u) =

              N

              }

              if (v==N-1){

              (v) = 1

              N

              }

              else {

              (v) = 2

              N

            2. Algorithm 3

              IDT(dst [] [] [])

              { width=width of image, N=8;

              height=height of image; q=(width/8)*(height/8) for ( i=0;i < q; i++)

              { for ( x=0; x< N; x++) { for ( y=0; y< N; y++)

              {

              sum=0;

              for( u=0;u<N;u++)

              { for( v=0;b<N;v++)

              { if (u==N-1) {

              (u) = 1

              N

              }

              else{

              2

              (u) =

              N

              }

              if (v==N-1){

              (v) = 1

              N

              } }

              sum=0; else {

              for( x=0;x<N;x++)

              { for( y=0;y<N;y++)

              { sum= sum + block[i][x][y]*

              (v) = 2

              N

              sin 2x 1(u 1) *

              2 y 1(v 1) ; }

              sin

              2N

              2N

              sum= sum + (u) * (v) *dst[i][u][v] *

              sin 2x 1(u 1) * sin 2 y 1(v 1) ;

              NCETRASECT-2015 Conference Proceedings

              2N

              2N

      2. Transform Input image into equivalent DST image

} // end of for loop v } // end of for loop u

idst[i][j][k]=sum;

} // end of for loop y }

// end of for loop x

} // end of for loop i }

// end of IDST

      1. Algorithm 4

        Get_Image(idst [] [] []) { k=0;

        width=width of image; height=height of image;

        for ( i=0; i < width; i++) { for ( j=0; j < height; j++) {

        xpos = i * n; ypos = j * n;

        for (a=0; a < n; a++)

        { for (b=0; b < n; b++)

        {

        color=(int)idst[k][a][b];

        set color at position (xpos+a, ypos+b);

        }// end of loop b

        } // end of loop a k++;

        } // end of loop j

        } // end of loop i

        }// end of Get_Image

    1. Outputs

8*8 block 1

8*8 block 2

8*8 block 3

8*8 block 4

8*8 block 1

8*8 block 2

8*8 block 3

8*8 block 4

1. Convert pixel matrix into blocks of size 8*8

=

Input Image of size Output blocks of size 16*16 8*8

FDST

Input Image of size Output DST Image of 16*16 size 16*16

  1. Get modified image from DST image

    IDST

    Input DCT Image of Output Image of size size 16*16 16*16

    3. MODIFICATION IN ORIGINAL DST

    1. Using cosine operator rather than sine

      There is a difference of /2 between sine & cosine operator hence using cosine rather than sine operator in DST may loss pixel data

      FDST

      Using cosine operator

      Input Image of size Output DST Image of 16*16 size 16*16

      IDST

      Using cosine operator

      Input Image of size Output Image of size 16*16 16*16

    2. Change in block size

      All shading of black & white image can be expressed in 8 bit of blocks hence we use block size 8*8 to perform DST on it. But in color image each color value of a pixel can be expressed into 24 bit of block which contain 8 bit red + 8 bit green + 8 bit blue. To transform a color image into its equivalent DST format we extract each 8 bit color component from 24 bit of block & then perform 8*8 DST on each color component rather than using 24*24 DST for 24 bit block. The main reason is that if use 24*24 DST rather than 8*8 DST the time complexity of DST is increases in a very large amount.

      For example

      For an image of size 48*48

      1. If 8*8 DST used

        Total no of blocks q=(48/8)*(48/8)=36 For FDST

        for ( i=0;i < q; i++) // loop runs 36 times

        { for ( u=0; u< 8; u++) // loop runs 36*8 times

        { for ( v=0; v < 8; v++)// loop runs 36*8*8 times

        {

        }

        for( x=0;x<8;x++)// loop runs 36*8*8*8 times

        { for( y=0;y<8;y++) // loop runs 36*8*8*8*8

        times

        {

        } // end of for loop y } // end of for loop x

        } // end of for loop v

        } // end of for loop u

        } // end of for loop i

        Total no. of iteration = 36*8*8*8*8= 147456

      2. If 24*24 DST used

      Total no of blocks q=(48/24)*(48/24)=4 For FDST

      for ( i=0;i < q; i++) // loop runs 4 times

      { for ( u=0; u< 24; u++) // loop runs 4*24 times

      { for ( v=0; v < 24; v++)// loop runs 4*24*24 times

      {

      }

      for( x=0;x<24;x++)// loop runs 4*24*24*24 times

      {

      for( y=0;y<24;y++) // loop runs 4*24*24*24*24 times

      {

      } // end of for loop y } // end of for loop x

      } // end of for loop v

      } // end of for loop u

      } // end of for loop i

      Total no. of iteration =4*24*24*24*24= 1327104

      Total no. of iteration =4*24*24*24*24= 1327104

      Hence 24 * 24 DST required 1327104-147456=1179648 extra iteration to preform DST which increases time complexity in large amount hence DST used with block size 8*8.

    3. MSE & PSNR

      Following table gives a comparative analysis of quality of transformed image using DST, DST with cosine operator & original DCT.

      MSE

      PSNR

      2D DST

      0.38

      52.38

      2DDST with cosine

      2239.17

      14.63

      2D DCT

      0.28

      53.64

      Table 1: MSE & PSNR value of input image

      after transformation

      As the table show 2D DCT has minimum MSE & maximum PSNR value hence it is best transformation technique & we cannot use DST with cosine operator in normal way because it gives an image with high amount of noises in the pixel of image.

  2. CONCLUSION

The result presented in this document shows that

  1. It is very easy to implement DST rather than other transformation on image except DCT.

  2. If DST used with cosine operator rather than sin pixel data may lose.

  3. If DST used with block size 24*24 rather than block size 8*8 then time complexity of DST is increases in very large amount.

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  2. Anjali Kapoor and Dr. Renu Dhir, Image Compression Using Fast 2- D DCT Technique, International Journal on

Computer Science and Engineering (IJCSE), vol. 3 pp. 2415-2419, 6

  1. Maneesha Gupta and Dr.Amit Kumar Garg, Analysis Of Image Compression Algorithm Using DCT International

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  2. N.Ahmed, T.Natatarajan, and K.R. Rao, Discrete Cosine Transform,

June 2011.

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